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

Answer: (a)

Let the numbers be x and (x + 2)

Then, x × (x + 2) = 582168

$x^2 + 2x - 582168 = 0$

$x^2 + 764x - 762x - 582168 = 0$

(x + 764) (x - 762) = 0

x = 762

Answer: (d)

$(m - n)^2 = {4mn}/(m + n -1)$

⇒ $(m - n)^2$ (m + n - 1 ) = 4mn

⇒ $(m - n)^2 (m + n - 1 ) = (m + n)^2 - (m - n)^2$

⇒ $(m - n)^2 ( m + n) = (m + n)^2$

⇒ $(m - n)^2$ = (m + n)

(Since, m + n ≠‚ 0)

The above equation has infinitely many solutions where m and n are positive integers.

We can put m + n = v and m - n = u, and re-write the equation as $u^2$ = v and then plug in different values of u and v to get different pairs of (m, n).

Answer: (a)

Suppose the number is x

∵ $x × 1/5 = 81$

x = 81 × 5 = 405

∴ ${x × 68}/100 = {405 × 68}/100$

x = 275.4

Answer: (a)

As p, q, r are non-negative integers, the maximum will be achieved when the value of each variable is closed to each other.

i.e., p, q, r are 3, 3, 4 (not necessarily in the same order).

Hence the value of

pq + qr + pr + pqr = 3 × 3 + 3 × 4 + 3 × 4 + 3 × 3 × 4

= 9 + 12 + 12 + 36 = 69

Answer: (b)

On dividing 6709 by 9,

we get remainder = 4.

∴ Required number to be subtracted = 4.