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Question 1 of 10
1. Question
1 pointsCategory: Quantitative AptitudeA man takes 4 hours and 25 minutes in walking a distance and riding back to the starting place. He could walk both ways in 6 hours. The time taken by him to ride both ways is
Correct
Explanation:
A man takes ‘a’ hrs to walk to a certain distance and ride back. If walking both ways takes him b hrs, then
Riding both ways takes him [a – (ba)] hrs
So 4 hours and 25 minutes – (6 hours – 4 hours and 25 minutes)
= 4 hours and 25 minutes – 1 hr 35 minutesIncorrect
Explanation:
A man takes ‘a’ hrs to walk to a certain distance and ride back. If walking both ways takes him b hrs, then
Riding both ways takes him [a – (ba)] hrs
So 4 hours and 25 minutes – (6 hours – 4 hours and 25 minutes)
= 4 hours and 25 minutes – 1 hr 35 minutes 
Question 2 of 10
2. Question
1 pointsCategory: Quantitative AptitudeA man calculated his loss% on selling at article to be 25% on the basis of selling price. What is the loss% on the basis of cost price?
Correct
Explanation:
Shortcut:
(25/(100+25))*100 = 20%
OR
25 = [CPSP]/SP * 100
So SP = 4CP – 4SP
So SP = (4/5)*CP
Now actual loss% on CP = [CPSP]/CP * 100
Put SP = (4/5)*CP
So loss% = [CP(4CP/5)]/CP * 100 = 100/4Incorrect
Explanation:
Shortcut:
(25/(100+25))*100 = 20%
OR
25 = [CPSP]/SP * 100
So SP = 4CP – 4SP
So SP = (4/5)*CP
Now actual loss% on CP = [CPSP]/CP * 100
Put SP = (4/5)*CP
So loss% = [CP(4CP/5)]/CP * 100 = 100/4 
Question 3 of 10
3. Question
1 pointsCategory: Quantitative AptitudeA and B can complete a work in 12 days and 20 days respectively. They started work but alternately starting with B. In how many days will the work get completed this way?
Correct
Explanation:
Total work = LCM[12,20] = 60
So efficiency of A = 60/12 = 5, and efficiency of B = 60/20 = 3
So 2 days work of A and B = 5+3 = 8
Multiply by 7 both sides
So 14 days work of A and B = 56
Remaining work = 6056 = 4
Now Bs turn, he can do work =3 in 1 day
So now after 15 days, work left = 43 = 1
Now As turn, he can do 5 work in 1 day, so 1 work in 1/5 day
So total 15 1/5 daysIncorrect
Explanation:
Total work = LCM[12,20] = 60
So efficiency of A = 60/12 = 5, and efficiency of B = 60/20 = 3
So 2 days work of A and B = 5+3 = 8
Multiply by 7 both sides
So 14 days work of A and B = 56
Remaining work = 6056 = 4
Now Bs turn, he can do work =3 in 1 day
So now after 15 days, work left = 43 = 1
Now As turn, he can do 5 work in 1 day, so 1 work in 1/5 day
So total 15 1/5 days 
Question 4 of 10
4. Question
1 pointsCategory: Quantitative AptitudeThe ratio of base to corresponding altitude of a triangle is 3 : 4. If the area of the triangle is 1176 sq. cm, what is the length of the base of triangle?
Correct
Explanation:
3x and 4x
So (1/2) * 3x * 4x = 1176
Solve, x = 14 cm
So base = 3*14Incorrect
Explanation:
3x and 4x
So (1/2) * 3x * 4x = 1176
Solve, x = 14 cm
So base = 3*14 
Question 5 of 10
5. Question
1 pointsCategory: Quantitative AptitudeRs 158 is divided between A and B. A gets Rs 23 more than C who gets Rs 101 less than both A and B together. The amount that B got is
Correct
Explanation:
A+B = 158
So C = 158101 = 57
So A got = 57+23 = 80
So B got 15880Incorrect
Explanation:
A+B = 158
So C = 158101 = 57
So A got = 57+23 = 80
So B got 15880 
Question 6 of 10
6. Question
1 pointsCategory: Quantitative AptitudeThe expenditure on lightning 6 bulbs for 8 days at 6 hours each day is Rs 450. How many bulbs can be lightened for 10 days at 5 hours each day spending Rs 625?
Correct
Explanation:
6*8*6*625 = x*10*5*450
Solve, x = 8Incorrect
Explanation:
6*8*6*625 = x*10*5*450
Solve, x = 8 
Question 7 of 10
7. Question
1 pointsCategory: Quantitative AptitudeThe incomes of A and B is in the ratio 5 : 6. Their expenditures are in the ratio 5 : 8. A saved Rs 2000 and B saved Rs 1400. What is the income of B?
Correct
Explanation:
5x, 6x
5y, 8y
So 5x – 5y = 2000, gives x – y = 400
And 6x – 8y = 1400, gives 3x – 4y = 700
Solve both equations, x = 900
So income of B = 6*900Incorrect
Explanation:
5x, 6x
5y, 8y
So 5x – 5y = 2000, gives x – y = 400
And 6x – 8y = 1400, gives 3x – 4y = 700
Solve both equations, x = 900
So income of B = 6*900 
Question 8 of 10
8. Question
1 pointsCategory: Quantitative AptitudeA sum of money invested at compound interest amounts to Rs 3146 at the end of 2 years and amounts to Rs 3460.60 at the end of 3 years. The rate per cent per annum is
Correct
Explanation:
P [1 + r/100]^3 = 3460.60
And P [1 + r/100]^2 = 3146
Divide both equation
[1 + r/100] = 3460.60/3146 gives [1 + r/100] = 34606/31460
So r/100 = (34606/31460) – 1
So r/100 = 3146/31460 = 1/10
So r = 10Incorrect
Explanation:
P [1 + r/100]^3 = 3460.60
And P [1 + r/100]^2 = 3146
Divide both equation
[1 + r/100] = 3460.60/3146 gives [1 + r/100] = 34606/31460
So r/100 = (34606/31460) – 1
So r/100 = 3146/31460 = 1/10
So r = 10 
Question 9 of 10
9. Question
1 pointsCategory: Quantitative AptitudeA dishonest seller claims to sell his articles at cost price but he uses false weight of 750 gm for 1 kg. What is his gain percent?
Correct
Explanation:
(1000g – 750g)/750g * 100 = 100/3%Incorrect
Explanation:
(1000g – 750g)/750g * 100 = 100/3% 
Question 10 of 10
10. Question
1 pointsCategory: Quantitative AptitudeThe sum of digits of a two digit number is 8. The original number is 17 less than twice the number obtained by reversing the digits. What is the difference between the digits of the number?
Correct
Explanation:
Original number – 10x+y
x+y = 8
And 10x + y = 2(10y + x) – 17, so 19y – 8x = 17
Solve both equations, y = 3 and x = 5
So difference = 53Incorrect
Explanation:
Original number – 10x+y
x+y = 8
And 10x + y = 2(10y + x) – 17, so 19y – 8x = 17
Solve both equations, y = 3 and x = 5
So difference = 53
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