CBSE Class 7 Mathematics Comparing Quantities Worksheet Set 02

Question 1. Show that in an isosceles triangle, angles opposite to equal sides are equal.
Answer: Let \( \Delta \text{ABC} \) be an isosceles triangle where side \( \text{AB} = \text{AC} \).
Draw the angle bisector of \( \angle \text{A} \) so that it meets the base \( \text{BC} \) at point D.
Now, compare \( \Delta \text{ABD} \) and \( \Delta \text{ACD} \):
- \( \text{AB} = \text{AC} \) (Given equal sides)
- \( \angle \text{BAD} = \angle \text{CAD} \) (AD bisects \( \angle \text{A} \))
- \( \text{AD} = \text{AD} \) (Common side shared by both triangles)
By the SAS (Side-Angle-Side) congruence rule:
\( \Delta \text{ABD} \cong \Delta \text{ACD} \).
Because corresponding parts of congruent triangles are equal (CPCT):
\( \angle \text{B} = \angle \text{C} \).
Hence, the angles opposite to the equal sides are equal.
In simple words: Draw a line down the middle from top corner A. This splits the triangle into two matching halves, which makes corner angle B equal to corner angle C.

Exam Tip: State your construction step clearly (drawing the bisector of angle A) before proving the two triangles congruent using SAS.

Question 2. If ∆ABC ≅ ∆FED under the correspondence ABC ↔ FED, write all the corresponding congruent parts of the triangles.
Answer: When two triangles are congruent, their matching vertices line up in order.
Here, the correspondence is \( \text{A} \leftrightarrow \text{F} \), \( \text{B} \leftrightarrow \text{E} \), and \( \text{C} \leftrightarrow \text{D} \).
1. Corresponding congruent sides:
- Side \( \text{AB} \leftrightarrow \text{Side FE} \) (or \( \text{AB} = \text{FE} \))
- Side \( \text{BC} \leftrightarrow \text{Side ED} \) (or \( \text{BC} = \text{ED} \))
- Side \( \text{AC} \leftrightarrow \text{Side FD} \) (or \( \text{AC} = \text{FD} \))
2. Corresponding congruent angles:
- \( \angle \text{A} \leftrightarrow \angle \text{F} \) (or \( \angle \text{A} = \angle \text{F} \))
- \( \angle \text{B} \leftrightarrow \angle \text{E} \) (or \( \angle \text{B} = \angle \text{E} \))
- \( \angle \text{C} \leftrightarrow \angle \text{D} \) (or \( \angle \text{C} = \angle \text{D} \))
In simple words: Match the letters in the same positions. A goes with F, B goes with E, and C goes with D for all three sides and all three angles.

Exam Tip: List both groups of parts - all three matching sides and all three matching angles - to get full credit.

Question 3. Arun bought a car for Rs 3,50,000. The next year, the price went upto Rs 3,70,000. What was the percentage of price increase.
Answer: Initial price of the car = Rs 3,50,000.
New price of the car = Rs 3,70,000.
Increase in price = \( 3,70,000 - 3,50,000 = \text{Rs } 20,000 \).
Percentage increase formula:
\( \text{Percentage increase} = \frac{\text{Increase in price}}{\text{Original price}} \times 100\% \)
\( = \frac{20000}{350000} \times 100\% \)
\( = \frac{2}{35} \times 100\% = \frac{200}{35}\% = \frac{40}{7}\% = 5\frac{5}{7}\% \approx 5.71\% \).
Thus, the percentage increase in price is \( 5\frac{5}{7}\% \) (or approximately 5.71%).
In simple words: The car price rose by Rs 20,000. Divide 20,000 by the starting price of 3,50,000 and multiply by 100 to get 5 5/7%.

Exam Tip: Always divide the rise in price by the original starting cost, never by the new price.

