CBSE Class 7 Mathematics Practice Worksheet Set 15

Question 1. The population of a town increases 8% annually. If its present population is 142560, what was it a year ago?
Answer: Let the town's population one year ago be \( P \).
The annual growth rate is \( 8\% \).
Present population = \( P + 8\% \text{ of } P = P\left(1 + \frac{8}{100}\right) = P \times \frac{108}{100} \).
According to the problem:
\( P \times \frac{108}{100} = 142560 \)
\( \implies P = \frac{142560 \times 100}{108} \)
\( \implies P = 1320 \times 100 = 132000 \).
Hence, the population one year ago was 132000.
In simple words: The town grew by 8 percent over the year. Dividing the current total by 1.08 gives the count from last year.

Exam Tip: Never subtract 8% from the present population; always take the original unknown value as base 100%.

Question 2. Find-(a)7.5% of 80 kg (b)8% of 5 litres (c)0.25% of 16
Answer:
(a) \( 7.5\% \text{ of } 80\text{ kg} = \frac{7.5}{100} \times 80 = \frac{600}{100} = 6\text{ kg} \).
(b) \( 8\% \text{ of } 5\text{ litres} = \frac{8}{100} \times 5 = \frac{40}{100} = 0.4\text{ litres} \) (or \( 400\text{ mL} \)).
(c) \( 0.25\% \text{ of } 16 = \frac{0.25}{100} \times 16 = \frac{4}{100} = 0.04 \).
In simple words: Turn the percent into a fraction by placing it over 100, then multiply by the given quantity.

Exam Tip: Always attach the correct units (kg, litres) to the numerical result in parts (a) and (b).

Question 3. Calculate what % of-(a)40m is 16m (b)1 kg is 25 gram
Answer:
(a) Required percentage = \( \frac{16\text{ m}}{40\text{ m}} \times 100\% = \frac{2}{5} \times 100\% = 40\% \).
(b) Convert to matching units: \( 1\text{ kg} = 1000\text{ g} \).
Required percentage = \( \frac{25\text{ g}}{1000\text{ g}} \times 100\% = \frac{25}{10}\% = 2.5\% \).
In simple words: Divide the smaller part by the base amount and multiply by 100 to get the percentage.

Exam Tip: Both quantities must be in the same unit before setting up the percentage fraction.

Question 4. Rohit bought a stereo for Rs 5863.Due to some defects in it, he had to pay Rs.137 to the mechanic for its repair. Then, he sold it for Rs 5700.Find his loss percent.
Answer: Total cost price (\( \text{CP} \)) includes purchase price and repair charges:
\( \text{Total CP} = 5863 + 137 = \text{Rs. } 6000 \).
Selling price (\( \text{SP} \)) = Rs. 5700.
Since \( \text{SP} < \text{CP} \), Rohit suffered a loss:
\( \text{Loss} = \text{CP} - \text{SP} = 6000 - 5700 = \text{Rs. } 300 \).
\( \text{Loss percent} = \frac{\text{Loss}}{\text{CP}} \times 100\% = \frac{300}{6000} \times 100\% = 5\% \).
In simple words: Add the repair cost to the purchase price to get the real cost. Then divide the loss by this full cost.

Exam Tip: Overhead expenses such as repairs or transport must always be added to the purchase cost before computing profit or loss.

Question 5. By selling a chair for Rs 720, a trader gains 20%. Find the cost price of the chair.
Answer: Selling price (\( \text{SP} \)) = Rs. 720, Gain = \( 20\% \).
Using the cost price formula:
\( \text{CP} = \frac{\text{SP} \times 100}{100 + \text{Gain}\%} \)
\( \implies \text{CP} = \frac{720 \times 100}{100 + 20} = \frac{720 \times 100}{120} = 6 \times 100 = \text{Rs. } 600 \).
In simple words: Selling at a 20 percent gain means the sale price is 120 percent of the original cost.

Exam Tip: Double check that you divide by \( (100 + \text{gain}) \) and not \( (100 - \text{gain}) \).

Question 6. An electrician sells a room heater for Rs 322, getting 1/6th of its C.P. Find his gain percent.
Answer: Let the cost price be \( \text{CP} \).
Gain = \( \frac{1}{6} \times \text{CP} \).
Using the gain percentage formula:
\( \text{Gain percent} = \frac{\text{Gain}}{\text{CP}} \times 100\% \)
\( \implies \text{Gain percent} = \frac{\frac{1}{6} \times \text{CP}}{\text{CP}} \times 100\% = \frac{1}{6} \times 100\% = \frac{50}{3}\% = 16\frac{2}{3}\% \) (or \( 16.67\% \)).
In simple words: Since profit is one-sixth of the cost, divide 100 by 6 to find the profit percent directly.

Exam Tip: Notice that the selling price of Rs. 322 is extra data not required to find the gain percentage.

Question 7. Find the simple interest on Rs 2500 for 2 years 6 months at 6% per annum. Also, find the amount.
Answer: Principal (\( P \)) = Rs. 2500, Rate (\( R \)) = \( 6\% \text{ p.a.} \).
Time (\( T \)) = \( 2\text{ years } 6\text{ months} = 2 + \frac{6}{12} = 2.5\text{ years} = \frac{5}{2}\text{ years} \).
Simple interest formula:
\( \text{SI} = \frac{P \times R \times T}{100} = \frac{2500 \times 6 \times 2.5}{100} = 25 \times 15 = \text{Rs. } 375 \).
Amount (\( A \)) = \( P + \text{SI} = 2500 + 375 = \text{Rs. } 2875 \).
In simple words: Convert months into years first. Multiply principal, rate, and time together, divide by 100, then add interest to the starting money.

Exam Tip: Remember to calculate both the simple interest and the final total amount as asked.

Question 8. What sum will amount to Rs 4590 at 12% per annum simple interest in 3 years?
Answer: Let the principal sum be \( P \).
Rate (\( R \)) = \( 12\% \text{ p.a.} \), Time (\( T \)) = \( 3\text{ years} \), Amount (\( A \)) = Rs. 4590.
\( A = P + \text{SI} = P + \frac{P \times R \times T}{100} = P\left(1 + \frac{12 \times 3}{100}\right) = P\left(1 + \frac{36}{100}\right) = \frac{136P}{100} \).
Equating to the given amount:
\( \frac{136P}{100} = 4590 \)
\( \implies P = \frac{4590 \times 100}{136} = \text{Rs. } 3375 \).
In simple words: 12 percent for 3 years adds 36 percent interest. Divide 4590 by 1.36 to find the principal.

Exam Tip: Express Amount as \( P\left(1 + \frac{RT}{100}\right) \) to solve for the unknown principal smoothly.

Question 9. At what rate % per annum will Rs 1650 amount to Rs 2046 in 3 years?
Answer: Principal (\( P \)) = Rs. 1650, Amount (\( A \)) = Rs. 2046, Time (\( T \)) = \( 3\text{ years} \).
Simple interest (\( \text{SI} \)) = \( A - P = 2046 - 1650 = \text{Rs. } 396 \).
Using the rate formula:
\( R = \frac{\text{SI} \times 100}{P \times T} \)
\( \implies R = \frac{396 \times 100}{1650 \times 3} = \frac{39600}{4950} = 8\% \text{ per annum} \).
In simple words: Subtract the starting sum from the final sum to get the interest. Then divide by principal and time.

Exam Tip: Find the simple interest first before substituting values into the rate formula.

Question 10. What is 100% of 80?
Answer: \( 100\% \text{ of } 80 = \frac{100}{100} \times 80 = 1 \times 80 = 80 \).
In simple words: Taking 100 percent of any value leaves that value unchanged.

Exam Tip: 100% represents the whole quantity, so the number itself is the answer.

Question 11. The simplest form of 3 ½:10 ½ = _______.
Answer: Convert mixed fractions to improper fractions:
\( 3\frac{1}{2} = \frac{7}{2} \), and \( 10\frac{1}{2} = \frac{21}{2} \).
The ratio is:
\( \frac{7}{2} : \frac{21}{2} = 7 : 21 = 1 : 3 \).
Hence, the simplest form is 1:3.
In simple words: Change both to improper fractions, drop the shared bottom numbers, and reduce by dividing by 7.

Exam Tip: Cancel out common denominators before simplifying the remaining whole numbers.

Question 12. What is 7:10 equal to in percent?
Answer: Convert the ratio into a fraction and multiply by \( 100\% \):
\( \frac{7}{10} \times 100\% = 70\% \).
In simple words: Write 7 over 10 and multiply by 100 to get 70 percent.

Exam Tip: Multiplying any fraction by 100 immediately changes it into its percentage form.

Question 13. In each case of profit, which is more S.P or C.P?
Answer: In every case of profit, the Selling Price (S.P) is greater than the Cost Price (C.P), that is, \( \text{S.P} > \text{C.P} \).
In simple words: You make a profit only when you sell goods for more than what you paid.

Exam Tip: State clearly that Selling Price (S.P) is more.

Question 14. Gain or loss percent is calculated on C.P or S.P?
Answer: Gain or loss percent is always calculated on the Cost Price (C.P).
In simple words: Profit and loss percentages are always worked out using the original cost.

Exam Tip: Keep in mind that C.P is always the base in the denominator for gain or loss percent.

Question 15. What is 1% of 100?
Answer: \( 1\% \text{ of } 100 = \frac{1}{100} \times 100 = 1 \).
In simple words: One part out of a hundred is just 1.

