CBSE Class 7 Mathematics Algebraic Expression Worksheet Set 04

Algebraic Expressions

Question 1. Find the like terms of the following.
a. x²y - 3xy² + 4xy² + y² - 3y² + x²y
b. a² - 2ab + b² - 4ab + b² - c² + 2a² - b² + 2c²
Answer:
a. Grouping terms with identical variable factors:
- Terms with \( x^2y \): \( x^2y \) and \( x^2y \)
- Terms with \( xy^2 \): \( -3xy^2 \) and \( 4xy^2 \)
- Terms with \( y^2 \): \( y^2 \) and \( -3y^2 \)

b. Grouping matching literal terms:
- Terms with \( a^2 \): \( a^2 \) and \( 2a^2 \)
- Terms with \( ab \): \( -2ab \) and \( -4ab \)
- Terms with \( b^2 \): \( b^2 \), \( b^2 \), and \( -b^2 \)
- Terms with \( c^2 \): \( -c^2 \) and \( 2c^2 \)
In simple words: Like terms have the exact same letters and powers. Group them together by matching their letter parts.

Exam Tip: Ignore the numbers in front and look only at the variable letters and their powers to spot like terms.

Question 2. Add:
a. 21b-32+8b-10b+10
b. 3p²q²-4pq+5-10p²q²+7p²q²
c. 14x+10y-12xy-13; 18-7x-10y+8xy; 4xy
d. ab-4a; 4b-ab; 4a-4b
e. (3y²+6y-10)+(8y-2y²-8)
Answer:
a. Combine like terms:
\[ (21b + 8b - 10b) + (-32 + 10) = 19b - 22 \]

b. Combine like terms:
\[ (3p^2q^2 - 10p^2q^2 + 7p^2q^2) - 4pq + 5 = 0p^2q^2 - 4pq + 5 = -4pq + 5 \]

c. Add all three expressions together:
\[ (14x + 10y - 12xy - 13) + (18 - 7x - 10y + 8xy) + 4xy \]
Group like terms:
\[ = (14x - 7x) + (10y - 10y) + (-12xy + 8xy + 4xy) + (-13 + 18) \]
\[ = 7x + 0y + 0xy + 5 = 7x + 5 \]

d. Add the given expressions:
\[ (ab - 4a) + (4b - ab) + (4a - 4b) \]
\[ = (ab - ab) + (-4a + 4a) + (4b - 4b) = 0 \]

e. Add the polynomials:
\[ (3y^2 - 2y^2) + (6y + 8y) + (-10 - 8) = y^2 + 14y - 18 \]
In simple words: Put matching terms side by side and add their numbers up. Terms that cancel each other become zero.

Exam Tip: Be careful with positive and negative signs when adding constant terms.

Question 3. Subtract:
a. -m²+6mn from 4m²-2mn+7
b. 3pq-6q²-2p² from 6p²+2q²-2pq
c. 6a²-3ab+2b² from 10ab-5a²-2b²
d. -3x²+4y²+3 from 4y²-5
e. 3x(y-5) from y(x+5)
f. 2x³-y³+3y-3 from x³+y³+2y+1
Answer:
a. Subtract the first expression from the second:
\[ (4m^2 - 2mn + 7) - (-m^2 + 6mn) \]
\[ = 4m^2 - 2mn + 7 + m^2 - 6mn = 5m^2 - 8mn + 7 \]

b. Subtract the first expression from the second:
\[ (6p^2 + 2q^2 - 2pq) - (3pq - 6q^2 - 2p^2) \]
\[ = 6p^2 + 2q^2 - 2pq - 3pq + 6q^2 + 2p^2 = 8p^2 + 8q^2 - 5pq \]

c. Subtract the first expression from the second:
\[ (10ab - 5a^2 - 2b^2) - (6a^2 - 3ab + 2b^2) \]
\[ = 10ab - 5a^2 - 2b^2 - 6a^2 + 3ab - 2b^2 = -11a^2 - 4b^2 + 13ab \]

