CBSE Class 7 Mathematics Practice Worksheet Set 13

Question. From the adjoining figure, find the values of x and y.
Answer: At the bottom-right corner, the interior angle \( x \) and the given outer angle \( 130^\circ \) sit on a straight line, making a linear pair:
\( x + 130^\circ = 180^\circ \)
\( \implies x = 180^\circ - 130^\circ = 50^\circ \)
Inside the triangle, the sum of all three angles is always \( 180^\circ \). Let the top inside angle be \( \angle A \):
\( \angle A + 55^\circ + x = 180^\circ \)
\( \angle A + 55^\circ + 50^\circ = 180^\circ \)
\( \implies \angle A + 105^\circ = 180^\circ \)
\( \implies \angle A = 180^\circ - 105^\circ = 75^\circ \)
Because the two straight lines cross at the top point, angle \( y \) and the top inside angle are vertically opposite:
\( y = \angle A = 75^\circ \).
Thus, \( x = 50^\circ \) and \( y = 75^\circ \).
In simple words: The straight line gives x by taking 130 away from 180. Then the three triangle corners add up to 180, and crossing lines make y equal to the top corner.

Exam Tip: State the geometric reason clearly in words (such as "linear pair" and "vertically opposite angles") beside each calculation step.

Question. One of the angles of a triangle is equal to the sum of the other two. Find the measure of that angle. What type of triangle is it?
Answer: Let the three interior angles of the triangle be \( \angle A \), \( \angle B \), and \( \angle C \).
According to the given condition:
\( \angle A = \angle B + \angle C \)
By the angle sum property of a triangle:
\( \angle A + \angle B + \angle C = 180^\circ \)
Substitute \( \angle A \) in place of \( (\angle B + \angle C) \):
\( \angle A + \angle A = 180^\circ \)
\( 2\angle A = 180^\circ \)
\( \implies \angle A = \frac{180^\circ}{2} = 90^\circ \)
Since one of the angles is \( 90^\circ \), the triangle is a right-angled triangle.
In simple words: When one corner equals the other two added together, it must take half of the full 180 degrees, which makes it a 90-degree right triangle.

Exam Tip: Both parts must be answered: state the numerical value \( 90^\circ \) as well as the name "right-angled triangle".

Question. Evaluate: 3 x 9 x 3⁰
Answer: Any non-zero number raised to the power 0 is equal to 1, so \( 3^0 = 1 \).
\( 3 \times 9 \times 3^0 = 3 \times 9 \times 1 = 27 \).
In simple words: Any number with a power of zero turns into 1, so you just multiply 3 by 9.

Exam Tip: Do not mistake \( 3^0 \) for 0; remember the basic exponent law \( a^0 = 1 \).

Question. Evaluate: 2⁰+ 2 +2²
Answer: Calculate the value of each term individually:
\( 2^0 = 1 \)
\( 2^1 = 2 \)
\( 2^2 = 4 \)
Add the calculated terms:
\( 1 + 2 + 4 = 7 \).
In simple words: Work out each power first to get 1, 2, and 4, then add them up to reach 7.

Exam Tip: Write down the numerical value of each power before doing the final addition.

Question. A television set is sold for Rs 10,200 at a loss of 15%. Find its cost price.
Answer: Given that:
Selling Price (\( \text{SP} \)) = Rs. 10,200
Loss percentage = \( 15\% \)
Using the cost price formula:
\( \text{CP} = \frac{\text{SP} \times 100}{100 - \text{Loss}\%} \)
\( \implies \text{CP} = \frac{10200 \times 100}{100 - 15} = \frac{10200 \times 100}{85} \)
Divide 10200 by 85:
\( \implies \text{CP} = 120 \times 100 = \text{Rs. } 12000 \).
Thus, the cost price of the television set is Rs. 12,000.
In simple words: Since selling at a 15 percent loss means getting 85 percent of the original price, divide 10,200 by 85 and multiply by 100.

Exam Tip: Never calculate 15% of the selling price; loss percent is always based on the unknown cost price.

Question. In how many years will Rs 2500 yield a simple interest of Rs 675 at 9% per annum?
Answer: Given data:
Principal (\( P \)) = Rs. 2500
Simple Interest (\( \text{SI} \)) = Rs. 675
Rate of interest (\( R \)) = \( 9\% \text{ per annum} \)
Using the simple interest formula:
\( \text{SI} = \frac{P \times R \times T}{100} \)
Rearranging to find Time (\( T \)):
\( T = \frac{\text{SI} \times 100}{P \times R} \)
\( \implies T = \frac{675 \times 100}{2500 \times 9} = \frac{675}{25 \times 9} = \frac{675}{225} = 3\text{ years} \).
Hence, it will take 3 years.
In simple words: Multiply interest by 100, then divide by the principal and the rate to find the number of years.

Exam Tip: Cancel out zeros between 100 and 2500 first to simplify the fraction quickly.

Question. 15 kg of rice costs Rs 112.50. Find the cost of 24 kg of rice.
Answer: Using the unitary method:
Cost of \( 15\text{ kg} \) of rice = Rs. 112.50
Cost of \( 1\text{ kg} \) of rice = \( \frac{112.50}{15} = \text{Rs. } 7.50 \)
Cost of \( 24\text{ kg} \) of rice = \( 24 \times 7.50 = \text{Rs. } 180 \).
Therefore, \( 24\text{ kg} \) of rice costs Rs. 180.
In simple words: Find the price of one single kilogram first, then multiply that price by 24.

