Home Mathematics Mcqs Ratio and Proportion

# A and B together have Rs. 1210. If $$\frac{4}{5}$$ of A’s amount is equal to $$\frac{2}{5}$$ of B’s amount, how much amount does B have?

A. RS. 400
B. RS. 423
C. RS. 467
D. RS. 484

Solution:

$\left. \begin{array} { l } { \frac { 4 } { 15 } A = \frac { 2 } { 5 } B } \\ { \Rightarrow A = ( \frac { 2 } { 5 } \times \frac { 15 } { 4 } ) B } \\ { \Rightarrow A = \frac { 3 } { 2 } B } \\ { \Rightarrow \frac { A } { B } = \frac { 3 } { 2 } } \\ { \Rightarrow A : B = 3 : 2 } \\ { \therefore B ^ { \prime } s \text { share } = Rs \cdot ( 1210 \times \frac { 2 } { 5 } )}\\{ = Rs .484 } \end{array} \right.$

# Two numbers are respectively 20% and 50% more than a third number. The ratio of the two numbers is:

A. 4:5
B. 5:4
C. 1:2
D. 3:4

Solution:

$\left. \begin{array} { l } { \text { Let the third number be } x \text { . } } \\ { \text { Then, first number } = 120 \% \text { of } x = \frac { 120 x } { 100 } }\\ {= \frac { 6 x } { 5 } } \\ { \text { Second number } = 150 \% \text { of } x = \frac { 150 x } { 100 } = \frac { 3 x } { 2 } } \\ { \therefore \text { Ratio of first two numbers } = ( \frac { 6 x } { 5 } : \frac { 3 x } { 2 } )}\\ {= 12 x : 15 x = 4 : 5 } \end{array} \right.$

# 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. 7000
B. 2000
C. 3000
D. 4000

Solution:

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

$\left. \begin{array} { l } { \text { Then, } 4 x – 3 x = 1000 } \\ { \Rightarrow x = 1000 } \\ { \therefore B ^ { \prime } s \text { share } = R s .2 x = \text { Rs. } ( 2 \times 1000 )}\\ {= \text { Rs. } 2000 } \end{array} \right.$

# 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. 7 : 6 : 1
B. 2 : 6 : 7
C. 4 : 5 : 2
D. 2 : 3 : 4

Solution:

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

$\left. \begin{array} { l } { \text { Number of increased seats are } ( 140 \% \text { of } 5 x ) ,}\\ {( 150 \% \text { of } 7 x ) \text { and } ( 175 \% \text { of } 8 x ) } \\ { \Rightarrow ( \frac { 140 } { 100 } \times 5 x ) , ( \frac { 150 } { 100 } \times 7 x ) \text { and } ( \frac { 175 } { 100 } \times 8 x ) } \\ { \Rightarrow 7 x , \frac { 21 x } { 2 } \text { and } 14 x } \\ { \therefore \text { The required ratio } = 7 x : \frac { 21 x } { 2 } : 14 x } \\ { \Rightarrow 14 x : 21 x : 28 x } \\ { \Rightarrow 2 : 3 : 4 } \end{array} \right.$

# In a mixture 60 litres, the ratio of milk and water 2 : 1. If this ratio is to be 1 : 2, then the quantity of water to be further added is:

A. 40 litres
B. 20 litres
C. 60 litres
D. 70 litres

Solution:

$\left. \begin{array} { l } { \text { Quantity of milk } = ( 60 \times \frac { 2 } { 3 } ) \text { litres } }\\ {= 40 \text { litres. } } \\ { \text { Quantity of water in it } = ( 60 – 40 ) \text { litres } }\\ {= 20 \text { litres. } } \\ { \text { New ratio } = 1 : 2 } \end{array} \right.$

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

$\left. \begin{array} { l } { \text { Then, milk: water } = ( \frac { 40 } { 20 + x } ) } \\ { \text { Now, } ( \frac { 40 } { 20 + x } ) = \frac { 1 } { 2 } } \\ { \Rightarrow 20 + x = 80 } \\ { \Rightarrow x = 60 } \\ { \therefore \text { Quantity of water to be added } = 60 \text { litres } } \end{array} \right.$

# 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. 31 : 12
B. 13 : 43
C. 42 : 32
D. 21 : 22

Solution:

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).

$\left. \begin{array} { l } { \Rightarrow ( \frac { 120 } { 100 } \times 7 x ) \text { and } ( \frac { 110 } { 100 } \times 8 x ) } \\ { \Rightarrow \frac { 42 x } { 5 } \text { and } \frac { 44 x } { 5 } } \\ { \therefore \text { The required ratio } = ( \frac { 42 x } { 5 } : \frac { 44 x } { 5 } )}\\{ = 21 : 22 } \end{array} \right.$

# 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. 38,000
B. RS. 12,000
C. RS. 19,000
D. RS. 27,000

Solution:

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

$\left. \begin{array} { l } { \text { Then, } \frac { 2 x + 4000 } { 3 x + 4000 } = \frac { 40 } { 57 } } \\ { \Rightarrow 57 ( 2 x + 4000 ) = 40 ( 3 x + 4000 ) } \\ { \Rightarrow 6 x = 68,000 } \\ { \Rightarrow 3 x = 34,000 } \\ { \text { Sumit’s present salary } = ( 3 x + 4000 ) }\\ {= Rs . ( 34000 + 4000 ) = Rs .38,000 } \end{array} \right.$

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