CBSE Class 7 Mathematics Practice Worksheet Set 12

Question 1. Write in exponential form:
a) 8 x p x p x p x p x y x y x k
b) 3 x a x a x a x 3 x b x b
Answer:
a) \( 8 \times p \times p \times p \times p \times y \times y \times k = 2^3 p^4 y^2 k \) (or \( 8p^4 y^2 k \))
b) \( 3 \times a \times a \times a \times 3 \times b \times b = (3 \times 3) \times (a \times a \times a) \times (b \times b) = 3^2 a^3 b^2 \) (or \( 9a^3 b^2 \))
In simple words: Count how many times each letter or number multiplies itself. Write that count as a small power at the top.

Exam Tip: Group identical numbers and variables together first before writing their powers.

Question 2. Write the coefficient of each term of the expression :
a) 7m -- 10m + 2/3m -- m + 5
b) -3k + 15k - 6k - 4
Answer:
a) For the expression \( 7m - 10m + \frac{2}{3}m - m + 5 \):
- Coefficient of term \( 7m \) is \( 7 \)
- Coefficient of term \( -10m \) is \( -10 \)
- Coefficient of term \( \frac{2}{3}m \) is \( \frac{2}{3} \)
- Coefficient of term \( -m \) is \( -1 \)
- The constant term is \( 5 \)

b) For the expression \( -3k + 15k - 6k - 4 \):
- Coefficient of term \( -3k \) is \( -3 \)
- Coefficient of term \( 15k \) is \( 15 \)
- Coefficient of term \( -6k \) is \( -6 \)
- The constant term is \( -4 \)
In simple words: The coefficient is the number sitting right in front of a variable. If there is a minus sign, keep it with the number.

Exam Tip: Never forget the negative sign when writing coefficients of negative terms.

Question 3. Identify monomials, binomials , trinomials from the following:
a) 8b + 3a -2
b) -5a + 3b + ab - 1
c) 2/3 xyz
d) 1/4 m - 1/3 n
Answer:
a) \( 8b + 3a - 2 \) contains 3 terms, so it is a trinomial.
b) \( -5a + 3b + ab - 1 \) contains 4 terms, so it is a quadrinomial (or simply a polynomial).
c) \( \frac{2}{3}xyz \) contains only 1 term, so it is a monomial.
d) \( \frac{1}{4}m - \frac{1}{3}n \) contains 2 terms, so it is a binomial.
In simple words: One term is a monomial, two terms make a binomial, and three terms make a trinomial.

Exam Tip: Terms are separated by plus or minus signs. Count the parts to classify the expression easily.

Question 4. Preethi took a loan of Rs 6000 from a money lender ,who charged interest at the rate of 15% per annum . After 2 years Preethi paid him Rs 7400 and a calculator to clear the debt . What is the price of the calculator ?
Answer:
Given:
Principal (\( P \)) = Rs. 6000
Rate (\( R \)) = 15% per annum
Time (\( T \)) = 2 years

Simple Interest (\( \text{SI} \)):
\[ \text{SI} = \frac{P \times R \times T}{100} = \frac{6000 \times 15 \times 2}{100} = \text{Rs. } 1800 \]

Total amount due to clear the loan:
\[ A = P + \text{SI} = 6000 + 1800 = \text{Rs. } 7800 \]

Preethi paid Rs. 7400 in cash plus a calculator.
Price of the calculator = Total debt - Cash paid
\[ = 7800 - 7400 = \text{Rs. } 400 \]
In simple words: Find the full money owed using the interest formula. Then take away the cash paid to find the calculator cost.

Exam Tip: Always add simple interest to the principal first to get the total amount due before subtracting payments.

Question 5. From the sum of a + 3b - 9ab , 2a - b + 8ab , b + 8 and a - 3ab, subtract the sum of -3a + 4b - 8ab and a – b + 3 + 7ab.
Answer:
Step 1: Find the first sum:
\[ (a + 3b - 9ab) + (2a - b + 8ab) + (b + 8) + (a - 3ab) \]
Group like terms:
\[ = (a + 2a + a) + (3b - b + b) + (-9ab + 8ab - 3ab) + 8 \]
\[ = 4a + 3b - 4ab + 8 \]

Step 2: Find the second sum:
\[ (-3a + 4b - 8ab) + (a - b + 3 + 7ab) \]
Group like terms:
\[ = (-3a + a) + (4b - b) + (-8ab + 7ab) + 3 \]
\[ = -2a + 3b - ab + 3 \]

Step 3: Subtract the second sum from the first sum:
\[ (4a + 3b - 4ab + 8) - (-2a + 3b - ab + 3) \]
\[ = 4a + 3b - 4ab + 8 + 2a - 3b + ab - 3 \]
Group like terms:
\[ = (4a + 2a) + (3b - 3b) + (-4ab + ab) + (8 - 3) \]
\[ = 6a - 3ab + 5 \]
In simple words: Add the first four expressions together. Then add the next two expressions together. Finally, subtract the second result from the first.

