__Ratio and Proportion Practice Problems – Set 2__**1. If sum of two numbers is 1210 and if 4/15 of one number is 2/5 of the other. Then one of the two numbers is:**

a) 284

b) 362

c) 482

**Answer:**

**D)**

**Explanation:**

Let A and B be two numbers.

Then 4/15 of A = 2/5 of B

A = (2/5 x 15/4)B

A = 3B/2

A : B = 3 : 2

Given, A + B = 1210

Then, 3X + 2X = 1210

X = 242

Therefore, 2X = 2(242) = 484

And 3X = 3(242) = 726

**2. Let a, b and c be three numbers such that b is 505 more than c and a is 20% more than c then a:b is**

a) 2:3

b) 1:3

c) 4:5

d) 3:7

**Answer:**

**C)**

**Explanation:**

Given that,

a is 20% more than c then a = 120% of c = 120c/100 =
6c/5

And b is 50% more than c then b = 150% of c =
150c/100 = 3c/2

Required ratio = a : b = 6c/5 : 3c/2 = 6/5 : 3/2 =
4:5

**3. The salary of two friends Ramu and Raju are in the ratio 4:5. If the salary of each one increases by Rs.6000, then the new ratio becomes 48:55. What is Raju's present salary?**

a) Rs.10500

b) Rs.10500

c) Rs.11500

d) Rs.12500

**Answer:**

**B)**

**Explanation:**

Ratio their salary is 4:5

Let the original salary of Ramu and Raju be 4k and
5k respectively.

After increasing Rs.6000, the ratio becomes 48:55

That is,

(4k+6000)/(5k+6000) = 48/55

55(4k+6000) = 48(5k+6000)

220k+330000 = 240k+288000

20k= 42000

We have to find the original salary of Raju; that
is, 5k.

If 20k = 42000 then 5k = 10500.

**4. The number of candidates writing three different entrance exams is in the ratio 4:5:6. There is a proposal to increase these numbers of candidates by 40%, 60% and 85% respectively. What will be the ratio of increased numbers?**

a) 14:15:16

b) 12:15:19

c)13:19:21

d) none of these

**Answer:**

**D)**

**Explanation:**

Given ratio of number of candidates is 4:5:6

Let the number of candidates for 3 exams be 4k, 5k
and 6k respectively.

After increasing, number of candidates become (140%
of 4k), (160% of 5k) & (185% of 6k)

That is, (140x4k)/100, (160x5k)/100 and (185x6k)/100

= 56k/10, 80k/10 and 111k/10

Now, the required new ratio is: 56k/100 : 80k/10 :
111k/10

= 56 : 80 : 111

**5. The ratio of salary of two persons X and Y is 5:8. If the salary of X increases by 60% and that of Y decreases by 35% then the new ratio of their salaries become 40:27. What is X's salary?**

a) Rs.15000

b) Rs.12000

c) Rs.19500

d) data inadequate.

**Answer:**

**D)**

**Explanation:**

Ratio of salary of X and Y is 5:8

Let the original salary of X and Y be Rs.5k and
Rs.8k respectively.

After increasing 60%, new salary of X = 160% of 5k =
160x5k/100 = 80k/10 ...(1)

After decreasing 35%, new salary of Y = (100-35)% of
8k = 65% of 8k = 52k/10 ...(2)

Given that, new ratio is 40:27

That is, 80k/10 : 52k/10 = 40/27

This does not give the value of k; so that we cannot
find X's exact salary.

Hence the answer is data inadequate.

**6. A factory employs skilled workers, unskilled workers and clerks in the proportion 8 : 5 : 1 and the wage of a skilled worker, an unskilled worker and a clerk are in the ratio 5 : 2 : 3. When 20 unskilled workers are employed, the total daily wages of all amount to Rs. 3180. Find the daily wages paid to each category of employees.**

a) Rs. 2100, Rs. 800, Rs. 280.

b) Rs. 2400, Rs. 480, Rs. 300.

c) Rs. 2400, Rs. 600, Rs. 180.

d) Rs. 2200, Rs. 560, Rs. 420.

**Answer:**

**C)**

**Explanation:**

Skilled workers, unskilled workers and clerks are in
the proportion 8 : 5 : 1

Given 20 unskilled workers, So 5/14 x K = 20, K =
56,

Therefore there are 32 skilled workers, 20 unskilled
workers and 4 clerks

Ratio of amount of 32 skilled workers, 20 unskilled
workers and 4 clerks

= 5 x 32 : 2 x 20 : 3 x 4

= 160 : 40 : 12 or 40 : 10 : 3

Now, divide Rs 3,180 in the ratio 40 : 10 : 3

We get, Rs. 2400, Rs. 600, Rs. 180.

**7. The soldiers in two armies when they met in a battle were in the ratio of 10 : 3. Their respective losses were as 20 : 3 and the survivors as 40 : 13. If the number of survivors in the larger army be 24,000, find the original number of soldiers in each army ?**

a) 28000, 8400

b) 25000, 7500

c) 29000, 2750

d) 26000, 7800

**Answer:**

**A)**

**Explanation:**

Let the soldiers in the two armies be 10X, 3X and
losses be 20Y, 3Y,

Then we have,

10X - 20Y = 24000 ...(i)

And 3X - 3Y = 24000 x 13/40 = 7800

Solving, we have 10X = 28000, 3X = 8400

**8.**

**What must be added to each of the numbers 7, 11 and 19, so that the resulting numbers may be in continued proportion?**

a)
3

b)
5

c)
4

d)
-3

**Answer:**

**D)**

**Explanation:**

Let X be the required number, then

(7 + X) : (11 + X) :: (11 +X) : (19 + X)

(7 + X) (19 + X) = (11 + X)^2

X^2 + 26X + 133 = X^2 + 22X + 121

4X = - 12 or X = - 3

**9. The mean proportional between 45 and a certain number is three times the mean proportional between 5 and 22. The number is ?**

a) 24

b) 49

c) 22

d) 9

**Answer:**

**C)**

**Explanation:**

If X be the required number, then

(45 x X)

^{1/2}= 3 x (5 x 22)^{1/2}
45X = 9 x 110

X = 22

**10.**

**A**

**and B are two alloys of argentum and brass prepared by mixing metals in proportions 7 : 2 and 7 : 11 respectively. If equal quantities of the two alloys are melted to form a third alloy C, the proportion of argentum and brass in C will be ?**

a)
5 : 9

b)
5 : 7

c)
7 : 5

d)
9 : 5

**Answer:**

**C)**

**Explanation:**

The first alloy is prepared by mixing metals in
proportions 7 : 2,

Then the weight of the first alloy will be = 7x + 2x

Assume ’x’ = 1Kg

The weight of the first alloy will be 9kg

The second alloy is prepared by mixing metals in
proportions 7 : 11,

The weight of the second alloy will be 18kg

If equal quantities of the two alloys are melted to
form a third alloy,

Then 18Kg of first alloy should be used, then its
ratio will become 14 : 4,

Now, the proportion of argentum and brass in third
alloy = 7 + 14 : 11 + 4 = 21 : 15 = 7 : 5