Question 4. Amina buys a book for Rs 275 and sells it at a loss of 15%. How much does she sell it for?
Answer: Cost price of the book (CP) = Rs 275.
Loss percentage = 15%.
Calculate the amount of loss:
\( \text{Loss} = 15\% \text{ of } 275 = \frac{15}{100} \times 275 = \frac{4125}{100} = \text{Rs } 41.25 \).
Now find the selling price (SP):
\( \text{SP} = \text{CP} - \text{Loss} \)
\( = 275 - 41.25 = \text{Rs } 233.75 \).
Therefore, Amina sells the book for Rs 233.75.
In simple words: 15% of Rs 275 is a loss of Rs 41.25. Take Rs 41.25 away from Rs 275 to get Rs 233.75.

Exam Tip: Be careful with subtraction involving decimals: write 275 as 275.00 before subtracting 41.25.

Question 5. What rate gives Rs 280 as interest on a sum of Rs 56,000 in 2 years?
Answer: Simple Interest (SI) = Rs 280.
Principal (P) = Rs 56,000.
Time (T) = 2 years.
Formula for interest rate (R):
\( R = \frac{\text{SI} \times 100}{P \times T} \)
\( R = \frac{280 \times 100}{56000 \times 2} \)
\( R = \frac{28000}{112000} = \frac{28}{112} = \frac{1}{4}\% = 0.25\% \).
Hence, the required rate of interest is 0.25% per annum (or \( \frac{1}{4}\% \)).
In simple words: Put the given values into the rate formula. Simplifying gives an annual rate of 0.25%.

Exam Tip: Cancel out zeroes first: \( \frac{280}{560 \times 2} = \frac{280}{1120} = \frac{1}{4} = 0.25\% \).

Question 6. How many lines of symmetry are there in:-
(a) Rectangle
(b) Square
(c) Isosceles triangle
(d) Circle.
Answer:
(a) Rectangle: 2 lines of symmetry (connecting the midpoints of opposite sides).
(b) Square: 4 lines of symmetry (2 along its diagonals and 2 connecting midpoints of opposite sides).
(c) Isosceles triangle: 1 line of symmetry (passing through the vertex between the equal sides).
(d) Circle: Infinitely many lines of symmetry (any straight line passing through its centre).
In simple words: A rectangle has 2 fold lines, a square has 4, an isosceles triangle has 1, and a circle has endlessly many lines of symmetry.

Exam Tip: Remember that a rectangle does not have diagonal lines of symmetry - only its vertical and horizontal bisectors work.

Question 7. Find the order of rotational symmetry of :-
(a) A square
(b) An equilateral triangle
(c) A parallelogram.
Answer:
(a) A square: Order of rotational symmetry is 4 (it looks identical at rotations of 90°, 180°, 270°, and 360°).
(b) An equilateral triangle: Order of rotational symmetry is 3 (it looks identical at rotations of 120°, 240°, and 360°).
(c) A parallelogram: Order of rotational symmetry is 2 (it looks identical at rotations of 180° and 360°).
In simple words: During a full turn, a square matches its original shape 4 times, an equilateral triangle matches 3 times, and a parallelogram matches 2 times.

Exam Tip: The order of rotational symmetry is the number of times a shape fits onto itself in one full 360-degree rotation.

Question 8. Draw a net of i) a cuboid of dimension 3cm X 2cm X 1cm. ii) a cube of edge 3cm.
Answer:
(i) Net of a cuboid (3 cm × 2 cm × 1 cm):
A flat pattern made of 6 rectangular faces that fold together into a cuboid.
(ii) Net of a cube (edge 3 cm):
A flat pattern made of 6 identical square faces of side 3 cm that fold to form a cube.
In simple words: A net is an open 2D pattern made of 6 faces that can be folded into a solid 3D shape.

Exam Tip: Make sure every net has exactly 6 connected faces that fold together without overlapping.

