Aptitude & reasoning

 

Pipes and Cistern

 

1. Inlet:

A pipe connected with a tank or a cistern or a reservoir, that fills it, is known as an inlet.

 

Outlet:

A pipe connected with a tank or cistern or reservoir, emptying it, is known as an outlet.

 

2. If a pipe can fill a tank in xhours, then:

part filled in 1 hour =	1	.
	x

3. If a pipe can empty a tank in yhours, then:

part emptied in 1 hour =	1	.
	y

4. If a pipe can fill a tank in xhours and another pipe can empty the full tank in y hours (where y > x), then on opening both the pipes, then

the net part filled in 1 hour =		1	–	1		.
		x		y

5. If a pipe can fill a tank in xhours and another pipe can empty the full tank in y hours (where x > y), then on opening both the pipes, then

the net part emptied in 1 hour =		1	–	1		.
		y		x

 

 

Questions:

 

Level-I:

 

1.	Three pipes A, B and C can fill a tank from empty to full in 30 minutes, 20 minutes, and 10 minutes respectively. When the tank is empty, all the three pipes are opened. A, B and C discharge chemical solutions P,Q and R respectively. What is the proportion of the solution R in the liquid in the tank after 3 minutes?
	A. 5 11 B. 6 11 C. 7 11 D. 8 11

 

2.	Pipes A and B can fill a tank in 5 and 6 hours respectively. Pipe C can empty it in 12 hours. If all the three pipes are opened together, then the tank will be filled in:
	A. 1 13 hours 17 B. 2 8 hours 11 C. 3 9 hours 17 D. 4 1 hours 2

 

3.	A pump can fill a tank with water in 2 hours. Because of a leak, it took 2 hours to fill the tank. The leak can drain all the water of the tank in:
	A. 4 1 hours 3 B. 7 hours C. 8 hours D. 14 hours
4.	Two pipes A and B can fill a cistern in 37 minutes and 45 minutes respectively. Both pipes are opened. The cistern will be filled in just half an hour, if the B is turned off after:
	A. 5 min. B. 9 min. C. 10 min. D. 15 min.

 

5.	A tank is filled by three pipes with uniform flow. The first two pipes operating simultaneously fill the tank in the same time during which the tank is filled by the third pipe alone. The second pipe fills the tank 5 hours faster than the first pipe and 4 hours slower than the third pipe. The time required by the first pipe is:
	A. 6 hours B. 10 hours C. 15 hours D. 30 hours
6.	Two pipes can fill a tank in 20 and 24 minutes respectively and a waste pipe can empty 3 gallons per minute. All the three pipes working together can fill the tank in 15 minutes. The capacity of the tank is:
	A. 60 gallons B. 100 gallons C. 120 gallons D. 180 gallons

 

7.	A tank is filled in 5 hours by three pipes A, B and C. The pipe C is twice as fast as B and B is twice as fast as A. How much time will pipe A alone take to fill the tank?
	A. 20 hours B. 25 hours C. 35 hours D. Cannot be determined E. None of these

 

8.	Two pipes A and B together can fill a cistern in 4 hours. Had they been opened separately, then B would have taken 6 hours more than A to fill the cistern. How much time will be taken by A to fill the cistern separately?
	A. 1 hour B. 2 hours C. 6 hours D. 8 hours

 

9.	Two pipes A and B can fill a tank in 20 and 30 minutes respectively. If both the pipes are used together, then how long will it take to fill the tank?
	A. 12 min B. 15 min C. 25 min D. 50 min

 

10.	Two pipes A and B can fill a tank in 15 minutes and 20 minutes respectively. Both the pipes are opened together but after 4 minutes, pipe A is turned off. What is the total time required to fill the tank?
	A. 10 min. 20 sec. B. 11 min. 45 sec. C. 12 min. 30 sec. D. 14 min. 40 sec.
11.	Level-II:   One pipe can fill a tank three times as fast as another pipe. If together the two pipes can fill the tank in 36 minutes, then the slower pipe alone will be able to fill the tank in:
	A. 81 min. B. 108 min. C. 144 min. D. 192 min.

 

12.	A large tanker can be filled by two pipes A and B in 60 minutes and 40 minutes respectively. How many minutes will it take to fill the tanker from empty state if B is used for half the time and A and B fill it together for the other half?
	A. 15 min B. 20 min C. 27.5 min D. 30 min

 

13.	A tap can fill a tank in 6 hours. After half the tank is filled, three more similar taps are opened. What is the total time taken to fill the tank completely?
	A. 3 hrs 15 min B. 3 hrs 45 min C. 4 hrs D. 4 hrs 15 min

 

14.	Three taps A, B and C can fill a tank in 12, 15 and 20 hours respectively. If A is open all the time and B and C are open for one hour each alternately, the tank will be full in:
	A. 6 hours B. 6 2 hours 3 C. 7 hours D. 7 1 hours 2

 

15.	Three pipes A, B and C can fill a tank in 6 hours. After working at it together for 2 hours, C is closed and A and B can fill the remaining part in 7 hours. The number of hours taken by C alone to fill the tank is:
	A. 10 B. 12 C. 14 D. 16
16.	How much time will the leak take to empty the full cistern? I. The cistern is normally filled in 9 hours.  II. It takes one hour more than the usual time to fill the cistern because of la leak in the bottom.
	A. I alone sufficient while II alone not sufficient to answer B. II alone sufficient while I alone not sufficient to answer C. Either I or II alone sufficient to answer D. Both I and II are not sufficient to answer E. Both I and II are necessary to answer

 

17.	How long will it take to empty the tank if both the inlet pipe A and the outlet pipe B are opened simultaneously? I. A can fill the tank in 16 minutes.  II. B can empty the full tank in 8 minutes.
	A. I alone sufficient while II alone not sufficient to answer B. II alone sufficient while I alone not sufficient to answer C. Either I or II alone sufficient to answer D. Both I and II are not sufficient to answer E. Both I and II are necessary to answer

 

18.	If both the pipes are opened, how many hours will be taken to fill the tank? I. The capacity of the tank is 400 litres. II. The pipe A fills the tank in 4 hours.  III. The pipe B fills the tank in 6 hours.
	A. Only I and II B. Only II and III C. All I, II and III D. Any two of the three E. Even with all the three statements, answer cannot be given.

 

 

Answers:

 

Level-I:

 

Answer:1 Option B

 

Explanation:

Part filled by (A + B + C) in 3 minutes = 3		1	+	1	+	1		=		3 x	11		=	11	.
		30		20		10					60			20

 

Part filled by C in 3 minutes =	3	.
	10

 

Required ratio =		3	x	20		=	6	.
		10		11			11

 

Answer:2 Option C

 

Explanation:

Net part filled in 1 hour		1	+	1	–	1		=	17	.
		5		6		12			60

 

The tank will be full in	60	hours i.e., 3	9	hours.
	17		17

 

 

Answer:3 Option D

 

Explanation:

Work done by the leak in 1 hour =		1	–	3		=	1	.
		2		7			14

Leak will empty the tank in 14 hrs.

 

 

Answer:4 Option B

 

Explanation:

Let B be turned off after x minutes. Then,

Part filled by (A + B) in x min. + Part filled by A in (30 –x) min. = 1.

x		2	+	1		+ (30 – x).	2	= 1
		75		45			75

 

	11x	+	(60 -2x)	= 1
	225		75

11x + 180 – 6x = 225.

x = 9.

 

 

Answer:5 Option C

 

Explanation:

Suppose, first pipe alone takes x hours to fill the tank .

Then, second and third pipes will take (x -5) and (x – 9) hours respectively to fill the tank.

	1	+	1	=	1
	x		(x – 5)		(x – 9)

 

	x – 5 + x	=	1
	x(x – 5)		(x – 9)

(2x – 5)(x – 9) = x(x – 5)

x² – 18x + 45 = 0

(x – 15)(x – 3) = 0

x = 15.    [neglecting x = 3]

 

 

Answer:6 Option C

 

Explanation:

Work done by the waste pipe in 1 minute =	1	–		1	+	1
	15			20		24

 

=		1	–	11
		15		120

 

= –	1	.    [-ve sign means emptying]
	40

 

Volume of	1	part = 3 gallons.
	40

Volume of whole = (3 x 40) gallons = 120 gallon

 

 

Answer:7 Option C

 

Explanation:

Suppose pipe A alone takes x hours to fill the tank.

Then, pipes B and C will take	x	and	x	hours respectively to fill the tank.
	2		4

 

	1	+	2	+	4	=	1
	x		x		x		5

 

	7	=	1
	x		5

x = 35 hrs.

 

Answer:8 Option C

 

Explanation:

Let the cistern be filled by pipe A alone in x hours.

Then, pipe B will fill it in (x + 6) hours.

