  # IBPS RRB PO Quantitative Aptitude Quiz 9 5 Steps - 3 Clicks

# IBPS RRB PO Quantitative Aptitude Quiz 9

### Introduction

What is Quantitative Aptitude test?
Quantitative Aptitude is one of the prominent competitive aptitude subjects which evaluates numerical ability and problem solving skills of candidates. This test forms the major part of a number of important entrance and recruitment exams for different fields. The Quantitative Aptitude section primarily has questions related to the Simplification, Numbering Series, and Compound Interest, etc.

A candidate with quantitative aptitude knowledge will be in a better position to analyse and make sense of the given data. Quantitative Aptitude knowledge is an important measure for a prospective business executive’s abilities.

The article IBPS RRB PO Quantitative Aptitude Quiz 9 provides Quantitative Aptitude questions with answers useful to the candidates preparing for Competitive exams, Entrance exams, Interviews etc. IBPS RRB has released IBPS RRB Officer 2019 Official Notification for Officer Scale(I, II, and II). Quantitative Aptitude plays major role to qualify examination. The article IBPS RRB PO Quantitative Aptitude Quiz 9 will assist the students to know the expected questions from Quantitative Aptitude.

### Quiz

Q1. Average age of a group of 40 students is 37 yrs. If a teacher is included then the average age rises by 1,then find the age of the teacher after 2 years ?

A. 77
B. 80
C. 78
D. 81

Explanation:
X = 37*40 = 1480
X = 38*41 = 1558
Teacher age = 1558 – 1480 = 78
Aftr 2 yrs = 78 + 2 = 80

2. At present Priya is twice Sariga’s age. 8 yrs hence the respective ratio between Priya and Sariga’s age will be 22 :13, What is Sariga’s present age ?

A. 26
B. 36
C. 18
D. 20

Explanation:

P = 2S
$$\frac{2s+8}{s+8}$$ = $$\frac{22}{13}$$
26s + 104 = 22s + 176
4s = 72
S= 18yr

3. The ratio between the present ages of X and Y is 4:9 respectively. 2yrs from now, Y’s age will become twice the age of X.What was the difference between their ages five yrs ago ?

A. 10
B. 7
C. 12
D. 9

Explanation:
$$\frac{4x+2}{7x+2}$$ = $$\frac{1}{2}$$
8x + 4 = 9x + 2
x = 2
x:y = 8:18
5 yrs ago = 3:13
D = 13-3 = 10

4. The ratio of present age of Kaviya and Lakshmi is 6:7, the ratio of their ages after 8yrs would be 8:9, What is Kaviya’s present age?

A. 20
B. 24
C. 28
D. 30

Explanation:
$$\frac{6x+8}{7x+8}$$ = $$\frac{8}{9}$$
54x + 72 = 56x + 64
2x = 8
x = 4
K = 6*4 = 24yrs

5. The ratio between the present age of Maha and Deepa is 5:X. Maha is 9yrs younger than Parveen. Parveen’s age after 9yrs will be 33yrs. The difference between the Deepa and Maha age is same as the present age of Parveen. Find X

A. 13
B. 10
C. 11
D. 11

Explanation:
P present age = 33 – 9 = 24
M = 24 – 9 =15
D – M = 24
D – 15 = 24
D = 24 + 15 = 39
15:39 = 5:13
X = 13

Q1. The present age of Akil is 3 times that of Nirmal. 3 yrs hence the age of Akil will become 4yrs more than Nirmal. age The present age of Akil is

A. 1
B. 2
C. 4
D. 6

Explanation:
A = 3N
A + 3 = N + 4 + 3
3N + 3 = N + 7
2N = 4
N = $$\frac{4}{2}$$
N = 2yrs
A = 3*2 = 6yrs

2. The arithmetic mean of the scores of a group of students in a test was 52. The brightest 20% of them secured a mean score of 80 and the dullest 25% a mean score of 31. The mean score of remaining 55% is:

A. 51.4
B. 52.6
C. 56.1
D. 55.3

Explanation:

Let the required mean score be a
Then, 20 x 80 + 25 x 31 + 55 x a = 52 x 100
1600 + 775 + 55a = 5200
55a = 2825
a = 51.4

3. The average of a non-zero number and its square is 5 times the number. The number is:

A. 0, 7
B. 0, 6
C. 5, 7
D. 0, 9

Explanation:
Let the number be x.
Then,
$$\frac{(x + x^2) }{2}$$= 5x
x^2 – 9x = 0
x (x – 9) = 0
x = 0 or x = 9.

4. The average weight of 8 persons increases by 2.5 kg when a new person comes in place of one them weighing 65 kg. What might be the weight of the new person?

A. 65 kg
B. 70 kg
C. 85 kg
D. 92 kg

Explanation:
Total weight increased = (8 x 2.5) kg = 20 kg
Weight of new person = (65 + 20) kg = 85 kg

5. The average age of the boys in a class is 16 years and that of the girls is 15 years. The average age for the whole class is:

A. 15
B. 16
C. 17

Explanation:
Clearly, to find the average, we ought to know the number of boys, girls or students in the class, neither of which has been given
So, the data provided is inadequate

1. A rectangular box has dimensions 1.6 m × 1 m × 0.6 m. How many cubical boxes each of side 20 cm can be fit inside the rectangular box?

A. 220
B. 205
C. 120
D. 165

Explanation: a + b = 6
20 cm = 0.2 m
So number of boxes that can fit = 1.6 × 1 × $$\frac{0.6 }{0.2}$$ × 0.2 × 0.2

2. What is the volume of a cylinder whose curved surface area is 704 c$${m }^{2}$$ and height is 14 cm?

A. 2878 c$${m }^{3$$
B. 2914 c$${m }^{3}$$
C. 2396 c$${m }^{3}$$
D. 2816 c$${m }^{3}$$

Explanation:
2ᴨrh = 704, h = 14
Solve both, so r = 8
Volume = ᴨ$${r}^{2$$h

3. The total cost of painting the walls of a room is Rs 475. Find the cost of painting the walls of another room whose length, breadth and height each are double than the dimensions of the previous room.

A. Rs 1780
B. Rs 1900
C. Rs 1846
D. Rs 1960

Area of first room = 2(l+b)*h
After all dimensions doubled, new area = 2(2l + 2b)*2h = 4[2(l+b)*h ] = 4 times previous area, so cost of painting = 4*475

4. A circular wire of diameter 84 cm is bent into a rectangle with sides ratio 6 : 5. What are the respective sides of the rectangle?

A. 60 cm, 72 cm
B. 78 cm, 65 cm
C. 72 cm, 60 cm
D. 72 cm, 60 cm

Explanation:
Length of wire = 2ᴨr = 2($$\frac{22}{7}$$)*42 = 264 cm
Perimeter of rectangle = 2(6x + 5x) = 264
Solve, x = 12
So dimensions – 12*6, 12*5

5. The ratio of the outer and the inner perimeters of a circular path is 9 : 8. The path is 3 metres wide. What is the diameter of the outer circle?

A. 20 m
B. 27 m
C. 35 m
D. 39 m

$$\frac{2ᴨr}{2ᴨR }$$ = $$\frac{9}{8}$$
So $$\frac{r}{R}$$ = $$\frac{9}{8}$$, so r = ($$\frac{9}{8}$$)R
r – R = 3, so ($$\frac{9}{8}$$)R – R = 3
R = 24, so r = ($$\frac{9}{8}$$)*24 = 27