Sasi Math (Revised for 1024)

Scroll:Distance, Time and Speed >> Distance, Time & Speed >> ps (1891)

Written Instructions:

For each Multiple Choice Question (MCQ), four options are given. One of them is the correct answer. Make your choice (1,2,3 or 4). Write your answers in the brackets provided..

For each Short Answer Question(SAQ) and Long Answer Question(LAQ), write your answers in the blanks provided.

Leave your answers in the simplest form or correct to two decimal places.

Murugan cycles to work every day. If he cycles at a speed of 27 $\frac{km}{h}$ , he would reach his office at 8.50 a.m. If he cycles at 21 $\frac{km}{h}$ , he would reach office 20 minutes later. What should his average cycling speed be if he wants to arrive at his office at 9.00 a.m. Give your answer correct to 2 decimal places.

$\frac{km}{h}$

Answer:_______________

Selvan cycles to work every day. If he cycles at a speed of 29 $\frac{km}{h}$ , he would reach his office at 8.50 a.m. If he cycles at 24 $\frac{km}{h}$ , he would reach office 20 minutes later. What should his average cycling speed be if he wants to arrive at his office at 9.00 a.m. Give your answer correct to 2 decimal places.

$\frac{km}{h}$

Answer:_______________

Kalai cycles to work every day. If he cycles at a speed of 37 $\frac{km}{h}$ , he would reach his office at 8.50 a.m. If he cycles at 30 $\frac{km}{h}$ , he would reach office 20 minutes later. What should his average cycling speed be if he wants to arrive at his office at 9.00 a.m. Give your answer correct to 2 decimal places.

$\frac{km}{h}$

Answer:_______________

Sriram cycles to work every day. If he cycles at a speed of 31 $\frac{km}{h}$ , he would reach his office at 8.50 a.m. If he cycles at 24 $\frac{km}{h}$ , he would reach office 20 minutes later. What should his average cycling speed be if he wants to arrive at his office at 9.00 a.m. Give your answer correct to 2 decimal places.

$\frac{km}{h}$

Answer:_______________

Selvan cycles to work every day. If he cycles at a speed of 19 $\frac{km}{h}$ , he would reach his office at 8.50 a.m. If he cycles at 15 $\frac{km}{h}$ , he would reach office 20 minutes later. What should his average cycling speed be if he wants to arrive at his office at 9.00 a.m. Give your answer correct to 2 decimal places.

$\frac{km}{h}$

Answer:_______________

Karna cycles to work every day. If he cycles at a speed of 33 $\frac{km}{h}$ , he would reach his office at 8.50 a.m. If he cycles at 27 $\frac{km}{h}$ , he would reach office 20 minutes later. What should his average cycling speed be if he wants to arrive at his office at 9.00 a.m. Give your answer correct to 2 decimal places.

$\frac{km}{h}$

Answer:_______________

Moorthi cycles to work every day. If he cycles at a speed of 29 $\frac{km}{h}$ , he would reach his office at 8.50 a.m. If he cycles at 24 $\frac{km}{h}$ , he would reach office 20 minutes later. What should his average cycling speed be if he wants to arrive at his office at 9.00 a.m. Give your answer correct to 2 decimal places.

$\frac{km}{h}$

Answer:_______________

Barathan cycles to work every day. If he cycles at a speed of 36 $\frac{km}{h}$ , he would reach his office at 8.50 a.m. If he cycles at 30 $\frac{km}{h}$ , he would reach office 20 minutes later. What should his average cycling speed be if he wants to arrive at his office at 9.00 a.m. Give your answer correct to 2 decimal places.

$\frac{km}{h}$

Answer:_______________

Narayanan cycles to work every day. If he cycles at a speed of 37 $\frac{km}{h}$ , he would reach his office at 8.50 a.m. If he cycles at 30 $\frac{km}{h}$ , he would reach office 20 minutes later. What should his average cycling speed be if he wants to arrive at his office at 9.00 a.m. Give your answer correct to 2 decimal places.

$\frac{km}{h}$

Answer:_______________

10)

Prem cycles to work every day. If he cycles at a speed of 38 $\frac{km}{h}$ , he would reach his office at 8.50 a.m. If he cycles at 33 $\frac{km}{h}$ , he would reach office 20 minutes later. What should his average cycling speed be if he wants to arrive at his office at 9.00 a.m. Give your answer correct to 2 decimal places.

