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Question 1 of 10

1. Question

1 points

Category: Quantitative Aptitude

Upstream speed of boat E is 9 km/hr. How many more hours will it take to cover a distance of 378 km upstream than same distance downstream?

Correct

Solution: Upstream speed = 9, speed of boat = 15 km/hr, so speed of stream = 15-9 = 6 km/hr, Downstream speed = 15+ 6 = 21 km/hr So, the required time = 378/9(upstream) – 378/21(downstream) = 24 hours

Incorrect

Solution: Upstream speed = 9, speed of boat = 15 km/hr, so speed of stream = 15-9 = 6 km/hr, Downstream speed = 15+ 6 = 21 km/hr So, the required time = 378/9(upstream) – 378/21(downstream) = 24 hours

Question 2 of 10

2. Question

1 points

Category: Quantitative Aptitude

On a particular day, ratio of upstream speed to downstream speed of boat D is 3 : 7. It took 20 hours more to cover a distance upstream than same distance downstream by boat D. On that particular day, boat B covered same distance in how much time?

Correct

Solution: Let speed of stream on that day = x km/hr
So (20-x)/(20+x) = 3/7
Solve, x = 8 km/hr
So y/(20-8) – y/(20+8) = 20
Solve, y = 420 km
So required time = 420/(16+8) = 420/24 = 17.5 hours

Incorrect

Solution: Let speed of stream on that day = x km/hr
So (20-x)/(20+x) = 3/7
Solve, x = 8 km/hr
So y/(20-8) – y/(20+8) = 20
Solve, y = 420 km
So required time = 420/(16+8) = 420/24 = 17.5 hours

Question 3 of 10

3. Question

1 points

Category: Quantitative Aptitude

For boat C, its upstream speed is 6 km/hr on a particular day. Find the difference in time in covering 360 km by boats A and C on that particular day.

Correct

Solution: Upstream speed of C = 6 km/hr, so speed of stream: 9 – x = 6, x = 3 km/hr
Downstream speed of boat C = 9+3 = 12 km/hr
Downstream speed of boat A = 12+3 = 15 km/hr
So difference in timings = 360/12 – 360/15 = 30 – 24 = 6 hours

Incorrect

Solution: Upstream speed of C = 6 km/hr, so speed of stream: 9 – x = 6, x = 3 km/hr
Downstream speed of boat C = 9+3 = 12 km/hr
Downstream speed of boat A = 12+3 = 15 km/hr
So difference in timings = 360/12 – 360/15 = 30 – 24 = 6 hours

Question 4 of 10

4. Question

1 points

Category: Quantitative Aptitude

Boat B covered 756 km downstream on Monday for which it took 6 hours less than that in which it covered half distance upstream on Tuesday. On Tuesday, speed of stream was 2 km/hr more than that on Monday. Find the downstream speed of boat B on Tuesday.

Correct

Solution: Let on Monday, speed of stream is x km/hr, then on Tuesday it is (x+2) km/hr, On Monday it covered 756 km, so on Tuesday it covered 756/2 = 378 km
So 756/(16+x) = 378/(16-(x+2)) – 6 ………………(1)
126/(16+x) = 63/(14-x) – 1
OR Solve with options in
378/[16 -(y+2 )] – 756/(16+y )=6
63 [1/(14-y) – 2/(16+y) ] = 1
Solve, x = 5 km/hr, So speed of stream on Tuesday = (5+2) = 7 km/hr
So downstream speed of boat B = (16+7) = 23 km/hr

Incorrect

Solution: Let on Monday, speed of stream is x km/hr, then on Tuesday it is (x+2) km/hr, On Monday it covered 756 km, so on Tuesday it covered 756/2 = 378 km
So 756/(16+x) = 378/(16-(x+2)) – 6 ………………(1)
126/(16+x) = 63/(14-x) – 1
OR Solve with options in
378/[16 -(y+2 )] – 756/(16+y )=6
63 [1/(14-y) – 2/(16+y) ] = 1
Solve, x = 5 km/hr, So speed of stream on Tuesday = (5+2) = 7 km/hr
So downstream speed of boat B = (16+7) = 23 km/hr

Question 5 of 10

5. Question

1 points

Category: Quantitative Aptitude

On a particular day, the time taken by boat A to cover 255 km downstream is 2 hours more than the time taken by boat D to cover 195 km upstream. Find the ratio of upstream speed to downstream speed in case of boat A on that particular day.

Correct

Solution: Let x km/hr is the speed of stream on that particular day.
So 255/(12+x) = 195/(20-x) + 2
Solve, x = 5 km/hr
So in case of boat A, ratio of speed upstream : downstream = (12-5) : (12+5) = 7 : 17

Incorrect

Solution: Let x km/hr is the speed of stream on that particular day.
So 255/(12+x) = 195/(20-x) + 2
Solve, x = 5 km/hr
So in case of boat A, ratio of speed upstream : downstream = (12-5) : (12+5) = 7 : 17

Question 6 of 10

6. Question

1 points

Category: Quantitative Aptitude

Directions for Question 6-8: The questions consist of two statements numbered “I, II and III” given below it. You have to decide whether the data provided in the statement are sufficient to answer the questions?

