Fractions
Any unit can be divided into any numbers of equal parts, one or more of this parts is called fraction of that unit. e.g. one-forth (1/4), one-third (1/3), three-seventh (3/7) etc.

The lower part indicates the number of equal parts into which the unit is divided, is called denominator. The upper part, which indicates the number of parts taken from the fraction is called the numerator. The numerator and the denominator of a fraction are called its terms.

• A fraction is unity, when its numerator and denominator are equal.
• A fraction is equal to zero if its numerator is zero.
• The denominator of a fraction can never be zero.
• The value of a fraction is not altered by multiplying or dividing the numerator and the denominator by the same number.e.g. 2/3 = 2/6 = 8/12 = (2/4)/(3/4)
• When there is no common factor between numerator and denominator it is called in its lowest terms.e.g. 15/25 = 3/5
• When a fraction is reduced to its lowest term, its numerator and denominator are prime to each other.
• When the numerator and denominator are divided by its HCF, fraction reduces to its lowest term.

Proper fraction: A fraction in which numerator is less than the denominator. e.g. 1/4, 3/4, 11/12 etc.

 

Improper Fraction:  A fraction in which numerator is equal to or more than the denominator. e.g. 5/4, 7/4, 13/12 etc.

 

Like fraction: Fractions in which denominators are same is called like fractions.

e.g. 1/12, 5/12, 7/12, 13/12 etc.

 

Unlike fraction: Fractions in which denominators are not same is called, unlike fractions.

e.g. 1/12, 5/7, 7/9 13/11 etc.

 

Compound Fraction: Fraction of a fraction is called a compound fraction.

e.g. 1/2 of 3/4 is a compound fraction.

 

Complex Fractions: Fractions in which numerator or denominator or both are fractions, are called complex fractions.

 

Continued fraction: Fraction that contain additional fraction is called continued fraction.

e.g.

 

 

 

Rule: To simplify a continued fraction, begin from the bottom and move upwards.

 

Decimal Fractions: Fractions in which denominators are 10 or multiples of 10 is called, decimal fractions. e.g. 1/10, 3/100, 2221/10000 etc.

 

Recurring Decimal: If in a decimal fraction a digit or a set of digits is repeated continuously, then such a number is called a recurring decimal. It is expressed by putting a dot or bar over the digits. e.g.

 

 

Pure recurring decimal: A decimal fraction in which all the figures after the decimal point is repeated is called a pure recurring decimal.

 

Mixed recurring decimal: A decimal fraction in which only some of the figures after the decimal point is repeated is called a mixed recurring decimal.

 

Conversion of recurring decimal into proper fraction: 

CASE I: Pure recurring decimal

 

Write the repeated digit only once in the numerator and put as many nines as in the denominator as the number of repeating figures. e.g.

 

CASE II: Mixed recurring decimal

In the numerator, take the difference between the number formed by all the digits after the decimal point and that formed by the digits which are not repeated. In the denominator, take the number formed as many nines as there are repeating digits followed by as many zeros as is the number of non-repeating digits. e.g.

 
Questions

Level-I

1.	Evaluate : (2.39)2 – (1.61)2 2.39 – 1.61
	A. 2 B. 4 C. 6 D. 8

 

2.	What decimal of an hour is a second ?
	A. .0025 B. .0256 C. .00027 D. .000126

 

3.	The value of (0.96)3 – (0.1)3 is: (0.96)2 + 0.096 + (0.1)2
	A. 0.86 B. 0.95 C. 0.97 D. 1.06

 

4.	The value of 0.1 x 0.1 x 0.1 + 0.02 x 0.02 x 0.02 is: 0.2 x 0.2 x 0.2 + 0.04 x 0.04 x 0.04
	A. 0.0125 B. 0.125 C. 0.25 D. 0.5

 

5.	If 2994 ÷ 14.5 = 172, then 29.94 ÷ 1.45 = ?
	A. 0.172 B. 1.72 C. 17.2 D. 172
6.	When 0.232323….. is converted into a fraction, then the result is:
	A. 1 5 B. 2 9 C. 23 99 D. 23 100

 

7.	.009 = .01 ?
	A. .0009 B. .09 C. .9 D. 9

 

8.	The expression (11.98 x 11.98 + 11.98 x x + 0.02 x 0.02) will be a perfect square for x equal to:
	A. 0.02 B. 0.2 C. 0.04 D. 0.4

 

9.	(0.1667)(0.8333)(0.3333) is approximately equal to: (0.2222)(0.6667)(0.1250)
	A. 2 B. 2.40 C. 2.43 D. 2.50

 

10.	3889 + 12.952 – ? = 3854.002
	A. 47.095 B. 47.752 C. 47.932 D. 47.95
11.	Level-II     0.04 x 0.0162 is equal to:
	A. 6.48 x 10-3 B. 6.48 x 10-4 C. 6.48 x 10-5 D. 6.48 x 10-6

 

12.	4.2 x 4.2 – 1.9 x 1.9 is equal to: 2.3 x 6.1
	A. 0.5 B. 1.0 C. 20 D. 22

 

13.	If 144 = 14.4 , then the value of x is: 0.144 x
	A. 0.0144 B. 1.44 C. 14.4 D. 144

 

 

 

14.	The price of commodity X increases by 40 paise every year, while the price of commodity Y increases by 15 paise every year. If in 2001, the price of commodity X was Rs. 4.20 and that of Y was Rs. 6.30, in which year commodity X will cost 40 paise more than the commodity Y ?
	A. 2010 B. 2011 C. 2012 D. 2013

 

15.	Which of the following are in descending order of their value ?
	A. 1 , 2 , 3 , 4 , 5 , 6 3 5 7 5 6 7 B. 1 , 2 , 3 , 4 , 5 , 6 3 5 5 7 6 7 C. 1 , 2 , 3 , 4 , 5 , 6 3 5 5 6 7 7 D. 6 , 5 , 4 , 3 , 2 , 1 7 6 5 7 5 3
16.	Which of the following fractions is greater than 3 and less than 5 ? 4 6
	A. 1 2 B. 2 3 C. 4 5 D. 9 10

 

17.	The rational number for recurring decimal 0.125125…. is:
	A. 63 487 B. 119 993 C. 125 999 D. None of these

 

18.	617 + 6.017 + 0.617 + 6.0017 = ?
	A. 6.2963 B. 62.965 C. 629.6357 D. None of these

 

19.	The value of 489.1375 x 0.0483 x 1.956 is closest to: 0.0873 x 92.581 x 99.749
	A. 0.006 B. 0.06 C. 0.6 D. 6

 

20.	0.002 x 0.5 = ?
	A. 0.0001 B. 0.001 C. 0.01 D. 0.1

 

 

 

 

 

Answers

Level-I

Answer:1 Option B

 

Explanation:

Given Expression =	a2 – b2	=	(a + b)(a – b)	= (a + b) = (2.39 + 1.61) = 4.
	a – b		(a – b)

 

Answer:2 Option C

 

Explanation:

Required decimal =	1	=	1	= .00027
	60 x 60		3600

 

 

Answer:3 Option A

 

Explanation:

Given expression	= (0.96)3 – (0.1)3 (0.96)2 + (0.96 x 0.1) + (0.1)2
	= a3 – b3 a2 + ab + b2
	= (a – b)   = (0.96 – 0.1)   = 0.86

Answer:4 Option B

 

Explanation:

Given expression =	(0.1)3 + (0.02)3	=	1	= 0.125
	23 [(0.1)3 + (0.02)3]		8

 

 

 

 

Answer:5 Option C

 

Explanation:

29.94	=	299.4
1.45		14.5

 

=		2994	x	1		[ Here, Substitute 172 in the place of 2994/14.5 ]
		14.5		10

 

=	172
	10

= 17.2

 

 

Answer:6 Option C

 

Explanation:

0.232323… = 0.23 =	23
	99

 

Answer:7 Option C

 

Explanation:

Let	.009	= .01;     Then x =	.009	=	.9	= .9
	x		.01		1

 

 

Answer:8 Option C

 

Explanation:

Given expression = (11.98)² + (0.02)² + 11.98 x x.

For the given expression to be a perfect square, we must have

11.98 x x = 2 x 11.98 x 0.02 or x   = 0.04

 

Answer:9 Option D

 

Explanation:

Given expression	= (0.3333) x (0.1667)(0.8333) (0.2222) (0.6667)(0.1250)
	= 3333 x 1 x 5 6 6 2222 2 x 125 3 1000
	= 3 x 1 x 5 x 3 x 8 2 6 6 2
	= 5 2
	= 2.50

 

Answer:10 Option D

 

Explanation:

Let 3889 + 12.952 – x = 3854.002.

Then x = (3889 + 12.952) – 3854.002

= 3901.952 – 3854.002

= 47.95.

 

Level-II

Answer:11 Option B

 

Explanation:

4 x 162 = 648. Sum of decimal places = 6.
So, 0.04 x 0.0162 = 0.000648 = 6.48 x 10^-4

 

Answer:12 Option B

 

Explanation:

Given Expression =	(a2 – b2)	=	(a2 – b2)	= 1.
	(a + b)(a – b)		(a2 – b2)

 

 

Answer:13 Option A

 

Explanation:

144	=	14.4
0.144		x

 

	144 x 1000	=	14.4
	144		x

 

x =	14.4	= 0.0144
	1000

 

 

Answer:14 Option B

 

Explanation:

Suppose commodity X will cost 40 paise more than Y after z years.

Then, (4.20 + 0.40z) – (6.30 + 0.15z) = 0.40

0.25z = 0.40 + 2.10

z =	2.50	=	250	= 10.
	0.25		25

X will cost 40 paise more than Y 10 years after 2001 i.e., 2011.

 

 

 

Answer:15 Option D

Answer:16 Option C

 

Explanation:

3	= 0.75,	5	= 0.833,	1	= 0.5,	2	= 0.66,	4	= 0.8,	9	= 0.9.
4		6		2		3		5		10

Clearly, 0.8 lies between 0.75 and 0.833.

	4	lies between	3	and	5	.
	5		4		6

 

 

 

Answer:17 Option C

 

Explanation:

0.125125… = 0.125 =	125
	999

 

 

Answer:18 Option C

 

Explanation:

617.00

6.017

0.617

+  6.0017

——–

629.6357

———

 

Answer:19 Option B

 

Explanation:

489.1375 x 0.0483 x 1.956		489 x 0.05 x 2
0.0873 x 92.581 x 99.749		0.09 x 93 x 100

 

=	489
	9 x 93 x 10

 

=	163	x	1
	279		10

 

=	0.58
	10

= 0.058  0.06.

 

Answer:20 Option B

 

Explanation:

2 x 5 = 10.

Sum of decimal places = 4

0.002 x 0.5 = 0.001

Discount

 

The discount is referred to the reduction in the price of some commodity or service. It may anywhere appear in the distribution channel in the form of modifications in marked price (printed on the item) or in retail price (set by retailer usually by pasting a sticker on the item) or in list price (quoted for the buyer). The discount is provided for the purpose of increasing sales, to clear out old stock, to encourage distributors, to reward potential customer etc. In short, the discount can serve as a way to attract customers for a particular item or service.

In math, discount is one of the easiest way to raise the customers of particular product. Discounts are a significant element of your online merchandising plan. You build discounts so that you can force sales on items or collection of products to your customers who convene particular conditions. In math, the discount problems can be solved by using discount formula.

The “discount rate” means the interest rate. Discount rate is based on the simple interest rate. To calculate simple interest rate, just find out the interest rate for one period (multiply by amount, interest rate, period) but calculate the discount rate, just multiply by the amount and an interest rate. This is called the define discount rate.

To calculate the discount rate, just multiply the amount by an interest rate. By using the Formula Discount rate DR = pr (p = principal amount, r = interest rate).

 

What is Discount Rate?

Discount rate is one of the simple ways to increase the customers of particular product. Discounts are a important element of your online merchandising strategy. You make discounts so that you can force sales on products or collection of products to your customers who meet certain particular conditions.

 

 

The formula used to calculate the discount is discount = marked price – selling price.

Here,

 

Selling price is what you actually pay for the item.

 

Marked price is the normal price of the item without a discount.

 

Discount is either a dollar rate or a percentage of the marked cost.

 

Discount Rate Definition

Discount Rate is the cost of the total amount generally less than its original value is called . In other words, a total bill will generally sell at a discount, and the discount rate is annualized percentage of this discount, that is percentage is adjusted to give an annual percentage.

 

Discount Rate Formula

Formula of the Discount Rate is:

 

Discount rate DR = pr

where,

• p = principal amount
• r = interest rate

 

 

 

Questions:

Level-I

1: Ricky purchase the dress. That dress rate was Rs1000 at 10% discount . Find discount rate? And then ricky how many dollars give to cashier?

2: Kalvin purchased land for 50000 dollars at 20% in 2000th year. Then 2004th year that land sales 3000 dollars. How many dollars he loss?

3. The marked price of a ceiling fan is $ 1250 and the shopkeeper allows a discount of 6% on it. Find the selling price of the fan.
4. A trader marks his goods at 40% above the cost price and allows a discount of 25%. What is his gain percent?
5. A dealer purchased a washing machine for $ 7660. He allows a discount of 12% on its marked price and still gains 10%. Find the marked price of the machine.
6. How much per cent above the cost price should a shopkeeper mark his goods so that after allowing a discount of 25% on the marked price, he gains 20%?
7. Find the single discount equivalent to two successive discounts of 20% and 10%.
8. A merchant who marked his goods up by 50% subsequently offered a discount of 20% on the marked price. What is the percentage profit that the merchant make after offering the discount?

 

9. Applied to a bill for Rs. 1,00,000 the difference between a discount of 40% and two successive discounts of 36% and 4% is:
10. On a 20% discount sale, an article costs Rs. 596. What was the original price of the article?

Level-II:

11. A discount of 15% on one article is the same as discount of 20% on a second article. The costs of the
12. A discount of 2 ½% is given to the customer on marked price of an article. A man bought the article for Rs. 39. The marked price of article is:
13. Printed price of an article is Rs. 900 but the retailer gets a discount of 40%. He sells the article for Rs. 900. Retailer’s gain percent is:
14. The marked price of a watch was Rs. 720. A man bought the same watch for Rs. 550.80, after getting two successive discounts. If the first discount was 10%, what was the second discount rate?
15. A shopkeeper marks his goods 20% above cost price, but allows 30% discount for cash. His net loss is:
16. A retailer buys 40 pens at the marked price of 36 pens from a wholesaler. If he sells these pens giving a discount of 1%, what is the profit percent?
17. A pizzeria has a coupon that reads, “Getoff a $9.00 cheese pizza.” What is the discount? What is the sale price of the cheese pizza?

18.In a video store, a DVD that sells for $15 is marked, “10% off.” What is the sale price of the DVD?

 

Answers:

Level-I:

 

Solution:1
Here,

Principal amount p = 1000 rs

Interest rate r = 10%

Discount rate DR = pr

DR = 1000*

= 100

The discount amount for the dress is 100.

Discount rate DR = 100.

Dress rate = principal amount – discount rate

= 1000 – 100

=900

Ricky gives 900 rs to cashier

 

 

Solution:2
Principal amount p = 50000 dollars

Interest rate r = 20%

Discount rate DR = pr

DR = 50000 x 2010020100 in 2000th year

= 10000

Discount rate DR = 1000 dollars in 2000th year.

The discount amount is 10000 dollars.

Discount rate DR = 50000*30/100 in 2004th year

Discount rate =15000 dollars.

The discount amount is 15000 dollars.

Loss Discount rate in 2004th year – Discount rate in 2000th year

=15000 dollars – 10000 dollars

=5000 dollars

Kalvin 5000 dollars losses in that land.
 

Solution:3

Marked price = $ 1250 and discount = 6%.

