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.

Karna cycles to work every day. If he cycles at a speed of 26 $\frac{km}{h}$ , 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:_______________

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 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:_______________

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 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:_______________

Kalai 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:_______________

Moorthi 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:_______________

Selvan 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:_______________

Karna 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:_______________

Balaji 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:_______________

Narayanan cycles to work every day. If he cycles at a speed of 41 $\frac{km}{h}$ , 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)

Pugal 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:_______________

Answer: 23.23 $\frac{km}{h}$

SOLUTION 1 :

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

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

So at 26 $\frac{km}{h}$ , Karna is ahead by 7 km at 8.50 a.m., compared to if Karna cycles at a speed of 21 $\frac{km}{h}$ .

Step 2: 26 km - 21 km = 5 km

At 26 $\frac{km}{h}$ , Karna is ahead by 5 km every hour.

Step 3: 7 km ÷ 5 $\frac{km}{h}$ = 1.4 h.

At 26 $\frac{km}{h}$ , Karna has been cycling for 1.4 h to be 7 km ahead at 8.50 a.m.

So at 26 $\frac{km}{h}$ , Karna takes 1.4 h to reach the office.

Step 4:	1h	→	26 km
	1.4h	→	1.4 x 26 km
		=	36.4 km

Total distance travelled is 36.4 km.

Step 5:	The time when Karna started cycling	→	8.50 a.m. - 1.4 h
		=	8.50 a.m. - 1 h 24 min
		=	7.26 a.m.

Step 6:	Cycling time if Karna wants to reach school at 9.00 a.m.	→	9.00 a.m. - 7.26 a.m.
		=	1 h 34 min
		=	1 h + $\frac{34}{60}$ h
		≈	1.56667 h

Step 7:	Average speed	→	36.4 km ÷ 1.56667 h
		=	23.23399 $\frac{km}{h}$
		≈	23.23 $\frac{km}{h}$

Answer: 34.89 $\frac{km}{h}$

SOLUTION 1 :

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

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

So at 37 $\frac{km}{h}$ , Narayanan is ahead by 11 km at 8.50 a.m., compared to if Narayanan cycles at a speed of 33 $\frac{km}{h}$ .

Step 2: 37 km - 33 km = 4 km

At 37 $\frac{km}{h}$ , Narayanan is ahead by 4 km every hour.

Step 3: 11 km ÷ 4 $\frac{km}{h}$ = 2.75 h.

At 37 $\frac{km}{h}$ , Narayanan has been cycling for 2.75 h to be 11 km ahead at 8.50 a.m.

So at 37 $\frac{km}{h}$ , Narayanan takes 2.75 h to reach the office.

Step 4:	1h	→	37 km
	2.75h	→	2.75 x 37 km
		=	101.75 km

Total distance travelled is 101.75 km.

Step 5:	The time when Narayanan started cycling	→	8.50 a.m. - 2.75 h
		=	8.50 a.m. - 2 h 45 min
		=	6.05 a.m.

Step 6:	Cycling time if Narayanan wants to reach school at 9.00 a.m.	→	9.00 a.m. - 6.05 a.m.
		=	2 h 55 min
		=	2 h + $\frac{55}{60}$ h
		≈	2.91667 h

Step 7:	Average speed	→	101.75 km ÷ 2.91667 h
		=	34.88567 $\frac{km}{h}$
		≈	34.89 $\frac{km}{h}$

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

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

So at 22 $\frac{km}{h}$ , Pugal is ahead by 5 km at 8.50 a.m., compared to if Pugal cycles at a speed of 15 $\frac{km}{h}$ .

Step 2: 22 km - 15 km = 7 km

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

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

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

So at 22 $\frac{km}{h}$ , Pugal 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 Pugal 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 Pugal 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: 26.67 $\frac{km}{h}$

SOLUTION 1 :

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

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

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

Step 2: 30 km - 24 km = 6 km

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

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

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

So at 30 $\frac{km}{h}$ , Kalai 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 Kalai 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 Kalai 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: 35.32 $\frac{km}{h}$

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 38 $\frac{km}{h}$ , 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: 38 km - 33 km = 5 km

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

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

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

So at 38 $\frac{km}{h}$ , Moorthi 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 Moorthi 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 Moorthi 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 $\frac{km}{h}$

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 Selvan cycles at 33 $\frac{km}{h}$ :

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

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

Step 2: 40 km - 33 km = 7 km

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

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

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

So at 40 $\frac{km}{h}$ , Selvan 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 Selvan 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 Selvan 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: 23.63 $\frac{km}{h}$

SOLUTION 1 :

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

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

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

Step 2: 27 km - 21 km = 6 km

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

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

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

So at 27 $\frac{km}{h}$ , Karna 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 Karna 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 Karna 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: 26.67 $\frac{km}{h}$

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 30 $\frac{km}{h}$ , 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: 30 km - 24 km = 6 km

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

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

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

So at 30 $\frac{km}{h}$ , Balaji 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 Balaji 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 Balaji 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: 38.34 $\frac{km}{h}$

SOLUTION 1 :

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

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

So at 41 $\frac{km}{h}$ , Narayanan is ahead by 12 km at 8.50 a.m., compared to if Narayanan cycles at a speed of 36 $\frac{km}{h}$ .

Step 2: 41 km - 36 km = 5 km

At 41 $\frac{km}{h}$ , Narayanan is ahead by 5 km every hour.

Step 3: 12 km ÷ 5 $\frac{km}{h}$ = 2.4 h.

At 41 $\frac{km}{h}$ , Narayanan has been cycling for 2.4 h to be 12 km ahead at 8.50 a.m.

So at 41 $\frac{km}{h}$ , Narayanan takes 2.4 h to reach the office.

Step 4:	1h	→	41 km
	2.4h	→	2.4 x 41 km
		=	98.4 km

Total distance travelled is 98.4 km.

Step 5:	The time when Narayanan started cycling	→	8.50 a.m. - 2.4 h
		=	8.50 a.m. - 2 h 24 min
		=	6.26 a.m.

Step 6:	Cycling time if Narayanan wants to reach school at 9.00 a.m.	→	9.00 a.m. - 6.26 a.m.
		=	2 h 34 min
		=	2 h + $\frac{34}{60}$ h
		≈	2.56667 h

Step 7:	Average speed	→	98.4 km ÷ 2.56667 h
		=	38.33761 $\frac{km}{h}$
		≈	38.34 $\frac{km}{h}$

10)

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

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

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

Step 2: 19 km - 15 km = 4 km

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

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

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

So at 19 $\frac{km}{h}$ , Pugal 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 Pugal 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 Pugal 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 $\frac{km}{h}$