Punjab State Board PSEB 8th Class Maths Book Solutions Chapter 13 Direct and Inverse Proportions InText Questions and Answers.

## PSEB 8th Class Maths Solutions Chapter 13 Direct and Inverse Proportions InText Questions

Try These : [Textbook Page No. 204]

1. Observe the following tables and find if x and y are directly proportional.

Question (i)

X | 20 | 17 | 14 | 11 | 8 | 5 | 2 |

y | 40 | 34 | 28 | 22 | 16 | 10 | 4 |

Solution:

\(\begin{aligned}

&\frac{20}{40}=\frac{1}{2}, \frac{17}{34}=\frac{1}{2}, \frac{14}{28}=\frac{1}{2}, \frac{11}{22}=\frac{1}{2}, \frac{8}{16}=\frac{1}{2} \\

&\frac{5}{10}=\frac{1}{2}, \frac{2}{4}=\frac{1}{2}

\end{aligned}\)

The value of \(\frac{\text {x}}{\text {y}}\) is same for different values of x and y. So these values x and y are directly proportional.

Question (ii)

X | 6 | 10 | 14 | 18 | 22 | 26 | 30 |

y | 4 | 8 | 12 | 16 | 20 | 24 | 28 |

Solution:

\(\begin{aligned}

&\frac{6}{4}=\frac{3}{2}, \frac{10}{8}=\frac{5}{4}, \frac{14}{12}=\frac{7}{6}, \frac{18}{16}=\frac{9}{8}, \frac{22}{20}=\frac{11}{10}, \\

&\frac{26}{24}=\frac{13}{12}, \frac{30}{28}=\frac{15}{14}

\end{aligned}\)

The values of \(\frac{\text {x}}{\text {y}}\) are different for different values of x and y respectively. So these values of x and y are not directly proportional.

Question (iii)

X | 5 | 8 | 12 | 15 | 18 | 20 |

y | 15 | 24 | 36 | 60 | 72 | 100 |

Solution:

The values of \(\frac{\text {x}}{\text {y}}\) are different for different values of x and y respectively.

So these values of x and y are not directly proportional.

2. Principal = ₹ 1000, Rate = 8 % per annum. Fill in the following table and find which type of interest (simple or compound) changes in direct proportion with time period.

Time Period | 1 year | 2 years | 3 years |

Simple Interest (in ₹) | |||

Compound Interest (in ₹) |

Solution:

Simple interest : SI = \(\frac{P \times R \times T}{100}\)

For calculation:

P = ₹ 1000, R = 8 %, T = …………….

Time (T) → | 1 year: T = 1 | 2 years : T = 2 | 3 years : T = 3 |

Simple interest SI = \(\frac{P \times R \times T}{100}\) | ₹ \(\frac{1000 \times 8 \times 1}{100}\) = ₹ 80 |
₹ \(\frac{1000 \times 8 \times 2}{100}\) = ₹ 160 |
₹ \(\frac{1000 \times 8 \times 3}{100}\) = ₹ 240 |

\(\frac{\text { SI }}{\text { T }}\) | \(\frac {80}{1}\) = 80 | \(\frac {160}{2}\) = 80 | \(\frac {240}{3}\) = 80 |

Here, the ratio of simple interest with time period is same for every year.

Hence, simple interest changes in direct proportion with time period.

Compound interest:

For calculation:

P = ₹ 1000, R = 8 %, T = ……………

Time → | 1 year : n = 1 |

A = P(1 + \(\frac {R}{100}\))^{n
}CI = A – P |
A = 1000(1 + \(\frac {8}{100}\))^{1
}= 1000 × \(\frac {108}{100}\) = 1080∴ CI = 1080 – 1000 = ₹ 80 |

\(\frac{\text { CI }}{\text { T }}\) | \(\frac {80}{1}\) |

Time → | 2 years : n = 2 |

A = P(1 + \(\frac {R}{100}\))^{n
}CI = A – P |
A = 1000(1 + \(\frac {8}{100}\))^{2
}= 1000 × \(\frac {108}{100}\) × \(\frac {108}{100}\) = ₹ 1166.40∴ CI = ₹ 1166.40 – 1000 = ₹ 166.40 |

\(\frac{\text { CI }}{\text { T }}\) | \(\frac {166.40}{2}\) |

Time → | 3 years : n = 3 |

A = P(1 + \(\frac {R}{100}\))^{n
}CI = A – P |
A = 1000(1 + \(\frac {8}{100}\))^{3
}= 1000 × \(\frac {108}{100}\) × \(\frac {108}{100}\) × \(\frac {108}{100}\) = ₹ 1259.712∴ CI = ₹ 1259.712 – ₹ 1000 = ₹ 259.712 |

\(\frac{\text { CI }}{\text { T }}\) | \(\frac {259.712}{3}\) |

Here, the ratio of CI and T is not same.

