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
