CBSE Class 7 Mathematics Practice Worksheet Set 10

Question 1. Divide Rs. 45000 among P, Q and R in the ratio 2 : 3 : 4. What will it be in percentages.
Answer: First, find the sum of all ratio terms:
Total parts = \( 2 + 3 + 4 = 9 \).
Now, divide the total amount of Rs. 45,000 into these parts:
P's share = \( \frac{2}{9} \times 45000 = \text{Rs. } 10000 \)
Q's share = \( \frac{3}{9} \times 45000 = \text{Rs. } 15000 \)
R's share = \( \frac{4}{9} \times 45000 = \text{Rs. } 20000 \)
Next, convert each part into a percentage:
Percentage for P = \( \frac{2}{9} \times 100\% = \frac{200}{9}\% = 22\frac{2}{9}\% \) (or \( 22.22\% \))
Percentage for Q = \( \frac{3}{9} \times 100\% = \frac{1}{3} \times 100\% = 33\frac{1}{3}\% \) (or \( 33.33\% \))
Percentage for R = \( \frac{4}{9} \times 100\% = \frac{400}{9}\% = 44\frac{4}{9}\% \) (or \( 44.44\% \)).
In simple words: Add the ratio numbers to get 9 parts. Find each person's money share, then multiply their fraction by 100 to get percentages.

Exam Tip: Add up your calculated rupee values to ensure they total Rs. 45,000, and verify the percentages sum to 100%.

Question 2. The three angles of a triangle are in the ration 1 : 2 : 3. Find their measures. Is it right-triangle?
Answer: Let the three angles of the triangle be \( 1x \), \( 2x \), and \( 3x \).
The total sum of angles inside any triangle is \( 180^\circ \):
\( x + 2x + 3x = 180^\circ \)
\( 6x = 180^\circ \)
\( \implies x = \frac{180^\circ}{6} = 30^\circ \)
Now, calculate each angle:
First angle = \( 1 \times 30^\circ = 30^\circ \)
Second angle = \( 2 \times 30^\circ = 60^\circ \)
Third angle = \( 3 \times 30^\circ = 90^\circ \)
Yes, it is a right-angled triangle because one of its angles measures exactly \( 90^\circ \).
In simple words: Combine the parts to find that x equals 30 degrees. The angles are 30, 60, and 90 degrees, making it a right triangle.

Exam Tip: Always state both the angle values and the concluding yes/no confirmation with its geometric reason.

Question 3. Find the S.P. given,
i) C.P. = Rs. 700, gain = 12%
ii) C.P. = Rs. 960, gain = 7 ½ %
iii) C.P. = Rs. 2000 and loss = 4 ½ %
Answer:
i) \( \text{Gain} = 12\% \text{ of } 700 = \frac{12}{100} \times 700 = \text{Rs. } 84 \).
\( \text{S.P.} = \text{C.P.} + \text{Gain} = 700 + 84 = \text{Rs. } 784 \).

ii) Convert the mixed fraction: \( 7\frac{1}{2}\% = \frac{15}{2}\% \).
\( \text{Gain} = \frac{15}{2 \times 100} \times 960 = \frac{15 \times 96}{20} = \frac{15 \times 24}{5} = 3 \times 24 = \text{Rs. } 72 \).
\( \text{S.P.} = \text{C.P.} + \text{Gain} = 960 + 72 = \text{Rs. } 1032 \).

iii) Convert the mixed fraction: \( 4\frac{1}{2}\% = \frac{9}{2}\% \).
\( \text{Loss} = \frac{9}{2 \times 100} \times 2000 = \frac{9}{200} \times 2000 = 9 \times 10 = \text{Rs. } 90 \).
\( \text{S.P.} = \text{C.P.} - \text{Loss} = 2000 - 90 = \text{Rs. } 1910 \).
In simple words: Work out the profit or loss in rupees first. Add profit to the cost price, or subtract loss from it.

Exam Tip: Convert mixed fractional percentages into improper fractions before multiplying to avoid arithmetic errors.

