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.

Balaji 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:_______________

Sriram 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:_______________

Murugan 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:_______________

Kumar 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:_______________

Sriram 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 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 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:_______________

Karna 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:_______________

Chidambaram 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:_______________

Karna 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:_______________

10)

Kumar 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:_______________

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

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

So at 39 $\frac{km}{h}$ , 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: 39 km - 33 km = 6 km

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

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

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

So at 39 $\frac{km}{h}$ , Balaji 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 Balaji 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 Balaji 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: 17.50 $\frac{km}{h}$

SOLUTION 1 :

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

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

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

Step 2: 21 km - 15 km = 6 km

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

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

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

So at 21 $\frac{km}{h}$ , Sriram 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 Sriram 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 Sriram 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: 33.04 $\frac{km}{h}$

SOLUTION 1 :

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

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

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

Step 2: 37 km - 30 km = 7 km

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

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

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

So at 37 $\frac{km}{h}$ , Murugan 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 Murugan 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 Murugan 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: 17.79 $\frac{km}{h}$

SOLUTION 1 :

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

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

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

Step 2: 22 km - 15 km = 7 km

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

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

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

So at 22 $\frac{km}{h}$ , Kumar 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 Kumar 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 Kumar 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: 21.08 $\frac{km}{h}$

SOLUTION 1 :

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

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

So at 25 $\frac{km}{h}$ , Sriram is ahead by 6 km at 8.50 a.m., compared to if Sriram cycles at a speed of 18 $\frac{km}{h}$ .

Step 2: 25 km - 18 km = 7 km

At 25 $\frac{km}{h}$ , Sriram is ahead by 7 km every hour.

Step 3: 6 km ÷ 7 $\frac{km}{h}$ = 0.85714 h.

At 25 $\frac{km}{h}$ , Sriram has been cycling for 0.85714 h to be 6 km ahead at 8.50 a.m.

So at 25 $\frac{km}{h}$ , Sriram takes 0.85714 h to reach the office.

Step 4:	1h	→	25 km
	0.85714h	→	0.85714 x 25 km
		=	21.4285 km

Total distance travelled is 21.4285 km.

Step 5:	The time when Sriram started cycling	→	8.50 a.m. - 0.85714 h
		=	8.50 a.m. - 0 h 51 min
		=	7.99 a.m.

Step 6:	Cycling time if Sriram wants to reach school at 9.00 a.m.	→	9.00 a.m. - 7.99 a.m.
		=	1 h 0 min
		=	1 h + $\frac{0}{60}$ h
		≈	1.01667 h

Step 7:	Average speed	→	21.4285 km ÷ 1.01667 h
		=	21.07714 $\frac{km}{h}$
		≈	21.08 $\frac{km}{h}$

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

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

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

Step 2: 38 km - 33 km = 5 km

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

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

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

So at 38 $\frac{km}{h}$ , Aravindh 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 Aravindh 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 Aravindh 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: 17.14 $\frac{km}{h}$

SOLUTION 1 :

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

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

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

Step 2: 20 km - 15 km = 5 km

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

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

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

So at 20 $\frac{km}{h}$ , Karna 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 Karna 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 Karna 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: 23.23 $\frac{km}{h}$

SOLUTION 1 :

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

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

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

Step 2: 26 km - 21 km = 5 km

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

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

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

So at 26 $\frac{km}{h}$ , Chidambaram 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 Chidambaram 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 Chidambaram 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: 36.26 $\frac{km}{h}$

SOLUTION 1 :

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

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

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

Step 2: 40 km - 33 km = 7 km

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

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

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

So at 40 $\frac{km}{h}$ , Karna 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 Karna 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 Karna 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 $\frac{km}{h}$

10)

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 Kumar cycles at 27 $\frac{km}{h}$ :

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

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

Step 2: 33 km - 27 km = 6 km

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

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

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

So at 33 $\frac{km}{h}$ , Kumar 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 Kumar 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 Kumar 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 $\frac{km}{h}$