d. Subtract the first expression from the second:
\[ (4y^2 - 5) - (-3x^2 + 4y^2 + 3) \]
\[ = 4y^2 - 5 + 3x^2 - 4y^2 - 3 = 3x^2 - 8 \]

e. Expand brackets first:
\( y(x + 5) = xy + 5y \)
\( 3x(y - 5) = 3xy - 15x \)
Now subtract:
\[ (xy + 5y) - (3xy - 15x) = xy + 5y - 3xy + 15x = -2xy + 15x + 5y \]

f. Subtract the first expression from the second:
\[ (x^3 + y^3 + 2y + 1) - (2x^3 - y^3 + 3y - 3) \]
\[ = x^3 + y^3 + 2y + 1 - 2x^3 + y^3 - 3y + 3 = -x^3 + 2y^3 - y + 4 \]
In simple words: Write the term that comes after "from" first. Reverse the signs of the second expression and combine like terms.

Exam Tip: In subtraction problems, always enclose the expression being subtracted inside brackets so you remember to switch all its signs.

Question 4. Simplify if x=-2, a=-3, b=-1
a. 3x-5-x+9
b. 2a-5b-4+9
c. 2x+3b²-2
d. a²-2a+11
e. 2a-4b+x+2
f. \frac{3x+4+4}{2}
g. \frac{5b+4}{2}
h. \frac{a+b-1}{2}
Answer:
Given: \( x = -2 \), \( a = -3 \), and \( b = -1 \).

a. Simplify: \( 3x - x - 5 + 9 = 2x + 4 \)
Substitute \( x = -2 \):
\[ 2(-2) + 4 = -4 + 4 = 0 \]

b. Simplify: \( 2a - 5b + 5 \)
Substitute \( a = -3 \) and \( b = -1 \):
\[ 2(-3) - 5(-1) + 5 = -6 + 5 + 5 = 4 \]

c. Substitute \( x = -2 \) and \( b = -1 \):
\[ 2(-2) + 3(-1)^2 - 2 = -4 + 3(1) - 2 = -4 + 3 - 2 = -3 \]

d. Substitute \( a = -3 \):
\[ (-3)^2 - 2(-3) + 11 = 9 + 6 + 11 = 26 \]

e. Substitute \( a = -3 \), \( b = -1 \), and \( x = -2 \):
\[ 2(-3) - 4(-1) + (-2) + 2 = -6 + 4 - 2 + 2 = -2 \]

f. Substitute \( x = -2 \):
\[ \frac{3(-2) + 4 + 4}{2} = \frac{-6 + 8}{2} = \frac{2}{2} = 1 \]

g. Substitute \( b = -1 \):
\[ \frac{5(-1) + 4}{2} = \frac{-5 + 4}{2} = -\frac{1}{2} \]

h. Substitute \( a = -3 \) and \( b = -1 \):
\[ \frac{(-3) + (-1) - 1}{2} = \frac{-5}{2} = -\frac{5}{2} \]
In simple words: Replace every letter with its number. Be extra careful with minus signs when multiplying and squaring.

Exam Tip: Remember that squaring a negative number always results in a positive value, such as \( (-3)^2 = 9 \).

Perimeter and Area

Question 1. Find the breadth of a rectangular plot of land if its area is 880sq.m and the length is 44m. Also find its perimeter.
Answer:
Given: Area = \( 880\text{ m}^2 \), Length (\( l \)) = \( 44\text{ m} \).

Formula for breadth (\( b \)):
\[ b = \frac{\text{Area}}{\text{Length}} = \frac{880}{44} = 20\text{ m} \]

Formula for perimeter:
\[ \text{Perimeter} = 2(l + b) = 2(44 + 20) = 2(64) = 128\text{ m} \]
The breadth is \( 20\text{ m} \) and the perimeter is \( 128\text{ m} \).
In simple words: Divide area by length to get breadth 20 m. Add length and breadth, then double the sum to get perimeter 128 m.

Exam Tip: State the units clearly: meters (m) for length, breadth, and perimeter.