Exam Tip: Change every sign inside the bracket when subtracting an algebraic expression.

Question 6. The angles of a triangle are in the ratio 5:6:7.Calculate the measures of all the angles of the triangle.
Answer:
Let the angles of the triangle be \( 5x \), \( 6x \), and \( 7x \).
The sum of all three interior angles in a triangle is \( 180^\circ \).
\[ 5x + 6x + 7x = 180^\circ \]

\( \implies 18x = 180^\circ \)

\( \implies x = \frac{180^\circ}{18} = 10^\circ \)

Now, find each angle:
First angle = \( 5 \times 10^\circ = 50^\circ \)
Second angle = \( 6 \times 10^\circ = 60^\circ \)
Third angle = \( 7 \times 10^\circ = 70^\circ \)
In simple words: Add the ratio parts to get 18. Divide 180 by 18 to find one part, then multiply by each ratio number.

Exam Tip: Check your final answer by adding all three angles to see if they sum up to \( 180^\circ \).

Question 7. The angles of a triangle are in the ratio 2:3:5. Calculate the smallest and the largest angles of the triangle.
Answer:
Let the angles be \( 2x \), \( 3x \), and \( 5x \).
Sum of angles of a triangle = \( 180^\circ \).
\[ 2x + 3x + 5x = 180^\circ \]

\( \implies 10x = 180^\circ \)

\( \implies x = \frac{180^\circ}{10} = 18^\circ \)

Smallest angle = \( 2x = 2 \times 18^\circ = 36^\circ \)
Largest angle = \( 5x = 5 \times 18^\circ = 90^\circ \)
In simple words: Divide 180 by the sum of parts, 10. Multiply 18 by 2 for the smallest angle and by 5 for the largest.

Exam Tip: The question asks only for the smallest and largest angles, so state both clearly in your final step.

Question 8. The measures of two angles of a triangle are 65 ⁰ and 35 ⁰ .Find the third angle.
Answer:
Let the third angle be \( x \).
The sum of angles in a triangle is \( 180^\circ \).
\[ 65^\circ + 35^\circ + x = 180^\circ \]
\[ 100^\circ + x = 180^\circ \]

\( \implies x = 180^\circ - 100^\circ = 80^\circ \)
Hence, the third angle measures \( 80^\circ \).
In simple words: Add the two given angles together. Subtract that number from 180 to find the missing angle.

Exam Tip: Mention the angle sum property of a triangle to ensure full credit.

Question 9. Find the simple intrest , if Principal = Rs 5000 , Rate = 15% per annum , Time =5 yrs
Answer:
Given:
Principal (\( P \)) = Rs. 5000
Rate of interest (\( R \)) = 15% p.a.
Time (\( T \)) = 5 years

Formula for simple interest:
\[ \text{SI} = \frac{P \times R \times T}{100} \]
\[ \text{SI} = \frac{5000 \times 15 \times 5}{100} = 50 \times 75 = \text{Rs. } 3750 \]
In simple words: Multiply the principal, rate, and time together. Then divide by 100 to get the interest.

Exam Tip: Write the units (Rs.) clearly next to your final answer.

Question 10. Find the gain or loss per cent if cost price = Rs 800, selling price = Rs 720
Answer:
Given:
Cost Price (\( \text{CP} \)) = Rs. 800
Selling Price (\( \text{SP} \)) = Rs. 720

Since \( \text{SP} < \text{CP} \), there is a loss.
\[ \text{Loss} = \text{CP} - \text{SP} = 800 - 720 = \text{Rs. } 80 \]

Loss percentage:
\[ \text{Loss } \% = \left( \frac{\text{Loss}}{\text{CP}} \times 100 \right) \% \]
\[ = \left( \frac{80}{800} \times 100 \right) \% = 10\% \]
In simple words: Subtract the selling price from the cost price to get the loss. Divide that loss by the cost price and multiply by 100.

