Ratios and Proportions (114 Questions with Explanation)

1.

If two numbers are in the ratio 2:3. If 10 is added to both of the numbers then the ratio becomes 3:4 then find the smallest number?

   A.) 10
   B.) 20
   C.) 30
   D.) 40

Answer: Option 'B'

2:3 
2x + 10 : 3x + 10 = 3 : 4 
4[2x + 10] = 3[3x + 10] 
8x + 40 = 9x + 30 
9x - 8x = 40 - 30 
x = 10 
Then smallest number is = 2 
2x = 20 
Short cut method: 
a:b = 2:3 
c:d = 3:4 
1.Cross multiplication with both ratios a × d ~ b × c = 2 * 4 ~ 3 × 3 = 8 ~ 9 = 1 
2. If 10 is added both the number means 10 × 3 = 30 and 10 × 4 = 40, 
Then 30 ~ 40 = 10 
=> 1 ---> 10 
=> 2 ---> 20 

2.

For every 2 chocolate Ram gets Shyam is given 3 chocolate. If the total number of chocolate distributed to them is 65, then find the number of chocolates Shyam received.

   A.) 30
   B.) 39
   C.) 34
   D.) 45

Answer: Option 'B'

39

3.

A total amount of Rs. 6900 is to be divided among A,B and C in such a way that one-third of A's part, half of B's part and three-fourth of C's part is equal. Then B's part is:

   A.) Rs. 2178.95
   B.) Rs. 2861.76
   C.) Rs. 2239.14
   D.) Rs. 2356.85

Answer: Option 'A'

Given, A + B + C = 6900 ----> (1)
and(1/3) A = (1/2) B = (3/4) C
=> A = (3/2) B -----> (2)
and C = [4/(2 × 3)] B
=> C = (2/3) B -----> (3)
Substitute (2) and (3) in (1)
=> (3/2)B+ B + (2/3)B = 6900
=> B [(3/2) + 1 + (2/3)] = 6900
On taking L.C.M we get,
=> B [(9 + 6 + 4) / 6] = 6900
=> B [19 / 6] = 6900
=> B = 6900 * (6/19)
=> B = 2178.95
Then B's Part = Rs.2178.95

4.

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.) One-ninth
   B.) One-seventh
   C.) One-eighth
   D.) Seven-eighths

Answer: Option 'B'

When there are 8 people, the share of each person is 1/8 => (Original Share)
When there are 7 people, the share of each person is 1/7
Increase in the share of each person is 1/7 – 1/8 = 1 / 56
1 / 56 ==> Which is 1/7 of 1/8 ==> 1/7 of the original share of each person.

5.

Two numbers are in the ratio 2:3. If eight is added to both the numbers, the ratio becomes 3:4. The numbers are 

   A.) 15 and 20
   B.) 13 and 17
   C.) 16 and 24
   D.) 17 and 9

Answer: Option 'C'

Let Numbers be m and n,
m/n = 2/3
Adding 8 on both sides,we get
=> (m+8) / (n+8) = 3 / 4
=> 4m + 32 = 3n + 24
By solving each of them seperately we get the value of m , n
=> 4m + 32 =0
=> 4m = 32
=> m = 32 / 4
=> m = 8
3n + 24 = 0
=> 3n = 24
=> n = 24 / 3
=> n = 8
So according to the given ratios,
The numbers are in 2 : 3
m = 8 × 2 = 16
n = 8 × 3 = 24
So m =16 , n = 24

6.

If Rs.900/- Rupees are divided among a,b and c in such a way that A’s share 3 times that of B and B’s share is 2 times that of C. The A’s share is? 

   A.) Rs.600/-
   B.) Rs.100/-
   C.) Rs.200/-
   D.) Rs.700/-

Answer: Option 'A'

A:B:C = 6:2:1 
Total parts = 9 
A's share is = 6 parts
9 -----> Rs.900/- 
6 -----> Rs.600/-

7.