Question 2. Perimeter of a rectangular sheet is 400cm.If the length is 50cm.Find its breadth and also find its area.
Answer:
Given: Perimeter = \( 400\text{ cm} \), Length (\( l \)) = \( 50\text{ cm} \).

Formula for perimeter:
\[ 2(l + b) = 400 \]
\[ 50 + b = \frac{400}{2} = 200 \]

\( \implies b = 200 - 50 = 150\text{ cm} \)

Now, find the area:
\[ \text{Area} = l \times b = 50 \times 150 = 7500\text{ cm}^2 \]
The breadth is \( 150\text{ cm} \) and the area is \( 7500\text{ cm}^2 \).
In simple words: Half of the perimeter is 200 cm. Take away 50 cm to get breadth 150 cm. Multiply them to get area 7500 sq cm.

Exam Tip: Divide perimeter by 2 first to get the sum of length and breadth.

Question 3. Find the area of a square whose perimeter is 160m.
Answer:
Given: Perimeter of square = \( 160\text{ m} \).
Formula for side (\( s \)):
\[ s = \frac{\text{Perimeter}}{4} = \frac{160}{4} = 40\text{ m} \]

Area of the square:
\[ \text{Area} = s^2 = 40 \times 40 = 1600\text{ m}^2 \]
The side length is \( 40\text{ m} \) and the area is \( 1600\text{ m}^2 \).
In simple words: Divide perimeter 160 by 4 to get side 40 m. Multiply 40 by 40 to find the area of 1600 sq m.

Exam Tip: A square has 4 equal sides, so always divide perimeter by 4 to find one side.

Question 4. The area of a square 1600sq.m.Find the area of a rectangle whose breadth is the same as that of the side of the square and length is 10m more than its breadth.
Answer:
Area of the square = \( 1600\text{ m}^2 \).
Side of the square (\( s \)):
\[ s = \sqrt{1600} = 40\text{ m} \]

For the rectangle:
Breadth (\( b \)) = side of square = \( 40\text{ m} \)
Length (\( l \)) = \( b + 10 = 40 + 10 = 50\text{ m} \)

Area of the rectangle:
\[ \text{Area} = l \times b = 50 \times 40 = 2000\text{ m}^2 \]
The area of the rectangle is \( 2000\text{ m}^2 \).
In simple words: The square has a side of 40 m. The rectangle has width 40 m and length 50 m. Multiplying them gives 2000 sq m.

Exam Tip: Read the relation between sides carefully: length is 10 m more than the breadth.

Question 5. The length and breadth of a rectangular plot are in the ratio of 4:3.If the perimeter is 140m.
Answer:
Let length (\( l \)) = \( 4x \) and breadth (\( b \)) = \( 3x \).
Perimeter of the rectangle = \( 140\text{ m} \).
\[ 2(l + b) = 140 \]
\[ 2(4x + 3x) = 140 \]
\[ 2(7x) = 140 \]
\[ 14x = 140 \]

\( \implies x = \frac{140}{14} = 10 \)

Now, find the dimensions:
Length = \( 4x = 4 \times 10 = 40\text{ m} \)
Breadth = \( 3x = 3 \times 10 = 30\text{ m} \)
In simple words: Set the sides as 4x and 3x. Use the perimeter formula to find x equals 10, giving sides of 40 m and 30 m.

Exam Tip: Check your answer: \( 2(40 + 30) = 2(70) = 140\text{ m} \).

Question 6. The perimeter of square is 96m.The area of rectangle is 2sq.m less than the area of given square .If the breadth of the rectangle is 14m.Find its length.
Answer:
Side of the square (\( s \)):
\[ s = \frac{\text{Perimeter}}{4} = \frac{96}{4} = 24\text{ m} \]
Area of the square:
\[ \text{Area}_{\text{square}} = s^2 = 24 \times 24 = 576\text{ m}^2 \]

Area of the rectangle is \( 2\text{ m}^2 \) less than the square's area:
\[ \text{Area}_{\text{rectangle}} = 576 - 2 = 574\text{ m}^2 \]

Given breadth of rectangle = \( 14\text{ m} \):
\[ \text{Length} = \frac{\text{Area}}{\text{Breadth}} = \frac{574}{14} = 41\text{ m} \]
The length of the rectangle is \( 41\text{ m} \).
In simple words: The square has area 576 sq m. The rectangle has area 574 sq m. Divide 574 by breadth 14 to find length 41 m.

