# IBPS RRB Clerk 2019 Prelims: Quant Test Day 10

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- Question 1 of 10
##### 1. Question

1 pointsCategory: 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 = 55Incorrect**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 pointsCategory: 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.**Correct**Answer: 4) 3400**

Explanation:

A’s investment*time of investment = (X-1000)*12

B’s investment*time of investment = (X+3000)*6

ATQ,

(Total annual profit):(A’s profit) = 7:3

{(X-1000)*12 + (X+3000)*6} / (X-1000)*12} = 7/3

{(X-1000)*2 + (X+3000)}/(X-1000)*2} = 7/3

(2X-2000+X+3000)/(2X-2000) = 7/3

(3X+1000)/(2X-2000) = 7/3

9X+3000 = 14X-14000

5X = 17000

X = 3400Incorrect**Answer: 4) 3400**

Explanation:

A’s investment*time of investment = (X-1000)*12

B’s investment*time of investment = (X+3000)*6

ATQ,

(Total annual profit):(A’s profit) = 7:3

{(X-1000)*12 + (X+3000)*6} / (X-1000)*12} = 7/3

{(X-1000)*2 + (X+3000)}/(X-1000)*2} = 7/3

(2X-2000+X+3000)/(2X-2000) = 7/3

(3X+1000)/(2X-2000) = 7/3

9X+3000 = 14X-14000

5X = 17000

X = 3400 - Question 3 of 10
##### 3. Question

1 pointsCategory: Quantitative Aptitude**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. 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 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 = 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 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 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) = 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 pointsCategory: 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 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

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 pointsCategory: 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 pointsCategory: 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 = 937.5Incorrect**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 pointsCategory: 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 pointsCategory: 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 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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