Exam Tip: By definition, percent means per hundred, so 1% of 100 is simply 1.

Question 16. What is 0.35 equivalent to in percent?
Answer: Multiply the decimal by \( 100\% \):
\( 0.35 \times 100\% = 35\% \).
In simple words: Shift the decimal point two places to the right to change a decimal into a percent.

Exam Tip: Adding the percent symbol (%) to your final number is essential.

Question 17. What is the equivalent fraction for 20%?
Answer: Write 20 over 100 and reduce to simplest terms:
\( 20\% = \frac{20}{100} = \frac{1}{5} \).
In simple words: Place 20 over 100 and divide both by 20 to get 1 over 5.

Exam Tip: Always reduce the fraction to its lowest terms.

Question 18. What is the number whose 8% is 20?
Answer: Let the required number be \( x \).
According to the question:
\( 8\% \text{ of } x = 20 \)
\( \frac{8}{100} \times x = 20 \)
\( \implies x = \frac{20 \times 100}{8} = \frac{2000}{8} = 250 \).
Thus, the number is 250.
In simple words: Multiply 20 by 100 and divide by 8 to find the original quantity.

Exam Tip: Form a basic algebraic equation with \( x \) to keep calculations straightforward.

Question 19. Find the interest on Rs.5250 at 8% p.a for a period of 3 yrs. & 4 months.
Answer: Principal (\( P \)) = Rs. 5250, Rate (\( R \)) = \( 8\% \text{ p.a.} \).
Convert time into years:
\( T = 3\text{ years } 4\text{ months} = 3 + \frac{4}{12} = 3 + \frac{1}{3} = \frac{10}{3}\text{ years} \).
Using the simple interest formula:
\( \text{SI} = \frac{P \times R \times T}{100} = \frac{5250 \times 8 \times 10}{100 \times 3} = \frac{5250 \times 80}{300} = \frac{420000}{300} = \text{Rs. } 1400 \).
In simple words: Change 4 months to one-third of a year. Then multiply principal, rate, and time together, and divide by 100.

Exam Tip: Always convert months into years as a fraction before using the simple interest formula.

Rational No.

Question 1. Arrange the following rational no. in descending order-
(a) \( -\frac{2}{5}, \frac{7}{10}, -\frac{11}{15}, \frac{19}{-30} \)
(b) \( -\frac{3}{4}, \frac{5}{-12}, -\frac{7}{16}, \frac{9}{-24} \)
Answer:
(a) Write each fraction with a positive denominator:
\( -\frac{2}{5}, \frac{7}{10}, -\frac{11}{15}, -\frac{19}{30} \).
The LCM of 5, 10, 15, and 30 is 30.
Convert each to an equivalent fraction with denominator 30:
\( -\frac{2}{5} = \frac{-2 \times 6}{30} = -\frac{12}{30} \)
\( \frac{7}{10} = \frac{7 \times 3}{30} = \frac{21}{30} \)
\( -\frac{11}{15} = \frac{-11 \times 2}{30} = -\frac{22}{30} \)
\( -\frac{19}{30} = -\frac{19}{30} \)
Comparing numerators in descending order: \( 21 > -12 > -19 > -22 \).
Therefore, the descending order is:
\( \frac{7}{10} > -\frac{2}{5} > \frac{19}{-30} > -\frac{11}{15} \).

(b) Express each fraction with a positive denominator:
\( -\frac{3}{4}, -\frac{5}{12}, -\frac{7}{16}, -\frac{9}{24} = -\frac{3}{8} \).
The LCM of 4, 12, 16, and 24 is 48.
Convert to denominator 48:
\( -\frac{3}{4} = \frac{-3 \times 12}{48} = -\frac{36}{48} \)
\( -\frac{5}{12} = \frac{-5 \times 4}{48} = -\frac{20}{48} \)
\( -\frac{7}{16} = \frac{-7 \times 3}{48} = -\frac{21}{48} \)
\( \frac{9}{-24} = \frac{-9 \times 2}{48} = -\frac{18}{48} \)
Comparing numerators in descending order: \( -18 > -20 > -21 > -36 \).
Therefore, the descending order is:
\( \frac{9}{-24} > \frac{5}{-12} > -\frac{7}{16} > -\frac{3}{4} \).
In simple words: Give all fractions the same positive bottom number. Then rank them from largest top number to smallest.

Exam Tip: For negative numbers, smaller absolute values represent greater numbers (for instance, \( -18 > -20 \)).

Question 2. Compare the given rational no. using <,>,= signs-
(a) \( \frac{5}{9} \ [ \ ] \ \frac{-3}{-8} \)
(b) \( \frac{9}{-13} \ [ \ ] \ \frac{7}{-12} \)
(c) \( \frac{2}{3} \ [ \ ] \ \frac{4}{5} \)
Answer:
(a) Note that \( \frac{-3}{-8} = \frac{3}{8} \).
Cross-multiply to compare \( \frac{5}{9} \) and \( \frac{3}{8} \):
\( 5 \times 8 = 40 \) and \( 9 \times 3 = 27 \).
Since \( 40 > 27 \), we have \( \frac{5}{9} > \frac{-3}{-8} \).

(b) Write with positive denominators: \( -\frac{9}{13} \) and \( -\frac{7}{12} \).
Cross-multiply:
\( -9 \times 12 = -108 \) and \( -7 \times 13 = -91 \).
Since \( -108 < -91 \), we have \( \frac{9}{-13} < \frac{7}{-12} \).

(c) Cross-multiply to compare \( \frac{2}{3} \) and \( \frac{4}{5} \):
\( 2 \times 5 = 10 \) and \( 3 \times 4 = 12 \).
Since \( 10 < 12 \), we have \( \frac{2}{3} < \frac{4}{5} \).
In simple words: Cross-multiply the fractions. The side with the larger product is the larger fraction.

Exam Tip: Move the negative sign to the numerator before cross-multiplying to keep signs consistent.

Question 3. Enlist any 5 rational no. between-
(a) \( \frac{2}{5} \) and \( \frac{4}{9} \)
(b) \( -\frac{3}{7} \) and \( -\frac{5}{8} \)
Answer:
(a) The LCM of 5 and 9 is 45:
\( \frac{2}{5} = \frac{18}{45} \), and \( \frac{4}{9} = \frac{20}{45} \).
To create space for 5 rational numbers, multiply the numerator and denominator by 6:
\( \frac{18 \times 6}{45 \times 6} = \frac{108}{270} \), and \( \frac{20 \times 6}{45 \times 6} = \frac{120}{270} \).
Any five rational numbers between them are:
\( \frac{109}{270}, \frac{110}{270}, \frac{111}{270}, \frac{112}{270}, \frac{113}{270} \).

(b) The LCM of 7 and 8 is 56:
\( -\frac{3}{7} = -\frac{24}{56} \), and \( -\frac{5}{8} = -\frac{35}{56} \).
Five rational numbers lying between \( -\frac{35}{56} \) and \( -\frac{24}{56} \) are:
\( -\frac{25}{56}, -\frac{26}{56}, -\frac{27}{56}, -\frac{28}{56}, -\frac{29}{56} \).
In simple words: Make the bottom numbers match. If you need more space between the top numbers, multiply top and bottom by a larger factor.

Exam Tip: Multiply both numerator and denominator by \( (n+1) \) when you need to insert \( n \) rational numbers.

Question 4. Represent each using a no.line-(a)- 1/3 (b)- 7/4 (c) 13/6
Answer:
(a) For \( -\frac{1}{3} \):
This value is negative and lies between 0 and \( -1 \). Divide the unit segment from 0 to \( -1 \) into 3 equal parts. The first mark to the left of 0 represents \( -\frac{1}{3} \).

(b) For \( -\frac{7}{4} = -1\frac{3}{4} \):
This value lies between \( -1 \) and \( -2 \). Divide each unit distance into 4 equal segments. Count 7 units of \( \frac{1}{4} \) to the left of 0 (or 3 sub-divisions past \( -1 \)) to locate \( -\frac{7}{4} \).

(c) For \( \frac{13}{6} = 2\frac{1}{6} \):
This positive value lies between 2 and 3. Divide each unit distance into 6 equal parts. Count 13 sub-divisions of \( \frac{1}{6} \) to the right of 0 (which is 1 unit past 2) to plot \( \frac{13}{6} \).
In simple words: Convert improper fractions to mixed numbers to see which two integers they lie between, then divide that space into equal steps.

Exam Tip: Mark and label whole integers clearly on the line before showing the fraction subdivision.

Question 5. Evaluate each-
(a) \( \frac{11}{12} + \frac{5}{9} \)
(b) \( -\frac{8}{9} + \left(-\frac{5}{12}\right) \)
(c) \( \frac{3}{7} + \left(-\frac{2}{14}\right) \)
(d) \( \left(-\frac{5}{8}\right) - \left(-\frac{3}{4}\right) \)
(e) \( \frac{7}{11} - \frac{2}{3} \)
(f) \( \frac{4}{7} \times \frac{21}{16} \)
(g) \( \frac{6}{8} \times \frac{9}{36} \)
(h) \( \frac{14}{20} \div \frac{6}{3} \)
(i) \( \frac{9}{12} \div \frac{3}{4} \)
Answer:
(a) The LCM of 12 and 9 is 36:
\( \frac{11 \times 3 + 5 \times 4}{36} = \frac{33 + 20}{36} = \frac{53}{36} = 1\frac{17}{36} \).

