number system Model Questions & Answers, Practice Test for ibps clerk prelims

Answer: (c)

Given, p = 2 × 3 × 5 ... 59 × 61 = ... 0

Also, 2 ≤ n ≤ 59

Now, we check the sequence p + n

Since, unit digit of p is zero.

Therefore, for every even value of n, (p + n) is always divisible.

For odd value of n = 2, 5, ... 59

Take n = 3

∴ p + n = p + 3 = (2 . 3 . 5 ... 59 . 61 + 3)

= 3(2 . 5 ... 59 . 61 + 1) which is divisible.

Similarly, for even value of n, p + n is divisible.

Hence, it is clear always p + n is that divisible by any number.

So, there is no prime number exist in this sequence.

Answer: (d)

Amount received by each person

= $172850/25$ = Rs.6914

Answer: (b)

No.s divided by 7 leaing 3 as reminder are -

7 + 3 = 10

14 + 3 = 17

21 + 3 = 24

28 + 3 = 31

35 + 3 = 38

42 + 3 = 45

49 + 3 = 52

56 + 3 = 59

63 + 3 = 66

70 + 3 = 73

77 + 3 = 80

84 + 3 = 87

91 + 3 = 94

Total 13 terms.

Sum = 3 + 13 + 7 [1 + 2 + 3 + -----+13]

= 39 + $(7 × {13 × 14}/2)$ = 39 + 637 = 676

So, option (c) is correct.

Answer: (d)

Given equation, $1/m + 4/n = 1/12$

⇒ 12(n + 4m) = mn

⇒ 12n + 48m = mn

⇒ m(48 – n) = –12n

⇒ m(n – 48) = 12n

∴ m = ${12n}/{n - 48}$...(i)

Here, as m and n are positive integers, therefore > 48.

But n is an odd integer less than 60, therefore possible values of n = 49, 51, 53, 55, 57 and 59.

But on putting n = 53, 55, and 59 in Eq. (i),

we get the non-integer values of m

On putting n = 49, 51 and 57 , we get the value of m = 588, 204 and 76, respectively.

Hence, there are three possible pairs of m and n that satisfy the equation.

Answer: (b)

92 = $2^2$ × 23

85 = 5 × 17

Since there is no common factor in 92 and 85, therefore they are relatively prime.