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**.

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**Answer** : Option B

**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:**

**Answer** : Option C

**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:**

**Answer** : Option A

**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:**

**Answer** : Option B

**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:**

**Answer** : Option B

**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

**Answer** : Option C

**Explanation**: 101 \(\frac{27}{100000}\)

= 101 + \(\frac{27}{100000}\)

= 101 + .00027 = 101.00027

**2. \(\frac{5 × 1.6 – 2 × 1.4 }{1.3}\) = ?**

**Answer** : Option D

**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 ?**

**Answer** : Option B

**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:**

**Answer** : Option B

**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 ?**

**Answer** : Option B

**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

**Answer** : Option B

**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)?**

**Answer** : Option B

**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 :**

**Answer** : Option B

**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 :**

**Answer** : Option C

**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:**

**Answer** : Option D

**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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