number systems Model Questions & Answers, Practice Test for ibps clerk mains

Answer: (b)

Any number that are divisible by 3 only when sum of its digits is divisible by 3

∴ (x + y + 2 + 3 + 5) = (10 + x + y)

As x + y ≤ 5,

then (10 + x + y) = 12 or 15

∴ (x, y) = (2, 0), (1, 1) (2, 3), (3, 2), (1, 4), (4, 1) and (5, 0)

Hence, number of possible pairs = 7

Answer: (d)

${5 + √10}/{5√5 - 2√20 - √32 + √50}$

= ${5 + √10}/{5√5 - 4√5 - 4√2 + 5√2}$

= ${5 + √10}/{√5 + √2} = {(5 + √10)(√5 - √2)}/{(√5 + √2)(√5 - √2)}$

= ${5√5 + √50 - 5√2 - √20}/{5 - 2}$

= ${5√5 + 5√2 - 5√2 - 2√5}/3$

= ${3√5}/3 =√5$

Answer: (c)

Let 'r' be the remainder

⇒ 221 – r, 116 – r, 356 – r are exactly divisible by that number.

Now, if two numbers are divisible by a number, then so is their difference

⇒ [(221 – r) – (116 – r)], [(356 – r) – (116 – r)].

and [(356 – r) – (221 – r)] are divisible by that number

⇒ 105, 135, 240 are divisible by that number

= HCF of 105, 135, 140 = 15.

Answer: (b)

From the given alternatives,

112 × 114 = 12768

∴ Larger number = 114

Answer: (c)

We have to find the least number which is divisible by 3, 5, 6 and 9 and is also a perfect square.

The LCM of 3, 5, 6 and 9 is 3 × 3 × 2 × 5 = 90.

Hence, the required number is 90 × 2 × 5 = 900.