NEET PG (NTA)-National Eligibility cum Entrance Test Post Graduate (Medical) NEET-PG Coaching Programs

🎯 36 Numeric, 2996 MCQs (& PYQs) with Full Explanations (2024-2025 Exam)

Click Here to View & Get Complete Material

Rs. 650.00

3 Year Validity (Multiple Devices)

Application of Linear Graphs: Variation of Linear Graph Applications

Variation in the value of different quantities depend upon the variation in values of other quantities.

For example: if the number of persons visiting a hotel increases, earning of the hotel increases and vice versa if a number of people are employed, time taken to accomplish a job decreases.

Sometimes value of one quantity increases with the increase in the value of other quantity while in other cases the value of one quantity decreases with an increase in the value of other quantity.

Hence, two quantities can either exist in direct proportion or indirect proportion. Relationship between these two quantities can be represented in an arithmetic manner or graphical manner.

Sometimes these two quantities exhibit a linear dependence, in other words, variation in the value of one quantity is proportional to the first power of variation in the value of other quantity. This is done by linear graphs.

Linear Graph Applications

Problem: Miti can ride a scooter constantly at a speed of 20 km/hour. Draw a distance-time graph for this situation. With the help of the linear graph, calculate

  1. The time taken by Miti to ride 100 km.
  2. The distance covered by Miti in 3 hours.

Solution: let՚s form a table of values correlating the time of travel and the distance travelled by Miti in the particular time interval.

Let՚s Form a Table of Values Correlating the Time of Travel and the Distance Travelled by Miti in the Particular Time Interval
Time (hr.)123456
Distance (km)20406080100120

Let us draw a graph with the help of these tabulated values,

Illustration: Linear Graph Applications

With the help of the above graph, we can calculate the required values,

1. Time taken to cover 100 km by Miti:

  • Here observing the X-coordinate of the graph corresponding to the Y-coordinate of the graph at 100 km.
  • Y – coordinate of 100 km is related to X – coordinate of 5 hr.
  • Hence, Miti covered 100 km in 5 hrs.

2. Distance covered by Miti in 3 hours:

Using same thing of question 1,

Y-coordinate of the graph corresponding to the X-coordinate of the graph at 3 hr is 60 km.

Hence, distance covered by Miti in 3 hours = 60 km

Example:

Study the following graph carefully and answer the questions given below.

Exports from Three Companies Over the Years (in ₹ crore)

Illustration: Linear Graph Applications

1. For which of the following pairs of years the total exports from the three companies together are equal?

(A) 1995 and 1998

(B) 1996 and 1998

(C) 1997 and 1998

(D) 1995 and 1996

Answer: (D)

Total exports of the three companies X, Y and Z together, during various years are:

In 1993,

Exports of companies

Exports of companies

Exports of companies

Total exports in 1993

In 1994,

Exports of companies

Exports of companies

Exports of companies

Total exports in 1993

Same thing using other years,

In 1995 = ₹ (40 + 60 + 120) crores = ₹ 220 crores.

In 1996 = ₹ (70 + 60 + 90) crores = ₹ 220 crores.

In 1997 = ₹ (100 + 80 + 60) crores = ₹ 240 crores.

In 1998 = ₹ (50 + 100 + 80) crores = ₹ 230 crores.

In 1999 = ₹ (120 + 140 + 100) crores = ₹ 360 crores.

Here clearly seen that the total exports of the three companies X, Y and Z together are same during the years 1995 and 1996.