PIPES & CISTERN

 

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.

 

  1. If a pipe can fill a tank in xhours, then:
part filled in 1 hour = 1 .
x
  1. If a pipe can empty a tank in yhours, then:
part emptied in 1 hour = 1 .
y
  1. 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
  1. 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 ROOTS

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 3To 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 9is 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
  1. 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

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. aman = am + n
am am – n
an
   
  • (am)namn

 

  1. (ab)nanbn

 

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

 

  1. If 5√5 * 53÷ 5-3/2= 5a+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:

xz = y2        10(0.48z) = 10(2 x 0.70) = 101.40

0.48z = 1.40

 z = 140 = 35 = 2.9 (approx.)
48 12

 

Answer:4 Option A

 

Explanation:

5a = 3125        5a = 55

a = 5.

5(a – 3) = 5(5 – 3) = 52 = 25.

 

 

Answer:5 Option C

 

Explanation:

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

3x 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)

= 44(1/4)

= 41

= 4

Answer:7 Option B

 

Explanation:

(10)150 ÷ (10)146 = 10150
10146

= 10150 – 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 = 5x.

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

5x = 5(15 + 2.5 – 4.5)

5x = 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)  .   xa(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

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 = ( divisorquotient ) + 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.

  1. 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/n(n+1)

(12+22+32+………+n2) = 1/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

 

  1. Find the value of 1/(3+1/(3+1/(3-1/3)))
A.) 3/10 B.) 10/3
C.) 27/89 D.) 89/27

  1. 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

 

 

 

 

  1. 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
   
  1. 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., 30th 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 & LOSS

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.
  1. Loss Percentage: (Loss %)
    Loss % = Loss x 100
C.P.
  1. Selling Price: (S.P.)
    SP = (100 + Gain %) x C.P
100
         
  1. Selling Price: (S.P.)
    SP = (100 – Loss %) x C.P.
100
  1. Cost Price: (C.P.)
    C.P. = 100 x S.P.
(100 + Gain %)
  1. Cost Price: (C.P.)
    C.P. = 100 x S.P.
(100 – Loss %)
  1. If an article is sold at a gain of say 35%, then S.P. = 135% of C.P.
  2. If an article is sold at a loss of say, 35% then S.P. = 65% of C.P.
  3. 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
  1. 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

 

  1. A man buys an article for Rs. 27.50 and sells it for Rs 28.60. Find his gain percent
  2. 1%
  3. 2%
  4. 3%
  5. 4%

 

 

  1. A TV is purchased at Rs. 5000 and sold at Rs. 4000, find the lost percent.
  2. 10%
  3. 20%
  4. 25%
  5. 28%

 

 

  1. 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

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 = xn.
  3. If the current age is x, then age nyears ago = x– n.
  4. The ages in a ratio abwill 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).

FRACTIONS

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

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 amountr = 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?

  1. 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.
  2. A trader marks his goods at 40% above the cost price and allows a discount of 25%. What is his gain percent?
  3. 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.
  4. 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%?
  5. Find the single discount equivalent to two successive discounts of 20% and 10%.
  6. 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?

 

  1. 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:
  2. On a 20% discount sale, an article costs Rs. 596. What was the original price of the article?

Level-II:

  1. A discount of 15% on one article is the same as discount of 20% on a second article. The costs of the
  2. 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:
  3. 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:
  4. 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?
  5. A shopkeeper marks his goods 20% above cost price, but allows 30% discount for cash. His net loss is:
  6. 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?
  7. 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 :

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

 

 

 

 

 

 

  1. 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

 

 

 

 

 

 

 

  1. 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

 

 

  1. 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.

 

 

 

  1. 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

 

 

 

 

 

  1. 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

Mixed ratio and proportion

 

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 1st part = 7 ×465=3255

The 2nd 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: 1st rice cost = 120, 2nd rice cost = 145 and 3rd 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.