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# Ratio and Proportion MCQ Question

#### Ratio and Proportion MCQ Question with Details Solution

Q21. Three containers have their volumes in the ratio 3 : 4 : 5. They are full of mixtures of milk and water. The mixtures contain milk and water in the ratio of (4: 1), ( 3 : 1) and (5 : 2) respectively. The contents of all these three containers are poured into a fourth container. The ratio of milk and water in the fourth container is:

(a) 4 : 1

(b) 151 : 48

(c) 157 : 53

(d) 5 : 2

 Answer: (c) 157 : 53Solution: Let the three containers contain 3k, 4k and 5k litres of mixtures, respectively. k = ratio constant.Milk in 1st mix. = $\frac{3\mathrm{k}\times 4}{5}=\frac{12\mathrm{k}}{5}$Water in 1st mix. = $\frac{3\mathrm{k}\times 1}{5}=\frac{3\mathrm{k}}{5}$Milk in 2nd mix. = $\frac{4\mathrm{k}\times 3}{4}=3\mathrm{k}$Water in 2nd mix. = $\frac{4\mathrm{k}\times 1}{4}=\mathrm{k}$Milk in 3rd mix. = $\frac{5\mathrm{k}\times 5}{7}=\frac{25\mathrm{k}}{7}$Water in 3rd mix. =$\frac{5\mathrm{k}\times 2}{7}=\frac{10\mathrm{k}}{7}$Total Milk = $\frac{12\mathrm{k}}{5}+3\mathrm{k}+\frac{25\mathrm{k}}{7}=\frac{314\mathrm{k}}{35}$Total Water = $\frac{3\mathrm{k}}{5}+\mathrm{k}+\frac{10\mathrm{k}}{7}=\frac{106\mathrm{k}}{35}$The ratio of milk and water in the fourth container = $\frac{314\mathrm{k}}{35}:\frac{106\mathrm{k}}{35}=157:53$

Q22. The prize money of Rs. 1,800 is divided among 3 students A, B and C in such a way that 4 times the share of A is equal to 6 times the share of B, which is equal to 3 times the share of C. Then A’s share is

(a) Rs. 400

(b) Rs. 600

(c) Rs. 700

(d) Rs. 800

 Answer: (b) Rs. 600Solution: 4A = 6B ⇒2A = 3B ⇒ A : B = 3 : 2B = 3C ⇒2 B = C ⇒B : C = 1 : 2Now A : B : C = 3 : 2 : 4A’s share = $1800\times \frac{3}{9}=600$

Q23. A bag contains Rs. 216 in the form of one rupee, 50 paisa and 25 paisa coins in the ratio of 2: 3: 4. The number of 50 paisa coins is:

(a) 96

(b) 144

(c) 114

(d) 141

 Answer: (b) 144Solution: ratio of coin =  2 : 3 : 4 ∴ ratio if their value = 2 : 1.5 : 1 = 4 : 3 : 2Value of 50 paisa coins = $216\times \frac{3}{9}=72$∴ The number of 50 paisa coins = 72 × 2 = 144

Q24. Two numbers are in the ratio 2 : 3. If 9 is added to each number, they will be in the ratio 3 : 4. What is the product of the two numbers?

(a) 360

(b) 480

(c) 486

(d) 512

 Answer: (c) 486Solution: 1st number 2k and 2nd Number = 3kNow $\frac{2\mathrm{k}+9}{3\mathrm{k}+9}=\frac{3}{4}$⇒9k + 27 = 8k + 36⇒ k = 36 – 27 = 91st number = 18 and 2nd Number = 27Product of the two numbers = 18 × 27 = 486Using trick: 1st number = $\frac{2\times 9\left( 3-4 \right)}{\left( 2\times 4-3\times 3 \right)}=18$2nd Number = $\frac{3\times 9\left( 3-4 \right)}{\left( 2\times 4-3\times 3 \right)}=27$Product of the two numbers = 18 × 27 = 486

Q25. In a school, the ratio of boys and girls is 4:5. When 100 girls leave the school, the ratio becomes 6:7. How many boys are there in the school?

(a) 1800

(b) 1200

(c) 1000

(d) 1500

(e) None of the above

 Answer: (b) 1200Solution:  Let boys = 4k and girls = 5kAccording to question,$\frac{4\mathrm{k}}{5\mathrm{k}-100}=\frac{6}{7}$⇒ 30k – 600 = 28k⇒ 2k = 600 or k = 300∴ Boys are there in the school = 300 × 4 = 1200

Q26. 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) 25Solution: Let boys = 4k and girls = 5kAccording to question,$\frac{4\mathrm{k}+10}{5\mathrm{k}}=\frac{6}{5}$⇒ 30k = 20k + 50⇒ 10k = 50 or k = 5∴ Girls are there in the class = 5 × 5 = 25

Q27. The speeds of three cars are in the ratio of 2:3:4. Find the ratio between the times taken by these cars to cover the same distance.

(a) 2 : 3 : 4

(b) 4 : 3 : 2

(c) 4 : 3 : 6

(d) 6 : 4 : 3

(e) None of the above

 Answer: (d) 6 : 4 : 3Solution: Speed and time are invers respectively.So, Ratio of time = $\frac{1}{2}:\frac{1}{3}:\frac{1}{4}=6:4:3$

Q28. Mr. Amit inherits 2505 gold coins and divides them among his three sons; Bharat, Parat and Marat; in a certain ratio. Out of the total coins received by each of them, Bharat sells 30 coins, Parat donates his 30 coins and Marat losses 25 coins. Now, the ratio of gold coins with them is 46 : 41 : 34, respectively. How many coins did Parat receive from his father?

(a) 705

(b) 950

(c) 800

(d) 850

(e) None of the above

 Answer: (d) 850Solution: Let gold coin have they now 46k, 41k, and 34k respectively.According question,(46k + 30) + (41k + 30) + (34k+25) = 2505⇒121k + 85 = 2505⇒ 121k = 2420 or k = 20∴ Parat receive coin from his father = 41 × 20 + 30 = 850

Q29. The ratio of three number is 3 : 4 : 7 and their product is 18144. Find the greatest number?

(a) 18

(b) 24

(c) 42

(d) 36

 Answer: (c) 42Solution:  let numbers are 3k, 4k, and 7kAccording to question,3k × 4k × 7k = 18144⇒ 84k3 = 18144⇒k3 = 216 or k = 6Greatest number = 7 × 6 = 42Using trick:Greatest number =$7\times \sqrt[3]{\frac{18144}{3\times 4\times 7}}=42$Rules: The ratio of three number is x : y : z and their product is ‘P’. The numbers are:1st Number = $x\times \sqrt[3]{\frac{P}{xyz}}$2nd Number = $y\times \sqrt[3]{\frac{P}{xyz}}$3rd Number = $z\times \sqrt[3]{\frac{P}{xyz}}$

Q30. If the ratio of milk and water in the allegation of 200 liter is 3:1 then water must be added in it so that ratio of milk and water would be 5:2

(a) 10

(b) 20

(c) 30

(d) 40

 Answer: (a) 10Solution: Milk = $200\times \frac{3}{4}=150$ and Water= $200\times \frac{1}{4}=50$Let x liter water must be added∴ $\frac{150}{50+\mathrm{x}}=\frac{5}{2}$⇒ 250 + 5x = 300⇒ 5x = 50 or x = 10Using trick:Required amount of water = $\frac{200\left( 3\times 2-1\times 5 \right)}{5\left( 3+1 \right)}=10$

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