	1	+	1	=	1
	x		(x + 6)		4

 

	x + 6 + x	=	1
	x(x + 6)		4

x² – 2x – 24 = 0

(x -6)(x + 4) = 0

x = 6.     [neglecting the negative value of x]

 

 

Answer:9 Option A

 

Explanation:

Part filled by A in 1 min =	1	.
	20

 

Part filled by B in 1 min =	1	.
	30

 

Part filled by (A + B) in 1 min =		1	+	1		=	1	.
		20		30			12

Both pipes can fill the tank in 12 minutes.

 

 

Answer:10 Option D

 

Explanation:

Part filled in 4 minutes = 4		1	+	1		=	7	.
		15		20			15

 

Remaining part =		1 –	7		=	8	.
			15			15

 

Part filled by B in 1 minute =	1
	20

 

	1	:	8	:: 1 : x
	20		15

 

x =		8	x 1 x 20		= 10	2	min = 10 min. 40 sec.
		15				3

The tank will be full in (4 min. + 10 min. + 40 sec.) = 14 min. 40 sec.

 

Level-II:

 

Answer:11 Option C

 

Explanation:

Let the slower pipe alone fill the tank in x minutes.

Then, faster pipe will fill it in	x	minutes.
	3

 

	1	+	3	=	1
	x		x		36

 

	4	=	1
	x		36

x = 144 min.

 

 

 

Answer:12 Option D

 

Explanation:

Part filled by (A + B) in 1 minute =		1	+	1		=	1	.
		60		40			24

Suppose the tank is filled in x minutes.

Then,	x		1	+	1		= 1
	2		24		40

 

	x	x	1	= 1
	2		15

x = 30 min.

 

Answer:13 Option B

 

Explanation:

Time taken by one tap to fill half of the tank = 3 hrs.

Part filled by the four taps in 1 hour =		4 x	1		=	2	.
			6			3

 

Remaining part =		1 –	1		=	1	.
			2			2

 

	2	:	1	:: 1 : x
	3		2

 

x =		1	x 1 x	3		=	3	hours i.e., 45 mins.
		2		2			4

So, total time taken = 3 hrs. 45 mins.

 

Answer:14 Option C

 

Explanation:

(A + B)’s 1 hour’s work =		1	+	1		=	9	=	3	.
		12		15			60		20

 

(A + C)’s hour’s work =		1	+	1		=	8	=	2	.
		12		20			60		15

 

Part filled in 2 hrs =		3	+	2		=	17	.
		20		15			60

 

Part filled in 6 hrs =		3 x	17		=	17	.
			60			20

 

Remaining part =		1 –	17		=	3	.
			20			20

 

Now, it is the turn of A and B and	3	part is filled by A and B in 1 hour.
	20

Total time taken to fill the tank = (6 + 1) hrs = 7 hrs.

 

Answer:15 Option C

 

Explanation:

Part filled in 2 hours =	2	=	1
	6		3

 

Remaining part =		1 –	1		=	2	.
			3			3

 

(A + B)’s 7 hour’s work =	2
	3

 

(A + B)’s 1 hour’s work =	2
	21

C’s 1 hour’s work = { (A + B + C)’s 1 hour’s work } – { (A + B)’s 1 hour’s work }

=		1	–	2		=	1
		6		21			14

C alone can fill the tank in 14 hours.

 

Answer:16 Option E

 

Explanation:

1. Time taken to fill the cistern without leak = 9 hours.

Part of cistern filled without leak in 1 hour =	1
	9

1. Time taken to fill the cistern in presence of leak = 10 hours.

Net filling in 1 hour =	1
	10

 

Work done by leak in 1 hour =		1	–	1		=	1
		9		10			90

Leak will empty the full cistern in 90 hours.

Clearly, both I and II are necessary to answer the question.

Correct answer is (E).

 

 

 

 

Answer:17 Option E

 

Explanation:

I. A’s 1 minute’s filling work =	1
	16

 

II. B’s 1 minute’s filling work =	1
	8

 

(A + B)’s 1 minute’s emptying work =		1	–	1		=	1
		8		16			16

Tank will be emptied in 16 minutes.

Thus, both I and II are necessary to answer the question.

Correct answer is (E).

 

Answer:18 Option B

 

Explanation:

II. Part of the tank filled by A in 1 hour =	1
	4

 

III. Part of the tank filled by B in 1 hour =	1
	6

 

(A + B)’s 1 hour’s work =		1	+	1		=	5
		4		6			12

 

A and B will fill the tank in	12	hrs = 2 hrs 24 min.
	5

So, II and III are needed.

Correct answer is (B).

Square Root & Cube Root

 

Step 1: First of all group the number in pairs of 2 starting from the right.

 

Step 2: To get the ten’s place digit, Find the nearest square (equivalent or greater than or less than) to the first grouped pair from left and put the square root of the square.

 

Step 3: To get the unit’s place digit of the square root

 

Remember the following

If number ends in	Unit’s place digit of the square root
1	1 or 9(10-1)
4	2 or 8(10-2)
9	3 or 7(10-3)
6	4or 6(10-4)
5	5
0	0

 

Lets see the logic behind this for a better understanding

We know,

1²=1

2²=4

3²=9

4²=16

5²=25

6²=36

7²=49

8²=64

9²=81

10²=100

 

Now, observe the unit’s place digit of all the squares.

Do you find anything common?

 

We notice that,

Unit’s place digit of both 1² and 9² is 1.

Unit’s place digit of both 2² and 8² is 4

Unit’s place digit of both 3² and 7² is 9

Unit’s place digit of both 4² and 6² is 6.

Step 4: Multiply the ten’s place digit (found in step 1) with its consecutive number and compare the result obtained with the first pair of the original number from left.

 

Remember,

If first pair of the original number > Result obtained on multiplication then  select the greater number  out of the two numbers as the unit’s place digit of the square root.

 

If firstpair of the original number < the result obtained on multiplication,then select the lesser number out of the two numbers as the unit’s place digit of the square root.

 

 

Let us consider an example to get a better understanding of the method

 

 

Example 1: √784=?

Step 1: We start by grouping the numbers in pairs of two from right as follows

7 84

 

Step 2: To get the ten’s place digit,

We find that nearest square to first group (7) is 4 and √4=2

Therefore ten’s place digit=2

 

Step 3: To get the unit’s place digit,

We notice that the number ends with 4, So the unit’s place digit of the square root should be either 2 or 8(Refer table).

 

Step 4: Multiplying the ten’s place digit of the square root that we arrived at in step 1(2) and its consecutive number(3) we get,

2×3=6
ten’s place digit of original number > Multiplication result
7>6
So we need to select the greater number (8) as the unit’s place digit of the square root.
Unit’s place digit =8

Ans:√784=28

 

 

 

Cube roots of perfect cubes

It may take two-three minutes to find out cube root of a perfect cube by using conventional method. However we can find out cube roots of perfect cubes very fast, say in one-two seconds using Vedic Mathematics.

We need to remember some interesting properties of numbers to do these quick mental calculations which are given below.

 

Points to remember  for speedy  calculation of cube roots

1. To calculate cube root of any perfect cube quickly, we need to remember the cubes of 1 to 10 which is given below.

13	=	1
23	=	8
33	=	27
43	=	64
53	=	125
63	=	216
73	=	343
83	=	512
93	=	729
103	=	1000

2. From the above cubes of 1 to 10, we need to remember an interesting property.

13 = 1	=>	If last digit of the perfect cube = 1, last digit of the cube root = 1
23 = 8	=>	If last digit of the perfect cube = 8, last digit of the cube root = 2
33 = 27	=>	If last digit of the perfect cube = 7, last digit of the cube root = 3
43 = 64	=>	If last digit of the perfect cube = 4, last digit of the cube root = 4
53 = 125	=>	If last digit of the perfect cube =5, last digit of the cube root = 5
63 = 216	=>	If last digit of the perfect cube = 6, last digit of the cube root = 6
73 = 343	=>	If last digit of the perfect cube = 3, last digit of the cube root = 7
83 = 512	=>	If last digit of the perfect cube = 2, last digit of the cube root = 8
93 = 729	=>	If last digit of the perfect cube = 9, last digit of the cube root = 9
103 = 1000	=>	If last digit of the perfect cube = 0, last digit of the cube root = 0

 

It’s very easy to remember the relations given above because

1	->	1	(Same numbers)
8	->	2	(10’s complement of 8 is 2 and 8+2 = 10)
7	->	3	(10’s complement of 7 is 3 and 7+3 = 10)
4	->	4	(Same numbers)
5	->	5	(Same numbers)
6	->	6	(Same numbers)
3	->	7	(10’s complement of 3 is 7 and 3+7 = 10)
2	->	8	(10’s complement of 2 is 8 and 2+8 = 10)
9	->	9	(Same numbers)
0	->	0	(Same numbers)

