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# SSC-CGL Maths Questions and Answers PDF

Download SSC CGL Maths Questions with answers PDF based on previous papers very useful for SSC CGL exams. Top-15 Very Important Questions for SSC Exams.

Question 1: The greatest number, which when subtracted from 5834, gives a number exactly divisible by each of 20, 28, 32 and 35, is

a) 1120

b) 4714

c) 5200

d) 5600

Question 2: A number, when divided by 114, leaves remainder 21. If the same number is divided by 19, then the remainder will be

a) 1

b) 2

c) 7

d) 17

Question 3: A manufacturer marked an article at Rs. 50 and sold it allowing 20% discount. If his profit was 25%, then the cost price of the article was

a) Rs. 40

b) Rs. 35

c) Rs. 32

d) Rs. 30

Question 4: At what rate percent per annum will a sum of Rs. 1,000 amount to Rs. 1,102.50 in 2 years at compound interest?

a) 5

b) 5.5

c) 6

d) 6.5

Question 5: In how many years will a sum of Rs. 800 at 10% per annum compound interest, compounded semiannually becomes Rs. 926.10 ?

a) 1$\frac{1}{2}$

b) 1$\frac{2}{3}$

c) 2$\frac{1}{3}$

d) 2$\frac{1}{2}$

Question 6: A copper wire of length 36 m and diameter 2 mm is melted to form a sphere. The radius of the sphere (in cm) is:

a) 2.5

b) 3

c) 3.5

d) 4

Question 7: $\sqrt{8 + \sqrt{57 + \sqrt{38 + \sqrt{108 + \sqrt{169}}}}}$

a) 4

b) 6

c) 8

d) 10

Question 8: The unit digit in the product $122^{173}$ is

a) 2

b) 4

c) 6

d) 8

Question 9: A copper wire is bent in the form of an equilateral triangle, and has an area $121\sqrt{3}$ cm. If the same wire is bent into the form of a circle, the area(in cm2) enclosed by the wire in(Take $\pi = \frac{22}{7}$)

a) 364.5

b) 693.5

c) 346.5

d) 639.5

Question 10: At present, the ratio of the ages of Maya and Chhaya is 6:5 and fifteen years from now, the ratio will get changed to 9:8. Maya’s present age is

a) 21 years

b) 24 years

c) 30 years

d) 40 years

Question 11: Which one of the following will completely divide 571 + 572 + 573 ?

a) 150

b) 160

c) 155

d) 30

Question 12: A student was asked to divide a number by 6 and add 12 to the quotient. He, however, first added 12 to the number and then divided it by 6, getting 112 as the answer. The correct answer should have been

a) 124

b) 122

c) 118

d) 114

Question 13: The difference between the compound interest and simple interest for the amount Rs. 5,000 in 2 years is Rs.32. The rate of interest is

a) 5%

b) 8%

c) 10%

d) 12%

InstructionsDirections: The following graph shows the production of cotton bales of 100 kg each in lakhs by different states A, B, C, D and E over the years. Study the graph and answer the following Questions. Question 14: In which State(s) is there a steady increase in the production of cotton during the given period?

a) A and B

b) B and D

c) A and C

d) D and E

Question 15: If $x = 1 + \sqrt{2} + \sqrt{3}$ , then the value of $(2x^4 – 8x^3 – 5x^2 + 26x- 28)$ is __?

a) $6\sqrt{6}$

b) $0$

c) $3\sqrt{6}$

d) $2\sqrt{6}$

Given: Numbers- First = 5834
Second = x (Suppose)
And number (5834 – x) is divisible by each of 20,28,32,35
Let’s say it is y
Hence 5834 – x = y
or x = 5834 – y
Now for x to be greatest y should be least
hence y should be least common multiple of 20,28,32,35
y = 1120
now x = 5834 – 1120
x = 4714

Let the given number be x
Let a be the quotient when x is divided by 114
So $\frac{x}{114}$ = a$\frac{21}{114}$
so x = 114a + 21
when x is divided by 19 it can be written as
$\frac{x}{19} = \frac{114a + 21}{19}$
114 is divisible by 19 and 21 leaves a remainder of 2.

Given: Marked Price = 50
Discount = 20%of 50 = $50\times0.5$ = 10
Hence sold price = 50 – 10 = 40
Let’s say cost price is $x$
Profit = 25% of x (Always remember profit and loss applicable only on cost price) = $\frac{x}{4}$
Hence sold price will be $x$ + $\frac{x}{4}$ = $\frac{5x}{4}$
or $\frac{5x}{4}$ = 40
$x$ = 32

Let’s say rate is r
hence $1000\times(1+\frac{r}{100})^{2}$ = 1102.5
now on solving the we will get r = 5

When we are compounding it semiannually its rate becomes 5% and number of years will 2n
so for compound interest:
926.10=$800\times(1+\frac{10}{100})^{2n}$
solve for n.

since we know volume will remain same while melting
$\pi r_{1}^{2}h= \frac{4}{3}\pi r_{2}^{3}$
where $r_{1}$ is radius of cylinderical wire and $r_{2}$ is radius of sphere and h is length of wire
putting values we will get $r_{2}$ = 3 cm.

Start from the root of 169 then second root will reduce to 11, thrid root will reduce to 7, fourth root will reduce to 8, and finally it reduce to value 4

As we know a number with unit digit 2 have repeating cycle of 2,4,8,6 after every fourth power
as power is 173 or (172+1) where till 172 , 43rd cycle will get complete and next unit digit will be 2.

Area of equilateral triangle is $\frac{\sqrt{3}}{4} a^{2}$ where a is side of triangle
which is equals to $121{\sqrt{3}}$
or a = 22 and whole length of wire will be 66
from here when it is bend to make a circle, circumference will be $2\pi r$ = 66
r = 10.5
hence area of circle will be $\pi r^{2}$ = 346.5

Let’s say maya’s age is $6x$ and chaya’s age is $5x$.
after 15 years ages will be $6x+15$ and $5x+15$.
New ratio will be $\frac{6x+15}{5x+15} = \frac{9}{8}$.
After solving above equation we will get $x$ equals to 5
So maya’s age will be 30.

Among all options only option C has unit digit 5, and in given equation unit digit will also be 5.
So only 155 can divide the given equation completely.

Let’s say number is N
So according to student result is $112= \frac{N+12}{6}$
or N = 660
Correct answer will be = $\frac{660}{6} +12 = 110+12 = 122$

Difference between compound interest and simple interest for 2 years will be
= $(P((1+\frac{r}{100})^2) – P) – 2P \frac{r}{100} = 32$ (where P is principal amount 5000 and r is rate )
after solving above equation we will get r = 8%

Only in A and C there is a steady increment in production of cotton as in D and E , It is decresed and in B production is equal for two years. Hence answer will be C).

x = 1+ $\sqrt {2} + \sqrt {3}$
$(x-1)^{2}$ = $(\sqrt {2} + \sqrt {3}) ^ {2}$
$x^{2} +1 – 2x = 5 + 2 \sqrt {6}$
$x^{2} – 2x = 4 + 2 \sqrt {6}$ ( eq. (1) )
$(x^{2} – 2x)^{2} = x^{4} + 4x^{2} – 4x^{3} = 40 + 16\sqrt{6}$ eq (2)
Now in $2x^{4} – 8x^{3} – 5x^{2} + 26x – 28$
or $2(x^{4} – 4x^{3}) – 5x^{2} + 26x – 28$ ( putting values from eq (1) and eq (2) )
After solving we will get it reduced to $6\sqrt{6}$