Questions and Answers Type: | MCQ (Multiple Choice Questions). |

Main Topic: | Quantitative Aptitude. |

Quantitative Aptitude Sub-topic: | Time Speed and Distance Aptitude Questions and Answers. |

Number of Questions: | 10 Questions with Solutions. |

- If a man swims \(10 \ km\) upstream and \(12 \ km\) downstream, taking \(2 \ hours\) each time, then find the speed of stream?
- \(0.3 \ km/hr\)
- \(0.4 \ km/hr\)
- \(0.5 \ km/hr\)
- \(1.0 \ km/hr\)

Answer: (c) \(0.5 \ km/hr\)

Solution: Given, speed of upstream \((U_s) = \frac{10}{2} = 5 \ km/hr\)

speed of downstream \((D_s) = \frac{12}{2} = 6 \ km/hr\)

then speed of stream, $$ S_s = \frac{D_s - U_s}{2} $$ $$ = \frac{6 - 5}{2} = 0.5 \ km/hr $$

Solution: Given, speed of upstream \((U_s) = \frac{10}{2} = 5 \ km/hr\)

speed of downstream \((D_s) = \frac{12}{2} = 6 \ km/hr\)

then speed of stream, $$ S_s = \frac{D_s - U_s}{2} $$ $$ = \frac{6 - 5}{2} = 0.5 \ km/hr $$

- If a girl can swim \(18 \ km\) upstream and \(26 \ km\) downstream, taking \(2 \ hours\) each time, then find speed of the girl in still water?
- \(10 \ km/hr\)
- \(11 \ km/hr\)
- \(12 \ km/hr\)
- \(15 \ km/hr\)

Answer: (b) \(11 \ km/hr\)

Solution: Given, speed of upstream \((U_s) = \frac{18}{2} = 9 \ km/hr\)

speed of downstream \((D_s) = \frac{26}{2} = 13 \ km/hr\)

then speed of the girl in still water, $$ B_s = \frac{D_s + U_s}{2} $$ $$ = \frac{9 + 13}{2} = 11 \ km/hr $$

Solution: Given, speed of upstream \((U_s) = \frac{18}{2} = 9 \ km/hr\)

speed of downstream \((D_s) = \frac{26}{2} = 13 \ km/hr\)

then speed of the girl in still water, $$ B_s = \frac{D_s + U_s}{2} $$ $$ = \frac{9 + 13}{2} = 11 \ km/hr $$

- If a boat goes \(30 \ km\) upstream and \(45 \ km\) downstream, taking \(3 \ hours\) each time, then find the speed of stream and speed of the boat in still water?
- \(S_s = 2.5 \ km/hr\), \(B_s = 12.5 \ km/hr\)
- \(S_s = 3.5 \ km/hr\), \(B_s = 13.5 \ km/hr\)
- \(S_s = 2.5 \ km/hr\), \(B_s = 15.5 \ km/hr\)
- \(S_s = 5.5 \ km/hr\), \(B_s = 12.5 \ km/hr\)

Answer: (a) \(S_s = 2.5 \ km/hr\), \(B_s = 12.5 \ km/hr\)

Solution: Given, speed of upstream \((U_s) = \frac{30}{3} = 10 \ km/hr\)

speed of downstream \((D_s) = \frac{45}{3} = 15 \ km/hr\)then speed of stream, $$ S_s = \frac{D_s - U_s}{2} $$ $$ = \frac{15 - 10}{2} = 2.5 \ km/hr $$ then speed of the boat in still water, $$ B_s = \frac{D_s + U_s}{2} $$ $$ = \frac{15 + 10}{2} = 12.5 \ km/hr $$

Solution: Given, speed of upstream \((U_s) = \frac{30}{3} = 10 \ km/hr\)

speed of downstream \((D_s) = \frac{45}{3} = 15 \ km/hr\)then speed of stream, $$ S_s = \frac{D_s - U_s}{2} $$ $$ = \frac{15 - 10}{2} = 2.5 \ km/hr $$ then speed of the boat in still water, $$ B_s = \frac{D_s + U_s}{2} $$ $$ = \frac{15 + 10}{2} = 12.5 \ km/hr $$