Question 9. Construct isosceles right-angled triangle ABC, where m∠ACB = 900 and AC = 6cm.
Answer: In an isosceles right-angled triangle with \( \angle \text{C} = 90^\circ \), the two perpendicular sides must be equal.
Therefore, side \( \text{BC} = \text{AC} = 6\text{ cm} \).
Steps of construction:
1. Draw a line segment \( \text{AC} = 6\text{ cm} \) using a ruler.
2. At point C, construct a right angle of \( 90^\circ \) using a compass or protractor, drawing ray CX.
3. Set your compass to a radius of 6 cm. With C as the centre, draw an arc cutting ray CX at point B.
4. Join point A and point B with a straight line.
\( \Delta \text{ABC} \) is the required isosceles right-angled triangle.
In simple words: Draw a 6 cm baseline AC, build a 90-degree angle at corner C, mark 6 cm up to point B, and join A to B.

Exam Tip: In an isosceles right triangle, the hypotenuse is the longest side, so the two legs meeting at the 90-degree corner must be the equal sides.

Question 10. Construct ∆ABC in which AB= 3cm, BC= 5cm and ∠B = 600 . Also, write the steps of construction.
Answer: Steps of construction:
1. Draw a base line segment \( \text{BC} = 5\text{ cm} \) using a ruler.
2. At vertex B, draw an angle of \( 60^\circ \) using a compass, creating ray BX.
3. With B as the centre and a compass radius of 3 cm, cut an arc across ray BX to mark point A.
4. Connect point A to point C using a straight line.
\( \Delta \text{ABC} \) is the required triangle.
In simple words: Draw a 5 cm base line BC. At corner B, make a 60-degree angle, measure 3 cm up to A, and join A to C.

Exam Tip: Construct the 60-degree angle using an arc with a compass rather than a protractor to earn full practical geometry marks.

Question 11. Find the mean, median and mode of the following data:-
13, 16, 12, 14, 19, 12, 14, 13, 14, 14
Answer: Total number of observations (\( n \)) = 10.
1. Mean:
Sum of all observations = \( 13 + 16 + 12 + 14 + 19 + 12 + 14 + 13 + 14 + 14 = 141 \).
\( \text{Mean} = \frac{\text{Sum of observations}}{n} = \frac{141}{10} = 14.1 \).
2. Median:
Arrange the data values in ascending order:
12, 12, 13, 13, 14, 14, 14, 14, 16, 19.
Since \( n = 10 \) is an even number, the median is the average of the 5th and 6th values:
5th value = 14, 6th value = 14.
\( \text{Median} = \frac{14 + 14}{2} = 14 \).
3. Mode:
The number 14 occurs 4 times, which is more frequent than any other value.
\( \text{Mode} = 14 \).
In simple words: The mean is 14.1, found by dividing the sum by 10. The middle value is 14, and the most common number is also 14.

Exam Tip: Always sort the numbers from smallest to largest before calculating the median.

Question 12. The following table gives the number of students classified according to their favourite subjects:-

Construct a bar graph to represent this data.
Answer: Choose a suitable scale along the vertical axis (y-axis): let 1 unit = 5 students.
Plot the subjects along the horizontal axis (x-axis) with equal spacing between bars of uniform width:
- English: height = 30 students
- Hindi: height = 35 students
- Maths: height = 40 students
- Science: height = 25 students
- Social Science: height = 30 students
In simple words: Draw bars of equal width on a grid. Let each grid unit represent 5 students, and raise each subject's bar to its given student count.

Exam Tip: State your chosen scale (e.g., 1 unit = 5 students) clearly in the top-right corner of your graph.

Question 13. A bag contains 75 red balls, 25 green balls and 90 blue balls. A ball is taken out from the bag at random. What is the probability that this ball is not a blue ball?
Answer: Total number of balls in the bag:
\( 75 + 25 + 90 = 190 \).
Number of balls that are not blue (red balls + green balls):
\( 75 + 25 = 100 \).
Probability formula:
\( \text{Probability} = \frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}} \)
\( = \frac{100}{190} = \frac{10}{19} \).
Hence, the probability of not picking a blue ball is \( \frac{10}{19} \).
In simple words: There are 100 balls that are not blue out of 190 total balls. This gives a probability of 10/19.

