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Q #6
:

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.

6.How many sweets did I distribute to the two of them?

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26 + 37 = 63. Choice (3)

Q #7
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Directions : Select the correct alternative from the given choices.

When a number is either doubled or tripled the number of factors of the number increase by 60. If the number is divisible by 9 but not by 8, how many factors does the number have?

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Let the number be N.

    Given that 2N has 60 more factors than N

    ⇒ The number of factors of N, which are not the multiples of 2 is 60 also given that 3N has 60 more factors than N.

    ⇒ The number of factors of N, which are not the multiples of 3 are 60.

    ⇒ Both the powers of 2 and 3 in N are the same.

    If N is a multiple of 9 (i.e 32), then it is also a multiple of 2^2 (i.e. 4).

    If N is not a multiple of 8 (i.e. 23 ), than it is not a multiple of 3^3 (i.e. 27),

    ∴ N is of the form, N = 22 x P or 32 x Q

    ∴ The number of factors of N is (2+1) x 60 =180   


Q #8
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Directions : Select the correct alternative from the given choices.

The registration number of Mr. Dutt’s bike is a unique one. It gets increased by 7155 when seen upside down. What is the registration number of Mr. Dutt’s bike?

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Solution. Only 0, 1, 6, 8 and 9 make sense, when seen upside down. From these numbers we find that Arvind's bike 1906 => 9061 – 1906 = 7155. Choice (2)

Q #9
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Directions : Select the correct alternative from the given choices.

A number when successively divided by 4, 5, 6 and 7 leaves remainders of 1, 4, 5 and 6 respectively. Which of the following can be the remainder when the above number is divided by 93?

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Q #10
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Directions : These questions are based on the data given below.

A set of natural numbers, P, is formed using some of the first 1000 natural numbers. The set contains the maximum number of elements such that each element satisfies the following two conditions. I. No numbers of the set P is prime. II. Any two numbers of the set P are prime to each other.

What could be the greatest possible natural number in the set P?

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Consider the square root of 1000. The largest prime less than this is 31. So the set of all prime numbers less than or equal to 31 i.e., { 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31} must be represented in set A.

 

The above set itself cannot be set A, as the members of set A are non-prime.

 

Therefore, each member of set A must represent exactly one of these prime numbers as, set A is such that maximum number of elements. If, any member of the set is a combination of 2 or more primes from the above set, then the number of elements in set A will get reduced as the members of this set should be co-prime to each other.

Set A could be  {22, 32, 52, 112, 132, 172, 192, 232, 292, 312,}

 

It can be noted that there are more possibilities for set A. A number of set A could t, a product of exactly one prime number from the original set and another prime number which is not from the original set.

 

If  none of the prime numbers is used on more than one occasion then the set of numbers will be co-primes and hence satisfy the coordination for set A.

 

Consider 999

    999 – 99 x 11 = 27 x 37 = 33 x 37

    The prime number 3 from the original set is represented only once even though it is multiplied three times.

    ∴ 999 can be a member of the set and it is obviously the larges