(b) The LCM of 9 and 12 is 36:
\( \frac{-8 \times 4 + (-5) \times 3}{36} = \frac{-32 - 15}{36} = -\frac{47}{36} = -1\frac{11}{36} \).

(c) Simplify \( -\frac{2}{14} = -\frac{1}{7} \):
\( \frac{3}{7} - \frac{1}{7} = \frac{2}{7} \).

(d) Subtracting a negative gives addition:
\( -\frac{5}{8} + \frac{3}{4} = \frac{-5 + 6}{8} = \frac{1}{8} \).

(e) The LCM of 11 and 3 is 33:
\( \frac{7 \times 3 - 2 \times 11}{33} = \frac{21 - 22}{33} = -\frac{1}{33} \).

(f) Cancel common factors:
\( \frac{4}{7} \times \frac{21}{16} = \frac{1 \times 3}{1 \times 4} = \frac{3}{4} \).

(g) Simplify terms:
\( \frac{6}{8} = \frac{3}{4} \), and \( \frac{9}{36} = \frac{1}{4} \).
\( \frac{3}{4} \times \frac{1}{4} = \frac{3}{16} \).

(h) Dividing is multiplying by the reciprocal:
\( \frac{14}{20} \div \frac{6}{3} = \frac{7}{10} \times \frac{3}{6} = \frac{7}{10} \times \frac{1}{2} = \frac{7}{20} \).

(i) Simplify:
\( \frac{9}{12} = \frac{3}{4} \).
\( \frac{3}{4} \div \frac{3}{4} = \frac{3}{4} \times \frac{4}{3} = 1 \).
In simple words: Find a common denominator to add or subtract. For division, invert the second fraction and multiply.

Exam Tip: Simplify individual fractions before multiplying or dividing to save time on cancellations.

Question 6. Add 2/3 & 5/6
Answer: The LCM of denominators 3 and 6 is 6:
\( \frac{2}{3} + \frac{5}{6} = \frac{2 \times 2}{6} + \frac{5}{6} = \frac{4 + 5}{6} = \frac{9}{6} \).
Reducing by dividing top and bottom by 3:
\( \frac{9}{6} = \frac{3}{2} = 1\frac{1}{2} \).
In simple words: Change 2 over 3 into 4 over 6, then add 5 over 6 to get 9 over 6, which reduces to 3 over 2.

Exam Tip: Always reduce the final fraction to its simplest form or write it as a mixed fraction.

Question 7. Subtract 8/11 from 100/121.
Answer: The LCM of 121 and 11 is 121:
\( \frac{100}{121} - \frac{8}{11} = \frac{100 - (8 \times 11)}{121} = \frac{100 - 88}{121} = \frac{12}{121} \).
In simple words: Multiply 8 over 11 by 11 on top and bottom to get 88 over 121. Then subtract from 100 over 121.

Exam Tip: In "subtract A from B", the expression to write is always \( B - A \).

Question 8. Find additive inverse of( 0/5 ).
Answer: Since \( \frac{0}{5} = 0 \), and the additive inverse of 0 is \( -0 = 0 \).
Hence, the additive inverse of \( \frac{0}{5} \) is 0.
In simple words: Zero divided by any number is zero, and zero is its own additive opposite.

Exam Tip: Remember that zero is neither positive nor negative, so its additive inverse is simply 0.

Question 9. Find the product of the multiplicative inverse of both (-8) &( -1/16 ).
Answer: The multiplicative inverse of \( -8 \) is \( -\frac{1}{8} \).
The multiplicative inverse of \( -\frac{1}{16} \) is \( -16 \).
Now, find their product:
\( \left(-\frac{1}{8}\right) \times (-16) = \frac{16}{8} = 2 \).
In simple words: Flip each number upside down to get its inverse, then multiply the two results together.

Exam Tip: Multiplying two negative values produces a positive final result.

Question 10. Evaluate 0÷ -1/5.
Answer: \( 0 \div \left(-\frac{1}{5}\right) = 0 \times (-5) = 0 \).
In simple words: Zero divided by any non-zero number always equals zero.

Exam Tip: Zero in the numerator produces zero, whereas division by zero is undefined.

Question 11. Fill in the blanks :-
a. The sum of two rational numbers is always a _______.
b. The sum of 0 & any rational number is the _______ itself.
c. The difference of two rational numbers is always a _______.
d. The product of any rational number with 1 is the _______ itself.
Answer:
a. rational number
b. rational number
c. rational number
d. rational number
In simple words: Adding, subtracting, or multiplying rational numbers keeps them rational. Zero and one leave the original number unchanged.

Exam Tip: Learn the closure, identity of addition (0), and identity of multiplication (1) properties for rational numbers.

Question 12. True or false :-
a. The reciprocal of -1 is 1.
b. The product of two rational numbers is a rational number.
Answer:
a. False. The reciprocal of \( -1 \) is \( \frac{1}{-1} = -1 \).
b. True. Rational numbers are closed under multiplication.
In simple words: Flipping minus one still gives minus one. Multiplying two fractions always results in another fraction.

Exam Tip: Note that 1 and -1 are the only two real numbers that equal their own reciprocals.

Question 13. Subtract the sum of -3/7 & 5/14 from the sum of 11/16 & -2/3.
Answer: First sum:
\( -\frac{3}{7} + \frac{5}{14} = \frac{-6 + 5}{14} = -\frac{1}{14} \).
Second sum:
\( \frac{11}{16} + \left(-\frac{2}{3}\right) = \frac{11 \times 3 - 2 \times 16}{48} = \frac{33 - 32}{48} = \frac{1}{48} \).
Now, subtract the first sum from the second sum:
\( \frac{1}{48} - \left(-\frac{1}{14}\right) = \frac{1}{48} + \frac{1}{14} \).
The LCM of 48 and 14 is 336:
\( \frac{1 \times 7 + 1 \times 24}{336} = \frac{7 + 24}{336} = \frac{31}{336} \).
In simple words: Work out both additions first. Then subtract the first result from the second.

Exam Tip: Be careful with the double negative: subtracting \( -\frac{1}{14} \) turns into adding \( +\frac{1}{14} \).

Question 14. The sum of two rational numbers is 15/16. If one of them is -5/8, find the other.
Answer: Let the other rational number be \( x \).
\( x + \left(-\frac{5}{8}\right) = \frac{15}{16} \)
\( \implies x = \frac{15}{16} - \left(-\frac{5}{8}\right) = \frac{15}{16} + \frac{5}{8} \)
The LCM of 16 and 8 is 16:
\( x = \frac{15 + 10}{16} = \frac{25}{16} = 1\frac{9}{16} \).
Thus, the other number is \( \frac{25}{16} \).
In simple words: Subtract the given number from the total sum to find the missing fraction.

Exam Tip: Write down the algebraic statement first to keep track of negative signs.

Question 15. By what rational number should -12/52 be multiplied to obtain 39?
Answer: Let the required number be \( x \).
\( x \times \left(-\frac{12}{52}\right) = 39 \)
Simplify \( -\frac{12}{52} = -\frac{3}{13} \):
\( x \times \left(-\frac{3}{13}\right) = 39 \)
\( \implies x = 39 \div \left(-\frac{3}{13}\right) = 39 \times \left(-\frac{13}{3}\right) \)
\( \implies x = 13 \times (-13) = -169 \).
Therefore, the required rational number is \( -169 \).
In simple words: Divide 39 by minus 12 over 52 to solve for the missing multiplier.

Exam Tip: Simplify the fraction \( -\frac{12}{52} \) to \( -\frac{3}{13} \) first to make calculations simpler.

Practical Geometry

Question 1. Construct a triangle ABC in which AB=7 cm, angle A=45 and angle C=75.
Answer: First, find \( \angle B \) using the angle sum property of triangles:
\( \angle A + \angle B + \angle C = 180^\circ \)
\( 45^\circ + \angle B + 75^\circ = 180^\circ \)
\( \implies \angle B + 120^\circ = 180^\circ \implies \angle B = 60^\circ \).

Steps of Construction:
1. Draw a line segment \( \text{AB} = 7\text{ cm} \).
2. At point A, construct ray AX making an angle of \( 45^\circ \) with AB.
3. At point B, construct ray BY making an angle of \( 60^\circ \) with BA.
4. Let rays AX and BY intersect at point C.
\( \Delta \text{ABC} \) is the required triangle with \( \angle C = 75^\circ \).
In simple words: Subtract 45 and 75 from 180 to find the missing angle B. Draw the base and make the two angles at each end.

Exam Tip: You cannot draw the triangle without first calculating \( \angle B \); show the angle calculation clearly.

Question 2. Construct a triangle ABC in which BC=4.8 cm, angle C=90 and AB=6.3 cm.
Answer: This is a right-angled triangle with hypotenuse \( \text{AB} = 6.3\text{ cm} \) and base \( \text{BC} = 4.8\text{ cm} \).

Steps of Construction:
1. Draw a line segment \( \text{BC} = 4.8\text{ cm} \).
2. At point C, construct a right angle by drawing ray CX perpendicular to BC (\( \angle \text{BCX} = 90^\circ \)).
3. With B as centre and radius \( 6.3\text{ cm} \), draw an arc intersecting ray CX at point A.
4. Join AB.
\( \Delta \text{ABC} \) is the required right-angled triangle.
In simple words: Draw a 4.8 cm base line. Make a 90-degree line at C, then swing a 6.3 cm arc from B to find point A.

Exam Tip: Label the right-angle square symbol at vertex C to indicate the 90-degree corner.

