Mathematics

Class 92025–26 SNC28 Questions SolvedEnglish Medium

Exercise 6.1 — Class 9 Mathematics Solutions

All 28 questions below are worked step by step — every step clearly worked out with mathematical reasons — for the Punjab Board 9th class Mathematics textbook.

Exercise 6.1

9 Questions (19 Sub-parts)

Quick Solution Step by Step

Every step explained, with hints and the method used.

Question 1

EasyUnder Review

Find in which quadrant the following angles lie. Write a co-terminal angle for each.

Hint

Locate the terminal side by comparing the angle with the quadrant boundaries, then add or subtract $360^{\circ}$ to get a co-terminal angle.

Solution

(This question consists of multiple subparts. See solutions below.)

Quick Solution

Q1Part (i)

EasyUnder Review

$65^{\circ}$

Hint

Compare the angle with the boundaries $0^{\circ}, \pm 90^{\circ}, \pm 180^{\circ}, \pm 270^{\circ}$ , then add $360^{\circ}$ .

Solution

Locate the terminal side.

A positive angle is measured anticlockwise from the positive $x$ -axis. Quadrant I covers angles between $0^{\circ}$ and $90^{\circ}$ . Its terminal side lies between the positive $x$ -axis and the positive $y$ -axis.

$0^{\circ} < 65^{\circ} < 90^{\circ}$
Name the quadrant.

So $65^{\circ}$ lies in Quadrant I.
Write a co-terminal angle.

Co-terminal angles share the same terminal side, so adding $360^{\circ}$ (one full turn) gives one.
$\begin{matrix} 65^{\circ} + 360^{\circ} & = 425^{\circ} \end{matrix}$

Answer

$65^{\circ} \in Quadrant I, co-terminal angle = 425^{\circ}$

Quick Solution

Q1Part (ii)

EasyUnder Review

$135^{\circ}$

Hint

Compare the angle with the boundaries $0^{\circ}, \pm 90^{\circ}, \pm 180^{\circ}, \pm 270^{\circ}$ , then add $360^{\circ}$ .

Solution

Locate the terminal side.

A positive angle is measured anticlockwise. Quadrant II covers angles between $90^{\circ}$ and $180^{\circ}$ . Its terminal side lies between the positive $y$ -axis and the negative $x$ -axis.

$90^{\circ} < 135^{\circ} < 180^{\circ}$
Name the quadrant.

So $135^{\circ}$ lies in Quadrant II.
Write a co-terminal angle.

Co-terminal angles share the same terminal side, so adding $360^{\circ}$ (one full turn) gives one.
$\begin{matrix} 135^{\circ} + 360^{\circ} & = 495^{\circ} \end{matrix}$

Answer

$135^{\circ} \in Quadrant II, co-terminal angle = 495^{\circ}$

Quick Solution

Q1Part (iii)

EasyUnder Review

$- 40^{\circ}$

Hint

Compare the angle with the boundaries $0^{\circ}, \pm 90^{\circ}, \pm 180^{\circ}, \pm 270^{\circ}$ , then add $360^{\circ}$ .

Solution

Locate the terminal side.

A negative angle is measured clockwise from the positive $x$ -axis. Turning clockwise by less than $90^{\circ}$ lands in Quadrant IV. Its terminal side lies between the positive $x$ -axis and the negative $y$ -axis.

$- 90^{\circ} < - 40^{\circ} < 0^{\circ}$
Name the quadrant.

So $- 40^{\circ}$ lies in Quadrant IV.
Write a co-terminal angle.

Co-terminal angles share the same terminal side, so adding $360^{\circ}$ (one full turn) gives one.
$\begin{matrix} - 40^{\circ} + 360^{\circ} & = 320^{\circ} \end{matrix}$

Answer

$- 40^{\circ} \in Quadrant IV, co-terminal angle = 320^{\circ}$

Quick Solution

Q1Part (iv)

EasyUnder Review

$210^{\circ}$

Hint

Compare the angle with the boundaries $0^{\circ}, \pm 90^{\circ}, \pm 180^{\circ}, \pm 270^{\circ}$ , then add $360^{\circ}$ .

Solution

Locate the terminal side.

A positive angle is measured anticlockwise. Quadrant III covers angles between $180^{\circ}$ and $270^{\circ}$ . Its terminal side lies between the negative $x$ -axis and the negative $y$ -axis.