The ages of Manoj and Amit are in the ratio 2:3. After 12 years, their ages will be in the ratio 11:15. The age of Amit is: 

   A.) 48 years
   B.) 56 years
   C.) 40 years
   D.) 32 years

Answer: Option 'A'

Given the present age ratio of Manoj and Amit = 2 : 3
=> present age of Manoj = 2x
present age of Amit = 3x
Given after 12years, their ages will be in the ratio 11:15.
thus, after 12 years the age of Manoj will be = 2x + 12
and the age of Amit will be= 3x + 12
=> (age of Manoj after 12 years / age of Amitafter 12 years) = 11 / 15
=> (2x + 12) / (3x + 12) = 11 / 15
=> 15(2x + 12) = 11(3x + 12)
=> 30x + 180 = 33x + 132
=>33x - 30x = 180 - 132
=> 3x = 48
Thus,present age of Amit = 3x = 48 years.

8.

If a:b=1:2, b:c=3:4 and c:d = 2:3 find a:b:c:d? 

   A.) 3:6:8:24
   B.) 6:12:16:24
   C.) 6:18:24:16
   D.) 3:6:2:3

Answer: Option 'B'

a:b = 1:2, b:c = 3:4, c:d = 2:3
1:2 
3:4 
( a = 1 × 3 = 3, b = 2 × 3 = 6 and c = 2 × 4 = 8)
(a = a × b, b= b × b and c= b × c) 
a:b:c = 3:6:8 
a:b:c = 3:6:8 and c:d = 2:3
(Note: First a,b,c multiplication with c means 2 and last c means 8
multiplication with d means 3
a:b:c:d = 6:12:16:24 

9.

Salaries of Robert and Sam are in the ratio 2 : 3. If the salary of each is increased by Rs. 4000, the new ratio becomes 40 : 57. What is Sam's present salary?

   A.) 39000
   B.) 38000
   C.) 34000
   D.) 36000

Answer: Option 'B'

Let the original salary of Robert and Sam be 2k and 3k respectively.
Then (2k+4000)/(3k+4000)=40/57 => 57(2k+4000) = 40(3k+4000) => 114k + 2,28,000 = 120k + 1,60,000 => 68000 = 6k => 3k = 34000
Sam's present salary = 3k+4000 = 34000 + 4000 = 38000

10.

A mixture contains milk and water in the ratio 3:2. On adding 10 liters of water, the ratio of milk to water becomes 2:3. Total quantity of milk & water before adding water to it?

   A.) 10
   B.) 50
   C.) 20
   D.) 40

Answer: Option 'C'

milk:water = 3:2 
after adding 10 liters of water
milk:water = 2:3 
Olny water patrs increase when mixture of water
milk:wate = 3:2 = 2*(3:2) = 6:4 
after adding 10 liters of water 
milk:water = 2:3 = 3*(2:3) = 6:9 
milk parts always same 
Short cut method: 
milk:water = 6 : 4  
after adding 10 liters of water 
milk:water = 6 : 9  
milk is same but water increse 10liters then the water ratio is increse 5 parts  
5 part ---> 10 liters  
The quantity of milk in the original mixture is = 6 : 4 = 6 + 4 = 10 
10 parts ---> 20 liters  (Answer is = 20) 
Short cut method - 2 : for Only milk problems 
milk : water 
6 : 4 
6 : 9 
milk ratio same but water ratio 5 parts incress per 10 liters 
5 part of ratio ----> 10 liters 
10 part of ratio ---> 20 liters 

11.

If Rs.1440/- are divided among A,B and C so that A receives 1/3rd as much as B and B receives 1/4th as much as C. The amount B received is: 

   A.) Rs.90/-
   B.) Rs.270/-
   C.) Rs.1080/-
   D.) Rs.27/-

Answer: Option 'B'

A:B:C = 1:3:12 
Total parts = 16 
B's share is = 3 parts 
16 -----> 1440 
1 -----> 90 
3 -----> 270 (B's share is 270) 

12.

Rs.780 is divided among 2 men, 6 women and 8 boys so that the share of a man, a woman and a boy are in the ratio 3: 2: 1. Then, how much does a boy get?