Discount = 6% of Marked Price

= (6% of $ 1250)

= $ {1250 × (6/100)}

= $ 75

Selling price = (Marked Price) – (discount) 

= $ (1250 – 75)

= $ 1175.

Hence, the selling price of the fan is $ 1175.

 

Solution:4

Let the cost price be $ 100.

Then, marked price = $ 140.

Discount = 25% of Marked Price 

= (25% of $ 140)

= $ {140 × (25/100)

= $ 35.

Selling price = (marked price) – (discount) 

= $ (140 – 35)

= $ 105.

Gain% = (105 – 100) % = 5%.

Hence, the trader gains 5%.

 

Solution:5

Cost price of the machine = $ 7660, Gain% = 10%.

Therefore, selling price = [{(100 + gain%)/100} × CP]

= $ [{(100 + 10)/100} × 7660]

= $ [(110/100) × 7660]

= $ 8426.

Let the marked price be $ x.

Then, the discount = 12% of $x

= $ {x × (12/100)}

= $ 3x/25

Therefore, SP = (Marked Price) – (discount)

= $ (x – 3x/25)

= $ 22x/25.

But, the SP = $ 8426.

Therefore, 22x/25 = 8426

⇒ x = (8426 × 25/22)

⇒ x = 9575.

Hence, the marked price of the washing machine is $ 9575

 

Solution:6

Let the cost price be $ 100.

Gain required = 20%.

Therefore, selling price = $ 120.

Let the marked price be $x.

Then, discount = 25% of $x

= $ (x × 25/100)

= $ x/4

Therefore, selling price = (Marked Price) – (discount)

= $ {x – (x/4)

= $ 3x/4

Therefore, 3x/4 = 120

⇔ x = {120 × (4/3)} = 160

Therefore, marked price = $ 160.

Hence, the marked price is 60% above cost price.

 

Solution:7

Let the marked price of an article be $ 100.

Then, first discount on it = $ 20.

Price after first discount = $ (100 – 20) = $ 80.

Second discount on it = 10% of $ 80

= $ {80 × (10/100)} = $ 8.

Price after second discount = $ (80 – 8) = $ 72.
Net selling price = $ 72.

Single discount equivalent to given successive discounts = (100 – 72)% = 28%

 

Solution:8 The easiest way to solve these kinds of problems is to assume a value for the merchant’s cost price.
To make calculations easy, it is best to assume the cost price to be $100.

The merchant marks his goods up by 50%.
Therefore, his marked price (quoted price) = cost price + mark up.
Marked price = $100 + 50% of $100 = 100 + 50 = $150.

The merchant offers a discount of 20% on his marked price.
Discount offered = 20% of 150 = $30.

Therefore, he finally sold his goods for $150 – $30 = $ 120.
We assumed his cost to be $100 and he sold it finally for $120.

Therefore, his profit = $20 on his cost of $ 100.
Hence, his % profit = profit/cost price * 100 = 20/100*100  = 20%.

 

Solution:9 40% of Rs. 1,00,000 = Rs. 40,000
36% of 1,00,000 = 36000
4% of 36,000 = Rs. 2,560.
Therefore, two successive discounts on Rs. 1,00,000 = 36,000 + 2560 = Rs. 38,560.
Difference between a discount of 40% and two successive discounts of 36% and 4%
= 40,000 – 38,560
= Rs. 1,440

Solution:10 If the selling price of the article is S, then
S – 20% of S = 596
S – S/5 = 596
4S/5 = 596
⇒ S = 596 x 5/4
⇒ S = 745

Level-II

Solution:11Let the prices of two articles be X and Y
From the question 15X/100 = 20Y/100
X/Y = 20/15
Thus the ratio of prices of two articles is 4 : 3
Any two amounts in the ratio 4 : 3 will satisfy the condition.
In the above instance, Rs. 80 and Rs. 60 is the answer.

Solution:12 Formula for Marked Price = 100 x SP/(100 – d%) = 100 x 39/(100 – 2.5%)
= 3900 / 97.5
= Rs. 40.
Marked Price of Article is Rs. 40.

Solution:13 Retailer gets a discount of 40% means he buys it at 60% of the price
60% x 900 = Rs. 540
Profit on selling it at Rs. 900 = 900 – 540 = Rs. 360.
Profit % = (Profit / C.P) x 100 = (360 / 540) x 100 = 66²/₃
Retailer’s Gain percent is 66²/₃

Solution:1410% discount on 720 = Rs. 72
Cost after 1st discount = 720 – 72 = Rs. 648.
Cost after 2nd discount = Rs. 550.80
Therefore 2nd discount = 648 – 550.80 = Rs. 97.20
Discount % = (97.2 x 100)/648 = 15%
Second discount rate = 15%.

Solution:15 Let the cost price be Rs. 100.
M.P. (which is 20% above C.P.) = Rs. 120.
30% discount on Rs. 120 = Rs. 36.
Selling Price = Rs. 120 – 36 = Rs. 84
Cost Price = Rs 100 and Selling Price = Rs 84 {since CP > SP, it is a loss}
Loss% = (16/100) x 100 = 16%.
His net loss percent is 16%.

Solution:16 Assuming the M.P. of each pen to be Rs. 10, the M.P. of 36 pens = Rs. 360
Cost price of 40 pens = Rs. 360 (from the question)
Cost price of each pen = 360/40 = Rs. 9
Selling Price of each pen at a discount of 1% on a marked price of Rs. 10 = 99% x 10 = Rs. 9.90
Profit = 9.90 – 9.00 = Rs. 0.90
Profit % = (0.90/9.00) x 100 = 10%
Profit % = 10%.

Solution:17 The discount is $3.00 and the sale price is $6.00

Solution:18 The rate is 10%. Thus, the customer is paying 90% for the DVD

The sale price is: 0.90 x $15.00 = $13.50

The sale price is $13.50.

Partnership : Partnership is an association of two or more parties, they put money for business.         Simple Partnership: Simple partnership is one in which the capitals of the partners are invested for the same time. The profit or losses are divided among the partners in the ratio of their investments.         Compound Partnership: Compound Partnership is one which the capitals of the partners are invested for different periods. In such cases equivalent capitals are calculated for a unit time by multiplying the capital with the number of units of time. The profits or losses are then divided in the ratio of these equivalent capitals. Tus the ratio of profits is directly proportional to both capital invested as time.         Working partner: A partner who participates in the working and manages the business is called a Working Partner.         Sleeping Partner: A partner who only invests capital but does not participate in the working of the business is called a Sleeping Partner.                 Division of Profit and Loss:     1. Rule :When investment of all partners are for the same time, the loss or profit is distributed among partners in the ratio of investment. Ex. Let P and Q invested Rs. a and b for one year in a business then share of profit and loss be , P’s share of profit : Q’s share profit = a : b 2.Rule : When investments are for different time period, then profit ratio is calculated as capital multiplied by length of investment Ex. P’s share of profit : Q’s share profit = a* t1 : b* t2     Questions with solutions Level-I

1. A, B and C enter into a partnership. They invest Rs. 40,000, Rs. 80,000 and Rs. 1,20,000 respectively. At the end of the first year, B withdrawns Rs. 40,000, while at the end of the second year, C withdraws Rs. 80,000. In what ratio will the profit be shared at the end of 3 years ?

 

 

 

Solution: A : B : C = (40,000 X 36) : (80,000 X 12 + 40,000 X 24) : (120,000 X 24 + 40,000 X 12)   =     3: 4: 16

 

 

 

 

 

 

2. A, B, C enter into a partnership investing Rs. 35,000, Rs.45,000 and Rs.55,000 respectively. The respective shares of A, B, C in an annual profit of Rs.40,500 are ?

 

 

 

Solution : A : B : C = 35000 : 45000 : 55000 = 7 : 9 : 11.

 

A’s share = Rs (40500 x 7/27) = Rs. 10500

 

B’s share = Rs.(40500× 9/27) = Rs. 13500

 

C’s share = Rs.(40500×11/27)= Rs. 16500

 

 

 

 

 

 

 

3. In a business, Lucky invests Rs. 35,000 for 8 months and manju invests Rs 42,000 for 10 months. Out of a profit of Rs. 31,570. Manju’s share is 😕

 

 

Solution :      lucky: Manju = (35000 X 8) : (42,000 X 10) = 2:3
Manju’s share = Rs.3/5×31570 = Rs. 18,942

 

 

4. Amar started a business investing Rs. 70,000. Ramki joined him after six months with an amount of Rs. 1,05,000 and Sagar joined them with Rs. 1.4 lakhs after another six months. The amount of profit earned should be distributed in what ratio among Aman, Rakhi and Sagar respectively, 3 years after Aman started the business ?

 

 

Solution: Amar : Ramki : Sagar =

 

(70000 X 36) : (105000 X 30) : (140000 X24)   = 12 : 15 : 16.

 

 

 

5 . A begins a business with Rs 450 and is joined afterwards by B with  Rs 300. After how many months does B join if the profits at the end of the year is divided in the ratio 2 : 1?

 

 

Solution.-.(B) Suppose B joins for x months.

Then,     450 ´12    =    2
300 ´ x           1

x =450× 6
300

 

x= 9 months

\B joins after (12 – 9) = 3 months.

 

 

 

6. Shekhar started a business investing Rs. 25,000 in 1999. In 2000, he invested an additional amount of Rs. 10,000 and Rajeev joined him with an amount of Rs. 35,000. In 2001, Shekhar invested another additional amount of Rs. 10,000 and Jatin joined them with an amount of Rs. 35,000. What will be Rajeev’s share in the profit of Rs. 1,50,000 earned at the end of 3 years from the start of the business in 1999?.

 
Solution : Shekhar : Rajeev : Jatin  =

(25000  X  12 + 35000  X  12 + 45000  X  12) : (35000  X 24) :   (35000  X  12)
= 1260000   :  840000  :  420000  =   3  :  2  :  1.
Rajeev’s share   =  Rs.(150000×26)  =   Rs. 50000

 

 

 

 

 

7. A,B and C started a business with Rs.15000, Rs.25000 and Rs.35000 respectively.  A was paid 10% of the total profit as a salary and the balance was divided in the ration of investment.  If A’s share is Rs.4,200, then C’s share is: ?

 

 

 

Solution : A, B and C must divide their salaries in the ratio :

15,000 : 25,000:35,000 = 3:5:7
Assume total Profit = 100X.

then A share is 10% of 100X for managing business and 3/15 part of 90X for his investment (as the remaining profit is   (100X – 10X = 90X)
So total A’s share  =  10X  + 315 × 90X =  4,200
⇒X = 150
Substituting X  = 150 in 90X we get remaining profit for sharing. That is Rs.13,500
Now C’s share  = 715×13,500  =  Rs.6,300

 

 

Level-II

1.	A and B invest in a business in the ratio 3 : 2. If 5% of the total profit goes to charity and A’s share is Rs. 855, the total profit is:
	A. Rs. 1425 B. Rs. 1500 C. Rs. 1537.50 D. Rs. 1576 Answer:1 Option B Explanation: Let the total profit be Rs. 100. After paying to charity, A’s share = Rs. 95 x 3 = Rs. 57. 5 If A’s share is Rs. 57, total profit = Rs. 100. If A’s share Rs. 855, total profit = 100 x 855 = 1500
2.	A, B and C jointly thought of engaging themselves in a business venture. It was agreed that A would invest Rs. 6500 for 6 months, B, Rs. 8400 for 5 months and C, Rs. 10,000 for 3 months. A wants to be the working member for which, he was to receive 5% of the profits. The profit earned was Rs. 7400. Calculate the share of B in the profit.
	A. Rs. 1900 B. Rs. 2660 C. Rs. 2800 D. Rs. 2840 Answer: 2 Option B   Explanation: For managing, A received = 5% of Rs. 7400 = Rs. 370. Balance = Rs. (7400 – 370) = Rs. 7030. Ratio of their investments = (6500 x 6) : (8400 x 5) : (10000 x 3) = 39000 : 42000 : 30000 = 13 : 14 : 10  B’s share = Rs. 7030 x 14 = Rs. 2660. 37

	3 .A, B and C enter into a partnership in the ratio  :  : . After 4 months, A increases his share 50%. If the total profit at the end of one year be Rs. 21,600, then B’s share in the profit is:
	A. Rs. 2100 B. Rs. 2400 C. Rs. 3600 D.   Rs. 4000 Answer:3 Option D   Explanation: Ratio of initial investments = 7 : 4 : 6 = 105 : 40 : 36. 2 3 5 Let the initial investments be 105x, 40x and 36x.  A : B : C = 105x x 4 + 150 x 105x x 8 : (40x x 12) : (36x x 12) 100 = 1680x : 480x : 432x = 35 : 10 : 9. Hence, B’s share = Rs. 21600 x 10 = Rs. 4000. 54
4.	A, B, C subscribe Rs. 50,000 for a business. A subscribes Rs. 4000 more than B and B Rs. 5000 more than C. Out of a total profit of Rs. 35,000, A receives:
	A. Rs. 8400 B. Rs. 11,900 C. Rs. 13,600 D. Rs. 14,700   Answer:4 Option D   Explanation: Let C = x. Then, B = x + 5000 and A = x + 5000 + 4000 = x + 9000. So, x + x + 5000 + x + 9000 = 50000 3x = 36000 x = 12000 A : B : C = 21000 : 17000 : 12000 = 21 : 17 : 12.  A’s share = Rs. 35000 x 21 = Rs. 14,700. 50

5.	Three partners shared the profit in a business in the ratio 5 : 7 : 8. They had partnered for 14 months, 8 months and 7 months respectively. What was the ratio of their investments?
	A. 5 : 7 : 8 B. 20 : 49 : 64 C. 38 : 28 : 21 D. None of these   Answer:5 Option B   Explanation: Let their investments be Rs. x for 14 months, Rs. y for 8 months and Rs. z for 7 months respectively. Then, 14x : 8y : 7z = 5 : 7 : 8. Now, 14x = 5        98x = 40y        y = 49 x 8y 7 20   And, 14x = 5        112x = 35z        z = 112 x = 16 x. 7z 8 35 5    x : y : z = x : 49 x : 16 x = 20 : 49 : 64. 20 5
6.	A starts business with Rs. 3500 and after 5 months, B joins with A as his partner. After a year, the profit is divided in the ratio 2 : 3. What is B’s contribution in the capital?
	A. Rs. 7500 B. Rs. 8000 C. Rs. 8500 D. Rs. 9000 Answer:6 Option D   Explanation: Let B’s capital be Rs. x. Then, 3500 x 12 = 2 7x 3 14x = 126000 x = 9000.

7.	A and B entered into partnership with capitals in the ratio 4 : 5. After 3 months, A withdrew  of his capital and B withdrew  of his capital. The gain at the end of 10 months was Rs. 760. A’s share in this profit is:
	A. Rs. 330 B. Rs. 360 C. Rs. 380 D. Rs. 430 Answer:7 Option A       Explanation: A : B = 4x x 3 + 4x – 1 x 4x x 7 : 5x x 3 + 5x – 1 x 5x x 7 4 5 = (12x + 21x) : (15x + 28x) = 33x :43x = 33 : 43.    A’s share = Rs. 760 x 33 = Rs. 330. 76

8.	A and B started a partnership business investing some amount in the ratio of 3 : 5. C joined then after six months with an amount equal to that of B. In what proportion should the profit at the end of one year be distributed among A, B and C?
	A. 3 : 5 : 2 B. 3 : 5 : 5 C. 6 : 10 : 5 D. Data inadequate Answer:8 Option C   Explanation: Let the initial investments of A and B be 3x and 5x. A : B : C = (3x x 12) : (5x x 12) : (5x x 6) = 36 : 60 : 30 = 6 : 10 : 5.

