SSC CHSL 4th March 2018 Quant Asked Questions With Video Explanations (Last year paper)

0
1082
SSC CHSL Important Quant Questions
SSC CHSL Important Quant Questions

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.

Download Quant Questions for of SSC CHSL 4th March 2018 PDF

10 SSC CHSL Mocks for Rs . 117

Take a free mock test for SSC CHSL

Download SSC CHSL Previous Papers

More SSC CHSL Important Questions and Answers PDF

Watch the video here (If not loaded below)

 

Download SSC CHSL 4 March 2018 Asked Questions

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

a) 117549/343

b) 117550/343

c) 117649/343

d) 117650/343

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

a) 30

b) 31

c) 33

d) 32

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

a) $5\sqrt{5}-6$

b) $5\sqrt{5}-30$

c) $15\sqrt{5}-30$

d) $45\sqrt{5}-630$

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

a) 1

b) 4

c) 79

d) 1789

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

a) The corresponding sides are proportional to each other.

b) The corresponding angles are equal.

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

d) All option are correct.#

SSC CHSL PREVIOUS PAPERS

SSC CHSL Study Material (FREE Tests)

Question 6: If the height of the equilateral triangle is 2√3 cm, then determine the area (in $cm^{2}$) of the equilateral triangle.

a) 6

b) 2√3

c) 4√3

d) 12

Question 7: 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) 30%

b) 32%

c) 28%

d) 33%

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

a) 96

b) 72

c) 144

d) 68

Question 9: 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) 7

b) 6

c) 8

d) 4

Question 10: 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) 10

b) 12.5

c) 15

d) 17.5

SSC CHSL FREE MOCK TEST

Answers & Solutions:

1) Answer (C)

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

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

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

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

= $\frac{117649}{343}$

=> Ans – (C)

2) Answer (B)

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

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

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

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

=> Ans – (B)

3) Answer (C)

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

Rationalizing the denominator, we get :

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

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

=> Ans – (C)

4) Answer (A)

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

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) Answer (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)

FREE SSC CHSL MOCK TEST SERIES

FREE SSC STUDY MATERIAL

6) Answer (C)

Let side of the equilateral triangle = $s$ cm

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

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

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

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

= $4\sqrt3$ $cm^2$

=> Ans – (C)

7) Answer (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) Answer (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) Answer (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) Answer (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)

DOWNLOAD APP TO ACESSES DIRECTLY ON MOBILE

SSC CHSL Important Q&A PDF

LEAVE A REPLY

Please enter your comment!
Please enter your name here