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.

Narayanan 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:_______________

Barathan 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:_______________

Selvan 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:_______________

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:_______________

Narayanan 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 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:_______________

Barathan 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:_______________

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:_______________

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 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:_______________

Barathan 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:_______________

10)

Aravindh 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:_______________

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 Narayanan cycles at 36 $\frac{km}{h}$ :

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

So at 43 $\frac{km}{h}$ , 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: 43 km - 36 km = 7 km

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

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

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

So at 43 $\frac{km}{h}$ , Narayanan 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 Narayanan 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 Narayanan 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 $\frac{km}{h}$

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

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

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

Step 2: 28 km - 24 km = 4 km

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

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

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

So at 28 $\frac{km}{h}$ , Barathan 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 Barathan 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 Barathan 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: 30.15 $\frac{km}{h}$

SOLUTION 1 :

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

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

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

Step 2: 34 km - 27 km = 7 km

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

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

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

So at 34 $\frac{km}{h}$ , Selvan 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 Selvan 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 Selvan 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: 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: 26.91 $\frac{km}{h}$

SOLUTION 1 :

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

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

So at 31 $\frac{km}{h}$ , Narayanan is ahead by 8 km at 8.50 a.m., compared to if Narayanan cycles at a speed of 24 $\frac{km}{h}$ .

Step 2: 31 km - 24 km = 7 km

At 31 $\frac{km}{h}$ , Narayanan is ahead by 7 km every hour.

Step 3: 8 km ÷ 7 $\frac{km}{h}$ = 1.14286 h.

At 31 $\frac{km}{h}$ , Narayanan has been cycling for 1.14286 h to be 8 km ahead at 8.50 a.m.

So at 31 $\frac{km}{h}$ , Narayanan takes 1.14286 h to reach the office.

Step 4:	1h	→	31 km
	1.14286h	→	1.14286 x 31 km
		=	35.42866 km

Total distance travelled is 35.42866 km.

Step 5:	The time when Narayanan started cycling	→	8.50 a.m. - 1.14286 h
		=	8.50 a.m. - 1 h 9 min
		=	7.41 a.m.

Step 6:	Cycling time if Narayanan wants to reach school at 9.00 a.m.	→	9.00 a.m. - 7.41 a.m.
		=	1 h 19 min
		=	1 h + $\frac{19}{60}$ h
		≈	1.31667 h

Step 7:	Average speed	→	35.42866 km ÷ 1.31667 h
		=	26.90777 $\frac{km}{h}$
		≈	26.91 $\frac{km}{h}$

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

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

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

Step 2: 39 km - 33 km = 6 km

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

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

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

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

Answer: 26.91 $\frac{km}{h}$

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 31 $\frac{km}{h}$ , 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: 31 km - 24 km = 7 km

At 31 $\frac{km}{h}$ , Balaji is ahead by 7 km every hour.

Step 3: 8 km ÷ 7 $\frac{km}{h}$ = 1.14286 h.

At 31 $\frac{km}{h}$ , Balaji has been cycling for 1.14286 h to be 8 km ahead at 8.50 a.m.

So at 31 $\frac{km}{h}$ , Balaji takes 1.14286 h to reach the office.

Step 4:	1h	→	31 km
	1.14286h	→	1.14286 x 31 km
		=	35.42866 km

Total distance travelled is 35.42866 km.

Step 5:	The time when Balaji started cycling	→	8.50 a.m. - 1.14286 h
		=	8.50 a.m. - 1 h 9 min
		=	7.41 a.m.

Step 6:	Cycling time if Balaji wants to reach school at 9.00 a.m.	→	9.00 a.m. - 7.41 a.m.
		=	1 h 19 min
		=	1 h + $\frac{19}{60}$ h
		≈	1.31667 h

Step 7:	Average speed	→	35.42866 km ÷ 1.31667 h
		=	26.90777 $\frac{km}{h}$
		≈	26.91 $\frac{km}{h}$

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

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

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

Step 2: 21 km - 15 km = 6 km

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

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

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

So at 21 $\frac{km}{h}$ , Barathan 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 Barathan 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 Barathan 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 $\frac{km}{h}$

10)

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

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

So at 21 $\frac{km}{h}$ , 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: 21 km - 15 km = 6 km

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

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

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

So at 21 $\frac{km}{h}$ , Aravindh 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 Aravindh 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 Aravindh 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 $\frac{km}{h}$