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Question 1 of 10

1. Question

1 points

Category: Quantitative Aptitude

The 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 = 8x-x = 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

Incorrect

Answer: 3) 55
Explanation:
Let the speed of boat in still water = 8x and speed of current = x
Upstream speed of the boat = 8x-x = 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 points

Category: Quantitative Aptitude

“A” invest “X-1000” 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.

Book 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. 1665

Incorrect

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 points

Category: Quantitative Aptitude

28 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 = 38

Incorrect

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 points

Category: Quantitative Aptitude

A 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) = 82x-16.4
Quantity of water left = 18x – (20-16.4) = 18x-3.6
ATQ,
Milk left – water left = 64
82x-16.4 – (18x-3.6) = 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

Incorrect

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) = 82x-16.4
Quantity of water left = 18x – (20-16.4) = 18x-3.6
ATQ,
Milk left – water left = 64
82x-16.4 – (18x-3.6) = 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 points

Category: Quantitative Aptitude

6 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
x-y = -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

Incorrect

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
x-y = -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 points

Category: Quantitative Aptitude

The 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)
L-B = 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)
L-B = 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 points

Category: Quantitative Aptitude

Rs. 1800 at x% on the compound interest for 2 years amount to Rs. 2592. Principal P at (X-5)% 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 = 2592-1800
{(2x + x^{2}/100)/100}*18 = 792
{(2x + x^{2}/100)} = 792/18
{(2x + x^{2}/100)} = 44
200x+x^{2} = 4400
x^{2}+220x-20x-4400 = 0
x^{2}+220x-20x -4400 = 0
x(x+220) – 20(x+220) = 0
(x+220)(x-20) = 0
x = 20, -220 (-ve value is not possible)
x = 20
Therefore, rate of interest = 20%
ATQ,
P + {P*(20-5)*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

Incorrect

Answer: 2) Rs. 937.5 Explanation:
{(2x + x^{2}/100)/100}*1800 = 2592-1800
{(2x + x^{2}/100)/100}*18 = 792
{(2x + x^{2}/100)} = 792/18
{(2x + x^{2}/100)} = 44
200x+x^{2} = 4400
x^{2}+220x-20x-4400 = 0
x^{2}+220x-20x -4400 = 0
x(x+220) – 20(x+220) = 0
(x+220)(x-20) = 0
x = 20, -220 (-ve value is not possible)
x = 20
Therefore, rate of interest = 20%
ATQ,
P + {P*(20-5)*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 points

Category: Quantitative Aptitude

If 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 = 35-21 = 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 = 35-21 = 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 points

Category: Quantitative Aptitude

If 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 years

Incorrect

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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