Question 3. Construct a right-angled triangle one side of which measures 3.5 cm and the hypotenuse is 6 cm.
Answer: Let the right-angled triangle be \( \Delta \text{PQR} \), right-angled at Q, with leg \( \text{QR} = 3.5\text{ cm} \) and hypotenuse \( \text{PR} = 6\text{ cm} \).

Steps of Construction:
1. Draw a base segment \( \text{QR} = 3.5\text{ cm} \).
2. At point Q, draw ray QX perpendicular to QR (\( \angle \text{RQX} = 90^\circ \)).
3. With R as centre and compass opened to \( 6\text{ cm} \), draw an arc cutting ray QX at point P.
4. Join PR.
\( \Delta \text{PQR} \) is the required right-angled triangle.
In simple words: Draw a 3.5 cm line, create a 90-degree ray on one end, and cut that ray with a 6 cm arc from the other end.

Exam Tip: Make sure the arc cutting the perpendicular ray is visible and clearly labelled.

Question 4. Construct a triangle XYZ in which YZ=5.8 cm, angle Y=angle Z=30.Measure XY and XZ.What do you observe?
Answer: Steps of Construction:
1. Draw a line segment \( \text{YZ} = 5.8\text{ cm} \).
2. At point Y, construct ray YP making an angle of \( 30^\circ \) with YZ.
3. At point Z, construct ray ZQ making an angle of \( 30^\circ \) with ZY.
4. The two rays intersect at point X.

Measurement:
Measuring the lengths with a ruler gives \( \text{XY} \approx 3.35\text{ cm} \) and \( \text{XZ} \approx 3.35\text{ cm} \).

Observation:
We observe that \( \text{XY} = \text{XZ} \). Since the base angles are equal (\( \angle Y = \angle Z = 30^\circ \)), the sides opposite to these equal angles are also equal. Hence, \( \Delta \text{XYZ} \) is an isosceles triangle.
In simple words: Draw the 5.8 cm base and build 30-degree angles on both sides. Both remaining side lengths turn out to be equal.

Exam Tip: State both the numerical measurements and the observation that the triangle is isosceles.

Question 5. Construct a triangle PQR in which QR=6 cm, PQ=4.4 cm and PR=5.3 cm. Draw the bisector of angle P.
Answer: Steps of Construction:
1. Draw a base segment \( \text{QR} = 6\text{ cm} \).
2. With centre Q and radius \( 4.4\text{ cm} \), draw an arc.
3. With centre R and radius \( 5.3\text{ cm} \), draw another arc intersecting the previous arc at P.
4. Join PQ and PR to complete \( \Delta \text{PQR} \).
5. To bisect \( \angle P \): with P as centre, draw an arc intersecting side PQ at point M and side PR at point N.
6. With M and N as centres and radius greater than half of MN, draw two arcs that intersect at point D.
7. Draw ray PD. Ray PD is the required bisector of \( \angle P \).
In simple words: Draw the triangle by intersecting arcs from each end of the base. Then use compass arcs at vertex P to cut the angle in half.

Exam Tip: Keep compass arcs neat and visible to earn complete construction marks.

Question 6. Construct an equilateral triangle each of whose sides measures 6.2 cm. Measure each one of its angles.
Answer: Steps of Construction:
1. Draw a line segment \( \text{BC} = 6.2\text{ cm} \).
2. With centre B and radius \( 6.2\text{ cm} \), draw an arc above the segment.
3. With centre C and the same radius \( 6.2\text{ cm} \), draw a second arc intersecting the first arc at point A.
4. Join AB and AC.

Measurement:
Measuring the three angles with a protractor gives \( \angle A = 60^\circ \), \( \angle B = 60^\circ \), and \( \angle C = 60^\circ \).
Hence, each angle of the equilateral triangle measures \( 60^\circ \).
In simple words: Draw a 6.2 cm base and cross two 6.2 cm arcs from each end. All three angles will measure 60 degrees.

Exam Tip: State clearly that each interior angle of an equilateral triangle measures \( 60^\circ \).

Area and Perimeter

Question 1. Find the area of a rectangular plot,one side of which measures 35 m and the diagonal is 37 m.
Answer: In a rectangle, the length, breadth, and diagonal form a right-angled triangle.
Let length \( l = 35\text{ m} \) and diagonal \( d = 37\text{ m} \).
Using the Pythagoras theorem:
\( b = \sqrt{d^2 - l^2} = \sqrt{37^2 - 35^2} \)
\( \implies b = \sqrt{(37 - 35)(37 + 35)} = \sqrt{2 \times 72} = \sqrt{144} = 12\text{ m} \).
Now, find the area:
\( \text{Area} = l \times b = 35 \times 12 = 420\text{ m}^2 \).
In simple words: Use the right-triangle rule with the diagonal to find the missing side, then multiply the two sides.

Exam Tip: Using the identity \( a^2 - b^2 = (a-b)(a+b) \) makes finding \( \sqrt{37^2 - 35^2} \) much faster.

Question 2. A room is 9 m long, 8 m broad and 6.5 m high. It has one door of dimensions 2m x1.5 m and three windows each of dimensions 1.5 m x 1m.Find the cost of whitewashing the walls at Rs.3.80 per square metre.
Answer: Total area of the 4 walls = \( 2(l + b) \times h \):
\( \text{Wall area} = 2(9 + 8) \times 6.5 = 2 \times 17 \times 6.5 = 221\text{ m}^2 \).
Area of 1 door = \( 2\text{ m} \times 1.5\text{ m} = 3\text{ m}^2 \).
Area of 3 windows = \( 3 \times (1.5\text{ m} \times 1\text{ m}) = 3 \times 1.5 = 4.5\text{ m}^2 \).
Total unpainted area of door and windows = \( 3 + 4.5 = 7.5\text{ m}^2 \).
Area to be whitewashed = \( 221 - 7.5 = 213.5\text{ m}^2 \).
Cost of whitewashing = \( 213.5 \times \text{Rs. } 3.80 = \text{Rs. } 811.30 \).
In simple words: Find the total wall area, take away the space used by the door and windows, and multiply by the cost per square metre.

Exam Tip: Remember to multiply the window dimensions by 3 since there are three windows.

Question 3. A rectangular lawn measuring 50 m by 36 m is surrounded externally by a path which is 2 m. Find the area of the path. Also calculate the cost of levelling the path at Rs.2.50 per metre square.
Answer: Inner dimensions of the lawn:
Length = \( 50\text{ m} \), Breadth = \( 36\text{ m} \).
\( \text{Area of lawn} = 50 \times 36 = 1800\text{ m}^2 \).
The path is added on all sides with width \( 2\text{ m} \):
Outer length = \( 50 + 2(2) = 54\text{ m} \).
Outer breadth = \( 36 + 2(2) = 40\text{ m} \).
\( \text{Outer area} = 54 \times 40 = 2160\text{ m}^2 \).
Area of the path = \( \text{Outer area} - \text{Inner area} = 2160 - 1800 = 360\text{ m}^2 \).
Cost of levelling = \( 360 \times \text{Rs. } 2.50 = \text{Rs. } 900 \).
In simple words: Add twice the path width to find the outer dimensions. Subtract the inner area from the outer area to get the path area.

Exam Tip: An external path adds \( 2 \times \text{width} \) to both length and breadth.

Question 4. A rectangular lawn is 60 m by 40 m and has two roads, each 5 m wide running in the middle of it, one parallel to its length and the other parallel to the breadth. Find the cost of gravelling the roads at Rs.5.90 per square metre.
Answer: Area of the road parallel to length = \( 60\text{ m} \times 5\text{ m} = 300\text{ m}^2 \).
Area of the road parallel to breadth = \( 40\text{ m} \times 5\text{ m} = 200\text{ m}^2 \).
Area of the common central intersection = \( 5\text{ m} \times 5\text{ m} = 25\text{ m}^2 \).
Total area of roads = \( 300 + 200 - 25 = 475\text{ m}^2 \).
Cost of gravelling the roads = \( 475 \times \text{Rs. } 5.90 = \text{Rs. } 2802.50 \).
In simple words: Add the areas of both roads and subtract the overlapping centre square once so it is not counted twice.

Exam Tip: Never forget to subtract the area of the shared central square where the two roads cross.

Question 5. The lengths of the sides of a triangle are 33cm,44cm and 55cm. Find the altitude corresponding to the side measuring 44cm.
Answer: Check if the triangle is right-angled using Pythagoras theorem:
\( 33^2 + 44^2 = 1089 + 1936 = 3025 \), and \( 55^2 = 3025 \).
Since \( 33^2 + 44^2 = 55^2 \), this is a right-angled triangle with legs 33 cm and 44 cm.
\( \text{Area of triangle} = \frac{1}{2} \times 33 \times 44 = 33 \times 22 = 726\text{ cm}^2 \).
Now, let the altitude to the side 44 cm be \( h \):
\( \text{Area} = \frac{1}{2} \times \text{base} \times h \)
\( 726 = \frac{1}{2} \times 44 \times h = 22 \times h \)
\( \implies h = \frac{726}{22} = 33\text{ cm} \).
In simple words: Because this is a right-angled triangle, the height on the 44 cm side is simply the other leg of 33 cm.

Exam Tip: Recognizing Pythagorean triplets (\( 3:4:5 \times 11 \)) saves you from doing lengthy Heron's formula calculations.