$\frac{km}{h}$

Answer:_______________

Answer: 23.63 $\frac{km}{h}$

SOLUTION 1 :

Step 1: Distance to be covered from 8.50 a.m. to 9.10 a.m. if Murugan cycles at 21 $\frac{km}{h}$ :

1h	→	21 km
20 min	→	$\frac{20}{60}$ x 21 km
	=	7 km.

So at 27 $\frac{km}{h}$ , Murugan is ahead by 7 km at 8.50 a.m., compared to if Murugan cycles at a speed of 21 $\frac{km}{h}$ .

Step 2: 27 km - 21 km = 6 km

At 27 $\frac{km}{h}$ , Murugan is ahead by 6 km every hour.

Step 3: 7 km ÷ 6 $\frac{km}{h}$ = 1.16667 h.

At 27 $\frac{km}{h}$ , Murugan has been cycling for 1.16667 h to be 7 km ahead at 8.50 a.m.

So at 27 $\frac{km}{h}$ , Murugan takes 1.16667 h to reach the office.

Step 4:	1h	→	27 km
	1.16667h	→	1.16667 x 27 km
		=	31.50009 km

Total distance travelled is 31.50009 km.

Step 5:	The time when Murugan started cycling	→	8.50 a.m. - 1.16667 h
		=	8.50 a.m. - 1 h 10 min
		=	7.40 a.m.

Step 6:	Cycling time if Murugan wants to reach school at 9.00 a.m.	→	9.00 a.m. - 7.40 a.m.
		=	1 h 20 min
		=	1 h + $\frac{20}{60}$ h
		≈	1.33333 h

Step 7:	Average speed	→	31.50009 km ÷ 1.33333 h
		=	23.62513 $\frac{km}{h}$
		≈	23.63 $\frac{km}{h}$

Answer: 26.26 $\frac{km}{h}$

SOLUTION 1 :

Step 1: Distance to be covered from 8.50 a.m. to 9.10 a.m. if Selvan cycles at 24 $\frac{km}{h}$ :

1h	→	24 km
20 min	→	$\frac{20}{60}$ x 24 km
	=	8 km.

So at 29 $\frac{km}{h}$ , Selvan is ahead by 8 km at 8.50 a.m., compared to if Selvan cycles at a speed of 24 $\frac{km}{h}$ .

Step 2: 29 km - 24 km = 5 km

At 29 $\frac{km}{h}$ , Selvan is ahead by 5 km every hour.

Step 3: 8 km ÷ 5 $\frac{km}{h}$ = 1.6 h.

At 29 $\frac{km}{h}$ , Selvan has been cycling for 1.6 h to be 8 km ahead at 8.50 a.m.

So at 29 $\frac{km}{h}$ , Selvan takes 1.6 h to reach the office.

Step 4:	1h	→	29 km
	1.6h	→	1.6 x 29 km
		=	46.4 km

Total distance travelled is 46.4 km.

Step 5:	The time when Selvan started cycling	→	8.50 a.m. - 1.6 h
		=	8.50 a.m. - 1 h 36 min
		=	7.14 a.m.

Step 6:	Cycling time if Selvan wants to reach school at 9.00 a.m.	→	9.00 a.m. - 7.14 a.m.
		=	1 h 46 min
		=	1 h + $\frac{46}{60}$ h
		≈	1.76667 h

Step 7:	Average speed	→	46.4 km ÷ 1.76667 h
		=	26.2641 $\frac{km}{h}$
		≈	26.26 $\frac{km}{h}$

Answer: 33.04 $\frac{km}{h}$

SOLUTION 1 :

Step 1: Distance to be covered from 8.50 a.m. to 9.10 a.m. if Kalai cycles at 30 $\frac{km}{h}$ :

1h	→	30 km
20 min	→	$\frac{20}{60}$ x 30 km
	=	10 km.

So at 37 $\frac{km}{h}$ , Kalai is ahead by 10 km at 8.50 a.m., compared to if Kalai cycles at a speed of 30 $\frac{km}{h}$ .

Step 2: 37 km - 30 km = 7 km

At 37 $\frac{km}{h}$ , Kalai is ahead by 7 km every hour.

Step 3: 10 km ÷ 7 $\frac{km}{h}$ = 1.42857 h.

