number systems Model Questions & Answers, Practice Test for ssc chsl tier 1

Answer: (c)

Let the 3 digits of number A be x, y and z

Hence A = 100x + 10y + z

On reversing the digits of number A,

we get the number B i.e., z y x.

∴ B = 100z + 10y + x

As B > A ⇒ z > x ...(i)

B – A = 99z – 99x = 99(z – x)

As 99 is not divisible by 7

so (z – x) has to be divisible by 7. ...(ii)

Using (i) & (ii), the only possible values of z and x are (8, 1) and (9, 2)

So the minimum and maximum range of A are 108 and 299, which Î106 < A < 305

Answer: (d)

Since the 7, 11 and 13 all are the factors of such a number so (d) is the correct answer.

Answer: (c)

Clearly, (2272 – 875) = 1397, is exactly divisible by N.

Now, 1397 = 11 × 127

∴ The required 3-digit number is 127, the sum of whose digits is 10.

Answer: (a)

Sum of digits = (5 + 1 + 7 + x + 3 + 2 + 4) = (22 + x), which must be divisible by 3.

∴ x = 2.

Answer: (b)

Let the number be (10x + y)

According to question

⇒ x + y = 7 --- (i)

(10y + x) – (10x + y) = 27

⇒ 9(y – x) = 27

⇒ y – x = 3 --- (ii)

From eq (i) and (ii)

x = 2; y = 5

so x × y = 2 × 5 = 10

So, option (c) is correct.