 

Also see
8 ->  2 and 2 ->  8
7 -> 3 and 3-> 7

 

 

 

 

 

Questions

Level-I

1.	The cube root of .000216 is:
	A. .6 B. .06 C. 77 D. 87

 

2.	What should come in place of both x in the equation x = 162 . 128 x
	A. 12 B. 14 C. 144 D. 196

 

3.	The least perfect square, which is divisible by each of 21, 36 and 66 is:
	A. 213444 B. 214344 C. 214434 D. 231444

 

4.	1.5625 = ?
	A. 1.05 B. 1.25 C. 1.45 D. 1.55

 

5.	If 35 + 125 = 17.88, then what will be the value of 80 + 65 ?
	A. 13.41 B. 20.46 C. 21.66 D. 22.35
6.	If a = 0.1039, then the value of 4a2 – 4a + 1 + 3a is:
	A. 0.1039 B. 0.2078 C. 1.1039 D. 2.1039

 

7.	If x = 3 + 1 and y = 3 – 1 , then the value of (x2 + y2) is: 3 – 1 3 + 1
	A. 10 B. 13 C. 14 D. 15

 

8.	A group of students decided to collect as many paise from each member of group as is the number of members. If the total collection amounts to Rs. 59.29, the number of the member is the group is:
	A. 57 B. 67 C. 77 D. 87

 

9.	The square root of (7 + 35) (7 – 35) is
	A. 5 B. 2 C. 4 D. 35

 

 

 

10.	If 5 = 2.236, then the value of 5 – 10 + 125 is equal to: 2 5
	A. 5.59 B. 7.826 C. 8.944 D. 10.062       Level-II
11.	625 x 14 x 11 is equal to: 11 25 196
	A. 5 B. 6 C. 8 D. 11

 

12.	0.0169 x ? = 1.3
	A. 10 B. 100 C. 1000 D. None of these

 

13.	3 – 1 2 simplifies to: 3
	A. 3 4 B. 4 3 C. 4 3 D. None of these

 

14.	How many two-digit numbers satisfy this property.: The last digit (unit’s digit) of the square of the two-digit number is 8 ?
	A. 1 B. 2 C. 3 D. None of these

 

15.	The square root of 64009 is:
	A. 253 B. 347 C. 363 D. 803

 

 

16. √29929 = ?

A. 173 B. 163 C. 196 D. 186

 

 

 

 

 

 

17. √106.09 = ?
A.	10.6
B.	10.5
C.	10.3
D.	10.2

 

18.  ?/√196 = 5
A.	76
B.	72
C.	70
D.	75

 

Answers

Level-I

 

Answer:1 Option B

 

Explanation:

(.000216)1/3	=		216		1/3
			106

 

=		6 x 6 x 6		1/3
		102 x 102 x 102

 

=	6
	102

 

=	6
	100

= 0.06

 

Answer:2 Option A

 

Explanation:

Let	x	=	162
	128		x

Then x² = 128 x 162

= 64 x 2 x 18 x 9

= 8² x 6² x 3²

= 8 x 6 x 3

= 144.

x = 144 = 12.

 

Answer:3 Option A

 

Explanation:

L.C.M. of 21, 36, 66 = 2772.

Now, 2772 = 2 x 2 x 3 x 3 x 7 x 11

To make it a perfect square, it must be multiplied by 7 x 11.

So, required number = 2² x 3² x 7² x 11² = 213444

 

Answer:4 Option B

 

Explanation:

1|1.5625( 1.25

|1

|——-

22| 56

| 44

|——-

245| 1225

| 1225

|——-

|    X

|——-

1.5625 = 1.25.

 

 

Answer:5 Option D

 

Explanation:

35 + 125 = 17.88

35 + 25 x 5 = 17.88

35 + 55 = 17.88

85 = 17.88

5 = 2.235

80 + 65 = 16 x 5 + 65

= 45 + 65

= 105 = (10 x 2.235) = 22.35

 

 

 

Answer:6 Option C

 

Explanation:

4a² – 4a + 1 + 3a = (1)² + (2a)² – 2 x 1 x 2a + 3a

= (1 – 2a)² + 3a

= (1 – 2a) + 3a

= (1 + a)

= (1 + 0.1039)

= 1.1039

 

Answer:7 Option C

 

Explanation:

x =	(3 + 1)	x	(3 + 1)	=	(3 + 1)2	=	3 + 1 + 23	= 2 + 3.
	(3 – 1)		(3 + 1)		(3 – 1)		2

 

y =	(3 – 1)	x	(3 – 1)	=	(3 – 1)2	=	3 + 1 – 23	= 2 – 3.
	(3 + 1)		(3 – 1)		(3 – 1)		2

x² + y² = (2 + 3)² + (2 – 3)²

= 2(4 + 3)

= 14

 

Answer:8 Option C

 

Explanation:

Money collected = (59.29 x 100) paise = 5929 paise.

Number of members = 5929 = 77

 

 

Answer:9 Option B

 

Explanation:

(7 + 35)(7 – 35)

=

(7)2 – (35)2

= 49 – 45  = 4  = 2

 

 

Answer:10 Option B

 

Explanation:

5	–	10	+ 125	=	(5)2 – 20 + 25 x 55
2		5			25

 

=	5 – 20 + 50
	25

 

=	35	x	5
	25		5

 

=	355
	10

 

=	7 x 2.236
	2

 

= 7 x 1.118

 

= 7.826

 

 

Level-II

Answer:11 Option A

 

Explanation:

Given Expression =	25	x	14	x	11	= 5.
	11		5		14

 

 

 

Answer:12 Option B

 

Explanation:

Let 0.0169 x x = 1.3.

Then, 0.0169x = (1.3)² = 1.69

x =	1.69	= 100
	0.0169

 

 

 

Answer:13 Option C

 

Explanation:

	3 –	1		2	= (3)2 +		1		2	– 2 x 3 x	1
		3					3				3

 

= 3 +	1	– 2
	3

 

= 1 +	1
	3

 

=	4
	3

 

 

 

Answer:14 Option D

 

Explanation:

A number ending in 8 can never be a perfect square.

 

 

Answer:15 Option A

 

Explanation:

2 |64009( 253      |4      |———-45  |240      |225      |———-503| 1509      |  1509      |———-      |     X      |———-

64009 = 253.

 

 

Answer:16 Option A

 

Explanation:
√29929 = So, √29929 = 173

 

 

Answer:17 Option C

 

Answer:18 Option C

Surds

A surd is a square root which cannot be reduced to a rational number.

For example,  is not a surd.

However  is a surd.

If you use a calculator, you will see that  and we will need to round the answer correct to a few decimal places. This makes it less accurate.

If it is left as , then the answer has not been rounded, which keeps it exact.

Here are some general rules when simplifying expressions involving surds.

 

 

 

1. a^mx aⁿ = a^m + n

am	= am – n
an

• (a^m)ⁿ= a^mn

 

1. (ab)ⁿ= aⁿbⁿ

 

	a		n	=	an
	b				bn

1. a⁰= 1

 

 

Questions

Level-I

 

1.	(17)3.5 x (17)? = 178
	A. 2.29 B. 2.75 C. 4.25 D. 4.5

 

2.	If a x – 1 = b x – 3 , then the value of x is: b a
	A. 1 2 B. 1 C. 2 D. 7 2

 

3.	Given that 100.48 = x, 100.70 = y and xz = y2, then the value of z is close to:
	A. 1.45 B. 1.88 C. 2.9 D. 3.7

 

4.	If 5a = 3125, then the value of 5(a – 3) is:
	A. 25 B. 125 C. 625 D. 1625

 

5.	If 3(x – y) = 27 and 3(x + y) = 243, then x is equal to:
	A. 0 B. 2 C. 4 D. 6

 

.6.	(256)0.16 x (256)0.09 = ?
	A. 4 B. 16 C. 64 D. 256.25

 

7.	The value of [(10)150 ÷ (10)146]
	A. 1000 B. 10000 C. 100000 D. 106

 

8.	1  + 1 + 1 = ? 1 + x(b – a) + x(c – a) 1 + x(a – b) + x(c – b) 1 + x(b – c) + x(a – c)
	A. 0 B. 1 C. xa – b – c D. None of these

 

9.	(25)7.5 x (5)2.5 ÷ (125)1.5 = 5?
	A. 8.5 B. 13 C. 16 D. 17.5 E. None of these

 

10.	(0.04)-1.5 = ?
	A. 25 B. 125 C. 250 D. 625

 

 

Level-II

 

11.	(243)n/5 x 32n + 1 = ? 9n x 3n – 1
	A. 1 B. 2 C. 9 D. 3n

 

12.	1 + 1 = ? 1 + a(n – m) 1 + a(m – n)
	A. 0 B. 1 2 C. 1 D. am + n

 

13.	If m and n are whole numbers such that mn = 121, the value of (m – 1)n + 1 is:
	A. 1 B. 10 C. 121 D. 1000

 

14.	xb (b + c – a) . xc (c + a – b) . xa (a + b – c) = ? xc xa xb
	A. xabc B. 1 C. xab + bc + ca D. xa + b + c

 

15. If 5√5 * 5³÷ 5^-3/2= 5^a+2 , the value of a is:
    A. 4
    B. 5
    C. 6
    D. 8

16.(132)7 ×(132)? =(132)11.5. A. 3 B. 3.5 C. 4 D. 4.5     17. (ab)x−2=(ba)x−7. What is the value   of x ?   A. 3 B. 4 C. 3.5 D. 4.5       18. (0.04)-2.5 = ?   A. 125 B. 25 C. 3125 D. 625

Answers

Level-I

Answer:1 Option D

 

Explanation:

Let (17)^3.5 x (17)^x = 17⁸.

