Questions & Answers: "Variation, Time and Work"

Complete guide to "Variation, Time and Work" for Math students. Below you will find important questions and model answers to help you prepare.

5 Questions

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Question 1

1 Mark

A work is completed by some workers in a few days. If the number of workers is made 2/3 rd times the original number of workers, then how many times will be the days required to be increased than the original to complete the same work?

Options

Option A Diagram for Option A
Option B Diagram for Option B
Option C Diagram for Option C
Option D is correctDiagram for Option D

Question 2

1 Mark

To complete a job, ‘A’ requires 12 days. For the same job, ‘B’ requires 20 days to complete the job. Both of them worked together for 3 days and then A left the job. To complete the remaining job, how many days ‘B’ will require?

Options

Option A

18

Option B

15

Option C is correct

12

Option D

10

Question 3

1 Mark

If m = 30 then n = 5. Write the equation of variation.

Diagram for Question 3

Options

Option A

m = 6 n

Option B is correct

mn = 150

Option C

n = 150 m

Option D

mn = 6

Question 4

1 Mark

If x varies directly with y and the value of x = 35, and y = 14, then find which of the following is the equation of variation?

Options

Option A is correct

2x = 5y

Option B

2y = 5x

Option C

2x + 5y = 0

Option D

2y – 5x = 0

Question 5

1 Mark

A train of length 1000 m. is running at a speed of 72 km/hr. How much time will it take to pass a tunnel of length 800 m? (Choose two correct alternatives.)

Options

Option A is correct

90 seconds

Option B is correct

1121\frac{1}{2} minutes

Option C

2 minutes

Option D

120 seconds

Explanation

To pass a tunnel, the train must cover a distance equal to its own length plus the length of the tunnel.The total distance (DD) to be covered is the sum of the length of the train (LTL_T) and the length of the tunnel (LTunnelL_{Tunnel}):D=LT+LTunnel=1000 m+800 m=1800 mD = L_T + L_{Tunnel} = 1000\text{ m} + 800\text{ m} = 1800\text{ m}The speed of the train (STS_T) is given as 7272 km/hr. We need to convert this speed to meters per second (m/s):ST=72 km/hr=72×1000 m3600 s=72×518 m/s=4×5 m/s=20 m/sS_T = 72\text{ km/hr} = 72 \times \frac{1000\text{ m}}{3600\text{ s}} = 72 \times \frac{5}{18}\text{ m/s} = 4 \times 5\text{ m/s} = 20\text{ m/s}Now, we can calculate the time (TT) taken using the formula T=DSTT = \frac{D}{S_T}:T=1800 m20 m/s=90 secondsT = \frac{1800\text{ m}}{20\text{ m/s}} = 90\text{ seconds}We also need to check the options in minutes. 9090 seconds is equal to 9060\frac{90}{60} minutes, which is 1.51.5 minutes or 1121\frac{1}{2} minutes.Therefore, both 9090 seconds and 1121\frac{1}{2} minutes are correct alternatives.