Simplification Problems for SSC CGL

1.

The population of a city A which is 68000 decreases at the rate of 1200/year. Population of city B which is 42000, increases at the rate of 800 per year. Find in how many years, the population of cities A and B are equal? 

   A.) 9 years
   B.) 13 years
   C.) 15 years
   D.) 10 years

Answer: Option 'B'

We have to find the population of cities A and B after x years. 
Step 1: Population of city A = 68000, decreases at the rate of 1200/year
68000 – 1200x
Step 2: Population of city B = 42000, increases at the rate of 800/year
42000 + 800x
Step 3: Find after how many population of cities A and B are equal. 
Population of city A = Population of city B 
68000 – 1200x = 42000 + 800x
68000 – 42000 = 1200x + 800x 
26000 = 2000x
x = 13

2.

6156  ÷ √? × 53 = 4028

   A.) 6889
   B.) 6561
   C.) 6995
   D.) 6235

Answer: Option 'B'

6153 × 53/√? = 4028
⇒ √? = 80.96 ≈ 81
⇒ ? = 6561

3.

Simplify: 13.05 × 4.5 ÷ 0.5

   A.) 119.75
   B.) 117.45 
   C.) 115.50
   D.) 140.60 

Answer: Option 'B'

Follow the rule of ‘VBODMAS’ 
No bar, No bracket. Hence, first step starts from ‘DMAS’ 
Step 1: Divide 4.5 ÷ 0.5=9
Step 2: Multiply 13.05 × 9 = 117.45

4.

98643 – 21748 = 51212 + ?

   A.) 25683
   B.) 26687
   C.) 27893
   D.) 26874

Answer: Option 'A'

Given eqn,
98643 – 21748 = 51212 + ? 
⇒ ? = 98643 – 21748 – 51212 
⇒ ? = 25683

5.

If x/y = 3/4 and 8x + 5y = 22, then find the value of x.

   A.) 1.5
   B.) 2.3
   C.) 4.5
   D.) 3.6

Answer: Option 'A'

y = 4/3 x
Substitute this value in 8x+5y=22
8x + 5(4/3) x = 22
44x=66
x=1.5

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