Ratio and Proportion MCQ Question

Solution: 2A = 3B = 4C = k (let)

A = $\frac{\mathrm{k}}{2}$ ; B =$\frac{\mathrm{k}}{3}$; C = $\frac{\mathrm{k}}{4}$

A : B : C = $\frac{\mathrm{k}}{2}:\frac{\mathrm{k}}{3}:\frac{\mathrm{k}}{4}=12:8:6=6:4:3$

Using Trick:

A : B : C = $\frac{1}{2}:\frac{1}{3}:\frac{1}{4}=6:4:3$

Trick: If xA = yB = zC, then find A : B : C = $\frac{1}{\mathrm{x}}:\frac{1}{\mathrm{y}}:\frac{1}{\mathrm{z}}$

Solution: Total number of student = 8670

Boys = 4545

∴ Girls = 8670 – 4545 = 4125

Ratio of the total number of boys to the total number of girls in the school = 4545:4125 = 303:275

Solution: Boys : Girls = 4:5

∴ Boys = 4k, Girls = 5k, k = ratio constant

10 more boys join the class

Now, $\frac{4\mathrm{k}+10}{5\mathrm{k}}=\frac{6}{5}$

⇒ 30k = 20k+50

⇒10k = 50 or k = 5

Numbers of girl = 25

Solution: let x will added both.

Now, $\frac{5+\mathrm{x}}{9+\mathrm{x}}=\frac{7}{10}$

⇒ 50 + 10x = 63 + 7x

⇒ 3x = 13 or x = $\frac{13}{3}$

Since x is the least integer.

$\therefore \mathrm{x}\geqslant \frac{13}{3}\,\,\mathrm{or} \mathrm{x}=5$

Solution:

Ration of the coins = 5 : 6 : 4

Number of 1 rupee coin = $465\times \frac{5}{15}=155$

Number of 50 paisa coin = $465\times \frac{6}{15}=186$

Number of 25 paisa coin = $465\times \frac{4}{15}=124$

Rules: x coins consists of 1 rupee, 50 paisa and 25 paisa coins. Their values are in the ratio a:b:c. The number of each type of coins respectively is

First we convert ratio from values to number

Ratio of the coins = a : 2b : 4c

Number of 1 rupee coin = $\mathrm{x}\times \frac{\mathrm{a}}{\mathrm{a}+2\mathrm{b}+4\mathrm{c}}$

Number of 50 paisa coin = $\mathrm{x}\times \frac{2\mathrm{b}}{\mathrm{a}+2\mathrm{b}+4\mathrm{c}}$

Number of 25 paisa coin =$\mathrm{x}\times \frac{4\mathrm{c}}{\mathrm{a}+2\mathrm{b}+4\mathrm{c}}$

Solution: a : b : c : d = 2 : 3 : 4 : 5

Trick: If A:B = m : n, B:C = n : o and C : D = o : p then A:B:C :D= m : n : o : q

Solution: Let A’s share be Rs. x, B’s share be Rs. y. Then, C’s share = Rs. [671 – (x + y)]

Now, x + 3 : y  + 7 : 671 – (x + y) + 9 = 1: 2 : 3

x + 3 : y + 7 : 680 – (x + y) = 1: 2 : 3

⇒ x + 3 = $\frac{1}{6}\times 690=115$ or x = 112

⇒ y + 7 = $\frac{2}{6}\times 690=230$ = or y = 223

∴ C’s share = Rs. [671 – (x + y)] = 671 – 335 = 336

Using trick: After increased total share = 671 + 3 + 7 + 9 = 690

C’s share = $\frac{3}{6}\times 690 -9=336$

Solution:

From ratio we can say C’s share is maximum

or

A’s share = $\frac{105}{533}\times 1066=210$

B’s share = $\frac{140}{533}\times 1066=280$

C’s share = $\frac{168}{533}\times 1066=336$ (Maximum)

D’s share = $\frac{120}{533}\times 1066=240$

C’s share is maximum

Solution: Let income of A = Rs. 3m, income of B = Rs. 2m and expenditure of A = Rs. 5p, expenditure of B = Rs. 3p

Now, saving = income – expenditure

∴ 3m – 5p = 2m – 3p = 200

⇒ m = 2p and p = 200

∴ m = 400

∴ A’s income = Rs. 1200

Using Trick: Monthly income of A = $\frac{200\times 3\left( 3-5 \right)}{3\times 3-2\times 5}=1200$

Solution: Gold in C in one unit = $\left( \frac{7}{9}+\frac{7}{18} \right) \,\,=\frac{7}{6}$

Copper in C in one unit = $\left( \frac{2}{9}+\frac{11}{18} \right) \,\,=\frac{5}{6}$

Ratio of gold and copper in C = $\frac{7}{6}:\frac{5}{6}=7:5$