Exam Tip: Carry out division step by step: \( 574 \div 14 = 41 \).

Question 7. Find the area of the parallelogram.
(i) b=6cm h=5cm
(ii) b=7cm h=4cm
(iii) b=8.4cm h=2.1cm
Answer:
Formula for area of a parallelogram: \( \text{Area} = \text{base} \times \text{height} \)

(i) Given: \( b = 6\text{ cm} \), \( h = 5\text{ cm} \)
\[ \text{Area} = 6 \times 5 = 30\text{ cm}^2 \]

(ii) Given: \( b = 7\text{ cm} \), \( h = 4\text{ cm} \)
\[ \text{Area} = 7 \times 4 = 28\text{ cm}^2 \]

(iii) Given: \( b = 8.4\text{ cm} \), \( h = 2.1\text{ cm} \)
\[ \text{Area} = 8.4 \times 2.1 = 17.64\text{ cm}^2 \]
In simple words: Multiply base by height to get the area of each parallelogram.

Exam Tip: Be careful with decimal places in part (iii): \( 8.4 \times 2.1 \) has two decimal digits.

Question 8. Find the area of the triangle.
(i) b=6cm h=9cm
(ii) b=12cm h=5cm
Answer:
Formula for area of a triangle: \( \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} \)

(i) Given: \( b = 6\text{ cm} \), \( h = 9\text{ cm} \)
\[ \text{Area} = \frac{1}{2} \times 6 \times 9 = 3 \times 9 = 27\text{ cm}^2 \]

(ii) Given: \( b = 12\text{ cm} \), \( h = 5\text{ cm} \)
\[ \text{Area} = \frac{1}{2} \times 12 \times 5 = 6 \times 5 = 30\text{ cm}^2 \]
In simple words: Multiply base and height together, then divide by 2 to get the area of the triangle.

Exam Tip: Do not forget to multiply by \( \frac{1}{2} \) when calculating the area of a triangle.

Question 9. Find the missing values of the following.

Answer:
Calculations row by row:
- Row 1 (Triangle): \( \text{Height} = \frac{2 \times \text{Area}}{\text{Base}} = \frac{2 \times 97.44}{8.4} = \frac{194.88}{8.4} = 23.2\text{ cm} \)
- Row 2 (Triangle): \( \text{Base} = \frac{2 \times \text{Area}}{\text{Height}} = \frac{2 \times 216}{18} = \frac{432}{18} = 24\text{ cm} \)
- Row 3 (Triangle): \( \text{Height} = \frac{2 \times \text{Area}}{\text{Base}} = \frac{2 \times 32.76}{7.8} = \frac{65.52}{7.8} = 8.4\text{ cm} \)
- Row 4 (Parallelogram): \( \text{Height} = \frac{\text{Area}}{\text{Base}} = \frac{170.5}{22} = 7.75\text{ cm} \)
- Row 5 (Parallelogram): \( \text{Base} = \frac{\text{Area}}{\text{Height}} = \frac{2512}{31.4} = 80\text{ cm} \)
- Row 6 (Parallelogram): \( \text{Base} = \frac{\text{Area}}{\text{Height}} = \frac{240}{8} = 30\text{ cm} \)

Completed Table:

In simple words: For triangles, double the area before dividing by the known side. For parallelograms, divide area directly by the known side.

Exam Tip: Remember: triangle area uses \( \frac{1}{2} \times b \times h \), while parallelogram area is simply \( b \times h \).

