Question

Draw the graph for the following table of values, with suitable scales on the axes. Interest on deposits for a year:

Deposit in (in rs) ⠀SI (in rs)
5000 ⠀⠀⠀⠀⠀⠀⠀325
10000 ⠀⠀⠀⠀⠀⠀650⠀
15000 ⠀⠀⠀⠀⠀⠀975⠀
20000⠀⠀⠀⠀⠀⠀1300⠀
25000⠀⠀⠀⠀⠀⠀1625

a. Does the graph pass through the origin?

b. Use the graph to find the interest on 12500 for a year.

c. To get an interest of rs 1462.50 per year, how much money should be deposited?​

Answer 1

QUESTION

Draw the graph for the following table of values, with suitable scales on the axes. Interest on deposits for a year:

Deposit in (in rs) ⠀⠀SI (in rs)

5000 ⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀325

10000 ⠀⠀⠀⠀⠀⠀⠀⠀⠀650⠀

15000 ⠀⠀⠀⠀⠀⠀⠀⠀⠀975⠀

20000⠀⠀⠀⠀⠀⠀⠀⠀⠀1300⠀

25000⠀⠀⠀⠀⠀⠀⠀⠀⠀1625

Answer :

Along the x-axis:

Take 1 small square = 5000 units

Along the y-axis:

Take 2 small square = 325 units

Plot the points A(5000, 325), B(10000, 650), C(15000, 975), D(20000, 1300) and E(25000, 1625).

Join them successively to obtain the required graph. Extend the graph line to meet the coordinate axes.

[ Note - See The Attachment ]

_______________________

a. Does the graph pass through the origin?

Ans. Yes, the graph line passes through the origin.

b. Use the graph to find the interest on 12500 for a year.

Ans. On the x-axis take a point P (midway between 10000 and 15000) at x = 12500. Draw PQ ⊥ x-axis, meeting the graph line at Q.

Clearly QP = $\sf(650 + \dfrac{325}{2} )$ Units

= ( 650 + 162.5 ) units = 812.5 units

Thus , the interest on rs 12500 for a year is rs 812.50

c. To get an interest of rs 1462.50 per year, how much money should be deposited?

Ans. On the y-axis take a point L (midway between 1300 and 1625) at y = 1462.50.

Draw LF ⊥ y-axis and let it meet the graph line at F. From F draw a vertical line which meets the x-axis at a point. This point is 22500.

Thus, to get an interest of 1462.50, the money to be deposited is rs 22500.

_______________________

ADDITIONAL INFORMATION

LINE GRAPH

• A line graph is a type of graph which displays information as a series of data points connected consecutively by line segments, thereby showing changes in the data over time.

• It helps us to visualise a trend in data over intervals of time. The adjacent line graph shows the yearly sales figures for a manufacturing company.

LINEAR GRAPH

• If you recall, a line graph is made up of line segments joined one after another.

• There can be instances when the line graph is a one continuous line.

• In such cases the graph is called a linear graph.

Answer 2

Answer:

A. Does the graph pass through the origin?

Does the graph pass through the origin?Ans. Yes, the graph line passes through the origin.

B. Use the graph to find the interest on 12500 for a year.

Use the graph to find the interest on 12500 for a year.Ans. On the x-axis take a point P (midway between 10000 and 15000) at x = 12500. Draw PQ ⊥ x-axis, meeting the graph line at Q.

Use the graph to find the interest on 12500 for a year.Ans. On the x-axis take a point P (midway between 10000 and 15000) at x = 12500. Draw PQ ⊥ x-axis, meeting the graph line at Q.Clearly QP = \sf(650 + \dfrac{325}{2} )(650+

Use the graph to find the interest on 12500 for a year.Ans. On the x-axis take a point P (midway between 10000 and 15000) at x = 12500. Draw PQ ⊥ x-axis, meeting the graph line at Q.Clearly QP = \sf(650 + \dfrac{325}{2} )(650+ 2

Use the graph to find the interest on 12500 for a year.Ans. On the x-axis take a point P (midway between 10000 and 15000) at x = 12500. Draw PQ ⊥ x-axis, meeting the graph line at Q.Clearly QP = \sf(650 + \dfrac{325}{2} )(650+ 2325

Use the graph to find the interest on 12500 for a year.Ans. On the x-axis take a point P (midway between 10000 and 15000) at x = 12500. Draw PQ ⊥ x-axis, meeting the graph line at Q.Clearly QP = \sf(650 + \dfrac{325}{2} )(650+ 2325

Use the graph to find the interest on 12500 for a year.Ans. On the x-axis take a point P (midway between 10000 and 15000) at x = 12500. Draw PQ ⊥ x-axis, meeting the graph line at Q.Clearly QP = \sf(650 + \dfrac{325}{2} )(650+ 2325 ) Units

Use the graph to find the interest on 12500 for a year.Ans. On the x-axis take a point P (midway between 10000 and 15000) at x = 12500. Draw PQ ⊥ x-axis, meeting the graph line at Q.Clearly QP = \sf(650 + \dfrac{325}{2} )(650+ 2325 ) Units= ( 650 + 162.5 ) units = 812.5 units

Use the graph to find the interest on 12500 for a year.Ans. On the x-axis take a point P (midway between 10000 and 15000) at x = 12500. Draw PQ ⊥ x-axis, meeting the graph line at Q.Clearly QP = \sf(650 + \dfrac{325}{2} )(650+ 2325 ) Units= ( 650 + 162.5 ) units = 812.5 unitsThus , the interest on rs 12500 for a year is rs 812.50

C.To get an interest of rs 1462.50 per year, how much money should be deposited?

To get an interest of rs 1462.50 per year, how much money should be deposited?Ans. On the y-axis take a point L (midway between 1300 and 1625) at y = 1462.50.

To get an interest of rs 1462.50 per year, how much money should be deposited?Ans. On the y-axis take a point L (midway between 1300 and 1625) at y = 1462.50.Draw LF ⊥ y-axis and let it meet the graph line at F. From F draw a vertical line which meets the x-axis at a point. This point is 22500.

To get an interest of rs 1462.50 per year, how much money should be deposited?Ans. On the y-axis take a point L (midway between 1300 and 1625) at y = 1462.50.Draw LF ⊥ y-axis and let it meet the graph line at F. From F draw a vertical line which meets the x-axis at a point. This point is 22500.Thus, to get an interest of 1462.50, the money to be deposited is rs 22500.