$180^{\circ} < 210^{\circ} < 270^{\circ}$
Name the quadrant.

So $210^{\circ}$ lies in Quadrant III.
Write a co-terminal angle.

Co-terminal angles share the same terminal side, so adding $360^{\circ}$ (one full turn) gives one.
$\begin{matrix} 210^{\circ} + 360^{\circ} & = 570^{\circ} \end{matrix}$

Answer

$210^{\circ} \in Quadrant III, co-terminal angle = 570^{\circ}$

Quick Solution

Q1Part (v)

EasyUnder Review

$- 150^{\circ}$

Hint

Compare the angle with the boundaries $0^{\circ}, \pm 90^{\circ}, \pm 180^{\circ}, \pm 270^{\circ}$ , then add $360^{\circ}$ .

Solution

Locate the terminal side.

A negative angle is measured clockwise. Turning clockwise past $- 90^{\circ}$ but not past $- 180^{\circ}$ lands in Quadrant III. Its terminal side lies between the negative $y$ -axis and the negative $x$ -axis.

$- 180^{\circ} < - 150^{\circ} < - 90^{\circ}$
Name the quadrant.

So $- 150^{\circ}$ lies in Quadrant III.
Write a co-terminal angle.

Co-terminal angles share the same terminal side, so adding $360^{\circ}$ (one full turn) gives one.
$\begin{matrix} - 150^{\circ} + 360^{\circ} & = 210^{\circ} \end{matrix}$

Answer

$- 150^{\circ} \in Quadrant III, co-terminal angle = 210^{\circ}$

Quick Solution

Question 2

EasyUnder Review

Convert the following into degrees, minutes and seconds.

Hint

Multiply the decimal part of the degrees by 60 to get minutes, then the decimal part of the minutes by 60 to get seconds.

Solution

(This question consists of multiple subparts. See solutions below.)

Quick Solution

Q2Part (i)

EasyUnder Review

${123.456}^{\circ}$

Hint

Use $1^{\circ} = 60^{'}$ and $1^{'} = 60^{''}$ on the decimal parts.

Solution

Separate the whole degrees.

Split the angle into its whole number of degrees and its decimal part.
$\begin{matrix} {123.456}^{\circ} & = 123^{\circ} + {0.456}^{\circ} \end{matrix}$
Convert the decimal degrees to minutes.

Since $1^{\circ} = 60^{'}$ , multiply the decimal part by 60.
$\begin{matrix} {0.456}^{\circ} & = (0.456 \times 60)^{'} \\ = {27.36}^{'} \\ = 27^{'} + {0.36}^{'} \end{matrix}$
Convert the decimal minutes to seconds.

Since $1^{'} = 60^{''}$ , multiply the remaining decimal part by 60.
$\begin{matrix} {0.36}^{'} & = (0.36 \times 60)^{''} \\ = {21.6}^{''} \end{matrix}$
Combine.

Put the degrees, minutes and seconds together.
$\begin{matrix} {123.456}^{\circ} & = 123^{\circ} 27^{'} {21.6}^{''} \end{matrix}$

Answer

${123.456}^{\circ} = 123^{\circ} 27^{'} {21.6}^{''}$

Quick Solution

Q2Part (ii)

EasyUnder Review

${58.7891}^{\circ}$

Hint

Use $1^{\circ} = 60^{'}$ and $1^{'} = 60^{''}$ on the decimal parts.

Solution

Separate the whole degrees.

Split the angle into its whole number of degrees and its decimal part.
$\begin{matrix} {58.7891}^{\circ} & = 58^{\circ} + {0.7891}^{\circ} \end{matrix}$
Convert the decimal degrees to minutes.

Since $1^{\circ} = 60^{'}$ , multiply the decimal part by 60.
$\begin{matrix} {0.7891}^{\circ} & = (0.7891 \times 60)^{'} \\ = {47.346}^{'} \\ = 47^{'} + {0.346}^{'} \end{matrix}$
Convert the decimal minutes to seconds.

Since $1^{'} = 60^{''}$ , multiply the remaining decimal part by 60.
$\begin{matrix} {0.346}^{'} & = (0.346 \times 60)^{''} \\ = {20.76}^{''} \end{matrix}$
Combine.

