Q 31.

If n is a whole number greater than 1 then, n^{2}(n^{2} - 1) is always divisible by.

Options:

60

24

48

12

Put n = 2,

n^{2}(n^{2} - 1) = 12 which is divisible by 12.

Put n = 3,

n^{2}(n^{2} - 1) = 72 which is divisible by 12.

Put n = 4,

n^{2}(n^{2} - 1) = 240 which is divisible by 12.

So, we can say that n^{2}(n^{2} - 1) is always divisible by 12.

Q 32.

Find the remainder when 9^{19} + 6 is divided by 8.

Options:

4

5

6

7

Given,

N = 9^{19} + 6

= (1+8)^{19} + 6

N/8 = [(1+8)^{19} + 6]/8

= (1 + 6)/8

= 7/8

So, remainder = 7

Q 33.

If N is divided by 56 it gives 29 as remainder. Find the remainder when N is divided by 8.

Options:

3

4

5

6

As per question,

N = 56q + 29

= (8x7q) + (8x3) + 5

= 8(7q + 3) + 5

So, remainder = 5

Q 34.

Find the sum of first 6 multiples of 5.

Options:

105

100

95

90

Given, n = 6 and x = 5

Sum of first n multiples of x = x[n(n+1)/2]

= 5[6(6+1)/2]

= 105

Q 35.

Find the sum of first 5 odd numbers.

Options:

21

23

25

27

Given, n = 5

Sum of first n odd numbers = n^{2}

= 5^{2}

= 25

Q 36.

Find the sum of first 5 even numbers.

Options:

28

30

32

34

Given, n = 5

Sum of first n even numbers = n(n+1)

= 5(5+1)

= 30

Q 37.

Find the sum of the cubes of the first 5 natural numbers.

Options:

223

224

225

226

Given, n = 5

Sum of the cubes of first n natural numbers = [n(n+1)/2]^{2}

= [5(5+1)/2]^{2}

= 225

Q 38.

Find the sum of the squares of first 10 natural numbers.

Options:

375

385

395

365

Given, n = 10

Sum of the sqaures of first n natural numbers = n(n+1)(2n+1)/6

= 10(10+1)(20+1)/6

= 385

Q 39.

Find the remainder when 4^{1000} is divided by 7.

Options:

1

2

3

4

4^{1000}/7

= ((4^{2})^{500})/7

= ((16)^{500})/7

= ((14 + 2)^{500})/7

= ((2)^{500})/7

= (2^{2} x (2^{3})^{166})/7

= (4 x (8)^{166})/7

= (4 x (7 + 1)^{166})/7

= (4 x (1)^{166})/7

= 4

Q 40.

If the sum and difference of the digits of a two digit number is 14 and 2 respectively. Find the product of the digits.

Options:

46

48

50

Insufficient data

Let the two digits be x and y.

Given, x + y = 14 and x - y = 2

Therefore, x = 8 and y = 6

So, product = xy = 8x6 = 48

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