Exam Tip: Always calculate profit or loss percentage on the Cost Price, never on the Selling Price.

Question 11. One of the acute angles of a right –angled triangle is 40⁰ .Find the other one.
Answer:
In a right-angled triangle, one angle is \( 90^\circ \).
The sum of the two acute angles is \( 90^\circ \).
Let the unknown acute angle be \( x \).
\[ 40^\circ + x = 90^\circ \]

\( \implies x = 90^\circ - 40^\circ = 50^\circ \)
The measure of the other acute angle is \( 50^\circ \).
In simple words: The two sharp angles in a right triangle always add up to 90 degrees. Subtract 40 from 90 to find the answer.

Exam Tip: Using the fact that acute angles in a right triangle are complementary saves time during exams.

Question 12. The measures of two angles of a triangle are 68 ⁰ and 72⁰ .Find the third angle.
Answer:
Let the third angle be \( \angle 3 \).
Sum of angles in a triangle = \( 180^\circ \).
\[ 68^\circ + 72^\circ + \angle 3 = 180^\circ \]
\[ 140^\circ + \angle 3 = 180^\circ \]

\( \implies \angle 3 = 180^\circ - 140^\circ = 40^\circ \)
The measure of the third angle is \( 40^\circ \).
In simple words: Combine 68 and 72 to get 140. Take 140 away from 180 to find the last angle.

Exam Tip: Show the addition step clearly before subtracting from 180 degrees.

Question 13. One of the acute angles of a right triangle is 48⁰ .Find other acute angle.
Answer:
In a right-angled triangle, the sum of both acute angles equals \( 90^\circ \).
Let the other acute angle be \( y \).
\[ 48^\circ + y = 90^\circ \]

\( \implies y = 90^\circ - 48^\circ = 42^\circ \)
Therefore, the other acute angle is \( 42^\circ \).
In simple words: Subtract 48 from 90 to get 42 degrees.

Exam Tip: Remember that the right angle is already \( 90^\circ \), so the remaining two angles must share the other \( 90^\circ \).

Question 14. One of the angles of a triangle is 110 and the other two angles are equal. What is the measure of each of these equal angles ?
Answer:
Let each of the two equal angles be \( x \).
Sum of all three angles = \( 180^\circ \).
\[ 110^\circ + x + x = 180^\circ \]
\[ 110^\circ + 2x = 180^\circ \]

\( \implies 2x = 180^\circ - 110^\circ = 70^\circ \)

\( \implies x = \frac{70^\circ}{2} = 35^\circ \)
Each of the equal angles measures \( 35^\circ \).
In simple words: Take 110 away from 180 to get 70. Share that 70 equally between the two angles to get 35 each.

Exam Tip: Set up an equation with \( 2x \) to show examiner each step of your calculation.

Question 15. In the figure A B II QR , find the value of < A PQ and <B PR , given that <PRQ =55⁰ and <PQR = 40⁰.
Answer:
Given that line \( AB \parallel QR \) and transversal lines are \( PQ \) and \( PR \).
- For line \( AB \parallel QR \) with transversal \( PQ \):
\( \angle APQ \) and \( \angle PQR \) are alternate interior angles.
\[ \angle APQ = \angle PQR = 40^\circ \]
- For line \( AB \parallel QR \) with transversal \( PR \):
\( \angle BPR \) and \( \angle PRQ \) are alternate interior angles.
\[ \angle BPR = \angle PRQ = 55^\circ \] In simple words: When two lines are parallel, the z-shaped alternate angles inside are equal.

Exam Tip: State "alternate interior angles" as the geometric reason to secure full marks.

Question 16. Find a number whose 8% is 6.
Answer:
Let the unknown number be \( x \).
According to the problem:
\[ 8\% \text{ of } x = 6 \]
\[ \frac{8}{100} \times x = 6 \]

\( \implies x = \frac{6 \times 100}{8} = \frac{600}{8} = 75 \)
The required number is \( 75 \).
In simple words: Multiply 6 by 100 and then divide by 8 to find the original number.

Exam Tip: Set up \( \frac{8}{100} \times x = 6 \) first rather than guessing the number.

Question 17. Add: 5x + 7y – 6z, 4y + 3x , 9x + 2z - 9y and 2y -2x.
Answer:
Write down the sum of all expressions:
\[ (5x + 7y - 6z) + (3x + 4y) + (9x - 9y + 2z) + (-2x + 2y) \]
Rearrange and combine like terms:
- Terms with \( x \): \( 5x + 3x + 9x - 2x = 15x \)
- Terms with \( y \): \( 7y + 4y - 9y + 2y = 4y \)
- Terms with \( z \): \( -6z + 2z = -4z \)

Sum = \( 15x + 4y - 4z \)
In simple words: Put matching letters together and add their numbers up.