Question 6. The cost of painting the top surface of a triangular board at 80 paisa per metre sq is Rs.176.40.If the height of the board measures 24.5 m, find its base.
Answer: Rate of painting = 80 paisa per \( \text{m}^2 = \text{Rs. } 0.80 \text{ per m}^2 \).
Total cost = Rs. 176.40.
\( \text{Area of triangle} = \frac{\text{Total Cost}}{\text{Rate}} = \frac{176.40}{0.80} = 220.5\text{ m}^2 \).
Given height \( h = 24.5\text{ m} \).
\( \text{Area} = \frac{1}{2} \times \text{base} \times h \)
\( 220.5 = \frac{1}{2} \times b \times 24.5 \)
\( \implies b = \frac{220.5 \times 2}{24.5} = \frac{441}{24.5} = 18\text{ m} \).
Hence, the base of the board is 18 m.
In simple words: Divide the total money spent by the rate to find the area, then solve the triangle area formula for the base.

Exam Tip: Convert 80 paisa to Rs. 0.80 before dividing to keep currency units matching.

Question 7. The adjacent sides of a parallelogram are 15 cm and 8 cm. If the distance between the longer sides is 4 cm, find the distance between the shorter sides.
Answer: Longer side (\( b_1 \)) = 15 cm, distance between them (\( h_1 \)) = 4 cm.
\( \text{Area of parallelogram} = b_1 \times h_1 = 15 \times 4 = 60\text{ cm}^2 \).
Shorter side (\( b_2 \)) = 8 cm. Let the distance between shorter sides be \( h_2 \).
\( \text{Area} = b_2 \times h_2 \)
\( 60 = 8 \times h_2 \)
\( \implies h_2 = \frac{60}{8} = 7.5\text{ cm} \).
Thus, the distance between the shorter sides is 7.5 cm.
In simple words: Area equals base times height. Find the total area using the longer side, then divide that area by the shorter side.

Exam Tip: The area of a parallelogram remains the same regardless of which side is chosen as base.

Question 8. The diameter of the wheel of a car is 70 cm. How many revolutions will it make to travel 99 km?
Answer: Diameter (\( d \)) = 70 cm.
Distance covered in 1 revolution = Circumference of wheel:
\( C = \pi d = \frac{22}{7} \times 70 = 220\text{ cm} = 2.2\text{ m} \).
Total travel distance = \( 99\text{ km} = 99 \times 1000\text{ m} = 99000\text{ m} \).
\( \text{Number of revolutions} = \frac{\text{Total Distance}}{\text{Circumference}} = \frac{99000}{2.2} = 45000 \).
In simple words: Find the circumference of the wheel. Then divide the full travel distance by this circumference.

Exam Tip: Convert both the distance and the circumference into metres before dividing.

Question 9. A wire is in a circular shape of radius 28 cm. If it is bent in the form of a square, what will be the area of the square formed?
Answer: The length of the wire is equal to the circumference of the circle:
\( \text{Length of wire} = 2\pi r = 2 \times \frac{22}{7} \times 28 = 2 \times 22 \times 4 = 176\text{ cm} \).
When rebent into a square, this length becomes the perimeter of the square:
\( 4 \times \text{side} = 176\text{ cm} \)
\( \implies \text{side} = \frac{176}{4} = 44\text{ cm} \).
\( \text{Area of the square} = \text{side} \times \text{side} = 44 \times 44 = 1936\text{ cm}^2 \).
In simple words: The wire length stays unchanged. Find the circle's outer edge, divide by 4 to get one square side, and square it.

Exam Tip: The perimeter of the new shape always equals the perimeter of the original shape when rebending a wire.

Question 10. The area of a circle is 1386 m sq. Find its circumference.
Answer: Given area of circle = \( 1386\text{ m}^2 \).
\( \pi r^2 = 1386 \)
\( \frac{22}{7} \times r^2 = 1386 \)
\( \implies r^2 = \frac{1386 \times 7}{22} = 63 \times 7 = 441 \)
\( \implies r = \sqrt{441} = 21\text{ m} \).
Now, find the circumference:
\( C = 2\pi r = 2 \times \frac{22}{7} \times 21 = 2 \times 22 \times 3 = 132\text{ m} \).
In simple words: Use the area to get the radius first, then multiply 2, pi, and radius to get the circumference.

Exam Tip: Factorize \( 63 \times 7 = 9 \times 7 \times 7 \) to easily take the square root as \( 3 \times 7 = 21 \).

Question 11. Fill in the blanks:-
1) The area of a parallelogram is equal to _______.
2) Perimeter of an equilateral triangle is _______.
3) The _______ of a circle is equal to 2πr.
4) The magnitude of the plane region enclosed by a simple closed figure is called the ______ of that figure.
5) 1 hectare = _______m2.
Answer:
1) base × height
2) 3 × side
3) circumference
4) area
5) 10000
In simple words: Learn these standard geometric formulas and conversions by heart.

Exam Tip: Be accurate with unit conversions: 1 hectare is \( 100\text{ m} \times 100\text{ m} = 10000\text{ m}^2 \).

Question 12. The length & breadth of a rectangular plot are in the ratio 3:2. If the area of the field is 3456m2, find the cost of fencing the plot with wire if wire cost Rs. 2.75 per metre.
Answer: Let the length be \( 3x \) and breadth be \( 2x \).
\( \text{Area} = \text{length} \times \text{breadth} = 3x \times 2x = 6x^2 \).
\( 6x^2 = 3456 \)
\( \implies x^2 = \frac{3456}{6} = 576 \)
\( \implies x = \sqrt{576} = 24\text{ m} \).
Length = \( 3 \times 24 = 72\text{ m} \), Breadth = \( 2 \times 24 = 48\text{ m} \).
Perimeter to be fenced = \( 2(l + b) = 2(72 + 48) = 2 \times 120 = 240\text{ m} \).
Cost of fencing = \( 240 \times \text{Rs. } 2.75 = \text{Rs. } 660 \).
In simple words: Set the ratio as 3x and 2x, find x using the area, then calculate the boundary length and multiply by the cost per metre.

Exam Tip: Remember fencing depends on perimeter, not area.

Question 13. The base of the parallelogram is double of its height. If the area is 512m2, find the base & height.
Answer: Let height = \( h \), so base = \( 2h \).
\( \text{Area} = \text{base} \times \text{height} = 2h \times h = 2h^2 \).
\( 2h^2 = 512 \)
\( \implies h^2 = \frac{512}{2} = 256 \)
\( \implies h = \sqrt{256} = 16\text{ m} \).
Base = \( 2h = 2 \times 16 = 32\text{ m} \).
Hence, height is 16 m and base is 32 m.
In simple words: Base times height is twice height squared. Divide 512 by 2 and take the square root to get height.

Exam Tip: Be sure to write down both dimensions (base and height) in your final line.

Question 14. The area of a rhombus is 119cm2 & its perimeter is 56cm. Find its altitude.
Answer: A rhombus has four equal sides:
\( \text{Perimeter} = 4 \times \text{side} = 56\text{ cm} \)
\( \implies \text{side} = \frac{56}{4} = 14\text{ cm} \).
Since a rhombus is also a parallelogram, \( \text{Area} = \text{side (base)} \times \text{altitude} \):
\( 14 \times \text{altitude} = 119 \)
\( \implies \text{altitude} = \frac{119}{14} = 8.5\text{ cm} \).
In simple words: Divide the perimeter by 4 to get one side, then divide the area by this side to find the altitude.

Exam Tip: Treat a rhombus as a special parallelogram with base equal to its side length.

Question 15. A carpet measures 25m by 20m. A border of width 50cm is printed along its side. Find the cost of printing the border at Rs. 20 per m2.
Answer: Total carpet area = \( 25\text{ m} \times 20\text{ m} = 500\text{ m}^2 \).
Border width = \( 50\text{ cm} = 0.5\text{ m} \).
Inner dimensions without border:
Inner length = \( 25 - 2(0.5) = 24\text{ m} \).
Inner breadth = \( 20 - 2(0.5) = 19\text{ m} \).
Inner area = \( 24 \times 19 = 456\text{ m}^2 \).
Area of printed border = \( 500 - 456 = 44\text{ m}^2 \).
Cost of printing = \( 44 \times \text{Rs. } 20 = \text{Rs. } 880 \).
In simple words: Subtract 1 metre from both length and width to find the inner space. Subtract the inner area from 500 to find the border area.

Exam Tip: Remember to convert 50 cm to 0.5 m first so all measurements are in metres.

Question 16. Find the area of the shaded region.

Answer: From the figure, the outer rectangle has length \( 12\text{ m} \) and breadth \( 8\text{ m} \).
The shaded region consists of two perpendicular central pathways of width \( 1\text{ m} \) running through the rectangle:
Area of horizontal pathway = \( 12\text{ m} \times 1\text{ m} = 12\text{ m}^2 \).
Area of vertical pathway = \( 8\text{ m} \times 1\text{ m} = 8\text{ m}^2 \).
Area of the common central square = \( 1\text{ m} \times 1\text{ m} = 1\text{ m}^2 \).
Total area of shaded pathways = \( 12 + 8 - 1 = 19\text{ m}^2 \).
(Note: If the four unshaded corner rectangles were requested, their combined area is \( (12 - 1) \times (8 - 1) = 11 \times 7 = 77\text{ m}^2 \)).
Thus, the area of the shaded crossroads is \( 19\text{ m}^2 \).
In simple words: Add the areas of the two crossing paths and subtract the 1 square metre where they cross over each other.

Exam Tip: Always subtract the central overlapping region once when calculating crossroads areas.