9.	A, B, C rent a pasture. A puts 10 oxen for 7 months, B puts 12 oxen for 5 months and C puts 15 oxen for 3 months for grazing. If the rent of the pasture is Rs. 175, how much must C pay as his share of rent?
	A. Rs. 45 B. Rs. 50 C. Rs. 55 D. Rs. 60 Answer:9 Option A   Explanation: A : B : C = (10 x 7) : (12 x 5) : (15 x 3) = 70 : 60 : 45 = 14 : 12 : 9.  C’s rent = Rs. 175 x 9 = Rs. 45. 35

10.	A and B started a business in partnership investing Rs. 20,000 and Rs. 15,000 respectively. After six months, C joined them with Rs. 20,000. What will be B’s share in total profit of Rs. 25,000 earned at the end of 2 years from the starting of the business?
	A. Rs. 7500 B. Rs. 9000 C. Rs. 9500 D. Rs. 10,000 Answer: 10 Option A   Explanation: A : B : C = (20,000 x 24) : (15,000 x 24) : (20,000 x 18) = 4 : 3 : 3.  B’s share = Rs. 25000 x 3 = Rs. 7,500. 10

 

Ratio

Introduction:

Ratio is the relation which one quantity bears to another of the same kind. The ratio of two quantities a and b is the fraction a/b and we write it as a: b.

In the ratio a: b, we call a as the first term or antecedent and b, the second term or consequent.

 

Note: The multiplication or division of each term of a ratio by the same non- zero number does not affect the ratio.

 

Compound Ratio: – It is obtained by multiplying together the numerators for new numerator and denominators for new denominator.

 

 

Example 1. If the ratios are 4:3, 15:20, 2:6 and 3:5 find the compound ratio?

 

 

 

 

 

 

 

 

Example2. If we divide 4185 into two parts such that they are in ratio 7:2, then find the values of both the parts?

Sol 2. Let the actual variable be 7x and 2x.

So, the 1^st part = 7 ×465=3255

The 2^nd part = 2 ×465=930

 

 

Note:

The ratio of first , second and third quantities is given by

ac : bc : bd

 

If the ratio between first and second quantity is a:b and third and fourth is c:d .

Similarly, the ratio of first, second, third and fourth quantities is given by
ace : bce : bde : bdf
If the ratio between first and second quantity is a: b and third and fourth is c:d.

 

                                                 Proportion

 
Introduction:-
Four quantities are said to be proportional if the two ratios are equal i.e.  the A, B, C and D are proportion. It is denoted by “::” it is written as A : B : C : D where A and D are extremes and B and C are called means .
                             Product of the extreme = Product of the means

 

 

Direct proportion: – The two given quantities are so related that if one quantity increases (or decreases) then the other quantity also increases (or decreases).

Example 1. If 5 pens cost Rs 10 then 15 pen cost?

Sol 1. It is seen that if number of pens increases then cost also increases. So,

5 pens: 15 pens:: Rs 10 : required cost

 

 

Inverse proportion: – The two given quantities are so related that if one quantity increases (or decreases) then the other quantity also decreases (or increases).

Example 2.If 10 men can do a work in 20 days then in how many days 20 men can do that work?

Sol 2. Here if men increase then days should decrease, so this is a case of inverse proportion, so

10 men: 20 men :: required days : 20 days

 

 

Rule of three: It Is the method of finding 4th term of a proportion if all the other three are given, if ratio is a:b :: c:d then ,

 

 

 

                                             ALLIGATION

Introduction:-

The word allegation means linking. It is used to find:

1. The proportion in which the ingredients of given price are mixed to produce a new mixture at a given price.
2. The mean or average value of mixture when the price of the two or more ingredients and the proportion in which they are mixed are given.

Mathematical Formula:

 

For two ingredient:-

 

 

Example 1: If the rice at Rs 3.20 per kg and the rice at Rs 3.50 per kg be mixed then what should be their proportion so that the new mixture be worth Rs 3.35 per kg ?

Sol 1: CP of 1 kg of cheaper rice                          CP of 1 kg of dearer rice

Hence they must be mixed in equal proportion i.e. 1:1

 

 

Example 2: Find out the ratio of new mixture so that it will cost Rs 1.40 per kg from the given three kinds of rice costing Rs 1.20, Rs 1.45 and Rs 1.74?

 

Sol 2: 1^st rice cost = 120, 2^nd rice cost = 145 and 3^rd rice cost = 174 paisa.

From the above rule: we have,

Therefore, three rice must be mixed in 39: 20: 20 ratios to have a new mixture of rice.

 

 

Questions

Level-I

..	1.   A  and B together have Rs. 1210. If  of A’s amount is equal to  of B’s amount, how much amount does B have?
	A. Rs. 460 B. Rs. 484 C. Rs. 550 D. Rs. 664

 

2.	Two numbers are respectively 20% and 50% more than a third number. The ratio of the two numbers is:
	A. 2 : 5 B. 3 : 5 C. 4 : 5 D. 6 : 7

 

 

3.	A sum of money is to be distributed among A, B, C, D in the proportion of 5 : 2 : 4 : 3. If C gets Rs. 1000 more than D, what is B’s share?
	A. Rs. 500 B. Rs. 1500 C. Rs. 2000 D. None of these

 

 

 

4.	Seats for Mathematics, Physics and Biology in a school are in the ratio 5 : 7 : 8. There is a proposal to increase these seats by 40%, 50% and 75% respectively. What will be the ratio of increased seats?
	A. 2 : 3 : 4 B. 6 : 7 : 8 C. 6 : 8 : 9 D. None of these

 

 

5.	In a mixture 60 litres, the ratio of milk and water 2 : 1. If this ratio is to be 1 : 2, then the quanity of water to be further added is:
	A. 20 litres B. 30 litres C. 40 litres D. 60 litres
6.	The ratio of the number of boys and girls in a college is 7 : 8. If the percentage increase in the number of boys and girls be 20% and 10% respectively, what will be the new ratio?
	A. 8 : 9 B. 17 : 18 C. 21 : 22 D. Cannot be determined

 

7.	Salaries of Ravi and Sumit are in the ratio 2 : 3. If the salary of each is increased by Rs. 4000, the new ratio becomes 40 : 57. What is Sumit’s salary?
	A. Rs. 17,000 B. Rs. 20,000 C. Rs. 25,500 D. Rs. 38,000

 

8.	If 0.75 : x :: 5 : 8, then x is equal to:
	A. 1.12 B. 1.2 C. 1.25 D. 1.30

 

 

9.	The sum of three numbers is 98. If the ratio of the first to second is 2 :3 and that of the second to the third is 5 : 8, then the second number is:
	A. 20 B. 30 C. 48 D. 58

 

 

	10 .If Rs. 782 be divided into three parts, proportional to  :  : , then the first part is:
	A. Rs. 182 B. Rs. 190 C. Rs. 196 D. Rs. 204

 

 

 

 

Answers

1. Answer:Option B

 

Explanation:

	4	A	=	2	B
	15			5

 

A =		2	x	15	B
		5		4

 

A =	3	B
	2

 

	A	=	3
	B		2

A : B = 3 : 2.

B’s share = Rs.		1210 x	2		= Rs. 484.
			5

 

 

 

 

2 .Answer: Option C

 

Explanation:

Let the third number be x.

Then, first number = 120% of x =	120x	=	6x
	100		5

 

Second number = 150% of x =	150x	=	3x
	100		2

 

Ratio of first two numbers =		6x	:	3x		= 12x : 15x = 4 : 5.

 

 

3 .Answer: Option C

Explanation:

Let the shares of A, B, C and D be Rs. 5x, Rs. 2x, Rs. 4x and Rs. 3x respectively.

Then, 4x – 3x = 1000

x = 1000.

B’s share = Rs. 2x = Rs. (2 x 1000) = Rs. 2000.

 

 

4 .Answer: Option A

 

Explanation:

 

Originally, let the number of seats for Mathematics, Physics and Biology be 5x, 7x and 8x respectively.

 

Number of increased seats are (140% of 5x), (150% of 7x) and (175% of 8x).

 

		140	x 5x		,		150	x 7x		and		175	x 8x
		100					100					100

 

7x,	21x	and 14x.
	2

 

The required ratio = 7x :	21x	: 14x
	2

 

14x : 21x : 28x

 

2 : 3 : 4.

 

 

 

 

 

 

 

 

 

 

 

5 .Answer: Option D

 

Explanation:

Quantity of milk =		60 x	2	litres = 40 litres.
			3

Quantity of water in it = (60- 40) litres = 20 litres.

New ratio = 1 : 2

Let quantity of water to be added further be x litres.

 

 

Then, milk : water =		40		.
		20 + x

 

Now,		40		=	1
		20 + x			2

 

20 + x = 80

 

x = 60.

Quantity of water to be added = 60 litres.

 

6 .Answer: Option C

 

Explanation:

 

Originally, let the number of boys and girls in the college be 7x and 8x respectively.

 

Their increased number is (120% of 7x) and (110% of 8x).

 

		120	x 7x		and		110	x 8x
		100					100

 

	42x	and	44x
	5		5

 

The required ratio =		42x	:	44x		= 21 : 22

 

7 .Answer: Option D

 

Explanation:

Let the original salaries of Ravi and Sumit be Rs. 2x and Rs. 3x respectively.

 

Then,	2x + 4000	=	40
	3x + 4000		57

57(2x + 4000) = 40(3x + 4000)

 

6x = 68,000

 

3x = 34,000

 

Sumit’s present salary = (3x + 4000) = Rs.(34000 + 4000) = Rs. 38,000.

 

 

 

8 .Answer: Option B

 

Explanation:

(x x 5) = (0.75 x 8)    x =		6		= 1.20
		5

 

 

 

 

 

 

 

 

9 .Answer: Option B

 

Explanation:

Let the three parts be A, B, C. Then,

 

A : B = 2 : 3 and B : C = 5 : 8 =		5 x	3		:		8 x	3		= 3 :	24
			5					5			5

 

A : B : C = 2 : 3 :	24	= 10 : 15 : 24
	5

 

B =		98 x	15		= 30.
			49

 

 

 

10 .Answer: Option D

 

 

Explanation:

 

Given ratio =  :  :  = 6 : 8 : 9.

 

1st part = Rs.		782 x	6		= Rs. 204

 

 

 

 

Level-II

11.	The salaries A, B, C are in the ratio 2 : 3 : 5. If the increments of 15%, 10% and 20% are allowed respectively in their salaries, then what will be new ratio of their salaries?
	A. 3 : 3 : 10 B. 10 : 11 : 20 C. 23 : 33 : 60 D. Cannot be determined     Answer: Option C   Explanation: Let A = 2k, B = 3k and C = 5k. A’s new salary = 115 of 2k = 115 x 2k = 23k 100 100 10   B’s new salary = 110 of 3k = 110 x 3k = 33k 100 100 10   C’s new salary = 120 of 5k = 120 x 5k = 6k 100 100    New ratio 23k : 33k : 6k = 23 : 33 : 60 10 10
12.	If 40% of a number is equal to two-third of another number, what is the ratio of first number to the second number?
	A. 2 : 5 B. 3 : 7 C. 5 : 3 D. 7 : 3 Answer: Option C   Explanation: Let 40% of A = 2 B 3   Then, 40A = 2B 100 3   2A = 2B 5 3   A = 2 x 5 = 5 B 3 2 3 A : B = 5 : 3.

 

13.	The fourth proportional to 5, 8, 15 is:
	A. 18 B. 24 C. 19 D. 20     Answer: Option B   Explanation: Let the fourth proportional to 5, 8, 15 be x. Then, 5 : 8 : 15 : x 5x = (8 x 15)   x = (8 x 15) = 24. 5

 

14.	Two number are in the ratio 3 : 5. If 9 is subtracted from each, the new numbers are in the ratio 12 : 23. The smaller number is:
	A. 27 B. 33 C. 49 D. 55 Answer: Option B   Explanation: Let the numbers be 3x and 5x. Then, 3x – 9 = 12 5x – 9 23 23(3x – 9) = 12(5x – 9) 9x = 99 x = 11. The smaller number = (3 x 11) = 33.

 

15.	In a bag, there are coins of 25 p, 10 p and 5 p in the ratio of 1 : 2 : 3. If there is Rs. 30 in all, how many 5 p coins are there?
	A. 50 B. 100 C. 150 D. 200 Answer: Option C   Explanation: Let the number of 25 p, 10 p and 5 p coins be x, 2x, 3x respectively. Then, sum of their values = Rs. 25x + 10 x 2x + 5 x 3x = Rs. 60x 100 100 100 100   60x = 30     x = 30 x 100 = 50. 100 60 Hence, the number of 5 p coins = (3 x 50) = 150.

 

 

 

Introduction:

There are four main directions – East, West, North and South as shown below:

 

 

 

 

There are four cardinal directions – North-East (N-E), North-West (N-W), South-East (S-E), and South-West (S-W) as shown below:

 

 

 

Key points

 

1. At the time of sunrise if a man stands facing the east, his shadow will be towards west.
2. At the time of sunset the shadow of an object is always in the east.
3. If a man stands facing the North, at the time of sunrise his shadow will be towards his left and at the time of sunset it will be towards his right.
4. At 12:00 noon, the rays of the sun are vertically downward hence there will be no shadow

 

 

 

 

 

 

 

 

 

Practice Questions

Type 1:

Siva starting from his house, goes 5 km in the East, then he turns to his left and goes 4 km. Finally he turns to his left and goes 5 km. Now how far is he from his house and in what direction?

Solution:

From third position it is clear he is 4 km from his house and is in North direction.

 

 

 

 

 

 

Type 2:

Suresh starting from his house, goes 4 km in the East, then he turns to his right and goes 3 km. What minimum distance will be covered by him to come back to his house?

Solution:

 

Type 3:

One morning after sunrise Juhi while going to school met Lalli at Boring road crossing. Lalli’s shadow was exactly to the right of Juhi. If they were face to face, which direction was Juhi facing?

 

Solution: In the morning sunrises in the east.

So in morning the shadow falls towards the west.

Now Lalli’s shadow falls to the right of the Juhi. Hence Juhi is facing South.

 

 

 

 

Type 4: Hema starting from her house walked 5 km to reach the crossing of Palace. In which direction she was going, a road opposite to this direction goes to Hospital. The road to the right goes to station. If the road which goes to station is just opposite to the road which IT-Park, then in which direction to Hema is the road which goes to IT-Park?

Solution:

From II it is clear that the road which goes to IT-Park is left to Hema.

 

 

 

 

 

 

 

 

Questions

 

Level-1

 

1.	One morning Udai and Vishal were talking to each other face to face at a crossing. If Vishal’s shadow was exactly to the left of Udai, which direction was Udai facing?
	A. East B. West C. North D. South

2.	Y is in the East of X which is in the North of Z. If P is in the South of Z, then in which direction of Y, is P?
	A. North B. South C. South-East D. None of these

3.	If South-East becomes North, North-East becomes West and so on. What will West become?
	A. North-East B. North-West C. South-East D. South-West

4.	A man walks 5 km toward south and then turns to the right. After walking 3 km he turns to the left and walks 5 km. Now in which direction is he from the starting place?
	A. West B. South C. North-East D. South-West

5.	Rahul put his timepiece on the table in such a way that at 6 P.M. hour hand points to North. In which direction the minute hand will point at 9.15 P.M. ?
	A. South-East   B. South   C. North D. West
6.	Rasik walked 20 m towards north. Then he turned right and walks 30 m. Then he turns right and walks 35 m. Then he turns left and walks 15 m. Finally he turns left and walks 15 m. In which direction and how many metres is he from the starting position?
	A. 15 m West B. 30 m East C. 30 m West D. 45 m East

7.

