Choose the appropriate alternative :

59.  then which of the following must be true?

(a) X=2, Y=1

(b) X>0, Y<0

(c) X,Y both positive

(d) X,Y both negative

60. (BE)2 = MPB, where B, E, M & P are distinct integers, then M = ?

(a) 2

(b) 3

(c) 9

(d) None of these.

61. Five digit numbers are formed using only 0,1,2,3, 4 exactly once. What is the difference between the maximum and minimum number that can be formed?

(a) 19800

(b) 41976

(c) 32976

(d) None of these.

62. How many numbers can be formed from 1,2, 3, 4, 5 (without repetition), when the digit at the units place must be greater than that in the tens place?

(a) 54

(b) 60

(c) 51/3

(d)

63. Distance between A and B is 72 km. Two men started walking from A and B at the same time towards each other. The person who started from A traveled uniformly with average speed 4 kmph. While the other man traveled with varying speed as follows: In first hour his speed was 2 kmph, in the second hour it was 2.5 kmph, in the third hour it was 3 kmph, and so on. When will they meet each other?

(a) 7 hours

(b) 10 hours

(c) 35 km from A

(d) midway between A & B

64. P,Q,R,S are four statements. Relation between these statements is as follows:

I.    If P is true then Q must be true.

II.   If Q is true then R must be true.

III.  If S is true then either Q is false or R is false.

Then which of the following must be true.

(a) If P is true then S is false.

(b) If S is false then Q must be true.

(c) If Q is true then P must be true.

(d) If R is true then Q must be true.

 59 (b) 60 (b)
 61 (c) 62 (b) 63 (d) 64 (a)

59. The best possible way to solve this to check each of the given answer choices. In options (a), (c) and (d), either both X & Y are positive or both X & Y are negative. Since we have (-Y) in the numerator of our expression and (-X) in the denominator, X & Y will never be both positive and neither will XY be positive. Hence both the numerator and the denominator of our expression will be 1 and the value will always be 1. Hence the only possible answer choice is (b).

60. Since MPB is a three digit number, and also the square of a two digit number, it can have a maximum value of 961 viz.312. This means that the number BE should be less than or equal to 31. So B can only take the values 0, 1, 2 and 3. Since last digit of MPB is also B, it can only be 0 or 1 (as none of the squares end in 2 or 3). The only squares that end in 0 are 100, 400 and 900. But for this to occur the last digit of BE also has to be 0. Since E & B are distinct integers, both of them cannot be 0. Hence B has to be 1. BE can be a number between 11 and 19 (as we have also ruled out 10), with its square also ending in 1. Hence the number BE can only be 11 or 19. 112 = 121. This is not possible as this will mean that M is also equal to 1. Hence our actual numbers are 192 = 361. Hence M = 3.

61. The maximum and the minimum five digit number that can be formed using only 0, 1, 2, 3, 4 exactly once are 43210 and 10234 respectively. The difference between them is 43210 – 10234 = 32976.

62. The digit in the units place must be greater than that in the tens place. So if we have 5 in the units place, the remaining four digits need not be any particular order. So we will have 4! numbers. However if we have 4 in the units place, we cannot have 5 in the tens place. Hence the tens place has to be one among 1,2 or 3. This can be done in 3 ways. The remaining 3 digits can be filled in the remaining three places in 3! ways. Hence total we will have (3 x 3!) numbers ending in 4. Similarly if we have 3 in the units place, the tens place can only be 1 or 2. This can be done in 2 ways. The remaining 3 digits can be arranged in the remaining 3 places in 3! Ways. Hence we will have (2 x 3!) numbers ending in 3. Similarly we can find that there will be (3!) numbers ending in 2 and no number ending in 1. So total number of numbers satisfying the given condition = 4! + (3 x 3!) + (2 x 3!) + 3! = 4! + 6 x 3! = 24 + (6 x 6) = 60.

63. Since A & B are moving in opposite directions, we will add their speeds to calculate the effective speeds. In other words in the first hour they would effectively cover a distance of (4+2) = 6 kms. towards each other. In the second hours they would effectively cover a distance of (4+2.5) = 6.5 kms. towards each other. And in the third hour (4+3) = 7 kms., in the fourth hour (4+3.5) = 7.5 kms. and so on. We can see that the distances that they cover in each hour is in AP viz. 6, 6.5, 7, 7.5 ….with a = 6 and d = 0.5. Since they have to effectivel cover a distance of 72 kms., the time taken to cover this much distance would be the time taken to meet each other. So, the sum of the first n terms of our AP has to be 72. If we are to express this in our equation of sum of first n terms of the AP we will get : Sn = n/2 x [2a + (n – 1)d]. Substituting our values we will get, 72 = n/2 x [12 + 0.5(n – 1)]. Solving this we get n = 9 hours. In 9 hours A would have covered (9 x 4) = 36 kms. So B would also have covered (72 – 36) = 36 kms. Hence they would meet midway between A & B.

64. If P is true then both Q and R have to be true. For S to be true, either Q or R must be false. Hence if P is true, S cannot be true. Hence the answer is (a).

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