Exam Tip: Clearly state the price for one unit before multiplying to earn full method marks.

Question. What per cent of 350 is 84?
Answer: Let the required percentage be \( x\% \).
\( \frac{x}{100} \times 350 = 84 \)
\( \implies x = \frac{84 \times 100}{350} \)
\( \implies x = \frac{84 \times 2}{7} = 12 \times 2 = 24\% \).
Hence, 84 is \( 24\% \) of 350.
In simple words: Put 84 over 350 and multiply by 100 to change it into a percent.

Exam Tip: Divide 84 and 7 directly after canceling common factors of 10 and 5 to save calculation time.

Question. A sum of money amounts to Rs 1550 in two years at 12% per annum. Find the sum.
Answer: Let the principal sum be \( P \).
Given:
Time (\( T \)) = 2 years
Rate (\( R \)) = \( 12\% \text{ per annum} \)
Amount (\( A \)) = Rs. 1550
The total amount is:
\( A = P + \text{SI} = P + \frac{P \times R \times T}{100} \)
\( 1550 = P\left(1 + \frac{12 \times 2}{100}\right) = P\left(1 + \frac{24}{100}\right) = P\left(\frac{124}{100}\right) \)
\( \implies P = \frac{1550 \times 100}{124} = \frac{1550 \times 25}{31} \)
Since \( 1550 \div 31 = 50 \):
\( \implies P = 50 \times 25 = \text{Rs. } 1250 \).
Therefore, the required sum of money is Rs. 1250.
In simple words: With 24 percent total interest, the final amount is 124 percent of the starting sum. Divide by 124 and multiply by 100.

Exam Tip: Express Amount as \( P\left(1 + \frac{RT}{100}\right) \) to keep the algebraic steps clean and simple.

Question. By selling a toy at Rs 51.75, a dealer gains 15%. Find the cost price of the toy.
Answer: Given data:
Selling Price (\( \text{SP} \)) = Rs. 51.75
Gain percentage = \( 15\% \)
Using the cost price formula:
\( \text{CP} = \frac{\text{SP} \times 100}{100 + \text{Gain}\%} \)
\( \implies \text{CP} = \frac{51.75 \times 100}{100 + 15} = \frac{5175}{115} \)
Dividing 5175 by 115 gives:
\( \text{CP} = \text{Rs. } 45 \).
Thus, the cost price of the toy is Rs. 45.
In simple words: A 15 percent gain means the toy sold for 115 percent of its original cost. Divide 5175 by 115.

Exam Tip: Be sure to write \( (100 + \text{gain}) \) in the denominator when profit is made.

Question. 35%of a number is 224. Find the number.
Answer: Let the unknown number be \( x \).
According to the question:
\( \frac{35}{100} \times x = 224 \)
\( \implies x = \frac{224 \times 100}{35} \)
Simplify by dividing 224 and 35 by 7:
\( \implies x = \frac{32 \times 100}{5} = 32 \times 20 = 640 \).
Thus, the number is 640.
In simple words: Divide 224 by 35 to find one percent, then multiply by 100 to get the full number.

Exam Tip: Notice that \( 224 = 32 \times 7 \) and \( 35 = 5 \times 7 \) to cancel the factor of 7 quickly.

Question. Find the value of m such that (3/5)³ X (3/5) ᵐ⁺¹ = (3/5)⁸
Answer: Using the product law of exponents, \( a^p \times a^q = a^{p+q} \):
\( \left(\frac{3}{5}\right)^{3 + (m + 1)} = \left(\frac{3}{5}\right)^8 \)
\( \left(\frac{3}{5}\right)^{m + 4} = \left(\frac{3}{5}\right)^8 \)
Since the bases are identical on both sides, equate the exponents:
\( m + 4 = 8 \)
\( \implies m = 8 - 4 = 4 \).
Therefore, the value of \( m \) is 4.
In simple words: Add the powers on the left side together, then set the sum equal to the power on the right side.

Exam Tip: Explicitly mention that bases are equal on both sides before setting the exponents equal.

Question. Simplify: (5¹ X 3¹) ÷ 6¹
Answer: Evaluate the numbers in powers of 1:
\( 5^1 = 5 \), \( 3^1 = 3 \), and \( 6^1 = 6 \).
\( (5 \times 3) \div 6 = 15 \div 6 = \frac{15}{6} \).
Dividing numerator and denominator by 3 gives the simplest form:
\( \frac{15}{6} = \frac{5}{2} \) (or \( 2.5 \)).
In simple words: Multiply 5 by 3 to get 15, then divide by 6 and simplify the fraction to 5 over 2.

Exam Tip: Always reduce fractional answers to their lowest terms.

Question.The standard form of -10 / -15 is ................................
Answer: Both numerator and denominator have negative signs, which cancel out:
\( \frac{-10}{-15} = \frac{10}{15} \)
Divide both numbers by their greatest common factor, 5:
\( \frac{10 \div 5}{15 \div 5} = \frac{2}{3} \).
Hence, the standard form is \( \frac{2}{3} \).
In simple words: Negative divided by negative becomes positive, and dividing top and bottom by 5 leaves 2 over 3.

Exam Tip: A rational number is in standard form only when its denominator is positive and its terms share no common factor other than 1.