Put the degrees, minutes and seconds together.
$\begin{matrix} {58.7891}^{\circ} & = 58^{\circ} 47^{'} {20.76}^{''} \end{matrix}$

Answer

${58.7891}^{\circ} = 58^{\circ} 47^{'} {20.76}^{''} \approx 58^{\circ} 47^{'} 21^{''}$

Quick Solution

Q2Part (iii)

EasyUnder Review

${90.5678}^{\circ}$

Hint

Use $1^{\circ} = 60^{'}$ and $1^{'} = 60^{''}$ on the decimal parts.

Solution

Separate the whole degrees.

Split the angle into its whole number of degrees and its decimal part.
$\begin{matrix} {90.5678}^{\circ} & = 90^{\circ} + {0.5678}^{\circ} \end{matrix}$
Convert the decimal degrees to minutes.

Since $1^{\circ} = 60^{'}$ , multiply the decimal part by 60.
$\begin{matrix} {0.5678}^{\circ} & = (0.5678 \times 60)^{'} \\ = {34.068}^{'} \\ = 34^{'} + {0.068}^{'} \end{matrix}$
Convert the decimal minutes to seconds.

Since $1^{'} = 60^{''}$ , multiply the remaining decimal part by 60.
$\begin{matrix} {0.068}^{'} & = (0.068 \times 60)^{''} \\ = {4.08}^{''} \end{matrix}$
Combine.

Put the degrees, minutes and seconds together.
$\begin{matrix} {90.5678}^{\circ} & = 90^{\circ} 34^{'} {4.08}^{''} \end{matrix}$

Answer

${90.5678}^{\circ} = 90^{\circ} 34^{'} {4.08}^{''}$

Quick Solution

Question 3

EasyUnder Review

Convert the following into decimal degrees.

Hint

Divide the minutes by 60 and the seconds by 3600, then add them to the degrees.

Solution

(This question consists of multiple subparts. See solutions below.)

Quick Solution

Q3Part (i)

EasyUnder Review

$65^{\circ} 32^{'} 15^{''}$

Hint

Use $1^{'} = {\frac{1}{60}}^{\circ}$ and $1^{''} = {\frac{1}{3600}}^{\circ}$ .

Solution

Convert minutes and seconds to degrees.

Since $1^{'} = {(\frac{1}{60})}^{\circ}$ and $1^{''} = {(\frac{1}{3600})}^{\circ}$ , divide the minutes by 60 and the seconds by 3600.
$\begin{matrix} 32^{'} & = {(\frac{32}{60})}^{\circ} \\ = {0.5333}^{\circ} \end{matrix}$ $\begin{matrix} 15^{''} & = {(\frac{15}{3600})}^{\circ} \\ = {0.0042}^{\circ} \end{matrix}$
Add to the whole degrees.

Add both decimal parts to the degrees. Keeping the exact fractions, $\frac{32}{60} + \frac{15}{3600} = \frac{1935}{3600} = 0.5375$ , so the rounded values add correctly.
$\begin{matrix} 65^{\circ} 32^{'} 15^{''} & = 65^{\circ} + {0.5333}^{\circ} + {0.0042}^{\circ} \\ = {65.5375}^{\circ} \end{matrix}$

Answer

$65^{\circ} 32^{'} 15^{''} = {65.5375}^{\circ}$

Quick Solution

Q3Part (ii)

EasyUnder Review

$42^{\circ} 18^{'} 45^{''}$

Hint

Use $1^{'} = {\frac{1}{60}}^{\circ}$ and $1^{''} = {\frac{1}{3600}}^{\circ}$ .

Solution

Convert minutes and seconds to degrees.

Since $1^{'} = {(\frac{1}{60})}^{\circ}$ and $1^{''} = {(\frac{1}{3600})}^{\circ}$ , divide the minutes by 60 and the seconds by 3600.
$\begin{matrix} 18^{'} & = {(\frac{18}{60})}^{\circ} \\ = {0.3}^{\circ} \end{matrix}$ $\begin{matrix} 45^{''} & = {(\frac{45}{3600})}^{\circ} \\ = {0.0125}^{\circ} \end{matrix}$
Add to the whole degrees.

Add both decimal parts to the degrees.
$\begin{matrix} 42^{\circ} 18^{'} 45^{''} & = 42^{\circ} + {0.3}^{\circ} + {0.0125}^{\circ} \\ = {42.3125}^{\circ} \end{matrix}$

Answer

$42^{\circ} 18^{'} 45^{''} = {42.3125}^{\circ}$

Quick Solution

Q3Part (iii)

EasyUnder Review

$78^{\circ} 45^{'} 36^{''}$

Hint

Use $1^{'} = {\frac{1}{60}}^{\circ}$ and $1^{''} = {\frac{1}{3600}}^{\circ}$ .