Exam Tip: Collect like terms in alphabetical order so you do not miss any term.

Question 18. From the figure write down: a) each linear pair b) each pair of vertically opposite angles
Answer:
Based on the lines \( AB \) and \( CD \) intersecting at point \( O \):
a) Linear pairs (adjacent angles on a straight line that sum to \( 180^\circ \)):
1. \( \angle AOC \) and \( \angle COB \)
2. \( \angle COB \) and \( \angle BOD \)
3. \( \angle BOD \) and \( \angle AOD \)
4. \( \angle AOD \) and \( \angle AOC \)

b) Pairs of vertically opposite angles:
1. \( \angle AOC \) and \( \angle BOD \)
2. \( \angle AOD \) and \( \angle BOC \) In simple words: Angles next to each other on a line form a linear pair. Angles sitting across from each other form vertically opposite angles.

Exam Tip: There are four distinct linear pairs at a single intersection of two straight lines.

Question 19. Find the perimeter and area of a square whose side is 0.15 m.
Answer:
Given side of the square (\( s \)) = \( 0.15\text{ m} \)

1. Perimeter of square:
\[ \text{Perimeter} = 4 \times s = 4 \times 0.15 = 0.6\text{ m} \]

2. Area of square:
\[ \text{Area} = s \times s = 0.15 \times 0.15 = 0.0225\text{ m}^2 \]
In simple words: Multiply the side by 4 to get distance around the edge. Multiply the side by itself to find the flat space inside.

Exam Tip: Be careful with the decimal point in multiplication; \( 0.15 \times 0.15 \) has four decimal places.

Question 20. Each side of a square is doubled. What is the change in its area?
Answer:
Let the original side of the square be \( s \).
Original area (\( A_1 \)) = \( s^2 \)

When each side is doubled:
New side = \( 2s \)
New area (\( A_2 \)) = \( (2s)^2 = 4s^2 = 4 \times A_1 \)

Therefore, the area becomes \( 4 \) times its original value (an increase of 300%).
In simple words: When you double the side length, the space inside grows by two times two, which is four times larger.

Exam Tip: Remember that squaring \( 2s \) gives \( 4s^2 \), not \( 2s^2 \).

Question 21. The area of a rectangle is 144 m² and its length is 16 m. find its breadth.
Answer:
Given:
Area of rectangle = \( 144\text{ m}^2 \)
Length (\( l \)) = \( 16\text{ m} \)

Formula:
\[ \text{Area} = \text{length} \times \text{breadth} \]
\[ 144 = 16 \times b \]

\( \implies b = \frac{144}{16} = 9\text{ m} \)
The breadth of the rectangle is \( 9\text{ m} \).
In simple words: Divide the total area by length to get the width.

Exam Tip: Always state the correct unit of measurement (m) with your final answer.

Question 22. Express the following per cents as fractions in simplest forms:
a) 26%
b) 55%
c) 120%
Answer:
a) \( 26\% = \frac{26}{100} = \frac{13}{50} \)
b) \( 55\% = \frac{55}{100} = \frac{11}{20} \)
c) \( 120\% = \frac{120}{100} = \frac{6}{5} = 1\frac{1}{5} \)
In simple words: Put the number over 100 and cancel common factors until you cannot simplify any more.

Exam Tip: Reduce fractions completely to lowest terms to avoid losing half a mark.

Question 23. Express each of the following as per cents:
a) 1.21
b) 0.051
c) 14/25
Answer:
a) \( 1.21 \times 100\% = 121\% \)
b) \( 0.051 \times 100\% = 5.1\% \)
c) \( \frac{14}{25} \times 100\% = 14 \times 4\% = 56\% \)
In simple words: Multiply any decimal or fraction by 100 and add the percent symbol.

Exam Tip: Moving the decimal point two places to the right converts a decimal directly to a percentage.

Question 24. What is the number whose 8% is 480?
Answer:
Let the number be \( N \).
\[ 8\% \text{ of } N = 480 \]
\[ \frac{8}{100} \times N = 480 \]

\( \implies N = \frac{480 \times 100}{8} = 60 \times 100 = 6000 \)
The required number is \( 6000 \).
In simple words: Divide 480 by 8 to find one percent, then multiply by 100 to get the full number.

