number system Model Questions & Answers, Practice Test for ibps so prelims 2023

Answer: (b)

$7^174 = (7)^{43 × 4} × 7^2$

= 1 × 9 = 9

2. Let n is a add number then (n+ 2) is also a odd number,

Now, $(n + 2)^2 – (n)^2$ = (2n + 2).2

= 4(n + 1)

As n is odd, so (n + 1) is even number and must be divisible by 2.

Hence difference of square of two odd number is away divisible by 8.

3. Let n and (n + 2) are two odd number.

Now, n(n + 2) + 1 = $n^2 + 2n + 1 = (n + 1)^2$

This is perfect square

Hence, statement 2 and 3 are true

Answer: (b)

Factors of 88 = 11 × 8

2784936

= (Sum of odd place digit) – (Sum of even place digit)

= 25 – 14 = 11 (This is divisible by 11)

2784936:

Last three digit is 936 is divisible by 8.

So this number is also divisible by 8.

Now, 2784936 is divisible by 88.

Answer: (d)

Let n = 6

Therefore $√n = √6$ ≈ 2.4

Now, the divisor of 6 are 1, 2, 3

If we take 2 as divisor then $√n$ > 2 > 1.

Statement I is true.

If we take 3 as divisor then 6 > 3 > 2.4,

i.e. n > $√n$

Therefore statement II is true

Answer: (b)

From options,

(a) $14/(1+ 4) = 14/5$ = 4 (Rem) and

$14/{1 × 4} = 14/4$ = 2 (Rem)

Since, remainder is not same.

(b) $23/(2 + 3) = 23/5$ = 3 (Rem) and

$23/{2 × 3} = 23/6$ = 5 (Rem)

Since, remainder is not same.

(c) $32/(3 + 2) = 32/5$ = 2 (Rem) and

$32/(3 × 2) = 32/6$ = 2 (Rem)

Since, remainder is same.

∴ Difference of quotients = 6 – 5 = 1

(d) $41/(4 + 1) = 41/5$ = 1 (Rem) and

$41/(4 × 1) = 41/4$ = 1 (Rem)

Since, remainder is same.

But difference of quotients = 10 – 8 = 2 ≠ 1

Answer: (d)

$2^{12n} – 6^{4n} = (2^12)^n – (6^4)^n= (4096)^n – (1296)^n$

= $(4096 – 1296) [(4096)^{n – 1} + (4096)^{n – 2}(1296) + ... + (1296)^{n – 1}]$ = 2800(k)

Hence, last two digits are always be zero.