Thus, compound interest is not proportional with time period.

Think, Discuss and Write: [Textbook Page No. 204]

1. If we fix time period and the rate of interest, simple interest changes proportionally with principal. Would there be a similar relationship for compound interest? Why?

Solution:

Time period (T) and rate of interest (R) are fixed, then

Simple interest = \(\frac {PRT}{100}\) = P × Constant

So simple interest depends on principal. Simple interest changes proportionally with principal.

Now, compound interest = P(1 + \(\frac {R}{100}\))^{T} – P

i.e., A – P

= P [(1 + \(\frac {R}{100}\)^{T} – 1]

= P × Constant

So compound interest depends on principal.

If principal increases or decreases, then compound interest will also increases or decreases.

Thus, compound interest changes with principal.

Think, Discuss and Write : [Textbook Page No. 209]

1. Take a few problems discussed so far under ‘direct variation’. Do you think that they can be solved by ‘unitary method’?

Solution:

Yes, each problem can be solved by unitary method.

e.g. Question 4 of Exercise: 13.1

Number of bottles filled in 6 hours = 840

∴ The number of bottles filled in 1 hour = \(\frac {840}{6}\) = 140

The number of bottles filled in 5 hours = 140 × 5 = 700

Thus, 700 bottles will it fill in five hours.

Try These : [Textbook Page No. 211]

1. Observe the following tables and find which pair of variables (here x and y) are in inverse proportion.

Question (i)

X | 50 | 40 | 30 | 20 |

y | 5 | 6 | 7 | 8 |

Solution:

x_{1} = 50 and y_{1} = 5

∴ x_{1}y_{1} = 50 × 5

∴ x_{1}y_{1} = 250

x_{2} = 40 and y_{2} = 6

∴ x_{2}y_{2} = 40 × 6

∴ x_{2}y_{2} = 240

x_{3} = 30 and y_{3} = 7

∴ x_{3}y_{3} = 30 × 7

∴ x_{3}y_{3} = 210

x_{4} = 20 and y_{4} = 8

∴ x_{4}y_{4} = 20 × 8

∴ x_{4}y_{4} = 160

Now 250 ≠ 240 ≠ 210 ≠ 160

∴ x_{1}y_{1} ≠ x_{2}y_{2} ≠ x_{3}y_{3} ≠ x_{4}y_{4}

∴ x and y are not in inverse proportion.

Question (ii)

X | 100 | 200 | 300 | 400 |

y | 60 | 30 | 20 | 15 |

Solution:

x_{1} = 100 and y_{1} = 60

∴ x_{1}y_{1} = 100 × 60

∴ x_{1}y_{1} = 6000

x_{2} = 200 and y_{2} = 30

∴ x_{2}y_{2} = 200 × 30

∴ x_{2}y_{2} = 6000

x_{3} = 300 and y_{3} = 20

∴ x_{3}y_{3} = 300 × 20

∴ x_{3}y_{3} = 6000

x_{4} = 400 and y_{4} = 15

∴ x_{4}y_{4} = 400 × 15

∴ x_{4}y_{4} = 6000

Now x_{1}y_{1} = x_{2}y_{2} = x_{3}y_{3} = x_{4}y_{4}

∴ x and y are in inverse proportion.

Question (iii)

X | 90 | 60 | 45 | 30 | 20 | 5 |

y | 10 | 15 | 20 | 25 | 30 | 35 |

Solution:

x_{1} = 90 and y_{1} = 10

∴ x_{1}y_{1} = 90 × 10

∴ x_{1}y_{1} = 900

x_{2} = 60 and y_{2} = 15

∴ x_{2}y_{2} = 60 × 15

∴ x_{2}y_{2} = 900

x_{3} = 45 and y_{3} = 20

∴ x_{3}y_{3} = 45 × 20

∴ x_{3}y_{3} = 900

x_{4} = 30 and y_{4} = 25

∴ x_{4}y_{4} = 30 × 25

∴ x_{4}y_{4} = 750

x_{5} = 20 and y_{5} = 30

∴ x_{5}y_{5} = 20 × 30

∴ x_{5}y_{5} = 600

x_{6} = 30 and y_{6} = 35

∴ x_{6}y_{6} = 5 × 35

∴ x_{6}y_{6} = 175

Now x_{1}y_{1} = x_{2}y_{2} = x_{3}y_{3} ≠ x_{4}y_{4} ≠ x_{5}y_{5} ≠ x_{6}y_{6}

∴ x and y are in inverse proportion.