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If one student walk from his house to school at 5 km/ph, he late by 30 minutes. However, if he walks at 6 km/ph, he is late by 5 minutes only. The distance of his school from his house is……

A.

2.5 km

B.

3.6 km

C.

5.5 km

D.

12.5 km

Answer with explanation

Answer: Option DExplanation

Workspace

A.

4

B.

6

C.

8

D.

10

Answer with explanation

Answer: Option CExplanation

Amy can travel clockwise or anticlockwise on the diagram.

Clockwise, she has no choice of route from A to B, a choice of one out of two routes from B to C, and a choice of one out of two routes from C back to A. This gives four possible routes.

Similarly, anticlockwise she has four different routes.

Total routes = 8

Workspace

A.

15 km/hr

B.

20 km/hr

C.

D.

30 km/hr

Answer with explanation

Answer: Option BExplanation

Let the second racer takes ‘x’ hr with speed S_{2}

and the first racer takes [x – (5/6)] hr with speed S_{1}

Given, Total distance = 50 km

Then, S_{1} = 50/[x – (5/6)]

S_{2} = 50/x

Total speed = S_{1} + S_{2}

= {50/[x – (5/6)]} + [50/x]

= 50 {1/[x – (5/6)] + [1/x]}

= 50 {x + [x – (5/6)] / [x – (5/6)]x}

= 50 {[2x – (5/6)] / [x^{2} – (5x/6)]}

= 50 {[(12x – 5)/6] / [(6x^{2} – 5x)/6]}

= 50 {[12x – 5] / [6x^{2} – 5x]}

As they cross each other in 1hr,

Now, Time = Distance/ Total Speed

—> 1 = 50 / (50 {[12x – 5] / [6x^{2} – 5x]})

—> 1 = 1 / {[12x – 5] / [6x^{2} – 5x]}

—> 1 = [6x^{2} – 5x] / [12x – 5]

—> [12x – 5] = [6x^{2} – 5x]

—> 6x^{2} – 5x – 12x + 5 = 0

—> 6x^{2} – 17x + 5 = 0

—> 6x^{2} – 15x – 2x + 5 = 0

—> 3x(2x – 5) – 1(2x – 5) = 0

—> (2x – 5) (3x – 1) = 0

—> x = 5/2, 1/3

[Neglecting x = 1/3]

Now, We have to find the speed of the slower racer ie., second racer.

Put x = 5/2 in S_{2}, we get

S_{2} = 50/(5/2)

= (50 * 2) / 5

= 20 km/hr

Thus, the speed of the slower racer = 20 km/hr.

Workspace

Sravan drove from home to a neighboring town at the speed of 50 km/h and on his returning journey, he drove at the speed of 45 km/h and also took an hour longer to reach home. What distance did he cover?

A.

350 kms

B.

450 kms

C.

700 kms

D.

900 kms

Answer with explanation

Answer: Option DExplanation

Let the distance he covered each way = d kms

According to the question,

d/45 – d/50 = 1

=> d = 450 kms.

Hence, the total distance he covered in his way = d + d = 2 d = 2 x 450 = 900 kms.

Workspace

Jennifer travels first 4 hours of her journey at a speed of 80 miles/hr and the remaining distance in 6 hours at a speed of 30 miles/hr. What is her average speed in miles/hr?

A.

50 miles / hr

B.

60 miles / hr

C.

75 miles / hr

D.

92 miles / hr

Answer with explanation

Answer: Option AExplanation

Workspace

A man leaves a point P at 6 a.m. and reaches the point Q at 10 a.m. another man leaves the point give at 8 a.m. and reaches the point P at 12 noon. At what time do they meet?

A.

8 a.m.

B.

8.30 a.m

C.

9 a.m.

D.

9.30 a.m.

Answer with explanation

Answer: Option CExplanation

let total distance is 2s

first person takes 4hr to cover the distance

then speed= 2s/4= s/2 km/hr

second person also takes 4hr to cover the distance

then speed= 2s/4= s/2 km/hr

so their speeds are same

when second person starts the journey while first person completes the half of the total journey

so next 1hr they met each other

8+1=9a.m

Workspace

A.

30 km/h

B.

48 km/h

C.

42 km/h

D.

35 km/h

Answer with explanation

Answer: Option AExplanation

Workspace

An aeroplane flying 1000 km covers the first 200 km at the rate of 200 km/hr, the second 200 km at 400 km/hr, the third 200 km at 600 km / hr & last 200 km at the rate of 800 km/hr. Determine the average speed of the aeroplane?

A.

250 km/hr

B.

300 km/hr

C.

480 km/hr

D.

600 km/hr

Answer with explanation

Answer: Option CExplanation

Workspace

Two girls move in opposite directions, one from A to B and other from B to A. The girl from A reaches the destination in 16 hrs and girl from B reaches her destination in 25 hrs, after having met. If former’s speed is 25 km/hr, what will be the speed of latter?

A.

10 km/hr

B.

12 km/hr

C.

16 km/hr

D.

20 km/hr

Answer with explanation

Answer: Option DExplanation

Workspace

Ramesh says, “Driving at an average speed of 60 kmph, I reach office 10 minutes early. However, if I drive at a speed 10 kmph lesser than the earlier, I get late by half an hour”. Find the distance between Ramesh’s office and home.

A.

60 km

B.

200 km

C.

90 km

D.

100 km

Answer with explanation

Answer: Option BExplanation

Let distance be D

With speed 50km/hr (10 kmph less than the earlier 60 kmph), he is 30 minutes late

With speed 60 km/hr he is 10 minutes early

Difference between two times = 30+10 = 40min = [40 / 60]hours.

Note: Difference between two given times can also be easily measured or checked by looking at a watch or imagining a watch.

D = S x T

==> T = D / S (or) S = D / T

Also, time = T = D / S

==> [ (D / 50) – (D / 60) ] = [ 40 / 60 ]

==> [ 60D – 50D / 3000] (L.C.M) = [ 40 / 60 ]

==> [10D / 3000] = [ 40 / 60 ]

==> D = 200 km.

Workspace

A man goes to his office from his house at a speed of 3 km/hr and returns at a speed of 2 km/hr. If he takes 5 hours in going and coming, what is the distance between his house and office?

A.

3 km

B.

5km

C.

6 km

D.

4 km

Answer with explanation

Answer: Option CExplanation

Workspace

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