Question 4. Find the C.P. given,
i) S.P. = Rs. 2725, gain = 9%
ii) S.P. = Rs. 819, loss = 9%
Answer:
i) Using the cost price formula when there is a gain:
\( \text{C.P.} = \frac{\text{S.P.} \times 100}{100 + \text{Gain}\%} = \frac{2725 \times 100}{100 + 9} = \frac{272500}{109} \)
Dividing 2725 by 109 gives 25:
\( \implies \text{C.P.} = 25 \times 100 = \text{Rs. } 2500 \).

ii) Using the cost price formula when there is a loss:
\( \text{C.P.} = \frac{\text{S.P.} \times 100}{100 - \text{Loss}\%} = \frac{819 \times 100}{100 - 9} = \frac{819 \times 100}{91} \)
Dividing 819 by 91 gives 9:
\( \implies \text{C.P.} = 9 \times 100 = \text{Rs. } 900 \).
In simple words: Divide the sale price by 109 for profit or by 91 for loss, then multiply by 100 to get the original cost.

Exam Tip: Never calculate percentage directly on the selling price to find cost price; always divide by \( (100 \pm \text{rate}) \).

Question 5. Find the Amount in each case.
i) P = Rs. 1500, R = 9% p.a., T = 3 years
ii) P = Rs. 2250, R = 8 ⅓% p.a., T = 2 years
iii) P = Rs. 5000, R = 12 ½% p.a., T = 2 years
Answer:
i) Simple Interest \( I = \frac{P \times R \times T}{100} = \frac{1500 \times 9 \times 3}{100} = 15 \times 27 = \text{Rs. } 405 \).
\( \text{Amount} = P + I = 1500 + 405 = \text{Rs. } 1905 \).

ii) Rate \( R = 8\frac{1}{3}\% = \frac{25}{3}\% \).
\( I = \frac{2250 \times \frac{25}{3} \times 2}{100} = \frac{2250 \times 50}{300} = \frac{2250}{6} = \text{Rs. } 375 \).
\( \text{Amount} = P + I = 2250 + 375 = \text{Rs. } 2625 \).

iii) Rate \( R = 12\frac{1}{2}\% = \frac{25}{2}\% \).
\( I = \frac{5000 \times \frac{25}{2} \times 2}{100} = \frac{5000 \times 25}{100} = 50 \times 25 = \text{Rs. } 1250 \).
\( \text{Amount} = P + I = 5000 + 1250 = \text{Rs. } 6250 \).
In simple words: Find the simple interest using the standard rule, then add that interest back to the starting money.

Exam Tip: Be sure to compute the total Amount (\( A = P + I \)) and not just stop at Simple Interest.

Question 6. Find the sum borrowed (Principal) in each case.
i) I = Rs. 1080, R = 9% p.a., T = 3 years
ii) I = Rs. 800, R = 8% p.a., T = 2 years
iii) I = Rs. 1050, R = 5% p.a., T = 3 years
Answer: The principal is given by \( P = \frac{I \times 100}{R \times T} \).

i) \( P = \frac{1080 \times 100}{9 \times 3} = \frac{108000}{27} = \text{Rs. } 4000 \).

ii) \( P = \frac{800 \times 100}{8 \times 2} = \frac{80000}{16} = \text{Rs. } 5000 \).

iii) \( P = \frac{1050 \times 100}{5 \times 3} = \frac{105000}{15} = \text{Rs. } 7000 \).
In simple words: Multiply interest by 100, then divide by rate times time to find the starting borrowed sum.

Exam Tip: Rearrange \( I = \frac{PRT}{100} \) to write \( P = \frac{100 \times I}{RT} \) clearly before substituting values.

Question 7. Find the Rate of Interest, given
i) Int = Rs. 825, P = Rs. 3300, T = 3 years
ii) P = Rs. 6000, Int= Rs. 780, T = 2 years
iii) P = Rs. 3500, Int= Rs. 490, T = 2 years
Answer: The formula for the interest rate is \( R = \frac{\text{Int} \times 100}{P \times T} \).

i) \( R = \frac{825 \times 100}{3300 \times 3} = \frac{825}{33 \times 3} = \frac{825}{99} = \frac{25}{3}\% = 8\frac{1}{3}\% \text{ p.a.} \) (or \( 8.33\% \text{ p.a.} \)).

ii) \( R = \frac{780 \times 100}{6000 \times 2} = \frac{780}{60 \times 2} = \frac{78}{12} = 6.5\% \text{ p.a.} \) (or \( 6\frac{1}{2}\% \text{ p.a.} \)).

iii) \( R = \frac{490 \times 100}{3500 \times 2} = \frac{490}{35 \times 2} = \frac{490}{70} = 7\% \text{ p.a.} \).
In simple words: Multiply the interest by 100, then divide by principal times years to get the yearly percent.

