Quantitative Aptitude
SPEED TIME AND DISTANCE MCQs
Time & Distance
Total Questions : 422
| Page 7 of 43 pages
Answer: Option A. -> 40 km/hr
Answer: (a)Total distance covered = 400 km.Total time = $25/2$ hours$3/4$th of total journey= $3/4$ × 400 = 300 km.Time taken = $\text"Distance"/ \text"Speed"$= $300/30$ = 10 hoursRemaining time = $25/2$ –10= ${25 - 20}/2 = 5/2$ hoursRemaining distance = 100 km.∴ Required speed of car= $100/{5/2} = {100 × 2}/5$ = 40 kmph.
Answer: (a)Total distance covered = 400 km.Total time = $25/2$ hours$3/4$th of total journey= $3/4$ × 400 = 300 km.Time taken = $\text"Distance"/ \text"Speed"$= $300/30$ = 10 hoursRemaining time = $25/2$ –10= ${25 - 20}/2 = 5/2$ hoursRemaining distance = 100 km.∴ Required speed of car= $100/{5/2} = {100 × 2}/5$ = 40 kmph.
Answer: Option A. -> 70 km
Answer: (a)Let the length of journey be x km, then$x/35 - x/40 = 15/60 = 1/4$${8x - 7x}/280 = 1/4$$x = 280/4 = 70$ km
Answer: (a)Let the length of journey be x km, then$x/35 - x/40 = 15/60 = 1/4$${8x - 7x}/280 = 1/4$$x = 280/4 = 70$ km
Answer: Option A. -> 6
Answer: (a)Using Rule 5,If two equal distances are covered at two unequal speed of x kmph and y kmph,then average speed = $({2xy}/{x + y})$= ${2 × 12 × 4}/{12 + 4} = 96/16$ = 6 kmph
Answer: (a)Using Rule 5,If two equal distances are covered at two unequal speed of x kmph and y kmph,then average speed = $({2xy}/{x + y})$= ${2 × 12 × 4}/{12 + 4} = 96/16$ = 6 kmph
Answer: Option C. -> 14$2/5$ km/hr
Answer: (c)Using Rule 5,Average speed= $({2xy}/{x + y})$ kmph= $({2 × 12 × 18}/{12 + 18})$ kmph= $({2 × 12 × 18}/30)$ kmph= 14$2/5$ kmph
Answer: (c)Using Rule 5,Average speed= $({2xy}/{x + y})$ kmph= $({2 × 12 × 18}/{12 + 18})$ kmph= $({2 × 12 × 18}/30)$ kmph= 14$2/5$ kmph
Answer: Option D. -> 36 kmph
Answer: (d)Using Rule 3,Average speed= $\text"Total distance"/\text"time taken"$= ${30 × 12/60 + 45 × 8/60}/{12/60 + 8/60}$= 12 × 3 = 36 kmph
Answer: (d)Using Rule 3,Average speed= $\text"Total distance"/\text"time taken"$= ${30 × 12/60 + 45 × 8/60}/{12/60 + 8/60}$= 12 × 3 = 36 kmph
Answer: Option A. -> 9.00 a.m.
Answer: (a)Using Rule 11,Time taken by 1st man to reach B after meeting 2nd man at C is '$t_1$' and time taken by 2nd man to reach A after meeting 1st man at C is '$t_2$' then:${\text"Speed of 1st man"(s_1)}/{\text"Speed of 2nd man"(s_2)} = √{t_2/t_1}$Distance from A to B = $s_1t_1 + S_2t_2$
Answer: (a)Using Rule 11,Time taken by 1st man to reach B after meeting 2nd man at C is '$t_1$' and time taken by 2nd man to reach A after meeting 1st man at C is '$t_2$' then:${\text"Speed of 1st man"(s_1)}/{\text"Speed of 2nd man"(s_2)} = √{t_2/t_1}$Distance from A to B = $s_1t_1 + S_2t_2$
Answer: Option A. -> 20 km/hr
Answer: (a)Using Rule 3,If a man travels different distances $d_1,d_2,d_3$, and so on with different speeds $s_1,s_2,s_3$, respectively then,Average speed = $({d_1 + d_2 + d_3 + ...})/{d_1/S_1 + d_2/S_2 + d_3/S_3 + ...}$
Answer: (a)Using Rule 3,If a man travels different distances $d_1,d_2,d_3$, and so on with different speeds $s_1,s_2,s_3$, respectively then,Average speed = $({d_1 + d_2 + d_3 + ...})/{d_1/S_1 + d_2/S_2 + d_3/S_3 + ...}$
Answer: Option D. -> 54 km/hr.
Answer: (d)Total distance covered by the bus= 150 km. + 2 × 60 km.= (150 + 120) km.= 270 km.∴ Average speed = $\text"Total distance"/ \text"Time taken"$= $270/5$ = 54 kmph.
Answer: (d)Total distance covered by the bus= 150 km. + 2 × 60 km.= (150 + 120) km.= 270 km.∴ Average speed = $\text"Total distance"/ \text"Time taken"$= $270/5$ = 54 kmph.
Answer: Option C. -> 14.4 kmph
Answer: (c)Time taken by Kamal= $100/{18 × 5/18}$ = 20 secondsTime taken by Bimal= 20 + 5 = 25 secondsBimal's speed= $100/25$ =4 m/sec = ${4 × 18}/5$ kmph= 14.4 kmph.
Answer: (c)Time taken by Kamal= $100/{18 × 5/18}$ = 20 secondsTime taken by Bimal= 20 + 5 = 25 secondsBimal's speed= $100/25$ =4 m/sec = ${4 × 18}/5$ kmph= 14.4 kmph.
Answer: Option D. -> 2 km.
Answer: (d)Distance between starting point and multiplex = x metreTime = $\text"Distance"/ \text"Speed"$According to the question,$x/3 - x/4 = {5 + 5}/60$${4x - 3x}/12 = 1/6$$x/12 = 1/6$$x = 12/6$ = 2 km.
Answer: (d)Distance between starting point and multiplex = x metreTime = $\text"Distance"/ \text"Speed"$According to the question,$x/3 - x/4 = {5 + 5}/60$${4x - 3x}/12 = 1/6$$x/12 = 1/6$$x = 12/6$ = 2 km.
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