CMAT Speed Time Distance Work Questions

Let the efficiency of Bhushan be $$x$$ units/hr.

So, efficiency of Amit be $$1.5x$$ units/hr.

Given, Bhushan alone can do a piece of work in 30 hours.

So, total work = $$30x$$ units

Given, Bhushan starts working and had already worked for 12 hours.

So, work done by Bhushan = $$12x$$ units

So, remaining work = $$30x-12x=18x$$ units

Now, this remaining work will be completed by both Amit and Bhushan.

So, required number of hours = $$\dfrac{18x}{x+1.5x}=\dfrac{18x}{2.5x}=\dfrac{180}{25}=7.2$$ hours.

So, correct answer is option (D).

Let $$T_w$$ be the time taken to walk one way and $$T_r$$ be the time taken to ride one way.

So, $$T_w+T_r=$$ $$4$$ hours $$20$$ minutes

or, $$T_w+T_r=260$$ ---->(1)

Now, it is given, if he walks on both sides, he loses 1 hour

so, $$2T_w=260+60=320$$ ---->(2)

Multiplying equation (1) by 2 and subtracting equation (2) from it,

$$2T_r=520-320=200$$

So, time taken in riding both ways = 200 minutes = 3 hours and 20 minutes.

So, correct answer is option (C).

Let say length of race be $$D$$ metres and speed of $$A$$, $$B$$ and $$C$$ be $$S_A,\ S_B,\ S_c$$ m/s.

Now, A beats B by 30 meters

So, $$\dfrac{S_A}{S_B}=\dfrac{D}{D-30}$$ --->(1)

Also, B beats C by 20 meters

So, $$\dfrac{S_B}{S_C}=\dfrac{D}{D-20}$$ ----->(2)

Also, A beats C by 48 meters.

So, $$\dfrac{S_A}{S_C}=\dfrac{D}{D-48}$$ ---->(3)

Multiplying (1) and (2) and equating with option (3),

$$\dfrac{D}{D-30}\times\ \dfrac{D}{D-20}=\dfrac{D}{D-48}$$

or, $$D\left(D-48\right)=\left(D-30\right)\left(D-20\right)$$

or, $$D^2-48D=D^2-50D+600$$

or, $$50D-48D=600$$

or, $$2D=600$$

or, $$D=300$$ meters

Putting the value of $$D$$ in (1) and (2),

$$\dfrac{S_A}{S_B}=\dfrac{300}{300-30}=\dfrac{300}{270}=\dfrac{10}{9}=\dfrac{50}{45}$$

$$\dfrac{S_B}{S_C}=\dfrac{300}{280}=\dfrac{15}{14}=\dfrac{45}{42}$$

The speed of A, B and C are in the ratio 50 : 45 : 42.

So, the correct answer is option C, statement I is true but statement II is false.

Let the distance travelled by leopard in one step be $$x$$ units.

It is given, distance traversed by leopard in 5 steps is equal to the distance traversed by tiger in 4 steps

So, distance travelled by tiger in one step = $$\dfrac{5x}{4}$$ units.

Now, leopard moved 3 steps and tiger moved 5 steps in the same time.

So, in the same time, leopard is moving $$3\times\ x=3x$$ units.

And, distance travelled by the tiger = $$5\times\ \dfrac{5x}{4}=\dfrac{25x}{4}$$ units.

So, speed ratios of leopard and tiger

= $$\dfrac{3x}{\dfrac{25x}{4}}=\dfrac{12x}{25x}=\dfrac{12}{25}$$

So, number of rounds covered by leopard = $$\dfrac{12}{25}\times\ 100=48$$

Let the work done by 1 man and 1 woman in 1 day, m and w, respectively.

It is given that 3 men and 18 women together take 2 days to complete a piece of work.

Total work = (3m+18w)*2

6 men alone can complete the piece of work in 3 days = 6m*3 = 18m

Work done is the same for both cases.

So, 18m = 6m+36w

12m = 36w

m = 3w

Total work = 18m = 18*3w = 54w

Therefore, 9 women will complete the work in x days. = 9w*x = 54w

Thus, x = 6 days.

Statement I :

If Shyam can finish the task by working alone in 8 days, so he will do $$\dfrac{6}{8}\times100=75\%$$ os the work in 6 days.

The remaining 25% is done by Ram is 6 days.

So, if Ram does 25% of the work in 6 days, he will take $$\dfrac{100}{25}\times6=24$$ days to complete the work alone.

Hence, the given statement is true.

Statement II :

Assume 1 person does 1 man hour work in one hour

So, in one week, working 8 hours a day, 

The total work done by 6 person is $$6\times8\times7=336$$ man hours

Assume the payment to be Rs. $$x$$ per man-hour

So, $$336x=8400$$

or, $$x=25$$

The total man hours for 9 persons working 6 hours a day for a week will be

$$9\times6\times7=378$$

So, the money earned is $$378x=378\times25=9450$$

Hence, the given statement is true.

Assume the length of the first train to be $$x$$ m

So, the length of the second train is $$0.5x$$ m

The total length to be covered = $$1.5x$$ m

As they are travelling in the opposite directions, their speed will add in relative frame

So, the combined speed is $$48+42\ =\ 90\ $$kmph or $$\ 90\times\dfrac{5}{18}=25$$ m/s

So, $$\dfrac{1.5x}{25}=12$$

$$1.5x=300$$

or, $$x=200$$ m

Given it passes the station in 45 sec

Let the length of the station be $$l$$ m

So, $$\dfrac{l+x}{48\times\dfrac{5}{18}}=45$$

$$l+x=600$$

$$l=400$$ m

Hence, the answer is 400 m

Let us assume Seeta and Geeta are at A and B initially.

Distance between A and B = 200 m.

Seeta and Geeta have the same speed.

So, their travel distance will be the same since they start together.

Seeta travels = 40+25 = 65 meters in the horizontal direction from Point A and 15+15 = 30 meters in the vertical direction.

Total distance travelled by Seeta = 65+30 = 95 meters.

Hence, Geeta also travels = 95 meters.

Since Seeta travels only 65 meters on line AB, and Geeta travels = 95 meters on line AB.

Distance between them = 200-65-95 = 40 meters.