Exam Tip: Include "p.a." (per annum) with the final rate percentage to earn full marks.

Question 8. The price of Yamaha Motor Cycle in a particular year was Rs. 69,000. The next year the price went up to Rs. 73,830. What was the percentage of increase ?
Answer: Original price = Rs. 69,000
New price = Rs. 73,830
Increase in price = \( 73830 - 69000 = \text{Rs. } 4830 \).
Formula for percentage increase:
\( \text{Percentage increase} = \frac{\text{Increase in price}}{\text{Original price}} \times 100\% \)
\( \implies \text{Percentage increase} = \frac{4830}{69000} \times 100\% = \frac{483}{69}\% = 7\% \).
Hence, the price increased by \( 7\% \).
In simple words: Subtract the two prices to get 4830 rupees. Divide that by the starting price and multiply by 100.

Exam Tip: Always place the initial (original) price in the denominator when calculating percentage change.

Question 9. Convert each part of the ratio to percentage.
i) 7:13 ii) 3 : 4 : 5 iii) 12 : 13
Answer:
i) Total parts = \( 7 + 13 = 20 \).
First part = \( \frac{7}{20} \times 100\% = 7 \times 5\% = 35\% \)
Second part = \( \frac{13}{20} \times 100\% = 13 \times 5\% = 65\% \).

ii) Total parts = \( 3 + 4 + 5 = 12 \).
First part = \( \frac{3}{12} \times 100\% = \frac{1}{4} \times 100\% = 25\% \)
Second part = \( \frac{4}{12} \times 100\% = \frac{1}{3} \times 100\% = 33\frac{1}{3}\% \) (or \( 33.33\% \))
Third part = \( \frac{5}{12} \times 100\% = \frac{125}{3}\% = 41\frac{2}{3}\% \) (or \( 41.67\% \)).

iii) Total parts = \( 12 + 13 = 25 \).
First part = \( \frac{12}{25} \times 100\% = 12 \times 4\% = 48\% \)
Second part = \( \frac{13}{25} \times 100\% = 13 \times 4\% = 52\% \).
In simple words: Add up the ratio terms. Divide each part by this total and multiply by 100 to get percentages.

Exam Tip: Quick self-check: verify that the percentages in each sub-part sum up to exactly 100%.

Ch. 10 : Practical Geometry

Question 1. Construct ∆ ABC , given AB= 6cm, BC=8cm and AC=10cm Measure ∠ B.
Answer: Steps of Construction:
1. Draw a baseline segment \( \text{BC} = 8\text{ cm} \).
2. Using a compass set to a radius of \( 6\text{ cm} \), draw an arc with B as the centre.
3. With C as the centre, draw an arc of radius \( 10\text{ cm} \) cutting the first arc at point A.
4. Connect A to B and A to C to form triangle ABC.

Measurement:
Measuring \( \angle B \) with a protractor gives \( \angle B = 90^\circ \).
(Note: Since \( 6^2 + 8^2 = 36 + 64 = 100 = 10^2 \), triangle ABC is a right-angled triangle by the converse of Pythagoras theorem). In simple words: Draw the 8 cm base. Swing a 6 cm arc from one end and a 10 cm arc from the other. The angle between the two shorter sides is 90 degrees.

Exam Tip: Keep construction arcs clear and visible; do not erase them, as examiners check for them.

Question 2. Construct an isosceles triangle ABC with AB = AC. BC = 5.5cm. Perimeter = 17.5 cm.
Answer: Perimeter = \( \text{AB} + \text{AC} + \text{BC} = 17.5\text{ cm} \).
Since \( \text{AB} = \text{AC} \):
\( 2\text{AB} + 5.5 = 17.5 \)
\( 2\text{AB} = 17.5 - 5.5 = 12\text{ cm} \)
\( \implies \text{AB} = \text{AC} = 6\text{ cm} \).

Steps of Construction:
1. Draw a line segment \( \text{BC} = 5.5\text{ cm} \).
2. Set your compass to \( 6\text{ cm} \). With B as the centre, swing an arc above BC.
3. Keeping the radius at \( 6\text{ cm} \), swing an arc from C cutting the earlier arc at point A.
4. Draw line segments AB and AC to complete the isosceles triangle ABC. In simple words: Take 5.5 cm away from the perimeter to get 12 cm, so the two equal sides are 6 cm each. Use compass arcs to make the triangle.

