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.

Raj cycles to work every day. If he cycles at a speed of 34 $\frac{km}{h}$ , 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:_______________

Sriram cycles to work every day. If he cycles at a speed of 28 $\frac{km}{h}$ , 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 28 $\frac{km}{h}$ , 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:_______________

Sriram cycles to work every day. If he cycles at a speed of 34 $\frac{km}{h}$ , 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:_______________

Balaji 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 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:_______________

Pugal cycles to work every day. If he cycles at a speed of 25 $\frac{km}{h}$ , 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:_______________

Balaji cycles to work every day. If he cycles at a speed of 40 $\frac{km}{h}$ , 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:_______________

Murugan cycles to work every day. If he cycles at a speed of 25 $\frac{km}{h}$ , 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:_______________

Aravindh cycles to work every day. If he cycles at a speed of 43 $\frac{km}{h}$ , he would reach his office at 8.50 a.m. If he cycles at 36 $\frac{km}{h}$ , 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)

Chidambaram cycles to work every day. If he cycles at a speed of 28 $\frac{km}{h}$ , 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:_______________

Answer: 30.15 $\frac{km}{h}$

SOLUTION 1 :

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

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

So at 34 $\frac{km}{h}$ , Raj is ahead by 9 km at 8.50 a.m., compared to if Raj cycles at a speed of 27 $\frac{km}{h}$ .

Step 2: 34 km - 27 km = 7 km

At 34 $\frac{km}{h}$ , Raj is ahead by 7 km every hour.

Step 3: 9 km ÷ 7 $\frac{km}{h}$ = 1.28571 h.

At 34 $\frac{km}{h}$ , Raj has been cycling for 1.28571 h to be 9 km ahead at 8.50 a.m.

So at 34 $\frac{km}{h}$ , Raj takes 1.28571 h to reach the office.

Step 4:	1h	→	34 km
	1.28571h	→	1.28571 x 34 km
		=	43.71414 km

Total distance travelled is 43.71414 km.

Step 5:	The time when Raj started cycling	→	8.50 a.m. - 1.28571 h
		=	8.50 a.m. - 1 h 17 min
		=	7.33 a.m.

Step 6:	Cycling time if Raj wants to reach school at 9.00 a.m.	→	9.00 a.m. - 7.33 a.m.
		=	1 h 27 min
		=	1 h + $\frac{27}{60}$ h
		≈	1.45 h

Step 7:	Average speed	→	43.71414 km ÷ 1.45 h
		=	30.14768 $\frac{km}{h}$
		≈	30.15 $\frac{km}{h}$

Answer: 24.00 $\frac{km}{h}$

SOLUTION 1 :

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

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

So at 28 $\frac{km}{h}$ , Sriram is ahead by 7 km at 8.50 a.m., compared to if Sriram cycles at a speed of 21 $\frac{km}{h}$ .

Step 2: 28 km - 21 km = 7 km

At 28 $\frac{km}{h}$ , Sriram is ahead by 7 km every hour.

Step 3: 7 km ÷ 7 $\frac{km}{h}$ = 1 h.

At 28 $\frac{km}{h}$ , Sriram has been cycling for 1 h to be 7 km ahead at 8.50 a.m.

So at 28 $\frac{km}{h}$ , Sriram takes 1 h to reach the office.

Step 4:	1h	→	28 km
	1h	→	1 x 28 km
		=	28 km

Total distance travelled is 28 km.

Step 5:	The time when Sriram started cycling	→	8.50 a.m. - 1 h
		=	8.50 a.m. - 1 h 0 min
		=	7.50 a.m.

Step 6:	Cycling time if Sriram wants to reach school at 9.00 a.m.	→	9.00 a.m. - 7.50 a.m.
		=	1 h 10 min
		=	1 h + $\frac{10}{60}$ h
		≈	1.16667 h

Step 7:	Average speed	→	28 km ÷ 1.16667 h
		=	23.99993 $\frac{km}{h}$
		≈	24.00 $\frac{km}{h}$

Answer: 25.85 $\frac{km}{h}$

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 28 $\frac{km}{h}$ , 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: 28 km - 24 km = 4 km

At 28 $\frac{km}{h}$ , Selvan is ahead by 4 km every hour.

