Problems on boats and streams

1. If a man can swim downstream at 6kmph and upstream at 2km ph his speed in still water is : ans :4 km/hr

2. A man can row upstream at 8kmph and downstream at 12kmph the speed of the stream is Ans:2.5km/hr

3. If anshul rows 15km upstream and 21km downstream taking 3 hours each time, then the speed of the stream is Ans:1km/hr

4. A man rows 750m in 675 seconds against the stream and returns in 7 ½ minutes. How rowing speed in still water is Ans:5km/hr

5. A man rows 13km upstream in 5 hours and also 28km downstream in 5 hours. The velocity of the sream is Ans: 1.5 km/hr

6. If a boat goes 7km upstream in 42 minutes and the speed of the stream is 3kmph, then the speed of the boat in still water is Ans: 13km/hr

7. A man can row 9 1/3 kmph in still water and finds that it takes him thrice as much time to row up than as to row down the same distance in the river. The speed of the current is ans: 4 2/3 km/hr

8. A man can row a boat at 10kmph in still water. IF the speed of the stream is 6kmph, the time taken to row a distance of 80km down the stream is ans:5hours

9. A boat takes 4hours for traveling downstream from point A to point B and coming back to point A upstream. If the velocity of the stream is 2km ph and the speed of the boat in still water is 4kmph, what is the distance between A and B? Ans:6km

10. IF a man rows at 6kmph is still water and 4:5kmph against the current, then his along the current is : Ans: 7.5km/hr

11. If a man’s rate with the current is 11kmph and the rate of the current is 1.5kmph, then the man’s rate against the current is Ans:8lm/hr

12. Speed of a boat in standing water is 9kmph and the speed of the stream is 1:5kmph. A man rows to a place at a distance of 10.5 km and comes back to the starting point. The total time taken by him is ans: 24 hours

13. Aboat moves upstream at the rate of 1km in 10 minutes and downstream at the rate of 1km in 6 minutes. The speed of the current is ans:2km/hr

14. River is running at 2kmphr. IF takes a man twice as long to row ap as to row down the river. The rate of the man is still water is ans:6km/hr

15. A man rows to a place 48km distant and back in 14 hours. He finds that he can row 4km with the stream in the same time as 3km against the stream. The rate of the stream is :ans:1km/hr

16. The current of stream runs at 1kmph. A motor baot goes 35km upstream and back again to the starting point in 12hours. The speed of the motor boat in still water is ans:6km/hr

17. A baot covers 24km upstream and 36km downstream in 6 hours while it cover 35km upstream and 24km downstream in 6 ½ hours. The velocity of the current is ans:2km/hr

18. A man can row three quarters of a kilometer against the stream in 11 ¼ minutes and returns in 7 ¼ mintues. The speed of the man is still water is ans:5km/hr

19. The speed of abaot in still water is 15km/hr and the rate of current is 3 km/hr. the distance traveld downstream in 12 minutes is ans:3.6km

20. A man can row 5kmph in still water. If the river is running at 1kmph, it makes him 75minutes to row to a place and back. How far is the place? Ans:3km

21. If a man rows at the rate of 5kmph in still water and his rate against the current is 3.5kmph then the man’s rate along the current is ans: 6.5km/hr

All numbers used are real numbers.

1. Larry is driving at 70 miles per hour. How many minutes does it take him to travel 28 miles ?Ο 2½ Ο 3½ Ο 4 Ο 6 Ο 24Answer: 24Speed = Distance / Time. So, Time = Distance / Speed = (28/70) hr = 0.4 hr = 0.4 x 60 minutes = 24 minutes.

3. A ship sails 3 nautical miles due south, then 12 nautical miles due west, and then another 12 nautical miles due north. How far, in nautical miles, is the ship from its starting point ?Ο 15Ο 18 Ο 21 Ο 24 Ο 27 Answer: 15Effectively, the ship sails 12 nautical miles due west and 9 nautical miles due north. These form the sides of a right-angled triangle, whose hypotenuse is to be determined. By the Pythagorean Theorem, Hypotenuse = (122 + 92)½= (144 + 81)½ = √225 = 15.

