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## Arithmetic: Time Speed Distance Test-5

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*Arithmetic: Time Speed Distance Test-5*.You scored %%SCORE%% out of %%TOTAL%%.You correct answer percentage: %%PERCENTAGE%% .Your performance has been rated as %%RATING%% Your answers are highlighted below.

Question 1 |

If a man walks 20 km at 5 km/hr, he will be late by 40 minutes. If he walks at 8 km per hr, how early from the fixed time will he reach?

A | 15 minutes |

B | 25 minutes |

C | 50 minutes |

D | $ \displaystyle 1\frac{1}{2}$Minutes |

Question 1 Explanation:

The time taken by the man is 20 /5 = 4 hours.

If he is 40 mins late he should have reached by 3 hours 20 mins. = 200 mins

When he travels at 8 Km/hr, the person would reach at 20/8 = 2 hours 30 mins. = 150 mins

He has reached 200-150 = 50 mins earlier.

If he is 40 mins late he should have reached by 3 hours 20 mins. = 200 mins

When he travels at 8 Km/hr, the person would reach at 20/8 = 2 hours 30 mins. = 150 mins

He has reached 200-150 = 50 mins earlier.

Question 2 |

If a man reduces his speed to 2/3, he takes 1 hour more in walking a certain distance. The time (in hours) to cover the distance with his normal speed is:

A | 2 |

B | 1 |

C | 3 |

D | 1.5 |

Question 2 Explanation:

According to the question
A reduction of speed to 2/3 or original means the time taken will be in the ratio of 2: (3-2)= 2:1 .
The time taken will be double.
Therefore, now he will take 2 hours to walk the same distance.

Question 3 |

A student rides on bicycle at 8 km/hour and reaches his school 2.5 minutes late. The next day he increases his speed to 10 km/hour and reaches school 5 minutes early. How far is the school from his house?

A | 5/8 km |

B | 8 km |

C | 5 km |

D | 10 km |

Question 3 Explanation:

Let t be the correct time to reach the school and all the speed be considered unit-less as the multiplying factor for conversion will cancelled on both sides of equation,

(t+2.5)8=10(t-5)

8t + 20 = 10t - 50

2t = 70

t=35 mins.

Distance from school =$ 10\times \frac{35-5}{60}=5km$

(t+2.5)8=10(t-5)

8t + 20 = 10t - 50

2t = 70

t=35 mins.

Distance from school =$ 10\times \frac{35-5}{60}=5km$

Question 4 |

A man covered a certain distance at some speed. Had he moved 3 km per hour faster, he would have taken 40 minutes less. If he had moved 2 km per hour slower, he would have taken 40 minutes more. The distance (in km) is:

A | 20 |

B | 35 |

C | 36 ^{2}/_{3} |

D | 40 |

Question 4 Explanation:

Let the distance be x km and initial speed be y kmph.

Acc. to the given statements

$ \displaystyle \begin{array}{l}\frac{x}{y}-\frac{x}{y+3}=\frac{40}{60}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...............\left( i \right)\\and.\\\frac{x}{y-2}-\frac{x}{y}=\frac{40}{60}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,....................\left( ii \right)\\From\,\,\,\,equations\,\,\,\left( i \right)\,\,\,\,and\,\,\,\left( ii \right).\\\frac{x}{y}-\frac{x}{y+3}=\frac{x}{y-2}-\frac{x}{y}\\\Rightarrow \frac{1}{y}-\frac{1}{y+3}=\frac{1}{y-2}-\frac{1}{y}\\\Rightarrow \frac{y+3-y}{y\,\left( y+3 \right)}=\frac{y-y+2}{y\,\left( y-2 \right)}\\\Rightarrow 3\,\left( y-2 \right)=2\left( y+3 \right)\\\Rightarrow 3y-6=2y+6\\\Rightarrow y=12\\From\,\,\,\,equation\,\,\,\left( i \right).\\\frac{x}{12}-\frac{x}{15}=\frac{40}{60}\Rightarrow \frac{5x-4x}{60}=\frac{2}{3}\\\Rightarrow x=\frac{2}{3}\times 60=40\\\therefore \,\,\,Dis\tan ce=40\,km.\end{array}$

Acc. to the given statements

$ \displaystyle \begin{array}{l}\frac{x}{y}-\frac{x}{y+3}=\frac{40}{60}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...............\left( i \right)\\and.\\\frac{x}{y-2}-\frac{x}{y}=\frac{40}{60}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,....................\left( ii \right)\\From\,\,\,\,equations\,\,\,\left( i \right)\,\,\,\,and\,\,\,\left( ii \right).\\\frac{x}{y}-\frac{x}{y+3}=\frac{x}{y-2}-\frac{x}{y}\\\Rightarrow \frac{1}{y}-\frac{1}{y+3}=\frac{1}{y-2}-\frac{1}{y}\\\Rightarrow \frac{y+3-y}{y\,\left( y+3 \right)}=\frac{y-y+2}{y\,\left( y-2 \right)}\\\Rightarrow 3\,\left( y-2 \right)=2\left( y+3 \right)\\\Rightarrow 3y-6=2y+6\\\Rightarrow y=12\\From\,\,\,\,equation\,\,\,\left( i \right).\\\frac{x}{12}-\frac{x}{15}=\frac{40}{60}\Rightarrow \frac{5x-4x}{60}=\frac{2}{3}\\\Rightarrow x=\frac{2}{3}\times 60=40\\\therefore \,\,\,Dis\tan ce=40\,km.\end{array}$

Question 5 |

Starting from his house one day, a student walks at a speed of 2

^{1}/_{2}km/h, and reaches his school 6 minutes late. Next day at the same time he increases his speed by 1 kmph. and reaches the school 6 minutes early. How far is the school from his house?A | (a) 2 km |

B | (b) 1 ^{1}/_{2 }km |

C | (c) 1 km |

D | (d) 1 ^{3}/_{4 }km |

Question 5 Explanation:

Let the speed be s kmph and t mins =appropriate time.

5/2(t+6) = (7/2) (t-6)

5t+30 = 7t -42

2t=72

t= 36 mins

s=7/2 t= 30 mins.=1/2 hour

d= 7/2 x 1/2= 1

5/2(t+6) = (7/2) (t-6)

5t+30 = 7t -42

2t=72

t= 36 mins

s=7/2 t= 30 mins.=1/2 hour

d= 7/2 x 1/2= 1

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