Step 3: 8 km ÷ 4 $\frac{km}{h}$ = 2 h.

At 28 $\frac{km}{h}$ , Selvan has been cycling for 2 h to be 8 km ahead at 8.50 a.m.

So at 28 $\frac{km}{h}$ , Selvan takes 2 h to reach the office.

Step 4:	1h	→	28 km
	2h	→	2 x 28 km
		=	56 km

Total distance travelled is 56 km.

Step 5:	The time when Selvan started cycling	→	8.50 a.m. - 2 h
		=	8.50 a.m. - 2 h 0 min
		=	6.50 a.m.

Step 6:	Cycling time if Selvan wants to reach school at 9.00 a.m.	→	9.00 a.m. - 6.50 a.m.
		=	2 h 10 min
		=	2 h + $\frac{10}{60}$ h
		≈	2.16667 h

Step 7:	Average speed	→	56 km ÷ 2.16667 h
		=	25.84611 $\frac{km}{h}$
		≈	25.85 $\frac{km}{h}$

Answer: 31.87 $\frac{km}{h}$

SOLUTION 1 :

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

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

So at 34 $\frac{km}{h}$ , Sriram is ahead by 10 km at 8.50 a.m., compared to if Sriram cycles at a speed of 30 $\frac{km}{h}$ .

Step 2: 34 km - 30 km = 4 km

At 34 $\frac{km}{h}$ , Sriram is ahead by 4 km every hour.

Step 3: 10 km ÷ 4 $\frac{km}{h}$ = 2.5 h.

At 34 $\frac{km}{h}$ , Sriram has been cycling for 2.5 h to be 10 km ahead at 8.50 a.m.

So at 34 $\frac{km}{h}$ , Sriram takes 2.5 h to reach the office.

Step 4:	1h	→	34 km
	2.5h	→	2.5 x 34 km
		=	85 km

Total distance travelled is 85 km.

Step 5:	The time when Sriram started cycling	→	8.50 a.m. - 2.5 h
		=	8.50 a.m. - 2 h 30 min
		=	6.20 a.m.

Step 6:	Cycling time if Sriram wants to reach school at 9.00 a.m.	→	9.00 a.m. - 6.20 a.m.
		=	2 h 40 min
		=	2 h + $\frac{40}{60}$ h
		≈	2.66667 h

Step 7:	Average speed	→	85 km ÷ 2.66667 h
		=	31.87496 $\frac{km}{h}$
		≈	31.87 $\frac{km}{h}$

Answer: 28.86 $\frac{km}{h}$

SOLUTION 1 :

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

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

So at 31 $\frac{km}{h}$ , Balaji is ahead by 9 km at 8.50 a.m., compared to if Balaji cycles at a speed of 27 $\frac{km}{h}$ .

Step 2: 31 km - 27 km = 4 km

At 31 $\frac{km}{h}$ , Balaji is ahead by 4 km every hour.

Step 3: 9 km ÷ 4 $\frac{km}{h}$ = 2.25 h.

At 31 $\frac{km}{h}$ , Balaji has been cycling for 2.25 h to be 9 km ahead at 8.50 a.m.

So at 31 $\frac{km}{h}$ , Balaji takes 2.25 h to reach the office.

Step 4:	1h	→	31 km
	2.25h	→	2.25 x 31 km
		=	69.75 km

Total distance travelled is 69.75 km.

Step 5:	The time when Balaji started cycling	→	8.50 a.m. - 2.25 h
		=	8.50 a.m. - 2 h 15 min
		=	6.35 a.m.

