number systems Model Questions & Answers, Practice Test for ssc mts paper 1

Answer: (b)

Let the smaller number be x

∵ x × (x + 2) = 4488

$x^2 + 2x – 4488 = 0$

(x + 68) (x – 66) = 0

∴ x = 66

Answer: (d)

Average score for the first

4 tests = ${3N + N + 20}/4 = {4N + 20}/4$ = N + 5

Answer: (b)

$7^4/2400$ gives us a remainder of 1.

Thus, the remainder of $7^99/2400$ would depend on the remainder of $7^3/2400$ → remainder = 343.

Answer: (e)

Suppose the greater number is x

x(x – 2) = 6888 or $x^2 – 2x – 6888$ = 0

$x^2 – 84x + 82x – 6888$ = 0

or x(x – 84) + 82(x – 84) = 0

(x – 84) (x + 82) = 0

or x = 84 or – 82

So, the greater number is 84.

Answer: (a)

$2^2 + 22^2 + 222^2 + 2222^2 + ……+ (2222 ……49 twos)^2$

= $2^2 + (2)^2 + 2^2 + …… + 2^2$ (49 twos)

= 4 + 4 + 4 + …… + 4 (49 twos)

= 4 × 49 = last digit is 6.