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.

Kalai cycles to work every day. If he cycles at a speed of 21 $\frac{km}{h}$ , 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:_______________

Pugal cycles to work every day. If he cycles at a speed of 24 $\frac{km}{h}$ , 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:_______________

Raj cycles to work every day. If he cycles at a speed of 20 $\frac{km}{h}$ , 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:_______________

Pugal 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:_______________

Barathan cycles to work every day. If he cycles at a speed of 24 $\frac{km}{h}$ , 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:_______________

Aravindh cycles to work every day. If he cycles at a speed of 35 $\frac{km}{h}$ , 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:_______________

Aravindh 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 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 35 $\frac{km}{h}$ , 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:_______________

Kumar 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:_______________

10)

Balaji 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:_______________

Answer: 17.50 $\frac{km}{h}$

SOLUTION 1 :

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

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

So at 21 $\frac{km}{h}$ , Kalai is ahead by 5 km at 8.50 a.m., compared to if Kalai cycles at a speed of 15 $\frac{km}{h}$ .

Step 2: 21 km - 15 km = 6 km

At 21 $\frac{km}{h}$ , Kalai is ahead by 6 km every hour.

Step 3: 5 km ÷ 6 $\frac{km}{h}$ = 0.83333 h.

At 21 $\frac{km}{h}$ , Kalai has been cycling for 0.83333 h to be 5 km ahead at 8.50 a.m.

So at 21 $\frac{km}{h}$ , Kalai takes 0.83333 h to reach the office.

Step 4:	1h	→	21 km
	0.83333h	→	0.83333 x 21 km
		=	17.49993 km

Total distance travelled is 17.49993 km.

Step 5:	The time when Kalai started cycling	→	8.50 a.m. - 0.83333 h
		=	8.50 a.m. - 0 h 50 min
		=	8.00 a.m.

Step 6:	Cycling time if Kalai wants to reach school at 9.00 a.m.	→	9.00 a.m. - 8.00 a.m.
		=	1 h 0 min
		=	1 h + $\frac{0}{60}$ h
		≈	1 h

Step 7:	Average speed	→	17.49993 km ÷ 1 h
		=	17.49993 $\frac{km}{h}$
		≈	17.50 $\frac{km}{h}$

Answer: 20.57 $\frac{km}{h}$

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 24 $\frac{km}{h}$ , 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: 24 km - 18 km = 6 km

At 24 $\frac{km}{h}$ , Pugal is ahead by 6 km every hour.

Step 3: 6 km ÷ 6 $\frac{km}{h}$ = 1 h.

At 24 $\frac{km}{h}$ , Pugal has been cycling for 1 h to be 6 km ahead at 8.50 a.m.

So at 24 $\frac{km}{h}$ , Pugal takes 1 h to reach the office.

Step 4:	1h	→	24 km
	1h	→	1 x 24 km
		=	24 km

Total distance travelled is 24 km.

Step 5:	The time when Pugal 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 Pugal 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	→	24 km ÷ 1.16667 h
		=	20.57137 $\frac{km}{h}$
		≈	20.57 $\frac{km}{h}$

Answer: 17.14 $\frac{km}{h}$

SOLUTION 1 :

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

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

So at 20 $\frac{km}{h}$ , Raj is ahead by 5 km at 8.50 a.m., compared to if Raj cycles at a speed of 15 $\frac{km}{h}$ .

Step 2: 20 km - 15 km = 5 km

At 20 $\frac{km}{h}$ , Raj is ahead by 5 km every hour.

Step 3: 5 km ÷ 5 $\frac{km}{h}$ = 1 h.

At 20 $\frac{km}{h}$ , Raj has been cycling for 1 h to be 5 km ahead at 8.50 a.m.

So at 20 $\frac{km}{h}$ , Raj takes 1 h to reach the office.

Step 4:	1h	→	20 km
	1h	→	1 x 20 km
		=	20 km

Total distance travelled is 20 km.

Step 5:	The time when Raj 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 Raj 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	→	20 km ÷ 1.16667 h
		=	17.14281 $\frac{km}{h}$
		≈	17.14 $\frac{km}{h}$

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

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

So at 28 $\frac{km}{h}$ , 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: 28 km - 21 km = 7 km

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

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

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

So at 28 $\frac{km}{h}$ , Pugal 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 Pugal 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 Pugal 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: 20.57 $\frac{km}{h}$

SOLUTION 1 :

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

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

So at 24 $\frac{km}{h}$ , Barathan is ahead by 6 km at 8.50 a.m., compared to if Barathan cycles at a speed of 18 $\frac{km}{h}$ .

