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

Answer: (b)

$x + 100/x > 50$

⇒ $x^2 - 50x + 100 > 0$

$x = {50 ± √{(50)^2 - 4 × 100 × 1}}/2$

using $x = {b ± √{b^2 - 4ac}}/{2a}$

$x = 25 ± 5√{21}$

so, $x^2 - 50x + 100 > 0$

{x - (25 + 5$√21$)}{x - (25 - 5$√21$)} > 0

As x ≤ 100 (Natural number)

So domain of x is = {1, 2, 48, 49, 50, 51------, 100}

Hence no. of values of x is 55.

Option (c) is correct.

Answer: (d)

Let p be any positive number of the form 2m, 2m + 1 for any whole number m.

Case I: p = 2m

$p^2 = 4m^2 = 4(m^2) = 4q$ where q = $m^2$

Case II: p = 2m + 1

$p^2 = (2m + 1)^2 = 4m^2 + 4m + 1$

= $4(m^2 + m) + 1$

= 4q + 1 where $q = m^2 + m$

From above we see that square of any positive integer is either of the form 4q or (4q + 1) for some integer q.

Answer: (d)

$94^3 - 23^3$ is divisible by 94 - 23 = 71

$94^3 - 71^3$ is divisible by 94 - 71 = 23

$23^3 - 71^3$ is divisible by 23 + 71 = 94

∴ $94^3 - 23^3 - 71^3$ is divisible by 23, 71 and 94

Answer: (c)

According to the question,

x + x + 2 + x + 4 + x + 6 + x + 8 = 140

or, 5x + 20 = 140

or, 5x = 120

∴ $x = 120/5$ = 24

∴ x + 8 = 24 + 8 = 32

The next set of five consecutive even number will start with = 34

∴ Required sum = 34 + 36 + 38 + 40 + 42 = 190

Answer: (b)

$3^{2n + 1} + 2^{2n + 1} = 3 × (3)^2n + 2 × (2)^2n$

= $3 × (9)^n + 2 × (4)^n$

Thus, when n is odd, then unit digit of $(9)^n$ = 9 and $(4)^n$ = 4

and when n is even, then unit digit of $(9)^n$ = 1 and $(4)^n$ = 6

Hence, when n is odd positive integer, then 3 × (unit digit of 9) + 2 × (unit digit of 4)

= 3 × 9 + 2 × 4 = 35

Hence, unit digit of $(3)^{2n + 1} + (2)^{2n + 1}$ = 5

Also, when n is even positive integer, then

3 × (unit digit of 9) + 2 × (unit digit of 4)

= 3 × 1 + 2 × 6 = 15

Hence, unit digit of $(3)^{2n + 1} + (2)^{2n + 1}$ = 5