Question 10. Find the circumference of a circle whose diameter is (a) 10cm (b) 7.7cm
Answer:
Formula for circumference: \( C = \pi d \)

(a) For diameter \( d = 10\text{ cm} \):
Using \( \pi = 3.14 \):
\[ C = 3.14 \times 10 = 31.4\text{ cm} \]
(Or using \( \pi = \frac{22}{7} \): \( C = \frac{220}{7}\text{ cm} \approx 31.43\text{ cm} \)).

(b) For diameter \( d = 7.7\text{ cm} \):
Using \( \pi = \frac{22}{7} \):
\[ C = \frac{22}{7} \times 7.7 = 22 \times 1.1 = 24.2\text{ cm} \]
In simple words: Multiply diameter by pi to get the distance around the circle.

Exam Tip: When the diameter is a multiple of 7 or 1.1, using \( \pi = \frac{22}{7} \) gives clean cancellations.

Question 11. The diameter of tyre is 77cm.Find the distance covered by it in 10 rounds.
Answer:
Given: Diameter (\( d \)) = \( 77\text{ cm} \).
Distance covered in 1 round = Circumference of tyre (\( C \)):
\[ C = \pi d = \frac{22}{7} \times 77 = 22 \times 11 = 242\text{ cm} \]

Distance covered in 10 rounds:
\[ = 10 \times 242 = 2420\text{ cm} \text{ (or } 24.2\text{ m)} \]
The distance covered is \( 2420\text{ cm} \).
In simple words: One full turn covers 242 cm. In 10 turns, the tyre travels 2420 cm.

Exam Tip: Distance covered in one revolution always equals the circumference of the wheel.

Question 12. The diameter of a car wheel is 49cm.Find how many times will it take to cover a distance of 3080cm.
Answer:
Given: Diameter (\( d \)) = \( 49\text{ cm} \), Total distance = \( 3080\text{ cm} \).
Circumference of wheel:
\[ C = \pi d = \frac{22}{7} \times 49 = 22 \times 7 = 154\text{ cm} \]

Number of rotations required:
\[ \text{Number of rotations} = \frac{\text{Total distance}}{\text{Circumference}} = \frac{3080}{154} = 20 \]
The wheel will rotate 20 times.
In simple words: One turn moves the wheel 154 cm. Divide 3080 by 154 to get 20 complete turns.

Exam Tip: Make sure total distance and circumference are in the same unit (cm) before dividing.

Question 13. A 7m wide road runs around a circular park whose circumference is 176m.Find the fencing the outer boundary of the road at the rate of Rs.21 per metre.
Answer:
Let the radius of the circular park be \( r \).
Inner circumference = \( 176\text{ m} \):
\[ 2\pi r = 176 \]
\[ 2 \times \frac{22}{7} \times r = 176 \]
\[ \frac{44}{7} \times r = 176 \]

\( \implies r = \frac{176 \times 7}{44} = 4 \times 7 = 28\text{ m} \)

Width of road = \( 7\text{ m} \).
Outer radius (\( R \)) = \( r + 7 = 28 + 7 = 35\text{ m} \).

Outer circumference = \( 2\pi R \):
\[ = 2 \times \frac{22}{7} \times 35 = 2 \times 22 \times 5 = 220\text{ m} \]

Cost of fencing outer boundary at Rs. 21 per metre:
\[ \text{Cost} = 220 \times 21 = \text{Rs. } 4620 \]
The cost of fencing the outer boundary is Rs. 4620.
In simple words: Find inner radius 28 m from circumference 176 m. Add 7 m road to get outer radius 35 m. Find outer circumference 220 m and multiply by 21 rupees.

Exam Tip: Add the path width to the inner radius to obtain the outer radius before finding the outer circumference.

Question 14. A wire is in the shape of a square with each side as 16.5cm is bent in the shape of a circle. Find the circumference and diameter of the circle.
Answer:
Side of square = \( 16.5\text{ cm} \).
Length of wire = Perimeter of square:
\[ \text{Perimeter} = 4 \times 16.5 = 66\text{ cm} \]

Since the same wire is bent into a circle, Circumference of circle = Length of wire = \( 66\text{ cm} \).
Now find diameter (\( d \)):
\[ \pi d = 66 \]
\[ \frac{22}{7} \times d = 66 \]

\( \implies d = \frac{66 \times 7}{22} = 3 \times 7 = 21\text{ cm} \)
The circumference is \( 66\text{ cm} \) and the diameter is \( 21\text{ cm} \).
In simple words: The wire is 66 cm long, so the circle has circumference 66 cm. Dividing by pi gives a diameter of 21 cm.