Exam Tip: Show the calculation finding the two equal sides (\( \text{AB} = \text{AC} = 6\text{ cm} \)) before listing construction steps.

Question 3. Construct ∆ PQR with m ∠Q = 120° and PQ = 7 cm = QR.
Answer: Steps of Construction:
1. Draw a line segment \( \text{QR} = 7\text{ cm} \).
2. At point Q, construct an angle of \( 120^\circ \) using a compass or protractor, and extend ray QX.
3. Set your compass to \( 7\text{ cm} \). With centre Q, cut an arc along ray QX at point P.
4. Connect point P to point R to complete the triangle PQR. In simple words: Draw a 7 cm base. Open an angle of 120 degrees at Q, mark 7 cm along that line for P, and join P to R.

Exam Tip: An angle of \( 120^\circ \) is obtuse; ensure the drawn ray opens wider than a right angle.

Question 4. Construct ∆ XYZ given YZ = 10 cm, ∠Y = 60°, ∠X = 90°
Answer: First, find the missing angle at vertex Z using the angle sum property of triangles:
\( \angle X + \angle Y + \angle Z = 180^\circ \)
\( 90^\circ + 60^\circ + \angle Z = 180^\circ \)
\( \implies \angle Z = 180^\circ - 150^\circ = 30^\circ \).

Steps of Construction:
1. Draw a baseline segment \( \text{YZ} = 10\text{ cm} \).
2. At point Y, construct ray YP making an angle of \( 60^\circ \) with YZ.
3. At point Z, construct ray ZQ making an angle of \( 30^\circ \) with ZY.
4. Let the rays YP and ZQ intersect at point X.
The angle at X will naturally measure \( 90^\circ \), completing right-angled triangle XYZ. In simple words: Take 90 and 60 away from 180 to find angle Z is 30 degrees. Draw the 10 cm base and create angles of 60 and 30 at the ends.

Exam Tip: You cannot draw the triangle without calculating \( \angle Z = 30^\circ \) first because the given baseline is YZ.

Question 5. Construct a right angled triangle with its legs 6 cm each.
Answer: Let the triangle be ABC, right-angled at B, with legs \( \text{AB} = 6\text{ cm} \) and \( \text{BC} = 6\text{ cm} \).

Steps of Construction:
1. Draw a baseline segment \( \text{BC} = 6\text{ cm} \).
2. At point B, construct a perpendicular ray BX such that \( \angle \text{CBX} = 90^\circ \).
3. With B as the centre and a radius of \( 6\text{ cm} \), cut an arc on ray BX at point A.
4. Connect point A to point C.
Triangle ABC is the required isosceles right-angled triangle. In simple words: Draw a 6 cm horizontal line. Make a 90-degree corner at one end, mark 6 cm up on it, and connect the ends.

Exam Tip: Draw the small square corner mark at the vertex to show clearly that it is a \( 90^\circ \) right angle.

Ch. 11 : Perimeter And Area

Question 1. Find the perimeter and area of a rectangle if its length = 13 cm and breadth = 5 cm.
Answer: For a rectangle with length \( l = 13\text{ cm} \) and breadth \( b = 5\text{ cm} \):
\( \text{Perimeter} = 2(l + b) = 2(13 + 5) = 2 \times 18 = 36\text{ cm} \).
\( \text{Area} = l \times b = 13 \times 5 = 65\text{ cm}^2 \).
In simple words: Add length and breadth, then double the sum for perimeter. Multiply length by breadth to get the area.

Exam Tip: Remember to write linear units (\( \text{cm} \)) for perimeter and square units (\( \text{cm}^2 \)) for area.

Question 2. Find the perimeter and area of a square whose side is 7 cm.
Answer: Given side of square \( s = 7\text{ cm} \):
\( \text{Perimeter} = 4 \times s = 4 \times 7 = 28\text{ cm} \).
\( \text{Area} = s \times s = 7 \times 7 = 49\text{ cm}^2 \).
In simple words: Multiply the side by 4 to find the outer edge, and multiply the side by itself to find the area.

Exam Tip: Keep the formulas distinct: perimeter is \( 4s \) while area is \( s^2 \).

Question 3. The perimeter of a square is 28cm. Find its area.
Answer: First, find the side length of the square:
\( \text{Side} = \frac{\text{Perimeter}}{4} = \frac{28}{4} = 7\text{ cm} \).
Now, calculate its area:
\( \text{Area} = \text{Side}^2 = 7 \times 7 = 49\text{ cm}^2 \).
In simple words: Divide the perimeter by 4 to get one side, then multiply that side by itself.

