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Question 1 of 10
1. Question
1 pointsCategory: Quantitative AptitudeThe ratio of the speed of a boat and current is 8:1. The ratio of time taken for distance “D+5” downstream and “D+15” upstream is 2:3. Find the distance D.
Correct
Answer: 3) 55
Explanation:
Let the speed of boat in still water = 8x and speed of current = x
Upstream speed of the boat = 8xx = 7x
And downstream speed of the boat = 8x+x = 9x
Time taken in travelling D+5 distance downstream = (D+5)/9x
And time taken in travelling D+15 distance upstream = (D+15)/7x
ATQ,
{(D+5)/9x)} : {(D+15)/7x} = 2:3
{(D+5)*7)} : {(D+15)*9} = 2/3
(7D+35)/(9D+135) = 2/3
21D+105 = 18D + 270
D = 55Incorrect
Answer: 3) 55
Explanation:
Let the speed of boat in still water = 8x and speed of current = x
Upstream speed of the boat = 8xx = 7x
And downstream speed of the boat = 8x+x = 9x
Time taken in travelling D+5 distance downstream = (D+5)/9x
And time taken in travelling D+15 distance upstream = (D+15)/7x
ATQ,
{(D+5)/9x)} : {(D+15)/7x} = 2:3
{(D+5)*7)} : {(D+15)*9} = 2/3
(7D+35)/(9D+135) = 2/3
21D+105 = 18D + 270
D = 55 
Question 2 of 10
2. Question
1 pointsCategory: Quantitative Aptitude“A” invest “X1000” and after 6 months B invest “X+3000”. The ratio of total annual profit and A’s profit is 7:3. Find the value of X.
Correct
Answer: 4) 3400
Explanation:
A’s investment*time of investment = (X1000)*12
B’s investment*time of investment = (X+3000)*6
ATQ,
(Total annual profit):(A’s profit) = 7:3
{(X1000)*12 + (X+3000)*6} / (X1000)*12} = 7/3
{(X1000)*2 + (X+3000)}/(X1000)*2} = 7/3
(2X2000+X+3000)/(2X2000) = 7/3
(3X+1000)/(2X2000) = 7/3
9X+3000 = 14X14000
5X = 17000
X = 3400Incorrect
Answer: 4) 3400
Explanation:
A’s investment*time of investment = (X1000)*12
B’s investment*time of investment = (X+3000)*6
ATQ,
(Total annual profit):(A’s profit) = 7:3
{(X1000)*12 + (X+3000)*6} / (X1000)*12} = 7/3
{(X1000)*2 + (X+3000)}/(X1000)*2} = 7/3
(2X2000+X+3000)/(2X2000) = 7/3
(3X+1000)/(2X2000) = 7/3
9X+3000 = 14X14000
5X = 17000
X = 3400 
Question 3 of 10
3. Question
1 pointsCategory: Quantitative AptitudeBook A is sold at 7% profit and Book B is sold at Rs. 165 more than the cost price of book A. If both the books have same cost price and the overall profit is 9%, find the selling price of book B.
Correct
Answer: 3) Rs. 1665
Explanation:
Let the cost price of book A = cost price of book B = 100x
Total cost price of both the books = 200x
Total selling price of both the books = {(100+9)/100}*200x = 218x
Selling price of book A = {(100+7)/100}*100 = 107x
Selling price of book B = 100x + 165
Therefore,
107x + 100x + 165 = 218x
207x + 165 = 218x
11x = 165
x = 15
Therefore, selling price of book B = 100x+165 = 100*15 + 165 = 1500+165 = Rs. 1665Incorrect
Answer: 3) Rs. 1665
Explanation:
Let the cost price of book A = cost price of book B = 100x
Total cost price of both the books = 200x
Total selling price of both the books = {(100+9)/100}*200x = 218x
Selling price of book A = {(100+7)/100}*100 = 107x
Selling price of book B = 100x + 165
Therefore,
107x + 100x + 165 = 218x
207x + 165 = 218x
11x = 165
x = 15
Therefore, selling price of book B = 100x+165 = 100*15 + 165 = 1500+165 = Rs. 1665 
Question 4 of 10
4. Question
1 pointsCategory: Quantitative Aptitude28 men can complete a work in 27 days, 12 men start work and worked for x days after that 8 men joined them and completed the remaining work in 15 days. Find the value of x.
