number systems Model Questions & Answers, Practice Test for ibps po prelims

Answer: (d)

Out of the given alternatives,

56 × 57 = 3192

Answer: (a)

a, a + 2, a + 4 are prime numbers.

Put value of 'a' starting from 3, we will have 3, 5 and 7 as the only set of prime numbers satisfying the given relationships.

Answer: (b)

Between 100 and 199, there will be 19 numbers which contain '2'.

They are as follows.

102, 112, 120 – 129 (10 numbers), 132, 142, 152, 162, 172, 182, 192.

Similar would be the case for 300 – 399, 400 – 499, 500 – 599, 600 – 699.

For 200 – 299, all 100 numbers will have 2.

∴ Total number of numbers containing

'2' = 19 × 6 + 100 = 114 + 100 = 214.

Answer: (d)

On dividing we find that when $11109999/1111$

Quotient is 9999 and remainder is 1110.

Answer: (d)

P = r s = r (1 – r) = $r – r^2$ (qr s + =1)

⇒ ${dP}/{dr}$ = 1 – 2r = 0 (For max. and min.)

⇒ $r = 1/2 ⇒ {d^2P}/{dr^2}$ = – 2

= Negative ⇒ Maximum

∴ Maximum value of rs = $1/4$