Step 6:	Cycling time if Balaji wants to reach school at 9.00 a.m.	→	9.00 a.m. - 6.35 a.m.
		=	2 h 25 min
		=	2 h + $\frac{25}{60}$ h
		≈	2.41667 h

Step 7:	Average speed	→	69.75 km ÷ 2.41667 h
		=	28.86203 $\frac{km}{h}$
		≈	28.86 $\frac{km}{h}$

Answer: 22.83 $\frac{km}{h}$

SOLUTION 1 :

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

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

So at 25 $\frac{km}{h}$ , Pugal is ahead by 7 km at 8.50 a.m., compared to if Pugal cycles at a speed of 21 $\frac{km}{h}$ .

Step 2: 25 km - 21 km = 4 km

At 25 $\frac{km}{h}$ , Pugal is ahead by 4 km every hour.

Step 3: 7 km ÷ 4 $\frac{km}{h}$ = 1.75 h.

At 25 $\frac{km}{h}$ , Pugal has been cycling for 1.75 h to be 7 km ahead at 8.50 a.m.

So at 25 $\frac{km}{h}$ , Pugal takes 1.75 h to reach the office.

Step 4:	1h	→	25 km
	1.75h	→	1.75 x 25 km
		=	43.75 km

Total distance travelled is 43.75 km.

Step 5:	The time when Pugal started cycling	→	8.50 a.m. - 1.75 h
		=	8.50 a.m. - 1 h 45 min
		=	7.05 a.m.

Step 6:	Cycling time if Pugal wants to reach school at 9.00 a.m.	→	9.00 a.m. - 7.05 a.m.
		=	1 h 55 min
		=	1 h + $\frac{55}{60}$ h
		≈	1.91667 h

Step 7:	Average speed	→	43.75 km ÷ 1.91667 h
		=	22.82605 $\frac{km}{h}$
		≈	22.83 $\frac{km}{h}$

Answer: 36.26 $\frac{km}{h}$

SOLUTION 1 :

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

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

So at 40 $\frac{km}{h}$ , Balaji is ahead by 11 km at 8.50 a.m., compared to if Balaji cycles at a speed of 33 $\frac{km}{h}$ .

Step 2: 40 km - 33 km = 7 km

At 40 $\frac{km}{h}$ , Balaji is ahead by 7 km every hour.

Step 3: 11 km ÷ 7 $\frac{km}{h}$ = 1.57143 h.

At 40 $\frac{km}{h}$ , Balaji has been cycling for 1.57143 h to be 11 km ahead at 8.50 a.m.

So at 40 $\frac{km}{h}$ , Balaji takes 1.57143 h to reach the office.

Step 4:	1h	→	40 km
	1.57143h	→	1.57143 x 40 km
		=	62.8572 km

Total distance travelled is 62.8572 km.

Step 5:	The time when Balaji started cycling	→	8.50 a.m. - 1.57143 h
		=	8.50 a.m. - 1 h 34 min
		=	7.16 a.m.

Step 6:	Cycling time if Balaji wants to reach school at 9.00 a.m.	→	9.00 a.m. - 7.16 a.m.
		=	1 h 44 min
		=	1 h + $\frac{44}{60}$ h
		≈	1.73333 h

Step 7:	Average speed	→	62.8572 km ÷ 1.73333 h
		=	36.26384 $\frac{km}{h}$
		≈	36.26 $\frac{km}{h}$

Answer: 22.83 $\frac{km}{h}$

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 25 $\frac{km}{h}$ , 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: 25 km - 21 km = 4 km

At 25 $\frac{km}{h}$ , Murugan is ahead by 4 km every hour.

Step 3: 7 km ÷ 4 $\frac{km}{h}$ = 1.75 h.

At 25 $\frac{km}{h}$ , Murugan has been cycling for 1.75 h to be 7 km ahead at 8.50 a.m.

So at 25 $\frac{km}{h}$ , Murugan takes 1.75 h to reach the office.