At 37 $\frac{km}{h}$ , Kalai has been cycling for 1.42857 h to be 10 km ahead at 8.50 a.m.

So at 37 $\frac{km}{h}$ , Kalai takes 1.42857 h to reach the office.

Step 4:	1h	→	37 km
	1.42857h	→	1.42857 x 37 km
		=	52.85709 km

Total distance travelled is 52.85709 km.

Step 5:	The time when Kalai started cycling	→	8.50 a.m. - 1.42857 h
		=	8.50 a.m. - 1 h 26 min
		=	7.24 a.m.

Step 6:	Cycling time if Kalai wants to reach school at 9.00 a.m.	→	9.00 a.m. - 7.24 a.m.
		=	1 h 36 min
		=	1 h + $\frac{36}{60}$ h
		≈	1.6 h

Step 7:	Average speed	→	52.85709 km ÷ 1.6 h
		=	33.03568 $\frac{km}{h}$
		≈	33.04 $\frac{km}{h}$

Answer: 26.91 $\frac{km}{h}$

SOLUTION 1 :

Step 1: Distance to be covered from 8.50 a.m. to 9.10 a.m. if Sriram cycles at 24 $\frac{km}{h}$ :

1h	→	24 km
20 min	→	$\frac{20}{60}$ x 24 km
	=	8 km.

So at 31 $\frac{km}{h}$ , Sriram is ahead by 8 km at 8.50 a.m., compared to if Sriram cycles at a speed of 24 $\frac{km}{h}$ .

Step 2: 31 km - 24 km = 7 km

At 31 $\frac{km}{h}$ , Sriram is ahead by 7 km every hour.

Step 3: 8 km ÷ 7 $\frac{km}{h}$ = 1.14286 h.

At 31 $\frac{km}{h}$ , Sriram has been cycling for 1.14286 h to be 8 km ahead at 8.50 a.m.

So at 31 $\frac{km}{h}$ , Sriram takes 1.14286 h to reach the office.

Step 4:	1h	→	31 km
	1.14286h	→	1.14286 x 31 km
		=	35.42866 km

Total distance travelled is 35.42866 km.

Step 5:	The time when Sriram started cycling	→	8.50 a.m. - 1.14286 h
		=	8.50 a.m. - 1 h 9 min
		=	7.41 a.m.

Step 6:	Cycling time if Sriram wants to reach school at 9.00 a.m.	→	9.00 a.m. - 7.41 a.m.
		=	1 h 19 min
		=	1 h + $\frac{19}{60}$ h
		≈	1.31667 h

Step 7:	Average speed	→	35.42866 km ÷ 1.31667 h
		=	26.90777 $\frac{km}{h}$
		≈	26.91 $\frac{km}{h}$

Answer: 16.76 $\frac{km}{h}$

SOLUTION 1 :

Step 1: Distance to be covered from 8.50 a.m. to 9.10 a.m. if Selvan cycles at 15 $\frac{km}{h}$ :

1h	→	15 km
20 min	→	$\frac{20}{60}$ x 15 km
	=	5 km.

So at 19 $\frac{km}{h}$ , Selvan is ahead by 5 km at 8.50 a.m., compared to if Selvan cycles at a speed of 15 $\frac{km}{h}$ .

Step 2: 19 km - 15 km = 4 km

At 19 $\frac{km}{h}$ , Selvan is ahead by 4 km every hour.

Step 3: 5 km ÷ 4 $\frac{km}{h}$ = 1.25 h.

At 19 $\frac{km}{h}$ , Selvan has been cycling for 1.25 h to be 5 km ahead at 8.50 a.m.

So at 19 $\frac{km}{h}$ , Selvan takes 1.25 h to reach the office.

Step 4:	1h	→	19 km
	1.25h	→	1.25 x 19 km
		=	23.75 km

Total distance travelled is 23.75 km.

Step 5:	The time when Selvan started cycling	→	8.50 a.m. - 1.25 h
		=	8.50 a.m. - 1 h 15 min
		=	7.35 a.m.

