CBSE Class 10 Mathematics Ratio and Proportion MCQs Set 29

Practice MCQs for Class 10 Mathematics Ratio and Proportion

Explore reliable objective questions for Ratio and Proportion tailored for Class 10 learners. Utilizing these Mathematics multiple-choice formats ensures thorough preparation and strengthens problem-solving speed for upcoming school assessments.

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Multiple Choice Questions

Question 1: The price of a necklace varies directly as the number of pearls in it and as the square root of the radius of a pearl. The price was Rs. 150 when it had 75 pearls each of radius 1 cm. Find the radius of a pearl of a necklace having 100 pearls whose cost is Rs. 600 (in cm).
(a) 2
(b) 9
(c) 3
(d) 4
Show Answer & Explanation

Answer: (b) 9

Explanation:
1. Price = k × n × √r: 150 = k × 75 × 1, so k = 2
2. 600 = 2 × 100 × √r, so √r = 3 and r = 9.

Teacher's Note:
1. Combine the direct variations into one equation with a constant k.

Question 2: The price of a book varies directly as the number of pages in it and inversely as the number of years that have elapsed since the date of purchase. Two books cost the same, but the number of pages in the first book is triple that of the second book. If the first book is sold 18 years ago, how long ago was the second book sold?
(a) 54 years
(b) 9 years
(c) 6 years
(d) 3 years
Show Answer & Explanation

Answer: (c) 6 years

Explanation:
1. Price = k × pages/years, so for equal prices pages/years is equal
2. 3p/18 = p/t, so t = 6 years.

Teacher's Note:
1. Equal prices mean equal values of pages divided by years.

Question 3: Akbar and Birbal purchased shares for the cost of their basic salaries which are in the ratio 5 : 6. Later the company gave them 40 additional shares each, due to which the ratio changed to 7 : 8. If the worth of each share is Rs. 75, what is the basic salary of the person who has fewer shares?
(a) Rs. 10500
(b) Rs. 7500
(c) Rs. 8800
(d) Rs. 9000
Show Answer & Explanation

Answer: (b) Rs. 7500

Explanation:
1. Let the shares be 5x and 6x: (5x + 40)/(6x + 40) = 7/8
2. 40x + 320 = 42x + 280, so x = 20
3. Fewer shares = 100, so the salary = 100 × 75 = Rs. 7500.

Teacher's Note:
1. Form an equation from the changed ratio.

Question 4: Every orphan receives bananas, oranges and grapes in the ratio of 3 : 2 : 7 in terms of dozens. The weight of a grape is 24 gm and the weights of a banana and an orange are in the ratio 4 : 5, while the weight of an orange is 150 gm. Find the ratio of the three fruits in terms of weight (banana : orange : grape).
(a) 90 : 75 : 42
(b) 180 : 150 : 82
(c) 75 : 42 : 90
(d) None of these
Show Answer & Explanation

Answer: (a) 90 : 75 : 42

Explanation:
1. Weight of a banana = 4/5 × 150 = 120 gm
2. Weights: banana 3 × 120 = 360, orange 2 × 150 = 300, grape 7 × 24 = 168
3. Ratio = 360 : 300 : 168 = 90 : 75 : 42.

Teacher's Note:
1. Multiply the number of fruits by the weight of one.

Question 5: In a zoo, there are rabbits and pigeons. If heads are counted, there are 340 heads and if legs are counted there are 1060 legs. How many pigeons are there?
(a) 120
(b) 150
(c) 180
(d) 170
Show Answer & Explanation

Answer: (b) 150

Explanation:
1. Let r rabbits and p pigeons: r + p = 340 and 4r + 2p = 1060
2. 2r = 1060 − 680 = 380, so r = 190
3. p = 340 − 190 = 150.

Teacher's Note:
1. Rabbits have 4 legs and pigeons have 2.

Question 6: In three vessels, each of 25 litres capacity, a mixture of milk and water is filled. The ratio of milk and water are 3 : 1, 2 : 3 and 4 : 3 in the respective vessels. If all three vessels are emptied into a single large vessel, what will be the ratio of water to milk in the resultant mixture?
(a) 179 : 241
(b) 197 : 214
(c) 219 : 117
(d) 179 : 234
Show Answer & Explanation

Answer: (a) 179 : 241

Explanation:
1. Water = 25(1/4 + 3/5 + 3/7) = 25 × 179/140
2. Milk = 25(3/4 + 2/5 + 4/7) = 25 × 241/140
3. Water : milk = 179 : 241.

