Problems on Trains

If the manufacturer gains 10 %, the wholesale dealer 15 % and the retailer 25 %, then find the cost of production of a table if the retail price was Rs. 1265.

Answer: (b)

Let the cost of Production = Rs. p

Then, as per question,

⇒ \(\left ( \frac{125}{100} \times \frac{115}{100} \times \frac{110}{100} \times p \right ) = 1265\)

⇒ \(\frac{253}{160}p = 1265\)

p = 800

A dealer sold two of his cattle for Rs. 500 each. On one of them he lost 10% on the other, he gained 10%. His gain or loss percent in the entire transaction was:

Answer: (c)

Loss% = (common gain or loss % / 10)2

= (10/10)² %

= 1%.

Two, trains, one from Howrah to Patna and the other from Patna to Howrah, start simultaneously. After they meet, the trains reach their destinations after 9 hours and 16 hours respectively. The ratio of their speeds is:

Answer: (b)

Let us name the trains as A and B. Then,

(A's speed) : (B's speed)

= \(\sqrt{b} : \sqrt{a}\)

= \(\sqrt{16} : \sqrt{9}\)

= 4 : 3.

A retailer buys a radio for Rs 225. His overhead expenses are Rs 15. He sell is the radio for Rs 300. The profit percent of the retailer is

Answer: (c)

Cost price = (225 + 15) = 240 sell price = 300

Gain = \(\frac{64}{240} \times 100\) = 25%

In what ratio should water and wine be mixed so that after selling the mixture at the cost price a profit of 33.33% is made?

Answer: (b)

33.33% profit means there is one part water and 3 part is pure wine. so the required ratio of water and wine in the mixture is 1 : 3

A train moves with a speed of 108 kmph. Its speed in metres per second is :

Answer: (c)

108 kmph = \(\left ( 108 \times \frac{5}{18} \right )\) m/sec = 30 m/s.

Two stations P and Q are 110 km apart on a straight track. One train starts from P at 7 a.m. and travels towards Q at 20 kmph. Another train starts from Q at 8 a.m. and travels towards P at a speed of 25 kmph. At what time will they meet?

Answer: (b)

Assume both trains meet after x hours after 7 am

Distance covered by train starting from P in x hours = 20x km

Distance covered by train starting from Q in (x-1) hours = 25(x-1)

Total distance = 110

⇒ 20x + 25(x-1) = 110

⇒ 45x = 135

x = 3

Means, they meet after 3 hours after 7 am, i.e., they meet at 10 am.

A train travelling at a speed of 75 mph enters a tunnel \(3\frac{1}{2}\) miles long. The train is \(\frac{1}{4}\) mile long. How long does it take for the train to pass through the tunnel from the moment the front enters to the moment the rear emerges?

Answer: (b)

Total distance covered = \(\left [ \frac{7}{2} \times \frac{1}{4} \right ]\) miles

= \(\frac{15}{4}\)  miles

Time taken = \(\left [ \frac{15}{4 \times 75} \right ]\) hrs

= \(\frac{1}{20}\) hrs

= \(\left [ \frac{1}{20} \times 60 \right ]\) min

= 3 min

Two trains having equal lengths, take 10 seconds and 15 seconds respectively to cross a post. If the length of each train is 120 meters, in what time (in seconds) will they cross each other when traveling in opposite direction?

Answer: (c)

Speed of train 1 = \(\frac{120}{10}\) m/sec = 12 m/sec

Speed of train 2 = \(\frac{120}{15}\) m/sec = 8 m/sec

if they travel in opposite direction, relative speed = 12 + 8 = 20 m/sec

distance covered = 120 + 120 = 240 m

\(time = \frac{distance}{speed} = \frac{240}{20}\) = 12 sec

Two trains 140 m and 160 m long run at the speed of 60 km/hr and 40 km/hr respectively in opposite directions on parallel tracks. The time (in seconds) which they take to cross each other, is:

Answer: (d)

Relative speed = (60 + 40) km/hr = \(\left [ 100 \times \frac{5}{18} \right ]\) m/sec = \(\frac{250}{9}\) m/sec.

Distance covered in crossing each other = (140 + 160) m = 300 m.

Required time = \(\left [ 300 \times \frac{9}{250}\) \right ][/latex] sec = \(\frac{54}{5}\) sec = 10.8 sec.