Step 6:	Cycling time if Selvan wants to reach school at 9.00 a.m.	→	9.00 a.m. - 7.35 a.m.
		=	1 h 25 min
		=	1 h + $\frac{25}{60}$ h
		≈	1.41667 h

Step 7:	Average speed	→	23.75 km ÷ 1.41667 h
		=	16.76467 $\frac{km}{h}$
		≈	16.76 $\frac{km}{h}$

Answer: 29.70 $\frac{km}{h}$

SOLUTION 1 :

Step 1: Distance to be covered from 8.50 a.m. to 9.10 a.m. if Karna cycles at 27 $\frac{km}{h}$ :

1h	→	27 km
20 min	→	$\frac{20}{60}$ x 27 km
	=	9 km.

So at 33 $\frac{km}{h}$ , Karna is ahead by 9 km at 8.50 a.m., compared to if Karna cycles at a speed of 27 $\frac{km}{h}$ .

Step 2: 33 km - 27 km = 6 km

At 33 $\frac{km}{h}$ , Karna is ahead by 6 km every hour.

Step 3: 9 km ÷ 6 $\frac{km}{h}$ = 1.5 h.

At 33 $\frac{km}{h}$ , Karna has been cycling for 1.5 h to be 9 km ahead at 8.50 a.m.

So at 33 $\frac{km}{h}$ , Karna takes 1.5 h to reach the office.

Step 4:	1h	→	33 km
	1.5h	→	1.5 x 33 km
		=	49.5 km

Total distance travelled is 49.5 km.

Step 5:	The time when Karna started cycling	→	8.50 a.m. - 1.5 h
		=	8.50 a.m. - 1 h 30 min
		=	7.20 a.m.

Step 6:	Cycling time if Karna wants to reach school at 9.00 a.m.	→	9.00 a.m. - 7.20 a.m.
		=	1 h 40 min
		=	1 h + $\frac{40}{60}$ h
		≈	1.66667 h

Step 7:	Average speed	→	49.5 km ÷ 1.66667 h
		=	29.69994 $\frac{km}{h}$
		≈	29.70 $\frac{km}{h}$

Answer: 26.26 $\frac{km}{h}$

SOLUTION 1 :

Step 1: Distance to be covered from 8.50 a.m. to 9.10 a.m. if Moorthi cycles at 24 $\frac{km}{h}$ :

1h	→	24 km
20 min	→	$\frac{20}{60}$ x 24 km
	=	8 km.

So at 29 $\frac{km}{h}$ , Moorthi is ahead by 8 km at 8.50 a.m., compared to if Moorthi cycles at a speed of 24 $\frac{km}{h}$ .

Step 2: 29 km - 24 km = 5 km

At 29 $\frac{km}{h}$ , Moorthi is ahead by 5 km every hour.

Step 3: 8 km ÷ 5 $\frac{km}{h}$ = 1.6 h.

At 29 $\frac{km}{h}$ , Moorthi has been cycling for 1.6 h to be 8 km ahead at 8.50 a.m.

So at 29 $\frac{km}{h}$ , Moorthi takes 1.6 h to reach the office.

Step 4:	1h	→	29 km
	1.6h	→	1.6 x 29 km
		=	46.4 km

Total distance travelled is 46.4 km.

Step 5:	The time when Moorthi started cycling	→	8.50 a.m. - 1.6 h
		=	8.50 a.m. - 1 h 36 min
		=	7.14 a.m.

Step 6:	Cycling time if Moorthi wants to reach school at 9.00 a.m.	→	9.00 a.m. - 7.14 a.m.
		=	1 h 46 min
		=	1 h + $\frac{46}{60}$ h
		≈	1.76667 h

Step 7:	Average speed	→	46.4 km ÷ 1.76667 h
		=	26.2641 $\frac{km}{h}$
		≈	26.26 $\frac{km}{h}$

Answer: 32.73 $\frac{km}{h}$

SOLUTION 1 :

Step 1: Distance to be covered from 8.50 a.m. to 9.10 a.m. if Barathan cycles at 30 $\frac{km}{h}$ :

1h	→	30 km
20 min	→	$\frac{20}{60}$ x 30 km
	=	10 km.

So at 36 $\frac{km}{h}$ , Barathan is ahead by 10 km at 8.50 a.m., compared to if Barathan cycles at a speed of 30 $\frac{km}{h}$ .

Step 2: 36 km - 30 km = 6 km

At 36 $\frac{km}{h}$ , Barathan is ahead by 6 km every hour.

Step 3: 10 km ÷ 6 $\frac{km}{h}$ = 1.66667 h.

At 36 $\frac{km}{h}$ , Barathan has been cycling for 1.66667 h to be 10 km ahead at 8.50 a.m.

