MBA Quantitative Aptitude Questions & Answers – Number System Set I

1. A Shopkeeper offers total 150 dairies to its customers. As per the scheme one dairy will be offered on the purchase of a “Financial Management” book. Out of 150 dairies the cost of some dairies is Rs. 30 and the cost of the rest dairies is Rs. 50. Maximum how many customers can avail a dairy of Rs. 50 as an offer from the shopkeeper if the total cost of the dairies cannot exceed Rs. 7450?

a) 45

b) 120

c) Can’t be determined

d) None of these

a) 12321

b) 1

c) 2197

d) 11

4. The value of

(1.2.3…….9).(11.12.13…….19).(21.22.23……29).(31.32.33……..39).

(41.42.43……49) (51.52.53……..59)… (81.82……89)

5. The last 4 digits of the following expression will be

(1!)⁴ + (2!)³ + (3!)² + (4!)¹ + (10!)⁴ + (100!)³ + (1000!)² + (10000!)¹

a) 5939

b) 0069

c) 0969

d) Can’t be determined

a) -1

b) 1

c) 0

d) Can’t be determined

a) 130

b) 2200

c) 1100

d) Data insufficient

9. Which of the following is divisible by 2⁷+1?

I. 2¹⁴ + 1

II. 2²⁸ - 1

III. 2⁴⁹ + 1

a) (I) and (II)

b) (II) and (III)

c) (I) and (III)

d) All of these

10. Each family in Model town has almost two adults and the total number of boys in locality is less than the number of girls. Similarly the number of girls is less than the number of adults in the locality. Sampat Singh, the Minister of the locality is the only adult, then minimum number of families in his locality is:

a) 2

b) 3

c) 4

d) None of these

Answers

Explanation

Explanation (1) :

Option (a) and (b) are wrong since the maximum cost of total dairies is nearly Rs. 7450. Now, to maximize the number of pens of Rs. 50, we have to minimize the number of pens of Rs. 30 and the total cost cannot exceed Rs. 7450. So by hit and trial get required result. As:

Hence the maximum number of dairies of Rs. 50 is 147

Explanation (2) :

Explanation (3) :

Explanation (4) :

Explanation (5) :

Explanation (6) :

Explanation (7) :

Explanation (8) :

Explanation (9) :

Explanation (10) :

Let F be the family, B, G and A are boys, girls and adults respectively.

Since F < B < G < A

Now, if F = 1, then 1 < 2 < 3 < 4, adults cannot be 4 in one family.

If F = 2, then 2 < 3 < 4 < 5, adults cannot be 5 in 2 families.

If F = 3, then 3 < 4 < 5 < 6, not possible since one family has only two adults.

Again, if F = 4 than 4 < 5 < 6 < 7, possible since

(3 X 2) + (1 X 1) = 7

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