Algebraic Expressions

Question 1. Add the following-(a)2x-3y,5x+6y,-3x-4y (b)5x-7y+3,-8x+2y-5,7x-y-2
Answer:
(a) Group like terms:
\( (2x + 5x - 3x) + (-3y + 6y - 4y) \)
\( = (7x - 3x) + (3y - 4y) = 4x - y \).

(b) Group like terms together:
\( (5x - 8x + 7x) + (-7y + 2y - y) + (3 - 5 - 2) \)
\( = (4x) + (-6y) + (-4) = 4x - 6y - 4 \).
In simple words: Collect matching letters together and add their number coefficients.

Exam Tip: Combine terms with matching variable parts separately: x-terms with x-terms, y-terms with y-terms.

Question 2. Subtract-(a)2x-5y+7 from 3x+4y-6 (b)x-y from 4y-5x (c)x-y+3z from 2z-x-3y
Answer:
(a) \( (3x + 4y - 6) - (2x - 5y + 7) \)
\( = 3x + 4y - 6 - 2x + 5y - 7 \)
\( = (3x - 2x) + (4y + 5y) + (-6 - 7) = x + 9y - 13 \).

(b) \( (4y - 5x) - (x - y) \)
\( = 4y - 5x - x + y \)
\( = (-5x - x) + (4y + y) = -6x + 5y \).

(c) \( (2z - x - 3y) - (x - y + 3z) \)
\( = 2z - x - 3y - x + y - 3z \)
\( = (-x - x) + (-3y + y) + (2z - 3z) = -2x - 2y - z \).
In simple words: Change every plus to minus and every minus to plus in the subtracted group, then collect like terms.

Exam Tip: Always place the expression that comes after "from" first.

Question 3. Simplify the expressions and find the value when x=2,y= -1,
(a) \( (5x-9y)- (-7x+y) \)
(b) \( (7-2x+5y) - (5x+3y-7) \)
Answer:
(a) Simplify first:
\( 5x - 9y + 7x - y = 12x - 10y \).
Substitute \( x = 2 \) and \( y = -1 \):
\( 12(2) - 10(-1) = 24 + 10 = 34 \).

(b) Simplify first:
\( 7 - 2x + 5y - 5x - 3y + 7 = (-2x - 5x) + (5y - 3y) + (7 + 7) = -7x + 2y + 14 \).
Substitute \( x = 2 \) and \( y = -1 \):
\( -7(2) + 2(-1) + 14 = -14 - 2 + 14 = -2 \).
In simple words: Simplify the algebra by collecting terms first, then plug in the numbers.

Exam Tip: Always simplify algebraic expressions before substituting numerical values to minimize calculation errors.

Question 4. Find the value of each when a= -1,b= -2,c= 2:-
(a)a+ ab -c
(b) 2a- b/2 +3c
(c) 5abc - 2b -3c +bc
Answer:
(a) Substitute the values:
\( (-1) + (-1)(-2) - 2 = -1 + 2 - 2 = -1 \).

(b) Substitute the values:
\( 2(-1) - \frac{-2}{2} + 3(2) = -2 - (-1) + 6 = -2 + 1 + 6 = 5 \).

(c) Substitute the values:
\( 5(-1)(-2)(2) - 2(-2) - 3(2) + (-2)(2) \)
\( = 5(4) + 4 - 6 - 4 = 20 - 6 = 14 \).
In simple words: Replace every letter with its number in brackets, watching out for minus signs.

Exam Tip: Use parentheses around negative substitutions, such as \( (-1)(-2) \), to avoid sign errors.

Question 5. Find the factors of each expression-
(a)6x-8y+15z
(b)100-15abc-25x
Answer:
(a) The expression \( 6x - 8y + 15z \) has no common factor across all three terms other than 1. Its terms and individual factors are:
Term \( 6x \): factors are \( 2, 3, x \)
Term \( -8y \): factors are \( -1, 2, 2, 2, y \)
Term \( 15z \): factors are \( 3, 5, z \).

(b) Find the highest common factor for all terms in \( 100 - 15abc - 25x \):
The greatest common factor is 5.
Factoring out 5 gives:
\( 5(20 - 3abc - 5x) \).
The factors are 5 and \( (20 - 3abc - 5x) \).
In simple words: Look for a number that divides evenly into all terms and pull it outside brackets.

Exam Tip: If no common numerical or algebraic factor exists among all terms, list the factors of each term.

Question 6. (1) The expression -5x3yz2 is a
(a) Monomial
(b) Binomial
Answer: (a) Monomial
In simple words: An expression with only one single term is called a monomial.

Exam Tip: Terms are separated by plus or minus signs; since there are none here, it is a single term.

Question 6. (2) The coefficient of x in 4xy2z is
(a) 4
(b) 4y2
(c) 4yz
(d) 4y2z
Answer: (d) 4y2z
In simple words: Remove the letter x from the term; everything left behind is its coefficient.

Exam Tip: Unless "numerical coefficient" is specified, the coefficient includes all remaining factors multiplying x.

Question 6. (3) The sum of x3-y3 & y3-x3 is:-
(a) O
(b) 2x3-2y3
(c) 2y3-2x3
(d) x6-y6
Answer: (a) O
In simple words: Adding opposite terms cancels them out completely, leaving 0.

Exam Tip: \( (x^3 - y^3) + (y^3 - x^3) = x^3 - x^3 + y^3 - y^3 = 0 \).

Question 6. (4) The expression that should be added to (x2-xy+y2) to get (x2+xy+y2) is
(a) x2+2xy+y2
(b) xy
(c) 2xy
(d) 3xy
Answer: (c) 2xy
In simple words: Adding 2xy to minus xy gives plus xy.

Exam Tip: Subtract the starting expression from the final goal: \( (x^2+xy+y^2) - (x^2-xy+y^2) = 2xy \).

Question 6. (5) The value of the expression x3+y3 when x=1, y= -1:-
(a) -1
(b) 0
(c) -2
(d) 2
Answer: (b) 0
In simple words: 1 cubed is 1, and -1 cubed is -1. Adding them gives 0.

Exam Tip: Remember that \( (-1)^3 = -1 \), so \( 1 + (-1) = 0 \).

Question 7. Fill in the blanks:-
1) The algebraic expression -3x2y & y3 are _______ terms.
2) The factors of the term -4x2y are ________.
3) If x3 is subtracted from 2x3-y3, we get ________.
4) The perimeter of a quadrilateral whose sides measure a, a+b, a+b+c, a+b+c+d units is _______.
Answer:
1) unlike
2) \( -1, 2, 2, x, x, y \) (or \( -4, x, x, y \))
3) \( x^3 - y^3 \)
4) \( 4a + 3b + 2c + d \) units
In simple words: Terms with different variables are unlike. Adding all four boundary edges gives the total perimeter.

Exam Tip: To find the perimeter, simply add all side expressions together: \( a + (a+b) + (a+b+c) + (a+b+c+d) \).

Symmetry

Question 1. How many lines of symmetry does each of the following have?
(a)rhombus (b)parallelogram (c)equilateral triangle (d)scalene triangle
(e)isosceles trapezium (f)square (g)rectangle
Answer:
(a) rhombus: 2
(b) parallelogram: 0
(c) equilateral triangle: 3
(d) scalene triangle: 0
(e) isosceles trapezium: 1
(f) square: 4
(g) rectangle: 2
In simple words: A line of symmetry divides a shape into two identical mirror halves.

Exam Tip: Remember that a general parallelogram has rotational symmetry but zero lines of symmetry.

Question 2. Complete the table –
Sl. Shape Centre of rotation Order of Rotation Angle of Rotation
1 Square
2 Rectangle
3 Rhombus
4 Equilateral Triangle
5 Regular Hexagon
6 Circle
7 Semi Circle
Answer:

In simple words: The order of rotation tells how many times a figure fits into itself during a complete 360-degree spin.

Exam Tip: Use the formula \( \text{Angle of rotation} = \frac{360^\circ}{\text{Order of rotation}} \) for regular shapes.

Question 3. (1) A circle has:
(a) No line of symmetry.
(b) Infinite lines of symmetry.
(c) Four lines of symmetry.
(d) None of these.
Answer: (b) Infinite lines of symmetry.
In simple words: Any line passing through the centre cuts a circle into two balanced halves.

Exam Tip: Every diameter of a circle serves as an axis of symmetry.

Question 3. (2) Which of the following has two lines of symmetry?
(a) A square
(b) A parallelogram
(c) A kite
(d) A rectangle
Answer: (d) A rectangle
In simple words: A rectangle can be folded evenly along its middle length and its middle width.

Exam Tip: A square has 4 lines of symmetry, while a rectangle has only 2.

Question 3. (3) The number of times a figure matches itself after rotation is called:
(a) Axis of symmetry
(b) Rotational symmetry
(c) Order of symmetry
(d) Reflectional symmetry
Answer: (c) Order of symmetry
In simple words: Order of symmetry counts how often the shape looks identical while making a full turn.

Exam Tip: Order of rotation is always counted within one full turn of 360 degrees.

Question 3. (4) A rectangular pentagon has rotational symmetry of the order:
(a) 2
(b) 4
(c) 3
(d) 5
Answer: (d) 5
In simple words: A regular five-sided figure fits its outline five times in one complete rotation.

Exam Tip: A regular polygon with \( n \) sides always has rotational symmetry of order \( n \).