So at 36 $\frac{km}{h}$ , Barathan takes 1.66667 h to reach the office.

Step 4:	1h	→	36 km
	1.66667h	→	1.66667 x 36 km
		=	60.00012 km

Total distance travelled is 60.00012 km.

Step 5:	The time when Barathan started cycling	→	8.50 a.m. - 1.66667 h
		=	8.50 a.m. - 1 h 40 min
		=	7.10 a.m.

Step 6:	Cycling time if Barathan wants to reach school at 9.00 a.m.	→	9.00 a.m. - 7.10 a.m.
		=	1 h 50 min
		=	1 h + $\frac{50}{60}$ h
		≈	1.83333 h

Step 7:	Average speed	→	60.00012 km ÷ 1.83333 h
		=	32.7274 $\frac{km}{h}$
		≈	32.73 $\frac{km}{h}$

Answer: 33.04 $\frac{km}{h}$

SOLUTION 1 :

Step 1: Distance to be covered from 8.50 a.m. to 9.10 a.m. if Narayanan cycles at 30 $\frac{km}{h}$ :

1h	→	30 km
20 min	→	$\frac{20}{60}$ x 30 km
	=	10 km.

So at 37 $\frac{km}{h}$ , Narayanan is ahead by 10 km at 8.50 a.m., compared to if Narayanan cycles at a speed of 30 $\frac{km}{h}$ .

Step 2: 37 km - 30 km = 7 km

At 37 $\frac{km}{h}$ , Narayanan is ahead by 7 km every hour.

Step 3: 10 km ÷ 7 $\frac{km}{h}$ = 1.42857 h.

At 37 $\frac{km}{h}$ , Narayanan has been cycling for 1.42857 h to be 10 km ahead at 8.50 a.m.

So at 37 $\frac{km}{h}$ , Narayanan takes 1.42857 h to reach the office.

Step 4:	1h	→	37 km
	1.42857h	→	1.42857 x 37 km
		=	52.85709 km

Total distance travelled is 52.85709 km.

Step 5:	The time when Narayanan started cycling	→	8.50 a.m. - 1.42857 h
		=	8.50 a.m. - 1 h 26 min
		=	7.24 a.m.

Step 6:	Cycling time if Narayanan wants to reach school at 9.00 a.m.	→	9.00 a.m. - 7.24 a.m.
		=	1 h 36 min
		=	1 h + $\frac{36}{60}$ h
		≈	1.6 h

Step 7:	Average speed	→	52.85709 km ÷ 1.6 h
		=	33.03568 $\frac{km}{h}$
		≈	33.04 $\frac{km}{h}$

10)

Answer: 35.32 $\frac{km}{h}$

SOLUTION 1 :

Step 1: Distance to be covered from 8.50 a.m. to 9.10 a.m. if Prem cycles at 33 $\frac{km}{h}$ :

1h	→	33 km
20 min	→	$\frac{20}{60}$ x 33 km
	=	11 km.

So at 38 $\frac{km}{h}$ , Prem is ahead by 11 km at 8.50 a.m., compared to if Prem cycles at a speed of 33 $\frac{km}{h}$ .

Step 2: 38 km - 33 km = 5 km

At 38 $\frac{km}{h}$ , Prem is ahead by 5 km every hour.

Step 3: 11 km ÷ 5 $\frac{km}{h}$ = 2.2 h.

At 38 $\frac{km}{h}$ , Prem has been cycling for 2.2 h to be 11 km ahead at 8.50 a.m.

So at 38 $\frac{km}{h}$ , Prem takes 2.2 h to reach the office.

Step 4:	1h	→	38 km
	2.2h	→	2.2 x 38 km
		=	83.6 km

Total distance travelled is 83.6 km.

Step 5:	The time when Prem started cycling	→	8.50 a.m. - 2.2 h
		=	8.50 a.m. - 2 h 12 min
		=	6.38 a.m.

Step 6:	Cycling time if Prem wants to reach school at 9.00 a.m.	→	9.00 a.m. - 6.38 a.m.
		=	2 h 22 min
		=	2 h + $\frac{22}{60}$ h
		≈	2.36667 h

Step 7:	Average speed	→	83.6 km ÷ 2.36667 h
		=	35.32389 $\frac{km}{h}$
		≈	35.32 $\frac{km}{h}$