## Ratio and Proportion – Aptitude Questions and Answers

**Ratio and Proportion aptitude questions and answers section with explanation. Practice online test for various interview, competitive and entrance exams.**

**1. Three persons A, B and C divide a certain amount of money such that A’s share is Rs. 4 less than half of the total amount, B’s share is Rs. 8 more than half of what is left and finally C takes the rest which is Rs. 14. Find the total amount they initially had with them?**

A.Rs. 61 B.Rs. 85 C.Rs. 80 D.Rs. 70 E.None of these__Answer:__ C**Explanation:**

Let the total amount be Rs. p.

Let shares of A and B be Rs. x and Rs. y respectively.

C’s share was Rs. 14

we have, x + y + 14 = p —– (1)

From the given data, x = (p/2) – 4 —– (2)

Remaining amount = p – (p/2 – 4) => p/2 + 4.

y = 1/2(p/2 + 4) + 8 => p/4 + 10 —– (3)

From (1), (2) and (3)

p/2 – 4 + p/4 + 10 + 14 = p

3p/4 + 20 = p

p/4 = 20 => p = Rs. 80.**2. Two men Amar and Bhuvan have the ratio of their monthly incomes as 6 : 5. The ratio of their monthly expenditures is 3 : 2. If Bhuvan saves one-fourth of his income, find the ratio of their monthly savings?**

A.3 : 5 B.3 : 10 C.3 : 8 D.1 : 2 E.None of these__Answer:__ B**Explanation:**

Let the monthly incomes of Amar and Bhuvan be 6x and 5x respectively.

Let the monthly expenditure of Amar and Bhuvan be 3y and 2y respectively.

Savings of Bhuvan every month = 1/4(5x)

= (His income) – (His expenditure) = 5x – 2y.

=> 5x = 20x – 8y => y = 15x/8.

Ratio of savings of Amar and Bhuvan

= 6x – 3y : 1/4(5x) = 6x – 3(15x/8) : 5x/4 = 3x/8 : 5x/4

= 3 : 10.**3. The weights of three boys are in the ratio 4 : 5 : 6. If the sum of the weights of the heaviest and the lightest boy is 45 kg more than the weight of the third boy, what is the weight of the lightest boy?**

A.32 kg B.36 kg C.40 kg D.44 kg E.None of these__Answer:__ B**Explanation:**

Let the weights of the three boys be 4k, 5k and 6k respectively.

4k + 6k = 5k + 45

=> 5k = 45 => k = 9

Therefore the weight of the lightest boy

= 4k = 4(9) = 36 kg.**4. If p, q and r are positive integers and satisfy x = (p + q -r)/r = (p – q + r)/q = (q + r – p)/p, then the value of x is?**

A.1/2 B.1 C.-1/2 D.-1 E.None of these__Answer:__ B**Explanation:**

When two or more ratios are equal, each of the ratios are equal to sum of the numerators divided by the sum of the denominators, provided sum of the denominators is non-zero.

Hence, x = (p + q -r)/r = (p – q + r)/q = (q + r – p)/p

=> x = (p + q – r + p – q + r + q + r – p) / (r + q + p)

=> x = (r + q + p) / (r + q + p) = 1

p + q + r is non-zero.**5. What is the least number to be subtracted from 11, 15, 21 and 30 each so that resultant numbers become proportional?**

A.1 B.2 C.3 D.4 E.None of these__Answer:__ C**Explanation:**

Let the least number to be subtracted be x, then 11 – x, 15 – x, 21 – x and 30 – x are in proportion.

<=> (11 – x) : (15 – x) = (21 – x) : (30 – x)

=> (11 – x)(30 – x) = (15 – x)(21 – x)

From the options, when x = 3

=> 8 * 27 = 12 * 18**6. A gardener wants to plant trees in his garden in rows in such a way that the number of trees in each row to be the same. If there are 24 rows the number of trees in each row is 42 if there are 12 more rows find the number of trees in each row?**

A.63 B.28 C.48 D.32 E.None of these__Answer:__ B**Explanation:**

Required number of trees

= 24/36 * 42 = 28.**7. In an exam, a candidate secured 504 marks our of the maximum mark of M. If the maximum mark M is converted into 800 marks, he would have secured 384 marks. What is the value of M?**

A.750 B.1200 C.1125 D.975 E.None of these__Answer:__ E**Explanation:**

504/M = 384/800

(504 * 800) / 384 = M

M = 1050**8. In a fort, there are 1200 soldiers. If each soldier consumes 3 kg per day, the provisions available in the fort will last for 30 days. If some more soldiers join, the provisions available will last for 25 days given each soldier consumes 2.5 kg per day. Find the number of soldiers joining the fort in that case.**

A.420 B.528 C.494 D.464 E.None of these__Answer:__ B**Explanation:**

Assume x soldiers join the fort. 1200 soldiers have provision for 1200 (days for which provisions last them)(rate of consumption of each soldier)

= (1200)(30)(3) kg.

Also provisions available for (1200 + x) soldiers is (1200 + x)(25)(2.5) k

As the same provisions are available

=> (1200)(30)(3) = (1200 + x)(25)(2.5)

x = [(1200)(30)(3)] / (25)(2.5) – 1200 => x = 528.**9. 15 binders can bind 1400 books in 21 days. How many binders will be required to bind 1600 books in 20 days?**

A.14 B.18 C.24 D.28 E.None of these__Answer:__ B**Explanation:**

Binders Books Days

15 1400 21

x 1600 20

x/15 = (1600/1400) * (21/20) => x = 18**10. 150 men consume 1050 kg of rice in 30 days. In how many days will 70 men consume 980 kg of rice?**

A.30 B.60 C.45 D.90 E.None of these__Answer:__ B**Explanation:**

Rate of consumption of each man = 1050/(150 * 30) = 7/30 kg/day

Let us say 70 men take x days to consume 150 kg.

Quantity consumed by each item in x days = (7x/30) kg

Quantity consumed by 70 men in x days = (7/30 x)(70) kg

= (7/30 x) * (70) = 960

x = 60 days.

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