Correct
Answer: 5) 38
Explanation:
Let the total work = 28*27 = 756 units
Work done by 12 men in x days = 12x units
And work done by (12+8) men in 15 days = 20*15 = 300 units
Since, the work is completed
12x + 300 = 756
12x = 456
x = 456/12
x = 38Incorrect
Answer: 5) 38
Explanation:
Let the total work = 28*27 = 756 units
Work done by 12 men in x days = 12x units
And work done by (12+8) men in 15 days = 20*15 = 300 units
Since, the work is completed
12x + 300 = 756
12x = 456
x = 456/12
x = 38 
Question 5 of 10
5. Question
1 pointsCategory: Quantitative AptitudeA mixture has 82% milk and 18% water, 20 litres of the mixture is removed, then milk becomes 64 litres more than water in the remaining mixture. Find the initial quantity of the mixture.
Correct
Answer: 3) 120 litres
Explanation:
Let the initial quantity of mixture = 100x
Quantity of milk = 82x and the quantity of water = 18x
When 20 litre of mixture is removed,
Quantity of milk left = 82x – 20*(82/100) = 82x16.4
Quantity of water left = 18x – (2016.4) = 18x3.6
ATQ,
Milk left – water left = 64
82x16.4 – (18x3.6) = 64
82x – 16.4 – 18x + 3.6 = 64
64x 12.8 = 64
64x = 64+12.8
64x = 76.8
x = 76.8/64 = 1.2
Initial quantity of mixture = 100x = 100*1.2 = 120 litreIncorrect
Answer: 3) 120 litres
Explanation:
Let the initial quantity of mixture = 100x
Quantity of milk = 82x and the quantity of water = 18x
When 20 litre of mixture is removed,
Quantity of milk left = 82x – 20*(82/100) = 82x16.4
Quantity of water left = 18x – (2016.4) = 18x3.6
ATQ,
Milk left – water left = 64
82x16.4 – (18x3.6) = 64
82x – 16.4 – 18x + 3.6 = 64
64x 12.8 = 64
64x = 64+12.8
64x = 76.8
x = 76.8/64 = 1.2
Initial quantity of mixture = 100x = 100*1.2 = 120 litre 
Question 6 of 10
6. Question
1 pointsCategory: Quantitative Aptitude6 years ago, the sum of the ages of A and B was 75 years. The age of A after 12 years and age of B after 9 years will be the same. Then what is the age of B, 10 years from now?
Correct
Answer: 4) 55 years
Explanation:
Given, 6 years ago, the sum of ages of A and B = 75
Sum of present age of A and B will be 75+12 = 87 years
Let the present age of A is x and present age of B is y
x + y = 87 (equation 1)
also,
x +12 y+9
xy = 3 (equation 2)
From equation 1 and equation 2, we get
Present age of B = y = 45 years
Therefore, age of B after 10 years will be 45 +10 = 55 yearsIncorrect
Answer: 4) 55 years
Explanation:
Given, 6 years ago, the sum of ages of A and B = 75
Sum of present age of A and B will be 75+12 = 87 years
Let the present age of A is x and present age of B is y
x + y = 87 (equation 1)
also,
x +12 y+9
xy = 3 (equation 2)
From equation 1 and equation 2, we get
Present age of B = y = 45 years
Therefore, age of B after 10 years will be 45 +10 = 55 years 
Question 7 of 10
7. Question
1 pointsCategory: Quantitative AptitudeThe perimeter of rectangle A is 84m, where the difference of length and breadth is 8m. Area of rectangle B is 20% more than the area of rectangle A. Find the area of rectangle B.
Correct
Answer: 3) 510m^{2}
Explanation:
Perimeter of rectangle A
2(L+B) = 84
L+B = 42 (equation 1)
LB = 8
From equation 1 and equation 2,
L = 25 and B = 17
Area of rectangle A = 25*17 = 425m^{2}
Area of rectangle B = (120/100)*425 = (6/5)*425 = 510m^{2}Incorrect
Answer: 3) 510m^{2}
Explanation:
Perimeter of rectangle A
2(L+B) = 84
L+B = 42 (equation 1)
LB = 8
From equation 1 and equation 2,
L = 25 and B = 17
Area of rectangle A = 25*17 = 425m^{2}
Area of rectangle B = (120/100)*425 = (6/5)*425 = 510m^{2} 
Question 8 of 10
8. Question
1 pointsCategory: Quantitative AptitudeRs. 1800 at x% on the compound interest for 2 years amount to Rs. 2592. Principal P at (X5)% on the simple interest for 4 years amount to Rs. 1500. Find P.