Then, (17)^{3.5 + x} = 17⁸.

3.5 + x = 8

x = (8 – 3.5)

x = 4.5

 

Answer:2 Option C

 

Explanation:

Given	a		x – 1	=		b		x – 3
	b					a

 

a

x – 1

=

a

-(x – 3)

=

a

(3 – x)

b

b

b

x – 1 = 3 – x

2x = 4

x = 2.

 

 

Answer:3 Option C

 

Explanation:

x^z = y²        10^(0.48z) = 10^{(2 x 0.70)} = 10^1.40

0.48z = 1.40

z =	140	=	35	= 2.9 (approx.)
	48		12

 

Answer:4 Option A

 

Explanation:

5^a = 3125        5^a = 5⁵

a = 5.

5^{(a – 3)} = 5^{(5 – 3)} = 5² = 25.

 

 

Answer:5 Option C

 

Explanation:

3^x – y = 27 = 3³        x – y = 3 ….(i)

3^x + y = 243 = 3⁵        x + y = 5 ….(ii)

On solving (i) and (ii), we get x = 4

 

 

Answer:6 Option A

 

Explanation:

(256)^0.16 x (256)^0.09 = (256)^{(0.16 + 0.09)}

= (256)^0.25

= (256)^(25/100)

= (256)^(1/4)

= (4⁴)^(1/4)

= 4^4(1/4)

= 4¹

= 4

Answer:7 Option B

 

Explanation:

(10)150 ÷ (10)146 =	10150
	10146

= 10^{150 – 146}

= 10⁴

= 10000.

 

Answer:8 Option B

 

Explanation:

Given Exp. =	1  + 1  + 1 1 + xb + xc xa xa 1 + xa + xc xb xb 1 + xb + xa xc xc

 

=	xa	+	xb	+	xc
	(xa + xb + xc)		(xa + xb + xc)		(xa + xb + xc)

 

=	(xa + xb + xc)
	(xa + xb + xc)

= 1.

 

Answer:9 Option B

 

Explanation:

Let (25)^7.5 x (5)^2.5 ÷ (125)^1.5 = 5^x.

Then,	(52)7.5 x (5)2.5	= 5x
	(53)1.5

 

	5(2 x 7.5) x 52.5	= 5x
	5(3 x 1.5)

 

	515 x 52.5	= 5x
	54.5

5^x = 5^{(15 + 2.5 – 4.5)}

5^x = 5¹³

x = 13.

 

Answer:10 Option B

 

Explanation:

(0.04)-1.5 =		4		-1.5
		100

 

=		1		-(3/2)
		25

= (25)^(3/2)

= (5²)^(3/2)

= (5)^{2 x (3/2)}

= 5³

= 125.

 

Level-II

 

Answer:11 Option C

 

Explanation:

Given Expression	= (243)(n/5) x 32n + 1 9n x 3n – 1
	= (35)(n/5) x 32n + 1 (32)n x 3n – 1
	= (35 x (n/5) x 32n + 1) (32n x 3n – 1)
	= 3n x 32n + 1 32n x 3n – 1
	= 3(n + 2n + 1) 3(2n + n – 1)
	= 33n + 1 33n – 1
	= 3(3n + 1 – 3n + 1)   = 32   = 9.

Answer:12 Option C

 

Explanation:

1	+	1	=	1  + 1 1 + an am 1 + am an
1 + a(n – m)		1 + a(m – n)

 

=	am	+	an
	(am + an)		(am + an)

 

=	(am + an)
	(am + an)

= 1.

 

Answer:13 Option D

 

Explanation:

We know that 11² = 121.

Putting m = 11 and n = 2, we get:

(m – 1)^{n + 1} = (11 – 1)^{(2 + 1)} = 10³ = 1000.

 

Answer:14 Option B

 

Explanation:

Given Exp.	= x(b – c)(b + c – a) . x(c – a)(c +a – b) . x(a – b)(a + b – c)
	= x(b – c)(b + c) – a(b – c)  .  x(c – a)(c + a) – b(c – a) .  x(a – b)(a + b) – c(a – b)
	= x(b2 – c2 + c2 – a2 + a2 – b2)  .   x–a(b – c) – b(c – a) – c(a – b)
	= (x0 x x0)
	= (1 x 1) = 1.

 

Answer:15 option C

 

Answer:16

Explanation

am.an=am+n

(132)⁷ × (132)^x = (132)^11.5

=> 7 + x = 11.5

=> x = 11.5 – 7 = 4.5

 

 

Answer:17

Explanation:

an=1a−n

(ab)x−2=(ba)x−7⇒(ab)x−2=(ab)−(x−7)⇒x−2=−(x−7)⇒x−2=−x+7⇒x−2=−x+7⇒2x=9⇒x=92=4.5

 

Answer:18

Explanation:

a−n=1/an

(0.04)−2.5=(1/.04)2.5=(100/4)2.5=(25)2.5=(52)2.5=(52)(5/2)=55=3125

 

Simplification

Simplification is one of the most important part of Quantitative Aptitude section of any competitive exam. Today I am sharing all the techniques to solve Simplification questions quickly.

Rules of Simplification

V → Vinculum

B → Remove Brackets – in the order ( ) , { }, [ ]

O → Of
D → Division

M → Multiplication

A → Addition

S → Subtraction

 

Classification

Types	Description
Natural Numbers:	all counting numbers ( 1,2,3,4,5….∞)
Whole Numbers:	natural number + zero( 0,1,2,3,4,5…∞)
Integers:	All whole numbers including Negative number + Positive number(∞……-4,-3,-2,-1,0,1,2,3,4,5….∞)
Even & Odd Numbers :	All whole number divisible by 2 is Even (0,2,4,6,8,10,12…..∞) and which does not divide by 2 are Odd (1,3,5,7,9,11,13,15,17,19….∞)
Prime Numbers:	It can be positive or negative except 1, if the number is not divisible by any number except the number itself.(2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61….∞)
Composite Numbers:	Natural numbers which are not prime
Co-Prime:	Two natural number a and b are said to be co-prime if their HCF is 1.

 

Divisibility

Numbers	IF A Number	Examples
Divisible by 2	End with 0,2,4,6,8 are divisible by 2	254,326,3546,4718 all are divisible by 2
Divisible by 3	Sum of its digits  is divisible by 3	375,4251,78123 all are divisible by 3.  [549=5+4+9][5+4+9=18]18 is divisible by 3  hence 549 is divisible by 3.
Divisible by 4	Last two digit divisible by 4	5648 here last 2 digits are 48 which is divisible by 4 hence 5648 is also divisible by 4.
Divisible by 5	Ends with 0 or 5	225 or 330 here last digit digit is 0 or 5 that mean both the numbers are divisible by 5.
Divisible by 6	Divides by Both 2 & 3	4536 here last digit is 6 so it divisible by 2 & sum of its digit (like 4+5+3+6=18) is 18 which is divisible by 3.Hence 4536 is divisible by 6.
Divisible by 8	Last 3 digit divide by 8	746848 here last 3 digit 848 is divisible by 8 hence 746848 is also divisible by 8.
Divisible by 10	End with 0	220,450,1450,8450 all numbers has a last digit zero it means all are divisible by 10.
Divisible by 11	[Sum of its digit in odd places-Sum of its digits in even places]= 0 or multiple of 11	Consider the number 39798847 (Sum of its digits at odd places)-(Sum of its digits at even places)(7+8+9+9)-(4+8+7+3) (23-12) 23-12=11, which is divisible by 11. So 39798847 is divisible by 11.

Division & Remainder Rules

Suppose we divide 45 by 6

 

hence ,represent it as:

dividend = ( divisor✘quotient ) + remainder

or

divisior= [(dividend)-(remainder] / quotient

could be write it as

x = kq + r where (x = dividend,k = divisor,q = quotient,r = remainder)

 

 

Rules

1. Modulus of a Real Number:

Modulus of a real number a is defined as

|a| =		a, if a > 0
		–a, if a < 0

Thus, |5| = 5 and |-5| = -(-5) = 5.

