 # IBPS RRB Quantitative Aptitude Quiz Day 2 5 Steps - 3 Clicks

# IBPS RRB Quantitative Aptitude Quiz Day 2

### 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 Quantitative Aptitude Quiz Day 2 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 Quantitative Aptitude Quiz Day 2 will assist the students to know the expected questions from Quantitative Aptitude.

### Quiz

Directions(1-5): Volume and Surface Area

1. In a shower, 5 cm of rain falls. The volume of water that falls on 1.5 hectares of ground is:

A. 75 cu. m
B. 750 cu. m
C. 7500 cu. m
D. 75000 cu. m

Explanation: 1 hectare = 10,000 $${m}^{2}$$ = 60
Depth = $$\frac{5}{100}$$m
= $$\frac{1}{20}$$m
∴ Volume = (Area x Depth) = 15000 x $$\frac{1}{20}$$$${m}^{3}$$
= 750 $${m}^{3}$$

2. A hall is 15 m long and 12 m broad. If the sum of the areas of the floor and the ceiling is equal to the sum of the areas of four walls, the volume of the hall is:

A. 720
B. 900
C. 1200
D. 1800

Explanation: 2(15 + 12) x h = 2(15 x 12)
⇒ h = $$\frac{180}{27}$$m
= $$\frac{20}{3}$$m

∴ Volume = [15 x 12 x $$\frac{20}{3}$$]$${m}^{3}$$
= 1200$${m}^{3}$$

3. 66 cubic centimetres of silver is drawn into a wire 1 mm in diameter. The length of the wire in metres will be:

A. 84
B. 90
C. 168
D. 336

Explanation: Let the length of the wire be h.

Radius = $$\frac{1}{2}$$mm = $$\frac{1}{20}$$cm
⇒$$\frac{22}{7}$$ x $$\frac{1}{20}$$ x $$\frac{1}{20}$$ x h = 66
⇒$$\frac{66 × 20 × 20 × 7}{22}$$
= 8400 cm = 84 m.

4. A boat having a length 3 m and breadth 2 m is floating on a lake. The boat sinks by 1 cm when a man gets on it. The mass of the man is:

A. 12 kg
B. 60 kg
C. 72 kg
D. 96 kg

Explanation: Volume of water displaced = (3 x 2 x 0.01)$${m}^{3}$$
= 0.06 $${m}^{3}$$
∴ Mass of man = Volume of water displaced x Density of water
= (0.06 x 1000) kg
= 60 kg.

5. 50 men took a dip in a water tank 40 m long and 20 m broad on a religious day. If the average displacement of water by a man is 4 $${m}^{3}$$, then the rise in the water level in the tank will be:

A. 20 cm
B. 25 cm
C. 35 cm
D. 50 cm

Explanation: Total volume of water displaced = (4 x 50) $${m}^{3}$$
= 200 $${m}^{3}$$
∴ Rise in water level = $$\frac{200}{40 × 20}$$m
= 0.25 m
= 25 cm

Directions(1-5): Fractions
1. The fraction 101 $$\frac{27}{100000}$$

A. .01027
B. .10127
C. 101.00027
D. 101.000027

Explanation: 101 $$\frac{27}{100000}$$
= 101 + $$\frac{27}{100000}$$
= 101 + .00027 = 101.00027

2. $$\frac{5 × 1.6 – 2 × 1.4 }{1.3}$$ = ?

A. 0.4
B. 1.2
C. 1.4
D. 4

Explanation: Given Expression = $$\frac{8 – 2.8}{1.3}$$
$$\frac{5.2}{1.3}$$
= $$\frac{52}{13}$$
= 4

3. How many digits will be there to the right of the decimal point in the product of 95.75 and .02554 ?

A. 5
B. 6
C. 7
D. None of these

Explanation: Sum of decimal places = 7.

Since the last digit to the extreme right will be zero (since 5 x 4 = 20), so there will be 6 significant digits to the right of the decimal point.

4. 0.04 x 0.0162 is equal to:

A. 6.48 x $${10}^{-3}$$
B. 6.48 x $${10}^{-4}$$
C. 6.48 x $${10}^{-5}$$
D. 6.48 x $${10}^{-6}$$

Explanation: 4 x 162 = 648. Sum of decimal places = 6.
So, 0.04 x 0.0162 = 0.000648 = 6.48 x $${10}^{-4}$$

5. The price of commodity X increases by 40 paise every year, while the price of commodity Y increases by 15 paise every year. If in 2001, the price of commodity X was Rs. 4.20 and that of Y was Rs. 6.30, in which year commodity X will cost 40 paise more than the commodity Y ?

A. 2010
B. 2011
C. 2012
D. 2013

Explanation: Suppose commodity X will cost 40 paise more than Y after z years.

Then, (4.20 + 0.40z) – (6.30 + 0.15z) = 0.40
⇒ 0.25z = 0.40 + 2.10
⇒z = $$\frac{2.50}{0.25}$$
= $$\frac{250}{20}$$
= 10
∴ x will cost 40 paise more than Y 10 years after 2001 i.e., 2011

Directions(1-5): Probability
1. A bag contains 6 black and 8 white balls. One ball is drawn at random. What is the probability that the ball drawn is white?

A. $$\frac{3}{4}$$
B. $$\frac{4}{7}$$
C. $$\frac{1}{8}$$
D. $$\frac{3}{7}$$

Explanation: Let number of balls = (6 + 8) = 14.
A number of white balls = 8.

P (drawing a white ball) = $$\frac{8}{14}$$
= $$\frac{4}{7}$$

2. One card is drawn at random from a pack of 52 cards. What is the probability that the card drawn is a face card (Jack, Queen and King only)?

A. $$\frac{1}{13}$$
B. $$\frac{3}{13}$$
C. $$\frac{1}{4}$$
D. $$\frac{9}{52}$$

Explanation: Clearly, there are 52 cards, out of which there are 12 face cards.
P (getting a face card) = $$\frac{12}{52}$$
∴ = $$\frac{3}{13}$$

Directions(1-5): Mensuration
3. A circular wire of radius 42 cm is cut and bent in the form of a rectangle whose sides are in the ratio of 6:5. The smaller side of the rectangle is :

A. 30 cm
B. 60 cm
C. 72 cm
D. 132 cm

Explanation: circumference = $$\frac{2*22}{7*42}$$ = 264 cm
length of rectangle sides are 6x, 5x. then circumference is 2*(6x+5x) = 264, x = 12

smaller side of rectangle = 5x = 60 cm

4. The length of a rectangle is twice its breadth. If its length is decreased by 5 cm and the breadth is increased by 5 cm, the area of the rectangle is increased by 75 cm2. Therefore, the length of the rectangle is :

A. 20 cm
B. 30 cm
C. 40 cm
D. 50 cm

Explanation: b=x and l= 2x
then (2x-5)(x+)-x*2x= 75
2$${x}^{2}$$ + 5x-25-2$${x}^{2}$$ = 75
x = 20
length = 40 cm

Directions(1-5): Averages
5. The average weight of A, B and C is 45 kg. If the average weight of A and B be 40 kg and that of B and C be 43 kg, then the weight of B is:

A. 17 kg
B. 20 kg
C. 26 kg
D. 31 kg

Explanation: Let A, B, C represent their respective weights. Then, we have:

A + B + C = (45 x 3) = 135 …. (i)

A + B = (40 x 2) = 80 …. (ii)

B + C = (43 x 2) = 86 ….(iii)

Adding (ii) and (iii), we get: A + 2B + C = 166 …. (iv)

Subtracting (i) from (iv), we get : B = 31.

B’s weight = 31 kg.

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