Question. Simplify: [-3/2 X 4/5] + [9/5 X -10/3]
Answer: Simplify each bracketed term first:
First bracket:
\( \left[-\frac{3}{2} \times \frac{4}{5}\right] = -\frac{3 \times 2}{1 \times 5} = -\frac{6}{5} \)
Second bracket:
\( \left[\frac{9}{5} \times \left(-\frac{10}{3}\right)\right] = \frac{3 \times (-2)}{1 \times 1} = -6 = -\frac{30}{5} \)
Now add the two results:
\( -\frac{6}{5} + \left(-\frac{30}{5}\right) = \frac{-6 - 30}{5} = -\frac{36}{5} \) (or \( -7\frac{1}{5} \)).
In simple words: Solve each multiplication separately by canceling numbers first, then add the two negative fractions.

Exam Tip: Cancel common terms diagonally before multiplying to avoid dealing with large numbers.

Question. Simplify: [ 2 / 9 ]⁶ X [ 2 / 9 ]⁴
Answer: Using the product law of exponents, \( a^m \times a^n = a^{m+n} \):
\( \left[\frac{2}{9}\right]^6 \times \left[\frac{2}{9}\right]^4 = \left[\frac{2}{9}\right]^{6 + 4} = \left[\frac{2}{9}\right]^{10} \).
In simple words: When multiplying the same fraction, keep the fraction the same and add the powers together.

Exam Tip: Unless requested to expand, leaving the final expression in power notation is the expected standard answer.

Question. Write the product in standard form: 9/ -42 X -14 / 15.
Answer: Write the given fractions:
\( \frac{9}{-42} \times \frac{-14}{15} \)
Since both fractions contain a negative sign, the product is positive:
\( = \frac{9 \times 14}{42 \times 15} \)
Cancel common factors:
\( \frac{14}{42} = \frac{1}{3} \)
\( = \frac{9 \times 1}{3 \times 15} = \frac{3}{15} = \frac{1}{5} \).
Therefore, the product in standard form is \( \frac{1}{5} \).
In simple words: Two negative signs cancel to positive. Reduce 14 over 42 to 1 over 3, and 9 over 15 to 3 over 5, leaving 1 over 5.

Exam Tip: Take care of signs first: a negative multiplied by a negative always gives a positive answer.

Question. A radio set costing Rs 750 is sold at a loss of 14%.What is the selling price?
Answer: Given data:
Cost Price (\( \text{CP} \)) = Rs. 750
Loss percentage = \( 14\% \)
Find the loss amount:
\( \text{Loss} = \frac{14}{100} \times 750 = \frac{14 \times 15}{2} = 7 \times 15 = \text{Rs. } 105 \)
Now, find the Selling Price:
\( \text{SP} = \text{CP} - \text{Loss} = 750 - 105 = \text{Rs. } 645 \).
Thus, the selling price of the radio set is Rs. 645.
In simple words: Work out 14 percent of 750, which is 105 rupees, and subtract it from the original price.

Exam Tip: You can also find the selling price directly by calculating \( 86\% \) of 750.

Question. Find the value of:
a) 25% of Rs 12,500
b) 12½% of Rs 8400
Answer:
a) \( 25\% \text{ of Rs. } 12500 = \frac{25}{100} \times 12500 = 25 \times 125 = \text{Rs. } 3125 \).

b) First convert the mixed percentage into an improper fraction:
\( 12\frac{1}{2}\% = \frac{25}{2}\% = \frac{25}{2 \times 100} = \frac{25}{200} = \frac{1}{8} \)
\( \frac{1}{8} \times 8400 = \text{Rs. } 1050 \).
In simple words: 25 percent means one quarter of the money, while 12 and a half percent means one eighth of the money.

Exam Tip: Remembering that \( 12\frac{1}{2}\% = \frac{1}{8} \) makes calculations much faster.

Question. Anis got 40 marks out of 50 in mathematics and 30 marks out of 40 in science. Which is a better score?
Answer: Compare both scores by converting them into percentages:
Percentage in Mathematics = \( \frac{40}{50} \times 100\% = 40 \times 2 = 80\% \)
Percentage in Science = \( \frac{30}{40} \times 100\% = \frac{3}{4} \times 100\% = 75\% \)
Comparing the percentages: \( 80\% > 75\% \).
Therefore, the score in mathematics is better.
In simple words: Turn both test marks into percentages. 80 percent in math is higher than 75 percent in science.

Exam Tip: Do not just compare the raw marks; always convert to percentages or common-denominator fractions when maximum marks differ.

Question. Simplify: (3/4)⁷ ÷ (3/4)⁵
Answer: Using the quotient law of exponents, \( a^m \div a^n = a^{m-n} \):
\( \left(\frac{3}{4}\right)^7 \div \left(\frac{3}{4}\right)^5 = \left(\frac{3}{4}\right)^{7 - 5} = \left(\frac{3}{4}\right)^2 \)
Evaluate the power:
\( \left(\frac{3}{4}\right)^2 = \frac{3^2}{4^2} = \frac{9}{16} \).
In simple words: Subtract the smaller power from the larger power, leaving 3 over 4 squared, which equals 9 over 16.

Exam Tip: State the law \( a^m \div a^n = a^{m-n} \) alongside your working.

Question. Simplify: 9 + (-8 +3/5)
Answer: Remove the grouping parentheses and combine the integer terms:
\( 9 + (-8) + \frac{3}{5} = (9 - 8) + \frac{3}{5} = 1 + \frac{3}{5} \)
Convert to a single fraction:
\( 1 + \frac{3}{5} = \frac{5 + 3}{5} = \frac{8}{5} \) (or \( 1\frac{3}{5} \)).
In simple words: Take 8 away from 9 to get 1, then add 3 over 5 to end up with 8 over 5.

