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**Maths Ace Class 8 Solutions Chapter 15 Introduction to Graphs**

Welcome to NCTB Solutions. Here with this post we are going to help 8th class students for the Solutions of Maths Ace Prime Class 8 Math Book, Chapter 15, Introduction to Graphs. Here students can easily find step by step solutions of all the problems for Introduction to Graphs, Exercise 15.1 and 15.2 Also here our mathematics teacher’s are solved all the problems with easily understandable methods with proper guidance so that all the students can understand easily. Here in this post students will get chapter 15 solutions.

**Introduction to Graphs Exercise 15.1 Solution :**

**Question no – (2) **

**Solution : **

**(a) (2, 7)**

= The abscissa of (2, 7) is 2

**(b)** **(9, 19)**

= The abscissa (9, 19) is 2

**(c)** **(10, 12)**

= The abscissa (10, 12) is 10

**(d)** **(4, 3)**

= The abscissa (4, 3) is 4

**Question no – (3) **

**Solution : **

**(a) (21, 7)**

= Coordinate = 7

**(b)** **(8, 29)**

= Coordinate = 29

**(c) (5, 9)**

= Coordinate = 9

**(d) (2, 8)**

= Coordinate = 8

**Question no – (4) **

**Solution : **

Coordinates of vertices are,

A = (2, 1)

B = (5, 1)

C = (6, 2)

D = (3, 2)

**Introduction to Graphs Exercise 15.2 Solution :**

**Question no – (1) **

**Solution : **

**Table – (a)**

**y** = 2x – 1

= 2 × 2 – 1

= 4 – 1

= 3

**y** = 2x – 1

= 2 × 4 – 1

= 8 – 1

= 7

**y** = 2x – 1

= 2 × 6 – 1

= 12 – 1

= 11

**y** = 2x – 1

= 2 × 8 – 1

= 16 – 1

= 15

**Table – (b)**

**y** = 4x

= 4 × 3

= 12

**y** = 4x

= 4 × 9

= 36

**y** = 4x

= 4 × 15

= 60

**y** = 4x

= 4 × 6

= 24

**y** = 4x

= 4 × 12

= 48

**Table – (c)**

**y** = 5 x – 5

= 5 × 1 – 5

= 5 – 5

= 0

**y** = 5x – 5

= 5 × 2 – 5

= 10 – 5

= 5

**y** = 5x – 5

= 5 × 3 – 5

= 15 – 5

= 10

**y** = 5x – 5

= 5 × 4 – 5

= 20 – 5

= 15

**y** = 5x – 5

= 5 × 5 – 5

= 25 – 5

= 20

**Question no – (2) **

**Solution : **

**Graph – (a)**

X |
1 | 2 | 3 | 4 | 5 |

Y |
2 | 6 | 9 | 13 | 16 |

**Graph – (b)**

X |
2 | 4 | 6 | 8 | 10 | 12 |

Y |
10 | 20 | 30 | 40 | 50 | 60 |

**Question no – (3) **

**Solution : **

**(a)** Find the amount of rainfall on Wednesday.

= 5 cm

**(b)** On which day of the week was the rainfall minimum?

= Thurs day

**(c)** Find the average rainfall during the week.

= 3. 71 cm

= (6.1 + 3 + 3 + 5 + 1 + 5.1 + 3/7)

= 3.71 cm

**(d)** Find the decrease or increase in rainfall from Thursday to Friday.

= Thurs day = 1 cm

Friday = 5 cm

(Increase = (5 – 1) cm,

= 4 cm

**(e)** On which days was the amount of rainfall the same?

The day is = Monday & Thursday

**Question no – (5) **

**Solution : **

**(a)** What is the simple interest on depositing 6000?

= 1st, P = 500

= S.I = 45 Rs

Difference in principal

= (1000 – 500)

= 500

Difference in S.I

= (90 – 45) RST

= 45 RST

**∴** S.I on depositing 6000 is

= (6000 – 4000) = 2000 Principal

S.I is = (4 × 45) RST

= 180

**∴** Total = (360 + 180)

= 540 RST

**(b)** What amount should be deposited to earn 450 as simple interest?

Amount deposit for S.I 360 Rs is 4000

So, To S.I 450 RST Amount deposit

= (4000 + 1000) RST

= 5000 RST

**Question no – (6) **

**Solution : **

**(a)** Find the distance covered if he is driving at a speed of 45 km/hr.

**∴** Distance cover at speed 45 km/h,

= (45 × 240/40)

= (45 × 6)

= 270 km

**(b)** Find the speed of the taxi if he travels a distance of 450 km.

**∴ **Speed,

= 450/6 km/h

= 75 km/h

**Question no – (7) **

**Solution : **

It is a liner graph between Distance and Time.

**Next Chapter Solution : **