Exam Tip: You can also find this by subtracting the probability of getting a blue ball from 1: \( 1 - \frac{90}{190} = \frac{100}{190} = \frac{10}{19} \).

Question 14. From a circular sheet of radius 4cm, a circle of radius 3cm is removed. Find the area of the remaining sheet. ( Take π = 3.14 )
Answer: Radius of the original circular sheet (\( R \)) = 4 cm.
Radius of the removed inner circle (\( r \)) = 3 cm.
Formula for the area of the remaining sheet:
\( \text{Area} = \pi R^2 - \pi r^2 = \pi(R^2 - r^2) \)
Substitute the given values:
\( = 3.14 \times (4^2 - 3^2) \)
\( = 3.14 \times (16 - 9) \)
\( = 3.14 \times 7 = 21.98\text{ cm}^2 \).
Therefore, the area of the remaining circular sheet is 21.98 cm².
In simple words: Find the area of the large circle and subtract the area of the smaller cutout circle to get 21.98 cm².

Exam Tip: Factor out \( \pi \) as \( \pi(R^2 - r^2) \) first to make your multiplication faster and less prone to errors.

Question 15. Two cross roads, each of width 10cm, cut at right angles through the centre of a rectangular park of length 700m and breadth 300m and parallel to its sides. Find the area of the roads. Also, find the area of the park excluding cross roads. Give the answer in hectares.
Answer: Taking the standard road width of 10 m across the park:
Dimensions of the park: Length = 700 m, Breadth = 300 m.
Total area of the park = \( 700 \times 300 = 210,000\text{ m}^2 \).
1. Area of the roads:
- Area of road parallel to length = \( 700\text{ m} \times 10\text{ m} = 7,000\text{ m}^2 \).
- Area of road parallel to breadth = \( 300\text{ m} \times 10\text{ m} = 3,000\text{ m}^2 \).
- Area of the common square at the crossroads centre = \( 10\text{ m} \times 10\text{ m} = 100\text{ m}^2 \).
Total area of cross roads:
\( 7000 + 3000 - 100 = 9,900\text{ m}^2 \).
Convert to hectares (\( 1\text{ hectare} = 10,000\text{ m}^2 \)):
\( \frac{9900}{10000} = 0.99\text{ hectares} \).
2. Area of the park excluding cross roads:
\( 210,000 - 9,900 = 200,100\text{ m}^2 \).
Convert to hectares:
\( \frac{200100}{10000} = 20.01\text{ hectares} \).
(Note: If calculated with a literal width of \( 10\text{ cm} = 0.1\text{ m} \), the area of roads is \( 99.99\text{ m}^2 \approx 0.01\text{ hectares} \), and the remaining park area is \( 209,900.01\text{ m}^2 \approx 20.99\text{ hectares} \)).
In simple words: Add the areas of the two crossing road strips and subtract the middle overlap square. The roads cover 0.99 hectares, leaving 20.01 hectares for the park.

Exam Tip: Remember to subtract the central square once, because it gets counted twice when adding the two crossing roads.

Question 16. Find the breadth of a rectangular plot of land, if its area is 440m2 and the length is 22m. Also, find its perimeter.
Answer: Area of the rectangular plot = \( 440\text{ m}^2 \).
Length of the plot (\( l \)) = 22 m.
1. Calculate breadth (\( b \)):
\( \text{Breadth} = \frac{\text{Area}}{\text{Length}} = \frac{440}{22} = 20\text{ m} \).
2. Calculate perimeter:
\( \text{Perimeter} = 2(l + b) \)
\( = 2(22 + 20) = 2 \times 42 = 84\text{ m} \).
Therefore, the breadth of the plot is 20 m and its perimeter is 84 m.
In simple words: Divide the area of 440 by the length of 22 to get a width of 20 m. Then add length and width and double it to get a perimeter of 84 m.

Exam Tip: Check your units: breadth and perimeter are measured in metres (m), while area is in square metres (m²).