- A man swims from point A to B and takes \(2 \ hours\) for traveling upstream from point A to B and coming back from point B to A downstream. If the speed of stream is \(10 \ km/hr\) and speed of the man in still water is \(12 \ km/hr\), then find the distance between point A and B?
- \(2.58 \ km\)
- \(2.67 \ km\)
- \(3.67 \ km\)
- \(3.58 \ km\)

Answer: (c) \(3.67 \ km\)

Solution: Given, speed of stream \((S_s) = 10 \ km/hr\)

speed of the man in still water \((B_s) = 12 \ km/hr\), then

speed of upstream \((U_s) = B_s - S_s\) \(= 12 - 10 = 2 \ km/hr\)

speed of downstream \((D_s) = B_s + S_s\) \(= 12 + 10 = 22 \ km/hr\)

Let the distance between point A and B = \(x \ km\), then $$ \frac{x}{22} + \frac{x}{2} = 2 $$ $$ \frac{12 \ x}{22} = 2 $$ $$ 12 \ x = 44 $$ $$ x = \frac{44}{12} = 3.67 \ km $$ Hence the distance between the point A and B = \(3.67 \ km\)

Solution: Given, speed of stream \((S_s) = 10 \ km/hr\)

speed of the man in still water \((B_s) = 12 \ km/hr\), then

speed of upstream \((U_s) = B_s - S_s\) \(= 12 - 10 = 2 \ km/hr\)

speed of downstream \((D_s) = B_s + S_s\) \(= 12 + 10 = 22 \ km/hr\)

Let the distance between point A and B = \(x \ km\), then $$ \frac{x}{22} + \frac{x}{2} = 2 $$ $$ \frac{12 \ x}{22} = 2 $$ $$ 12 \ x = 44 $$ $$ x = \frac{44}{12} = 3.67 \ km $$ Hence the distance between the point A and B = \(3.67 \ km\)

- If a boat goes downstream with the speed of \(16 \ km/hr\) and upstream with the speed of \(14 \ km/hr\), then find the speed of the boat in still water?
- \(12 \ km/hr\)
- \(13 \ km/hr\)
- \(14 \ km/hr\)
- \(15 \ km/hr\)

Answer: (d) \(15 \ km/hr\)

Solution: Given, speed of downstream \((D_s) = 16 \ km/hr\)

speed of upstream \((U_s) = 14 \ km/hr\)

then speed of the boat in still water, $$ B_s = \frac{U_s + D_s}{2} $$ $$ = \frac{14 + 16}{2} $$ $$ = \frac{30}{2} = 15 \ km/hr $$

Solution: Given, speed of downstream \((D_s) = 16 \ km/hr\)

speed of upstream \((U_s) = 14 \ km/hr\)

then speed of the boat in still water, $$ B_s = \frac{U_s + D_s}{2} $$ $$ = \frac{14 + 16}{2} $$ $$ = \frac{30}{2} = 15 \ km/hr $$

- If a boat goes downstream at the speed of \(9 \ km/hr\) and speed of stream is \(4 \ km/hr\), then find the speed of boat in still water?
- \(5 \ km/hr\)
- \(6 \ km/hr\)
- \(4 \ km/hr\)
- \(2 \ km/hr\)

Answer: (a) \(5 \ km/hr\)

Solution: Given, speed of downstream \((D_s) = 9 \ km/hr\)

speed of stream \((S_s) = 4 \ km/hr\)

then speed of the boat in still water, $$ B_s = D_s - S_s $$ $$ = 9 - 4 = 5 \ km/hr $$

Solution: Given, speed of downstream \((D_s) = 9 \ km/hr\)

speed of stream \((S_s) = 4 \ km/hr\)

then speed of the boat in still water, $$ B_s = D_s - S_s $$ $$ = 9 - 4 = 5 \ km/hr $$

- A girl can swim \(28 \ km\) downstream, taking \(4 \ hours\). If speed of stream is \(3 \ km/hr\), then find the speed of upstream?
- \(1 \ km/hr\)
- \(2 \ km/hr\)
- \(3 \ km/hr\)
- \(5 \ km/hr\)

Answer: (a) \(1 \ km/hr\)

Solution: Given, speed of downstream \((D_s) = \frac{28}{4} = 7 \ km/hr\)

speed of stream \((S_s) = 3 \ km/hr\), then $$ S_s = \frac{D_s - U_s}{2} $$ $$ 3 = \frac{7 - U_s}{2} $$ $$ 6 = 7 - U_s $$ $$ U_s = 1 \ km/hr $$