Exam Tip: Grouping whole numbers together first simplifies calculations significantly.

Question. Express -5/8 as a rational number whose numerator is 60.
Answer: To turn the numerator \( -5 \) into \( 60 \), find the multiplying factor:
\( 60 \div (-5) = -12 \)
Multiply both the numerator and the denominator by \( -12 \):
\( \frac{-5 \times (-12)}{8 \times (-12)} = \frac{60}{-96} \).
Thus, the required rational number is \( \frac{60}{-96} \).
In simple words: Multiply top and bottom by minus 12 so that the top number becomes positive 60.

Exam Tip: Remember to multiply both numerator and denominator by the exact same value to maintain equality.

Question. Express in standard form
a) -14 / -56
b) 25 / -45
Answer:
a) The negative signs cancel out:
\( \frac{-14}{-56} = \frac{14}{56} \)
Divide numerator and denominator by their greatest common factor, 14:
\( \frac{14 \div 14}{56 \div 14} = \frac{1}{4} \).

b) Shift the negative sign to the numerator and divide both terms by their common factor, 5:
\( \frac{25}{-45} = \frac{-25}{45} = \frac{-25 \div 5}{45 \div 5} = -\frac{5}{9} \).
In simple words: Divide out common factors and ensure the bottom number is positive.

Exam Tip: Standard form strictly requires the denominator to be a positive integer.

Question. Ajay secured 28 marks out of 40 in a class test. Express this as a per cent.
Answer: Write the score as a fraction and multiply by \( 100\% \):
Percentage = \( \frac{28}{40} \times 100\% \)
Reduce the fraction:
\( \frac{28}{40} = \frac{7}{10} \)
\( \implies \frac{7}{10} \times 100\% = 7 \times 10 = 70\% \).
Thus, Ajay secured \( 70\% \).
In simple words: Put 28 over 40, simplify to 7 over 10, and multiply by 100 to get 70 percent.

Exam Tip: Simplify the fraction by canceling common factors before multiplying by 100.

Question. The value of (-1)³ X (-5)⁴ X 20 is
Answer: Evaluate each exponential part:
\( (-1)^3 = -1 \) (an odd power preserves the negative sign)
\( (-5)^4 = (-5) \times (-5) \times (-5) \times (-5) = 625 \) (an even power yields a positive value)
Multiply the terms together:
\( (-1) \times 625 \times 20 = -1 \times 12500 = -12500 \).
In simple words: Power 3 keeps minus 1 negative, power 4 makes minus 5 positive 625, and multiplying by 20 gives minus 12,500.

Exam Tip: Check whether the exponent is even or odd first to determine the sign of the evaluated base.

Question. The simple interest on Rs 7500 at 12% for 2½ years is .............................
Answer: Given data:
Principal (\( P \)) = Rs. 7500
Rate (\( R \)) = \( 12\% \text{ per annum} \)
Time (\( T \)) = \( 2\frac{1}{2}\text{ years} = \frac{5}{2}\text{ years} = 2.5\text{ years} \)
Using the simple interest formula:
\( \text{SI} = \frac{P \times R \times T}{100} \)
\( \implies \text{SI} = \frac{7500 \times 12 \times 5}{100 \times 2} = 75 \times 6 \times 5 = 75 \times 30 = \text{Rs. } 2250 \).
In simple words: Change 2 and a half years into 5 over 2, then multiply by principal and rate, and divide by 100.

Exam Tip: Convert mixed numbers to improper fractions before inserting them into formulas.

Question. The ratio of 85 paisa to Rs 6.80 in simplest form is ...........................
Answer: First, convert both quantities into the same unit:
Rs. \( 6.80 = 6.80 \times 100\text{ paisa} = 680\text{ paisa} \).
The ratio is:
\( 85 : 680 \)
Divide both terms by 85 (since \( 85 \times 8 = 680 \)):
\( \frac{85 \div 85}{680 \div 85} = \frac{1}{8} \).
Hence, the simplest form is \( 1 : 8 \).
In simple words: Turn the rupees into 680 paisa so units match, then divide both numbers by 85 to get 1 to 8.

Exam Tip: Ratios cannot be computed between different units; always convert both quantities to the smaller unit first.

Question. The area of a rectangle is 92.5 cm². If the breadth is 7.4 cm find the length.
Answer: The formula for the area of a rectangle is:
\( \text{Area} = \text{Length} \times \text{Breadth} \)
\( \implies \text{Length} = \frac{\text{Area}}{\text{Breadth}} \)
\( \implies \text{Length} = \frac{92.5}{7.4} = \frac{925}{74} = 12.5\text{ cm} \).
Therefore, the length of the rectangle is \( 12.5\text{ cm} \).
In simple words: Divide the area by the breadth to find the missing length.

Exam Tip: Shift both decimals one spot to the right to divide whole numbers easily (\( 925 \div 74 \)).

Question. The area of a square is 25 cm². Find the length of a side of the square.
Answer: Let the side of the square be \( s \).
\( \text{Area} = s^2 = 25\text{ cm}^2 \)
\( \implies s = \sqrt{25} = 5\text{ cm} \).
Thus, the side length of the square is \( 5\text{ cm} \).
In simple words: Take the square root of 25 to find the length of one side.