2. Virnaculum (or Bar):

When an expression contains Virnaculum, before applying the ‘BODMAS’ rule, we simplify the expression under the Virnaculum.

 

 

Example:

On dividing a certain number by 342, we get 47 as remainder. If the same number is divided by 18, what will be the remainder ?

Number = 342k + 47

( 18 ✘19k ) + ( 18 ✘2 ) + 11

18 ✘( 19k + 2 ) +11.

Remainder = 11

 

Sum Rules

(1+2+3+………+n) = ¹/₂ n(n+1)

(1²+2²+3²+………+n²) = ¹/₆ n (n+1) (2n+1)

(1³+2³+3³+………+n³) = ¹/4 n² (n+1)²

 

Questions:

Level-I:

 

1.	A man has Rs. 480 in the denominations of one-rupee notes, five-rupee notes and ten-rupee notes. The number of notes of each denomination is equal. What is the total number of notes that he has ?
	A. 45 B. 60 C. 75 D. 90

 

2.	There are two examinations rooms A and B. If 10 students are sent from A to B, then the number of students in each room is the same. If 20 candidates are sent from B to A, then the number of students in A is double the number of students in B. The number of students in room A is:
	A. 20 B. 80 C. 100 D.   200

 

3. The price of 10 chairs is equal to that of 4 tables. The price of 15 chairs and 2 tables together is Rs. 4000. The total price of 12 chairs and 3 tables is:

A. Rs. 3500 B. Rs. 3750 C. Rs. 3840 D. Rs. 3900

 

4. If a – b = 3 and a2 + b2 = 29, find the value of ab.

A. 10 B. 12 C. 15 D. 18

 

5.	The price of 2 sarees and 4 shirts is Rs. 1600. With the same money one can buy 1 saree and 6 shirts. If one wants to buy 12 shirts, how much shall he have to pay ?
	A. Rs. 1200 B. Rs. 2400 C. Rs. 4800 D. Cannot be determined E. None of these

 

6.	A sum of Rs. 1360 has been divided among A, B and C such that A gets  of what B gets and B gets  of what C gets. B’s share is:
	A. Rs. 120 B. Rs. 160 C. Rs. 240 D. Rs. 300

 

7.	One-third of Rahul’s savings in National Savings Certificate is equal to one-half of his savings in Public Provident Fund. If he has Rs. 1,50,000 as total savings, how much has he saved in Public Provident Fund ?
	A. Rs. 30,000 B. Rs. 50,000 C. Rs. 60,000 D. Rs. 90,000

 

8.	A fires 5 shots to B’s 3 but A kills only once in 3 shots while B kills once in 2 shots. When B has missed 27 times, A has killed:
	A. 30 birds B. 60 birds C. 72 birds D. 90 birds

 

9.	Eight people are planning to share equally the cost of a rental car. If one person withdraws from the arrangement and the others share equally the entire cost of the car, then the share of each of the remaining persons increased by:
	A. 1 7 B. 1 8 C. 1 9 D. 7 8

 

10.	To fill a tank, 25 buckets of water is required. How many buckets of water will be required to fill the same tank if the capacity of the bucket is reduced to two-fifth of its present ?
	A. 10 B. 35 C. 62.5 D. Cannot be determined E. None of these

 

Level-II:

1.	In a regular week, there are 5 working days and for each day, the working hours are 8. A man gets Rs. 2.40 per hour for regular work and Rs. 3.20 per hours for overtime. If he earns Rs. 432 in 4 weeks, then how many hours does he work for ?
	A. 160 B. 175 C. 180 D. 195

 

12.	Free notebooks were distributed equally among children of a class. The number of notebooks each child got was one-eighth of the number of children. Had the number of children been half, each child would have got 16 notebooks. Total how many notebooks were distributed ?
	A. 256 B. 432 C. 512 D. 640 E. None of these

 

13.	A man has some hens and cows. If the number of heads be 48 and the number of feet equals 140, then the number of hens will be:
	A. 22 B. 23 C. 24 D. 26

 

14.	(469 + 174)2 – (469 – 174)2 = ? (469 x 174)
	A. 2 B. 4 C. 295 D. 643

 

15.	David gets on the elevator at the 11th floor of a building and rides up at the rate of 57 floors per minute. At the same time, Albert gets on an elevator at the 51st floor of the same building and rides down at the rate of 63 floors per minute. If they continue travelling at these rates, then at which floor will their paths cross ?
	A. 19 B. 28 C. 30 D. 37

 

16. Find the value of 1/(3+1/(3+1/(3-1/3)))

A.) 3/10	B.) 10/3
C.) 27/89	D.) 89/27

17. 
    Find the value of

A.) 3½ 99;	B.) 34/99
C.) 2.131313	D.) 3.141414

 

18.Find the value of

((0.1)³ + (0.6)³ + (0.7)³ − (0.3)(0.6)(0.7))/((0.1)² + (0.6)² + (0.7)² − 0.006 − 0.42 − 0.07)

 

A.) 14/10	B.) 1.35
C.) 13/10	D.) 0

 

 

 

 

19. Solve(0.76 × 0.76 × 0.76 − 0.008)/(0.76 × 0.76 + 0.76 × 0.2 + 0.04)

A.) 0.56	B.) 0.65
C.) 0.54	D.) 0.45

20. Find the value of

A.) 1.5	B.) -1.5
C.) 1	D.) 0

 

Answers:

Level-I

 

Answer:1 Option D

 

Explanation:

Let number of notes of each denomination be x.

Then x + 5x + 10x = 480

16x = 480

x = 30.

Hence, total number of notes = 3x = 90.

 

 

Answer:2 Option C

 

Explanation:

Let the number of students in rooms A and B be x and y respectively.

Then, x – 10 = y + 10      x – y = 20 …. (i)

and x + 20 = 2(y – 20)      x – 2y = -60 …. (ii)

Solving (i) and (ii) we get: x = 100 , y = 80.

The required answer A = 100.

 

 

Answer:3 Option D

 

Explanation:

Let the cost of a chair and that of a table be Rs. x and Rs. y respectively.

Then, 10x = 4y   or   y =	5	x.
	2

15x + 2y = 4000

15x + 2 x	5	x = 4000
	2

20x = 4000

x = 200.

So, y =		5	x 200		= 500.
		2

Hence, the cost of 12 chairs and 3 tables = 12x + 3y

= Rs. (2400 + 1500)

= Rs. 3900.

 

 

 

 

 

Answer:4 Option A

 

Explanation:

2ab = (a² + b²) – (a – b)²

= 29 – 9 = 20

ab = 10.

 

 

 

Answer:5 Option B

 

Explanation:

Let the price of a saree and a shirt be Rs. x and Rs. y respectively.

Then, 2x + 4y = 1600 …. (i)

and x + 6y = 1600 …. (ii)

 

Divide equation (i) by 2, we get the below equation.

 

=> x +  2y =  800. — (iii)

 

Now subtract (iii) from (ii)

 

x +  6y = 1600  (-)

x +  2y =  800

—————-

4y =  800

—————-

 

Therefore, y = 200.

 

Now apply value of y in (iii)

 

=>  x + 2 x 200 = 800

 

=>  x + 400 = 800

 

Therefore x = 400

 

Solving (i) and (ii) we get x = 400, y = 200.

Cost of 12 shirts = Rs. (12 x 200) = Rs. 2400.

 

 

Answer:6 Option C

 

Explanation:

Let C’s share = Rs. x

Then, B’s share = Rs.	x	,   A’s share = Rs.		2	x	x		= Rs.	x
	4			3		4			6

 

	x	+	x	+ x = 1360
	6		4

 

	17x	= 1360
	12

 

x =	1360 x 12	= Rs. 960
	17

 

Hence, B’s share = Rs.		960		= Rs. 240.

 

 

Answer:7 Option C

 

Explanation:

Let savings in N.S.C and P.P.F. be Rs. x and Rs. (150000 – x) respectively. Then,

1	x =	1	(150000 – x)
3		2

 

	x	+	x	= 75000
	3		2

 

	5x	= 75000
	6

 

x =	75000 x 6	= 90000
	5

Savings in Public Provident Fund = Rs. (150000 – 90000) = Rs. 60000

 

 

Answer:8 Option A

 

Explanation:

Let the total number of shots be x. Then,

Shots fired by A =	5	x
	8

 

Shots fired by B =	3	x
	8

 

Killing shots by A =	1	of	5	x	=	5	x
	3		8			24

 

Shots missed by B =	1	of	3	x	=	3	x
	2		8			16

 

	3x	= 27 or x =		27 x 16		= 144.
	16			3

 

Birds killed by A =	5x	=		5	x 144		= 30.
	24			24

 

 

 

 

 

 

 

Answer:9 Option A

 

Explanation:

Original share of 1 person =	1
	8

 

New share of 1 person =	1
	7

 

Increase =		1	–	1		=	1
		7		8			56

 

Required fraction =	(1/56)	=		1	x	8		=	1
	(1/8)			56		1			7

 

 

Answer:10 Option C

 

Explanation:

Let the capacity of 1 bucket = x.