Exam Tip: Simplify \( \frac{480}{8} = 60 \) before multiplying by 100 to keep calculations simple.

Question 25. In a fabric, cotton and synthetic fibres are in the ratio of 3:5. What is the percentage of cotton fibre in the fabric?
Answer:
Given ratio of cotton to synthetic fibres = \( 3 : 5 \)
Total number of parts = \( 3 + 5 = 8 \)
Cotton share = \( \frac{3}{8} \)

Percentage of cotton fibre:
\[ = \left( \frac{3}{8} \times 100 \right)\% = \frac{300}{8}\% = 37.5\% \]
In simple words: Add the ratio parts together to get 8. Cotton is 3 parts out of 8, which is 37.5 percent.

Exam Tip: Always divide the part by the total sum of parts before multiplying by 100.

Question 26. In an examination, Kavita obtained 72% marks. If the maximum marks are 650, find the marks obtained by Kavita.
Answer:
Maximum marks = 650
Percentage scored = 72%

Marks obtained by Kavita:
\[ = 72\% \text{ of } 650 \]
\[ = \frac{72}{100} \times 650 = \frac{72 \times 65}{10} = \frac{4680}{10} = 468 \]
Kavita obtained 468 marks.
In simple words: Multiply 650 by 72 and divide by 100 to find the scored marks.

Exam Tip: Cancel zeroes first to simplify the multiplication step.

Question 27. 15% of the employees in a factory are females and the number of male employees is 255. Find the total number of employees. Also, find the number of female employees.
Answer:
Percentage of female workers = 15%
Percentage of male workers = \( 100\% - 15\% = 85\% \)

Let the total number of employees be \( T \).
\[ 85\% \text{ of } T = 255 \]
\[ \frac{85}{100} \times T = 255 \]

\( \implies T = \frac{255 \times 100}{85} = 3 \times 100 = 300 \)

Total number of employees = 300
Number of female employees = Total employees - Male employees
\[ = 300 - 255 = 45 \]
In simple words: Since 15 percent are women, 85 percent must be men. Use the 255 men to find the whole staff of 300, leaving 45 women.

Exam Tip: Finding the male percentage first makes solving for the total straightforward.

Question 28. Sarika obtained 432 marks out of 600 and Rajeev obtained 525 marks out of 750 in annual examination, Whose performance is better?
Answer:
Calculate Sarika's percentage:
\[ = \left( \frac{432}{600} \times 100 \right)\% = \frac{432}{6}\% = 72\% \]

Calculate Rajeev's percentage:
\[ = \left( \frac{525}{750} \times 100 \right)\% = \frac{525}{7.5}\% = 70\% \]

Comparing the two scores: \( 72\% > 70\% \).
Therefore, Sarika's performance is better.
In simple words: Turn both marks into percentages. Sarika got 72 percent and Rajeev got 70 percent, so Sarika did better.

Exam Tip: Never compare raw marks directly when the maximum totals are different; always convert to percentages.

Question 29. After spending 80% of his income and giving 10% of the remainder to a charity, a man has Rs 446 left on him. Find his income.
Answer:
Let the man's total income be \( x \).
Amount spent = \( 80\% \text{ of } x \)
Remaining income = \( 100\% - 80\% = 20\% \text{ of } x = 0.20x \)

Amount given to charity = \( 10\% \text{ of remainder} = 10\% \text{ of } 0.20x = 0.02x \)
Amount left = \( 0.20x - 0.02x = 0.18x \) (or 18% of total income)

Given amount left = Rs. 446
\[ 0.18x = 446 \]

\( \implies x = \frac{446}{0.18} = \frac{44600}{18} = \frac{22300}{9} \approx \text{Rs. } 2477.78 \)
The man's income is Rs. 2477.78 (or Rs. \( 2477\frac{7}{9} \)).
In simple words: After spending 80 percent, 20 percent is left. Taking away 10 percent of that leaves 18 percent of his total income.

Exam Tip: Note that charity is taken from the remainder, not from the original full income.

Question 30. Rosita bought a cycle for Rs 1850. He sold it to Mohan at a profit of 8%. Find the selling price of the cycle.
Answer:
Given:
Cost Price (\( \text{CP} \)) = Rs. 1850
Profit percentage = 8%

Profit amount:
\[ = 8\% \text{ of } 1850 = \frac{8}{100} \times 1850 = \frac{14800}{100} = \text{Rs. } 148 \]

Selling Price (\( \text{SP} \)):
\[ \text{SP} = \text{CP} + \text{Profit} = 1850 + 148 = \text{Rs. } 1998 \]
In simple words: Work out 8 percent of 1850 to get 148 rupees profit. Add that to the cost price to get 1998 rupees.