Exam Tip: Remember to include the single linear unit (cm) rather than area units (\( \text{cm}^2 \)).

Question. The area of a rectangle is 126 cm². If the breadth is 9 cm. Find the perimeter of the rectangle.
Answer: First, calculate the length of the rectangle:
\( \text{Length} = \frac{\text{Area}}{\text{Breadth}} = \frac{126}{9} = 14\text{ cm} \)
Now, use the perimeter formula:
\( \text{Perimeter} = 2 \times (\text{Length} + \text{Breadth}) \)
\( \implies \text{Perimeter} = 2 \times (14 + 9) = 2 \times 23 = 46\text{ cm} \).
Hence, the perimeter of the rectangle is \( 46\text{ cm} \).
In simple words: Divide 126 by 9 to get length 14, then add 14 and 9 and double the result.

Exam Tip: Find the length first, and remember that perimeter requires doubling the sum of length and breadth.

Question. The perimeter of a square is 42 cm. Find the area of the square.
Answer: A square has four equal sides:
\( \text{Perimeter} = 4 \times \text{side} = 42\text{ cm} \)
\( \implies \text{side} = \frac{42}{4} = 10.5\text{ cm} \)
Now, find the area:
\( \text{Area} = \text{side} \times \text{side} = 10.5 \times 10.5 = 110.25\text{ cm}^2 \).
In simple words: Divide 42 by 4 to get a side of 10.5, then multiply 10.5 by itself.

Exam Tip: Keep decimal calculations accurate: \( 10.5 \times 10.5 = 110.25 \).

Question. Find the height of a triangle whose area is 56 cm² and base is 16 cm.
Answer: The formula for the area of a triangle is:
\( \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} \)
Substitute the given values:
\( 56 = \frac{1}{2} \times 16 \times \text{height} \)
\( 56 = 8 \times \text{height} \)
\( \implies \text{height} = \frac{56}{8} = 7\text{ cm} \).
Therefore, the height of the triangle is \( 7\text{ cm} \).
In simple words: Halve the base to get 8, then divide 56 by 8 to find the height.

Exam Tip: Do not forget the factor of \( \frac{1}{2} \) in the triangle area formula.

Question. Find the area of a parallelogram with base 8.5 cm and height 4.2 cm.
Answer: Using the area formula for a parallelogram:
\( \text{Area} = \text{base} \times \text{height} \)
\( \implies \text{Area} = 8.5 \times 4.2 = 35.7\text{ cm}^2 \).
Hence, the area of the parallelogram is \( 35.7\text{ cm}^2 \).
In simple words: Multiply base by height directly to get 35.7 square centimetres.

Exam Tip: The area of a parallelogram is simply base times height; do not divide by 2.

Question 80. Find the area of a parallelogram whose base is 18 cm and height is 54 mm.
Answer: Convert the height into centimetres so that units match:
\( 54\text{ mm} = \frac{54}{10}\text{ cm} = 5.4\text{ cm} \)
Now calculate the area:
\( \text{Area} = \text{base} \times \text{height} \)
\( \implies \text{Area} = 18 \times 5.4 = 97.2\text{ cm}^2 \).
In simple words: Change 54 millimetres into 5.4 centimetres first, then multiply by 18.

Exam Tip: Always make sure both dimensions share the exact same measurement unit before multiplying.

Question. Find the area of a triangle whose base is 7 cm and height is 4.8 cm.
Answer: Using the area formula:
\( \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} \)
\( \implies \text{Area} = \frac{1}{2} \times 7 \times 4.8 = 7 \times 2.4 = 16.8\text{ cm}^2 \).
Thus, the area of the triangle is \( 16.8\text{ cm}^2 \).
In simple words: Halve 4.8 to get 2.4, then multiply by 7 to get 16.8 square centimetres.

Exam Tip: Halving the even decimal number first makes mental multiplication easy.

Question. The area of a triangle is 216 cm². If its height is 12 cm find its base.
Answer: The area formula is:
\( \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} \)
\( 216 = \frac{1}{2} \times \text{base} \times 12 \)
\( 216 = 6 \times \text{base} \)
\( \implies \text{base} = \frac{216}{6} = 36\text{ cm} \).
Therefore, the base of the triangle is \( 36\text{ cm} \).
In simple words: Divide height by 2 to get 6, then divide the area 216 by 6.

Exam Tip: Multiply the area by 2 and then divide by height to isolate the base.

Question. The perimeter of a square is 26 cm. Find its area.
Answer: Find the side of the square:
\( \text{side} = \frac{\text{Perimeter}}{4} = \frac{26}{4} = 6.5\text{ cm} \)
Now compute the area:
\( \text{Area} = \text{side}^2 = 6.5 \times 6.5 = 42.25\text{ cm}^2 \).
In simple words: Divide the perimeter by 4 to get one side, then multiply that side by itself.

Exam Tip: Remember to write the area unit as \( \text{cm}^2 \).

Question. 4² -2³ =-----------------------------
Answer: Calculate each power:
\( 4^2 = 16 \)
\( 2^3 = 8 \)
Subtract the results:
\( 16 - 8 = 8 \).
In simple words: 4 times 4 is 16, and 2 cubed is 8. Subtracting 8 from 16 leaves 8.

Exam Tip: Work out powers before performing subtraction.