Two cars start from the opposite places of a main road, 150 km apart. First car runs for 25 km and takes a right turn and then runs 15 km. It then turns left and then runs for another 25 km and then takes the direction back to reach the main road. In the mean time, due to minor break down the other car has run only 35 km along the main road. What would be the distance between two cars at this point?

A. 65 km B. 75 km C. 80 km D. 85 km

8.	Starting from the point X, Jayant walked 15 m towards west. He turned left and walked 20 m. He then turned left and walked 15 m. After this he turned to his right and walked 12 m. How far and in which directions is now Jayant from X?
	A. 32 m, South B. 47 m, East C. 42 m, North D. 27 m, South

 

9.	One evening before sunset Rekha and Hema were talking to each other face to face. If Hema’s shadow was exactly to the right of Hema, which direction was Rekha facing?
	A. North B. South C. East D. Data is inadequate

10.	A boy rode his bicycle Northward, then turned left and rode 1 km and again turned left and rode 2 km. He found himself 1 km west of his starting point. How far did he ride northward initially?
	A. 1 km B. 2 km C. 3 km D. 5 km

 

 

Answers:

1Answer: Option C

Explanation:

 

2Answer: Option D

Explanation:

P is in South-West of Y.

 

3Answer: Option C

Explanation:

It is clear from the diagrams that new name of West will become South-East.

 

4Answer: Option D

Explanation:

Hence required direction is South-West.

 

5Answer: Option D

Explanation:

At 9.15 P.M., the minute hand will point towards west.

 

6Answer: Option D

Explanation:

 

7Answer: Option A

Explanation:

 

 

 

 

8Answer: Option A

Explanation:

 

9Answer: Option B

Explanation:

In the evening sun sets in West. Hence then any shadow falls in the East. Since Hema’s shadow was to the right of Hema. Hence Rekha was facing towards South.

 

10Answer: Option B

Explanation:

The boy rode 2 km. Northward

 

 

Level – 2

 

Dev, Kumar, Nilesh, Ankur and Pintu are standing facing to the North in a playground such as given below:

1. Kumar is at 40 m to the right of Ankur.
2. Dev is are 60 m in the south of Kumar.
3. Nilesh is at a distance of 25 m in the west of Ankur.
4. Pintu is at a distance of 90 m in the North of Dev

 

 

1.	Which one is in the North-East of the person who is to the left of Kumar?
	A. Dev B. Nilesh C. Ankur D. Pintu

2.	If a boy starting from Nilesh, met to Ankur and then to Kumar and after this he to Dev and then to Pintu and whole the time he walked in a straight line, then how much total distance did he cover?
	A. 215 m B. 155 m C. 245 m D.   185 m

 

 

Directions to Solve

Each of the following questions is based on the following information:

1. Six flats on a floor in two rows facing North and South are allotted to P, Q, R, S, T and U.
2. Q gets a North facing flat and is not next to S.
3. S and U get diagonally opposite flats.
4. R next to U, gets a south facing flat and T gets North facing flat.

 

 

3.	If the flats of P and T are interchanged then whose flat will be next to that of U?
	A. P B. Q C. R D. T

4.	Which of the following combination get south facing flats?
	A. QTS B. UPT C. URP D. Data is inadequate

5.	The flats of which of the other pair than SU, is diagonally opposite to each other?
	A. QP B. QR C. PT D. TS
6.	Whose flat is between Q and S?
	A. T B. U C. R D. P

 

 

Directions to Solve

Each of the following questions is based on the following information:

1. 8-trees → mango, guava, papaya, pomegranate, lemon, banana, raspberry and apple are in two rows 4 in each facing North and South.
2. Lemon is between mango and apple but just opposite to guava.
3. Banana is at one end of a line and is just next in the right of guava or either banana tree is just after guava tree.
4. Raspberry tree which at one end of a line, is just diagonally opposite to mango tree.

 

 

	7 .Which of the following statements is definitely true?
	A. Papaya tree is just near to apple tree. B. Apple tree is just next to lemon tree. C. Raspberry tree is either left to Pomegranate or after. D. Pomegranate tree is diagonally opposite to banana tree.

8	Which tree is just opposite to raspberry tree?
	A. Papaya B. Pomegranate C. Papaya or Pomegranate D. Data is inadequate
9	Which tree is just opposite to banana tree?
	A. Mango B. Pomegranate C. Papaya D. Data is inadequate

 

 

Answer: 1 Option D

Explanation:

Ankur is in the left of Kumar. Hence Pintu is in North-East of Ankur

 

 

Answer: 2 Option A

Explanation:

Required distance = 25 m + 40 m + 60 m + 90 m

Required distance = 215 m

 

 

Answer:3 Option C

Explanation:

Hence flat R will be next to U.

 

 

Answer:4 Option C

 

Explanation:

Hence URP flat combination get south facing flats.

 

Answer:5 Option A

 

Explanation:

Hence QP is diagonally opposite to each other.

 

 

 

 

 

 

Answer:6 Option A

 

Explanation:

Hence flat T is between Q and S.

 

Answer: 7 Option B

 

Explanation:

 

 

Answer:8 Option C

 

Explanation:

 

 

 

 

 

 

 

Answer:9 Option A

 

Explanation:

 

In this module we will deal with basic concepts of time and distance, speed, average speed, conversion from km/h to m/s and vice versa. This chapter will form the basis of further concept of relative speed which is used in train and boat problems.

Important Formulas

1. Speed=Distance/Time
2. Distance=Speed×Time
3. Time=Distance/Speed
4. To convert Kilometers per Hour(km/hr) to Meters per Second(m/s)
   x km/hr=(x×5)/18m/s
5. To convert Meters per Second(m/s) to Kilometers per Hour(km/hr)
   x m/s=(x×18)/5 km/hr
6. If a car covers a certain distance at x kmph and an equal distance at y kmph, the average speed of the whole journey = 2xy/(x+y) kmph
7. Speed and time are inversely proportional (when distance is constant) ⇒Speed ∝1/Time (when distance is constant)
8. If the ratio of the speeds of A and B is a : b, then the ratio of the times taken by them to cover the same distance is 1/a:1/b or b : a

Solved Examples

Level 1

1.A person crosses a 600 m long street in 5 minutes. What is his speed in km per hour?
A. 8.2	B. 4.2
C. 6.1	D. 7.2

 

Answer : Option D

Explanation :

Distance = 600 meter

time = 5 minutes = 5 x 60 seconds = 300 seconds

Speed = distance/time=600/300=2m/s=(2×18)/5 km/hr=36/5 km/hr=7.2 km/hr

2.Two boys starts from the same place walking at the rate of 5 kmph and 5.5 kmph respectively in the same direction. What time will they take to be 8.5 km apart?
A. 17 hr	B. 14 hr
C. 12 hr	D. 19 hr

 

Answer : Option A

Explanation :

Relative speed = 5.5 – 5 = .5 kmph (because they walk in the same direction)

distance = 8.5 km

Time = distance/speed=8.5/.5=17 hr.

3.Walking 6/7th of his usual speed, a man is 12 minutes too late. What is the usual time taken by him to cover that distance?
A. 1 hr 42 min	B. 1 hr
C. 2 hr	D. 1 hr 12 min

 

Answer : Option D

Explanation :

New speed = 6/7 of usual speed
Speed and time are inversely proportional.
Hence new time = 7/6 of usual time
Hence, 7/6 of usual time – usual time = 12 minutes
=>1/6 of usual time = 12 minutes => usual time = 12 x 6 = 72 minutes = 1 hour 12 minutes

4.A man goes to his office from his house at a speed of 3 km/hr and returns at a speed of 2 km/hr. If he takes 5 hours in going and coming, what is the distance between his house and office?
A. 3 km	B. 4 km
C. 5 km	D. 6 km

 

Answer : Option D

Explanation :

If a car covers a certain distance at x kmph and an equal distance at y kmph,the average speed of the whole journey = 2xy/(x+y) kmph

Hence, average speed = (2×3×2)/(2+3)=12/5 km/hr .

Total time taken = 5 hours

⇒Distance travelled=(12/5)×5=12 km

⇒Distance between his house and office =12/2=6 km

5.If a person walks at 14 km/hr instead of 10 km/hr, he would have walked 20 km more. What is the actual distance travelled by him?
A. 80 km	B. 70 km
C. 60 km	D. 50 km

 

Answer : Option D

Explanation :

Assume that the person would have covered x km if travelled at 10 km/hr

⇒Speed = Distance/Time=x/10….. (Equation1)

Give that the person would have covered (x + 20) km if travelled at 14 km/hr
⇒Speed = Distance/Time=(x+20)/14….. (Equation2)

From Equations 1 and 2,
X/10=(x+20)/14⇒14x=10x+200⇒4x=200⇒x=200/4=50

6.A car travels at an average of 50 miles per hour for 212 hours and then travels at a speed of 70 miles per hour for 112 hours. How far did the car travel in the entire 4 hours?
A. 210 miles	B. 230 miles
C. 250 miles	D. 260 miles

 

Answer : Option B

Explanation :

Speed1 = 50 miles/hour

Time1 = 2*(1/2) hour=5/2 hour

⇒Distance1 = Speed1 × Time1 = (50×5)/2=25×5=125 miles

⇒Speed2 = 70 miles/hour

Time2 = 1*1/2 hour=3/2 hour

Distance2 = Speed2 × Time2 = 70×3/2=35×3=105 miles

Total Distance = Distance1 + Distance2 =125+105=230 miles

7.Sound is said to travel in air at about 1100 feet per second. A man hears the axe striking the tree, 11/5 seconds after he sees it strike the tree. How far is the man from the wood chopper?
A. 1800 ft	B. 2810 ft
C. 3020 ft	D. 2420 ft

 

Answer : Option D

Explanation :

Speed of the sound = 1100 ft/s ⇒Time = ¹¹/₅ second

Distance = Speed × Time = 1100 ×11/5=220×11=2420 ft

8.A man walking at the rate of 5 km/hr crosses a bridge in 15 minutes. What is the length of the bridge (in meters)?
A. 1250	B. 1280
C. 1320	D. 1340

 

Answer : Option A

Explanation :

Speed = 5 km/hr

Time = 15 minutes = 1/4 hour

Length of the bridge = Distance Travelled by the man

= Speed × Time = 5×1/4 km

=5×1/4×1000 metre=1250 metre

Level 2

1.A man takes 5 hours 45 min in walking to a certain place and riding back. He would have gained 2 hours by riding both ways. The time he would take to walk both ways is
A. 11 hrs	B. 8 hrs 45 min
C. 7 hrs 45 min	D. 9 hts 20 min

 

Answer : Option C

Explanation :

Given that time taken for riding both ways will be 2 hours lesser than the time needed for waking one way and riding back From this, we can understand that time needed for riding one way = time needed for waking one way – 2 hours Given that time taken in walking one way and riding back = 5 hours 45 min Hence The time he would take to walk both ways = 5 hours 45 min + 2 hours = 7 hours 45 min In fact, you can do all these calculations mentally and save a lot of time which will be a real benefit for you. 2.A man complete a journey in 10 hours. He travels first half of the journey at the rate of 21 km/hr and second half at the rate of 24 km/hr. Find the total journey in km.
A. 121 km	B. 242 km
C. 224 km	D. 112 km

 

Answer : Option C

Explanation :

distance = speed x time
Let time taken to travel the first half = x hr
then time taken to travel the second half = (10 – x) hr
Distance covered in  the first half = 21x
Distance covered in  the second half = 24(10 – x)
But distance covered in  the first half = Distance covered in the second half
=> 21x = 24(10 – x) => 21x = 240 – 24x => 45x = 240 => 9x = 48 => 3x = 16⇒x=16/3

Hence Distance covered in the first half = 21x=21×16/3=7×16=112 km. Total distance = 2×112=224 km

3.A car traveling with 5/7 of its actual speed covers 42 km in 1 hr 40 min 48 sec. What is the actual speed of the car?
A. 30 km/hr	B. 35 km/hr
C. 25 km/hr	D. 40 km/hr

 

Answer : Option B

Explanation :

Time = 1 hr 40 min 48 sec = 1hr +40/60hr+48/3600hr=1+2/3+1/75=126/75hr

Distance = 42 kmSpeed=distance/time=42(126/75) =42×75/126

⇒5/7 of the actual speed = 42×75/126

⇒actual speed = 42×75/126×7/5=42×15/18=7×15/3=7×5=35 km/hr

4.A man covered a certain distance at some speed. If he had moved 3 kmph faster, he would have taken 40 minutes less. If he had moved 2 kmph slower, he would have taken 40 minutes more. What is the the distance in km?
A. 36	B. 38
C. 40	D. 42

 

Answer : Option C

Explanation :

Let the distance be x km , the speed in which he moved = v kmph

Time taken when moving at normal speed – time taken when moving 3 kmph faster = 40 minutes

⇒x/v−x/(v+3)=40/60⇒x[1/v−1/(v+3)]=2/3⇒x[(v+3−v)/v(v+3)]=2/3

⇒2v(v+3)=9x…………….(Equation1)

Time taken when moving 2 kmph slower – Time taken when moving at normal speed = 40 minutes
⇒x/(v−2)−x/v=40/60⇒x[1/(v−2)−1/v]=2/3

⇒x[(v−v+2)/v(v−2)]=2/3⇒x[2/v(v−2)]=2/3

⇒x[1/v(v−2)]=1/3⇒v(v−2)=3x…………….(Equation2)

Equation1/Equation2

⇒2(v+3)/(v−2)=3⇒2v+6=3v−6⇒v=12

Substituting this value of v inEquation1⇒2×12×15=9x

=>x= (2×12×15)/9= (2×4×15)/3=2×4×5=40. Hence distance = 40 km

5.In covering a distance of 30 km, Arun takes 2 hours more than Anil. If Arun doubles his speed, then he would take 1 hour less than Anil. What is Arun’s speed?
A. 8 kmph	B. 5 kmph
C. 4 kmph	D. 7 kmph

 

Answer : Option B

Explanation :

Let the speed of Arun = x kmph and the speed of Anil = y kmph
distance = 30 km

We know that distance/speed=time. Hence, 30/x−30/y=2………..(Equation1)

30/y−30/2x=1………..(Equation2)

Equation1 + Equation2⇒30/x−30/2x=3⇒30/2x=3⇒15/x=3⇒5/x=1⇒x=5. Hence Arun’s speed = 5 kmph

6.A car travels first 160 km at 64 km/hr and the next 160 km at 80 km/hr. What is the average speed for the first 320 km of the tour?
A. 70.24 km/hr	B. 74. 24 km/hr
C. 71.11 km/hr	D. 72.21 km/hr

 

Answer : Option C

Explanation :

If a car covers a certain distance at x kmph and an equal distance at y kmph,the average speed of the whole journey = 2xy/(x+y) kmph.

