Practice Ratio and proportion - quantitative aptitude Online Quiz (set-1) For All Competitive Exams
Q-1) What number should be subtracted from both the terms of the ratio 11 : 15 so as to make it as 2 : 3 ?
(a)
(b)
(c)
(d)
Required number = x
${11 - x}/{15 - x} = 2/3$
33 - 3x = 30 - 2x
3x - 2x = 33 - 30
x = 3
Q-2) In 80 litres mixture of milk and water the ratio of amount of milk to that of amount of water is 7 : 3. In order to make this ratio 2 : 1, how many litres of water should be added ?
(a)
(b)
(c)
(d)
Quantity of milk
= $7/10 × 80 = 56$ litres
Quantity of water
= $3/10 × 80$ = 24 litres
Let x litre water be added Then,
$56/{24 + x} = 2/1$
24 + x =28 ⇒ x = 4 litres
Q-3) The mean proportional between $(3 +√2)$ and $(12 - √32)$ is
(a)
(b)
(c)
(d)
Using Rule 14Mean Proportion - Let x be the mean proportion between a and b, then a:x::x:b (Real condition)$a/x = x/b ⇒ x^2 =ab$$x =√{ab}$So, mean proportion of a and b = $√{ab}$If x be the mean proportion between (x - a) and (x - b) then what will be the value of x ?$x = {ab}/{a+b}$
Mean proportional
=$√{(3+√2)(12-√{32})}$
= $√{(3+√2)4(3-√2)}$
= 2$√{9 - 2} = 2√7$
Q-4) The present age of two persons are 36 and 50 years respectively. If after n years the ratio of their age will be 3 : 4, then the value of n is
(a)
(b)
(c)
(d)
${36 + n}/{50 + n} = 3/4$
144 + 4n = 150 + 3n
4n - 3n = 150 - 144
n = 6
Q-5) The ratio between two numbers is 3 : 4. If each number is increased by 6, the ratio becomes 4 : 5. The difference between the numbers is
(a)
(b)
(c)
(d)
Let the numbers be 3x and 4x.
${3x + 6}/{4x + 6} = 4/5$
16x + 24 = 15x + 30
x = 30 - 24 = 6
Required difference = 6
Using Rule 34Two numbers are in the ratio a:b and if each number is increased by x, the ratio becomes c:d. Then the two numbers will be${xa(c-d)}/{ad-bc}$ and ${xb(c-d)}/{ad-bc}$
Here, a = 3, b= 4, x = 6
c = 4, d = 5
The numbers are = ${xa(c-d)}/{ad-bc}$
= ${6.3(4 - 5)}/{3 ×5 - 4 × 4}$
= ${18 × -1}/{15 - 16}$ = 18
= ${xb(c-d)}/{ad-bc}$
= ${6 × 4(4 - 5)}/{3 × 5 - 4 × 4}$
= ${24 × (-1)}/{15 - 16}$ = 24
Numbers are 24 and 18.
Their difference = 24 - 18 = 6
Q-6) The average of 11 numbers is 36, whereas average of 9 of them is 34. If the remaining two numbers are in the ratio of 2 : 3, find the value of the smaller number (between remaining two numbers).
(a)
(b)
(c)
(d)
According to the question,
Sum of remaining two numbers
= 11 × 36 - 9 × 34
= 396 - 306 = 90
Ratio of the remaining two numbers = 2 : 3
Smaller number
= $2/5$ × 90 = 36
Q-7) The total number of students in a school was 660. The ratio between boys and girls was 13 : 9. After some days, 30 girls joined the school and some boys left the school and new ratio between boys and girls became 6 : 5. The number of boys who left the school is :
(a)
(b)
(c)
(d)
In the first case,
Boys = $660 × 13/22$ = 390
Girls = $660 × 9/22$ = 270
If x boys leave the school, then
${390 - x}/{270 + 30} = 6/5$
390 - x = 360
x = 390 - 360 = 30
Q-8) The ratio of the number of ladies to that of gents at a party was 3 : 2. When 20 more gents joined the party, the ratio was reversed. The number of ladies present at the party was
(a)
(b)
(c)
(d)
Let the number of ladies and gents be 3x and 2x respectively.
