Ratio and
Proportion Practice Problems – Set 3
1. Population of two villages X and Y are in the
ratio of 5:7 respectively. If the population of village Y increase by 25000 and
the population of village x remains unchanged the respective ratio of their
populations becomes 25:36. What is the population of village X?
a) 625000
b) 675000
c) 875000
d) 900000
Answer: A)
Solution: Let the population of Village X be x
Let the population of
Village Y be y
According to question
⇒ x/y = 5/7
⇒ 7x = 5y
Multiplying both sides by
5
⇒ 35x = 25y .... (i)
Now, New Population of Y
= y' = y + 25000
And New Ratio = x/y' =
25/36
⇒ 36x = 25y'
⇒ 36x = 25 × (y + 25000)
⇒ 36x = 25y + 625000
From eq. (1)
⇒ 36x = 35x + 625000
⇒ x = 625000
Hence, Population of
Village ‘X’ is 625000.
2. In a fraction, twice the numerator is two more
than the denominator. If 3 is added to the numerator and the denominator each,
then the resultant fraction will be 2/3. What is the original fraction?
a) 5/16
b) 6/13
c) 14/6
d) 7/12
e) 7/13
Answer: D)
Solution: Let’s take a fraction m/n
Now, according to the
question,
2m – 2 = n -------------
[1]
Also, (m + 3)/(n + 3) =
2/3 ------------- [2]
Solving the above
equations we get m = 7 and n = 12
Hence our fraction is
7/12
3. Arun covers a total distance of 500 km by bus,
train and bike in the ratio of 3 : 4 : 2 respectively. The speed at which
Journey was covered by bus, train and bike was in the the ratio of 3 : 2 :
4. Find out the ratio of the time taken by Bus , Train and Bike to cover
the distance?
a) 2 : 4 : 1
b) 1 : 2 : 3
c) 1 : 1 : 1
d) 2 : 1 : 4
e) None of these
Answer: A)
Solution: Since, the total distance covered by
Bus, Train and Bike is in the ratio of 3 : 4 : 2,
So, lets assume the
distance covered by Bus = 3a,
Distance covered by Train
= 4a,
And the distance covered
by Bike = 2a,
Similarly,
The speed of Bus, Train
and Bike in the ratio of 3 : 2 : 4
So lets assume speed of
Bus = 3b,
Speed of Train = 2b
And the speed of Bike =
4b
As we know Time =
(Distance/speed)
So Time taken by Bus =
3a/3b = (a/b)
Similarly time taken by
Train = 4a/2b = (2a/b)
And tine taken by Bike is
= 2a/4b = (a/2b)
So the required ratio = 2
: 4 : 1
4. Three persons A, B, C rent the grazing of the
park for Rs 570. A puts 126 oxen in the park for 3 months, B puts in 162 oxen
for 5 months and C puts in 216 oxen in 4 months. What part of rent should B
pay?
a) 250
b) 225
c) 200
d) 100
e) 255
Answer: B)
Solution: Given, A puts 126 oxen in the park
for 3 months, B puts in 162 oxen for 5 months and C puts in 216 oxen in 4
months.
∴ For A, usage = 126 × 3 = 378
For B, usage = 162 × 5 =
810
For C, usage = 216 × 4 =
864
Ratio of A : B :C = 384 :
810 : 864
If the total rent is 570,
rent paid by B = 810/(378 + 810 + 864) × 570
⇒ Rent paid by B = 810/2052 × 570 = 225
5. A man divides his property so that his son’s
share to his wife’s and the wife’s share to his daughter are both in the ratio
3 : 1. If the daughter gets Rs. 10,000 less than the son, find the total worth
of the property.
a) Rs.16200
b) Rs.16250
c) Rs.16500
d) Rs.15300
e) None of these
Answer: B)
Solution: Let share of daughter be m.
Then share of wife = 3 ×
m = 3m
And share of son = 3 × 3m
= 9m
Total property = m + 3m +
9m = 13m
We have
⇒ 9m – m = 10000 ⇒ 8m = 10000 ⇒ m = 10000/8 ⇒ m = 1250
Total property = 13m = 13
× 1250 = 16250
6. The measures of the three angles in a triangle
are in the ratio 1 : 1 : 2. Which of the following must be true?
I. The triangle is equilateral.
II. The triangle is isosceles.
III. The triangle is right triangle.
a) Only I is true
b) Only II is true
c) only III is true
d) only II and III are true
e) None of these
Answer: D)
Solution: Let the 3 angles of the triangle are
m, m and 2m.