By using the same formula, we can find out the average speed quickly average speed = (2×64×80)/(64+80)=(2×64×80)/144⇒ (2×32×40)/36= (2×32×10)/9⇒ (64×10)/9=71.11 kmph

7.A man rides his bicycle 10 km at an average speed of 12 km/hr and again travels 12 km at an average speed of 10 km/hr. What is his average speed for the entire trip approximately?
A. 11.2 kmph	B. 10 kmph
C. 10.2 kmph	D. 10.8 kmph

 

Answer : Option D

Explanation :

Total distance travelled = 10 + 12 = 22 km

Time taken to travel 10 km at an average speed of 12 km/hr = distance/speed=10/12 hr

Time taken to travel 12 km at an average speed of 10 km/hr = distance/speed=12/10 hr

Total time taken =10/12+12/10 hr

Average speed = distance/time=22/(10/12+12/10)=(22×120)/{(10×10)+(12×12)}

(22×120)/244=(11×120)/122=(11×60)/61=660/61≈10.8 kmph

8.An airplane covers a certain distance at a speed of 240 kmph in 5 hours. To cover the same distance in 123 hours, it must travel at a speed of:
A. 660 km/hr	B. 680 km/hr
C. 700 km/hr	D. 720 km/hr

 

Answer : Option D

Explanation :

Speed and time are inversely proportional ⇒Speed ∝ 1/Time (when distance is constant)

Here distance is constant and Speed and time are inversely proportional

Speed ∝ 1/Time⇒Speed1/Speed2=Time2/Time1

⇒240/Speed2=(1*2/3)5⇒240/Speed2=(5/3)/5⇒240/Speed2=1/3⇒Speed2=240×3=720 km/hr

9.A train can travel 50% faster than a car. Both start from point A at the same time and reach point B 75 kms away from A at the same time. On the way, however, the train lost about 12.5 minutes while stopping at the stations. What is the speed of the car?
A. 80 kmph	B. 102 kmph
C. 120 kmph	D. 140 kmph

 

Answer : Option C

Explanation :

Let speed of the car = x kmph

Then speed of the train = x *(100+50)/100=150 x /100=3 x /2 kmph

Time taken by the car to travel from A to B=75/x hours

Time taken by the train to travel from A to B=75/(3 x /2)+12.5/60 hours

Since both start from A at the same time and reach point B at the same time

75/x=75/(3 x /2)+12.5/60⇒25/x=12.5/60⇒x=(25×60)/12.5=2×60=120

TIME AND WORK

In these problems the number of persons, quantity of work done and time taken are important factors. Also time taken by a person depends on the efficiency of that person which comes into picture when different people do the work such as women, children do the work alongside the men. The problems related to time and work can be solved by two major approaches – ratio & proportions and unitary method. Let us proceed to find some formulae related to these questions.

Important Formulas – Time and Work

• If A can do a piece of work in n days, work done by A in 1 day = 1/n

 

• If A does 1/n work in a day, A can finish the work in n days

 

• If M1 men can do W1 work in D1 days working H1 hours per day and M2 men can do W2 work in D2 days working H2 hours per day (where all men work at the same rate), then

M1 D1 H1 / W1 = M2 D2 H2 / W2

 

• If A can do a piece of work in p days and B can do the same in q days, A and B together can finish it in pq / (p+q) days

 

• If A is thrice as good as B in work, then

Ratio of work done by A and B = 3:1

Ratio of time taken to finish a work by A and B = 1: 3

 

SOLVED EXAMPLES

Level 1

1.P is able to do a piece of work in 15 days and Q can do the same work in 20 days. If they can work together for 4 days, what is the fraction of work left?
A. 8/15	B. 7/15
C. 11/15	D. 2/11

 

Answer : Option A

Explanation :

Amount of work P can do in 1 day = 1/15

Amount of work Q can do in 1 day = 1/20

Amount of work P and Q can do in 1 day = 1/15 + 1/20 = 7/60

Amount of work P and Q can together do in 4 days = 4 × (7/60) = 7/15

Fraction of work left = 1 – 7/15= 8/15

2.A can do a piece of work in 4 hours . A and C together can do it in just 2 hours, while B and C together need 3 hours to finish the same work. B alone can complete the work in — hours.
A. 12 hours	B. 6 hours
C. 8 hours	D. 10 hours

 

Answer : Option A

Explanation :

Work done by A in 1 hour = 1/4

Work done by B and C in 1 hour = 1/3

Work done by A and C in 1 hour = 1/2

Work done by A,B and C in 1 hour = 1/4+1/3 = 7/12

Work done by B in 1 hour = 7/12 – 1/2 = 1/12

=> B alone can complete the work in 12 hours

3.A completes 80% of a work in 20 days. Then B also joins and A and B together finish the remaining work in 3 days. How long does it need for B if he alone completes the work?
A. 37 ½ days	B. 22 days
C. 31 days	D. 22 days

 

Answer : Option A

Explanation :

Work done by A in 20 days = 80/100 = 8/10 = 4/5

Work done by A in 1 day = (4/5) / 20 = 4/100 = 1/25 — (1)

Work done by A and B in 3 days = 20/100 = 1/5 (Because remaining 20% is done in 3 days by A and B)

Work done by A and B in 1 day = 1/15 —(2)

Work done by B in 1 day = 1/15 – 1/25 = 2/75

=> B can complete the work in 75/2 days = 37 ½ days

4.P can finish a work in 18 days. Q can finish the same work in 15 days. Q worked for 10 days and left the job. How many days does P alone need to finish the remaining work?
A. 8	B. 5
C. 4	D. 6

 

Answer : Option D

Explanation :

Work done by P in 1 day = 1/18

Work done by Q in 1 day = 1/15

Work done by Q in 10 days = 10/15 = 2/3

Remaining work = 1 – 2/3 = 1/3

Number of days in which P can finish the remaining work = (1/3) / (1/18) = 6

5.Anil and Suresh are working on a special assignment. Anil needs 6 hours to type 32 pages on a computer and Suresh needs 5 hours to type 40 pages. If both of them work together on two different computers, how much time is needed to type an assignment of 110 pages?
A. 7 hour 15 minutes	B. 7 hour 30 minutes
C. 8 hour 15 minutes	D. 8 hour 30 minutes

 

Answer : Option C

Explanation :

Pages typed by Anil in 1 hour = 32/6 = 16/3

Pages typed by Suresh in 1 hour = 40/5 = 8

Pages typed by Anil and Suresh in 1 hour = 16/3 + 8 = 40/3

Time taken to type 110 pages when Anil and Suresh work together = 110 × 3 /40 = 33/4

= 8 ¼ hours = 8 hour 15 minutes

6.P works twice as fast as Q. If Q alone can complete a work in 12 days, P and Q can finish the work in — days
A. 1	B. 2
C. 3	D. 4

 

Answer : Option D

Explanation :

Work done by Q in 1 day = 1/12

Work done by P in 1 day = 2 × (1/12) = 1/6

Work done by P and Q in 1 day = 1/12 + 1/6 = ¼

=> P and Q can finish the work in 4 days

7.A work can be finished in 16 days by twenty women. The same work can be finished in fifteen days by sixteen men. The ratio between the capacity of a man and a woman is
A. 1:3	B. 4:3
C. 2:3	D. 2:1

 

Answer : Option B

Explanation :

Work done by 20 women in 1 day = 1/16 Work done by 1 woman in 1 day = 1/(16×20) Work done by 16 men in 1 day = 1/15 Work done by 1 man in 1 day = 1/(15×16) 8.P,Q and R together earn Rs.1620 in 9 days. P and R can earn Rs.600 in 5 days. Q and R in 7 days can earn Rs.910. How much amount does R can earn per day?
A. Rs.40	B. Rs.70
C. Rs.90	D. Rs.100

 

Answer : Option B

Explanation :

Amount Earned by P,Q and R in 1 day = 1620/9 = 180 —(1)

Amount Earned by P and R in 1 day = 600/5 = 120 —(2)

Amount Earned by Q and R in 1 day = 910/7 = 130 —(3)

(2)+(3)-(1) => Amount Earned by P , Q and 2R in 1 day

– Amount Earned by P,Q and R in 1 day = 120+130-180 = 70

=>Amount Earned by R in 1 day = 70
Ratio of the capacity of a man and woman =1/(15×16) : 1/(16×20) = 1/15 : 1/20

= 1/3 :1/4 = 4:3

Level 2

1.P, Q and R can do a work in 20, 30 and 60 days respectively. How many days does it need to complete the work if P does the work and he is assisted by Q and R on every third day?
A. 10 days	B. 14 days
C. 15 days	D. 9 days

 

Answer : Option C

Explanation :

Amount of work P can do in 1 day = 1/20

Amount of work Q can do in 1 day = 1/30

Amount of work R can do in 1 day = 1/60

P is working alone and every third day Q and R is helping him

Work completed in every three days = 2 × (1/20) + (1/20 + 1/30 + 1/60) = 1/5

So work completed in 15 days = 5 × 1/5 = 1

Ie, the work will be done in 15 days

2.A is thrice as good as B in work. A is able to finish a job in 60 days less than B. They can finish the work in – days if they work together.
A. 18 days	B. 22 ½ days
C. 24 days	D. 26 days

 

Answer : Option B

Explanation :

If A completes a work in 1 day, B completes the same work in 3 days

Hence, if the difference is 2 days, B can complete the work in 3 days

=> if the difference is 60 days, B can complete the work in 90 days

=> Amount of work B can do in 1 day= 1/90

Amount of work A can do in 1 day = 3 × (1/90) = 1/30

Amount of work A and B can together do in 1 day = 1/90 + 1/30 = 4/90 = 2/45

=> A and B together can do the work in 45/2 days = 22 ½ days

3.P can do a work in the same time in which Q and R together can do it. If P and Q work together, the work can be completed in 10 days. R alone needs 50 days to complete the same work. then Q alone can do it in A. 30 days B. 25 days C. 20 days D. 15 days   Answer : Option B Explanation : Work done by P and Q in 1 day = 1/10 Work done by R in 1 day = 1/50 Work done by P, Q and R in 1 day = 1/10 + 1/50 = 6/50 But Work done by P in 1 day = Work done by Q and R in 1 day . Hence the above equation can be written as Work done by P in 1 day × 2 = 6/50 => Work done by P in 1 day = 3/50 => Work done by Q and R in 1 day = 3/50 Hence work done by Q in 1 day = 3/50 – 1/50 = 2/50 = 1/25 So Q alone can do the work in 25 days 4.6 men and 8 women can complete a work in 10 days. 26 men and 48 women can finish the same work in 2 days. 15 men and 20 women can do the same work in – days.
A. 4 days	B. 6 days
C. 2 days	D. 8 days

 

Answer : Option A

Explanation :

Let work done by 1 man in 1 day = m and work done by 1 woman in 1 day = b

Work done by 6 men and 8 women in 1 day = 1/10

=> 6m + 8b = 1/10

=> 60m + 80b = 1 — (1)

Work done by 26 men and 48 women in 1 day = 1/2

=> 26m + 48b = ½

=> 52m + 96b = 1— (2)

Solving equation 1 and equation 2. We get m = 1/100 and b = 1/200

Work done by 15 men and 20 women in 1 day

= 15/100 + 20/200 =1/4

=> Time taken by 15 men and 20 women in doing the work = 4 days

5.Machine P can print one lakh books in 8 hours. Machine Q can print the same number of books in 10 hours while machine R can print the same in 12 hours. All the machines started printing at 9 A.M. Machine P is stopped at 11 A.M. and the remaining two machines complete work. Approximately at what time will the printing of one lakh books be completed?
A. 3 pm	B. 2 pm
C. 1:00 pm	D. 11 am

 

Answer : Option C

Explanation :

Work done by P in 1 hour = 1/8

Work done by Q in 1 hour = 1/10

Work done by R in 1 hour = 1/12

Work done by P,Q and R in 1 hour = 1/8 + 1/10 + 1/12 = 37/120

Work done by Q and R in 1 hour = 1/10 + 1/12 = 22/120 = 11/60

From 9 am to 11 am, all the machines were operating.

Ie, they all operated for 2 hours and work completed = 2 × (37/120) = 37/60.

6.A can complete a work in 12 days with a working of 8 hours per day. B can complete the same work in 8 days when working 10 hours a day. If A and B work together, working 8 hours a day, the work can be completed in — days.
A. 5 5⁄11	B. 4 5⁄11
C. 6 4⁄11	D. 6 5⁄11

 

Answer : Option A

Explanation :

A can complete the work in 12 days working 8 hours a day

=> Number of hours A can complete the work = 12×8 = 96 hours

=> Work done by A in 1 hour = 1/96

B can complete the work in 8 days working 10 hours a day

=> Number of hours B can complete the work = 8×10 = 80 hours => Work done by B in 1 hour = 1/80

Work done by A and B in 1 hour = 1/96 + 1/80 = 11/480 => A and B can complete the work in 480/11 hours. A and B works 8 hours a day.

Hence total days to complete the work with A and B working together = (480/11)/ (8) = 60/11 days = 5 ⁵⁄₁₁ days

Pending work = 1- 37/60 = 23/60

Hours taken by Q an R to complete the pending work = (23/60) / (11/60) = 23/11

which is approximately equal to 2. Hence the work will be completed approximately 2 hours after 11 am ; ie around 1 pm

7.If daily wages of a man is double to that of a woman, how many men should work for 25 days to earn Rs.14400? Given that wages for 40 women for 30 days are Rs.21600.
A. 12	B. 14
C. 16	D. 18

 

Answer : Option C

Explanation :

Wages of 1 woman for 1 day = 21600/(40×30)

Wages of 1 man for 1 day = (21600×2)/(40×30)

Wages of 1 man for 25 days = (21600×2×25)/(40×30)

Number of men = 14400/(21600×2×25)/(40×30)=144/(216×50)/40×30)=144/9=16

8.There is a group of persons each of whom can complete a piece of work in 16 days, when they are working individually. On the first day one person works, on the second day another person joins him, on the third day one more person joins them and this process continues till the work is completed. How many days are needed to complete the work?
A. 3 1⁄4 days	B. 4 1⁄3 days
C. 5 1⁄6 days	D. 6 1⁄5 days

 

Answer : Option C

Explanation :

Work completed in 1st day = 1/16

Work completed in 2nd day = (1/16) + (1/16) = 2/16

Work completed in 3rd day = (1/16) + (1/16) + (1/16) = 3/16

An easy way to attack such problems is from the choices. You can see the choices are

very close to each other. So just see one by one.

For instance, The first choice given in 3 ¹⁄₄

The work done in 3 days = 1/16 + 2/16 + 3/16 = (1+2+3)/16 = 6/16

The work done in 4 days = (1+2+3+4)/16 = 10/16

The work done in 5 days = (1+2+3+4+5)/16 = 15/16, almost close, isn’t it?

The work done in 6 days = (1+2+3+4+5+6)/16 > 1

Hence the answer is less than 6, but greater than 5. Hence the answer is 5 ¹⁄₆ days.

(Just for your reference, work done in 5 days = 15/16)

Pending work in 6th day = 1 – 15/16 = 1/16.

In 6th day, 6 people are working and work done = 6/16.

To complete the work 1/16, time required = (1/16) / (6/16) = 1/6 days.

Hence total time required = 5 + 1/6 = 5 ¹⁄₆ days

 

Introduction

Money is not free and it costs to borrow the money. Normally, the borrower has to pay an extra amount in addition to the amount he had borrowed. i.e, to repay the loan, the borrower has to pay the sum borrowed and the interest.

Lender and Borrower

The person giving the money is called the lender and the person taking the money is the borrower.

Principal (sum)

Principal (or the sum) is the money borrowed or lent out for a certain period. It is denoted by P.

Interest

Interest is the extra money paid by the borrower to the owner (lender) as a form of compensation for the use of the money borrowed.

Simple Interest (SI)

If the interest on a sum borrowed for certain period is calculated uniformly, it is called simple interest(SI).

Amount (A)

The total of the sum borrowed and the interest is called the amount and is denoted by A

• The statement “rate of interest 10% per annum” means that the interest for one year on a sum of Rs.100 is Rs.10. If not stated explicitly, rate of interest is assumed to be for one year.

