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.

Kumar 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:_______________

Prem 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:_______________

Barathan 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:_______________

Balaji 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:_______________

Prem 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:_______________

Kumar 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:_______________

Moorthi 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:_______________

Kalai 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 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)

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:_______________

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 Kumar cycles at 18 $\frac{km}{h}$ :

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

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

Step 2: 23 km - 18 km = 5 km

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

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

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

So at 23 $\frac{km}{h}$ , Kumar 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 Kumar 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 Kumar 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: 35.75 $\frac{km}{h}$

SOLUTION 1 :

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

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

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

Step 2: 39 km - 33 km = 6 km

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

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

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

So at 39 $\frac{km}{h}$ , Prem 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 Prem 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 Prem 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: 19.80 $\frac{km}{h}$

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 22 $\frac{km}{h}$ , 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: 22 km - 18 km = 4 km

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

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

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

So at 22 $\frac{km}{h}$ , Barathan 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 Barathan 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 Barathan 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 $\frac{km}{h}$

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

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

So at 25 $\frac{km}{h}$ , Balaji is ahead by 7 km at 8.50 a.m., compared to if Balaji cycles at a speed of 21 $\frac{km}{h}$ .

Step 2: 25 km - 21 km = 4 km

At 25 $\frac{km}{h}$ , Balaji is ahead by 4 km every hour.

Step 3: 7 km ÷ 4 $\frac{km}{h}$ = 1.75 h.

At 25 $\frac{km}{h}$ , Balaji has been cycling for 1.75 h to be 7 km ahead at 8.50 a.m.

So at 25 $\frac{km}{h}$ , Balaji 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 Balaji 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 Balaji 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 Prem cycles at 24 $\frac{km}{h}$ :

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

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

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

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

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

So at 31 $\frac{km}{h}$ , Prem 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 Prem 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 Prem 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: 32.31 $\frac{km}{h}$

SOLUTION 1 :

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

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

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

Step 2: 35 km - 30 km = 5 km

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

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

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

So at 35 $\frac{km}{h}$ , Kumar 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 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	→	70 km ÷ 2.16667 h
		=	32.30764 $\frac{km}{h}$
		≈	32.31 $\frac{km}{h}$

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

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

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

Step 2: 40 km - 36 km = 4 km

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

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

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

So at 40 $\frac{km}{h}$ , Moorthi 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 Moorthi 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 Moorthi 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: 20.20 $\frac{km}{h}$

SOLUTION 1 :

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

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

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

Step 2: 23 km - 18 km = 5 km

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

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

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

So at 23 $\frac{km}{h}$ , Kalai 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 Kalai 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 Kalai 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: 25.85 $\frac{km}{h}$

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 28 $\frac{km}{h}$ , 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: 28 km - 24 km = 4 km

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

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

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

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