How much did Sohil get as profit at the year-end in the business done by Animesh, Sohil and Akhilesh?
Statement I: Animesh and Sohil invested for one year and in the proportion 1:2 respectively.
Statement II: Akhilesh invested Rs.8000/- for nine months, his profit was 3/2 times that of Sohil’s and his investment was 4 times that of Animesh.
Statement III: The three together got Rs.1000/- as profit at the year end.

Correct

Solution: From statement one we get Animesh and Sohail investment ratio for 12 months as 1:2, Now from II we get Animesh’s investment = 8000/4 = 2000, Now we know about Sohil’s investment = 2*2000. Therefore ussin the above two statements we get the exact ratio of investment of Animesh, Sohil and Akhilesh = 2000*12 : 4000*12 : 8000*9 = 1:2:3. Therefore using statement III we can find the exact profit of Sohil i.e 2/6*1000 =333.33. Hence all the statements are required.

Incorrect

Solution: From statement one we get Animesh and Sohail investment ratio for 12 months as 1:2, Now from II we get Animesh’s investment = 8000/4 = 2000, Now we know about Sohil’s investment = 2*2000. Therefore ussin the above two statements we get the exact ratio of investment of Animesh, Sohil and Akhilesh = 2000*12 : 4000*12 : 8000*9 = 1:2:3. Therefore using statement III we can find the exact profit of Sohil i.e 2/6*1000 =333.33. Hence all the statements are required.

Question 7 of 10

7. Question

1 points

Category: Quantitative Aptitude

Directions for Question 6-8: The questions consist of two statements numbered “I, II and III” given below it. You have to decide whether the data provided in the statement are sufficient to answer the questions?

What is the cost of flooring the rectangular hall?
Statement I: Length of the rectangle is 6m
Statement II: Breadth of the rectangle is two-third of its length
Statement III: Cost of flooring the area of 100cm^2 is Rs.45/-

Correct

Solution: Statement I gives the length, II gives Breadth and III gives cost. Hence, all are required to answer the question.

Incorrect

Solution: Statement I gives the length, II gives Breadth and III gives cost. Hence, all are required to answer the question.

Question 8 of 10

8. Question

1 points

Category: Quantitative Aptitude

Directions for Question 6-8: The questions consist of two statements numbered “I, II and III” given below it. You have to decide whether the data provided in the statement are sufficient to answer the questions?

What is the cost of 7 pens?
Statement I: The cost of 4 pens and 3 notebooks is Rs.68
Statement II: The cost of one notebook, 3 staplers and 5 pens is Rs.147
Statement III: The cost of 1 pen and 2 notebooks is Rs.32

Correct

Solution: Using I and III, solving simultaneously we can solve for 7 pens.

Incorrect

Solution: Using I and III, solving simultaneously we can solve for 7 pens.

Question 9 of 10

9. Question

1 points

Category: Quantitative Aptitude

Directions for Questions 9-10: Refer to the information below to answer the following questions.
A batch of 50 transistors contains 3 defective ones. Two transistors are selected at random from the batch and put into a radio set.

What is the probability that both the transistors selected are defective?

Correct

Solution: As there are 3 defective out of 50 in first pick and for second pick only two are defective out of 49 as one defective has already been picked up. Hence, the probability of both being defective is: 3/50 * 2/49 = 3/1225.

Incorrect

Solution: As there are 3 defective out of 50 in first pick and for second pick only two are defective out of 49 as one defective has already been picked up. Hence, the probability of both being defective is: 3/50 * 2/49 = 3/1225.

Question 10 of 10

10. Question

1 points

Category: Quantitative Aptitude

Directions for Questions 9-10: Refer to the information below to answer the following questions.
A batch of 50 transistors contains 3 defective ones. Two transistors are selected at random from the batch and put into a radio set.

What is the probability that only one is defective?

Correct

Solution: For this, two cases are possible, either one the first pick is defective or in the second pick is defective. Hence probability for first pick to be defective = (3/50 * 47/49) and for second pick in which second one is defective, probability is (47/50 * 3/49) Since either of the case is possible so probability = (3/50 * 47/49) + (47/50 * 3/49) = (141/1225)

Incorrect

Solution: For this, two cases are possible, either one the first pick is defective or in the second pick is defective. Hence probability for first pick to be defective = (3/50 * 47/49) and for second pick in which second one is defective, probability is (47/50 * 3/49) Since either of the case is possible so probability = (3/50 * 47/49) + (47/50 * 3/49) = (141/1225)

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