SSC CGL Previous Year Questions: Simplification and Square & Cube Root- 1 - SSC CGL MCQ

2 of 3 ÷ 3 × 2 + {4 × 3 – (5 × 2 + 3)}
⇒ 2 × 1 × 2 + {12 – (13)}
⇒ 2 × 2 + {– 1}
⇒ 4 – 1
⇒ 3

(9 ÷ 30)² × 2.4 + 0.3 of 12 × (1 – 0.3)² + 9 × (0.3)² = ?
⇒ (0.3)² × 2.4 + 3.6 × (0.7)² + 9 × (0.09)
⇒ (0.09) × 2.4 + 1.764 + 9 × (0.09)
⇒ 1.98 + 0.81
⇒ 2.79

15.2 + 5.8 ÷ 2.9 x 2 - 3.5 x 2 ÷ 0.5
15.2 + 2 x 2 - 7 x 2
15.2 + 4 – 14 = 5.2

4.5 – (3.2 ÷ 0.8 x 5) + (3 x 4 ÷ 6)
= 4.5 – (4 x 5) + 3 x (2/3)
= 4.5 – 20 + 2 = –13.5

This is in the form of

Here a = 5.75, b = 3.25

= 9/100 = 0.09

According to question.
(2 + 1) (2² + 1) (2⁴ + 1) (2⁸ + 1) = ?
⇒ (2 + 1) (2⁸ + 1) (2²+ 1) (2⁴ + 1)
⇒ (2⁹ + 2 + 2⁸ + 1) (2⁶ + 2² + 2⁴ + 1)
⇒ (2¹⁵ + 2¹⁴ + 2¹³ + 2¹² + 2¹¹ + 2¹⁰ + 2⁹ + 2⁸ + 2⁷ + 2⁶ + 2⁵ + 2⁴ + 2³ + 2² + 2) + 1
Here, a = 2, n = 15 r = 2
∴

⇒ (2¹⁶ – 2)
∴(2¹⁶ – 2) + 1 = (2¹⁶ – 1)
∴ (2 + 1) (2² + 1) (2⁴ + 1) (2⁸ + 1) = (2¹⁶ – 1)

Here,

Now,

Here,
2x – 3 (2x – 2) > x – 1 < 2 + 2x
2x – 6x + 6 > x – 1
⇒ 2x – 6x – x > –7
⇒ –5x > –7
x < 7/5 ...(i)
(x – 1) < (2 + 2x)
x – 1 < 2 + 2x
– 3 < x ...(ii)
From (i) and (ii),
x = –2.

Let the numbers are x and y.
According to question
x + y = 46/3 ...(i)
x – y = 14/3 ...(ii)
Adding eqn. (i) & (ii)

Product of number 

Hence , a + b + c = 6

L.C.M of Numbers of Powers = 12
"

using 

• a³ + b³: (a + b) (a² - ab + b²)

⇒ 2.3 + 0.3 = 2.6

Let the number be x

 ∴ x = 12
Now, 5/3 of x = (5/3)x 12 = 20.

By squaring the given relations, we get (i) and (iii) are incorrect relations from the given statement.

 
 
 

⇒ 5