Then, the capacity of tank = 25x.

New capacity of bucket =	2	x
	5

 

Required number of buckets =	25x
	(2x/5)

 

=	25x	x	5
			2x

 

=	125
	2

= 62.5

 

Level-II:

Answer:11 Option B

 

Explanation:

Suppose the man works overtime for x hours.

Now, working hours in 4 weeks = (5 x 8 x 4) = 160.

160 x 2.40 + x x 3.20 = 432

3.20x = 432 – 384 = 48

x = 15.

Hence, total hours of work = (160 + 15) = 175.

 

Answer:12 Option C

 

Explanation:

Let total number of children be x.

Then, x x	1	x =	x	x 16     x = 64.
	8		2

 

Number of notebooks =	1	x2 =		1	x 64 x 64		= 512

 

Answer:13 Option D

 

Explanation:

Let the number of hens be x and the number of cows be y.

Then, x + y = 48 …. (i)

and 2x + 4y = 140      x + 2y = 70 …. (ii)

Solving (i) and (ii) we get: x = 26, y = 22.

The required answer = 26.

 

Answer:14 Option B

 

Explanation:

Given exp. =	(a + b)2 – (a – b)2
	ab

 

=	4ab
	ab

= 4 (where a = 469, b = 174.)

 

Answer:15 Option C

 

Explanation:

Suppose their paths cross after x minutes.

Then, 11 + 57x = 51 – 63x        120x = 40

x =	1
	3

 

Number of floors covered by David in (1/3) min. =		1	x 57		= 19.
		3

So, their paths cross at (11 +19) i.e., 30^th floor.

 

Answer:16 Option ‘C’

Explanation:

1/[3 + (1/(3+1/(3 – 1/3)))]

=> 1/[3 + 1/(3 + 1/(8/3))]

=> 1/[3 + 1/(3 + 3/8)]

=> 1/[3 + 8/27]

=> 1/(89/27)

=> 27/89

 

Answer:17 Option ‘D’

Explanation:

6/9 + 7/9 + 9/9 + 69/99

2/3 + 7/9 + 1 + 69/99

(66 + 77 + 99 + 69)/99

311/99 => 3.141414

 

Answer:18 Option ‘A’

Explanation:

((0.1)³ + (0.6)³ + (0.7)³ − (0.3)(0.6)(0.7))/((0.1)² + (0.6)² + (0.7)² − 0.006 − 0.42 − 0.07)

=> (0.1 + 0.6 + 0.7)³/(0.1 + 0.6 + 0.7)²

=> 0.1 + 0.6 + 0.7 => 1.4 = 14/10

 

Answer:19 Option ‘A’

 

Answer:20 Option ‘D’

11/30 − [1/6 + 1/5 + [7/12 − 7/12]]

11/30 − [1/6 + 1/5 + [0]]

11/30 − [(5 + 6)/30]

11/30 − 11/30 = 0.

Profit and loss

 

IMPORTANT FACTS

Cost Price:

The price, at which an article is purchased, is called its cost price, abbreviated as C.P.

 

Selling Price:

The price, at which an article is sold, is called its selling prices, abbreviated as S.P.

 

Profit or Gain:

If S.P. is greater than C.P., the seller is said to have a profit or gain.

 

Loss:

If S.P. is less than C.P., the seller is said to have incurred a loss.

 

IMPORTANT FORMULAE

1. Gain = (S.P.) – (C.P.)
2. Loss = (C.P.) – (S.P.)
3. Loss or gain is always reckoned on C.P.
4. Gain Percentage: (Gain %)

Gain % =		Gain x 100
		C.P.

5. Loss Percentage: (Loss %)

Loss % =		Loss x 100
		C.P.

6. Selling Price: (S.P.)

SP =		(100 + Gain %)	x C.P
		100

7. Selling Price: (S.P.)

SP =		(100 – Loss %)	x C.P.
		100

8. Cost Price: (C.P.)

C.P. =		100	x S.P.
		(100 + Gain %)

9. Cost Price: (C.P.)

C.P. =		100	x S.P.
		(100 – Loss %)

10. If an article is sold at a gain of say 35%, then S.P. = 135% of C.P.
11. If an article is sold at a loss of say, 35% then S.P. = 65% of C.P.
12. When a person sells two similar items, one at a gain of say x%, and the other at a loss of x%, then the seller always incurs a loss given by:

Loss % =		Common Loss and Gain %		2	=		x		2	.
		10					10

13. If a trader professes to sell his goods at cost price, but uses false weights, then

Gain % =		Error	x 100	%.
		(True Value) – (Error)

 

Questions:

Level-I:

 

 

1.	Alfred buys an old scooter for Rs. 4700 and spends Rs. 800 on its repairs. If he sells the scooter for Rs. 5800, his gain percent is:
	A. 4 4 % 7 B. 5 5 % 11 C. 10% D. 12%

 

2.	The cost price of 20 articles is the same as the selling price of x articles. If the profit is 25%, then the value of xis:
	A. 15 B. 16 C. 18 D. 25

 

3.	If selling price is doubled, the profit triples. Find the profit percent.
	A. 66 2 3 B. 100 C. 105 1 3 D. 120

 

4.	In a certain store, the profit is 320% of the cost. If the cost increases by 25% but the selling price remains constant, approximately what percentage of the selling price is the profit?
	A. 30% B. 70% C. 100% D. 250%

 

 

5.	A vendor bought toffees at 6 for a rupee. How many for a rupee must he sell to gain 20%?
	A. 3 B. 4 C. 5 D. 6

 

6.	The percentage profit earned by selling an article for Rs. 1920 is equal to the percentage loss incurred by selling the same article for Rs. 1280. At what price should the article be sold to make 25% profit?
	A. Rs. 2000 B. Rs. 2200 C. Rs. 2400 D. Data inadequate

 

7.	A shopkeeper expects a gain of 22.5% on his cost price. If in a week, his sale was of Rs. 392, what was his profit?
	A. Rs. 18.20 B. Rs. 70 C. Rs. 72 D. Rs. 88.25

 

8.	A man buys a cycle for Rs. 1400 and sells it at a loss of 15%. What is the selling price of the cycle?
	A. Rs. 1090 B. Rs. 1160 C. Rs. 1190 D. Rs. 1202

 

9.	Sam purchased 20 dozens of toys at the rate of Rs. 375 per dozen. He sold each one of them at the rate of Rs. 33. What was his percentage profit?
	A. 3.5 B. 4.5 C. 5.6 D. 6.5

 

10.	Some articles were bought at 6 articles for Rs. 5 and sold at 5 articles for Rs. 6. Gain percent is:
	A. 30% B. 33 1 % 3 C. 35% D. 44%
11.	Level-II:     On selling 17 balls at Rs. 720, there is a loss equal to the cost price of 5 balls. The cost price of a ball is:
	A. Rs. 45 B. Rs. 50 C. Rs. 55 D. Rs. 60

 

 

12.	When a plot is sold for Rs. 18,700, the owner loses 15%. At what price must that plot be sold in order to gain 15%?
	A. Rs. 21,000 B. Rs. 22,500 C. Rs. 25,300 D. Rs. 25,800

 

13.	100 oranges are bought at the rate of Rs. 350 and sold at the rate of Rs. 48 per dozen. The percentage of profit or loss is:
	A. 14 2 % gain 7 B. 15% gain C. 14 2 % loss 7 D. 15 % loss

 

14.	A shopkeeper sells one transistor for Rs. 840 at a gain of 20% and another for Rs. 960 at a loss of 4%. His total gain or loss percent is:
	A. 5 15 % loss 17 B. 5 15 % gain 17 C. 6 2 % gain 3 D. None of these

 

 

15.	A trader mixes 26 kg of rice at Rs. 20 per kg with 30 kg of rice of other variety at Rs. 36 per kg and sells the mixture at Rs. 30 per kg. His profit percent is:
	A. No profit, no loss B. 5% C. 8% D. 10% E. None of these

 

16. A man buys an article for Rs. 27.50 and sells it for Rs 28.60. Find his gain percent
17. 1%
18. 2%
19. 3%
20. 4%

 

 

17. A TV is purchased at Rs. 5000 and sold at Rs. 4000, find the lost percent.
18. 10%
19. 20%
20. 25%
21. 28%

 

 

18. In terms of percentage profit, which among following the best transaction.
    1. P. 36, Profit 17
    2. P. 50, Profit 24
    3. P. 40, Profit 19
    4. P. 60, Profit 29

 

 

 

 

Answer:1 Option B

 

Explanation:

Cost Price (C.P.) = Rs. (4700 + 800) = Rs. 5500.

