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
(b) 4
(c) 5
(d) 7
The correct answers to the above question in:
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.
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Read more question and answers set 1 Based Verbal Reasoning Questions and Answers
Question : 1
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 : 2
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 : 3
A girl counted in the following way on the fingers of her left hand : She started by calling the thumb 1, the index finger 2, middle finger 3, ring finger 4, little finger 5 and then reversed direction calling the ring finger 6, middle finger 7 and so on. She counted upto 1994. She ended counting on which finger?
a) Middle finger
b) Thumb
c) Index finger
d) Ring finger
Answer »Answer: (c)
Clearly, while counting, the numbers associated to the thumb will be : 1, 9,17, 25,.....
i.e. numbers of the form (8n + 1).
Since 1994 = 249 × 8 + 2, so 1993 shall correspond to the thumb and 1994 to the index finger.
Question : 4
A monkey climbs 30 feet at the beginning of each hour and rests for a while when he slips back 20 feet before he again starts climbing in the beginning of the next hour. If he begins his ascent at 8.00 a.m., at what will he first touch a flag at 120 feet from the ground?
a) 6 p.m.
b) 4 p.m
c) 5 p.m.
d) 8 p.m.
Answer »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.
Question : 5
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 : 6
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.
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