 

• Let Principal = P, Rate = R% per annum and Time = T years. Then

  Simple Interest, SI = PRT/100

 

• From the above formula , we can derive the followings

  P=100×SI/RT

  R=100×SI/PT

  T=100×SI/PR

 

Some Formulae

1. If a sum of money becomes n times in T years at simple interest, then the rate of interest per annum can be given be R = 100(n−1)/T %
2. The annual instalment which will discharge a debt of D due in T years at R% simple interest per annum =100D/ (100T+RT(T-1)/2)
3. If an amount P₁is lent out at simple interest of R₁% per annum and another amount P₂ at simple interest rate of R₂% per annum, then the rate of interest for the whole sum can be given by
   R=(P1R1+P2R2)/ (P1+P2)
4. If a certain sum of money is lent out in n parts in such a manner that equal sum of money is obtained at simple interest on each part where interest rates are R₁, R₂, … , R_nrespectively and time periods are T₁, T₂, … , T_n respectively, then the ratio in which the sum will be divided in n parts can be given by (1/R1T1):(1/R2T2):⋯(1/RnTn)
5. If a certain sum of money P lent out for a certain time T amounts to P₁at R₁% per annum and to P₂at R₂% per annum, then P = (P2R1−P1R2)/ (R1−R2) and T = (P1−P2) ×100 years / (P2R1−P1R2)

SOLVED EXAMPLES

LEVEL 1

1.       Arun took a loan of Rs. 1400 with simple interest for as many years as the rate of interest. If he paid Rs.686 as interest at the end of the loan period, what was the rate of interest?
A. 8%	B. 6%
C. 4%	D. 7%

Ans. Let rate = R%

Then, Time, T = R years

P = Rs.1400

SI = Rs.686

SI= PRT/100⇒686 = 1400 × R × R/100⇒686=14 Rx R ⇒49=Rx R ⇒R=7

i.e.,Rate of Interest was 7%. (D)

2.       How much time will it take for an amount of Rs. 900 to yield Rs. 81 as interest at 4.5% per annum of simple interest?
A. 2 years	B. 3 years
C. 1 year	D. 4 years

 

 

Ans. P = Rs.900

SI = Rs.81

T = ?

R = 4.5%

T= 100×SI/PR = 100×81/(900×4.5) = 2 years (A)

3.       A sum of money at simple interest amounts to Rs. 815 in 3 years and to Rs. 854 in 4 years. The sum is :
A. Rs. 700	B. Rs. 690
C. Rs. 650	D. Rs. 698

 

 

Ans. Simple Interest (SI) for 1 year = 854-815 = 39

Simple Interest (SI) for 3 years = 39 × 3 = 117

Principal = 815 – 117 = Rs.698 (D)

 

4.       A sum fetched a total simple interest of Rs. 929.20 at the rate of 8 p.a. in 5 years. What is the sum?
A. Rs. 2323	B. Rs. 1223
C. Rs. 2563	D. Rs. 2353

 

Ans. SI = Rs.929.20

P = ?

T = 5 years

R = 8%

P = 100×SI/RT=100×929.20/(8×5) = Rs.2323 (A)

5.       What will be the ratio of simple interest earned by certain amount at the same rate of interest for 5 years and that for 15 years?
A. 3 : 2	B. 1 : 3
C. 2 : 3	D. 3 : 1

 
Solution 1
Let Principal = P

Rate of Interest = R%

Required Ratio = (PR×5/100)/ (PR×15/100) =1:3 (B)
Solution 2

Simple Interest = PRT100

Here Principal(P) and Rate of Interest (R) are constants

Hence, Simple Interest ∝ T

Required Ratio = Simple Interest for 5 years Simple Interest for 15 years=T1T2=515=13=1:3 (B)

6.       A sum of money amounts to Rs.9800 after 5 years and Rs.12005 after 8 years at the same rate of simple interest. The rate of interest per annum is
A. 15%	B. 12%
C. 8%	D. 5%

 

 

Ans. Simple Interest for 3 years = (Rs.12005 – Rs.9800) = Rs.2205

Simple Interest for 5 years = 22053×5=Rs.3675

Principal (P) = (Rs.9800 – Rs.3675) = Rs.6125

R = 100×SI/PT=100×3675/(6125×5) =12% (B)

7.       A lent Rs. 5000 to B for 2 years and Rs. 3000 to C for 4 years on simple interest at the same rate of interest and received Rs. 2200 in all from both of them as interest. The rate of interest per annum is:
A. 5%	B. 10%
C. 7%	D. 8%

 

 

Ans. Let the rate of interest per annum be R%

Simple Interest for Rs. 5000 for 2 years at rate R% per annum +Simple Interest for Rs. 3000 for 4 years at rate R% per annum = Rs.2200

⇒5000×R×2/100+3000×R×4/100=2200

⇒100R + 120R=2200⇒220R=2200⇒R=10

i.e, Rate = 10%. (B)

8.       In how many years, Rs. 150 will produce the same interest at 6% as Rs. 800 produce in 2 years at 4½% ?
A. 4 years	B. 6 years
C. 8 years	D. 9 years

 

 

Ans. Let Simple Interest for Rs.150 at 6% for n years = Simple Interest for Rs.800 at 4½ % for 2 years

150×6×n/100=800×4.5×2/100

150×6×n=800×4.5×2

n=8 years (C)

 

LEVEL 2

1.        Mr. Thomas invested an amount of Rs. 13,900 divided in two different schemes A and B at the simple interest rate of 14% p.a. and 11% p.a. respectively. If the total amount of simple interest earned in 2 years be Rs. 3508, what was the amount invested in Scheme B?
A. Rs. 6400	B. Rs. 7200
C. Rs. 6500	D. Rs. 7500

 

 

Ans. Let the investment in scheme A be Rs.x

and the investment in scheme B be Rs. (13900 – x)

We know that SI = PRT/100

Simple Interest for Rs.x in 2 years at 14% p.a. = x×14×2100=28x100Simple Interest for Rs.(13900 – x) in 2 years at 11% p.a. = (13900−x)×11×2/100 =22(13900−x)/100

Total interest =Rs.3508

Thus, 28x/100+22(13900−x)/100 = 3508

28x+305800−22x=350800

6x = 45000

x=45000/6=7500

Investment in scheme B = 13900 – 7500 = Rs.6400 (A)

2.       A certain sum in invested for T years. It amounts to Rs. 400 at 10% per annum. But when invested at 4% per annum, it amounts to Rs. 200. Find the time (T).
A. 45 years	B. 60 years
C. 40 years	D. 50 Years

 
Solution 1
Let the principal = Rs.x

and time = y years

Principal,x amounts to Rs.400 at 10% per annum in y years

Simple Interest = (400-x)

Simple Interest = PRT/100

⇒ (400−x) = x×10×y/100

⇒ (400−x) = xy/10— (equation 1)

Principal,x amounts to Rs.200 at 4% per annum in y years

Simple Interest = (200-x)

Simple Interest = PRT/100

⇒ (200−x) = x×4×y/100

⇒ (200−x) = xy/25— (equation 2)

(equation 1)/(equation2)

⇒(400−x) / (200−x) = (xy/10)/(xy/25)

⇒ (400−x)/ (200−x) =25/10

⇒ (400−x)/ (200−x) =52

⇒800−2x = 1000−5x

⇒200=3x

⇒x =200/3 Substituting this value of x in Equation 1, we get,

(400−200/3) = (200y/3)/10

⇒ (400−200/3) = 20y/3

⇒1200−200=20y

⇒1000=20y

y=1000/20=50 years (D)

Solution 2
If a certain sum of money P lent out for a certain time T amounts to P₁ at R₁% per annum and to P₂ at R₂% per annum, then

P = (P2R1−P1R2)/ (R1−R2)

T = (P1−P2)x 100 years/(P2R1−P1R2)

R₁ = 10%, R₂ = 4%

P₁ = 400, P₂ = 200

T = (P1−P2)x 100 / (P2R1−P1R2) = (400−200)x 100 / (200×10−400×4)

=200 x 100/ (2000−1600) =200 ×100/400 = 12×100=50 years (D)

3.       Mr. Mani invested an amount of Rs. 12000 at the simple interest rate of 10% per annum and another amount at the simple interest rate of 20% per annum. The total interest earned at the end of one year on the total amount invested became 14% per annum. Find the total amount invested.
A. Rs. 25000	B. Rs. 15000
C. Rs. 10000	D. Rs. 20000

 

Ans. If an amount P₁ is lent out at simple interest of R₁% per annum and another amount P₂ at simple interest rate of R₂% per annum, then the rate of interest for the whole sum can be given by

R= (P1R1+P2R2)/(P1+P2)

P₁ = Rs. 12000, R₁ = 10%

P₂ =? R₂ = 20%

R = 14%

14 = (12000×10+P2×20)/ (12000+P2)

12000×14+14P2 =120000+20P2

6P2=14×12000−120000=48000

⇒P2=8000

Total amount invested = (P₁ + P₂) = (12000 + 8000) = Rs. 20000 (D)

4.       A sum of money is lent at S.I. for 6 years. If the same amount is paid at 4% higher, Arun would have got Rs. 120 more. Find the principal
A. Rs. 200	B. Rs. 600
C. Rs. 400	D. Rs. 500

 

 

Ans. This means, simple interest at 4% for that principal is Rs.120

P=100×SI/ RT=100×120/ (4×6) =100×30/6 = 100×5 = 500 (D)

5.       The simple interest on Rs. 1820 from March 9, 2003 to May 21, 2003 at 7 1⁄2% rate is
A. Rs. 27.30	B. Rs. 22.50
C. Rs. 28.80	D. Rs. 29

 

 

Ans. Time, T = (22 + 30 + 21) days = 73 days = 73/365 year=1/5 year

Rate, R = 7.5%=15/2%

SI = PRT/100 = 1820× (15/2) × (1/5)/100 = 1820 × (3/2)/100 = 910 × 3/100

= 2730/100 = 27.30 (A)

6.       A sum of Rs. 7700 is to be divided among three brothers Vikas, Vijay and Viraj in such a way that simple interest on each part at 5% per annum after 1, 2 and 3 years respectively remains equal. The Share of Vikas is more than that of Viraj by
A. Rs.1200	B. Rs.1400
C. Rs.2200	D. Rs.2800

Ans. If a certain sum of money is lent out in n parts in such a manner that equal sum of money is obtained at simple interest on each part where interest rates are R₁, R₂, … , R_n respectively and time periods are T₁, T₂, … , T_n respectively, then the ratio in which the sum will be divided in n parts can be given by

1/R1T1:1/R2T2:⋯1/RnTn

 

T₁ = 1 , T₂ = 2, T₃ = 3

R₁ = 5 , R₂ = 5, R₃ = 5

Share of Vikas : Share of Vijay : Share of Viraj

= (1/5×1) : (1/5×2) : (1/5×3) = 1/1:1/2:1/3 = 6:3:2

Total amount is Rs. 7700

Share of Vikas = 7700×6/11=700×6 = 4200

Share of Viraj = 7700×2/11=700×2=1400

Share of Vikas is greater than Share of Viraj by (4200 – 1400) = Rs. 2800 (D)

 

7.       David invested certain amount in three different schemes A, B and C with the rate of interest 10% p.a., 12% p.a. and 15% p.a. respectively. If the total interest accrued in one year was Rs. 3200 and the amount invested in Scheme C was 150% of the amount invested in Scheme A and 240% of the amount invested in Scheme B, what was the amount invested in Scheme B?
A. Rs.5000	B. Rs.2000
C. Rs.6000	D. Rs.3000

 

 

Ans. Let x, y and x be his investments in A, B and C respectively. Then

Then, Interest on x at 10% for 1 year

+ Interest on y at 12% for 1 year

+ Interest on z at 15% for 1 year

= 3200

x×10×1/100+y×12×1/100+z×15×1/100=3200

⇒10x+12y+15z=320000−−−(1)

Amount invested in Scheme C was 240% of the amount invested in Scheme B

=>z=240y/100 = 60y/25=12y/5−−−(2)

Amount invested in Scheme C was 150% of the amount invested in Scheme A

=>z=150x/100=3x/2

=>x=2z/3=2/3×12y/5=8y/5−−−(3)

From(1),(2) and (3),

10x + 12y + 15z = 320000

10(8y/5)+12y+15(12y/5)=320000

16y+12y+36y=320000

64y=320000

y=320000/64=10000/2=5000

i.e.,Amount invested in Scheme B = Rs.5000 (A)

 

 

Speed has no sense of direction unlike the velocity. Relative speed is the speed of one object as observed from another moving object. Questions on train are the classic examples of relative speed and in all these questions it is assumed that trains move parallel to each other – whether in the same direction or the opposite direction. Thus, we shall see how the relative speed is calculated and using it we come to know the time taken by the trains to cross each other and some other like aspects.

Important Formulas – Problems on Trains

1. x km/hr = (x×5)/18 m/s

 

2. y m/s = (y×18)/5 km/hr

 

3. Speed = distance/time, that is, s = d/t

 

4. velocity = displacement/time, that is, v = d/t

 

5. Time taken by a train x meters long to pass a pole or standing man or a post
   = Time taken by the train to travel x meters.

 

6. Time taken by a train x meters long to pass an object of length y meters

= Time taken by the train to travel (x + y) metres.

 

7. Suppose two trains or two objects are moving in the same direction at v1 m/s and v2 m/s where v1 > v2,

then their relative speed = (v1 – v2) m/s

 

8. Suppose two trains or two objects are moving in opposite directions at v1 m/s and v2 m/s ,

then their relative speed = (v1+ v2) m/s

 

9. Assume two trains of length x metres and y metres are moving in opposite directions at v1 m/s and v2 m/s, Then

The time taken by the trains to cross each other = (x+y) / (v1+v2) seconds

 

10. Assume two trains of length x metres and y metres are moving in the same direction at at v1 m/s and v2 m/s where v1 > v2, Then

The time taken by the faster train to cross the slower train = (x+y) / (v1-v2) seconds

 

11. Assume that two trains (objects) start from two points P and Q towards each other at the same time and after crossing they take p and q seconds to reach Q and P respectively. Then,

A’s speed: B’s speed = √q: √p

 

 

Solved Examples

Level 1

1.A train is running at a speed of 40 km/hr and it crosses a post in 18 seconds. What is the length of the train?
A. 190 metres	B. 160 metres
C. 200 metres Answer : Option C	D. 120 metres

 

Explanation :

Speed of the train, v = 40 km/hr = 40000/3600 m/s = 400/36 m/s

Time taken to cross, t = 18 s

Distance Covered, d = vt = (400/36)× 18 = 200 m

Distance covered is equal to the length of the train = 200 m

2.A train having a length of 240 m passes a post in 24 seconds. How long will it take to pass a platform having a length of 650 m?
A. 120 sec	B. 99 s
C. 89 s	D. 80 s

 

Answer : Option C

Explanation :

v = 240/24 (where v is the speed of the train) = 10 m/s

t = (240+650)/10 = 89 seconds

3.Two trains having length of 140 m and 160 m long run at the speed of 60 km/hr and 40 km/hr respectively in opposite directions (on parallel tracks). The time which they take to cross each other, is
A. 10.8 s	B. 12 s
C. 9.8 s	D. 8 s

 

Answer : Option A

Explanation :

Distance = 140+160 = 300 m

Relative speed = 60+40 = 100 km/hr = (100×10)/36 m/s

Time = distance/speed = 300 / (100×10)/36 = 300×36 / 1000 = 3×36/10 = 10.8 s

4.A train moves past a post and a platform 264 m long in 8 seconds and 20 seconds respectively. What is the speed of the train?
A. 79.2 km/hr	B. 69 km/hr
C. 74 km/hr	D. 61 km/hr

 

Answer : Option A

Explanation :

Let x is the length of the train and v is the speed

Time taken to move the post = 8 s

=> x/v = 8

=> x = 8v — (1)

Time taken to cross the platform 264 m long = 20 s

(x+264)/v = 20

=> x + 264 = 20v —(2)

Substituting equation 1 in equation 2, we get

8v +264 = 20v

=> v = 264/12 = 22 m/s

= 22×36/10 km/hr = 79.2 km/hr

5.Two trains, one from P to Q and the other from Q to P, start simultaneously. After they meet, the trains reach their destinations after 9 hours and 16 hours respectively. The ratio of their speeds is
A. 2 : 3	B. 2 :1
C. 4 : 3	D. 3 : 2

 

Answer : Option C

Explanation :

Ratio of their speeds = Speed of first train : Speed of second train

= √16: √ 9

= 4:3

6.Train having a length of 270 meter is running at the speed of 120 kmph . It crosses another train running in opposite direction at the speed of 80 kmph in 9 seconds. What is the length of the other train?
A. 320 m	B. 190 m
C. 210 m	D. 230 m

 

Answer : Option D

Explanation :

Relative speed = 120+80 = 200 kmph = 200×10/36 m/s = 500/9 m/s

time = 9s

Total distance covered = 270 + x where x is the length of other train

(270+x)/9 = 500/9

=> 270+x = 500

=> x = 500-270 = 230 meter

7.Two stations P and Q are 110 km apart on a straight track. One train starts from P at 7 a.m. and travels towards Q at 20 kmph. Another train starts from Q at 8 a.m. and travels towards P at a speed of 25 kmph. At what time will they meet?
A. 10.30 a.m	B. 10 a.m.
C. 9.10 a.m.	D. 11 a.m.