Exam Tip: Show the division step finding the side length to secure method marks.

Question 4. The area of a square is 144 cm. Find its Perimeter.
Answer: Given area of the square = \( 144\text{ cm}^2 \).
Find the side length:
\( \text{Side} = \sqrt{\text{Area}} = \sqrt{144} = 12\text{ cm} \).
Now, calculate the perimeter:
\( \text{Perimeter} = 4 \times \text{Side} = 4 \times 12 = 48\text{ cm} \).
In simple words: Find the square root of 144 to get a side of 12 cm. Multiply by 4 to find the perimeter.

Exam Tip: Memorize squares up to 20; recognizing \( \sqrt{144} = 12 \) saves exam time.

Question 5. The perimeter of rectangle is 80cm. If its length is 25cm, find its area.
Answer: Perimeter of rectangle = \( 2(l + b) = 80\text{ cm} \).
Given length \( l = 25\text{ cm} \):
\( 2(25 + b) = 80 \)
\( 25 + b = \frac{80}{2} = 40 \)
\( \implies b = 40 - 25 = 15\text{ cm} \).
Now, find the area:
\( \text{Area} = l \times b = 25 \times 15 = 375\text{ cm}^2 \).
In simple words: Halve 80 to get 40, subtract 25 to find breadth is 15 cm, then multiply 25 by 15.

Exam Tip: Do not forget to divide the perimeter by 2 before subtracting the length.

Question 6. The perimeter of a square is same as the circumference of a circle. If the area of the square is 121 cm, find the area of the circle.
Answer: Area of the square = \( 121\text{ cm}^2 \).
Side of the square \( s = \sqrt{121} = 11\text{ cm} \).
Perimeter of the square = \( 4 \times s = 4 \times 11 = 44\text{ cm} \).
Given that circumference of the circle equals the perimeter of the square:
\( 2\pi r = 44\text{ cm} \)
\( 2 \times \frac{22}{7} \times r = 44 \)
\( \implies r = \frac{44 \times 7}{44} = 7\text{ cm} \).
Now, compute the area of the circle:
\( \text{Area} = \pi r^2 = \frac{22}{7} \times 7 \times 7 = 22 \times 7 = 154\text{ cm}^2 \).
In simple words: The square has side 11 cm and boundary 44 cm. Setting circle circumference to 44 cm gives radius 7 cm and area 154 sq cm.

Exam Tip: When using \( \pi = \frac{22}{7} \), cancel out 7 with one of the radius factors before multiplying.

Question 7. In the figure, ABCD is a parallelogram. AE ⊥ BC, AF ⊥ DC. If AB = 8 cm, BC = 10 cm, AF = 6 cm Find AE
Answer: In parallelogram ABCD, opposite sides are equal:
\( \text{DC} = \text{AB} = 8\text{ cm} \), and \( \text{AD} = \text{BC} = 10\text{ cm} \).
The area of a parallelogram is given by \( \text{Base} \times \text{Height} \).
Taking DC as the base and AF as the height:
\( \text{Area} = \text{DC} \times \text{AF} = 8 \times 6 = 48\text{ cm}^2 \).
The area can also be expressed using BC as the base and AE as the height:
\( \text{Area} = \text{BC} \times \text{AE} \)
\( 48 = 10 \times \text{AE} \)
\( \implies \text{AE} = \frac{48}{10} = 4.8\text{ cm} \).
Hence, AE is \( 4.8\text{ cm} \).
In simple words: The area is 8 times 6, which is 48. Divide 48 by the other base 10 to find AE is 4.8 cm.

Exam Tip: Equate the two area formulas: \( \text{base}_1 \times \text{height}_1 = \text{base}_2 \times \text{height}_2 \).

Question 8. Area of triangle is 60 sq cm. If its base is 12 cm, Find its altitude.
Answer: Formula for area of a triangle:
\( \text{Area} = \frac{1}{2} \times \text{base} \times \text{altitude} \)
Substitute the given values:
\( 60 = \frac{1}{2} \times 12 \times \text{altitude} \)
\( 60 = 6 \times \text{altitude} \)
\( \implies \text{altitude} = \frac{60}{6} = 10\text{ cm} \).
Therefore, the altitude of the triangle is \( 10\text{ cm} \).
In simple words: Half of the base is 6. Divide the area 60 by 6 to get an altitude of 10 cm.

