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.

853 + ? ÷ 17 = 1000

   A.) 2499
   B.) 2566
   C.) 2965
   D.) 2864

Answer: Option 'A'

853 + ?/17 = 1000
?/17 = 1000 – 853 = 147
? = 17 × 147 = 2499

3.

(764 × ?) ÷ 250 = 382

   A.) 115
   B.) 125
   C.) 135
   D.) 145

Answer: Option 'B'

(764 × ?)/250 = 382
? = (382 × 250)/764 = 125

4.

√? + 416 = (60% of 920) – 110

   A.) 676
   B.) 751
   C.) 786
   D.) 657

Answer: Option 'A'

√? + 416 = (60% of 920) – 110
√? = (50% of 920 + 10% of 920) – 110 – 416
√? = (460 + 92) –526
√? = 552 – 526 = 26
⇒ ? = 262 = 676

5.

1035 ÷ [(3/4) of (71 + 65) – 15 ¾] = ?

   A.) 12
   B.) 16
   C.) 17
   D.) 20

Answer: Option 'A'

1035 ÷ [(3/4)× 136 – 63/4] = x
X = 1035 ÷ [102 – (63/4)]
X = 1035 ÷ [(408 – 63)/4]
X = 1035 ÷ (345/4)
X = 1035 × 4/345
X = 12

6.

Simplify: y – [y – (x + y) – {y –(y – x – y )} + 2x]

   A.) 0
   B.) x
   C.) 4x
   D.) x/2

Answer: Option 'C'

Follow the rule of ‘VBODMAS’ 
Step 1: Solve the expression below bar
y-[y-(x+y)-{y-(y-x+y) }+2x]
Step 2: Solve () bracket
y-[y-(x+y)-{y-(2y-x) }+2x]
y-[-x-{y-2y+x}+2x]
Step 2: Solve {} bracket
y-[-x-{y-2y+x+2x}]
y-[-x-{-y+3x}]
y-[-x+y-3x]
Step 3: Solve [ ] bracket
y+4x-y
4x
Therefore, correct answer is 4x i.e option (c)

7.

(45)2 + (21)2 = (?)2 + 257

   A.) 45
   B.) 48
   C.) 47
   D.) 58

Answer: Option 'C'

(45)2 + (21)2 = (?)2 + 257
or, (?)2 = (45)2 + (21)2 – 257
or, (?)2 = 2025 + 441 – 257 = 2209.
or, ? = 2209 = 47

8.

52 + 132 – 112 = (?)3 – 52

   A.) 4
   B.) 6
   C.) 5
   D.) 8

Answer: Option 'C'

From the given equation
⇒ 25 + 169 – 121 = ?3 – 52
⇒ 25 + 48 + 52 = ?3
⇒ ?3 = 125
⇒ ?3 = (5)3 ⇒ ? = 5.
Hence, option C is correct.

9.

98643 – 21748 = 51212 + ?

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

Answer: Option 'A'

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

10.

A contractor pays Rs. 20 to a worker for each day and the worker forfeits Rs. 10 for each day if he is idle. At the end of 60 days, the worker gets Rs. 300. Find for how many days the worker was idle?

   A.) 34 days
   B.) 30 days 
   C.) 28 days
   D.) 40 days

Answer: Option 'B'

Step 1: Number of days worked by the worker = 60 and he remained idle for x days. Therefore, number of days worked = (60 – x) 
Step 2: Each day he was getting paid Rs. 20. Therefore, the payment received for working days = (60 – x) 20
Step 3: After subtracting the amount which he forfeited, he receives Rs. 300. 
Therefore,
(60 – x) 20 – 10x = 300
1200 – 20x – 10x = 300 
900 = 30x 
x = 30 days

Simplification Problems for SSC CGL Download Pdf