Question. 7 x 10⁴ + 2 x 10² +3 x 10 is same as ------------------
Answer: Write each component in standard numerical form:
\( 7 \times 10^4 = 70000 \)
\( 2 \times 10^2 = 200 \)
\( 3 \times 10 = 30 \)
Add them together:
\( 70000 + 200 + 30 = 70230 \).
In simple words: Place 7 in the ten-thousands place, 2 in the hundreds place, and 3 in the tens place to form 70,230.

Exam Tip: Be careful to fill missing place values (thousands and ones) with zeros.

Question. The diameter of a circle is 7 cm . Then its circumference is -----------------.
Answer: The circumference formula in terms of diameter \( d \) is:
\( C = \pi d \)
Taking \( \pi = \frac{22}{7} \):
\( C = \frac{22}{7} \times 7 = 22\text{ cm} \).
In simple words: Multiply diameter by pi. The 7 cancels out, leaving 22 centimetres.

Exam Tip: Using \( C = \pi d \) is quicker than halving the diameter and using \( 2\pi r \).

Question. If the numbers x and y both squared and added we get ------------------------.
Answer: Squaring \( x \) gives \( x^2 \), and squaring \( y \) gives \( y^2 \).
Adding them gives the expression:
\( x^2 + y^2 \).
In simple words: Square both letters and put a plus sign between them.

Exam Tip: Do not confuse the sum of squares \( x^2 + y^2 \) with the square of the sum \( (x + y)^2 \).

Question. The value of 3x² + 2x + 1 when x is ( -1 ) is ------------------.
Answer: Substitute \( x = -1 \) into the algebraic expression:
\( 3(-1)^2 + 2(-1) + 1 \)
\( = 3(1) - 2 + 1 \)
\( = 3 - 2 + 1 = 2 \).
In simple words: Replace x with minus 1. Squaring makes it positive 1, so 3 minus 2 plus 1 gives 2.

Exam Tip: Remember that \( (-1)^2 = +1 \), not \( -1 \).

Question. Δ ABC ≅ Δ PQR. If ∠A = 70°, ∠B = 60°, then ∠ R = -----------------------.
Answer: In \( \Delta \text{ABC} \), find the third angle \( \angle C \):
\( \angle A + \angle B + \angle C = 180^\circ \)
\( 70^\circ + 60^\circ + \angle C = 180^\circ \)
\( \implies \angle C = 180^\circ - 130^\circ = 50^\circ \)
Since \( \Delta \text{ABC} \cong \Delta \text{PQR} \), their corresponding angles are equal:
\( \angle R = \angle C = 50^\circ \).
In simple words: Subtract 70 and 60 from 180 to find angle C, which matches angle R.

Exam Tip: Match corresponding vertices carefully: vertex C corresponds to vertex R.

Question. Express 1600 as the product of powers of its prime factors.
Answer: Factorize 1600 by prime numbers:
\( 1600 = 16 \times 100 = 2^4 \times (2^2 \times 5^2) = 2^{4+2} \times 5^2 = 2^6 \times 5^2 \).
Thus, the prime factorization of 1600 is \( 2^6 \times 5^2 \).
In simple words: Break 1600 into prime factors: 2 multiplies six times and 5 multiplies twice.

Exam Tip: Verify by expanding: \( 2^6 \times 5^2 = 64 \times 25 = 1600 \).

Question. The perimeter of a rectangle is 100 cm .If the length is 30 cm.Find its breadth.Also find its area.
Answer: Let length \( l = 30\text{ cm} \) and breadth be \( b \).
\( \text{Perimeter} = 2(l + b) = 100 \)
\( 30 + b = \frac{100}{2} = 50 \)
\( \implies b = 50 - 30 = 20\text{ cm} \).
Now, calculate the area:
\( \text{Area} = l \times b = 30 \times 20 = 600\text{ cm}^2 \).
Hence, the breadth is \( 20\text{ cm} \) and the area is \( 600\text{ cm}^2 \).
In simple words: Half the perimeter is 50, so taking away length 30 leaves breadth 20. Then multiply 30 by 20.

Exam Tip: Be sure to provide both required answers: breadth and area.

Question.  Express the numbers in the following statements in the standard form.
a) The diameter of the earth is 1, 27, 56, 000mm .
b) In a galaxy there are on an average 100, 000 ,000 , 000 stars.
Answer:
a) Shift the decimal point 7 places to the left:
\( 12,756,000\text{ mm} = 1.2756 \times 10^7\text{ mm} \).

b) Shift the decimal point 11 places to the left:
\( 100,000,000,000 = 1.0 \times 10^{11} \) (or \( 1 \times 10^{11} \)) stars.
In simple words: Place the decimal right after the first digit and count how many spots it moved to find the power of 10.

Exam Tip: Ensure the number in front of the power of 10 is between 1 and 10.

Question. If Δ ABC ≅ Δ STR then (i) AB = --------------- (ii) ∠C = --------------- (iii) AC = --------------.
Answer: Under the given correspondence \( \text{ABC} \leftrightarrow \text{STR} \):
(i) \( \text{AB} = \text{ST} \)
(ii) \( \angle C = \angle R \)
(iii) \( \text{AC} = \text{SR} \).
In simple words: Match the letters by their exact positions: first two letters match, and third letters match.

Exam Tip: Always follow the exact sequence of letters given in the congruence statement.