We also know that the sum
of all the angles of a triangle is 180
Hence we have
⇒ m + m + 2m = 180
⇒ 4m = 180
⇒ m = 180/4
⇒ m = 45
Hence we have the angles
of the triangle as 45, 45 and 90
Since, 2 angles of the
triangle are equal, we can say that the triangle is isosceles triangle.
Also, we have one angle
as 90, Hence we can say that the triangle is a right triangle
7. Suman has a certain number of green and blue
pens. Which of the following cannot be ratio of the green pens to the blue
pens, if she has a total of 15 pens?
a) 2 : 3
b) 1 : 4
c) 1 : 2
d) 4 : 3
e) 4 : 11
Answer: D)
Solution: The number of pens (whether green or
blue) has to be an integer. If number of green pens as 2x and blue pens as 3x,
then, 2x + 3x = 15
⇒ x = 3,
∴ Green pens = 6 and blue pens = 9.
Similarly, option 2, 3
and 5are also possible as 15 is divisible by (1 + 4 = 5), (1 + 2 = 3) and (7 +
8 = 15) but 15 is not divisible by (4 + 3 = 7) as in option (4), this would
give us no of green and blue pens as a fraction which is not possible.
8. A bag contains an equal number of one rupee, 50
paise and 25 paise coins respectively. If the total value is Rs. 35, how many
coins of each type are there?
a) 20 coins
b) 30 coins
c) 28 coins
d) 25 coins
e) 22 coins
Answer: A)
Solution: Let number of coins of each
denomination = m.
Then we have total worth
of one rupee coins = 1 × m = m
Total worth of 50 paise
coins = 0.5 × m = 0.5m
And total worth of 25
paise coins = 0.25 × m = 0.25m
Total worth = m + 0.5m +
0.25m
= 1.75m
We are given that
⇒1.75m = 35
⇒m = 35/1.75
⇒m = 20
Hence we have 20 coins of
each denomination
9. A dog takes 3 leaps for every 5 leaps of a
hare. If one leap of the dog is equal to 3 leaps of the hare, the ratio of the
speed of the dog to that of the hare is
a) 8 : 5
b) 9 : 5
c) 8 : 7
d) 9 : 7
e) 5 : 9
Answer: B)
Solution: Let the length of 1 leap of dog be
“D” and that of Hare be “H”
If one leap of the dog is
equal to 3 leaps of the hare
⇒ 1 × D = 3 × H ⇒ D/H =
3/1
According to this ratio
⇒ D = 3x and H = x (Where x = constant)
⇒ Also, A dog takes 3 leaps for every 5 leaps of a hare
⇒ Let the time be T in which a dog takes 3 leaps and hare
takes 5
∵ Speed = Distance/Time
⇒ Speed of dog (for the 3 leaps taken in time T) = 3D/T = (3x ×
3)/T = 9x/T
⇒ Speed of dog (for the 3 leaps taken in time T) = 5H/T = (x ×
5)/T = 5x/T
⇒ Ratio of the speed of the speed of the dog to that of the
hare is = 9x/T : 5x/T = 9 : 5
10. Two numbers are such that their difference, their
sum and their product are in the ratio of 2 : 8 : 15. The product of the
numbers is :
a) 8
b) 15
c) 24
d) 35
e) 20
Answer: B)
Solution: Let the numbers be ‘a’ and ‘b’
respectively.
Given, two numbers are
such that their difference, their sum and their product are in the ratio of 2 :
8 : 15.
∴ (a – b) : (a + b)
: ab = 2 : 8 : 15
⇒ (a – b)/(a + b) = 2/8 ------- eq (1)
And (a – b)/ab = 2/15
------- eq (2)
Simplifying equation 1 we
get:
4a – 4b = a + b ⇒ 3a = 5b ⇒ a = 5b/3
Substituting in eq2 we
get: b = 3
Then a = 5
Product of the two
numbers = 3 × 5 = 15
11. A man before his death writes his will. In it,
he mentions that three-fifth of his wealth to be given to his Son-in-law and
daughter. The ratio of amount to be given to the daughter and the son-in-law
must be 4 : 1 respectively. Half the amount that his daughter is getting must
be donated to his favourite NGO. One fourth of the remaining amount, he
entitles his wife to receive and the rest which amounts to Rs. 16800, he wants
to be burried along his grave. What amount he donated to the NGO mentioned in
the question?
a) 25200
b) 28400
c) 33600
d) 36800
e) None of these
Answer: C)
Solution: Let the total wealth be Rs. 100.