Exam Tip: Rearrange the formula as \( \text{altitude} = \frac{2 \times \text{Area}}{\text{base}} \) for direct evaluation.

Question 9. The area of a circle is 38.5 cm². Find its Circumference.
Answer: Given area of circle = \( 38.5\text{ cm}^2 \).
\( \pi r^2 = 38.5 \)
\( \frac{22}{7} \times r^2 = 38.5 \)
\( \implies r^2 = \frac{38.5 \times 7}{22} = \frac{269.5}{22} = 12.25 \)
\( \implies r = \sqrt{12.25} = 3.5\text{ cm} \).
Now, calculate the circumference:
\( C = 2\pi r = 2 \times \frac{22}{7} \times 3.5 = 2 \times 22 \times 0.5 = 22\text{ cm} \).
Thus, the circumference is \( 22\text{ cm} \).
In simple words: Use the area to find that the radius is 3.5 cm, then calculate the boundary length as 22 cm.

Exam Tip: Notice that \( 12.25 = \frac{49}{4} \), so \( r = \sqrt{\frac{49}{4}} = \frac{7}{2} = 3.5\text{ cm} \), which avoids manual decimal square roots.

Question 10. Find the area of a circular track if the outer radius is 14 m and the inner radius is 7m.
Answer: Let outer radius \( R = 14\text{ m} \) and inner radius \( r = 7\text{ m} \).
Area of the circular track = Area of outer circle - Area of inner circle:
\( \text{Area} = \pi R^2 - \pi r^2 = \pi(R^2 - r^2) \)
Using the identity \( R^2 - r^2 = (R - r)(R + r) \):
\( \text{Area} = \frac{22}{7} \times (14 - 7)(14 + 7) = \frac{22}{7} \times 7 \times 21 = 22 \times 21 = 462\text{ m}^2 \).
Hence, the area of the track is \( 462\text{ m}^2 \).
In simple words: Subtract the smaller inner circle area from the larger outer circle area to get 462 square metres.

Exam Tip: Use the algebraic identity \( \pi(R - r)(R + r) \) to simplify circular ring calculations.

Question 11. The diameter of a wheel of a car is 42 cm. How far will it go in 100 revolutions.
Answer: Radius of the wheel \( r = \frac{42}{2} = 21\text{ cm} \).
Distance covered in 1 revolution = Circumference of the wheel:
\( C = 2\pi r = 2 \times \frac{22}{7} \times 21 = 2 \times 22 \times 3 = 132\text{ cm} \).
Distance covered in 100 revolutions:
\( \text{Total distance} = 100 \times 132 = 13200\text{ cm} \)
Converting into metres:
\( \text{Total distance} = \frac{13200}{100} = 132\text{ m} \).
In simple words: One turn of the wheel covers 132 cm. In 100 turns, it covers 13,200 cm, which is 132 metres.

Exam Tip: Remember to convert your final answer into metres by dividing by 100.

Question 12. The minute hand of a wall clock is 21 cm long. Find the area swept by the minute hand in 30 minutes.
Answer: The length of the minute hand acts as the radius: \( r = 21\text{ cm} \).
In 60 minutes, the minute hand completes 1 full circle.
In 30 minutes, it sweeps half of a circle (a semicircle, angle = \( 180^\circ \)).
\( \text{Area swept} = \frac{1}{2} \times \pi r^2 \)
\( \implies \text{Area swept} = \frac{1}{2} \times \frac{22}{7} \times 21 \times 21 = 11 \times 3 \times 21 = 33 \times 21 = 693\text{ cm}^2 \).
Thus, the area swept in 30 minutes is \( 693\text{ cm}^2 \).
In simple words: In 30 minutes the minute hand moves across half the clock. Half of the full circle's area is 693 sq cm.

Exam Tip: Note that 30 minutes is exactly \( \frac{30}{60} = \frac{1}{2} \) of the circle.

Ch. 12 : Algebraic Expressions

Question 1. Add : 3x² + 4 xy - 2y² and x² - 5xy + y²
Answer: Group like terms together:
\( (3x^2 + 4xy - 2y^2) + (x^2 - 5xy + y^2) \)
\( = (3x^2 + x^2) + (4xy - 5xy) + (-2y^2 + y^2) \)
\( = 4x^2 - xy - y^2 \).
In simple words: Add matching terms: 3x² plus 1x² gives 4x², 4xy minus 5xy gives -xy, and -2y² plus y² gives -y².