 

Answer : Option B

Explanation :

Assume both trains meet after x hours after 7 am

Distance covered by train starting from P in x hours = 20x km

Distance covered by train starting from Q in (x-1) hours = 25(x-1)

Total distance = 110

=> 20x + 25(x-1) = 110

=> 45x = 135

=> x= 3 Means, they meet after 3 hours after 7 am, ie, they meet at 10 am

8.Two trains are running in opposite directions in the same speed. The length of each train is 120 meter. If they cross each other in 12 seconds, the speed of each train (in km/hr) is
A. 42	B. 36
C. 28	D. 20

 

Answer : Option B

Explanation :

Distance covered = 120+120 = 240 m

Time = 12 s

Let the speed of each train = v. Then relative speed = v+v = 2v

2v = distance/time = 240/12 = 20 m/s

Speed of each train = v = 20/2 = 10 m/s

= 10×36/10 km/hr = 36 km/hr

 

Level 2

1.A train, 130 meters long travels at a speed of 45 km/hr crosses a bridge in 30 seconds. The length of the bridge is
A. 270 m	B. 245 m
C. 235 m	D. 220 m

 

Answer : Option B

Explanation :

Assume the length of the bridge = x meter

Total distance covered = 130+x meter

total time taken = 30s

speed = Total distance covered /total time taken = (130+x)/30 m/s

=> 45 × (10/36) = (130+x)/30

=> 45 × 10 × 30 /36 = 130+x

=> 45 × 10 × 10 / 12 = 130+x

=> 15 × 10 × 10 / 4 = 130+x

=> 15 × 25 = 130+x = 375

=> x = 375-130 =245

2.A train has a length of 150 meters. It is passing a man who is moving at 2 km/hr in the same direction of the train, in 3 seconds. Find out the speed of the train.
A. 182 km/hr	B. 180 km/hr
C. 152 km/hr	D. 169 km/hr

 

Answer : Option A

Explanation :

Length of the train, l = 150m

Speed of the man, Vm= 2 km/hr

Relative speed, Vr = total distance/time = (150/3) m/s = (150/3) × (18/5) = 180 km/hr

Relative Speed = Speed of train, Vt – Speed of man (As both are moving in the same direction)

=> 180 = Vt – 2 => Vt = 180 + 2 = 182 km/hr

3.Two trains running in opposite directions cross a man standing on the platform in 27 seconds and 17 seconds respectively. If they cross each other in 23 seconds, what is the ratio of their speeds?
A. Insufficient data	B. 3 : 1
C. 1 : 3	D. 3 : 2

 

Answer : Option D

Explanation :

Let the speed of the trains be x and y respectively

length of train1 = 27x

length of train2 = 17y

Relative speed= x+ y

Time taken to cross each other = 23 s

=> (27x + 17 y)/(x+y) = 23 => (27x + 17 y)/ = 23(x+y)

=> 4x = 6y => x/y = 6/4 = 3/2

4.A jogger is running at 9 kmph alongside a railway track in 240 meters ahead of the engine of a 120 meters long train . The train is running at 45 kmph in the same direction. How much time does it take for the train to pass the jogger?
A. 46	B. 36
C. 18	D. 22

 

Answer : Option B

Explanation :

Distance to be covered = 240+ 120 = 360 m

Relative speed = 36 km/hr = 36×10/36 = 10 m/s

Time = distance/speed = 360/10 = 36 seconds

5.A train passes a platform in 36 seconds. The same train passes a man standing on the platform in 20 seconds. If the speed of the train is 54 km/hr, The length of the platform is
A. None of these	B. 280 meter
C. 240 meter	D. 200 meter

 

Answer : Option C

Explanation :

Speed of the train = 54 km/hr = (54×10)/36 m/s = 15 m/s

Length of the train = speed × time taken to cross the man = 15×20 = 300 m

Let the length of the platform = L

Time taken to cross the platform = (300+L)/15

=> (300+L)/15 = 36

=> 300+L = 15×36 = 540 => L = 540-300 = 240 meter

6.A train overtakes two persons who are walking in the same direction to that of the train at 2 kmph and 4 kmph and passes them completely in 9 and 10 seconds respectively. What is the length of the train?
A. 62 m	B. 54 m
C. 50 m	D. 55 m

 

Answer : Option C

Explanation :

Let x is the length of the train in meter and v is its speed in kmph

x/9 = (v-2) (10/36) — (1)

x/10 = (v-4) (10/36) — (2)

Dividing equation 1 with equation 2

10/9 = (v-2)/(v-4) => 10v – 40 = 9v – 18 => v = 22

Substituting in equation 1, x/9 = 200/36 => x = 9×200/36 = 50 m

7.A train is traveling at 48 kmph. It crosses another train having half of its length, traveling in opposite direction at 42 kmph, in 12 seconds. It also passes a railway platform in 45 seconds. What is the length of the platform?
A. 500 m	B. 360 m
C. 480 m	D. 400 m

 

Answer : Option D

Explanation :

Speed of train1 = 48 kmph

Let the length of train1 = 2x meter

Speed of train2 = 42 kmph

Length of train 2 = x meter (because it is half of train1’s length)

Distance = 2x + x = 3x

Relative speed= 48+42 = 90 kmph = 90×10/36 m/s = 25 m/s

Time = 12 s

Distance/time = speed => 3x/12 = 25

=> x = 25×12/3 = 100 meter

Length of the first train = 2x = 200 meter

Time taken to cross the platform= 45 s

Speed of train1 = 48 kmph = 480/36 = 40/3 m/s

Distance = 200 + y where y is the length of the platform

=> 200 + y = 45×40/3 = 600

=> y = 400 meter

8.A train, 800 meter long is running with a speed of 78 km/hr. It crosses a tunnel in 1 minute. What is the length of the tunnel (in meters)?
A. 440 m	B. 500 m
C. 260 m	D. 430 m

 

Answer : Option B

Explanation :

Distance = 800+x meter where x is the length of the tunnel

Time = 1 minute = 60 seconds

Speed = 78 km/hr = 78×10/36 m/s = 130/6 = 65/3 m/s

Distance/time = speed

(800+x)/60 = 65/3 => 800+x = 20×65 = 1300

=> x = 1300 – 800 = 500 meter

9.Two trains are running at 40 km/hr and 20 km/hr respectively in the same direction. If the fast train completely passes a man sitting in the slower train in 5 seconds, the length of the fast train is :
A. 19 m	B. 2779 m
C. 1329 m	D. 33 m

 

Answer : Option B

Explanation :

Relative speed = 40-20 = 20 km/hr = 200/36 m/s = 100/18 m/s

Time = 5 s

Distance = speed × time = (100/18) × 5 = 500/18 m = 250/9 = 2779 m = length of the fast train

 

• Percentage

  Percent means for every 100

  So, when percent is calculated for any value, it means that we calculate the value for every 100 of the reference value.

  percent is denoted by the symbol %. For example, x percent is denoted by x%

• x%=x/100

  Example : 25%=25/100=1/4

• To express x/y as a percent,we have x/y=(x/y×100)%

  Example : 1/4=(1/4×100)%=25%

• If the price of a commodity increases by R%, the reduction in consumptionso as not to increase the expenditure = [R/(100+R)×100]%
• If the price of a commodity decreases by R%, the increase in consumptionso as not to decrease the expenditure = [R/(100−R)×100]%
• If the population of a town = P and it increases at the rate of R% per annum, thenPopulation after n years = P((1+R)/100))n
• If the population of a town = P and it increases at the rate of R% per annum, thenPopulation before n years = P((1+R)/100))n
• If the present value of a machine = P and it depreciates at the rate of R% per annum,

ThenValue of the machine after n years = P((1-R)/100))n

• If the present value of a machine = P and it depreciates at the rate of R% per annum,

ThenValue of the machine before n years = P((1-R)/100))n

 

Solved Examples

Level 1

1.    If A = x% of y and B = y% of x, then which of the following is true?
A. None of these	B. A is smaller than B.
C. Relationship between A and B cannot be determined.	D. If x is smaller than y, then A is greater than B.
E. A is greater than B.

 

Answer : Option A

Explanation :

A = xy/100 ………….(Equation 1)

B = yx/100……………..(Equation 2)

From these equations, it is clear that A = B

 

 

2.If 20% of a = b, then b% of 20 is the same as:
A. None of these	B. 10% of a
C. 4% of a	D. 20% of a

 

Answer :Option C

Explanation :

20% of a = b

=> b = 20a/100

b% of 20 = 20b/100=(20a/100) × 20/100

=(20×20×a)/(100×100)=4a/100 = 4% of a

 

3.Two numbers A and B are such that the sum of 5% of A and 4% of B is two-third of the sum of 6% of A and 8% of B. Find the ratio of A : B.
A. 2 : 1	B. 1 : 2
C. 1 : 1	D. 4 : 3

 

 

Answer :Option D

Explanation :

5% of A + 4% of B = 2/3(6% of A + 8% of B)

5A/100+4B/100=2/3(6A/100+8B/100)

⇒5A+4B=2/3(6A+8B)

⇒15A+12B=12A+16B

⇒3A=4B

⇒AB=43⇒A:B=4:3

4.The population of a town increased from 1,75,000 to 2,62,500 in a decade. What is the average percent increase of population per year?
A. 4%	B. 6%
C. 5%	D. 50%

 

Answer :Option C

Explanation :

Increase in the population in 10 years = 2,62,500 – 1,75,000 = 87500

% increase in the population in 10 years = (87500/175000)×100=8750/175=50%

Average % increase of population per year = 50%/10=5%

 

5.Three candidates contested an election and received 1136, 7636 and 11628 votes respectively. What percentage of the total votes did the winning candidate get?
A. 57%	B. 50%
C. 52%	D. 60%

 

Answer :Option A

Explanation :

Votes received by the winning candidate = 11628

Total votes = 1136 + 7636 + 11628 = 20400

Required percentage = (11628/20400)×100=11628/204=2907/51=969/17=57%

 

6.A fruit seller had some oranges. He sells 40% oranges and still has 420 oranges. How many oranges he had originally?
A. 420	B. 700
C. 220	D. 400

 

Answer :Option B

Explanation :

He sells 40% of oranges and still there are 420 oranges remaining

=> 60% of oranges = 420

⇒(60×Total Oranges)/100=420

⇒Total Oranges/100=7

⇒ Total Oranges = 7×100=700

7.A batsman scored 110 runs which included 3 boundaries and 8 sixes. What percent of his total score did he make by running between the wickets?
A. 499/11 %	B. 45 %
C. 500/11 %	D. 489/11 %

 

Answer :Option C

Explanation :

Total runs scored = 110

Total runs scored from boundaries and sixes = 3 x 4 + 8 x 6 = 60

Total runs scored by running between the wickets = 110 – 60 = 50

Required % = (50/110)×100=500/11%

 

8.What percentage of numbers from 1 to 70 have 1 or 9 in the unit’s digit?
A. 2023%	B. 20%
C. 21%	D. 2223%

 

 

Answer :Option B

Explanation :

Total numbers = 70

Total numbers in 1 to 70 which has 1 in the unit digit = 7

Total numbers in 1 to 70 which has 9 in the unit digit = 7

Total numbers in 1 to 70 which has 1 or 9 in the unit digit = 7 + 7 = 14

Required percentage = (14/70)×100=140/7=20%

 

Level 2

 

1.In an election between two candidates, one got 55% of the total valid votes, 20% of the votes were invalid. If the total number of votes was 7500, what was the number of valid votes that the other candidate got?
A. 2800	B. 2700
C. 2100	D. 2500

 

Answer :Option B

Explanation :

Total number of votes = 7500

Given that 20% of Percentage votes were invalid

=> Valid votes = 80%

Total valid votes = (7500×80)/100

1^st candidate got 55% of the total valid votes.

Hence the 2^nd candidate should have got 45% of the total valid votes
=> Valid votes that 2nd candidate got = (total valid votes ×45)/100

=7500×(80/100)×(45/100)=75×(4/5)×45=75×4×9=300×9=2700

 

2.In a competitive examination in State A, 6% candidates got selected from the total appeared candidates. State B had an equal number of candidates appeared and 7% candidates got selected with 80 more candidates got selected than A. What was the number of candidates appeared from each State?
A. 8200	B. 7500
C. 7000	D. 8000

 

Answer :Option D

Explanation :

State A and State B had an equal number of candidates appeared.

In state A, 6% candidates got selected from the total appeared candidates

In state B, 7% candidates got selected from the total appeared candidates

But in State B, 80 more candidates got selected than State A

From these, it is clear that 1% of the total appeared candidates in State B = 80

=> total appeared candidates in State B = 80 x 100 = 8000

=> total appeared candidates in State A = total appeared candidates in State B = 8000

 

3.In a certain school, 20% of students are below 8 years of age. The number of students above 8 years of age is 2/3 of the number of students of 8 years of age which is 48. What is the total number of students in the school?
A. 100	B. 102
C. 110	D. 90

 

 

Answer :Option A

Explanation :

Let the total number of students = x

Given that 20% of students are below 8 years of age

then The number of students above or equal to 8 years of age = 80% of x —–(Equation 1)

Given that number of students of 8 years of age = 48 —–(Equation 2)

Given that number of students above 8 years of age = 2/3 of number of students of 8 years of age

=>number of students above 8 years of age = (2/3)×48=32—–(Equation 3)

From Equation 1,Equation 2 and Equation 3,
80% of x = 48 + 32 = 80

⇒80x/100=80

⇒x100=1⇒x=100

4.In an examination, 5% of the applicants were found ineligible and 85% of the eligible candidates belonged to the general category. If 4275 eligible candidates belonged to other categories, then how many candidates applied for the examination?
A. 28000	B. 30000
C. 32000	D. 33000

 

Answer :Option B

Explanation :

Let the number of candidates applied for the examination = x

Given that 5% of the applicants were found ineligible.

It means that 95% of the applicants were eligible (∴ 100% – 5% = 95%)

Hence total eligible candidates = 95x/100

Given that 85% of the eligible candidates belonged to the general category

It means 15% of the eligible candidates belonged to other categories(∴ 100% – 85% = 15%)
Hence Total eligible candidates belonged to other categories

=(total eligible candidates×15)/100=(95x/100)×(15/100)

=(95x×15)/(100×100)

Given that Total eligible candidates belonged to other categories = 4275

⇒(95x×15)/(100×100)=4275

⇒(19x×15)/(100×100)=855

⇒(19x×3)/(100×100)=171

⇒(x×3)/(100×100)=9

⇒x/(100×100)=3

⇒x=3×100×100=30000

 

5.A student multiplied a number by 3/5 instead of 5/3.What is the percentage error in the calculation?
A. 64%	B. 32%
C. 34%	D. 42%

 

Answer :Option A

Explanation :

Let the number = 1

Then, ideally he should have multiplied 1 by 5/3.