Selling Price (S.P.) = Rs. 5800.

Gain = (S.P.) – (C.P.) = Rs.(5800 – 5500) = Rs. 300.

Gain % =		300	x 100	%	= 5	5	%
		5500				11

 

Answer:2 Option B

 

Explanation:

Let C.P. of each article be Re. 1 C.P. of x articles = Rs. x.

S.P. of x articles = Rs. 20.

Profit = Rs. (20 – x).

		20 – x	x 100 = 25
		x

2000 – 100x = 25x

125x = 2000

x = 16.

 

 

Answer:3 Option B

 

Explanation:

Let C.P. be Rs. x and S.P. be Rs. y.

Then, 3(y – x) = (2y – x)    y = 2x.

Profit = Rs. (y – x) = Rs. (2x – x) = Rs. x.

Profit % =		x	x 100	% = 100%

 

 

Answer:4 Option B

 

Explanation:

Let C.P.= Rs. 100. Then, Profit = Rs. 320, S.P. = Rs. 420.

New C.P. = 125% of Rs. 100 = Rs. 125

New S.P. = Rs. 420.

Profit = Rs. (420 – 125) = Rs. 295.

Required percentage =		295	x 100	%	=	1475	% = 70% (approximately).
		420				21

 

 

Answer:5 Option C

 

Explanation:

C.P. of 6 toffees = Re. 1

S.P. of 6 toffees = 120% of Re. 1 = Rs.	6
	5

 

For Rs.	6	, toffees sold = 6.
	5

 

For Re. 1, toffees sold =		6 x	5		= 5.
			6

 

 

Answer:6 Option A   Explanation: Let C.P. be Rs. x. Then, 1920 – x x 100 = x – 1280 x 100 x x 1920 – x = x – 1280 2x = 3200 x = 1600  Required S.P. = 125% of Rs. 1600 = Rs. 125 x 1600 = Rs 2000. 100

 

 

Answer:7 Option C

 

Explanation:

C.P. = Rs.		100	x 392		= Rs.		1000	x 392		= Rs. 320
		122.5					1225

Profit = Rs. (392 – 320) = Rs. 72.

 

Answer:8 Option C

 

Explanation:

S.P. = 85% of Rs. 1400 = Rs.		85	x 1400		= Rs. 1190
		100

 

 

 

Answer:9 Option C

 

Explanation:

Cost Price of 1 toy = Rs.		375		= Rs. 31.25
		12

Selling Price of 1 toy = Rs. 33

So, Gain = Rs. (33 – 31.25) = Rs. 1.75

Profit % =		1.75	x 100	%	=	28	% = 5.6%
		31.25				5

 

 

 

Answer:10 Option D

 

Explanation:

Suppose, number of articles bought = L.C.M. of 6 and 5 = 30.

C.P. of 30 articles = Rs.		5	x 30		= Rs. 25.
		6

 

S.P. of 30 articles = Rs.		6	x 30		= Rs. 36.
		5

 

Gain % =		11	x 100	% = 44%.
		25

 

 

Answer:11 Option D

 

Explanation:

(C.P. of 17 balls) – (S.P. of 17 balls) = (C.P. of 5 balls)

C.P. of 12 balls = S.P. of 17 balls = Rs.720.

C.P. of 1 ball = Rs.		720		= Rs. 60.
		12

 

 

Answer:12 Option C

 

Explanation:

85 : 18700 = 115 : x

x =		18700 x 115		= 25300.
		85

Hence, S.P. = Rs. 25,300.

 

Answer:13 Option A

 

Explanation:

C.P. of 1 orange = Rs.		350		= Rs. 3.50
		100

 

S.P. of 1 orange = Rs.		48		= Rs. 4
		12

 

Gain% =		0.50	x 100	%	=	100	% = 14	2	%
		3.50				7		7

 

 

 

Answer:14 Option B

 

Explanation:

C.P. of 1st transistor = Rs.		100	x 840		= Rs. 700.
		120

 

C.P. of 2nd transistor = Rs.		100	x 960		= Rs. 1000
		96

So, total C.P. = Rs. (700 + 1000) = Rs. 1700.

Total S.P. = Rs. (840 + 960) = Rs. 1800.

Gain % =		100	x 100	%	= 5	15	%
		1700				17

 

 

 

Answer:15 Option B

 

Explanation:

C.P. of 56 kg rice = Rs. (26 x 20 + 30 x 36) = Rs. (520 + 1080) = Rs. 1600.

S.P. of 56 kg rice = Rs. (56 x 30) = Rs. 1680.

Gain =		80	x 100	% = 5%.
		1600

 

Answer:16 Option D

 

Explanation:

So we have C.P. = 27.50
S.P. = 28.60

Gain = 28.60 – 27.50 = Rs. 1.10

Gain%=(Gain/Cost∗100)%=(1.10/27.50∗100)%=4%

 

 

 

 

Answer:17 Option B

 

Explanation:

We know, C.P. = 5000
S.P. = 4000
Loss = 5000 – 4000 = 1000
Loss%=(Loss/Cost∗100)%=(1000/5000∗100)%=20%

 

 

Answer:18 Option D

 

Explanation:

Hint: Calculate profit percent as

Profit% = (profit/cost) * 100

Age Problems

 

Important Formulas on “Problems on Ages”:

 

1. If the current age is x, then ntimes the age is nx.
2. If the current age is x, then age nyears later/hence = x+ n.
3. If the current age is x, then age nyears ago = x– n.
4. The ages in a ratio a: bwill be ax and bx.

5. If the current age is x, then	1	of the age is	x	.
	n		n

Example:

A problem with one variable: How old is Al?

Many single-variable algebra word problems have to do with the relations between different people’s ages. For example:

Al’s father is 45. He is 15 years older than twice Al’s age. How old is Al?

We can begin by assigning a variable to what we’re asked to find. Here this is Al’s age, so let Al’s age = x.

We also know from the information given in the problem that 45 is 15 more than twice Al’s age. How can we translate this from words into mathematical symbols? What is twice Al’s age?

Well, Al’s age is x, so twice Al’s age is 2x, and 15 more than twice Al’s age is 15 + 2x.That equals 45, right? Now we have an equation in terms of one variable that we can solve for x: 45 = 15 + 2x.

original statement of the problem:	45 = 15 + 2x
subtract 15 from each side:	30 = 2x
divide both sides by 2:	15 = x

Since x is Al’s age and x = 15, this means that Al is 15 years old.

It’s always a good idea to check our answer:

twice Al’s age is 2 x 15:	30
15 more than 30 is 15 + 30:	45

This should be the age of Al’s father, and it is.

 

 

Questions:

Level-I:

 

1.	Father is aged three times more than his son Ronit. After 8 years, he would be two and a half times of Ronit’s age. After further 8 years, how many times would he be of Ronit’s age?
	A. 2 times B. 2 1 times 2 C. 2 3 times 4 D. 3 times

 

2.	The sum of ages of 5 children born at the intervals of 3 years each is 50 years. What is the age of the youngest child?
	A. 4 years B. 8 years C. 10 years D. None of these

 

3.	A father said to his son, “I was as old as you are at the present at the time of your birth”. If the father’s age is 38 years now, the son’s age five years back was:
	A. 14 years B. 19 years C. 33 years D. 38 years

 

4.	A is two years older than B who is twice as old as C. If the total of the ages of A, B and C be 27, the how old is B?
	A. 7 B. 8 C. 9 D. 10 E. 11

 

5.	Present ages of Sameer and Anand are in the ratio of 5 : 4 respectively. Three years hence, the ratio of their ages will become 11 : 9 respectively. What is Anand’s present age in years?
	A. 24 B. 27 C. 40 D. Cannot be determined E. None of these

 

6.	A man is 24 years older than his son. In two years, his age will be twice the age of his son. The present age of his son is:
	A. 14 years B. 18 years C. 20 years D. 22 years

 

7.	Six years ago, the ratio of the ages of Kunal and Sagar was 6 : 5. Four years hence, the ratio of their ages will be 11 : 10. What is Sagar’s age at present?
	A. 16 years B. 18 years C. 20 years D. Cannot be determined E. None of these

 

8.	The sum of the present ages of a father and his son is 60 years. Six years ago, father’s age was five times the age of the son. After 6 years, son’s age will be:
	A. 12 years B. 14 years C. 18 years D. 20 years

 

