Arithmetic Reasoning

Question: A, B, C, D and E play a game of cards. A says to B, “If you give me three cards, you will have as many as E has and if I give you three cards, you will have as many as D has.” A and B together have 10 cards more than what D and E together have. If B has two cards more than what C has and the total number of cards be 133, how many cards does B have ?
[A].

22

[B].

23

[C].

25

[D].

35

Answer: Option C

Explanation:

Clearly, we have :

B-3 = E …(i)

B + 3 = D …(ii)

A+B = D + E+10 …(iii)

B = C + 2 …(iv)

A+B + C + D + E= 133 …(v)

From (i) and (ii), we have : 2 B = D + E …(vi)

From (iii) and (vi), we have : A = B + 10 …(vii)

Using (iv), (vi) and (vii) in (v), we get:

(B + 10) + B + (B – 2) + 2B = 133 5B = 125 B = 25.

Question: If a 1 mm thick paper is folded so that the area is halved at every fold, then what would be the thickness of the pile after 50 folds ?
[A].

100 km

[B].

1000 km

[C].

1 million km

[D].

1 billion km

Answer: Option D

Explanation:

Question: 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].

Thumb

[B].

Index finger

[C].

Middle finger

[D].

Ring finger

Answer: Option B

Explanation:

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 x 8 + 2, so 1993 shall correspond to the thumb and 1994 to the index finger.

Question: Five bells begin to toll together and toll respectively at intervals of 6, 5, 7, 10 and 12 seconds. How many times will they toll together in one hour excluding the one at the start ?
[A].

7 times

[B].

8 times

[C].

9 times

[D].

11 times

Answer: Option B

Explanation:

L.C.M. of 6, 5, 7, 10 and 12 is 420.

So, the bells will toll together after every 420 seconds i.e. 7 minutes.

Now, 7 x 8 = 56 and 7 x 9 = 63.

Thus, in 1-hour (or 60 minutes), the bells will toll together 8 times, excluding the one at the start.

Question: 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].

£ 9

[B].

£ 11

[C].

£ 13

[D].

£ 15

Answer: Option C

Explanation:

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: Mr. Johnson was to earn £ 300 and a free holiday for seven weeks’ work. He worked for only 4 weeks and earned £ 30 and a free holiday. What was the value of the holiday?
[A].

£ 300

[B].

£ 330

[C].

£ 360

[D].

£ 420

Answer: Option B

Explanation:

Question: 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].

25

[B].

37

[C].

54

[D].

65

Answer: Option A

Explanation:

Clearly, the required number would 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.

Question: Between two book-ends in your study are displayed your five favourite puzzle books. If you decide to arrange the five books in every possible combination and moved just one book every minute, how long would it take you ?
[A].

1 hour

[B].

2 hours

[C].

3 hours

[D].

4 hours

Answer: Option B

Explanation:

Clearly, number of ways of arranging 5 books = 5 ! = 5 x 4 x 3 x 2 x 1 = 120.

So, total time taken = 120 minutes = 2 hours.

Question: 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 all ‘not 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].

18

[B].

22

[C].

42

[D].

43

Answer: Option D

Explanation:

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 3such 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: 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 hold ?
[A].

4

[B].

5

[C].

6

[D].

7

Answer: Option C

Explanation:

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 clubs = (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.