Amount to be given to the
son-in-law and the daughter = 3/5 × 100 = 60
The given ratio of amount
of the son-in-law and the daughter = 1 : 4
∴ Amt. to be given to
the daughter = 4/5 × 60 = 48
∴ Amt. to be donated to the NGO = 1/2 × 48 = 24
Now, remaining amount =
100 – 60 – 24 = 16
Amt. to be given to wife
= 1/4 × 16 = 4
The final remaining
amount = 16 – 4 = 12
Applying the rule of
proportion here, we get
Since, 12 ≡ 16800
∴ 24 (the amount donated to the NGO) ≡ X
∴ x = 16800/12 × 24 = 33600
12. A cylindrical vessel contains milk. When one
fourth of vessel has been leaked vessel is filled with water. From here, three
fourths of mixture is leaked and the vessel again filled with water and now
again half of the vessel has been leaked and it is again filled with water. How
much percent of milk does the vessel finally contain?
a) 10721
b) 9375
c) 8543
d) 10503
e) None of these
Answer: B)
Solution: Suppose, vessel initially contains x
litres of milk. Acc. to question given-
Remaining milk = x (1 –
1/4) (1 – 3/4) (1 – 1/2)
Remaining milk = x
(3/4)(1/4)(1/2)
Remaining milk = 3x/32
Remaining milk percentage
to the original solution = (3x/32)/x × 100 = 9.375
13. A vessel contains a solution of alcohol and
water in the ratio 3 : 2. The volume of the contents is increased by 50% by
adding water to this. From this resultant solution 30 L is withdrawn and then
replaced with water. The resultant ratio of alcohol to water in the final
solution is 3 : 7. Find the original volume of the solution.
a) 80
b) 90
c) 100
d) 105
e) 110
Answer: A)
Solution: Alcohol content = 3x
Water content = 2x
Mixture = 5x
As it's given that after
adding water to solution, the solution gets increased by 50% (because of the
water added),
Therefore, quantity
of Alcohol in new solution = as it is =
3x
But quantity of water
will be = 2x + 2.5x = 4.5x
Threfore, new ratio = 2 :
3,
After removing 30 L of
solution,
Remaining Alcohol = 3x –
12 L
Remaining Water = 4.5x –
18 L
After adding 30 L water
to the solution,
Alcohol = 3x – 12
Water = 4.5x – 18 + 30 =
4.5 + 12
Now according to the
question, (3x – 12)/(4.5x + 12) = 3/7
21x – 84 = 13.5x + 36
=> 7.5x = 120 => x = 16
Volume of original soln.
= 5x = 80.
14. 20 L of mixture contains alcohol and soda in
the ratio 6 : 2. if 4L of this mixture
be replaced by 4 L of alcohol the ratio of alcohol to soda in the new
mixture?
a) 4 : 3
b) 4 : 1
c) 8 : 5
d) 5 : 9
e) 7 : 5
Answer: B)
Solution: 4 L of mixture will remove alcohol
and soda in the same ratio in which they are present now.
i.e., 6/8 × 4L → alcohol
will be removed.
And 2/8 × 4L → soda will
be removed
Old concentration
6/8 × 20L -> alcholol;
2/8 × 20L -> soda
Alcholol 15L and Soda 5L
After removed
Alcholol => 15 – 6/8 ×
4 = 15 – 3 = 12L
Soda => 5 – 2/8 × 4 =
4L
After replacement
Alcholol => 12 + 4 =
16
Soda => 4 + 0 = 4
Ratio becomes = 16 : 4 =
4 : 1
15. A cylindrical container having milk is at a
level of 5/7 of its actual level from this x litres is removed, the container
is 1/4th full. Finally 15 litres is added to the container making it half full.
The value of x and the capacity of the container.
a) 27.85L
b) 32.67L
c) 38.9L
d) 41.9L
e) None of these
Answer: A)
Solution: Let the full container has a volume =
y litres
Then According to
question –
5/7y – x = 1/4y => x =
5/7y – 1/4y => x = (20 – 7)y/28 = 13y/28
Now, 15L is added to 1/4y
it becomes 1/2y
=> 1/4y + 15 = 1/2y
=> y = 15 × 4 = 60 L
=> x = 13/28 × 60 =
(15 × 13)/7 = 27.85L