Exam Tip: You can also find the selling price directly by calculating \( 108\% \) of Rs. 1850.

Question 31. A vendor purchased 40 dozen bananas for Rs 250. Out of these, 30 bananas were rotten and could not be sold. At what rate per dozen should he sell the remaining bananas to make a profit of 20%?
Answer:
Total bananas purchased = \( 40 \times 12 = 480 \text{ bananas} \)
Cost price (\( \text{CP} \)) = Rs. 250

Desired profit = \( 20\% \text{ of } 250 = \frac{20}{100} \times 250 = \text{Rs. } 50 \)
Required total selling price = \( 250 + 50 = \text{Rs. } 300 \)

Rotten bananas = 30
Good bananas remaining = \( 480 - 30 = 450 \text{ bananas} \)
Number of dozens remaining = \( \frac{450}{12} = 37.5 \text{ dozens} \)

Selling rate per dozen:
\[ = \frac{\text{Required Total SP}}{\text{Number of dozens}} = \frac{300}{37.5} = \text{Rs. } 8 \text{ per dozen} \]
In simple words: Find the total money needed, which is 300 rupees. Throw away bad bananas, count the rest in dozens, and divide 300 by that count.

Exam Tip: Convert the good bananas back into dozens before calculating the price per dozen.

Question 32. Equal sums of Rs 30000 were lent to Rajeev and Deepak at 10.5% p.a. for a period of 3 years and 5.5 years resp. .Find the diff. of interests paid by them.
Answer:
Given:
Principal (\( P \)) = Rs. 30000
Rate (\( R \)) = 10.5% p.a.
Time for Rajeev (\( T_1 \)) = 3 years
Time for Deepak (\( T_2 \)) = 5.5 years

Difference in time period = \( 5.5 - 3 = 2.5 \text{ years} \)

Difference in simple interest paid:
\[ = \frac{P \times R \times (T_2 - T_1)}{100} \]
\[ = \frac{30000 \times 10.5 \times 2.5}{100} = 300 \times 26.25 = \text{Rs. } 7875 \]
In simple words: Find the time difference of 2.5 years first. Then calculate the interest for that extra time.

Exam Tip: Subtracting the times first saves you from calculating two separate interest equations.

Question 33. Write the following statements algebraically:
i) The sum of numbers x, y and z.
ii) Product of 4 and 'x' divided by the difference of 5 and 'x'.
iii) Half of 'x' subtracted from 5 less than 'x'.
Answer:
i) \( x + y + z \)
ii) \( \frac{4x}{5 - x} \)
iii) "5 less than x" is \( (x - 5) \) and "Half of x" is \( \frac{x}{2} \).
Subtracting half of x gives:
\[ (x - 5) - \frac{x}{2} = \frac{x}{2} - 5 \]
In simple words: Change the English sentences into math symbols step by step.

Exam Tip: Remember that "5 less than x" means \( x - 5 \), not \( 5 - x \).

Question 34. If x = 4, y=3, z = 2, find the value of x + y /xyz
Answer:
Given values: \( x = 4 \), \( y = 3 \), and \( z = 2 \).
Substitute into \( \frac{x + y}{xyz} \):
Numerator = \( x + y = 4 + 3 = 7 \)
Denominator = \( xyz = 4 \times 3 \times 2 = 24 \)
\[ \frac{x + y}{xyz} = \frac{7}{24} \]

(Note: If read strictly with order of operations as \( x + \frac{y}{xyz} \), the result is \( 4 + \frac{3}{24} = 4 + \frac{1}{8} = \frac{33}{8} \)).
In simple words: Swap each letter with its given number and work out the fraction.

Exam Tip: Evaluate numerator and denominator separately before forming the final fraction.

Question 35. If a = 3, b = 2 and c = 5, find the value of 2ab – bc / 3ac
Answer:
Given: \( a = 3 \), \( b = 2 \), and \( c = 5 \).
Substitute into \( \frac{2ab - bc}{3ac} \):
Numerator:
\[ 2ab - bc = 2(3)(2) - (2)(5) = 12 - 10 = 2 \]
Denominator:
\[ 3ac = 3(3)(5) = 45 \]

Result:
\[ \frac{2ab - bc}{3ac} = \frac{2}{45} \]
In simple words: Put the given numbers into the formula and subtract on top, then multiply on bottom.

