MCQ: Square Roots and Cube Roots - 1 - SSC CGL MCQ

? = (√6 + 1)² − 2√6 = 6 + 1 + 2√6 − 2√6 = 7

? = √225 + √2304 + (12)² = 15 + 48 + 144 = 207

Let volume of cubes = a³ and b³

a/b = 7/11
or a : b = 7 : 11

(16)² − 5³ + √169 = (?)²

256 – 125 + 13 = (?)²

144 = (?)²

? = ± 12

(xⁿ - 1) is divisible by x + 1 if n is an even number. (49)¹⁵ - 1 = [(7²)¹⁵ - 1] = (7)³⁰ - 1 which is divisible by 7 + 1 ,8.
10-x=8  so x=2

81x²=(9x)²
64y²=(8y)²
 so term missing 2(9x)(8y)=144xy

First five prime numbers are 2, 3, 5, 7, 11. Sum - 28

Resolve 136 into prime factors and make group of two of each prime factor

136 = 2 × 2 × 2 × 17

136 = (2 × 2) × 2 × 17

We find that 2 and 17 doesn't appear in group of two. So, 136 has to be multiplied with 34 to make it a perfect square.

(64)² - (36)² = 20 y Using formulae (a + b)(a - b) = a^{2 - b2}
=> (64 + 36)(64 - 36) = 20 y => 100 x 28 = 20 y => y = 5 x 28 = 140

∴ 24136 < (156)2
24136 < 24336
∴ we add 24336 – 24136 = 200
so that it becomes a perfect square

9548 + 7314 = 8362 + x ∴ x = 9548 + 7314 - 8362 = 8500

Let one number = a
∴ Second number = 4a
⇒ 4a × a = 1936
⇒ a² = 1936/4 = 484
⇒ a² = 484
⇒ a = 2 × 11 = 22
and 4a = 4 × 22 = 88
∴ Numbers are 22 and 88.

3²⁵ + 3²⁶ + 3²⁷ + 3²⁸ = 3²⁴ x (3¹ + 3² + 3³ + 3⁴) = 3²⁴ x (3 + 9 + 27 + 81) = 4⁶⁰ x 120 which is divisible by 30.

Let the side of square field = 'a' m

∴ Area of square field = a² sq. m

a² = 22500 m²

⇒ a = 150 m

Speed of cycling = 15 km / hr

Now, total distance to be covered along the boundary

= 4 × 150 = 600 m

∵ 25/6 m is covered in 1 sec.

∴ 600 m is covered in 600/25×6

=144 sec = 2 min 24 sec.

999 is the largest three digit number.
On dividing 999 by 32, remainder is 7.
999 - 7 = 992
Therefore, 992 is the largest three digit multiple of 32.