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From four corners of a square sheet of side 5 cm, four pieces, each in the shape of arc of a circle with radius 3 cm are cut out. The area of the remaining portion is

Correct

Explanation:

Area of all 4sectors = ᴨr^{2} = 9ᴨ
Area of square = 5*5 = 25
Area of remaining portion = 25-9ᴨ

Incorrect

Explanation:

Area of all 4sectors = ᴨr^{2} = 9ᴨ
Area of square = 5*5 = 25
Area of remaining portion = 25-9ᴨ

Question 3 of 10

3. Question

1 points

Category: Quantitative Aptitude

The value of the expression x^{4} – 19x^{3} + 19x^{2} – 19x + 20 at x = 18 is

Two pipes A and B can fill the tank in 18 and 24 minutes respectively. Both are opened and at the end of 6 minutes pipe A is closed. How much time will be taken by both pipes to fill the tank?

Correct

Explanation:
(1/18)*6 + (1/24)*x = 1
Solve, x = 16

Incorrect

Explanation:
(1/18)*6 + (1/24)*x = 1
Solve, x = 16

Question 5 of 10

5. Question

1 points

Category: Quantitative Aptitude

The metal to be used for covering a cylinder having external radius 5 cm, height 21 cm and thickness 1 cm is to be cast from a cylinder. What should be the height of the cylinder of radius 3 cm from which this casting can be done?

Correct

Explanation:
Volume of metal to be casted = (22/7)*[(5)^{2} – (4)^{2}]*21
So (22/7)*3^{2}*h = (22/7)*[(5)^{2} – (4)^{2}]*21
Solve, h = 21

Incorrect

Explanation:
Volume of metal to be casted = (22/7)*[(5)^{2} – (4)^{2}]*21
So (22/7)*3^{2}*h = (22/7)*[(5)^{2} – (4)^{2}]*21
Solve, h = 21

Question 6 of 10

6. Question

1 points

Category: Quantitative Aptitude

Distance between point P and Q is 480 km. A train starts from point P at 6:00 AM with 60 km/hr towards Q. Another train starts from point Q towards P at 7:00 AM with 80 km/kr. At what time the trains will meet?

Correct

Explanation:
When 2nd train starts i.e. at 7 AM (1 hr after 6 AM), distance covered by first train is 60 km (i.e. in 1 hr).
Now 2nd train also starts and distance between them is now (480-60) = 420 km
Both coming in opposite direction, so relative speed = (60+80) = 140 km/hr
So time = 7:00 AM + (420/140) = 7:00 AM +3 hrs = 10:00 AM

Incorrect

Explanation:
When 2nd train starts i.e. at 7 AM (1 hr after 6 AM), distance covered by first train is 60 km (i.e. in 1 hr).
Now 2nd train also starts and distance between them is now (480-60) = 420 km
Both coming in opposite direction, so relative speed = (60+80) = 140 km/hr
So time = 7:00 AM + (420/140) = 7:00 AM +3 hrs = 10:00 AM

Question 7 of 10

7. Question

1 points

Category: Quantitative Aptitude

A and B started a business by investing Rs 5000 and Rs 6000 respectively. After 4 months C came with investment of Rs 7000 and B withdrew his whole amount. Now after another 4 months, B reinvested the money as before and A withdrew half of his money. If at the end of year the total profit is Rs 22,000, how much does C get?

Correct

Explanation:
A invested 5000 for 8 months and then 2500 for 4 months. B invested 6000 for initial 4 months and then for last 4 months i.e. for 8 months. C invested 7000 for (12-4)= 8 months.
So A : B : C is
5000*8 + 2500*4 : 6000*8 : 7000*8
50*8 + 25*4 : 60*8 : 70*8
50*2 + 25 : 60*2 : 70*2
10*2 + 5 : 12*2 : 14*2
25 : 24 : 28
So C got = [28/(25+24+28)] * 22,000 = 8000

Incorrect

Explanation:
A invested 5000 for 8 months and then 2500 for 4 months. B invested 6000 for initial 4 months and then for last 4 months i.e. for 8 months. C invested 7000 for (12-4)= 8 months.
So A : B : C is
5000*8 + 2500*4 : 6000*8 : 7000*8
50*8 + 25*4 : 60*8 : 70*8
50*2 + 25 : 60*2 : 70*2
10*2 + 5 : 12*2 : 14*2
25 : 24 : 28
So C got = [28/(25+24+28)] * 22,000 = 8000

Question 8 of 10

8. Question

1 points

Category: Quantitative Aptitude

The HCF of x^{4} – 1 and x^{4} – 2x^{3} – 2x^{2} – 2x – 3 is

Correct

Explanation:
x^{4} – 1 = (x^{2} – 1) (x^{2} + 1) = (x – 1)(x+1) (x^{2} + 1)
Now x^{4} – 2x^{3} – 2x^{2} – 2x – 3
Putting x = -1 in this equation gives 0, so (x+1) is a factor, divide x^{4} – 2x^{3} – 2x^{2} – 2x – 3 by (x+1) gives x^{3} – 3x^{2} + x – 3
Now put x = 3, gives 0, so another factor is (x-3), divide (x-3) gives x^{2} + 1 which cannot be further divided
So x^{4} – 2x^{3} – 2x^{2} – 2x – 3 = (x^{2} + 1)(x+1)(x-3)
Now common factors in both expressions are (x^{2} + 1)(x+1) which is the HCF.

Incorrect

Explanation:
x^{4} – 1 = (x^{2} – 1) (x^{2} + 1) = (x – 1)(x+1) (x^{2} + 1)
Now x^{4} – 2x^{3} – 2x^{2} – 2x – 3
Putting x = -1 in this equation gives 0, so (x+1) is a factor, divide x^{4} – 2x^{3} – 2x^{2} – 2x – 3 by (x+1) gives x^{3} – 3x^{2} + x – 3
Now put x = 3, gives 0, so another factor is (x-3), divide (x-3) gives x^{2} + 1 which cannot be further divided
So x^{4} – 2x^{3} – 2x^{2} – 2x – 3 = (x^{2} + 1)(x+1)(x-3)
Now common factors in both expressions are (x^{2} + 1)(x+1) which is the HCF.

Question 9 of 10

9. Question

1 points

Category: Quantitative Aptitude

Two medians AD and BE of ∆ABC intersect at O at right angles. If AD = 27 cm and BE = 18 cm, then the length of BD in cm is

Correct

Explanation:

GD = (1/3)*AD = (1/3)*27 = 9 cm
BG = (2/3)*BE = (2/3)*18 = 12 cm
In rt angled ∆BOD
So BD = √(9^2 + 12^2) = 15

Incorrect

Explanation:

GD = (1/3)*AD = (1/3)*27 = 9 cm
BG = (2/3)*BE = (2/3)*18 = 12 cm
In rt angled ∆BOD
So BD = √(9^2 + 12^2) = 15

Question 10 of 10

10. Question

1 points

Category: Quantitative Aptitude

A can contains 80 litres of milk. 8 litres of this milk is taken out and replaced with water. This process is repeated again. Find the amount of remaining milk in the mixture?