Step 4:	1h	→	25 km
	1.75h	→	1.75 x 25 km
		=	43.75 km

Total distance travelled is 43.75 km.

Step 5:	The time when Murugan started cycling	→	8.50 a.m. - 1.75 h
		=	8.50 a.m. - 1 h 45 min
		=	7.05 a.m.

Step 6:	Cycling time if Murugan wants to reach school at 9.00 a.m.	→	9.00 a.m. - 7.05 a.m.
		=	1 h 55 min
		=	1 h + $\frac{55}{60}$ h
		≈	1.91667 h

Step 7:	Average speed	→	43.75 km ÷ 1.91667 h
		=	22.82605 $\frac{km}{h}$
		≈	22.83 $\frac{km}{h}$

Answer: 39.14 $\frac{km}{h}$

SOLUTION 1 :

Step 1: Distance to be covered from 8.50 a.m. to 9.10 a.m. if Aravindh cycles at 36 $\frac{km}{h}$ :

1h	→	36 km
20 min	→	$\frac{20}{60}$ x 36 km
	=	12 km.

So at 43 $\frac{km}{h}$ , Aravindh is ahead by 12 km at 8.50 a.m., compared to if Aravindh cycles at a speed of 36 $\frac{km}{h}$ .

Step 2: 43 km - 36 km = 7 km

At 43 $\frac{km}{h}$ , Aravindh is ahead by 7 km every hour.

Step 3: 12 km ÷ 7 $\frac{km}{h}$ = 1.71429 h.

At 43 $\frac{km}{h}$ , Aravindh has been cycling for 1.71429 h to be 12 km ahead at 8.50 a.m.

So at 43 $\frac{km}{h}$ , Aravindh takes 1.71429 h to reach the office.

Step 4:	1h	→	43 km
	1.71429h	→	1.71429 x 43 km
		=	73.71447 km

Total distance travelled is 73.71447 km.

Step 5:	The time when Aravindh started cycling	→	8.50 a.m. - 1.71429 h
		=	8.50 a.m. - 1 h 43 min
		=	7.07 a.m.

Step 6:	Cycling time if Aravindh wants to reach school at 9.00 a.m.	→	9.00 a.m. - 7.07 a.m.
		=	1 h 53 min
		=	1 h + $\frac{53}{60}$ h
		≈	1.88333 h

Step 7:	Average speed	→	73.71447 km ÷ 1.88333 h
		=	39.1405 $\frac{km}{h}$
		≈	39.14 $\frac{km}{h}$

10)

Answer: 25.85 $\frac{km}{h}$

SOLUTION 1 :

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

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

So at 28 $\frac{km}{h}$ , Chidambaram is ahead by 8 km at 8.50 a.m., compared to if Chidambaram cycles at a speed of 24 $\frac{km}{h}$ .

Step 2: 28 km - 24 km = 4 km

At 28 $\frac{km}{h}$ , Chidambaram is ahead by 4 km every hour.

Step 3: 8 km ÷ 4 $\frac{km}{h}$ = 2 h.

At 28 $\frac{km}{h}$ , Chidambaram has been cycling for 2 h to be 8 km ahead at 8.50 a.m.

So at 28 $\frac{km}{h}$ , Chidambaram takes 2 h to reach the office.

Step 4:	1h	→	28 km
	2h	→	2 x 28 km
		=	56 km

Total distance travelled is 56 km.

Step 5:	The time when Chidambaram started cycling	→	8.50 a.m. - 2 h
		=	8.50 a.m. - 2 h 0 min
		=	6.50 a.m.

Step 6:	Cycling time if Chidambaram wants to reach school at 9.00 a.m.	→	9.00 a.m. - 6.50 a.m.
		=	2 h 10 min
		=	2 h + $\frac{10}{60}$ h
		≈	2.16667 h

Step 7:	Average speed	→	56 km ÷ 2.16667 h
		=	25.84611 $\frac{km}{h}$
		≈	25.85 $\frac{km}{h}$