**Question 11:**A vendor takes 10 hours to walk to a certain place and cycle back. He would have gained three hours if he had covered both routes on cycle. How long would he have taken to walk both ways?

[1] 6.5 hrs

[2] 26 hrs

[3] 33 hrs

[4] 13 hrs

**Answer:**

**Explanation: **

One way walk +one way cycling = 10 hrs. Now multiply both sides by 2.

We get 2-way walk + 2-way cycling = 20 hrs ....(1)

Also given: 2-way cycling = 7 hrs ....(2)

Subtract (2) from (1) to get: 2-way walk = 20 - 7 = 13 hrs.

**Question 12:**If Sita walks at 5 kmph, she misses her train by 10 minutes. If she walks at 7 kmph, she reaches the station 10 minutes early. How much distance does she walk to the station?

[1] 5.8 km

[2] 35.6 km

[3] 10.6 km

[4] 92 km

**Answer:**

**Explanation: **

Let the distance be D.

\(\frac{D}{5} - \frac{D}{7} = \frac{{10 + 10}}{{60}}\)

\( \Rightarrow D = \frac{{35}}{6} = 5.8km\)

**Question 13:**Rahul can cover the distance to his school in 15 minutes. He covers 2/3rd of it at the rate of 5 kmph and the remaining at the rate of 10 kmph. The total distance to the school is

[1] 15 km

[2] 1.5 km

[3] 3 km

[4] 6km

**Answer:**

**Explanation: **

Let the distance to his school be x km.

According to the given conditions,

\(\frac{{2x}}{3} \times \frac{1}{5} + \frac{x}{3} \times \frac{1}{{10}} = \frac{{15}}{{60}}\)

\( \Rightarrow \frac{{4x + x}}{{30}} = \frac{1}{4} \Rightarrow x = \frac{{30}}{{4 \times 5}} = \frac{3}{2} = 1.5km\)

**Question 14:**A friend is spotted by Lalloo at a distance of 200 m. When Lalloo starts to approach him, the friend also starts moving in the same direction as Lalloo. If the speed of his friend is 15 kmph, and that of Lalloo is 20 kmph, then how far will the friend have to walk before Lalloo meets him?

[1] 600 m

[2] 0.6 m

[3] 6km

[4] 900 m

**Answer:**

**Explanation: **

Lalloo is unfortunate that the friend is moving away from him.

(Because the friend moves in same direction as Lalloo).

relative speed= 20- 15= 5,kmph. distance= 200 m.

Thus, Lalloo will meet his friend when he gains 200 m over him.

=> time required = distance / speed = 0.2/5 = 1/25 hrs.

=>Distance travelled by the friend in 1/25 hrs. (when Lalloo catches up him)

=> Time x Speed = 1/25 x 15 = 3/5 km = 600 m.

**Question 15:**A distance is covered at a certain speed in a certain time. If the double of this distance is covered in four times the time, then what is the ratio of the two speeds?

[1] 1.5 : 0.7

[2] 1 : 1.9

[3] 4 : 2

[4] 6 : 1

**Answer:**

**Explanation: **

Case I : Distance D Speed \({S_1}\) Time \(D/{S_1}\)

Case II : Distance 2D Speed \({S_2}\) Time \(4\left( {D/{S_1}} \right)\)

=> Speed for case II = \({S_2}\) = Distance/Time = \(2D/\left( {4D/{S_1}} \right) = {S_1}/2\)

Hence, speed for case I : speed for case II. = \({S_1}:{S_1}/2 = 1:1/2 = 2:1\)