Solution

Convert minutes and seconds to degrees.

Since $1^{'} = {(\frac{1}{60})}^{\circ}$ and $1^{''} = {(\frac{1}{3600})}^{\circ}$ , divide the minutes by 60 and the seconds by 3600.
$\begin{matrix} 45^{'} & = {(\frac{45}{60})}^{\circ} \\ = {0.75}^{\circ} \end{matrix}$ $\begin{matrix} 36^{''} & = {(\frac{36}{3600})}^{\circ} \\ = {0.01}^{\circ} \end{matrix}$
Add to the whole degrees.

Add both decimal parts to the degrees.
$\begin{matrix} 78^{\circ} 45^{'} 36^{''} & = 78^{\circ} + {0.75}^{\circ} + {0.01}^{\circ} \\ = {78.76}^{\circ} \end{matrix}$

Answer

$78^{\circ} 45^{'} 36^{''} = {78.76}^{\circ}$

Quick Solution

Question 4

EasyUnder Review

Convert the following into radian.

Hint

Multiply the number of degrees by $\frac{π}{180}$ and simplify the fraction.

Solution

(This question consists of multiple subparts. See solutions below.)

Quick Solution

Q4Part (i)

EasyUnder Review

$36^{\circ}$

Hint

Multiply by $\frac{π}{180}$ .

Solution

Use the conversion factor.

Since $180^{\circ} = π$ rad, we have $1^{\circ} = \frac{π}{180}$ rad.
$\begin{matrix} 36^{\circ} & = 36 \times \frac{π}{180} rad \\ = \frac{36 π}{180} rad \end{matrix}$
Simplify.

Divide 36 and 180 by their HCF, 36.
$\begin{matrix} 36^{\circ} & = \frac{π}{5} rad \\ \approx 0.6283 rad \end{matrix}$

Answer

$36^{\circ} = \frac{π}{5} rad \approx 0.6283 rad$

Quick Solution

Q4Part (ii)

EasyUnder Review

${22.5}^{\circ}$

Hint

Multiply by $\frac{π}{180}$ .

Solution

Use the conversion factor.

Since $180^{\circ} = π$ rad, we have $1^{\circ} = \frac{π}{180}$ rad.
$\begin{matrix} {22.5}^{\circ} & = 22.5 \times \frac{π}{180} rad \\ = \frac{22.5 π}{180} rad \end{matrix}$
Simplify.

Since $180 \div 22.5 = 8$ , the fraction reduces to $\frac{1}{8}$ .
$\begin{matrix} {22.5}^{\circ} & = \frac{π}{8} rad \\ \approx 0.3927 rad \end{matrix}$

Answer

${22.5}^{\circ} = \frac{π}{8} rad \approx 0.3927 rad$

Quick Solution

Q4Part (iii)

EasyUnder Review

${67.5}^{\circ}$

Hint

Multiply by $\frac{π}{180}$ .

Solution

Use the conversion factor.

Since $180^{\circ} = π$ rad, we have $1^{\circ} = \frac{π}{180}$ rad.
$\begin{matrix} {67.5}^{\circ} & = 67.5 \times \frac{π}{180} rad \\ = \frac{67.5 π}{180} rad \end{matrix}$
Simplify.

Since $67.5 = 3 \times 22.5$ and $180 = 8 \times 22.5$ , the fraction reduces to $\frac{3}{8}$ .
$\begin{matrix} {67.5}^{\circ} & = \frac{3 π}{8} rad \\ \approx 1.1781 rad \end{matrix}$

Answer

${67.5}^{\circ} = \frac{3 π}{8} rad \approx 1.1781 rad$

Quick Solution

Question 5

EasyUnder Review

Convert the following into degrees.

Hint

Multiply the radian measure by $\frac{180^{\circ}}{π}$ ; the $π$ cancels.

Solution

(This question consists of multiple subparts. See solutions below.)

Quick Solution

Q5Part (i)

EasyUnder Review

$\frac{π}{16} rad$

Hint

Multiply by $\frac{180^{\circ}}{π}$ and cancel $π$ .

