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Question 1 of 10
1. Question
1 pointsCategory: Quantitative Aptitude24% of a number added to 16% of 450 gives 156
Correct
Explanation:
24/100 * x + 16/100 * 450 = 156Incorrect
Explanation:
24/100 * x + 16/100 * 450 = 156 
Question 2 of 10
2. Question
1 pointsCategory: Quantitative AptitudeA and B started a business with Rs 5,000 and Rs 6,000 respectively. After 9 months C also joined with Rs 12000. At the same time A and B withdrew Rs 1,000 each. If at the end of year B and C together received Rs 12,250 as their share from the total profit, then what is the total profit?
Correct
Explanation:
5000*9 + 4000*3 : 6000*9 + 5000*3 : 12000*3
19 : 23 : 12
So [(23+12)/54]*x = 12,250
Solve, x = 18,900Incorrect
Explanation:
5000*9 + 4000*3 : 6000*9 + 5000*3 : 12000*3
19 : 23 : 12
So [(23+12)/54]*x = 12,250
Solve, x = 18,900 
Question 3 of 10
3. Question
1 pointsCategory: Quantitative AptitudeThe CP and SP of article are increased by Rs 30 and Rs 12 respectively, then the profit% becomes 33 1/3%. If the actual profit % was 40%, what is the cost price of the article?
Correct
Explanation:
CP = x, then SP = (140/100)*x = 14x/10
New CP = (x+30), new SP = (14x/10 + 12), new profit% = 33 1/3 = 100/3%
So (14x/10 + 12) = [(100/3 + 100)/100] * (x+30)
Solve, x = 420Incorrect
Explanation:
CP = x, then SP = (140/100)*x = 14x/10
New CP = (x+30), new SP = (14x/10 + 12), new profit% = 33 1/3 = 100/3%
So (14x/10 + 12) = [(100/3 + 100)/100] * (x+30)
Solve, x = 420 
Question 4 of 10
4. Question
1 pointsCategory: Quantitative AptitudeA person lent out certain sum on simple interest and the same sum on compound interest at a certain rate of interest per annum. He noticed that the ratio between the difference of compound interest and simple interest of 3 years and 2 years is 47 : 15. The rate of interest per annum is
Correct
Explanation:
For 2 years, diff in CI and SI = Pr^{2}/100^{2}
For 3 years diff is Pr^{2}(r+300)/100^{3}
Pr^{2}/100^{2} / Pr^{2}(r+300)/100^{3}
= 15/47
So 100/(r+300) = 15/47
Solve, r = 13 1/3Incorrect
Explanation:
For 2 years, diff in CI and SI = Pr^{2}/100^{2}
For 3 years diff is Pr^{2}(r+300)/100^{3}
Pr^{2}/100^{2} / Pr^{2}(r+300)/100^{3}
= 15/47
So 100/(r+300) = 15/47
Solve, r = 13 1/3 
Question 5 of 10
5. Question
1 pointsCategory: Quantitative AptitudeThere are two trains of lengths 500 m and 550 m respectively. Both start at same time. If they travel in same direction they cross each other in 50 seconds and if they travel in opposite direction towards each other, they cross each other in 10 seconds. What is the speed of the faster train?
Correct
Explanation:
Let speed of 1st train = x m/s, of 2nd train = y m/s
Then when they travel in same direction, relative speed = xy
So (500+550) = (xy)*50. This gives x –y = 21
When they travel in opposite direction, relative speed = x+y
So (500+550) = (x+y)*10. This gives x + y = 105
Solve both equations, x = 63Incorrect
Explanation:
Let speed of 1st train = x m/s, of 2nd train = y m/s
Then when they travel in same direction, relative speed = xy
So (500+550) = (xy)*50. This gives x –y = 21
When they travel in opposite direction, relative speed = x+y
So (500+550) = (x+y)*10. This gives x + y = 105
Solve both equations, x = 63 
Question 6 of 10
6. Question
1 pointsCategory: Quantitative AptitudeA alone can do a piece of work in 60 days while A and B together can complete the same work in 20 days. What will be the number of day in which both will complete the work if it is given that B worked for only half a day daily till the end?
