Ratio and Proportion
Questions for SSC/RRB Exams (05 – 04 – 2018)
1. Find the ratio of b : a given
that 0.2a + 0.3b = 0.4 and 0.3a + 0.4b = 0.6
a) 2
b) 0
c) -2
Answer: B)
Solution:
Multiplying
the equations as follows:
0.2a
+ 0.3b = 0.4 -----(1) × 3
0.3a
+ 0.4b = 0.6 -----(2) × 2
⇒ 0.6a +
0.9b = 1.2
And,
0.6a + 0.8b = 1.2
Subtracting
the above 2 equations, we get: b = 0
Substituting
the value of b in any of the equations, we get: a = 2
∴ b : a = 0
2. Two numbers are in the ratio 3 :
4. If 4 be added to both of them, then their ratio becomes 5 : 6. Find the sum
of the numbers:
a) 14
b) 15
c) 20
d) 25
Answer: A)
Solution:
Given,
two numbers are in the ratio 3 : 4.
Let
the numbers be 3a and 4a respectively.
Given,
if 4 be added to both of them, then their ratio becomes 5 : 6.
⇒ (3a +
4)/(4a + 4) = 5/6
⇒ 18a + 24 =
20a + 20
⇒ 2a = 4
⇒ a = 2
Sum
of numbers = 3a + 4a = 7a = 14
3. In a school, the ratio of boys
to girls is 5 : 4 and the ratio of girls to teachers is 8 : 1. The ratio of
students to teachers is:
a) 18 : 1
b) 10 : 1
c) 8 : 3
d) 42 : 4
Answer: A)
Solution:
Let
the number of teachers be x.
∵ ratio of
girls : teachers = 8 : 1,
Number
of girls = (8/1) × x = 8x
Now,
since ratio of boys : girls = 5 : 4
Number
of boys = (5/4) × 8x = 10x
Number
of students = number of boys + number of girls = 10x + 8x = 18x
∴ Ratio of
student: teachers = 18x : x = 18 : 1
Hence,
the ratio of students and teachers is 18: 1
4. In an ornament the ratio of
silver and aluminium is 7 : 5. The percentage of aluminium in the ornament is:
a) 58 1/3
b) 41 2/3
c) 42.5
d) 57.5
Answer: B)
Solution:
The
percentage of aluminium in the ornament = 5/12 * 100 = 41 2/3
5. The number of students of a
class is 66. The ratio of the number of male students to the number of female
students is 5 : 6. The number of female students is
a) 18
b) 36
c) 30
b) 24
Answer: B)
Solution:
Let
the number of male and female students be 5x and 6x respectively.
Total
number of students = 66
∴ 5x + 6x =
66
⇒ x = 6
∴ number of
females = 6x = 6 × 6 = 36
6. The present ages of three
persons are in proportions 3: 7: 8. Six years ago, the sum of their ages was
36. Find their present ages (in years).
a) 8, 20,
28
b) 9, 21,
24
c) 20, 35,
45
d) Can’t be
determined
Answer: B)
Solution:
Let
the present ages of persons be 3x, 7x and 8x respectively.
Six
years ago,
Their
ages would be (3x – 6), (7x – 6) and (8x – 6) years respectively.
Given:
the sum of their ages was 36
Thus,
(3x – 6) + (7x – 6) + (8x – 6) = 36
⇒18x – 18 = 36
⇒18x = 54
∴ x = 3
Therefore,
their present years will be (3 × 3), (7 × 3) and (8 × 3) i.e., 9, 21 and 24
years respectively.
7. The ratio of third proportional
to 12 and 30 and the mean proportional between 9 and 25 would be –
a) 3 : 4
b) 5 : 1
c) 2 : 7
d) 1 : 3
Answer: B)
Solution: Let the
third proportional to 12 and 30 be x
Then,12
: 30 : : 30 : x
⇒12x = 30 *
30 ⇒ x =
30*30/12 = 75
∴ Third
proportional to 12 and 30 = 75
Mean
proportional between 9 and 25 = √9 * 25 = 15
∴ Required
ratio = 75 : 15
⇒ Required
ratio = 5 :1
8. The fourth proportional to 0.12,
0.24, 8 is
a) 8.9
b) 56
c) 16
d) 17
Answer: C)
Solution:
Mathematically
the question can be written as,
0.12:
0.24 ∷ 8: x
[Where x = the fourth proportional]
⇒ 0.12/0.24
= 8/x
⇒ ½ = 8/x
⇒ x = 16
9. The fourth proportional to 75,
192 and 200 is equal to fourth proportional to 90, 384 and Q. Find the value of
Q.
a) 100
b) 108
c) 120
d) 126
Answer: C)
Solution:
Fourth
proportional to 75, 192 and 200 will be R, if
75/192
= 200/R
⇒ R = (200 × 192)/75
Fourth
proportional to 90, 384 and Q will be R, if
90/384
= Q/R
⇒ R =
(384Q/90)
⇒ (384Q/90)
= (200 × 192)/75
⇒ Q = (100 × 90)/75 =
120
10. The angles in a quadrilateral
PQRS are such that angle S is fourth proportional of angles P, Q and R. If the
ratio of angles Q and P is 5 : 4, and angle S is 125°, find the measure of
angle R.
a) 90°
b) 100°
c) 108°
d) 120°
Answer: B)
Solution:
Angle
S is fourth proportional of angles P, Q and R.
⇒ P/Q = R/S
Ratio
of angles Q and P is 5 : 4, and angle S is 125°.
So,
P/Q = 4/5
⇒ 4/5 =
R/125
⇒ R = 4 × 25 = 100
∴ Measure of
angle R is 100°.