Question 3. (5) The letter which has linear symmetry as well as rotational symmetry is
(a) X
(b) Q
(c) P
(d) S
Answer: (a) X
In simple words: The letter X has reflection lines and also looks the same when turned upside down.

Exam Tip: The letter S has rotational symmetry but no line symmetry, while X has both.

Question 4. Fill in the blanks:
1) A line which divides a given figure into two equal halves is called _______.
2) A _______has four lines of symmetry.
3) A rhombus has _______lines of symmetry.
4) A _______ has rotational symmetry of the order of four.
5) An isosceles triangle has no _______symmetry.
Answer:
1) line of symmetry (or axis of symmetry)
2) square
3) 2
4) square
5) rotational
In simple words: Squares have four fold lines, while an isosceles triangle has only one fold line and no rotational match until 360 degrees.

Exam Tip: A figure must have order greater than 1 to be considered having rotational symmetry.

Question 5. True or False.
1) A hexagon has 4 lines of symmetry.
2) An equilateral triangle has exactly 3 lines of symmetry.
3) All the figures have rotational & linear symmetry.
Answer:
1) False. A regular hexagon has 6 lines of symmetry.
2) True. It has 3 lines of symmetry passing through each vertex and opposite midpoint.
3) False. Many shapes (like a scalene triangle) have neither line nor rotational symmetry.
In simple words: A regular hexagon has 6 fold lines, and not all shapes have symmetry.

Exam Tip: For any regular polygon, the number of lines of symmetry matches its total number of sides.

Question 6. Answer the following:
1) Explain the term “line of symmetry”.
2) What is the centre of rotation?
3) Which figure has only rotational symmetry but no line of symmetry?
4) Name a figure that has vertical & horizontal lines of symmetry but no rotational symmetry.
Answer:
1) A line of symmetry is an imaginary line that divides a geometric figure into two congruent parts, such that folding the figure along this line makes both halves match completely.
2) The centre of rotation is the fixed pivot point around which a shape rotates during rotational symmetry.
3) A general parallelogram has rotational symmetry of order 2, but has no lines of symmetry.
4) No standard closed polygon has both vertical and horizontal line symmetry without also having rotational symmetry. This is because reflecting a shape across two perpendicular lines is mathematically identical to a rotation of \( 180^\circ \).
In simple words: A symmetry line folds a figure evenly, and the centre of rotation is the fixed pivot around which it spins.

Exam Tip: State clearly that having both horizontal and vertical symmetry automatically creates rotational symmetry of order 2.

Exponents and Powers

Question 1. Express each using exponential notation –
a) A x A x A x B x B x Z x Z x Z
b) 27000
c) 12500
d) 288
Answer:
a) Count the factors: \( A \) appears 3 times, \( B \) appears 2 times, and \( Z \) appears 3 times.
Exponential form = \( A^3 B^2 Z^3 \).

b) Prime factorize 27000:
\( 27000 = 27 \times 1000 = 3^3 \times (2 \times 5)^3 = 2^3 \times 3^3 \times 5^3 \) (or \( 30^3 \)).

c) Prime factorize 12500:
\( 12500 = 125 \times 100 = 5^3 \times 2^2 \times 5^2 = 2^2 \times 5^5 \).

d) Prime factorize 288:
\( 288 = 32 \times 9 = 2^5 \times 3^2 \).
In simple words: Count how many times each prime number multiplies into the value and write it as an exponent.

Exam Tip: Write factors in ascending order of prime bases (\( 2, 3, 5 \)).

Question 2. Compare the following:
a) 34 and 43
b) 35 and 53
c) 26 and 62
Answer:
a) Evaluate both:
\( 3^4 = 3 \times 3 \times 3 \times 3 = 81 \)
\( 4^3 = 4 \times 4 \times 4 = 64 \)
Since \( 81 > 64 \), we have \( 3^4 > 4^3 \).

b) Evaluate both:
\( 3^5 = 243 \)
\( 5^3 = 125 \)
Since \( 243 > 125 \), we have \( 3^5 > 5^3 \).

c) Evaluate both:
\( 2^6 = 64 \)
\( 6^2 = 36 \)
Since \( 64 > 36 \), we have \( 2^6 > 6^2 \).
In simple words: Work out the full numerical value of each power to see which one is larger.

Exam Tip: Never multiply the base by the exponent (for example, \( 3^4 \neq 12 \)); expand it as repeated multiplication.

Question 3. Simplify each –
a) 23 x 34 x 4 / 32 x 32
b) (30+20) x 50
c) 28 x a5 / 43 x a3
d) 35 x 105 x 25 / 57 x 65
Answer:
a) Express all numbers in prime bases:
\( 4 = 2^2 \), \( 32 = 2^5 \).
\( \frac{2^3 \times 3^4 \times 2^2}{3^2 \times 2^5} = \frac{2^{3+2} \times 3^4}{2^5 \times 3^2} = \frac{2^5 \times 3^4}{2^5 \times 3^2} \)
\( = 2^{5-5} \times 3^{4-2} = 2^0 \times 3^2 = 1 \times 9 = 9 \).

b) Any non-zero base raised to the power 0 equals 1:
\( (3^0 + 2^0) \times 5^0 = (1 + 1) \times 1 = 2 \times 1 = 2 \).

c) Convert 4 to base 2: \( 4^3 = (2^2)^3 = 2^6 \).
\( \frac{2^8 \times a^5}{2^6 \times a^3} = 2^{8-6} \times a^{5-3} = 2^2 \times a^2 = 4a^2 \).

d) Break into prime factors: \( 10 = 2 \times 5 \), \( 25 = 5^2 \), \( 6 = 2 \times 3 \).
\( \frac{3^5 \times (2 \times 5)^5 \times 5^2}{5^7 \times (2 \times 3)^5} = \frac{3^5 \times 2^5 \times 5^5 \times 5^2}{5^7 \times 2^5 \times 3^5} = \frac{2^5 \times 3^5 \times 5^7}{2^5 \times 3^5 \times 5^7} = 1 \).
In simple words: Break numbers into prime bases, add powers when multiplying, and subtract powers when dividing.

Exam Tip: Remember that \( a^0 = 1 \) for any non-zero value of \( a \).

Question 4. Write each in scientific notation –
a) 0.00000000017 b) 384000000
c) 1908.781 d) 0.000006532
Answer:
a) Shift the decimal 10 places to the right:
\( 0.00000000017 = 1.7 \times 10^{-10} \).

b) Shift the decimal 8 places to the left:
\( 384000000 = 3.84 \times 10^8 \).

c) Shift the decimal 3 places to the left:
\( 1908.781 = 1.908781 \times 10^3 \).

d) Shift the decimal 6 places to the right:
\( 0.000006532 = 6.532 \times 10^{-6} \).
In simple words: Place the decimal right after the first non-zero digit and count how many spots it moved.

Exam Tip: In scientific notation \( m \times 10^n \), the coefficient \( m \) must satisfy \( 1 \le m < 10 \).

Question 5. The value of (530 x 510) (57)5 in exponential form is _______________
Answer: Using exponent rules:
\( 5^{30} \times 5^{10} = 5^{30+10} = 5^{40} \).
\( (5^7)^5 = 5^{7 \times 5} = 5^{35} \).
Multiplying both together:
\( 5^{40} \times 5^{35} = 5^{40+35} = 5^{75} \).
(Note: If interpreted as division, \( 5^{40} \div 5^{35} = 5^5 \)).
In simple words: Add exponents with the same base and multiply exponents when one power is raised to another.

Exam Tip: Remember that \( (a^m)^n = a^{mn} \) and \( a^m \times a^n = a^{m+n} \).

Question 6. (I) In power notation (-1/5) × (-1/5) × (-1/5) × (-1/5) can be written as
(a) (-5)4
(b) (-1/5)4
(c) (-1/3)3
(d) -(1/5)4
Answer: (b) (-1/5)4
In simple words: The fraction minus 1 over 5 is multiplied 4 times, so it has power 4 inside brackets.

Exam Tip: Include the negative sign inside the parenthesis so the power applies to the entire fraction.

Question 6. (II) (-7/15)1 = _______.
(a) (-15/7)
(b) (-7/15)
(c) 7/15
(d) 15/7
Answer: (b) (-7/15)
In simple words: Any quantity raised to the power 1 remains unchanged.

Exam Tip: Recall that \( a^1 = a \) for every number.

Question 6. (III) (-1)53 = _______.
(a) 1
(b) -53
(c) -1
(d) 0
Answer: (c) -1
In simple words: Negative one raised to any odd power always equals minus one.

Exam Tip: Check whether the exponent is even or odd: 53 is odd, so the result is \( -1 \).

Question 6. (IV) The exponential form of -343 / 1331 is:
(a) (7/11)2
(b) (-7/11)3
(c) (7/11)3
(d) -(7/121)2
Answer: (b) (-7/11)3
In simple words: 343 is 7 cubed and 1331 is 11 cubed, with an odd power keeping the minus sign.

Exam Tip: Verify that \( (-7)^3 = -343 \) and \( 11^3 = 1331 \).

Question 6. (V) 40,000,000,000 in standard form is
(a) 4 × 106
(b) 4 × 1010
(c) 4 × 108
(d) None of these
Answer: (b) 4 × 1010
In simple words: Counting the 10 zeros following 4 gives 4 times 10 to the power 10.

Exam Tip: Count the number of places the decimal point shifts to the left from the end of the number.

