MBA Questions of the day (29 Sep’13)

Choose the appropriate alternative :

Q67 – 69 :

67. A, B, C, D are to be seated in a row. But C and D cannot be together. Also B cannot be at the third place.

Which of the following must be false?

(a) A is at the first place.

(b) A is at the second place.

(c) A is at the third place.

(d) A is at the fourth place.

68. If A is not at the third place, then C has the following option only.

(a) the first place only.

(b) the third place only.

(c) the first and third place only.

(d) any of the places.

69. If A and B are together then which of the following must be necessarily false.

(a) C is not at the first place.

(b) A is at the third place.

(c) D is at the first place.

(d) C is at the first place.

70. One year payment to the servant is Rs. 90 plus one turban. The servant leaves after 9 months and receives Rs.65 and a turban. Then find the price of the turban.

(a) Rs.10

(b) Rs.15

(c) Rs.7.5

(d) Cannot be determined.

72. Three wheels can complete respectively 60, 36, 24 revolutions per minute. There is a red spot on each wheel that touches the ground at time zero. After how much time, all these spots will simultaneously touch the ground again?

(a) 5/2 seconds.

(b) 5/3 seconds

(c) 5 seconds.

(d) 7.5 seconds.

73. A certain number when divided by 899 leaves the remainder 63. Find the remainder when the same number is divided by 29.

(a) 5

(b) 4

(c) 1

(d) Cannot be determined.

74. A is the set of positive integers such that when divided by 2,3,4,5,6 leaves the remainders 1,2,3,4,5 respectively. How many integers between 0 and 100 belong to set A?

(a) 0

(b) 1

(c) 2

(d) None of these.

Answers:

67. Since C and D cannot be together, they can occupy either of the following seats : (1^st & 3^rd), (1^st & 4^th) or (2^nd & 4^th). In the last two cases, since B cannot be in the 3^rd place, A will have to be there. Thus we can see that A can never be in the 1^st place. Hence statement (a) is false.

68. Since both neither A nor B can be at 3^rd place, this place has to be occupied by either D or C. And if either of them occupies this place the other one has to occupy the 1^st place (since D & C cannot be together). Hence C can only occupy either 1^st or 3^rd place. Hence the correct option is (c).

69. If A & B are together, but C & D are not then the only places that A & B can occupy are 2^nd and 3^rd. And since B cannot be at 3^rd place, A has to be at 3^rd place. Hence the correct option is (b).

70. Let the cost of the turban be T. Hence total payment for one year = Rs.90 + T. So the payment for 9 months should be ¾ (90 + T). But this is equal to (65 + T). Equating the two, we get T = Rs.10.

71. Let R be he radius of each circle. The R² / 2R= 2R/R² which implies that R/2 = 2/R i.e. R² = 4, i.e. R = 2. Then the length of the square is 8. Thus the area of the square is 64, while the area covered by each coin is 2² = 4. Since there are four coins, the area covered by coins is 4(4) = 16. Thus the area not covered by the coins is 64 – 16= 16(4 – ).

72. The time taken by the white spots on all three wheels to simultaneously touch the ground again will be equal to the LCM of the times taken by the three wheels to complete 1 revolution. The first wheel complete 60 revolutions per minute. This means that to complete 1 revolution it takes (60/60) = 1 sec. The second wheel completes 36 revolutions per minute. So to complete 1 revolution it takes (36/60) = 3/5 sec. Similarly the third wheel takes (24/60) = 2/5 sec. to complete one revolution. Hence LCM of 1, 3/5, 2/5 = LCM(1,3,5)/HCF(1,5,5) = 5/1 = 5 seconds.

73. The best way to solve this question is the method of simulation. i.e. take a number which when divided by 899 gives a remainder of 63. The smallest such number is (899 + 63) = 972. 972, when divided by 29 gives a remainder of 5. Hence the answer is 5.

HINT : KITS students please note that 899 itself is divisible by 29. Hence required remainder is the same as obtained by dividing 63 by 29 i.e. 5.

74. Note that the difference between the divisors and the remainders is constant. 2-1 = 3-2 = 4-3 = 5-4 = 6-5 = 1. In such a case the required number will always be [a multiple of LCM of (2,3,4,5,6) – (the constant difference)]. LCM of (2,3,4,5,6) = 60. Hence the required number will be 60n – 1. Thus we can see that the smallest such number is (60 x 1) – 1 = 59. The second smallest is (60 x 2) – 1 = 119. So between 1 and 100 there is only one such number viz.59.