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MCQ on Speed Time and Distance for all Competitive exams

Amit and Sujit together can complete an assignment of data entry in five days. Sujit’s speed is 80% of Amit’s speed and the total key depressions in the assignment are 5,76,000. What is Amit’s speed in key depressions per hour if they work for 8 hours a day?

(a) 4800

(b) 6400

(c) 8000

(d) 7200

(e) None of these

Solution: (c)

Ratio of the work done by Sujit and Amit = 80 : 100 = 4 : 5

Total key depressions done by Amit = \(\displaystyle \frac{5}{9}\times 576000=320000\)

Amit’s speed in key depressions per hour = \(\displaystyle \frac{{320000}}{{8\times 5}}=8000\)

Alternate method:

Let Amit’s speed of key depressions per hour be X.

So, Sumit’s speed of key depressions is 80% of X.

In total, the number of key depressions per hour by both of them is 1.8 X

The total number of key depressions in a day by both of them together is 1.8 \(\displaystyle \times \)8 X= 14.4 X

This equals 576000/5 = 115200.

So, 14.4 \(\displaystyle \times \) X = 115200

or, X = 115200/14.4 = 8000

An aeroplane flies with an average speed of 756 km/h. A helicopter takes 48 hours to cover twice the distance covered by aeroplane in 9 h. How much distance will the helicopter cover in 18 h? (Assuming that flights are non-stop and moving with uniform speed.)

(a) 5010 km

(b) 4875 km

(c) 5760 km

(d) 5103 km

(e) None of these

Solution: (d)

Distance covered by the aeroplane in 9 h = \(\displaystyle 9\times 756=6804km\)

Speed of helicopter = \(\displaystyle \frac{{2\times 6804}}{{48}}=283.5km/h\)

Distance covered by helicopter in 18 h = \(\displaystyle 283.5\times 18=5103km\)

Wheels of diameters 7 cm and 14 cm start rolling simultaneously from X and Y, which are 1980 cm apart, towards each other in opposite directions. Both of them make the same number of revolutions per second. If both of them meet after 10 seconds, the speed of the smaller wheel is:

(a) 22 cm / sec

(b) 44 cm / sec

(c) 66 cm / sec

(d) 132 cm / sec

(e) None of these

Solution: (c)

Let each wheel make x revolutions per sec. Then,

\(\displaystyle [(2\pi \times \frac{7}{2}\times x)+(2\pi \times 7\times x)]\times 10=1980\)

\(\displaystyle (\frac{{22}}{7}\times 7\times x)+(2\times \frac{{22}}{7}\times 7\times x)=198\)

 66x = 198 \(\displaystyle \Rightarrow \) x = 3.

Distance moved by smaller wheel in 3 revolutions

\(\displaystyle (2\times \frac{{22}}{7}\times \frac{7}{2}\times 3)cm/\sec =66cm/\sec \)

An aeroplane takes off 30 minutes later than the scheduled time and in order to reach its destination 1500 km away in time, it has to increase its speed by 250 km/h from its usual speed. Find its usual speed.

(a) 1000 km/h

(b) 750 km/h

(c) 850 km/h

(d) 650 km/h

(e) None of these

Solution: (b)

Let the usual speed of the aeroplane be x km/h.

Then, \(\displaystyle \frac{{1500}}{x}-\frac{1}{2}=\frac{{1500}}{{(x+250)}}\)

\(\displaystyle \frac{{1500}}{x}-\frac{{1500}}{{x+250}}=\frac{1}{2}\Rightarrow \frac{{1500x+75000-1500x}}{{x(x+250)}}=\frac{1}{2}\)

\(\displaystyle 750000={{x}^{2}}+250xor{{x}^{2}}+250x-750000=0\)

\(\displaystyle {{x}^{2}}+1000x-750x-75000=0\)

 (x + 1000) (x – 750) = 0 \(\displaystyle \Rightarrow \) x = 750, – 1000

Speed cannot be negative

We get x = 750 km/h

A farmer travelled a distance of 61 km in 9 hrs. He travelled partly on foot at the rate of 4 km/hr and partly on bicycle at the rate of 9 km/hr. The distance travelled on foot is

(a) 17 km

(b) 16 km

(c) 15km

(d) 14 km

(e) None of these

Solution: (b)

Then, distance travelled by bicycle = (61 –x) km

So, \(\displaystyle \frac{x}{4}+\frac{{61-x}}{9}=9\)

\(\displaystyle 9x+4(61-x)=9\times 36\)

9x – 4x = 324 – 244

5x = 80

x = 16 km