number system Model Questions & Answers, Practice Test for ssc cgl tier 1 2023

Answer: (b)

I. Given, n is a prime number greater than 5.

Now, $n^4 – 1 = (n^2 – 1) (n^2 + 1)$

= $(n – 1) (n + 1) (n^2 + 1)$

Put n = 7 (prime number greater than 5)

$n^4$ – 1 = (7 – 1) (7 + 1)(49 + 1)

= 6 . 8 . 50 = 2400

So, statements I is true.

II. Given, n ∈ W (whole number)

i.e., n = 0, 1, 2, 3, 4, 5, ...

For n = 0,

5n, (5n – 1), (5n + 1) = 0, – 1, 1 = $(0)^2 , – 1, (1)^2$

For n = 1,

5n, (5n – 1), (5n + 1) = 5, 4, 6 = 5, $(2)^2$ , 6

For n = 2,

5n, (5n – 1 ), (5n + 1 ) = 10, 9, 11 = 10, $(3)^2$ , 11

For n = 3,

5n, (5n – 1), (5n + 1) = 15, 14, 16 = 15, 14, $(4)^2$

For n = 4,

5n, (5n – 1), (5n + 1) = 20, 19, 21

For n = 5,

5n, (5n – 1), (5n + 1) = 25, 24, 26 = $(5)^2$ , 24, 26 ... so on.

So, statements II is true.

Answer: (b)

Here b is the largest square divisor of c and $a^2$ divides c then, it is sure that a divides b.

Answer: (b)

By remainder theorem,

$9^6$ will have the remainder 1 as 9 has the remainder 1.

Also ${9^6 + 7}/8$ will have the same remainder as

${(1)^6 + 7}/8$ which has the remainder equal to 0.

Answer: (d)

Let a = 7p + 5 and b = 7q + 4

where, p and q are natural numbers.

∴ ab = (7p + 5) (7q + 4)

ab = 49pq + (4p + 5q) 7 + 20

= 7 (7pq + 4p + 5q) + 7 × 2 + 6

when ab is divided by 7, we get the remainder 6.

Answer: (d)

Interval after which the devices will beep together

= (L.C.M. of 30, 60, 90, 105) min = 1260 min. = 21 hrs.

So, the devices will again beep together 21 hrs. after 12 noon i.e., at 9 a.m.