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.

Prem 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:_______________

Moorthi cycles to work every day. If he cycles at a speed of 39 $\frac{km}{h}$ , 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:_______________

Moorthi cycles to work every day. If he cycles at a speed of 23 $\frac{km}{h}$ , he would reach his office at 8.50 a.m. If he cycles at 18 $\frac{km}{h}$ , 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 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:_______________

Murugan 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 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:_______________

Karna cycles to work every day. If he cycles at a speed of 30 $\frac{km}{h}$ , 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:_______________

Aravindh 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:_______________

Balaji 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:_______________

Pugal cycles to work every day. If he cycles at a speed of 22 $\frac{km}{h}$ , he would reach his office at 8.50 a.m. If he cycles at 18 $\frac{km}{h}$ , 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)

Sriram 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:_______________

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 Prem cycles at 24 $\frac{km}{h}$ :

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

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

Step 2: 29 km - 24 km = 5 km

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

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

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

So at 29 $\frac{km}{h}$ , Prem 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 Prem 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 Prem 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: 35.75 $\frac{km}{h}$

SOLUTION 1 :

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

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

So at 39 $\frac{km}{h}$ , Moorthi is ahead by 11 km at 8.50 a.m., compared to if Moorthi cycles at a speed of 33 $\frac{km}{h}$ .

Step 2: 39 km - 33 km = 6 km

At 39 $\frac{km}{h}$ , Moorthi is ahead by 6 km every hour.

Step 3: 11 km ÷ 6 $\frac{km}{h}$ = 1.83333 h.

At 39 $\frac{km}{h}$ , Moorthi has been cycling for 1.83333 h to be 11 km ahead at 8.50 a.m.

So at 39 $\frac{km}{h}$ , Moorthi takes 1.83333 h to reach the office.

Step 4:	1h	→	39 km
	1.83333h	→	1.83333 x 39 km
		=	71.49987 km

Total distance travelled is 71.49987 km.

Step 5:	The time when Moorthi started cycling	→	8.50 a.m. - 1.83333 h
		=	8.50 a.m. - 1 h 50 min
		=	7.00 a.m.

Step 6:	Cycling time if Moorthi wants to reach school at 9.00 a.m.	→	9.00 a.m. - 7.00 a.m.
		=	2 h 0 min
		=	2 h + $\frac{0}{60}$ h
		≈	2 h

Step 7:	Average speed	→	71.49987 km ÷ 2 h
		=	35.74994 $\frac{km}{h}$
		≈	35.75 $\frac{km}{h}$

Answer: 20.20 $\frac{km}{h}$

SOLUTION 1 :

Step 1: Distance to be covered from 8.50 a.m. to 9.10 a.m. if Moorthi cycles at 18 $\frac{km}{h}$ :

1h	→	18 km
20 min	→	$\frac{20}{60}$ x 18 km
	=	6 km.

So at 23 $\frac{km}{h}$ , Moorthi is ahead by 6 km at 8.50 a.m., compared to if Moorthi cycles at a speed of 18 $\frac{km}{h}$ .

Step 2: 23 km - 18 km = 5 km

At 23 $\frac{km}{h}$ , Moorthi is ahead by 5 km every hour.

Step 3: 6 km ÷ 5 $\frac{km}{h}$ = 1.2 h.

At 23 $\frac{km}{h}$ , Moorthi has been cycling for 1.2 h to be 6 km ahead at 8.50 a.m.

So at 23 $\frac{km}{h}$ , Moorthi takes 1.2 h to reach the office.

Step 4:	1h	→	23 km
	1.2h	→	1.2 x 23 km
		=	27.6 km

Total distance travelled is 27.6 km.

Step 5:	The time when Moorthi started cycling	→	8.50 a.m. - 1.2 h
		=	8.50 a.m. - 1 h 12 min
		=	7.38 a.m.

Step 6:	Cycling time if Moorthi wants to reach school at 9.00 a.m.	→	9.00 a.m. - 7.38 a.m.
		=	1 h 22 min
		=	1 h + $\frac{22}{60}$ h
		≈	1.36667 h

Step 7:	Average speed	→	27.6 km ÷ 1.36667 h
		=	20.19507 $\frac{km}{h}$
		≈	20.20 $\frac{km}{h}$

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 Kalai cycles at 21 $\frac{km}{h}$ :

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

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

Step 2: 28 km - 21 km = 7 km

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

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

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

So at 28 $\frac{km}{h}$ , Kalai 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 Kalai 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 Kalai 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: 37.89 $\frac{km}{h}$

SOLUTION 1 :

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

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

So at 40 $\frac{km}{h}$ , Murugan is ahead by 12 km at 8.50 a.m., compared to if Murugan cycles at a speed of 36 $\frac{km}{h}$ .