Exam Tip: When a shape is bent into another shape, the perimeter remains unchanged.

Question 15. If the area of a circle is 616sq.cm.Find the circumference.
Answer:
Given: Area of circle = \( 616\text{ cm}^2 \).
\[ \pi r^2 = 616 \]
\[ \frac{22}{7} \times r^2 = 616 \]
\[ r^2 = \frac{616 \times 7}{22} = 28 \times 7 = 196 \]

\( \implies r = \sqrt{196} = 14\text{ cm} \)

Now, find the circumference:
\[ C = 2\pi r = 2 \times \frac{22}{7} \times 14 = 2 \times 22 \times 2 = 88\text{ cm} \]
The circumference of the circle is \( 88\text{ cm} \).
In simple words: Use the area to find that the radius is 14 cm. Then calculate two times pi times 14 to get 88 cm.

Exam Tip: Remember \( 14^2 = 196 \); finding the square root gives the radius directly.

Congruence of Triangles

Question 1. In the given figure PR=QS; PQ=SR. State third matching pair and the state the CC.
Answer:
In \( \triangle PQR \) and \( \triangle SRQ \):
1. \( PR = QS \) (Given)
2. \( PQ = SR \) (Given)
3. \( QR = RQ \) (Common side)

Third matching pair: Side QR = Side RQ (Common side).
Congruence Criterion (CC): SSS (Side-Side-Side) Congruence Criterion, so \( \triangle PQR \cong \triangle SRQ \). In simple words: The shared base QR belongs to both triangles. Having all three matching sides equal proves congruence by SSS.

Exam Tip: Always state "common side" as the geometric reason for shared sides in congruence proofs.

Question 2. In the given figure AD is the Median. AB = AC. Is triangle ABD=triangle ACD. Is ∠B=∠C. Why?
Answer:
Given: \( AB = AC \) and \( AD \) is the median to base \( BC \).
Since \( AD \) is the median, \( D \) is the midpoint of \( BC \), which gives \( BD = CD \).

In \( \triangle ABD \) and \( \triangle ACD \):
1. \( AB = AC \) (Given)
2. \( BD = CD \) (\( AD \) is the median)
3. \( AD = AD \) (Common side)

Therefore, by SSS Congruence Criterion:
\[ \triangle ABD \cong \triangle ACD \]
Yes, triangle ABD is congruent to triangle ACD.

Also, yes, \( \angle B = \angle C \) because corresponding parts of congruent triangles are equal (CPCT). In simple words: The median splits the base in half, making all three sides equal in both triangles. Matching angles are equal by CPCT.

Exam Tip: State "CPCT" (Corresponding Parts of Congruent Triangles) as the reason for equal corresponding angles.

Question 3. State three pairs of equal parts and the CC.
Answer:
In \( \triangle PQR \) and \( \triangle SRQ \):
1. \( \angle PQR = \angle SRQ = 75^\circ \) (since \( 40^\circ + 35^\circ = 75^\circ \))
2. \( QR = RQ \) (Common side)
3. \( \angle PRQ = \angle SQR = 35^\circ \) (Given)

Three pairs of equal parts:
- \( \angle PQR = \angle SRQ \)
- Side \( QR = \text{Side } RQ \)
- \( \angle PRQ = \angle SQR \)

Congruence Criterion (CC): ASA (Angle-Side-Angle) Congruence Criterion, which proves \( \triangle PQR \cong \triangle SRQ \). In simple words: The two triangles share side QR between equal angles of 75 degrees and 35 degrees, making them congruent by ASA.

Exam Tip: For ASA, the equal side must be the included side located directly between the two known equal angles.