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If a man travels different distances $d_1,d_2,d_3$,... and so on in different time $t_1,t_2,t_3$ respectively then,

**Average speed** = $\text"total travelled distance"/\text"total time taken in travelling distance"$

= ${d_1 + d_2 + d_3 +...}/{t_1 + t_2 + t_3 +...}$

If a bus travels from A to B with the speed x km/h and returns from B to A with the speed y km/h,then the average speed will be **$({2xy}/{x + y})$**

If a man travels different distances $d_1,d_2,d_3$, and so on with different speeds $s_1,s_2,s_3$, respectively then,

**Average speed **= $({d_1 + d_2 + d_3 + ...})/{d_1/S_1 + d_2/S_2 + d_3/S_3 + ...}$

If a man travels at the speed of $s_1$, he reaches his destination $t_1$ late while he reaches $t_2$ before when he travels at $s_2$ speed, then the distance between the two places is

**D = ${(S_1 × S_2)(t_1 + t_2)}/{S_2 - S_1}$**

**Question-1) **

**A person crosses a 600 m long street in 5 minutes. What is his speed in km per hour? **

A) 3.6 km/hr

B) 7.2 km/hr

C) 8.4 km/hr

D) 10 km/hr

Ans- B

Solution:

Speed = ${600}/ {5} × 60$ m/sec. = 2 m/sec.

= 2 x ${18}/{5}$km/hr =7.2 km/hr

**Question-2) **

**Negesh walks from his home to Bus station. If he walks at the speed of 9 kmph, he reaches the bus station 3 minutes before the arrival of the bus. However, if he walks at the speed of 6 kmph, he misses the bus by 2 minutes. Find the distance covered by him to reach the bus station.**

A) 3.6 km

B) 1.5 km

C) 5.5 km

D) 12 km

Ans- B

Solution:

$d/6 - d/9$ = ${3+2}/60$

${3d-2d}/28 = 1/12$

Therefore, d = $18/12 = 1.5$ km

**Question-3) **

**A man riding his bicycle covers 150 metres in 25 seconds. What is his speed in km per hour ?.**

A) 21.6 km/hr

B) 15.5 km/hr

C) 25.5 km/hr

D) 12 km/hr

Ans- A

Solution:

Speed = $150/25$ = 6 m/sec

= $6 × 18/5 = 108/5$ = 21.6 km/hr

**Question-4) **

**Two men start together to walk a certain distance, one at 4 km/h and another at 3 km/h. The former arrives half an hour before the latter. Find the distance?**

A) 5 kms

B) 6 kms

C) 7 kms

D) 7.5 kms

Ans- B

Solution:

Here $S_1 = 4, t_1 = x, S_2 = 3, t_2 = x + 1/2$

$S_1t_1 = S_2t_2$

$4 × x = 3(x + 1/2)$

$4x - 3x = 3/2 x = 3/2$

Distance= $4 × 3/2$ = 6 kms

**Question-5) **

**You arrive at your school 5 minutes late if you walk with a speed of 4 km/h, but you arrive 10 minutes before the scheduled time if you walk with a speed of 5 km/h. The distance of your school from your house (in km) is**

A) 5 kms

B) 7 kms

C) 8 kms

D) 8.5 kms

Ans- A

Solution:

If a man travels at the speed of $S_1$, he reaches his destination $t_1$ late while he reaches $t_2$ before when he travels at $S_2$ speed,

then the distance between the two places is D = ${(S_1 × S_2)(t_1 + t_2)}/{S_2 - S_1}$

Here, $S_1 = 4, t_1 = 5, S_2 = 5, t_2$= 10

Distance =${(S_1 × S_2)(t_1 + t_2)}/{S_2 - S_1}$

=${(4 × 5)(5 + 10)}/{5 - 4}$

=20 × $15/60$ = 5 kms

- Total number of questions : 20
- Each question carries 1 mark
- Negative marks of Each Questions: 0.25
- Skiping questions No marks will deduct.
- Time allotted : 30 minutes