Welcome to Online Quant Test in AffairsCloud.com. We are starting SSC CGL 2016 Course and we are creating sample questions in Quant section, type of which will be asked in SSC CGL 2016 !!!

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Question 1 of 10

1. Question

1 points

Category: Quantitative Aptitude

Two parallel chords of a circle of diameter 10 cm lying on the opposite sides of the centre are of lengths 6 cm and 10 cm. the distance between the chords is

Correct

Explanation:

OA = OC = 5 cm
AB = 6 cm, so AP = PB = 3 cm
CD = 8 cm, so CQ = QD = 4 cm
In ∆OCQ,
OQ = √(5^{2} – 4^{2}) = 3
In ∆OAP,
OP = √(5^{2} – 3^{2}) = 4
So PQ = OP+OQ = 4+3

Incorrect

Explanation:

OA = OC = 5 cm
AB = 6 cm, so AP = PB = 3 cm
CD = 8 cm, so CQ = QD = 4 cm
In ∆OCQ,
OQ = √(5^{2} – 4^{2}) = 3
In ∆OAP,
OP = √(5^{2} – 3^{2}) = 4
So PQ = OP+OQ = 4+3

Question 2 of 10

2. Question

1 points

Category: Quantitative Aptitude

Evaluate 3 cos 70 cosec 20 + 5 cos 46 cosec 44

Correct

Explanation:
3 cos 70 cosec 20 + 5 cos 46 cosec 44
= 3 cos (90-20) cosec 20 + 5 cos (90-44) cosec 44
= 3 sin 20 cosec 20 + 5 sin 44 cosec 44
= 3 + 5

Incorrect

Explanation:
3 cos 70 cosec 20 + 5 cos 46 cosec 44
= 3 cos (90-20) cosec 20 + 5 cos (90-44) cosec 44
= 3 sin 20 cosec 20 + 5 sin 44 cosec 44
= 3 + 5

If angles of elevation of a tower from two consecutive kilometer-stones along the road are 30 degree and 60 degree respectively, then the height of tower above the ground will be?

Correct

Explanation:

See the figure
tan 60 = h/BC so BC = h/√3
and tan 30 = h/(BC+1), so BC = √3 h – 1
so h/√3 = √3 h – 1
solve, h = √3/2

Incorrect

Explanation:

See the figure
tan 60 = h/BC so BC = h/√3
and tan 30 = h/(BC+1), so BC = √3 h – 1
so h/√3 = √3 h – 1
solve, h = √3/2

Question 5 of 10

5. Question

1 points

Category: Quantitative Aptitude

A price of an article is first decreased by 10% and then increased by 25%. If the final price of the article becomes Rs 360, what was the original price?

Correct

Explanation:
x * (90/100) * (125/100) = 360
solve, x = 320

Incorrect

Explanation:
x * (90/100) * (125/100) = 360
solve, x = 320

Question 6 of 10

6. Question

1 points

Category: Quantitative Aptitude

Directions (6-9): Study the bar graph below and answer the following questions.
The graph shows the number of students and their evaluation for 3 years.

The percentage passed in 1st division in 2013 was

Correct

Explanation:
Total students in 2013 = 170
Students passed in 1st division = 20
So required % = (20/170) * 100

Incorrect

Explanation:
Total students in 2013 = 170
Students passed in 1st division = 20
So required % = (20/170) * 100

Question 7 of 10

7. Question

1 points

Category: Quantitative Aptitude

Directions (6-9): Study the bar graph below and answer the following questions.
The graph shows the number of students and their evaluation for 3 years.

In which year the school had the best result in respect of percentage of pass candidates?

Correct

Explanation:
% of passed candidates:
Year 2013: 82 6/11%
Year 2014: (140/190)*100 = 73.7%
Year 2015: (150/200)8100 = 75%

Incorrect

Explanation:
% of passed candidates:
Year 2013: 82 6/11%
Year 2014: (140/190)*100 = 73.7%
Year 2015: (150/200)8100 = 75%

Question 8 of 10

8. Question

1 points

Category: Quantitative Aptitude

Directions (6-9): Study the bar graph below and answer the following questions.
The graph shows the number of students and their evaluation for 3 years.

The number of students passed in third division in year 2014 was

Correct

Explanation:
Students passed in third division in 2014 = 190 – 140

Incorrect

Explanation:
Students passed in third division in 2014 = 190 – 140

Question 9 of 10

9. Question

1 points

Category: Quantitative Aptitude

Directions (6-9): Study the bar graph below and answer the following questions.
The graph shows the number of students and their evaluation for 3 years.

The pass percentage in 2013 was

Correct

Explanation:
Total passed students in 2013 is 140
So required% = (140/170)*100

Incorrect

Explanation:
Total passed students in 2013 is 140
So required% = (140/170)*100

Question 10 of 10

10. Question

1 points

Category: Quantitative Aptitude

The elimination of ɸ from x cosɸ – y sinɸ = 3 and x sinɸ + y sinɸ = 4 will give