# SSC CHSL Quant Previous year 4th March 2018 asked Questions With Video Explanations PDF

For Previous year Quant questions of SSC CHSL 2018 Tier I exam download PDF. Go through the video of Quant questions explanations of last year 4th march 2018.

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Download SSC CHSL 4 March 2018 Asked Questions

**Question 1:Â **Find the value $((49)^\frac{3}{2}+(49)^\frac{-3}{2})$

a)Â 117549/343

b)Â 117550/343

c)Â 117649/343

d)Â 117650/343

**Question 2:Â **Calculate the total numbers of prime factors in the expression $(4)^{11} \times (5)^{5} \times (3)^{2} \times (13)^{2}$

a)Â 30

b)Â 31

c)Â 33

d)Â 32

**Question 3:Â **Which of the following is the correct rationalize form ofÂ $\frac{15}{\sqrt{5}+2}=?$

a)Â $5\sqrt{5}-6$

b)Â $5\sqrt{5}-30$

c)Â $15\sqrt{5}-30$

d)Â $45\sqrt{5}-630$

**Question 4:Â **Determine the value of $’x’$ if $x=\frac{(943+864)^{2}-(943-864)^{2}}{(1886\times1728)}$

a)Â 1

b)Â 4

c)Â 79

d)Â 1789

**Question 5:Â **Which of the following options is/are CORRECT about the similarity of the two triangles?

a)Â The corresponding sides are proportional to each other.

b)Â The corresponding angles are equal.

c)Â The corresponding sides may or may not be equal to each other.

d)Â All option are correct.#

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**Question 6:Â **If the height of the equilateral triangle is 2âˆš3 cm, then determine the area (in $cm^{2}$) of the equilateral triangle.

a)Â 6

b)Â 2âˆš3

c)Â 4âˆš3

d)Â 12

**Question 7:Â **Length and breadth of a rectangle are increased by 10% and 20% respectively. What will be the percentage increase in the area of rectangle?

a)Â 30%

b)Â 32%

c)Â 28%

d)Â 33%

**Question 8:Â **The ratio of three numbers is 3 : 6 : 8. If their product is 9216, then what is the sum of the three numbers?

a)Â 96

b)Â 72

c)Â 144

d)Â 68

**Question 9:Â **Rohit started a business with Rs 75000 and after some months Simran joined him with Rs 60000. If the profit at the end of the year is divided in the ratio 3 : 1, then after how many months did Simran join Rohit?

a)Â 7

b)Â 6

c)Â 8

d)Â 4

**Question 10:Â **A boy bought 50 chocolates for Rs 1000. If the average price of 30 chocolates is Rs 25, then what is the average price (in Rs) of the remaining

chocolates?

a)Â 10

b)Â 12.5

c)Â 15

d)Â 17.5

**Answers & Solutions:**

**1)Â AnswerÂ (C)**

Expression =Â $((49)^\frac{3}{2}+(49)^\frac{-3}{2})$

=Â $((7^2)^{\frac{3}{2}}+(7^2)^{\frac{-3}{2}})$

= $7^3+7^{-3}$

= $343+\frac{1}{343}$

= $\frac{117649}{343}$

=> Ans – (C)

**2)Â AnswerÂ (B)**

ExpressionÂ :Â $(4)^{11} \times (5)^{5} \times (3)^{2} \times (13)^{2}$

Prime factorization =Â $(2)^{22} \times (5)^{5} \times (3)^{2} \times (13)^{2}$

Now, there are 4 distinct factors = $2,3,5,13$

Total number of prime factors = $22+5+2+2=31$

=> Ans – (B)

**3)Â AnswerÂ (C)**

ExpressionÂ : $\frac{15}{\sqrt{5}+2}=?$

Rationalizing the denominator, we getÂ :

= $\frac{15}{\sqrt{5}+2}\times\frac{(\sqrt5-2)}{(\sqrt5-2)}$

= $\frac{15(\sqrt5-2)}{5-4}=15\sqrt5-30$

=> Ans – (C)

**4)Â AnswerÂ (A)**

Expression :Â $x=\frac{(943+864)^{2}-(943-864)^{2}}{(1886\times1728)}$

Using, $a^2-b^2=(a+b)(a-b)$, where $a=(943+864)$ and $b=(943-864)$

= $\frac{[(943+864)+(943-864)]\times[(943+864)-(943-864)]}{(1886\times1728)}$

= $\frac{(943+943)\times(864+864)}{(1886\times1728)}$

= $\frac{(1886)\times(1728)}{(1886\times1728)}=1$

=> Ans – (A)

**5)Â AnswerÂ (D)**

If two triangles are similar, then the corresponding sides are proportional to each other. Also, the corresponding angles are equal, but the corresponding sides may or may not be equal to each other. Thus, all are correct.

=> Ans – (D)

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**6)Â AnswerÂ (C)**

Let side of the equilateral triangle = $s$ cm

Median = Height of triangle = $2\sqrt3=\frac{\sqrt3s}{2}$

=> $s=2\times2=4$ cm

$\therefore$ AreaÂ of the equilateral triangle = $\frac{\sqrt3}{4} (s)^2$

= $\frac{\sqrt3}{4}\times(4)^2$

= $4\sqrt3$ $cm^2$

=> Ans – (C)

**7)Â AnswerÂ (B)**

Let both length and breadth of the rectangle be $l=10$ cm and $b=10$ cm

Area of rectangle = $A=l\times b=10\times10=100$ $cm^2$

New length when it is increased by 10% = $l’=10+(\frac{10}{100}\times10)=11$ cm

Similarly, new breadth when it is increased by 20% = $b’=10+(\frac{20}{100}\times10)=12$ cm

=> New area = $A’=11\times12=132$ $cm^2$

$\therefore$ % increase = $\frac{(132-100)}{100}\times100=32\%$

=> Ans – (B)

**8)Â AnswerÂ (D)**

Let the numbers be $3x,6x$ and $8x$

Product = $3x\times6x\times8x=9216$

=> $x^3=\frac{9216}{144}=64$

=> $x=\sqrt[3]{64}=4$

$\therefore$ Sum of the numbers = $3x+6x+8x=17x$

= $17\times4=68$

=> Ans – (D)

**9)Â AnswerÂ (A)**

Rohit invested Rs. 75,000 for 12 months and Simran invested Rs. 60,000 for $x$ months

=> Ratio of profits of P:Q = $(75,000\times12):(60,000\times x)$

According to ques,

=> $\frac{75}{5x}=\frac{3}{1}$

=> $5x\times3=75$

=> $x=\frac{75}{15}=5$

Thus, Simran joined Rohit after = $12-5=7$ months

=> Ans – (A)

**10)Â AnswerÂ (B)**

Total price of 50 chocolates = Rs. 1000

Total price of 30 chocolates = $30\times25=Rs.$ $750$

=> Total price of remaining $(50-30=20)$ chocolates = $1000-750=Rs.$ $250$

$\therefore$ Average price of 20 chocolates = $\frac{250}{20}=Rs.$ $12.5$

=> Ans – (B)