Expected Quant Questions For SSC CHSL Set 3

Expected Quant Set- 3 Questions for SSC CHSL download PDF based on previous year question paper of SSC exams. 20 Very important Expected Quant questions for SSC CHSL Exam.

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Question 1: The red blood cells in a blood sample grows by 10% per hour in first two hours, decreases by 10% in next one hour, remains constant in next one hour and again increases by 5% per hour in next two hours. If the original count of the red blood cells in the sample is 40000, find the approximate red blood cell count at the end of 6 hours.

a) 40000

b) 45025

c) 48025

d) 50025

Question 2: A string of length 24 cm is bent first into a square and then into a right-angled triangle by keeping one side of the square fixed as its base. Then the area of triangle equals to:

a) 24 cm$^{2}$

b) 60 cm$^{2}$

c) 40 cm$^{2}$

d) 28 cm$^{2}$

Question 3: The ratio of the volume of a cube to that of a sphere which will fit inside the cube is

a) 4 : $\pi$

b) 4 : $3\pi$

c) 6: $\pi$

d) 2 : $\pi$

Question 4: Which of the following has to be added to $36x^{2}-87x+52$ to make it a perfect square ?

a) 3x+3

b) -3x+3

c) 3x-3

d) -3x-3

Question 5: Which of the following has to be added to $16x^{2}-38x+23$ to make it a perfect square ?

a) -2x+2

b) -x+3

c) -2x-2

d) -x-3

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Question 6: If a:b=6:5 and b:c=12:14 then what is the value of $a^{2}:b^{2}:bc$

a) 216:150:225

b) 108:150:175

c) 216:125:175

d) 216:150:175

Question 7: If a:b=7:3 and b:c=9:11 then what is the value of $a^{2}:b^{2}:bc$

a) 6:49:11

b) 9:49:11

c) 49:9:11

d) 49:9:13

Question 8: What is the least number that has to be added to $9x^{2}-30x+42$ to make it a perfect square ?

a) 17

b) -17

c) 7

d) -7

Question 9: A person travels one fourth of his distance with 20 km/hr,half of the distance with 40 km/hr and the rest of the distance with 10 km/hr.What is his average speed throughout the journey ?

a) 20 km/hr

b) 25 km/hr

c) 30 km/hr

d) 40 km/hr

Question 10: A person travels with 60% of his speed to reach his destination 40 min late.If he travels with 120% of his speed in what time will he reach the destination ?

a) 60 min

b) 50 min

c) 36 min

d) 48 min

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Question 11: A number is increased by 10% and then decreased by 10% then by what percent should the value be increased to get the original value ?

a) 0 %

b) 100/99 %

c) 1%

d) 99/100 %

Question 12: A 20 litre mixture contains 60% of alcohol and 4 litres of alcohol is added to it then what is the water percentage in the mixture ?l

a) 33.33%

b) 20%

c) 40%

d) 25%

Question 13: In an exam for 150 marks, Rahul fails in the exam by 5% and got 45 marks what is the pass marks in the exam ?

a) 47.5

b) 52.5

c) 45.5

d) 37.5

Question 14: A person sold a product for Rs.1200 and got 20% loss. If he needs to get 30% profit, then the price at which he needs to sell is?

a) Rs.2000

b) Rs.1500

c) Rs.1600

d) Rs.1950

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Question 15: Marked price of an article is Rs 1000 and three successive discounts of 20% each are given. What is the profit/loss if cost price=500 ?

a) Rs 14

b) Rs 12

c) Rs 13

d) Rs 11

Question 16: Marked price of an article is Rs 480 and cost price of article is marked up by 20%. If the profit percent is 20% then what is the discount offered ?

a) 10

b) 20

c) 30

d) 0

Question 17: A person bought 15 pens at Rs 10 each and sold 5 of them at Rs 8 and other 7 at Rs 12 and remaining at Rs 16. What is the profit/loss percentage ?

a) 13.66%

b) 12.33%

c) 14.66%

d) 12.66%

Question 18: Marked price of an article is Rs 400 and a discount of 10% is given. If profit percent is 30% then what is the cost price ?

a) Rs 260

b) Rs 270

c) Rs 285

d) Rs 277

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Answers & Solutions:

1) Answer (C)

Original count = 40,000

In the next 2 hours, it increases by 10%

=> Blood cell count after 2 hours = $40,000(1+\frac{10}{100})^2=40,000(\frac{11}{10})^2$

= $40,000\times\frac{121}{100}=48,400$

It decreases by 10% in next hour

=> Blood cell count after 3 hours = $48,400(1-\frac{10}{100})^1$

= $48,400\times\frac{9}{10}=43,560$

It remains constant in the next hour, => Blood cell count after 4 hours = $43,560$

In the next 2 hours, it increases by 5%

=> Blood cell count after 6 hours = $43,560(1+\frac{5}{100})^2=43,560(\frac{21}{20})^2$

= $43,560\times\frac{441}{400}=48,024.9\approx48,025$

=> Ans – (C)

2) Answer (A)