Question. The price of a maruthi car is Rs 4 ,00,000 in a particular year. The next year the price went up to 4,50, 000 .What was the percentage of increase.
Answer: Original price = Rs. 4,00,000
Increased price = Rs. 4,50,000
Increase in price = \( 450000 - 400000 = \text{Rs. } 50000 \).
Percentage increase formula:
\( \text{Percentage increase} = \frac{\text{Increase}}{\text{Original Price}} \times 100\% \)
\( \implies \frac{50000}{400000} \times 100\% = \frac{5}{40} \times 100\% = \frac{1}{8} \times 100\% = 12.5\% \).
Thus, the percentage increase is \( 12.5\% \).
In simple words: The car price rose by 50,000 rupees. Divide 50,000 by the starting price 400,000 and multiply by 100.

Exam Tip: The base for percentage increase must always be the original price, not the increased price.

Question. Simplify : i) 3⁵ x 10⁵ x 25 / 5⁷ x 6⁵
Answer: Express composite numbers as powers of prime factors:
\( 10^5 = (2 \times 5)^5 = 2^5 \times 5^5 \)
\( 25 = 5^2 \)
\( 6^5 = (2 \times 3)^5 = 2^5 \times 3^5 \)
Substitute into the expression:
\( \frac{3^5 \times (2^5 \times 5^5) \times 5^2}{5^7 \times (2^5 \times 3^5)} = \frac{2^5 \times 3^5 \times 5^{5+2}}{2^5 \times 3^5 \times 5^7} = \frac{2^5 \times 3^5 \times 5^7}{2^5 \times 3^5 \times 5^7} = 1 \).
In simple words: Break 10 and 6 into prime factors 2, 3, and 5. The top and bottom match completely, giving 1.

Exam Tip: Converting all terms to base 2, 3, and 5 is the cleanest way to simplify exponential fractions.

Question. Construct Δ ABC given BC = 7.2 cm ∠A =50° and ∠B=60°.
Answer: First, find the third angle \( \angle C \) using the angle sum property of triangles:
\( \angle A + \angle B + \angle C = 180^\circ \)
\( 50^\circ + 60^\circ + \angle C = 180^\circ \)
\( \implies \angle C = 180^\circ - 110^\circ = 70^\circ \)

Steps of Construction:
1. Draw a line segment \( \text{BC} = 7.2\text{ cm} \).
2. At point B, draw a ray making an angle of \( 60^\circ \) with BC using a protractor.
3. At point C, draw another ray making an angle of \( 70^\circ \) with CB.
4. Let the two rays intersect at point A.
\( \Delta \text{ABC} \) is the required triangle with \( \angle A = 50^\circ \).
In simple words: Calculate the missing corner C as 70 degrees first. Then draw the 7.2 cm base and build the two corner angles so their rays meet at A.

Exam Tip: You cannot draw the triangle from base BC without finding \( \angle C \) first; show this calculation clearly.

Question. Simplify the expressions and find the value if x = 2 ; (i) x²+9- 2 ( x - 5 )
ii) 5(3x - 2 )+ 4x + 15
Answer:
(i) Simplify first:
\( x^2 + 9 - 2(x - 5) = x^2 + 9 - 2x + 10 = x^2 - 2x + 19 \)
Now substitute \( x = 2 \):
\( (2)^2 - 2(2) + 19 = 4 - 4 + 19 = 19 \).

(ii) Simplify first:
\( 5(3x - 2) + 4x + 15 = 15x - 10 + 4x + 15 = 19x + 5 \)
Now substitute \( x = 2 \):
\( 19(2) + 5 = 38 + 5 = 43 \).
In simple words: Open the brackets and combine like terms first. Then put in 2 for x to find the final number.

Exam Tip: Watch signs when expanding: \( -2 \times (-5) = +10 \).

Question. In the figure AB and CD bisect each other at O
i) state the three pairs of equal parts in triangles AOC and BOD
ii) Is Δ AOC ≅ Δ BOD ? Give reason iii) Is AC = BD ? Give reason
Answer:
(i) Since AB and CD bisect each other at point O:
1. \( \text{AO} = \text{BO} \) (O is the midpoint of AB)
2. \( \text{CO} = \text{DO} \) (O is the midpoint of CD)
3. \( \angle \text{AOC} = \angle \text{BOD} \) (vertically opposite angles)

(ii) Yes, \( \Delta \text{AOC} \cong \Delta \text{BOD} \) by the SAS (Side-Angle-Side) congruence criterion, because two sides and the included angle of one triangle are equal to the corresponding parts of the other.

(iii) Yes, \( \text{AC} = \text{BD} \) by CPCT (Corresponding Parts of Congruent Triangles).
In simple words: Halving lines gives two equal pairs of sides, and crossing lines give equal angles in between. That makes the triangles identical.

Exam Tip: Mention CPCT (Corresponding Parts of Congruent Triangles) as the specific reason for part (iii).

Question. A wire of length 88 cm is bent into the shape of a circle. Find the radius of the circle and also find its area ? If the same wire is bent into the shape of a square what will be the length of each side ? ( π = 22/7 )
Answer: The length of the wire forms the circumference of the circle:
\( 2\pi r = 88\text{ cm} \)
\( 2 \times \frac{22}{7} \times r = 88 \)
\( \implies \frac{44}{7} r = 88 \)
\( \implies r = \frac{88 \times 7}{44} = 2 \times 7 = 14\text{ cm} \).
Now calculate the area of the circle:
\( \text{Area} = \pi r^2 = \frac{22}{7} \times 14 \times 14 = 22 \times 2 \times 14 = 616\text{ cm}^2 \).