   A.) Rs. 30
   B.) Rs.60
   C.) Rs.40
   D.) Rs.240

Answer: Option 'A'

From given data,
780 = 2m + 6w +8b
And m : w : b = 3 : 2 : 1
=> m = 3x Rs.
w = 2x Rs.,
b = x Rs.
Subs these m, w, b values in 780 = 2m + 6w +8b
=> 780 = 2 (3x) + 6 (2x) + 8 x
=> 780 = 6x + 12x + 8x
=> 780 = 26x
=> x = 780 / 26
=> x = 30 Rs.
Thus, a boy will get Rs. 30

13.

If Rs.540/- are divided among A,B and C in such a way that A’s share is ½nd of B share and B’s share is 1/3rd of C’s share. The share of A is? 

   A.) Rs.80/-
   B.) Rs.360/-
   C.) Rs.60/-
   D.) Rs.120/-

Answer: Option 'C'

A:B:C = 1:2:6 
Total parts = 9 
A's share is = 1 parts 
9 -----> Rs.540/- 
1 -----> Rs.60/- 

14.

Rs.630/- distributed among A,B and C such that on decreasing their shares by RS.10,RS.5 and RS.15 respectively, The balance money would be divided among them in the ratio 3:4:5. Then, A’s share is:? 

   A.) Rs.150/-
   B.) Rs.200/-
   C.) Rs.160/-
   D.) Rs.255/-

Answer: Option 'C'

A:B:C = 3:4:5 
Total parts = 12 
A's share is = 3 parts 
12 -----> Rs.600/- 
3 -----> Rs.150/- 
A's total = 150 + 10 = Rs.160/-

15.

Two numbers are in the ratio 11:12, if 14 is added to each they are in the ratio 69:74. Find the two numbers. 

   A.) 63 and 74
   B.) 55 and 80
   C.) 55 and 60
   D.) 88 and 97

Answer: Option 'C'

Numbers are in the ratio 11 : 12
Two numbers are 11k and 12k
Adding 14 to both the numbers, 11k+14 and 12k+14
Given that 11k+14 : 12k+14 = 69 : 74
=> 74(11k+14) = 69(12k+14)
=> (814-828)k = 14 x (-5)
=> k = 5
First Number, 11 x k = 55 Second Number 12 x k = 60
Numbers are 55 and 60

16.

A bag contains 5p,10p and Rs20p coins in the ratio of 1:2:4 respectively. If the total money in the bag is Rs.84/-. Find the number of 10p coins in that bag? 

   A.) 160 coins
   B.) 20 coins
   C.) 128 coins
   D.) None of these

Answer: Option 'A'

5paisa : 10paisa : 20paisa = 1 : 2 : 4  ---> coins ratio  
= 5 : 20 : 80 ---> money ratio 
5 paisa × 80 coins = Rs.4/- 
10 paisa × 160 coins = Rs.16/- 
Then 10 paisa coins in that bag = 160 cions

17.

Two numbers are in the ratio 11:12, if 14 is added to each they are in the ratio 69:74. Find the two numbers.

   A.) 55 and 80
   B.) 55 and 60
   C.) 63 and 74
   D.) 88 and 97

Answer: Option 'B'

Numbers are in the ratio 11 : 12
Two numbers are 11k and 12k
Adding 14 to both the numbers, 11k+14 and 12k+14
Given that 11k+14 : 12k+14 = 69 : 74
=> 74(11k+14) = 69(12k+14)
=> 814k + (74 × 14) = 828k + (69 × 14)
=> 814k - 828k = (69 × 14) - (74 × 14)
=> (814 - 828) k = 14 × (69 - 74)
=> (-14)k = 14 x (-5)
=> -k = -5
=> k = 5
First Number is 11 x k = 11 × 5 = 55
Second Number is 12 x k = 12 × 5 = 60
Numbers are 55 and 60

18.

If x:y=5:3 then (8x-5y) : (8x+5y)=? 

   A.) 5:13
   B.) 13:5
   C.) 5:11
   D.) 11:5

Answer: Option 'C'

x/y = 5/3 (Given) 
(8x − 5y)/(8x + 5y) 
8(x/y) − 5/8(x/y)+5 
( on dividing Nr and Dr by y) 
(8(5/3) − 5)/(8(5/3) + 5) 
(40/3 − 5/1)/(40/3 + 5/1) 
[(40 − 15)/3]/[(40 + 15)/3] 
25/55 = 5/11 
(8x-5y):(8x+5y) = 5:11 

19.