Hence the correct result was 1 x (5/3) = (5/3)

By mistake, he multiplied 1 by 3/5.

Hence the result with the error = 1 x (3/5) = (3/5)

Error = 5/3−3/5=(25−9)/15=16/15

percentage error = (Error/True Value)×100={(16/15)/(5/3)}×100

=(16×3×100)/(15×5)=(16×100)/(5×5)=16×4=64%

 

6.The price of a car is Rs. 3,25,000. It was insured to 85% of its price. The car was damaged completely in an accident and the insurance company paid 90% of the insurance. What was the difference between the price of the car and the amount received ?
A. Rs. 76,375	B. Rs. 34,000
C. Rs. 82,150	D. Rs. 70,000

 

Answer :Option A

Explanation :

Price of the car = Rs.3,25,000

Car insured to 85% of its price

=>Insured price=(325000×85)/100

Insurance company paid 90% of the insurance
⇒Amount paid by Insurance company =(Insured price×90)/100

=325000×(85/100)×(90/100)=325×85×9=Rs.248625

Difference between the price of the car and the amount received

= Rs.325000 – Rs.248625 = Rs.76375

7.If the price of petrol increases by 25% and Benson intends to spend only an additional 15% on petrol, by how much % will he reduce the quantity of petrol purchased?
A. 8%	B. 7%
C. 10%	D. 6%

 

 

Answer :Option A

Explanation :

Assume that the initial price of 1 Litre petrol = Rs.100 ,Benson spends Rs.100 for petrol,

such that Benson buys 1 litre of petrol

After the increase by 25%, price of 1 Litre petrol = (100×(100+25))/100=Rs.125

Since Benson spends additional 15% on petrol,

amount spent by Benson = (100×(100+15))/100=Rs.115

Hence Quantity of petrol that he can purchase = 115/125 Litre

Quantity of petrol reduced = (1−115/125) Litre

Percentage Quantity of reduction = ((1−115/125))/1×100=(10/125)/×100=(10/5)×4=2×4=8%

8.30% of the men are more than 25 years old and 80% of the men are less than or equal to 50 years old. 20% of all men play football. If 20% of the men above the age of 50 play football, what percentage of the football players are less than or equal to 50 years?
A. 60%	B. 70%
C. 80%	D. 90%

 

Answer :Option C

Explanation :

Let total number of men = 100

Then

80 men are less than or equal to 50 years old

(Since 80% of the men are less than or equal to 50 years old)

=> 20 men are above 50 years old (Since we assumed total number of men as 100)

20% of the men above the age of 50 play football

⇒Number of men above the age of 50 who play football = (20×20)/100=4

Number of men who play football = 20 (Since 20% of all men play football)
Percentage of men who play football above the age of 50 = (4/20)×100=20%

=>Percentage of men who play football less than or equal to the age 50 = 100%−20%=80%

 

1. Alligation

   It is the rule which enables us to find the ratio in which two or more ingredients at the given price must be mixed to produce a mixture of a specified price.

2. Mean Price

   Mean price is the cost price of a unit quantity of the mixture

 

3. Suppose a container contains x of liquid from which y units are taken out and replaced by water.After n operations, the quantity of pure liquid = [x(1−y/x)n]

 

4. Rule of Alligation

   If two ingredients are mixed, then

   (Quantity of cheaper/Quantity of dearer)=(P. of dearer – Mean Price)/(Mean price – C.P. of cheaper)

Cost Price(CP) of a unit quantity of cheaper (c)	Cost Price(CP) of a unit quantity of dearer (d)
Mean Price
(m)
(d – m)	(m – c)

5. => (Cheaper quantity) : (Dearer quantity) = (d – m) : (m – c)

 

Solved Examples

Level 1

1.A container contains 40 litres of milk. From this container 4 litres of milk was taken out and replaced by water. This process was repeated further two times. How much milk is now contained by the container?
A. 26 litres	B. 29.16 litres
C. 28 litres	D. 28.2 litres

 

 

Answer : Option B

Explanation :

Assume that a container contains x of liquid from which y units are taken out and replaced

by water. After n operations, the quantity of pure liquid

=[x(1−y/x)n]

Hence milk now contained by the container = 40(1−4/40)3=40(1−1/10)3

40×9/10×9/10×9/10=(4×9×9×9)/100=29.16

2.A jar full of whiskey contains 40% alcohol. A part of this whisky is replaced by another containing 19% alcohols and now the percentage of alcohol was found to be 26%. The quantity of whisky replaced is
A. 43	B. 34
C. 32	D. 23

 

 

Answer : Option D

Explanation :

Concentration of alcohol in 1st Jar = 40%

Concentration of alcohol in 2nd Jar = 19%

After the mixing, Concentration of alcohol in the mixture = 26%

By the rule of alligation,

Concentration of alcohol in 1st Jar	Concentration of alcohol in 2nd Jar
40%	19%
Mean
26%
7	14

Hence ratio of 1st and 2nd quantities = 7 : 14 = 1 : 2

 

3.In what ratio should rice at Rs.9.30 per Kg be mixed with rice at Rs. 10.80 per Kg so that the mixture be worth Rs.10 per Kg ?
A. 7 : 8	B. 8 : 7
C. 6 : 7	D. 7 ; 6

 

 

Answer : Option B

 

Explanation :

By the rule of alligation, we have

 

Cost of 1 kg rice of 1st kind	Cost of 1 kg rice of 2nd kind
9.3	10.80
Mean Price
10
10.8-10 = .8	10 – 9.3 = .7

Required ratio = .8 : .7 = 8 : 7.

 

 

 

4.In what ratio must water be mixed with milk costing Rs.12 per litre in order to get a mixture worth of Rs.8 per litre?
A. 1 : 3	B. 2 : 2
C. 1 : 2	D. 3 : 1

 

 

Answer : Option C

 

Explanation :

By the rule of alligation, we have

Cost Price of 1 litre of water	Cost Price of 1 litre of milk
0	12
Mean Price
8
12-8=4	8-0=8

Required Ratio = 4 : 8 = 1 : 2

 

 

 

5.In 1 kg mixture of iron and manganese 20% of manganese. How much iron should be added so that the proportion of manganese becomes 10%
A. 1.5 Kg	B. 2 Kg
C. .5 Kg	D. 1 Kg

 

 

 

 

Answer : Option D

 

Explanation :

By the rule of alligation, we have

Percentage concentration of manganese in the mixture : 20	Percentage concentration of manganese in pure iron : 0
Percentage concentration of manganese in the final mixture
10
10 – 0 = 10	20 – 10 = 10

=> Quantity of the mixture : Quantity of iron = 10 : 10 = 1 : 1

Given that Quantity of the mixture = 1 Kg

Hence Quantity of iron to be added = 1 Kg

 

6.A trader has 1600Kg of sugar. He sells a part at 8% profit and the rest at 12% profit. If he gains 11% on the whole , find the quantity sold at 12%.
A. 1200 Kg	B. 1400 Kg
C. 1600 Kg	D. 800 Kg

 

 

 

Answer : Option A

 

Explanation :

By the rule of alligation, we have

 

% Profit by selling part1	% Profit by selling part2
8	12
Net % Profit
11
12 – 11 = 1	11 – 8 = 3

=>Quantity of part1 : Quantity of part2 = 1 : 3

Given that total quantity = 1600 Kg

Hence quantity of part2 (quantity sold at 12%) = 1600×3/4=1200

 

 

7.How many litres of water must be added to 16 liters of milk and water contains 10% water to make it 20% water in it
A. 4 litre	B. 2 litre
C. 1 litre	D. 3 litre

 

 

Answer : Option B

 

Explanation :

By the rule of alligation, we have

% Concentration of water in pure water : 100	% Concentration of water in the given mixture : 10
Mean % concentration
20
20 – 10 = 10	100 – 20 = 80

=> Quantity of water : Quantity of the mixture = 10 : 80 = 1 : 8

Here Quantity of the mixture = 16 litres

=> Quantity of water : 16 = 1 : 8

Quantity of water = 16×1/8=2 litre

 

8.A merchant has 1000 kg of sugar part of which he sells at 8% profit and the rest at 18% profit. He gains 14% on the whole. The Quantity sold at 18% profit is
A. 300	B. 400
C. 600	D. 500

 

 

Answer : Option C

 

Explanation :

By the rule of alligation,we have

Profit% by selling 1st part	Profit% by selling 2nd part
8	18
Net % profit
14
18-14=4	14-8=6

=> Quantity of part1 : Quantity of part2 = 4 : 6 = 2 : 3

Total quantity is given as 1000Kg

So Quantity of part2 (Quantity sold at 18% profit) = 1000×3/5=600Kg

 

Level 2

 

1.Tea worth Rs. 126 per kg and Rs. 135 per kg are mixed with a third variety of tea in the ratio 1 : 1 : 2. If the mixture is worth Rs. 153 per kg, what is the price of the third variety per kg ?
A. Rs.182.50	B. Rs.170.5
C. Rs.175.50	D. Rs.180

 

 

Answer : Option C

Explanation :

Tea worth Rs. 126 per kg and Rs. 135 per kg are mixed in the ratio 1 : 1

So their average price = (126+135)2=130.5

Hence let’s consider that the mixture is formed by mixing two varieties of tea.

one at Rs. 130.50 per kg and the other at Rs. x per kg in the ratio 2 : 2, i.e.,

1 : 1. Now let’s find out x.

By the rule of alligation, we can write as

Cost of 1 kg of 1st kind of tea	Cost of 1 kg of 2nd kind of tea
130.50	x
Mean Price
153
(x – 153)	22.50

(x – 153) : 22.5 = 1 : 1

=>x – 153 = 22.50

=> x = 153 + 22.5 = 175.5

 

2.A milk vendor has 2 cans of milk. The first contains 25% water and the rest milk. The second contains 50% water. How much milk should he mix from each of the containers so as to get 12 litres of milk such that the ratio of water to milk is 3 : 5?
A. 5litres, 7 litres	B. 7litres, 4 litres
C. 6litres, 6 litres	D. 4litres, 8 litres

 

 

Answer : Option C

Explanation :

Let cost of 1 litre milk be Rs.1

Milk in 1 litre mix in 1st can = 3/4 litre

Cost Price(CP) of 1 litre mix in 1st can = Rs.3/4

Milk in 1 litre mix in 2nd can = 1/2 litre

Cost Price(CP) of 1 litre mix in 2nd can = Rs.1/2

Milk in 1 litre of the final mix=5/8

Cost Price(CP) of 1 litre final mix = Rs.5/8

=> Mean price = 5/8
By the rule of alligation, we can write as

CP of 1 litre mix in 2nd can	CP of 1 litre mix in 1st can
1/2	3/4
Mean Price
5/8
3/4 – 5/8 = 1/8	5/8 – 1/2 = 1/8

=> mix in 2nd can :mix in 1st can = 1/8 : 1/8 = 1:1

ie, from each can, 12×12=6 litre should be taken

 

3.Two vessels A and B contain spirit and water in the ratio 5 : 2 and 7 : 6 respectively. Find the ratio in which these mixture be mixed to obtain a new mixture in vessel C containing spirit and water in the ration 8 : 5 ?
A. 3: 4	B. 4 : 3
C. 9 : 7	D. 7 : 9

 

 

Answer : Option D

 

Explanation :

 

Let Cost Price(CP) of 1 litre spirit be Rs.1

Quantity of spirit in 1 litre mixture from vessel A = 5/7

Cost Price(CP) of 1 litre mixture from vessel A = Rs. 5/7

Quantity of spirit in 1 litre mixture from vessel B = 7/13

Cost Price(CP) of 1 litre mixture from vessel B = Rs. 7/13

Quantity of spirit to be obtained in 1 litre mixture from vessel C = 8/13

Cost Price(CP) of 1 litre mixture from vessel C = Rs. 8/13 = Mean Price

By the rule of alligation, we can write as

 

CP of 1 litre mixture from vessel A	CP of 1 litre mixture from vessel B
5/7	7/13
Mean Price
8/13
8/13 – 7/13 = 1/13	5/7 – 8/13 = 9/91

=> Mixture from Vessel A : Mixture from Vessel B = 1/13 : 9/91 = 7 : 9 = Required Ratio

 

4.How many kilograms of sugar costing Rs. 9 per kg must be mixed with 27 kg of sugar costing Rs. 7 per Kg so that there may be a gain of 10 % by selling the mixture at Rs. 9.24 per Kg ?
A. 60 Kg	B. 63 kg
C. 58 Kg	D. 56 Kg

 

 

Answer : Option B

Explanation :

Selling Price(SP) of 1 Kg mixture= Rs. 9.24

Profit = 10%

Cost Price(CP) of 1 Kg mixture= 100×SP /(100+Profit%) =100×9.24/(100+10)

=100×9.24/110=92.4/11=Rs.8.4

By the rule of alligation, we have

CP of 1 kg sugar of 1st kind	CP of 1 kg sugar of 2nd kind
Rs. 9	Rs. 7
Mean Price
Rs.8.4
8.4 – 7 = 1.4	9 – 8.4 = .6

ie, to get a cost price of 8.4, the sugars of kind1 and kind2 should be mixed in the

ratio 1.4 : .6 = 14 : 6 = 7 : 3

Let x Kg of kind1 sugar is mixed with 27 kg of kind2 sugar

then x : 27 = 7 : 3

⇒x/27=7/3

⇒x=(27×7)/3=63

5.A container contains a mixture of two liquids P and Q in the ratio 7 : 5. When 9 litres of mixture are drawn off and the container is filled with Q, the ratio of P and Q becomes 7 : 9. How many litres of liquid P was contained in the container initially?
A. 23	B. 21
C. 19	D. 17

 

 

Answer : Option B

Explanation :

Let’s initial quantity of P in the container be 7x

and initial quantity of Q in the container be 5x

Now 9 litres of mixture are drawn off from the container

Quantity of P in 9 litres of the mixtures drawn off = 9×7/12=63/12=21/ 4

Quantity of Q in 9 litres of the mixtures drawn off = 9×5/12=45/12=1/54

HenceQuantity of P remains in the mixtures after 9 litres is drawn off =7x−21/4

Quantity of Q remains in the mixtures after 9 litres is drawn off =5x−15/4

Since the container is filled with Q after 9 litres of mixture is drawn off,

Quantity of Q in the mixtures=5x−15/4+9=5x+21/4

Given that the ratio of P and Q becomes 7 : 9

⇒(7x−21/4):(5x+21/4)=7:9

⇒(7x−21/4)(5x+21/4)=7/9

63x−(9×21/4)=35x+(7×2/14)

28x=(16×21/4)

x=(16×21)/(4×28)

litres of P contained in the container initially = 7x=(7×16×21)/(4×28)=(16×21)/(4×4)=21

 

6.A dishonest milkman sells his milk at cost price but he mixes it with water and thereby gains 25%. What is the percentage of water in the mixture?
A. 25%	B. 20%
C. 22%	D. 24%

 

 

Answer : Option B

Explanation :

Let CP of 1 litre milk = Rs.1

Given that SP of 1 litre mixture = CP of 1 Litre milk = Rs.1

Given than Gain = 25%

Hence CP of 1 litre mixture = 100/(100+Gain%)×SP

=100(100+25)×1

=100/125=4/5

By the rule of alligation, we have

CP of 1 litre milk	CP of 1 litre water
1	0
CP of 1 litre mixture
4/5
4/5 – 0 = 4/5	1- 4/5 = 1/5

=> Quantity of milk : Quantity of water = 4/5 : 1/5 = 4 : 1

Hence percentage of water in the mixture = 1/5×100=20%