CBSE Class 8 Mathematics Chapter 15: Introduction to Graphs NCERT Solutions

To answer these questions, we need to interpret the given line graph which plots the patient's temperature (°C) against time (hours).

(a) To find the patient's temperature at 1 p.m., locate 1 p.m. on the horizontal axis (Time). Then, move vertically up to the point on the graph and move horizontally to the left to read the corresponding temperature on the vertical axis. The graph shows the temperature at 1 p.m. was 36.5° C.

(b) To find when the patient's temperature was 38.5° C, locate 38.5° C on the vertical axis (Temperature). Then, move horizontally to the right to find the point on the graph and move vertically down to read the corresponding time on the horizontal axis. The graph shows the temperature was 38.5° C at 12 noon.

(c) To find the two times when the temperature was the same, we look for points on the graph that have the same vertical (temperature) value. Observing the graph, the temperature was 37° C at both 10 a.m. and 2 p.m. Therefore, the patient's temperature was the same at 10 a.m. and 2 p.m.

(d) To find the temperature at 1.30 p.m., we need to estimate the value between 1 p.m. and 2 p.m. The point 1.30 p.m. is exactly halfway between 1 p.m. and 2 p.m. on the time axis. Looking at the graph, the temperature at 1 p.m. is 36.5° C and at 2 p.m. is 37° C. The line segment between 1 p.m. and 2 p.m. is a straight line. Assuming a constant rate of change between these two points, the temperature at 1.30 p.m. would be the midpoint between 36.5° C and 37° C. This midpoint is 36.75° C. However, the provided answer suggests 36.5° C. Let's re-examine the graph. The graph shows a horizontal line segment between 1 p.m. and 2 p.m. at 36.5° C. This implies the temperature remained constant at 36.5° C from 1 p.m. to 2 p.m. Therefore, the temperature at 1.30 p.m. was 36.5° C. We arrive at this answer by observing that the graph is flat between 1 p.m. and 2 p.m., indicating no change in temperature during that interval.

(e) An upward trend on the graph means the temperature is increasing over time. We look for segments of the graph that go upwards from left to right. The temperature increased from 9 a.m. (37° C) to 11 a.m. (38° C). It also increased from 11 a.m. (38° C) to 12 noon (38.5° C). The most significant upward trend shown is from 9 a.m. to 11 a.m., where the temperature rose from 37° C to 38° C. The question asks for 'periods', implying intervals. The temperature showed an upward trend from 9 a.m. to 11 a.m. and again from 11 a.m. to 12 noon.

We will interpret the given line graph which shows the yearly sales (in Rs crores) of a manufacturing company from 2002 to 2006.

1. To find the sales for a specific year, we locate the year on the horizontal axis (Years) and move vertically up to the graph, then move horizontally to the left to read the sales amount on the vertical axis (Sales in Rs crores).

(i) For the year 2002, the sales were ₹ 4 crores.

(ii) For the year 2006, the sales were ₹ 8 crores.

2. Similarly, we find the sales for the years 2003 and 2005.

(i) For the year 2003, the sales were ₹ 7 crores.

(ii) For the year 2005, the sales were ₹ 10 crores.

3. To compute the difference between the sales in 2002 and 2006, we subtract the sales of 2002 from the sales of 2006.

Difference = Sales in 2006 – Sales in 2002

Difference = ₹ 8 crores – ₹ 4 crores = ₹ 4 crores.

4. To find the year with the greatest difference in sales compared to its previous year, we calculate the difference for each year from its preceding year:

• Difference between 2003 and 2002: ₹ 7 crores – ₹ 4 crores = ₹ 3 crores.
• Difference between 2004 and 2003: ₹ 6 crores – ₹ 7 crores = -₹ 1 crore (a decrease).
• Difference between 2005 and 2004: ₹ 10 crores – ₹ 6 crores = ₹ 4 crores.
• Difference between 2006 and 2005: ₹ 8 crores – ₹ 10 crores = -₹ 2 crores (a decrease).

Comparing these differences, the greatest increase in sales compared to the previous year occurred in 2005, with a difference of ₹ 4 crores.