Simplification Problems for SSC CGL

1.

853 + ? ÷ 17 = 1000

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

Answer: Option 'A'

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

2.

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

3.

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.

4.

70% of 1680 + ?% of 1750 = 55% of 2820 – 886

   A.) –29.2
   B.) 30
   C.) 34
   D.) 54

Answer: Option 'A'

70% of 1680 + ?% of 1750 = 55% of 2820 – 886
70/100 × 1680 + ?/100 × 1750 = 55/100 × 2820 – 886
1176 + ?/10 × 175 = 1551 – 886 = 665
?/10 × 175 = 665 – 1176 = –511
? = (–511 × 10)/175 = –29.2

5.

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

6.

6156  ÷ √? × 53 = 4028

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

Answer: Option 'B'

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

7.

In a farm, along with 50 hens, there were 45 goats and 8 horses and some farmers. If total number of feet be 224 more than number of heads, then find the number of farmers. 

   A.) 16
   B.) 11
   C.) 15
   D.) 14

Answer: Option 'C'

Let’s the number of farmers be y. 
Step 1: Find number of heads
= (50 hens + 45 goats + 8 horses + y farmers) 
= (103 + y)
Step 2: Number of feet 
= [(Hens 2 × 50) + (45 × 4) + (8 × 4) + (y × 2)]
= [100 + 180 + 32 + 2y]
= 312 + 2y 
Step 3: Find number of farmers 
(312 + 2y) – (103 + y) = 224
312 + 2y – 103 – y = 224
y = 15 
Number of farmers = 15

8.

4/9 × 1701 + 2/11 × 1386 = ?

   A.) 180
   B.) 1008
   C.) 1080
   D.) 1800

Answer: Option 'B'

4/9 × 1701 + 2/11 × 1386 = ?
? = (4 × 189) + (2 × 126)
? = 756 + 252 = 1008.
Hence, option B is correct.

9.

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)

10.

√? + 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

11.

[(4)3 × (5)4] ÷ (4)5 = ?

   A.) 39.0625
   B.) 44.0625
   C.) 43.0625
   D.) 41.0625

Answer: Option 'A'

? = [43 × 54]/45 = 54/42 = 625/16 = 39.0625


Simplification Problems for SSC CGL Download Pdf