Arithmetical Reasoning section 2 MCQ Practice Questions Answers Test with Solutions & More Shortcuts

Arithmetical Reasoning PRACTICE TEST [2 - EXERCISES]

Arithmetical Reasoning section 1 MCQ

Arithmetical Reasoning section 2 MCQ

Question : 6

Six identical cards are placed on a table. Each card has number '1' marked on one side and number '2' marked on its other side. All the six cards are placed in such a manner that the number '1' is on the upper side. In one try, exactly four (neither more nor less) cards are turned upside down. In how many least number of tries can the cards be turned upside down such that all the six cards show number '2' on the upper side ?

a) 7

b) 3

c) 5

d) This cannot be achieved

Answer: (b)

arithmetical reasoning verbal reasoning d6 35a

Question : 7

David gets on the elevator at the 11th floor of a building and rides up at the rate of 57 floors per minute. At the same time. Albert gets on an elevator at the 51st floor of the same building and rides down at the rate of 63 floors per minute. If they continue travelling at these rates, then at which floor will their paths cross?

a) 30

b) 19

c) 28

d) 37

Answer: (a)

Suppose their paths cross after x minutes.

Then, 11 + 57x = 51 – 63 x ⇔ 120 x = 10 ⇔ x = $1/3$

Number of floors covered by David in $1/3$ min

= $(1/3 × 57)$ = 19.

So, their paths cross at (11 + 19)th i.e., 30th floor.

Question : 8

I have a few sweets to be distributed. If I keep 2, 3 or 4 in a pack, I am left with one sweet. If I keep 5 in a pack, I am left with none. What is the minimum number of sweets I have to pack and distribute ?

a) 54

b) 25

c) 37

d) 65

Answer: (b)

The required number will be such that it leaves a remainder of 1 when divided by 2, 3 or 4 and no remainder when divided by 5. Such a number is 25 among options

Question : 9

Three ducks can be arranged as shown above to satisfy all the three given conditions. A certain number of horses and an equal number of men are going somewhere. Half of the owners are on their horses' back while the remaining ones are walking along leading their horses. If the number of legs walking on the ground is 70, how many horses are there ?

a) 14

b) 10

c) 12

d) 16

Answer: (a)

Let number of horses = number of men = x.

Then, number of legs = 4x + 2 x (x/2) = 5x.

So, 5x = 70 or x = 14.

Question : 10

First bunch of bananas has (1/4) again as many bananas as a second bunch. If the second bunch has 3 bananas less than the first bunch, then the number of bananas in the first bunch is

a) 12

b) 9

c) 10

d) 15

Answer: (d)

Let the number of bananas in the second bunch be x

Then, number of bananas in the first bunch

= x + $1/4 x = 5/4 x$

So, $5/4$ x - x = 3 ⇔ 5x - 4x = 12 ⇔ x = 12

∴ Number of bananas in the first bunch

= $(5/4 × 12)$ = 15

IMPORTANT verbal reasoning EXERCISES

Arithmetical Reasoning section 2 MCQ Online Quiz

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