Solution: Given, speed of downstream \((D_s) = \frac{28}{4} = 7 \ km/hr\)

speed of stream \((S_s) = 3 \ km/hr\), then $$ S_s = \frac{D_s - U_s}{2} $$ $$ 3 = \frac{7 - U_s}{2} $$ $$ 6 = 7 - U_s $$ $$ U_s = 1 \ km/hr $$

- A boat goes \(16 \ km\) upstream, taking \(2 \ hours\). If speed of stream is \(2 \ km/hr\), then find the speed of downstream?
- \(15 \ km/hr\)
- \(13 \ km/hr\)
- \(12 \ km/hr\)
- \(10 \ km/hr\)

Answer: (c) \(12 \ km/hr\)

Solution: Given, speed of upstream \((U_s) = \frac{16}{2} = 8 \ km/hr\)

speed of stream \((S_s) = 2 \ km/hr\), then $$ S_s = \frac{D_s - U_s}{2} $$ $$ 2 = \frac{D_s - 8}{2} $$ $$ 4 = D_s - 8 $$ $$ D_s = 12 \ km/hr $$

Solution: Given, speed of upstream \((U_s) = \frac{16}{2} = 8 \ km/hr\)

speed of stream \((S_s) = 2 \ km/hr\), then $$ S_s = \frac{D_s - U_s}{2} $$ $$ 2 = \frac{D_s - 8}{2} $$ $$ 4 = D_s - 8 $$ $$ D_s = 12 \ km/hr $$

- A boat goes downstream with the speed of \(25 \ km/hr\). If the speed of the boat in still water is \(15 \ km/hr\), then find the speed of upstream?
- \(2 \ km/hr\)
- \(3 \ km/hr\)
- \(4 \ km/hr\)
- \(5 \ km/hr\)

Answer: (d) \(5 \ km/hr\)

Solution: Given, speed of downstream \((D_s) = 25 \ km/hr\)

speed of boat in still water \((B_s) = 15 \ km/hr\), then $$ B_s = \frac{D_s + U_s}{2} $$ $$ 15 = \frac{25 + U_s}{2} $$ $$ 30 = 25 + U_s $$ $$ U_s = 5 \ km/hr $$

Solution: Given, speed of downstream \((D_s) = 25 \ km/hr\)

speed of boat in still water \((B_s) = 15 \ km/hr\), then $$ B_s = \frac{D_s + U_s}{2} $$ $$ 15 = \frac{25 + U_s}{2} $$ $$ 30 = 25 + U_s $$ $$ U_s = 5 \ km/hr $$

- A boat goes upstream with the speed of \(22 \ km/hr\). If the speed of the boat in still water is \(8 \ km/hr\), then find the speed of downstream?
- \(12 \ km/hr\)
- \(13 \ km/hr\)
- \(14 \ km/hr\)
- \(15 \ km/hr\)

Answer: (c) \(14 \ km/hr\)

Solution: Given, speed of upstream \((U_s) = 22 \ km/hr\)

speed of the boat in still water \((B_s) = 18 \ km/hr\), then $$ B_s = \frac{D_s + U_s}{2} $$ $$ 18 = \frac{D_s + 22}{2} $$ $$ 36 = D_s + 22 $$ $$ D_s = 14 \ km/hr $$

Solution: Given, speed of upstream \((U_s) = 22 \ km/hr\)

speed of the boat in still water \((B_s) = 18 \ km/hr\), then $$ B_s = \frac{D_s + U_s}{2} $$ $$ 18 = \frac{D_s + 22}{2} $$ $$ 36 = D_s + 22 $$ $$ D_s = 14 \ km/hr $$

Lec 1: Introduction to Time, Speed and Distance
Exercise-1
Lec 2: Concept of Train and Platform Case (1)
Exercise-2
Lec 3: Concept of Train and Platform Case (2)
Exercise-3
Lec 4: Concept of Train and Platform Case (3)
Exercise-4
Lec 5: Concept of Train and Platform Case (4)
Exercise-5
Lec 6: Concept of Acceleration
Exercise-6
Lec 7: Concept of Boat and Stream Case (1) and Case (2)
Exercise-7
Lec 8: Concept of Boat and Stream Case (3)
Exercise-8
Exercise-9
Exercise-10