9.	At present, the ratio between the ages of Arun and Deepak is 4 : 3. After 6 years, Arun’s age will be 26 years. What is the age of Deepak at present ?
	A. 12 years B. 15 years C. 19 and half D. 21 years

 

10.	Sachin is younger than Rahul by 7 years. If their ages are in the respective ratio of 7 : 9, how old is Sachin?
	A. 16 years B. 18 years C. 28 years D. 24.5 years E. None of these

 

 

11.	Level-II:         The present ages of three persons in proportions 4 : 7 : 9. Eight years ago, the sum of their ages was 56. Find their present ages (in years).
	A. 8, 20, 28 B. 16, 28, 36 C. 20, 35, 45 D. None of these

 

12.	Ayesha’s father was 38 years of age when she was born while her mother was 36 years old when her brother four years younger to her was born. What is the difference between the ages of her parents?
	A. 2 years B. 4 years C. 6 years D. 8 years

 

13.	A person’s present age is two-fifth of the age of his mother. After 8 years, he will be one-half of the age of his mother. How old is the mother at present?
	A. 32 years B. 36 years C. 40 years D. 48 years

 

14.	Q is as much younger than R as he is older than T. If the sum of the ages of R and T is 50 years, what is definitely the difference between R and Q’s age?
	A. 1 year B. 2 years C. 25 years D. Data inadequate E. None of these

 

15.	The age of father 10 years ago was thrice the age of his son. Ten years hence, father’s age will be twice that of his son. The ratio of their present ages is:
	A. 5 : 2 B. 7 : 3 C. 9 : 2 D. 13 : 4

 

16.	What is Sonia’s present age? I. Sonia’s present age is five times Deepak’s present age.  II. Five years ago her age was twenty-five times Deepak’s age at that time.
	A. I alone sufficient while II alone not sufficient to answer B. II alone sufficient while I alone not sufficient to answer C. Either I or II alone sufficient to answer D. Both I and II are not sufficient to answer E. Both I and II are necessary to answer

 

17.	Average age of employees working in a department is 30 years. In the next year, ten workers will retire. What will be the average age in the next year? I. Retirement age is 60 years.  II. There are 50 employees in the department.
	A. I alone sufficient while II alone not sufficient to answer B. II alone sufficient while I alone not sufficient to answer C. Either I or II alone sufficient to answer D. Both I and II are not sufficient to answer E. Both I and II are necessary to answer

 

 

18.	Divya is twice as old as Shruti. What is the difference in their ages? I. Five years hence, the ratio of their ages would be 9 : 5.  II. Ten years back, the ratio of their ages was 3 : 1.
	A. I alone sufficient while II alone not sufficient to answer B. II alone sufficient while I alone not sufficient to answer C. Either I or II alone sufficient to answer D. Both I and II are not sufficient to answer E. Both I and II are necessary to answer

 

 

 

Answers:

Level-I:

 

Answer:1 Option A

 

Explanation:

Let Ronit’s present age be x years. Then, father’s present age =(x + 3x) years = 4x years.

	(4x + 8) =	5	(x + 8)
		2

8x + 16 = 5x + 40

3x = 24

x = 8.

Hence, required ratio =	(4x + 16)	=	48	= 2.
	(x + 16)		24

 

 

Answer:2 Option A

 

Explanation:

Let the ages of children be x, (x + 3), (x + 6), (x + 9) and (x + 12) years.

Then, x + (x + 3) + (x + 6) + (x + 9) + (x + 12) = 50

5x = 20

x = 4.

Age of the youngest child = x = 4 years.

 

 

 

Answer:3 Option A

 

Explanation:

Let the son’s present age be x years. Then, (38 – x) = x

2x = 38.

x = 19.

Son’s age 5 years back (19 – 5) = 14 years.

 

Answer:4 Option D

 

Explanation:

Let C’s age be x years. Then, B’s age = 2x years. A’s age = (2x + 2) years.

(2x + 2) + 2x + x = 27

5x = 25

x = 5.

Hence, B’s age = 2x = 10 years.

 

Answer:5 Option A

 

Explanation:

Let the present ages of Sameer and Anand be 5x years and 4x years respectively.

Then,	5x + 3	=	11
	4x + 3		9

9(5x + 3) = 11(4x + 3)

45x + 27 = 44x + 33

45x – 44x = 33 – 27

x = 6.

Anand’s present age = 4x = 24 years.

 

Answer:6 Option D

 

Explanation:

Let the son’s present age be x years. Then, man’s present age = (x + 24) years.

(x + 24) + 2 = 2(x + 2)

x + 26 = 2x + 4

x = 22.

 

Answer:7 Option A

 

Explanation:

Let the ages of Kunal and Sagar 6 years ago be 6x and 5x years respectively.

Then,	(6x + 6) + 4	=	11
	(5x + 6) + 4		10

10(6x + 10) = 11(5x + 10)

5x = 10

x = 2.

Sagar’s present age = (5x + 6) = 16 years.

 

Answer:8 Option D

 

Explanation:

Let the present ages of son and father be x and (60 –x) years respectively.

Then, (60 – x) – 6 = 5(x – 6)

54 – x = 5x – 30

6x = 84

x = 14.

Son’s age after 6 years = (x+ 6) = 20 years..

 

Answer:9 Option B

 

Explanation:

Let the present ages of Arun and Deepak be 4x years and 3x years respectively. Then,

4x + 6 = 26        4x = 20

x = 5.

Deepak’s age = 3x = 15 years.

 

Answer:10 Option D

 

Explanation:

Let Rahul’s age be x years.

Then, Sachin’s age = (x – 7) years.

	x – 7	=	7
	x		9

9x – 63 = 7x

2x = 63

x = 31.5

Hence, Sachin’s age =(x – 7) = 24.5 years.

 

Answer:11 Option B

 

Explanation:

Let their present ages be 4x, 7x and 9x years respectively.

Then, (4x – 8) + (7x – 8) + (9x – 8) = 56

20x = 80

x = 4.

Their present ages are 4x = 16 years, 7x = 28 years and 9x = 36 years respectively.

 

Answer:12 Option C

 

Explanation:

Mother’s age when Ayesha’s brother was born = 36 years.

Father’s age when Ayesha’s brother was born = (38 + 4) years = 42 years.

Required difference = (42 – 36) years = 6 years.

 

Answer:13 Option C

 

Explanation:

Let the mother’s present age be x years.

Then, the person’s present age =		2	x		years.
		5

 

		2	x + 8		=	1	(x + 8)
		5				2

2(2x + 40) = 5(x + 8)

x = 40.

 

Answer:14 Option D

 

Explanation:

Given that:

1. The difference of age b/w R and Q = The difference of age b/w Q and T.

2. Sum of age of R and T is 50 i.e. (R + T) = 50.

Question: R – Q = ?.

Explanation:

R – Q = Q – T

(R + T) = 2Q

Now given that, (R + T) = 50

So, 50 = 2Q and therefore Q = 25.

Question is (R – Q) = ?

Here we know the value(age) of Q (25), but we don’t know the age of R.

Therefore, (R-Q) cannot be determined.

 

Answer:15 Option B

 

Explanation:

Let the ages of father and son 10 years ago be 3x and x years respectively.

Then, (3x + 10) + 10 = 2[(x + 10) + 10]

3x + 20 = 2x + 40

x = 20.

Required ratio = (3x + 10) : (x + 10) = 70 : 30 = 7 : 3.

 

 

Answer:16  Option E

 

Explanation:

I. S = 5D     D =	S	….(i)
	5

1. S – 5 = 25 (D – 5)    S = 25D – 120 ….(ii)

Using (i) in (ii), we get S =		25 x	S		– 120
			5

4S = 120.

S = 30.

Thus, I and II both together give the answer. So, correct answer is (E).

 

Answer:17 Option E

 

Explanation:

1. Retirement age is 60 years.
2. There are 50 employees in the department.

Average age of 50 employees = 30 years.

Total age of 50 employees = (50 x 30) years = 1500 years.

Number of employees next year = 40.

Total age of 40 employees next year (1500 + 40 – 60 x 10) = 940.

Average age next year =	940	years = 23	1	years.
	40		2

Thus, I and II together give the answer. So, correct answer is (E).

 

Answer:18   Option C

 

Explanation:

Let Divya’s present age be D years and Shruti’s present age b S years

Then, D = 2 x S        D – 2S = 0 ….(i)

I.	D + 5	=	9	….(ii)
	S + 5		5

 

II.	D – 10	=	3	….(iii)
	S – 10		1

From (ii), we get : 5D + 25 = 9S + 45        5D – 9S = 20 ….(iv)

From (iii), we get : D – 10 = 3S – 30        D – 3S = -20 ….(v)

Thus, from (i) and (ii), we get the answer.

Also, from (i) and (iii), we get the answer.

I alone as well as II alone give the answer. Hence, the correct answer is (C).