Teacher's Note:
1. Use the fraction of water and milk in each vessel.

Question 7: Two alloys are made up of copper and tin. The ratio of copper and tin in the first alloy is 1 : 3 and in the second alloy it is 2 : 5. In what ratio should the two alloys be mixed to obtain a new alloy in which the ratio of tin and copper is 8 : 3?
(a) 3 : 5
(b) 4 : 7
(c) 3 : 8
(d) 5 : 11
Show Answer & Explanation

Answer: (b) 4 : 7

Explanation:
1. Take x of the first and y of the second: copper = x/4 + 2y/7 and tin = 3x/4 + 5y/7
2. 3(3x/4 + 5y/7) = 8(x/4 + 2y/7)
3. 9x/4 + 15y/7 = 2x + 16y/7, so x/4 = y/7 and x : y = 4 : 7.

Teacher's Note:
1. Write copper and tin in the mixture separately.

Question 8: A vessel of capacity 2 litres has 25% alcohol and another vessel of capacity 6 litres had 40% alcohol. The total liquid of 8 litres was poured out in a vessel of capacity 10 litres and thus the rest part of the vessel was filled with water. What is the new concentration of the mixture?
(a) 31%
(b) 71%
(c) 49%
(d) 29%
Show Answer & Explanation

Answer: (d) 29%

Explanation:
1. Alcohol = 25% of 2 + 40% of 6 = 0.5 + 2.4 = 2.9 litres
2. Total volume = 10 litres, so the concentration = 29%.

Teacher's Note:
1. Water adds volume but no alcohol.

Question 9: The ratio of copper and nickel by weight in two alloys X and Y are 2 : 7 and 5 : 4 respectively. How many kilograms of the alloys X and Y are required to make 42 kg of new alloy Z in which the copper and nickel are in equal amounts?
(a) 6 kg and 36 kg
(b) 10 kg and 32 kg
(c) 7 kg and 35 kg
(d) None of these
Show Answer & Explanation

Answer: (c) 7 kg and 35 kg

Explanation:
1. Copper in x kg of X = 2x/9 and nickel = 7x/9; for y kg of Y: 5y/9 and 4y/9
2. Equal copper and nickel: 2x + 5y = 7x + 4y, so y = 5x
3. x + y = 42 gives x = 7 and y = 35.

Teacher's Note:
1. Equal amounts means copper = nickel in the final alloy.

Question 10: Seven years ago the ratio of the ages of P and Q (in years) was 7 : 6. Which of the following cannot be the ratio of their ages 6 years from now?
(a) 13 : 11
(b) 15 : 14
(c) 13 : 12
(d) 16 : 15
Show Answer & Explanation

Answer: (a) 13 : 11

Explanation:
1. Let the ages 7 years ago be 7k and 6k: ratio after 13 years = (7k + 13)/(6k + 13)
2. This is always less than 7/6 = 1.1667 for k > 0
3. 13 : 11 = 1.1818 is greater than 7/6, so it is not possible.

Teacher's Note:
1. Adding the same number to both terms moves a ratio towards 1.

Question 11: The ratio of the monthly incomes of Ram and Shyam is 3 : 4 and the ratio of their monthly expenditures is 4 : 5. If Shyam saves Rs. 400 per month, which of the following cannot be the savings of Ram (in Rs./month)?
(a) 290
(b) 280
(c) 270
(d) 310
Show Answer & Explanation

Answer: (d) 310

Explanation:
1. Incomes 3a and 4a, expenditures 4b and 5b: 4a − 5b = 400
2. Ram's savings = 3a − 4b = 3(400 + 5b)/4 − 4b = 300 − b/4
3. Since b > 0, Ram's savings are less than Rs. 300, so Rs. 310 is impossible.

Teacher's Note:
1. Express Ram's savings in terms of one unknown.

Question 12: The ratio of the present ages of Anand and Bala is 8 : 3. When Anand was 30 years old, Bala was 5 years old. Find the present age of Bala (in years).
(a) 10
(b) 12
(c) 15
(d) 20
Show Answer & Explanation

Answer: (c) 15

Explanation:
1. The age difference = 30 − 5 = 25 years = 5 parts, so 1 part = 5
2. Bala is 3 parts = 15 years.