Step 2: 24 km - 18 km = 6 km

At 24 $\frac{km}{h}$ , Barathan is ahead by 6 km every hour.

Step 3: 6 km ÷ 6 $\frac{km}{h}$ = 1 h.

At 24 $\frac{km}{h}$ , Barathan has been cycling for 1 h to be 6 km ahead at 8.50 a.m.

So at 24 $\frac{km}{h}$ , Barathan takes 1 h to reach the office.

Step 4:	1h	→	24 km
	1h	→	1 x 24 km
		=	24 km

Total distance travelled is 24 km.

Step 5:	The time when Barathan 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 Barathan 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	→	24 km ÷ 1.16667 h
		=	20.57137 $\frac{km}{h}$
		≈	20.57 $\frac{km}{h}$

Answer: 32.31 $\frac{km}{h}$

SOLUTION 1 :

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

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

So at 35 $\frac{km}{h}$ , Aravindh is ahead by 10 km at 8.50 a.m., compared to if Aravindh cycles at a speed of 30 $\frac{km}{h}$ .

Step 2: 35 km - 30 km = 5 km

At 35 $\frac{km}{h}$ , Aravindh is ahead by 5 km every hour.

Step 3: 10 km ÷ 5 $\frac{km}{h}$ = 2 h.

At 35 $\frac{km}{h}$ , Aravindh has been cycling for 2 h to be 10 km ahead at 8.50 a.m.

So at 35 $\frac{km}{h}$ , Aravindh takes 2 h to reach the office.

Step 4:	1h	→	35 km
	2h	→	2 x 35 km
		=	70 km

Total distance travelled is 70 km.

Step 5:	The time when Aravindh 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 Aravindh 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	→	70 km ÷ 2.16667 h
		=	32.30764 $\frac{km}{h}$
		≈	32.31 $\frac{km}{h}$

Answer: 17.79 $\frac{km}{h}$

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 22 $\frac{km}{h}$ , 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: 22 km - 15 km = 7 km

At 22 $\frac{km}{h}$ , Aravindh is ahead by 7 km every hour.

Step 3: 5 km ÷ 7 $\frac{km}{h}$ = 0.71429 h.

At 22 $\frac{km}{h}$ , Aravindh has been cycling for 0.71429 h to be 5 km ahead at 8.50 a.m.

So at 22 $\frac{km}{h}$ , Aravindh takes 0.71429 h to reach the office.

Step 4:	1h	→	22 km
	0.71429h	→	0.71429 x 22 km
		=	15.71438 km

Total distance travelled is 15.71438 km.

Step 5:	The time when Aravindh started cycling	→	8.50 a.m. - 0.71429 h
		=	8.50 a.m. - 0 h 43 min
		=	8.07 a.m.

Step 6:	Cycling time if Aravindh wants to reach school at 9.00 a.m.	→	9.00 a.m. - 8.07 a.m.
		=	0 h 53 min
		=	0 h + $\frac{53}{60}$ h
		≈	0.88333 h

Step 7:	Average speed	→	15.71438 km ÷ 0.88333 h
		=	17.78993 $\frac{km}{h}$
		≈	17.79 $\frac{km}{h}$

Answer: 32.31 $\frac{km}{h}$

SOLUTION 1 :

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

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

So at 35 $\frac{km}{h}$ , Balaji is ahead by 10 km at 8.50 a.m., compared to if Balaji cycles at a speed of 30 $\frac{km}{h}$ .

Step 2: 35 km - 30 km = 5 km

At 35 $\frac{km}{h}$ , Balaji is ahead by 5 km every hour.

Step 3: 10 km ÷ 5 $\frac{km}{h}$ = 2 h.

At 35 $\frac{km}{h}$ , Balaji has been cycling for 2 h to be 10 km ahead at 8.50 a.m.

So at 35 $\frac{km}{h}$ , Balaji takes 2 h to reach the office.

Step 4:	1h	→	35 km
	2h	→	2 x 35 km
		=	70 km

Total distance travelled is 70 km.

Step 5:	The time when Balaji 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 Balaji 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	→	70 km ÷ 2.16667 h
		=	32.30764 $\frac{km}{h}$
		≈	32.31 $\frac{km}{h}$

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

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

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

Step 2: 28 km - 24 km = 4 km

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

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

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

So at 28 $\frac{km}{h}$ , Kumar 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 Kumar 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 Kumar 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 $\frac{km}{h}$

10)

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

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

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

Step 2: 29 km - 24 km = 5 km

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

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

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

So at 29 $\frac{km}{h}$ , Balaji 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 Balaji 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 Balaji 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 $\frac{km}{h}$