Exam Tip: Collect like terms inside separate parentheses before adding coefficients.

Question 2. Subtract : x² - xy + 3y² from 2x² - 2 xy + y²
Answer: Place the expression that comes after "from" first, then subtract:
\( (2x^2 - 2xy + y^2) - (x^2 - xy + 3y^2) \)
Change the signs of the terms being subtracted:
\( = 2x^2 - 2xy + y^2 - x^2 + xy - 3y^2 \)
Combine like terms:
\( = (2x^2 - x^2) + (-2xy + xy) + (y^2 - 3y^2) \)
\( = x^2 - xy - 2y^2 \).
In simple words: Reverse the signs of the subtracted expression, then group and combine the matching variables.

Exam Tip: Always reverse every sign inside the second set of brackets when subtracting.

Question 3. From the sum of a² + 2 ab - b² and ab + 2b² subtract the sum of 2a³ + 5ab and -ab + b²
Answer: First, find the first sum:
\( \text{Sum 1} = (a^2 + 2ab - b^2) + (ab + 2b^2) = a^2 + (2ab + ab) + (-b^2 + 2b^2) = a^2 + 3ab + b^2 \).
Next, find the second sum:
\( \text{Sum 2} = (2a^3 + 5ab) + (-ab + b^2) = 2a^3 + (5ab - ab) + b^2 = 2a^3 + 4ab + b^2 \).
Now subtract Sum 2 from Sum 1:
\( (a^2 + 3ab + b^2) - (2a^3 + 4ab + b^2) \)
\( = a^2 + 3ab + b^2 - 2a^3 - 4ab - b^2 \)
\( = -2a^3 + a^2 + (3ab - 4ab) + (b^2 - b^2) \)
\( = -2a^3 + a^2 - ab \).
In simple words: Add the first two expressions, add the second two expressions, then subtract the second total from the first.

Exam Tip: Keep unlike powers like \( 2a^3 \) and \( a^2 \) separate; they cannot be merged.

Question 4. Find the value of 2x³ - 3x² - 5x + 1 when
i) x = 1 ii) x = -1 iii) x = 2 iv) x = -2
Answer: Let \( P(x) = 2x^3 - 3x^2 - 5x + 1 \).

i) When \( x = 1 \):
\( 2(1)^3 - 3(1)^2 - 5(1) + 1 = 2(1) - 3(1) - 5 + 1 = 2 - 3 - 5 + 1 = -5 \).

ii) When \( x = -1 \):
\( 2(-1)^3 - 3(-1)^2 - 5(-1) + 1 = 2(-1) - 3(1) + 5 + 1 = -2 - 3 + 5 + 1 = 1 \).

iii) When \( x = 2 \):
\( 2(2)^3 - 3(2)^2 - 5(2) + 1 = 2(8) - 3(4) - 10 + 1 = 16 - 12 - 10 + 1 = -5 \).

iv) When \( x = -2 \):
\( 2(-2)^3 - 3(-2)^2 - 5(-2) + 1 = 2(-8) - 3(4) + 10 + 1 = -16 - 12 + 10 + 1 = -17 \).
In simple words: Put each number in place of x, calculate powers carefully with signs, and add up the values.

Exam Tip: Remember that a negative number raised to an odd power remains negative, while an even power becomes positive.

Ch. 13 : Exponents And Powers

Question 1. Prime factorise and write in exponential form.
i) 504 ii) 1512 iii) 3087 iv) 2058
Answer:
i) \( 504 = 2 \times 2 \times 2 \times 3 \times 3 \times 7 = 2^3 \times 3^2 \times 7 \).

ii) \( 1512 = 2 \times 2 \times 2 \times 3 \times 3 \times 3 \times 7 = 2^3 \times 3^3 \times 7 \).

iii) \( 3087 = 3 \times 3 \times 7 \times 7 \times 7 = 3^2 \times 7^3 \).

iv) \( 2058 = 2 \times 3 \times 7 \times 7 \times 7 = 2 \times 3 \times 7^3 \).
In simple words: Break each number into prime factors, then count how many times each factor repeats and write as a power.

Exam Tip: Write prime factors in ascending numerical order (e.g., \( 2, 3, 7 \)).