A bag contains 5p,10p and Rs20p coins in the ratio of 4:2:1 respectively. If the total money in the bag is Rs.30/-. Find the number of 5p coins in that bag? 

   A.) 100 coins
   B.) 150 coins
   C.) 200 coins
   D.) 250 coins

Answer: Option 'C'

5paisa : Rs.10/- : Rs.20/- = 4 : 2 : 1  ---> coins ratio   
=  1 : 1 : 1 ----> money ratio 
3 ---> 30 
1 ---> Rs.10/- × 20 coins ( 1/- = 5paisa × 20 cions ) 
Rs.10/-  = 200 coins(5paisa)

20.

If a dozen mirrors are fallen down. The ratio between broken and unbroken mirrors is: 

   A.) 2:3
   B.) 3:4
   C.) 5:7
   D.) 5:4

Answer: Option 'C'

Dozen means = 12 
12/(2+3) = 12/5 ==> wrong 
12/(3+4) = 12/7 ==> wrong 
12/(5+7) = 12/12 = 1 ==> Right (Answer = 5:7)

21.

Two numbers are in the ratio 1:2, if 20 is added to each they are in the ratio 11:18. Find the two numbers.

   A.) 35 and 70
   B.) 25 and 70
   C.) 44 and 72
   D.) 44 and 88

Answer: Option 'A'

Numbers are in the ratio 1 : 2 
Two numbers are 1k and 2k 
Adding 20 to both the numbers, 1k+20 and 2k+20 
Given that 1k+20 : 2k+20 = 11 : 18 => 18(1k+20) = 11(2k+20) => (18-22)k = 20 x (-7) => k = 35 
First Number , 1 x k = 35 Second Number 2 x k = 70 
Numbers are 35 and 70

22.

Two numbers are in the ratio 3:5, If 8 is subtracted from each, then they are in the ratio 1:3. Then, the second number is:

   A.) 15
   B.) 20
   C.) 12
   D.) 18

Answer: Option 'B'

Given, ratio of two numbers is 3:5 Let, the two numbers be 3x and 5x.
If 8 is subtracted from each, then => (3x - 8) / (5x - 8) = 1/3 => 3(3x - 8) = (5x - 8) => 9x - 24 = 5x - 8 => 4x = 16 => x = 4
Second number = 5x = 5 × 4 = 20

23.

Find the mean proportional between 49 & 81? 

   A.) 16
   B.) 10
   C.) 63
   D.) 12

Answer: Option 'C'

Formula = √a×b
A = 49 and B = 81 
√49×81 = 7 × 9 = 63

24.

One-third of the contents of a container evaporated on the 1st day, three-fourths of the remaining evaporated on the second day. What part of the contents of the container is left at the end of the second day?

   A.) One-fourth
   B.) One-half
   C.) One-eighteenths
   D.) One-sixth

Answer: Option 'D'

After first day, 2/3 rd of the contents remain (i.e., 1 - (1/3) = 2/3)
After second day 2/3 (3/4) x (2/3) = 1/6 of the content remains

25.

A mixture contains milk and water in the ratio 3:2. If 4 liters of water is added to the mixture, milk and water in the mixture becomes equal. The quantity of milk in the mixture in liter is? 

   A.) 6
   B.) 12
   C.) 24
   D.) 36

Answer: Option 'B'

Given, Milk and Water ratio = 3 : 2
=> quantity of milk= 3x liters
quantity of water = 2x liters
If 4 liters of water is added to the mixture,
New Milk and Water ratio = 1 : 1 (Since equal),
quantity of milk in the new mixture,= 3x liters
quantity of water in the new mixture = 2x + 4 liters.
=>quantity of Milk in new mixture/ quantity of water in new mixture = 1 / 1
=> 3x / (2x + 4) = 1 / 1
=> 3x = (2x + 4)
=> 3x - 2x = 4
=> x = 4
Thus,quantity of milk= 3x liters = 3 × 4 = 12 liters


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