When bent into a square, the length of the wire forms the perimeter:
\( 4 \times \text{side} = 88\text{ cm} \)
\( \implies \text{side} = \frac{88}{4} = 22\text{ cm} \).
Thus, the radius is \( 14\text{ cm} \), the area is \( 616\text{ cm}^2 \), and the side of the square is \( 22\text{ cm} \).
In simple words: The wire's length gives the circle's outer edge to find the radius and area, then dividing by 4 gives one side of the square.

Exam Tip: Note that wire length corresponds to perimeter (or circumference), not area.

Question. From the sum of 3x + 5 and 2x² - 4x +4, subtract the sum of 3x² - 5x and - x² + 2x +8
Answer: First, find the sum of the first two expressions:
\( (3x + 5) + (2x^2 - 4x + 4) = 2x^2 + (3x - 4x) + (5 + 4) = 2x^2 - x + 9 \)
Next, find the sum of the second two expressions:
\( (3x^2 - 5x) + (-x^2 + 2x + 8) = (3x^2 - x^2) + (-5x + 2x) + 8 = 2x^2 - 3x + 8 \)
Now subtract the second sum from the first sum:
\( (2x^2 - x + 9) - (2x^2 - 3x + 8) \)
\( = 2x^2 - x + 9 - 2x^2 + 3x - 8 \)
\( = (2x^2 - 2x^2) + (-x + 3x) + (9 - 8) = 2x + 1 \).
In simple words: Add the first two groups, add the second two groups, then subtract the second result from the first.

Exam Tip: Put parentheses around the second sum so that the minus sign applies to every single term inside.

Question. The standard form of 18 / -45 = .............................
Answer: Shift the negative sign to the numerator:
\( \frac{18}{-45} = \frac{-18}{45} \)
Divide numerator and denominator by their greatest common divisor, 9:
\( \frac{-18 \div 9}{45 \div 9} = -\frac{2}{5} \).
Thus, the standard form is \( -\frac{2}{5} \).
In simple words: Move the minus sign to the top and divide both numbers by 9 to get minus 2 over 5.

Exam Tip: A rational number in standard form must never have a negative denominator.

Question. Which is greater -2/5 or 7/-3 ?
Answer: Write both rational numbers with positive denominators:
\( -\frac{2}{5} \) and \( -\frac{7}{3} \)
The lowest common denominator of 5 and 3 is 15:
\( -\frac{2}{5} = \frac{-2 \times 3}{15} = -\frac{6}{15} \)
\( -\frac{7}{3} = \frac{-7 \times 5}{15} = -\frac{35}{15} \)
Compare the numerators: since \( -6 > -35 \), we have:
\( -\frac{6}{15} > -\frac{35}{15} \implies -\frac{2}{5} > \frac{7}{-3} \).
Therefore, \( -\frac{2}{5} \) is greater.
In simple words: Convert both to denominator 15. Minus 6 is closer to zero than minus 35, so minus 2 over 5 is larger.

Exam Tip: For negative numbers, the number with the smaller magnitude is the greater value.

Question. Draw a number line and represent -7/4 , 5/4 ,-1, 2
Answer: Convert the fractional numbers to mixed numbers:
\( -\frac{7}{4} = -1\frac{3}{4} \) (lies between \( -1 \) and \( -2 \))
\( \frac{5}{4} = 1\frac{1}{4} \) (lies between 1 and 2)
Divide each unit gap into 4 equal subdivisions of length \( \frac{1}{4} \):
- Locate \( -1 \) directly on its tick mark.
- Count 3 subdivisions to the left of \( -1 \) to plot \( -\frac{7}{4} \).
- Count 1 subdivision to the right of 1 to plot \( \frac{5}{4} \).
- Locate 2 directly on its tick mark.
In simple words: Break each step between whole numbers into 4 equal quarters, then mark each point at its proper fraction.

Exam Tip: Label 0 and the key whole numbers clearly so the fractional positions are easy to identify.

Question. Find the circumference of a circle whose radius is 14 cm.( take π = 22/7 )
Answer: Given radius \( r = 14\text{ cm} \).
The formula for the circumference of a circle is:
\( C = 2\pi r \)
\( \implies C = 2 \times \frac{22}{7} \times 14 = 2 \times 22 \times 2 = 88\text{ cm} \).
Hence, the circumference is \( 88\text{ cm} \).
In simple words: Multiply 2, pi, and 14 together. The 7 cancels with 14 to give 88 centimetres.

Exam Tip: State the circumference formula before performing numerical substitution.

Question. A gardener wants to fence a circular garden of diameter 21m. Find the length of the rope he needs to purchase, if he makes 3 rounds of fence. Also find the cost Rs 10 per meter. (Take π = 22/7 )
Answer: Given diameter \( d = 21\text{ m} \).
Circumference of the garden (length of 1 round of fence) = \( \pi d \):
\( C = \frac{22}{7} \times 21 = 22 \times 3 = 66\text{ m} \)
For 3 complete rounds of fencing:
\( \text{Total length of rope} = 3 \times 66 = 198\text{ m} \).
Cost of rope at Rs. 10 per metre:
\( \text{Total cost} = 198 \times \text{Rs. } 10 = \text{Rs. } 1980 \).
Thus, the gardener needs \( 198\text{ m} \) of rope, and the total cost is Rs. 1980.
In simple words: Find the circle's boundary once, multiply by 3 rounds to get 198 metres, then multiply by 10 rupees per metre.

Exam Tip: Do not forget to multiply by 3 for the 3 rounds before calculating the final cost.