# SSC CGL repeated aptitude questions in last year papers (set-9)

Jun 19, 2017 11:45 IST

SSC Quant study material

We are providing you the set of 10 questions, which are most repetitive questions of the quantitative aptitude came in last five years. Try all these questions and attempt similar questions too. All are provided with complete explanation through which you can easily see analyse the complete solution. An important thing is that SSC repeats its questions, so it is very helpful to you as you may get the same question in exam.

Directions (1-3) : The distribution of fruit consumption in a sample of 72 seventeen – year – old girls is given in the graph below. Study the graph and answer the questions.

Distribution of fruit consumption

Q1. How many of these girls ate fewer than two servings per day?

a) 15

b) 40

c) 25

d) none of these

Solution:

Required number of girls = 15 + 10 = 25

Q2. What percent of these girls ate six or more servings per day?

a) 12.5 %

b) 13 %

c) 10 %

d) 11 %

Solution:

Q3. How many of these girls ate more than two servings but less than six servings per day?

a) 26

b) 18

c) 23

d) 38

Solution:

Required number of girls = 10 + 8 + 5 = 23

Q4. The following pie-chart shows the monthly expenditure of a family on food, clothing, rent, miscellaneous expenses and savings. What is the central angle for savings?

a) 54o

b) 56o

c) 50o

d) 52o

Solution:

Q5. The pie – chart gives the expenditure (in percentage) on various items and savings of a family during a month. Monthly savings of the family is 3. 000. On which item is the expenditure maximum and how much is it?

a) Others, 2.000

b) Food, 3.000

c) Others, 5.000

d) Food, 5.000

Solution:

Q6. Solve for x:

3x – 3x-1 = 486.

a) 7

b) 9

c) 5

d) 6

Solution:

3x – 3x-1 = 486

Q7. What is the value of

a) 3

b) 3/2

c) 1

d) 1/2

Solution

a) 9

b) 16

c) 4

d) 10

Solution:

Q9. If x (x – 3) = – 1, than the value of x3 (x3 – 18) is

a) - 1

b) 2

c) 1

d) 0

Solution:

Q.10 If xsin260o – (3/2)sec60o.tan230o + (4/5)sin245o tan260o = 0, then x is

a) –(1/15)

b) - 4

c) – (4/15)

d) - 2

Solution:

xsin260o – (3/2)sec60o.tan230o + (4/5)sin245o tan260o = 0

All The Best!

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