Solution

Use the conversion factor.

Since $π$ rad $= 180^{\circ}$ , we have $1$ rad $= \frac{180^{\circ}}{π}$ .
$\begin{matrix} \frac{π}{16} rad & = \frac{π}{16} \times \frac{180^{\circ}}{π} \\ = \frac{180^{\circ}}{16} \\ = {11.25}^{\circ} \end{matrix}$
Express in degrees and minutes.

Since ${0.25}^{\circ} = 15^{'}$ , this is also $11^{\circ} 15^{'}$ .

Answer

$\frac{π}{16} rad = {11.25}^{\circ}$

Quick Solution

Q5Part (ii)

EasyUnder Review

$\frac{11 π}{5} rad$

Hint

Multiply by $\frac{180^{\circ}}{π}$ and cancel $π$ .

Solution

Use the conversion factor.

Since $π$ rad $= 180^{\circ}$ , we have $1$ rad $= \frac{180^{\circ}}{π}$ .

\begin{matrix} \frac{11 π}{5} rad & = \frac{11 π}{5} \times \frac{180^{\circ}}{π} \\ = 11 \times 36^{\circ} \\ = 396^{\circ} \end{matrix}

Answer

$\frac{11 π}{5} rad = 396^{\circ}$

Quick Solution

Q5Part (iii)

EasyUnder Review

$\frac{7 π}{6} rad$

Hint

Multiply by $\frac{180^{\circ}}{π}$ and cancel $π$ .

Solution

Use the conversion factor.

Since $π$ rad $= 180^{\circ}$ , we have $1$ rad $= \frac{180^{\circ}}{π}$ .

\begin{matrix} \frac{7 π}{6} rad & = \frac{7 π}{6} \times \frac{180^{\circ}}{π} \\ = 7 \times 30^{\circ} \\ = 210^{\circ} \end{matrix}

Answer

$\frac{7 π}{6} rad = 210^{\circ}$

Quick Solution

Question 6

MediumUnder Review

Find the arc length and area of a sector with:

Hint

Use $ℓ = r θ$ and $A = \frac{1}{2} r^{2} θ$ with $θ$ in radians.

Solution

(This question consists of multiple subparts. See solutions below.)

Quick Solution

Q6Part (i)

EasyUnder Review

$r = 6$ cm and $θ = \frac{π}{3}$ rad

Hint

Substitute directly into $ℓ = r θ$ and $A = \frac{1}{2} r^{2} θ$ .

Solution

Find the arc length.

With $θ$ in radians, the arc length is $ℓ = r θ$ .
$\begin{matrix} ℓ & = 6 \times \frac{π}{3} \\ = 2 π cm \\ \approx 6.28 cm \end{matrix}$
Find the area of the sector.

The area of a sector is $A = \frac{1}{2} r^{2} θ$ .
$\begin{matrix} A & = \frac{1}{2} \times 6^{2} \times \frac{π}{3} \\ = \frac{36 π}{6} \\ = 6 π {cm}^{2} \\ \approx 18.85 {cm}^{2} \end{matrix}$

Answer

$ℓ = 2 π \approx 6.28 cm, A = 6 π \approx 18.85 {cm}^{2}$

Quick Solution

Q6Part (ii)

MediumUnder Review

$r = \frac{4.8}{π}$ cm and central angle $θ = \frac{5 π}{6}$ radians

Hint

Find $ℓ = r θ$ first, then use $A = \frac{1}{2} r ℓ$ .

Solution

Find the arc length.

With $θ$ in radians, $ℓ = r θ$ ; the $π$ cancels.
$\begin{matrix} ℓ & = \frac{4.8}{π} \times \frac{5 π}{6} \\ = \frac{24}{6} \\ = 4 cm \end{matrix}$
Find the area of the sector.

Since $A = \frac{1}{2} r^{2} θ = \frac{1}{2} r (r θ)$ , we can use $A = \frac{1}{2} r ℓ$ with the arc length just found.
$\begin{matrix} A & = \frac{1}{2} \times \frac{4.8}{π} \times 4 \\ = \frac{9.6}{π} {cm}^{2} \\ \approx 3.056 {cm}^{2} \end{matrix}$

Answer

$ℓ = 4 cm, A = \frac{9.6}{π} \approx 3.056 {cm}^{2}$

Quick Solution

Question 7

MediumUnder Review

If the central angle of a sector is $60^{\circ}$ and the radius of the circle is $12$ cm, find the area of the sector and the percentage of the total area of the circle it represents.