Question 2. Simplify :
i) \( \frac{3 \times 2^4 \times 3^5}{4 \times 9^2} \)
ii) \( \frac{18^4 \times 4^3 \times 9}{6^3 \times 27^2 \times 8} \)
Answer:
i) Express all terms using prime bases 2 and 3:
Numerator = \( 2^4 \times 3^{1+5} = 2^4 \times 3^6 \).
Denominator = \( 4 \times 9^2 = 2^2 \times (3^2)^2 = 2^2 \times 3^4 \).
Applying the quotient law \( a^m \div a^n = a^{m-n} \):
\( = \frac{2^4 \times 3^6}{2^2 \times 3^4} = 2^{4-2} \times 3^{6-4} = 2^2 \times 3^2 = 4 \times 9 = 36 \).

ii) Express composite bases into powers of 2 and 3:
\( 18 = 2 \times 3^2 \implies 18^4 = (2 \times 3^2)^4 = 2^4 \times 3^8 \)
\( 4 = 2^2 \implies 4^3 = (2^2)^3 = 2^6 \)
\( 9 = 3^2 \)
\( 6 = 2 \times 3 \implies 6^3 = 2^3 \times 3^3 \)
\( 27 = 3^3 \implies 27^2 = (3^3)^2 = 3^6 \)
\( 8 = 2^3 \)
Numerator = \( (2^4 \times 3^8) \times 2^6 \times 3^2 = 2^{4+6} \times 3^{8+2} = 2^{10} \times 3^{10} \).
Denominator = \( (2^3 \times 3^3) \times 3^6 \times 2^3 = 2^{3+3} \times 3^{3+6} = 2^6 \times 3^9 \).
Now simplify:
\( = \frac{2^{10} \times 3^{10}}{2^6 \times 3^9} = 2^{10-6} \times 3^{10-9} = 2^4 \times 3^1 = 16 \times 3 = 48 \).
In simple words: Turn numbers into powers of 2 and 3. Subtract bottom powers from top powers to get the final answer.

Exam Tip: Rewrite bases into prime factor form first before using exponent addition and subtraction rules.

Question 3. Express in standard form :
i) 75,850 ii) 20,035,000,000
Answer: In standard scientific form \( k \times 10^n \), where \( 1 \le k < 10 \):

i) Move the decimal point 4 places to the left:
\( 75,850 = 7.585 \times 10^4 \).

ii) Move the decimal point 10 places to the left:
\( 20,035,000,000 = 2.0035 \times 10^{10} \).
In simple words: Place the decimal point after the first non-zero digit, then count how many places it shifted to find the power of 10.

Exam Tip: Check that the leading number is at least 1 and strictly less than 10.

Ch. 15 : Visualising Solid Shapes

Question 1. Find the dimensions of a cuboid obtained by joining 3 cubes end to end.
Answer: Let the side length of each identical cube be \( a \).
When three such cubes are placed end to end in a single line:
Length of the resulting cuboid = \( a + a + a = 3a \)
Breadth of the resulting cuboid = \( a \)
Height of the resulting cuboid = \( a \)
Thus, the dimensions of the cuboid are \( 3a \times a \times a \) (or length is 3 times the cube side, with breadth and height equal to the cube side).
In simple words: Placing three cubes in a line triples the length while keeping the width and height unchanged.

Exam Tip: Clearly state that only the length increases by a factor of 3, while breadth and height stay the same.

Question 2. Draw ‘NETS’ for the following 3-D shapes.
i) Cube ii) Cuboid iii) a square-pyramid
iv) Tetrahedron (triangle-pyramid)
Answer:
i) Net of a Cube: Consists of 6 congruent squares arranged in a cross layout. ii) Net of a Cuboid: Formed by 6 rectangles with matching opposite pairs. iii) Net of a Square Pyramid: One square base surrounded by 4 triangular faces. iv) Net of a Tetrahedron: One central equilateral triangle with 3 matching triangles attached to its sides. In simple words: A net is the flat pattern you can cut and fold up to make a three-dimensional box or pyramid.

Exam Tip: Make sure the total face count in your drawn net matches the 3D solid (6 for cube/cuboid, 5 for square pyramid, 4 for tetrahedron).

Question 3. Complete the following table.
Answer:

In simple words: Faces are flat walls, edges are the folded crease lines, and vertices are the sharp corner points.

Exam Tip: Verify your completed table entries using Euler's formula: \( F + V - E = 2 \).