According to the question,
${3x}/{2x + 20} = 2/3$
9x = 4x + 40
5x = 40 ⇒ x = 8
Number of ladies = 3x
= 3 × 8 = 24
Q-9) In a bag, there are three types of coins — 1-rupee, 50 paise and 25-paise in the ratio of 3 : 8 : 20. Their total value is 372. The total number of coins is
(a)
(b)
(c)
(d)
Ratio of the number of coins of Re. 1, 50 paise and 25 paise
= 3 : 8 : 20
Ratio of the values of these coins
= $3 : 8/2 : 20/4$ = 3 : 4 : 5
Value of 1 rupee coins
= $3/12 × $372 = Rs.93
Value of 50 paise coins
= $4/12 × 372$ = Rs.124
Value of 25 paise coins
= $5/12$ × 372 = Rs.155
Number of coins
= 93 + 124 × 2 + 155 × 4
= 93 + 248 + 620 = 961
Q-10) A bag contains 90 in coins of denominations of 50 paise, 25 paise and 10 paise. If coins of 50 paise, 25 paise and 10 paise are in the ratio 2 : 3 : 5, then the number of 25 paise coins in the bag is
(a)
(b)
(c)
(d)
Ratio of values of 50 paise, 25 paise and 10 paise coins
= $2/2 : 3/4 : 5/10$
= $1 : 3/4 : 1/2$ = 4 : 3 : 2
Sum of the ratios = 4 + 3 + 2 = 9 Value of 25 paise coins
= $3/9 × 90$ = Rs.30
Number of 25 paise coins
= 30 × 4 = 120
Q-11) My grandfather was 9 times older than me 16 years ago. He will be 3 times of my age 8 years from now. Eight years ago, the ratio of my age to that of my grandfather was
(a)
(b)
(c)
(d)
16 years ago,
My age = x years
My grandfather’s age = 9x years
After 8 years from the present,
9x + 16 + 8 = 3(x + 8 + 16)
9x + 24 = 3x + 24 + 48
9x + 24 = 3x + 72
9x - 3x = 72 - 24
6x = 48
$x = 48/6$ = 8
Required ratio 8 years ago,
= (x + 8) : (9x + 8)
= (8 + 8) : (9 × 8 + 8)
= 16 : 80 = 1 : 5
Q-12) The ratio of the ages of a father and his son 10 years hence will be 5 : 3, while 10 years ago, it was 3:1. The ratio of the age of the son to that of the father today, is
(a)
(b)
(c)
(d)
Let the age of father 10 years hence is 5x years,
then age of son 10 years hence will be 3x years.
According to the question,
${5x - 10 - 10}/{3x - 10 - 10} = 3/1$
${5x - 20}/{3x - 20} = 3/1$
5x - 20 = 9x - 60
4x = 40 or x = 10
Required ratio
= (3x - 10) : (5x - 10)
= 20 : 40 = 1 : 2
Q-13) The ratio of the present ages of two boys is 3:4. After 3 years, the ratio of their ages is equal to will be 4:5.The ratio of their ages after 21 years will be
(a)
(b)
(c)
(d)
Let the ages of boys be 3x and 4x years respectively.
According to the question,
After 3 years
${3x + 3}/{4x + 3} = 4/5$
16x + 12 = 15x + 15
16x - 15x = 15 - 12
x = 3
Required ratio after 21 years
= ${3x + 21}/{4x + 21}$
= ${3 ×3 + 21}/{4 × 3 + 21} = {9 + 21}/{12 + 21}$
= $30/33 = 10/11$
Q-14) If 4 years ago the ratio between the ages of P and Q was 5 : 6 and the sum of their ages at present is 52 years, what is the ratio of their present ages ?
(a)
(b)
(c)
(d)
4 years ago,
P’s age = 5x years
Q’s age = 6x years
According to the question,
5x + 4 + 6x + 4 = 52
11x = 52 - 8 = 44
$x = 44/11$ = 4
Required ratio
= (5x + 4) : (6x + 4)
= (5 × 4 + 4) : (6 × 4 + 4)
= 24 : 28 = 6 : 7
Q-15) The ratio between Sumit’s and Prakash’s age at present is 2 : 3. Sumit is 6 years younger than Prakash. The ratio of Sumit’s age to Prakash’s age after 6 years will be
(a)
(b)
(c)
(d)
Sumit’s present age
= 2x years
Prakash’s present age
= 3x years
3x - 2x = 6 ⇒ x = 6
Required ratio
= (2 × 6 + 6) : (3 × 6 + 6)
= 18 : 24 = 3 : 4
Q-16) A container contains 60 kg of milk. From this container 6 kg of milk was taken out and replaced by water. This process was repeated further two times. The amount of milk left in the container is
(a)
(b)
(c)
(d)
Amount of milk left
= Initial amount ×
$(1-\text”Amount taken out in each operation”/\text”Initial amount”)^3$
= $60(1 - 6/60)^3$
= $60(1 - 1/10)^3$
= $60 × 9/10 × 9/10 × 9/10$
= 43.74 kg.
Q-17) A mixture contains spirit and water in the ratio 3 : 2. If it contains 3 litres more spirit than water, the quantity of spirit in the mixture is
(a)
(b)
(c)
(d)
Let the amount of water be x litres.
Then, ${x + 3}/x = 3/2$
or 2x + 6 = 3x
or x = 6
The quantity of spirit in the mixture
= x + 3 = 6 + 3 = 9 litres
Q-18) Harsha is 40 years old and Ritu is 60 years old. How many years ago was the ratio of their ages 3:5?
(a)
(b)
(c)
(d)
Let x years ago the ratio of their age was 3 : 5
According to the question
${40 - x}/{60 - x}= 3/5$
200 - 5x = 180 - 3x
2x = 20
x = 10 years
Q-19) At present, the ratio of the age of Maya and Chhaya is 6 : 5 and fifteen years from now, the ratio will get changed to 9 : 8. Maya’s present age is
(a)
(b)
(c)
(d)
Let Maya’s present age be 6x years and Chhaya’s present age be 5x years.
After 15 years,
${6x + 15}/{5x + 15} = 9/8$
48x + 120 = 45x + 135
48x - 45x = 135 - 120
3x = 15 ⇒ x = 5
Maya’s present age = 6x
= 6 × 5 = 30 years
Q-20) The ratio of A’s age to B’s age is 4 : 3. ‘A’ will be 26 years old after 6 years. The age of B now is :
(a)
(b)
(c)
(d)
A’s present age
= 4x years (let).
According to the question,
4x + 6 = 26
4x = 26 - 6 = 20
$x = 20/4$ = 5
B’s present age
= 3x = 3 × 5 = 15 years