Correct
Answer: 2) Rs. 937.5
Explanation:
{(2x + x^{2}/100)/100}*1800 = 25921800
{(2x + x^{2}/100)/100}*18 = 792
{(2x + x^{2}/100)} = 792/18
{(2x + x^{2}/100)} = 44
200x+x^{2} = 4400
x^{2}+220x20x4400 = 0
x^{2}+220x20x 4400 = 0
x(x+220) – 20(x+220) = 0
(x+220)(x20) = 0
x = 20, 220 (ve value is not possible)
x = 20
Therefore, rate of interest = 20%
ATQ,
P + {P*(205)*4}/100 = 1500
P + (P*15*4)/100 = 1500
P + (P*60)/100 = 1500
P + 3P/5 = 1500
8P/5 = 1500
P = 7500/8
P = 937.5Incorrect
Answer: 2) Rs. 937.5
Explanation:
{(2x + x^{2}/100)/100}*1800 = 25921800
{(2x + x^{2}/100)/100}*18 = 792
{(2x + x^{2}/100)} = 792/18
{(2x + x^{2}/100)} = 44
200x+x^{2} = 4400
x^{2}+220x20x4400 = 0
x^{2}+220x20x 4400 = 0
x(x+220) – 20(x+220) = 0
(x+220)(x20) = 0
x = 20, 220 (ve value is not possible)
x = 20
Therefore, rate of interest = 20%
ATQ,
P + {P*(205)*4}/100 = 1500
P + (P*15*4)/100 = 1500
P + (P*60)/100 = 1500
P + 3P/5 = 1500
8P/5 = 1500
P = 7500/8
P = 937.5 
Question 9 of 10
9. Question
1 pointsCategory: Quantitative AptitudeIf the perimeter of a rectangle A is 112m and the ratio of its length and breadth is 5:3. Find the area of the square whose side is 150% more than the difference of length and breadth of the rectangle.
Correct
Answer: 3) 1225m^{2}
Explanation:
Given,
2*(L+B) = 112
L+B = 56 (equation 1)
L/B = 5/3 (equation 2)
From equation 1 and equation 2, we get
Length = 35m and breadth = 21m
Difference of length and breadth = 3521 = 14
Side of square = {(100+150)/100}*14 = (250/100)*14 = (5/2)*14 = 35
Area of square = 35*35 = 1225m^{2}Incorrect
Answer: 3) 1225m^{2}
Explanation:
Given,
2*(L+B) = 112
L+B = 56 (equation 1)
L/B = 5/3 (equation 2)
From equation 1 and equation 2, we get
Length = 35m and breadth = 21m
Difference of length and breadth = 3521 = 14
Side of square = {(100+150)/100}*14 = (250/100)*14 = (5/2)*14 = 35
Area of square = 35*35 = 1225m^{2} 
Question 10 of 10
10. Question
1 pointsCategory: Quantitative AptitudeIf the average age of 3 persons is 45 years and the sum of age of 1st and 2nd person is 85 years and the sum of age of 3rd person and 2nd person is 100 years. Find the sum of age of 1st and 3rd person.
Correct
Answer: 2) 85 years
Explanation:
Given, the average age of 3 person = 45 years
Therefore, sum of age of three person = 3*45 = 135
1st + 2nd + 3rd = 135 (equation 1)
1st + 2nd = 85 (equation 2)
From, equation 1 and equation 2, we get
3rd = 50 years
Also,
2nd and 3rd = 100 (equation 3)
From equation 1 and equation 3,
1st = 35
Required sum
1st + 3rd = 35+50 = 85 yearsIncorrect
Answer: 2) 85 years
Explanation:
Given, the average age of 3 person = 45 years
Therefore, sum of age of three person = 3*45 = 135
1st + 2nd + 3rd = 135 (equation 1)
1st + 2nd = 85 (equation 2)
From, equation 1 and equation 2, we get
3rd = 50 years
Also,
2nd and 3rd = 100 (equation 3)
From equation 1 and equation 3,
1st = 35
Required sum
1st + 3rd = 35+50 = 85 years
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