Teacher's Note:
1. The difference of ages never changes.

Question 13: There are three sections A, B and C in class VIII of a school. The ratio of the number of students in A, B and C is 2 : 3 : 4. The section which has neither the maximum number of students nor the minimum number of students has a strength of 30. Find the total strength of the three sections.
(a) 63
(b) 82
(c) 72
(d) 90
Show Answer & Explanation

Answer: (d) 90

Explanation:
1. Section B is the middle one: 3 parts = 30, so 1 part = 10
2. Total = 9 parts = 90.

Teacher's Note:
1. The middle section corresponds to 3 parts.

Question 14: Find the triplicate ratio of (2y − x) : (2x − y), if x : y = 4 : 3.
(a) 64 : 125
(b) 27 : 64
(c) 8 : 125
(d) 27 : 125
Show Answer & Explanation

Answer: (c) 8 : 125

Explanation:
1. Take x = 4 and y = 3: 2y − x = 2 and 2x − y = 5
2. The ratio is 2 : 5, so the triplicate ratio = 8 : 125.

Teacher's Note:
1. The triplicate ratio of a : b is a³ : b³.

Question 15: Nine years ago A's age and B's age were in the ratio 5 : 7. Which of the following cannot be the ratio of their ages 5 years from now?
(a) 11 : 13
(b) 13 : 19
(c) 21 : 25
(d) 15 : 16
Show Answer & Explanation

Answer: (b) 13 : 19

Explanation:
1. Ages 9 years ago: 5k and 7k, so the ratio after 14 years = (5k + 14)/(7k + 14)
2. It is always greater than 5/7 = 0.714 (and less than 1)
3. 13 : 19 = 0.684 is less than 5/7, so it is impossible.

Teacher's Note:
1. Adding the same number to both terms moves a ratio towards 1.

Question 16: The force applied on a stationary body varies directly with the acceleration with which it starts to move. If a force of 10 N is applied on a stationary body, it starts to move with an acceleration of 2 m/s². Find the force (in N) to be applied on the body at which it starts to move with an acceleration of 4 m/s².
(a) 30
(b) 20
(c) 40
(d) 50
Show Answer & Explanation

Answer: (b) 20

Explanation:
1. F ∝ a, so F = k × a with k = 10/2 = 5
2. F = 5 × 4 = 20 N.

Teacher's Note:
1. Direct variation means the ratio F/a is constant.

Question 17: The ratio of the present ages of Ram and Shyam is 3 : 2. Which of the following cannot be the ratio of their ages 20 years ago?
(a) 8 : 5
(b) 17 : 10
(c) 9 : 5
(d) 7 : 5
Show Answer & Explanation

Answer: (d) 7 : 5

Explanation:
1. Present ages 3k and 2k with k > 10, so the ratio 20 years ago = (3k − 20)/(2k − 20)
2. This is always greater than 3/2 = 1.5
3. 7 : 5 = 1.4 is less than 1.5, so it is not possible.

Teacher's Note:
1. Subtracting the same number from both terms moves the ratio away from 1.

Question 18: Ninety coins are to be distributed among P, Q and R such that P gets twice as many coins as Q gets and Q gets thrice as many coins as R gets. Find the number of coins R gets.
(a) 6
(b) 2
(c) 10
(d) 9
Show Answer & Explanation

Answer: (d) 9

Explanation:
1. Let R get x, Q = 3x and P = 6x
2. 10x = 90, so x = 9.

Teacher's Note:
1. Express all three in terms of R.

Question 19: P varies inversely with √y. If y = 2, then P = 40. If P = 20, then find y.
(a) √8
(b) 8
(c) 4
(d) √2
Show Answer & Explanation

Answer: (b) 8

Explanation:
1. P × √y is constant: 40 × √2
2. 20 × √y = 40√2, so √y = 2√2 and y = 8.

Teacher's Note:
1. Inverse variation means a constant product.

Question 20: A solution of 30 litres of milk and water has 70% milk. How many litres of water must be added so that the volumes of milk and water will be in the ratio 3 : 2?
(a) 6
(b) 3
(c) 5
(d) 4
Show Answer & Explanation

Answer: (c) 5

Explanation:
1. Milk = 21 litres and water = 9 litres
2. 21/(9 + w) = 3/2, so 9 + w = 14 and w = 5 litres.

