Questions On Quantitative Techniques SMQTQ043

Direction (1 to 5): Study the given chart carefully and answer the following questions.
Train X

Train Y

1. The number of passengers boarding Train X at Q is what percent of the number of passengers boarding Train Y at S?

A) 37.03%
B) 67.09%
C) 47.10%
D) 56.36%

Ans. A
Required percentage = 100/270 × 100 = 37.03%

2. What is the difference between the speed of Train X and that of Train Y?

A) 2.73kmph
B) 3.47kmph
C) 8.6kmph
D) 4.82kmph

Ans. A
Speed of Train X = 1280 / 5 pm – 10:20 am =
1280 / 17 hours 20 minutes
= 1280 x 3 / 52 = 73.84kmph
Speed of train Y = 1280 / 6:00 pm - 12:00 noon
= 1280/18 hours = 71.11kmph
So, difference between the speed of train A and train B = 73.84- 71.11 = 2.73kmph

3. What is the ratio of the total passengers of Train X to that of Train Y?

A) 111:80
B) 104: 87
C) 112: 57
D) 98:102

Ans. B
Total passengers in train X = 400 + 100 + 90 + 300 + 150 = 1040
Total passengers in train Y = 300 + 150 + 270 + 50 + 100 = 870
Required ratio = 1040: 870 = 104: 87

4. The total income of Train X is what percent of the total income of Train Y?

A) 150.32%
B) 162.04%
C) 114.21%
D) 135.23%

Ans. C
Total income of train X = (400 × 50) + (500 × 70) + (590 × 280) + (890 × 100) + (1040 × 120) = Rs.434000
Total income of train Y = (300 × 120) + (450 × 100) + (720 × 280) + (770 × 70) + (870 × 50) = Rs.380000
Required % = 434000 x 100 / 340000
= 114.21%

5. If the average speed of Train X increases by 10% then when will it reach to its destination?

A) 9:45 am
B) 6:45 am
C) 8:45 am
D) 5:45 am

Ans. C
If the average speed of train X increases by 10% then its new speed = 73.84 × 110/100 = 81.22kmph Time taken by train X during the journey = 1280/81.22 = 15.75 hours = 15 hours 45 minutes
The time when the train will reach its destination = 5 pm + 15 hours 45 minutes = 8:45 am