Exam Tip: Be careful with multiplication signs when replacing algebraic variables with numbers.

Question 36. If x = 1 and y = 3, find the value of x³ + y³ / x² –xy + y²
Answer:
Given: \( x = 1 \) and \( y = 3 \).
Substitute the values into \( \frac{x^3 + y^3}{x^2 - xy + y^2} \):
Numerator:
\[ x^3 + y^3 = 1^3 + 3^3 = 1 + 27 = 28 \]
Denominator:
\[ x^2 - xy + y^2 = 1^2 - (1)(3) + 3^2 = 1 - 3 + 9 = 7 \]

Value:
\[ \frac{28}{7} = 4 \]

(Note: Using the identity \( x^3 + y^3 = (x + y)(x^2 - xy + y^2) \), the expression simplifies directly to \( x + y = 1 + 3 = 4 \)).
In simple words: Calculate the top part to get 28 and the bottom part to get 7. Divide 28 by 7 to get 4.

Exam Tip: Recognizing the algebraic identity \( \frac{x^3+y^3}{x^2-xy+y^2} = x+y \) gives you an instant double check.

Question 37. In the figure AB // CD, find the values of <EOA, <OXC, <CXF, given that <EOB =50⁰
Answer:
Given: \( AB \parallel CD \) with transversal line \( EF \) intersecting \( AB \) at \( O \) and \( CD \) at \( X \), and \( \angle EOB = 50^\circ \).

1. \( \angle EOA \):
Angles \( \angle EOA \) and \( \angle EOB \) form a linear pair on straight line \( AB \).
\[ \angle EOA = 180^\circ - \angle EOB = 180^\circ - 50^\circ = 130^\circ \]

2. \( \angle OXC \):
Since \( AB \parallel CD \) and \( EF \) is the transversal, corresponding angles are equal:
\[ \angle OXC = \angle EOA = 130^\circ \]

3. \( \angle CXF \):
Angles \( \angle OXC \) and \( \angle CXF \) form a linear pair on straight line \( EF \):
\[ \angle CXF = 180^\circ - \angle OXC = 180^\circ - 130^\circ = 50^\circ \] In simple words: Use straight line angle rules to get 130 degrees. Then match matching corners across the parallel lines.

Exam Tip: State the angle relationships such as "linear pair" and "corresponding angles" beside each answer step.

Question 38. In ∆ABC, <A = 25°, <B = 50°, find <C.
Answer:
In \( \triangle ABC \), the sum of all interior angles is \( 180^\circ \).
\[ \angle A + \angle B + \angle C = 180^\circ \]
\[ 25^\circ + 50^\circ + \angle C = 180^\circ \]
\[ 75^\circ + \angle C = 180^\circ \]

\( \implies \angle C = 180^\circ - 75^\circ = 105^\circ \)
In simple words: Add 25 and 50 to get 75. Subtract 75 from 180 to find angle C.

Exam Tip: Mention the angle sum property of triangles in your proof.

Question 39. The length of a rectangle is 2 1/5 times its breadth.If the breadth is 4.5 cm .Find the perimeter of the rectangle.
Answer:
Given breadth (\( b \)) = \( 4.5\text{ cm} \)
Length (\( l \)):
\[ l = 2\frac{1}{5} \times b = \frac{11}{5} \times 4.5 = 11 \times 0.9 = 9.9\text{ cm} \]

Perimeter of the rectangle:
\[ \text{Perimeter} = 2(l + b) \]
\[ = 2(9.9 + 4.5) = 2(14.4) = 28.8\text{ cm} \]
In simple words: Multiply 4.5 by 11 over 5 to get the length. Add length and breadth, then double the answer.

Exam Tip: Convert mixed fractions to improper fractions first to prevent arithmetic errors.

Question 40. In the adjoining figure ,AB =DC and AC = DB .Prove that ∆ ABC ≅ ∆DCB.
Answer:
In \( \triangle ABC \) and \( \triangle DCB \):
1. \( AB = DC \) (Given)
2. \( AC = DB \) (Given)
3. \( BC = CB \) (Common side)

Therefore, by SSS (Side-Side-Side) congruence criterion:
\[ \triangle ABC \cong \triangle DCB \] In simple words: All three sides of the first triangle match all three sides of the second triangle, so the triangles are identical.