String of length 24 cm is bent into square, => Perimeter of square = 24 cm

Let side of square = $a$ cm

=> $4a=24$

=> $a=\frac{24}{4}=6$ cm

Let the other side of triangle be $b$ and hypotenuse be $c$ cm

=> Perimeter of triangle = $a+b+c=24$

=> $b+c=24-6=18$

=> $c=18-b$ ————(i)

Also, using Pythagoras Theorem

=> $6^2+b^2=c^2$

=> $c^2-b^2=36$ ———–(ii)

Solving equations (i) and (ii), we get : $b=8$ cm and $c=10$ cm

$\therefore$ Area of triangle = $\frac{1}{2} ab$

= $\frac{1}{2}\times6\times8=24$ $cm^2$

=> Ans – (A)

3) Answer (C)

Let edge of cube be $2a$ cm and thus diameter of sphere = $2a$ cm

=> Radius of sphere = $\frac{2a}{2}=a$ cm

Volume of cube = $(2a)^3=8a^3$ $cm^3$ ———–(i)

Volume of sphere = $\frac{4}{3}\pi r^3$

= $\frac{4}{3} \pi \times(a)^3=\frac{4a^3\pi}{3}$ $cm^3$ ———–(ii)

Dividing equation (i) by (ii), we get :

=> Required ratio = $\frac{8a^3}{\frac{4a^3\pi}{3}}$

= $\frac{8\times3}{4\pi}=\frac{6}{\pi}$

$\therefore$ Ratio of the volume of a cube to that of a sphere which will fit inside the cube = $6:\pi$

=> Ans – (C)

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4) Answer (C)

let us write the equation in the form of $a^{2}-2ab+b^{2}$
$(6x-7)^{2}$=$36x^{2}-2*6x*7+7^{2}$
But in the question it is given as -87x+52 so we should add 3x-3 to it in order to make it a perfect square.

5) Answer (A)

let us write the equation in the form of $a^{2}-2ab+b^{2}$
$(4x-5)^{2}$=$16x^{2}-2*4x*5+5^{2}$
But in the question it is given as 38x+23 so we should add -2x+2 to it in order to make it a perfect square.

6) Answer (D)

let a=6x,b=5x,b=12y,c=14y
5x=12y,x=12/5y
$a^{2}$=$36*x^{2}$
$b^{2}$=$25*x^{2}$
bc=$5x*5x*14/12$
bc=$175/6*x^{2}$
$a^{2}:b^{2}:bc$=216:150:175

7) Answer (C)

let a=7x,b=3x,b=9y,c=11y
3x=9y,x=3y,a=7x
b=3x,c=11x/3
$a^{2}$=$49*x^{2}$
$b^{2}$=$9*x^{2}$
bc=$3x*11x/3$
bc=$11*x^{2}$
$a^{2}:b^{2}:bc$=49:9:11

8) Answer (B)

let us write the equation in the form of $a^{2}-2ab+b^{2}$
$(3x-5)^{2}$=$3x^{2}-2*3x*5+5^{2}$
But in the question it is given as 42 so we should add -17 to it in order to make it a perfect square.

9) Answer (A)

Average speed=Total distance/Total time
Time=distance/speed
Let total distance=x
Time taken for one fourth distance=(x/4)/20=x/80
Time taken for half the distance=(x/2)/40=x/80
Time taken for the other one fourth=(x/4)/10=x/40
Total time=4x/80
=x/20
Average speed=(x)/(x/20)
=20 km/hr

10) Answer (B)

let his speed be ‘v’ and time be ‘t’
Therefore vt=(3v/5)*(t+40)
5t=3t+120
2t=120
t=60 min
Now vt=(6v/5)(t1)
t1=5*t/6
t1=50 min

11) Answer (B)

let the value be x
Increased by 10% it becomes 1.1x
Decreased by 10% it becomes 1.1x*0.9
=0.99x
Percentage to be increased to get the original value=(0.01x/0/0.99x)*100
=100/99 %

12) Answer (A)

In a 20 litre mixture alcohol percent =60%
Therefore 20*60/100=12 litres
Water =20-12=8 litres
Total alcohol=12+4=16 litres
Water percentage=(8/24)*100
=33.33%

13) Answer (B)

Total marks in the exam =150
5% of the total marks=5*150/100
=7.5
Given marks=45
Therefore pass marks=45+7.5
=52.5

14) Answer (D)

Given Selling Price = Rs.1200
Loss percentage = 20%
Then, Cost Price of the product = $1200 \times \dfrac{100}{100-20} = 1200 \times \dfrac{100}{80} = Rs.1500$
Required Profit percentage = 30%
Then, the required Selling Price = 130% of Rs.1500 = Rs.1950

15) Answer (B)

MP=1000
Three successive discounts of 20% each and so SP=1000*(80/100)(80/100)(80/100)
SP=512
Profit=512-500
=12

16) Answer (D)

MP=480
1.2*CP=480
CP=400
SP=MP-d
SP=480-d
$\frac{480-d-400}{400}\times100$=20
80-d=80
d=0

17) Answer (C)

Total cost price=15*10
=150
Total selling price=5*8+12*7+3*16
=40+84+48
=172
Profit percent=$\frac{172-150}{150}\times100$
=14.66%

18) Answer (D)

MP=400
SP=MP-discount
SP=400*90/100
SP=360
Cost price=360*10/13
Cost price=Rs 277

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