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Q #31
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Directions : These questions are based on the data given below.

I had some sweets (less than 80) with me. I distributed them among Sachin and Saurav. After distributing the sweets, it was found that both of them had got a different number of sweets, but both the numbers had the same unique properties. Both the numbers could be expressed as the sum of the squares of two different numbers and also as the difference of the cubes of two different numbers. If Saurav got less sweets than Sachin, then how many sweets did Sachin get?

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There are only three numbers less than 80, which satisfy both the conditions. They are : 26 = 52 + 12 = 33 – 13, 37 = 62 + 12 = 43 – 33 and 61 = 52 + 62 = 53 - 43 But, 61 + 26 > 80 ...Saurav got 26 sweets andSachin got 37 sweets

Q #32
:

Directions : These questions are based on the data given below.

I had some sweets (less than 80) with me. I distributed them among Sachin and Saurav. After distributing the sweets, it was found that both of them had got a different number of sweets, but both the numbers had the same unique properties. Both the numbers could be expressed as the sum of the squares of two different numbers and also as the difference of the cubes of two different numbers. How many sweets did I distribute to the two of them?

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26 + 37 = 63.

Q #33
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Directions : These questions are based on the data given below.

N! (N >= 2004) is divided by 10th to leave a remainder of r(r >= 0). If m is the maximum possible are r is the minimum possible, then the last digit of the quotient is

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Here, N = 2004! is divided by the largest power of 10 that perfectly divides it. Hence all the zeroes from the end are removed and the last digit then is what is required. Since the highest power of 2 that divides 2004! is definitely greater than the highest power of 5 that divides 2004! ( 10 = 2 x 5), even after removing all zeroes, the quotient will still be even. Therefore The last digit of the quotient is always even irrespective of the parity (even or odd) of m.

Q #34
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Directions : These questions are based on the data given below.

Imagine you have 26 constants labeled A to Z. Each constant is assigned a value in the following manner: A = 1 B = 2C (2 raised to the power of the position value of C in the alphabet, i.e., 23) C = 2D (2 raised to the power of the position value of D in the alphabet, i.e. 24) and so on till Z = 2A (2 raised to the power of the position value of A in the alphabet, i.e. 21) What is the exact numerical value of the following expression? (T – A)* (T – B)* …..(T – Y)* (T – Z)

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None of the above because (T – T) = 0

Q #35
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Directions : These questions are based on the data given below.

In a number system to the base 20, letters A, B, C, ….. to K of the English alphabet are sequentially used to digitally represent the values 10, 11, 12, ….. to 20 (to the base 10). Calculate the value of [CAKE(20) – BAKE(20)] to the base 10.

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