Correct
Explanation:
B’s 1 day’s work = 1/20 – 1/60 = 1/30
Now in 1day B completes 1/30 of the work, so in half day he will complete 1/2 * 1/30 = 1/60 of the work
Now A’s 1 day’s work is 1/60 and B’s 1/60
So their total 1 day’s work = 1/60 + 1/60 = 1/30Incorrect
Explanation:
B’s 1 day’s work = 1/20 – 1/60 = 1/30
Now in 1day B completes 1/30 of the work, so in half day he will complete 1/2 * 1/30 = 1/60 of the work
Now A’s 1 day’s work is 1/60 and B’s 1/60
So their total 1 day’s work = 1/60 + 1/60 = 1/30 
Question 7 of 10
7. Question
1 pointsCategory: Quantitative AptitudeA solution contains milk and water in the ratio 5 : 2. In 1400 gms of solution, 100 gms of water is added. Find the percentage of water in the solution thus formed?
Correct
Explanation:
In 1400 gm, water is 2/7 * 1400 = 400 gm
Now 100 gm of water added, so water become (400+100) = 500gm
Total solution becomes (1400+100) = 1500 gm
Required % = (500/1500) * 100Incorrect
Explanation:
In 1400 gm, water is 2/7 * 1400 = 400 gm
Now 100 gm of water added, so water become (400+100) = 500gm
Total solution becomes (1400+100) = 1500 gm
Required % = (500/1500) * 100 
Question 8 of 10
8. Question
1 pointsCategory: Quantitative AptitudeThe rectangle whose sides are 10 cm and 6cm has perimeter equal to half of the perimeter of a square. If the diameter of a circle is equal to the side of the square, find the approximate circumference of the semicircle?
Correct
Explanation:
Perimeter if square = 2 * peri. of rect = 2 * 2(10+6) = 64
So side of square = 64/4 = 16
So diameter of circle = 16, radius = 16/2 = 8
So circumference of circle = ᴨr + d = (22/7)*(8) + 16Incorrect
Explanation:
Perimeter if square = 2 * peri. of rect = 2 * 2(10+6) = 64
So side of square = 64/4 = 16
So diameter of circle = 16, radius = 16/2 = 8
So circumference of circle = ᴨr + d = (22/7)*(8) + 16 
Question 9 of 10
9. Question
1 pointsCategory: Quantitative AptitudeWhat is the HCF of 2/5, 4/25, 6/11, 10/22?
Correct
Explanation:
HCF will be HCF of numerators/LCM of denominators
So HCF = HCF of (2,4,6,10)/(LCM of 5,25, 11, 22)
= 2/550Incorrect
Explanation:
HCF will be HCF of numerators/LCM of denominators
So HCF = HCF of (2,4,6,10)/(LCM of 5,25, 11, 22)
= 2/550 
Question 10 of 10
10. Question
1 pointsCategory: Quantitative AptitudeB is 3 years older than C and A is thrice older than B. If the ratio of ages of A and C is 4 : 1, what will be total of the ages of A and C, 4 years hence?
Correct
Explanation:
Let C’s present age = x, then B’s = x+3, and A’s = 3(x+3) = 3x+9
Now (3x+9)/x = 4/1
Solve, x = 9
A’s present age = 3x+9 = 36
C’s present age = x = 9
4 years after (A+C)’s age = 36+9+4+4Incorrect
Explanation:
Let C’s present age = x, then B’s = x+3, and A’s = 3(x+3) = 3x+9
Now (3x+9)/x = 4/1
Solve, x = 9
A’s present age = 3x+9 = 36
C’s present age = x = 9
4 years after (A+C)’s age = 36+9+4+4
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