Arithmetical Reasoning section 1 MCQ Detailed Explanation And More Example
MOST IMPORTANT verbal reasoning - 2 EXERCISES
The following question based on Arithmetical Reasoning topic of verbal reasoning
(a) 6 p.m.
(b) 4 p.m
(c) 5 p.m.
(d) 8 p.m.
The correct answers to the above question in:
Answer: (a)
Let ascent of the monkey in 1 hour = ( 30 – 20 ) = 10 feet.
So, the monkey ascends 90 feet in 9 hours i.e., 5 p.m.
Clearly, in the next 1 hour i.e., till 6 p.m.
The monkey ascends remaining 30 feet to touch the flag.
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Read more question and answers set 1 Based Verbal Reasoning Questions and Answers
Question : 1
A player holds 13 cards of four suits, of which seven are black and six are red. There are twice as many diamonds as spades and twice as many hearts as diamonds. How many clubs does he holds?
a) 6
b) 4
c) 5
d) 7
Answer »Answer: (a)
Clearly, the black cards are either clubs or spades while the red cards are either diamonds or hearts.
Let the number of spades be x. Then, number of cubes = (7 – x).
Number of diamonds = 2 x number of spades = 2x
Number of hearts = 2 x number of diamonds = 4x
Total number of cards = x + 2x + 4x + 7 – x = 6x + 7.
Therefore 6x + 7 = 13 ⇔ 6x = 6 ⇔ x – 1.
Hence, number of clubs = (7 – x) = 6.
Question : 2
Mac has £ 3 more than Ken, but then Ken wins on the horses and trebles his money, so that he now has £ 2 more than the original amount of money that the two boys had between them. How much money did Mac and Ken have between them before Ken's win ?
a) £ 13
b) £ 9
c) £ 11
d) £ 15
Answer »Answer: (a)
Let money with Ken = x. Then, money with Mac = x + £ 3.
Now, 3x = (x x + £ 3) + £ 2 ⇔ x = £ 5.
Therefore total money with Mac and Ken = 2x + £ 3 = £ 13.
Question : 3
Mr. X, a mathematician, defines a number as 'connected with 6 if it is divisible by 6 or if the sum of its digits is 6, or if 6 is one of the digits of the number. Other numbers are allenot connected with 6'. As per this definition, the number of integers from 1 to 60 (both inclusive) which are not connected with 6 is
a) 42
b) 18
c) 22
d) 43
Answer »Answer: (d)
Numbers from 1 to 60, which are divisible by 6 are : 6, 12, 18, 24, 30, 36, 42, 48, 54, 60.
There are 10 such numbers.
Numbers from 1 to 60, the sum of whose digits is 6 are 6, 15, 24, 33, 42, 51, 60.
There are 7 such numbers of which 4 are common to the above ones. So, there are 3 such uncommon numbers.
Numbers from 1 to 60, which have 6 as one of the digits are 6, 16, 26, 36, 46, 56, 60.
Clearly, there are 4 such uncommon numbers.
So, numbers 'not connected with 6'
= 60 – (10 + 3 + 4) = 43.
Question : 4
A group of 1200 persons consisting of captains and soldiers is travelling in a train. For every 15 soldiers there is one captain. The number of captains in the group is ?
a) 80
b) 70
c) 75
d) 85
Answer »Answer: (c)
Generally we may commit mistake of dividing 1200/15. But out of 16 persons there is one captain. so, it will be 1200/16 = 75
Question : 5
Four persons, Alok, Bhupesh, Chander and Dinesh have a total of ` 100 among themselves. Alok and Bhupesh between them have as much money as Chander and Dinesh between them, but Alok has more money than Bhupesh; and Chander has only half the money that Dinesh has. Alok has in fact ` 5 more than Dinesh has. Who has the maximum amount of money ?
a) Chander
b) Alok
c) Bhupesh
d) Dinesh
Answer »Answer: (b)
a + b + c + d = 100
Also, a + b = c + d = 50
c = $d/2, ∴ d/2 + d$ = 50 ⇒ d = 33.3
∴ d > c
Also, a > b & between a and d, a = d + 5
∴ a > d
∴ Alok (b) has the maximum money.
Question : 6
Four children A, B, C and D are having some chocolates each. A gives B as many as he already has, he gives C twice of what C already has and he gives D thrice of what D already has. Now, D gives 1/8th of his own chocolates to B. Then A gives 10% chocolates he now owns to C and 20% to B. Finally, all of them have 35 chocolates each. What is the original number of chocolates each had in the beginning ?
a) A-70, B-25, C-25, D-20
b) A -110, B-10, C-10, D-10
c) A-90, B-20, C-20, D-10
d) A-125, B-5, C-5, D-5
Answer »Answer: (b)
Work with options we find that option (b) is correct.
| Option (b) | A | B | C | D |
| Beginning | 110 | 10 | 10 | 10 |
| A gave | 110 - 60 = 50 | 10 + 10 = 20 | 10 + 20 = 30 | 10 + 30 = 40 |
| D gave | 20 + 5 = 25 | 40 - 5 = 35 | ||
| A gave | 50 - 15 = 35 | 25 + 10 = 35 | 30 + 5 = 35 |
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