Teacher's Note:
1. Milk stays at 21 litres.

Question 21: If (a + b)/(xa + yb) = (b + c)/(xb + yc) = (c + a)/(xc + ya), where x + y ≠ 0 and a + b + c ≠ 0, then each of these ratios is equal to
(a) 1
(b) 1/(x + y)
(c) 2/(x + y)
(d) 2/(a + b)
Show Answer & Explanation

Answer: (c) 2/(x + y)

Explanation:
1. Each ratio equals (sum of numerators)/(sum of denominators)
2. = 2(a + b + c)/((x + y)(a + b + c)) = 2/(x + y).

Teacher's Note:
1. Equal ratios: add all numerators and all denominators.

Question 22: If x : y = 3 : 5, find the duplicate ratio of (3x + y) : (5x − y).
(a) 49 : 25
(b) 25 : 9
(c) 36 : 25
(d) 49 : 36
Show Answer & Explanation

Answer: (a) 49 : 25

Explanation:
1. Take x = 3 and y = 5: 3x + y = 14 and 5x − y = 10
2. The ratio is 14 : 10 = 7 : 5, so the duplicate ratio = 49 : 25.

Teacher's Note:
1. The duplicate ratio of a : b is a² : b².

Question 23: Amit is as much younger to Barkha as he is older than Chaman. If the sum of the ages of Barkha and Chaman is 48 years, what is the present age of Amit?
(a) 18 years
(b) 36 years
(c) 24 years
(d) 28 years
Show Answer & Explanation

Answer: (c) 24 years

Explanation:
1. Barkha − Amit = Amit − Chaman, so 2 × Amit = Barkha + Chaman = 48
2. Amit = 24 years.

Teacher's Note:
1. The middle of three equally spaced values is their average.

Question 24: Bipin is 6 times as old as Alok. Bipin's age will be twice of Chandan's age after 10 years. If Chandan's 7th birthday was celebrated 3 years ago, what is Alok's present age?
(a) 15 years
(b) 12 years
(c) 5 years
(d) None of these
Show Answer & Explanation

Answer: (c) 5 years

Explanation:
1. Chandan is 7 + 3 = 10 years old now
2. After 10 years Chandan = 20, so Bipin = 40 after 10 years, so Bipin is 30 now
3. Alok = 30/6 = 5 years.

Teacher's Note:
1. Count forward from the 7th birthday.

Question 25: Renuka got married 8 years ago. Today her age is 1 1/3 times her age at the time of marriage. Her daughter's age is 1/8 times her age. Her daughter's age is
(a) 3 years
(b) 4 years
(c) 6 years
(d) 8 years
Show Answer & Explanation

Answer: (b) 4 years

Explanation:
1. Let her age now be x: x = 4/3 (x − 8), so 3x = 4x − 32 and x = 32
2. Daughter's age = 32/8 = 4 years.

Teacher's Note:
1. Her age at marriage was x − 8.

Question 26: Ten years ago B was twice as old as A. If the ratio of their present ages is 4 : 3, what is the sum of their present ages?
(a) 25 years
(b) 30 years
(c) 40 years
(d) 35 years
Show Answer & Explanation

Answer: (d) 35 years

Explanation:
1. Let B = 4k and A = 3k now: 4k − 10 = 2(3k − 10)
2. 4k − 10 = 6k − 20, so k = 5
3. Sum = 7k = 35 years.

Teacher's Note:
1. Use the present-age ratio and the past-age relation together.

Question 27: The sum of the ages of Aryabhatta and Shridhar is 45 years. Five years ago the product of their ages was 4 times Aryabhatta's age at that time. The present ages of Aryabhatta and Shridhar respectively are
(a) 25 and 20
(b) 35 and 10
(c) 36 and 9
(d) 40 and 5
Show Answer & Explanation

Answer: (c) 36 and 9

Explanation:
1. Let the ages be A and S: (A − 5)(S − 5) = 4(A − 5), so S − 5 = 4 and S = 9
2. A = 45 − 9 = 36.

Teacher's Note:
1. Cancel the common factor A − 5.

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