6. Four carpenters can individually complete a particular task in 3, 4, 5, and 7 hours, respectively. What is the maximum fraction of the task that can be completed in forty-five minutes if three of the carpenters work together at their respective rates ?Ο 11/30 Ο 47/80Ο 3/5 Ο 11/15 Ο 5/6 Answer: 47/80In one hour, the four carpenters can individually complete one-third, one-fourth, one-fifth and one-seventh of the task. For completion of the maximum fraction of the task, the rates of the three quickest carpenters must be added. Thus, (1/3) + (1/4) + (1/5) = (20 + 15 + 12)/60 = 47/60. In 1 hour, three carpenters together can complete (47/60)th of the task. In ¾ hr (i.e., 45 minutes), they can complete (47/80)th of the task. Note that (47/60) x ¾ = 47/80.

8. The value of (251/3 x 251/3) / 251/2 isΟ 51/6 Ο 51/3

Ο 5 Ο 5 (51/3) Ο 5 (51/2) Answer: 51/3

(251/3 x 251/3) / 251/2 = 252/3/251/2 = 252/3 − 1/2 = 251/6 = (52)1/6 = 51/3

11. A portion of $8000 is invested at a 5% annual return, while the remainder is invested at a 3% annual return. If the annual income from each of the two portions is the same, then what is the total income from the two investments after one year?Ο $280 Ο $300Ο $320 Ο $360 

Ο $480 Answer: $300Let the $8000 amount be divided into two parts x and y such that x + y = 8000. The annual return on x is 5% and that on y is 3%; so, 5x / 100 = 3y / 100 or y = 5x / 3. On eliminating y, one obtains x + 5x / 3 = 8000 or 8x / 3 = 8000. Thus, x = 3000 and y = 5000. So, 5x / 100 = 150 and 3y / 100 = 150. ∴ Total income after one year is $150 + $150 = $300.

13. A student first reduced a certain number by 20% and then increased the resulting number by 25%. If the original number was 1.643 x 1012, then the value of the new number (after reduction and increment) isΟ 1.561 x 1012 Ο 1.587 x 1012 Ο 1.60 x 1012 Ο 1.62 x 1012 Ο 1.643 x 1012

Answer: 1.643 x 1012

Let y be the original number. After a reduction of 20%, the result is 0.8y. Now, this number is increased by 25%. The result is 1.25 x 0.8y. Note that 1.25 = 5/4 and 0.8 = 4/5; so, 1.25 x 0.8y = y. Since a reduction by 20% and a subsequent increment by 25% gives the original number, the correct answer is 1.643 x 1012.

16. x and y are two distinct positive integers divisible by 5. Which of the following is necessarily divisible by 10?Ο x + y Ο x − y Ο xy Ο 2xyΟ x2 + y2 Answer: 2xy2xy is necessarily divisible by 10 because its factors are 2 and 5. The other quantities are not necessarily divisible by 10 as shown by the examples below. If x = 15 and y = 10, then x + y = 25. If x = 15 and y = 10, then x − y = 5. If x = 15 and y = 5, then xy = 75. If x = 15 and y = 10, then x2 + y2 = 225 + 100 = 325.

18. A number x is chosen at random from the set of integers that satisfy |x| < 9. What is the probability that (4 / x) > x?Ο 1 / 5 Ο 1 / 8 Ο 7 / 8 Ο 2 / 17 Ο 7 / 17Answer: 7 / 17The values of x that satisfy |x| < 9 are −9 < x < 9. There are 17 integers (−8, −7, ..., −1, 0, 1, ..., 7, 8) in this interval. If both sides of (4 / x) > x are multiplied by a positive value of x, then x2 < 4. The only (positive) integer that satisfies −2 < x < 2 is 1. If both sides of (4 / x) > x are multiplied by a negative value of x, then x2 > 4. The six integers that satisfy x < −2 are −8, −7, −6, −5, −4 and −3. Since only 7 out of 17 integers satisfy (4 / x) > x, the probability is 7 / 17.

21. The smallest number, which when decreased by 3 is divisible by 14, 18 and 30 isΟ 633Ο 663 Ο 693 Ο 723 Ο 753 Answer: 63314 = 2 x 7; 18 = 2 x 3 x 3; 30 = 2 x 3 x 5. So, Least Common Multiple (LCM) of 14, 18 and 30 = 2 x 3 x 3 x 5 x 7 = 630. Required number = 630 + 3 = 633.