Step 2: 40 km - 36 km = 4 km

At 40 $\frac{km}{h}$ , Murugan is ahead by 4 km every hour.

Step 3: 12 km ÷ 4 $\frac{km}{h}$ = 3 h.

At 40 $\frac{km}{h}$ , Murugan has been cycling for 3 h to be 12 km ahead at 8.50 a.m.

So at 40 $\frac{km}{h}$ , Murugan takes 3 h to reach the office.

Step 4:	1h	→	40 km
	3h	→	3 x 40 km
		=	120 km

Total distance travelled is 120 km.

Step 5:	The time when Murugan started cycling	→	8.50 a.m. - 3 h
		=	8.50 a.m. - 3 h 0 min
		=	5.50 a.m.

Step 6:	Cycling time if Murugan wants to reach school at 9.00 a.m.	→	9.00 a.m. - 5.50 a.m.
		=	3 h 10 min
		=	3 h + $\frac{10}{60}$ h
		≈	3.16667 h

Step 7:	Average speed	→	120 km ÷ 3.16667 h
		=	37.8947 $\frac{km}{h}$
		≈	37.89 $\frac{km}{h}$

Answer: 26.67 $\frac{km}{h}$

SOLUTION 1 :

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

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

So at 30 $\frac{km}{h}$ , Karna is ahead by 8 km at 8.50 a.m., compared to if Karna cycles at a speed of 24 $\frac{km}{h}$ .

Step 2: 30 km - 24 km = 6 km

At 30 $\frac{km}{h}$ , Karna is ahead by 6 km every hour.

Step 3: 8 km ÷ 6 $\frac{km}{h}$ = 1.33333 h.

At 30 $\frac{km}{h}$ , Karna has been cycling for 1.33333 h to be 8 km ahead at 8.50 a.m.

So at 30 $\frac{km}{h}$ , Karna takes 1.33333 h to reach the office.

Step 4:	1h	→	30 km
	1.33333h	→	1.33333 x 30 km
		=	39.9999 km

Total distance travelled is 39.9999 km.

Step 5:	The time when Karna started cycling	→	8.50 a.m. - 1.33333 h
		=	8.50 a.m. - 1 h 20 min
		=	7.30 a.m.

Step 6:	Cycling time if Karna wants to reach school at 9.00 a.m.	→	9.00 a.m. - 7.30 a.m.
		=	1 h 30 min
		=	1 h + $\frac{30}{60}$ h
		≈	1.5 h

Step 7:	Average speed	→	39.9999 km ÷ 1.5 h
		=	26.6666 $\frac{km}{h}$
		≈	26.67 $\frac{km}{h}$

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 Aravindh cycles at 15 $\frac{km}{h}$ :

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

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

Step 2: 19 km - 15 km = 4 km

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

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

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

So at 19 $\frac{km}{h}$ , Aravindh 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 Aravindh 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 Aravindh 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: 30.15 $\frac{km}{h}$

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 34 $\frac{km}{h}$ , 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: 34 km - 27 km = 7 km

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

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

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

So at 34 $\frac{km}{h}$ , Balaji 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 Balaji 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 Balaji 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: 19.80 $\frac{km}{h}$

SOLUTION 1 :

Step 1: Distance to be covered from 8.50 a.m. to 9.10 a.m. if Pugal cycles at 18 $\frac{km}{h}$ :

1h	→	18 km
20 min	→	$\frac{20}{60}$ x 18 km
	=	6 km.

So at 22 $\frac{km}{h}$ , Pugal is ahead by 6 km at 8.50 a.m., compared to if Pugal cycles at a speed of 18 $\frac{km}{h}$ .

Step 2: 22 km - 18 km = 4 km

At 22 $\frac{km}{h}$ , Pugal is ahead by 4 km every hour.

Step 3: 6 km ÷ 4 $\frac{km}{h}$ = 1.5 h.

At 22 $\frac{km}{h}$ , Pugal has been cycling for 1.5 h to be 6 km ahead at 8.50 a.m.

So at 22 $\frac{km}{h}$ , Pugal takes 1.5 h to reach the office.

Step 4:	1h	→	22 km
	1.5h	→	1.5 x 22 km
		=	33 km

Total distance travelled is 33 km.

Step 5:	The time when Pugal 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 Pugal 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	→	33 km ÷ 1.66667 h
		=	19.79996 $\frac{km}{h}$
		≈	19.80 $\frac{km}{h}$

10)

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 Sriram cycles at 30 $\frac{km}{h}$ :

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

So at 37 $\frac{km}{h}$ , 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: 37 km - 30 km = 7 km

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

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

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

So at 37 $\frac{km}{h}$ , Sriram 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 Sriram 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 Sriram 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 $\frac{km}{h}$