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Question 1 of 10

1. Question

1 points

Category: Quantitative Aptitude

50 items having equal weights are bought. Now 5 more items weighing 105 kg each were later bought, making average weight of all the 55 items as 95 kg. Find the weight of each of the 50 items bought earlier.

Correct

Explanation:
(50x +5*105)/55 = 95
Solve, x =94

Incorrect

Explanation:
(50x +5*105)/55 = 95
Solve, x =94

Question 2 of 10

2. Question

1 points

Category: Quantitative Aptitude

A 90 l mixture of milk and water contains them in the ratio 5 : 4 respectively. If some amount of water is added to this mixture and ratio of milk to water becomes 10 : 9 in new mixture, what will be the amount of total mixture?

Correct

Explanation:
In 90l mixture, milk = 5/9 * 90 = 50l, so water = 40 l
Let x l of water to be added
So (40+x)/50 = 9/10
Solve, x = 5
This means 5 l of water is added in 90 l of mixture, thus making it 95 l

Incorrect

Explanation:
In 90l mixture, milk = 5/9 * 90 = 50l, so water = 40 l
Let x l of water to be added
So (40+x)/50 = 9/10
Solve, x = 5
This means 5 l of water is added in 90 l of mixture, thus making it 95 l

Question 3 of 10

3. Question

1 points

Category: Quantitative Aptitude

The difference between interests obtained by investing a sum of money at compound interest for 2 years and 20% per annum compounded annually and compounded semi annually is Rs 843.5. Find the sum.

A and B are pipes such that A can empty the tank in 40 minutes and B can fill in 15 minutes. The tank is full of water and pipe A is opened. If after 10 minutes, pipe B is also opened, then in how much total time the tank will be full again?

Correct

Explanation:
Emptying pipe A is opened first for 10 minutes, so in 10 minutes the part of tank it has emptied is (1/40)*10 = 1/4
Now filling pipe is also opened, now since only 1/4 of the tank is empty so 1/4 is only to be filled by both pipes, let it take now x minutes, so
(1/15 – 1/40)*x = 1/4
Solve, x= 6
So total = 10+6 = 16 minutes [total time is asked – 10 minutes when emptying pipe was only opened, 6 minutes when both were operating.]

Incorrect

Explanation:
Emptying pipe A is opened first for 10 minutes, so in 10 minutes the part of tank it has emptied is (1/40)*10 = 1/4
Now filling pipe is also opened, now since only 1/4 of the tank is empty so 1/4 is only to be filled by both pipes, let it take now x minutes, so
(1/15 – 1/40)*x = 1/4
Solve, x= 6
So total = 10+6 = 16 minutes [total time is asked – 10 minutes when emptying pipe was only opened, 6 minutes when both were operating.]

Question 5 of 10

5. Question

1 points

Category: Quantitative Aptitude

How many 4 digit numbers can be made from the digits 3, 8, 2, 1, 0, and 4 without repetition?

Correct

Explanation:
0 cannot be on first place for it to be a 4 digit number,
So for 1st digit 5 choices, for second also 5 (because 0 can be placed here), then 4 for third place, 3 for fourth place
Total numbers = 5*5*4*3

Incorrect

Explanation:
0 cannot be on first place for it to be a 4 digit number,
So for 1st digit 5 choices, for second also 5 (because 0 can be placed here), then 4 for third place, 3 for fourth place
Total numbers = 5*5*4*3

Question 6 of 10

6. Question

1 points

Category: Quantitative Aptitude

There is a loss of 10% on an article when it is sold at 3/5 of its Marked Price. How much percent less is Cost Price than Marked Price?

Correct

Explanation:
Let MP = x
Then SP = 3/5 * x = 3x/5
Loss is 10%, so CP = [100/(100-10)] * 3x/5 = 2x/3
So required % = [(x – 2x/3)/(x)] * 100

Incorrect

Explanation:
Let MP = x
Then SP = 3/5 * x = 3x/5
Loss is 10%, so CP = [100/(100-10)] * 3x/5 = 2x/3
So required % = [(x – 2x/3)/(x)] * 100

Question 7 of 10

7. Question

1 points

Category: Quantitative Aptitude

If P denotes the perimeter and S the sum of the distances of a point within a triangle from its angular points, the

Correct

Explanation:

BO is extended to D.
In ∆ABD. AB + AD > BD
So AB + AD > OB + OD
In ∆ODC. OD + DC > OC
Adding equations
AB + AD + OD + DC > OB + OD + OC
So AB + AC > OB + OC
Similarly
BC + BA > OA + OC
and CA + CB > OA + OB
Adding all three equations
2(AB + AC +CA) > 2(OB + OC + OA)
So AB + BC + CA > OA + OB + OC

Incorrect

Explanation:

BO is extended to D.
In ∆ABD. AB + AD > BD
So AB + AD > OB + OD
In ∆ODC. OD + DC > OC
Adding equations
AB + AD + OD + DC > OB + OD + OC
So AB + AC > OB + OC
Similarly
BC + BA > OA + OC
and CA + CB > OA + OB
Adding all three equations
2(AB + AC +CA) > 2(OB + OC + OA)
So AB + BC + CA > OA + OB + OC

Question 8 of 10

8. Question

1 points

Category: Quantitative Aptitude

In right angled ∆ABC right angled at B, AB = 12 cm and BC = 9 cm. If D, E, and F are mid-points of AB, BC and CA respectively, then find the area of ∆DEF

Correct

Explanation:

Ar. of ∆DEF = ¼ ar. of ∆ABC
So Ar. of ∆DEF = ¼ * ½ * 12 * 9

Incorrect

Explanation:

Ar. of ∆DEF = ¼ ar. of ∆ABC
So Ar. of ∆DEF = ¼ * ½ * 12 * 9

Question 9 of 10

9. Question

1 points

Category: Quantitative Aptitude

If 5 sin^{2}ɸ + 3 cos^{2}ɸ = 4. Find the value of tanɸ.