Ratio and Proportion MCQ Question

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Ratio and Proportion MCQ Question with Details Solution 

Read: Concept and Short Trick

Q11. If 2A = 3B = 4C, then find A : B : C

(a) 2 : 3 : 4

(b) 4 : 3 : 2

(c) 6 : 4 : 3

(d) 3 : 4 : 6

Answer: (c) 6 : 4 : 3

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 = $\frac{1}{\mathrm{x}}:\frac{1}{\mathrm{y}}:\frac{1}{\mathrm{z}}$

Q12. The total number of students in a school is 8670. If the number of boys in the school is 4545, then what will be the ratio of the total number of boys to the total number of girls in the school?

(a) 303 : 275

(b) 275 : 303

(c) 11 : 12

(d) 12 : 11

(e) None of the above

Answer: (a) 303 : 275

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

Q13. In a class, the number of boys and girls is in the ratio of 4 : 5. If 10 more boys join the class, the ratio of numbers of boys and girls becomes 6 : 5. How many girls are there in the class?

(a) 20

(b) 30

(c) 25

(d) Couldn’t be determined

(e) None of the above

Answer: (c) 25

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

Q14. What is the least integer which when added to both terms of the ratio 5 : 9 will make a ratio greater than 7 : 10?

(a) 6

(b) 8

(c) 5

(d) 7

Answer: (c) 5

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.

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

Q15. 465 coins consists of 1 rupee, 50 paisa and 25 paisa coins. Their values are in the ratio 5: 3 : 1. The number of each type of coins respectively is

(a) 155, 186, 124

(b) 154, 187, 124

(c) 154, 185, 126

(d) 150, 140, 175

Answer: (a) 155, 186, 124

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 =$\mathrm{x}\times \frac{4\mathrm{c}}{\mathrm{a}+2\mathrm{b}+4\mathrm{c}}$

Q16. If a : b = 2 : 3, b : c = 3 : 4, c : d = 4 : 5, find a : b : c : d.

(a) 5 : 4 : 3 : 2

(b) 30 : 20 : 15 : 12

(c) 2 : 3 : 4 : 6

(d) 2 : 3 : 4 : 5

Answer: (d) 2 : 3 : 4 : 5

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

Q17. Divide Rs. 671 among A, B, C such that if their shares be increased by Rs. 3, Rs. 7 and Rs. 9 respectively, the remainder shall be in the ratio 1 : 2 : 3.

(a) Rs. 112, Rs. 223, Rs. 336

(b) Rs. 114, Rs. 221, Rs. 336

(c) Rs. 112, Rs. 227, Rs. 332

(d) Rs. 114, Rs. 223, Rs. 334

Answer: (a) Rs. 112, Rs. 223, Rs. 336

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 = $\frac{3}{6}\times 690 -9=336$

Q18. If Rs. 1066 is divided among A, B, C and D such that A : B = 3 : 4, B : C = 5 : 6 and C : D = 7 : 5, who will get the maximum?

(a) B

(b) A

(c) C

(d) D

Answer: (c) C

Solution:

From ratio we can say C’s share is maximum

rp_s18

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

Q19. The income of A and B are in the ratio 3 : 2 and expenses are in the ratio 5 : 3. If both save Rs. 200, what is the income of A?

(a) Rs. 1000

(b) Rs. 1200

(c) Rs. 1500

(d) Rs. 1800

Answer: (b) Rs. 1200

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 = $\frac{200\times 3\left( 3-5 \right)}{3\times 3-2\times 5}=1200$

Q20. A and B are two alloys of gold and copper prepared by mixing metals in the ratio 7 : 2 and 7 : 11 respectively. If equal quantities of the alloys are melted to form a third alloy C, the ratio of gold and copper in C will be:

(a) 5 : 7

(b) 5 : 9

(c) 7 : 5

(d) 9 : 5

Answer: (c) 7 : 5

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 = $\frac{7}{6}:\frac{5}{6}=7:5$

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