Exam Tip: Always clearly state the three matching conditions along with the criterion name (SSS).

Question 41. In ∆ ABC, ∠ A = ∠B. Name the equal sides.
Answer:
In a triangle, sides opposite to equal angles are always equal in length.
- Side opposite to \( \angle A \) is \( BC \).
- Side opposite to \( \angle B \) is \( AC \).

Therefore, the equal sides are \( BC \) and \( AC \) (that is, \( BC = AC \)).
In simple words: The sides facing the equal angles have the exact same length.

Exam Tip: Draw a quick sketch to avoid confusing which side is opposite to which vertex.

Question 42. Three angles of a triangle measure ( 2x + 3)° , (5x -2)° and (3x – 1)° respectively . Find the measures of these angles
Answer:
The sum of angles in a triangle is \( 180^\circ \).
\[ (2x + 3)^\circ + (5x - 2)^\circ + (3x - 1)^\circ = 180^\circ \]
Combine like terms:
\[ (2x + 5x + 3x) + (3 - 2 - 1) = 180 \]
\[ 10x + 0 = 180 \]

\( \implies 10x = 180 \)

\( \implies x = \frac{180}{10} = 18 \)

Calculate each angle:
First angle = \( 2(18) + 3 = 36 + 3 = 39^\circ \)
Second angle = \( 5(18) - 2 = 90 - 2 = 88^\circ \)
Third angle = \( 3(18) - 1 = 54 - 1 = 53^\circ \)
In simple words: Add the expressions to get 10x equals 180. Divide to find x is 18, then put 18 into each angle expression.

Exam Tip: Verify by checking that \( 39^\circ + 88^\circ + 53^\circ = 180^\circ \).

Question 43. ABCD is a rectangle. Using Pythagoras theorem , Complete the following.
1. AB² = --------------------
2. AD² = --------------------
Answer:
In rectangle \( ABCD \), each vertex angle is \( 90^\circ \) and \( AC \) is the diagonal.
In right-angled triangle \( ABC \), using Pythagoras theorem:
\[ AC^2 = AB^2 + BC^2 \]
1. \( AB^2 = AC^2 - BC^2 \) (or \( AC^2 - AD^2 \), since \( BC = AD \))

In right-angled triangle \( ADC \), using Pythagoras theorem:
\[ AC^2 = AD^2 + CD^2 \]
2. \( AD^2 = AC^2 - CD^2 \) (or \( AC^2 - AB^2 \), since \( CD = AB \)) In simple words: In a right triangle, square of a short side equals square of the diagonal minus square of the other side.

Exam Tip: Remember opposite sides of a rectangle are equal: \( BC = AD \) and \( AB = CD \).

Question 44. ∆PQR is isosceles. S is the midpoint of the base QR. Prove that ∆PQS ≅ ∆PRS.
Answer:
Given: \( \triangle PQR \) is isosceles with base \( QR \), so \( PQ = PR \). \( S \) is the midpoint of \( QR \), which means \( QS = RS \).

In \( \triangle PQS \) and \( \triangle PRS \):
1. \( PQ = PR \) (Given, sides of isosceles triangle)
2. \( QS = RS \) (Given, \( S \) is midpoint of base \( QR \))
3. \( PS = PS \) (Common side)

Therefore, by SSS congruence criterion:
\[ \triangle PQS \cong \triangle PRS \] In simple words: The two side lengths are equal, the bottom halves are equal, and they share the middle line. So the two triangles are identical.

Exam Tip: Explicitly mention that \( PS \) is a common side shared by both triangles.

Question 45. ∆ABC is isosceles with AB = AC.If ∠B =54°, what is the measure of ∠A ?
Answer:
In \( \triangle ABC \), since \( AB = AC \), the angles opposite to these sides are equal:
\[ \angle C = \angle B = 54^\circ \]

Sum of interior angles of a triangle is \( 180^\circ \):
\[ \angle A + \angle B + \angle C = 180^\circ \]
\[ \angle A + 54^\circ + 54^\circ = 180^\circ \]
\[ \angle A + 108^\circ = 180^\circ \]

\( \implies \angle A = 180^\circ - 108^\circ = 72^\circ \)
The measure of \( \angle A \) is \( 72^\circ \).
In simple words: The two base angles are both 54 degrees, making 108 degrees together. Subtract 108 from 180 to get 72 degrees for angle A.

Exam Tip: State the property "angles opposite to equal sides are equal" as your first reason.