MBA Exam

Given “a” is a positive integer find the smallest positive integer “b” such that 23a divides b3113 +1.

1) 8<sup>a</sup> – 3
2) 4<sup>a</sup> – 5
3) 2<sup>a</sup> – 3
4) 2<sup>3a</sup> –2
5) 2<sup>3a</sup> – 1

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MBA Other Question

In the figure given below ABCD is a square. It is also given that AL : LB = BM : MA = DR : RC = CS : SD = 1 : 2. What is the ratio of the area of the Δ EFG to the area of the Δ ERC?

1) 1:3
2) 1:4
3) 1:16
4) 1:25
Suppose for any real number x, [x] denotes the greatest integer less than or
equal to x. Let L(x, y) = [x] + [y] + [x + y] and R(x, y) = [2x] + [2y]. Then it
is impossible to find any two positive real numbers x and y for whichPlease explain the solution

1) L(x, y) = R(x, y)
2) L(x, y) not equal R(x, y)
3) L(x, y) < R(x, y)
4) L(x, y) > R(x, y)