Hint

Change $60^{\circ}$ to radians, use $A = \frac{1}{2} r^{2} θ$ , then compare with $π r^{2}$ .

Solution

Convert the angle to radians.

The sector formula needs $θ$ in radians, so multiply by $\frac{π}{180}$ .
$\begin{matrix} θ & = 60 \times \frac{π}{180} \\ = \frac{π}{3} rad \end{matrix}$
Find the area of the sector.

Use $A_{s} = \frac{1}{2} r^{2} θ$ .
$\begin{matrix} A_{s} & = \frac{1}{2} \times 12^{2} \times \frac{π}{3} \\ = \frac{144 π}{6} \\ = 24 π {cm}^{2} \\ \approx 75.40 {cm}^{2} \end{matrix}$
Find the area of the circle.

The whole circle has area $π r^{2}$ .
$\begin{matrix} A_{c} & = π \times 12^{2} \\ = 144 π {cm}^{2} \end{matrix}$
Find the percentage.

Divide the sector's area by the circle's area and multiply by 100. Working with the exact values keeps $π$ out of the answer.
$\begin{matrix} Percentage & = \frac{24 π}{144 π} \times 100 % \\ = \frac{1}{6} \times 100 % \\ = 16 \frac{2}{3} % \\ \approx 16.67 % \end{matrix}$

Answer

$A = 24 π \approx 75.40 {cm}^{2}, Percentage \approx 16.67 %$

Quick Solution

Question 8

MediumUnder Review

Find the percentage of the area of a sector subtending an angle of $\frac{π}{8}$ radians.

Hint

The fraction of the circle is $\frac{θ}{2 π}$ ; the radius cancels.

Solution

Write the sector area in terms of the radius.

Let the radius be $r$ . The sector area is $\frac{1}{2} r^{2} θ$ .
$\begin{matrix} A_{s} & = \frac{1}{2} r^{2} \times \frac{π}{8} \\ = \frac{π r^{2}}{16} \end{matrix}$
Compare with the area of the circle.

The circle's area is $π r^{2}$ . The radius cancels, so the answer does not depend on the size of the circle.
$\begin{matrix} Percentage & = \frac{π r^{2} / 16}{π r^{2}} \times 100 % \\ = \frac{100}{16} % \\ = 6.25 % \end{matrix}$

Answer

$6.25 %$

Quick Solution

Question 9

HardUnder Review

A circular sector of radius $r = 12$ cm has an angle of $150^{\circ}$ . This sector is cut out and then bent to form a cone. What is the slant height and radius of the base of the cone?

(Figure: a sector with centre $O$ , both bounding radii $12$ cm and central angle $150^{\circ}$ .)

Hint

The sector's radius becomes the slant height, and its arc length becomes the base circumference.

Solution

Find the slant height.

When the sector is rolled into a cone, its two straight edges (the radii) run from the tip of the cone to the base edge, so the sector's radius becomes the cone's slant height.

$ℓ = 12 cm$
Find the arc length of the sector.

Convert $150^{\circ}$ to radians, then use $s = r θ$ .
$\begin{matrix} θ & = 150 \times \frac{π}{180} \\ = \frac{5 π}{6} rad \end{matrix}$ $\begin{matrix} s & = 12 \times \frac{5 π}{6} \\ = 10 π cm \end{matrix}$
Find the base radius.

The curved arc wraps round to become the circumference of the cone's base, so $2 π R = s$ .
$\begin{matrix} 2 π R & = 10 π \\ R & = \frac{10 π}{2 π} \\ = 5 cm \end{matrix}$

Answer

$Slant height = 12 cm, base radius = 5 cm$

Quick Solution

The exercises and question numbering reproduced on these pages are from Mathematics for Class 9 (National Curriculum of Pakistan 2023), published by the Punjab Curriculum and Textbook Board (PCTB), Lahore, authored by Muhammad Akhtar Shirani, Madiha Mahmood, and Ghulam Murtaza. PCTB holds the copyright in the original textbook. PrepSure is not affiliated with, endorsed by, or sponsored by PCTB. The worked solutions, explanations, hints and method notes are PrepSure's own original work, written and reviewed by our team. If you hold rights in this material and believe anything here exceeds fair use, write to us and we will take it down.

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