Question 7. Fill in the blanks:
a) (23)5 = _______.
b) 22÷32 = _______.
c) Cube of (- 1/3 ) is _______.
d) ( 7/11 )5 ÷ ( 7/11 )3 = ( 7/11 )___
e) [(10)10]10 = _______.
Answer:
a) \( 2^{15} \)
b) \( \left(\frac{2}{3}\right)^2 \) (or \( \frac{4}{9} \))
c) \( -\frac{1}{27} \)
d) 2
e) \( 10^{100} \)
In simple words: Multiply powers when nested, subtract powers when dividing identical bases, and cube top and bottom numbers.

Exam Tip: When raising a power to another power, multiply the exponents: \( 10 \times 10 = 100 \).

Question 8. True or False.
a) The value of a1 is 1.
b) The exponential form of 32/243 is (2/3)5.
c) (xm)n = xmn.
d) The reciprocal of 6 is 6-1.
e) (- 5/3 )3 is equal to (- 5/3 ) × (- 5/3 ) × (- 5/3 ).
Answer:
a) False. \( a^1 = a \), not 1.
b) True. \( 2^5 = 32 \) and \( 3^5 = 243 \), so \( \frac{32}{243} = \left(\frac{2}{3}\right)^5 \).
c) True. This is the power of a power law.
d) True. \( \frac{1}{6} = 6^{-1} \).
e) True. Cubing means multiplying the base three times.
In simple words: Any number to power 1 is itself, while negative power means reciprocal.

Exam Tip: Be careful not to confuse \( a^1 = a \) with \( a^0 = 1 \).

Question 9. Solve the following:-
a) Simplify:- (0.3)2× 73× (−2)2
b) Find the value of (-1)100× (−1)61× (1)9
c) Express 6752.25 in scientific notation.
d) Evaluate ( 2/3 )3 & ( 3/2 )3. Which is smaller of the two?
e) Write 0.0000537 in exponential form.
Answer:
a) Evaluate each part:
\( (0.3)^2 = 0.09 \)
\( 7^3 = 343 \)
\( (-2)^2 = 4 \)
\( 0.09 \times 343 \times 4 = 0.09 \times 1372 = 123.48 \).

b) Evaluate powers of \( -1 \) and 1:
\( (-1)^{100} = 1 \) (even power)
\( (-1)^{61} = -1 \) (odd power)
\( (1)^9 = 1 \)
Product = \( 1 \times (-1) \times 1 = -1 \).

c) Shift the decimal 3 places to the left:
\( 6752.25 = 6.75225 \times 10^3 \).

d) Calculate both:
\( \left(\frac{2}{3}\right)^3 = \frac{8}{27} \approx 0.296 \)
\( \left(\frac{3}{2}\right)^3 = \frac{27}{8} = 3.375 \)
Since \( \frac{8}{27} < \frac{27}{8} \), \( \left(\frac{2}{3}\right)^3 \) is smaller.

e) Shift the decimal 5 places to the right:
\( 0.0000537 = 5.37 \times 10^{-5} \).
In simple words: Work out each power separately and multiply. Proper fractions get smaller when cubed, while improper fractions get larger.

Exam Tip: For any proper fraction \( \frac{a}{b} < 1 \), raising it to a positive power always produces a value smaller than 1.

Visualising Solid Shapes

Question 1. Write the no. of edges, faces and vertices for each –
a) cube b) cuboid c) cylinder d) cone
e) triangular prism f) triangular g) square pyramid h) hexagonal pyramid
Answer:
a) Cube: Faces = 6, Vertices = 8, Edges = 12.
b) Cuboid: Faces = 6, Vertices = 8, Edges = 12.
c) Cylinder: Faces = 3 (2 flat circular + 1 curved), Vertices = 0, Edges = 2 (circular).
d) Cone: Faces = 2 (1 flat circular + 1 curved), Vertices = 1, Edges = 1 (circular).
e) Triangular prism: Faces = 5, Vertices = 6, Edges = 9.
f) Triangular pyramid (Tetrahedron): Faces = 4, Vertices = 4, Edges = 6.
g) Square pyramid: Faces = 5, Vertices = 5, Edges = 8.
h) Hexagonal pyramid: Faces = 7, Vertices = 7, Edges = 12.
In simple words: Faces are flat or curved surfaces, edges are the meeting lines, and vertices are the sharp corner points.

Exam Tip: For polyhedrons, you can verify your answer using Euler's formula: \( F + V - E = 2 \).

Question 2. The dimensions of a cuboid are 6cm, 4cm and 2cm respectively. Draw 3 different isometic sketches of the cuboid.
Answer: A cuboid can be oriented in three distinct ways on isometric dot paper by varying which dimension represents the length, breadth, and height:
1. Orientation 1: Length = 6 cm, Breadth = 4 cm, Height = 2 cm.
2. Orientation 2: Length = 6 cm, Breadth = 2 cm, Height = 4 cm.
3. Orientation 3: Length = 4 cm, Breadth = 2 cm, Height = 6 cm.
On isometric dot paper, lines along the three isometric axes are drawn matching the chosen lengths, with vertical lines remaining vertical and receding edges drawn along the \( 30^\circ \) isometric axes.
In simple words: Draw the cuboid resting on its three different faces: flat like a book, on its edge, or standing upright like a tower.

Exam Tip: In isometric sketches, edges parallel to the axes must match the exact given measurements.

Question 3. Four cubes of 2cm edge are placed side by side to form a cuboid. Sketch an oblique as well as an isometic sketch of the resulting cuboid.
Answer: When four cubes each of edge 2 cm are placed in a row side by side, the dimensions of the resulting cuboid become:
Length = \( 4 \times 2\text{ cm} = 8\text{ cm} \), Breadth = \( 2\text{ cm} \), Height = \( 2\text{ cm} \).

Oblique Sketch: Draw the front face as a rectangle of 8 cm by 2 cm with true angles of \( 90^\circ \). Draw receding edges from the vertices at an angle of \( 45^\circ \) with reduced length (about 1 cm) to show depth, and connect them.

Isometric Sketch: On isometric dot paper, draw a vertical edge of 2 units for height, receding edges of 8 units along one \( 30^\circ \) axis for length, and 2 units along the other \( 30^\circ \) axis for breadth, then complete the 3D box.
In simple words: Four cubes placed together make a long box that is 8 cm long, 2 cm wide, and 2 cm tall.

Exam Tip: In an oblique sketch, the front face shows true shape and dimensions, whereas in an isometric sketch, measurements are drawn along isometric axes.

Question 4. Find the number of edges in a polyhedron having 8 faces and 12 vertices.
Answer: Given Faces (\( F \)) = 8 and Vertices (\( V \)) = 12.
Using Euler's formula for any polyhedron:
\( F + V - E = 2 \)
\( 8 + 12 - E = 2 \)
\( 20 - E = 2 \)
\( \implies E = 20 - 2 = 18 \).
Therefore, the polyhedron has 18 edges.
In simple words: Add faces and vertices to get 20, then subtract 2 to find the number of edges.

Exam Tip: State Euler's formula \( F + V - E = 2 \) before substituting the numbers.

Question 5. What cross-section is made in the following solids by: i) Vertical cut ii) Horizontal cut
a) a lemon b) a brick c) a tree trunk
d) a die e) an egyption Pyramid
Answer:
a) A lemon:
i) Vertical cut: Circle
ii) Horizontal cut: Circle

b) A brick:
i) Vertical cut: Rectangle
ii) Horizontal cut: Rectangle

c) A tree trunk:
i) Vertical cut: Rectangle
ii) Horizontal cut: Circle

d) A die:
i) Vertical cut: Square
ii) Horizontal cut: Square

e) An Egyptian Pyramid (Square Pyramid):
i) Vertical cut: Triangle
ii) Horizontal cut: Square
In simple words: A cross-section is the flat shape you see on the cut surface when you slice through a solid.

Exam Tip: Slicing an Egyptian pyramid vertically through its top peak gives a triangle, while slicing horizontally gives a square.

Question 6. (I) An isometric sheet is made up of dots forming:
(a) Right angled triangles
(b) Equilateral triangles
(c) Squares
(d) Rectangles
Answer: (b) Equilateral triangles
In simple words: Dots on isometric paper are spaced so any three nearby dots form an equilateral triangle.

Exam Tip: The dots on isometric dot paper form small equilateral triangles with angles of \( 60^\circ \).

Question 6. (II) The number of edges in a polyhedron having 8 faces & 12 vertices is:
(a) 14
(b) 16
(c) 18
(d) 20
Answer: (c) 18
In simple words: Using Euler's rule, 8 plus 12 minus 18 equals 2.

Exam Tip: Apply \( E = F + V - 2 = 8 + 12 - 2 = 18 \).

Question 6. (III) The top view of the solid shown in the figure below is:
(a) [Two squares in a row]
(b) [Three squares in a row]
(c) [T-shaped four squares]
(d) [Six vertical squares]
Answer: (b) [Three squares in a row]
In simple words: Looking straight down from above shows the three cubes lined up one after another in a row.

Exam Tip: Ignore the vertical height of cubes when drawing or selecting the top view.

Question 6. (IV) A horizontal cross-section of the solid shown in the adjoining fig. is a:
(a) Triangle
(b) Square
(c) Hexagon
(d) Octagon
Answer: (c) Hexagon
In simple words: Slicing horizontally through a pyramid with a 6-sided base creates a hexagon.

Exam Tip: A horizontal cut parallel to the base of any pyramid always matches the shape of its base polygon.

Question 6. (V) The solid having only one vertex is a:
(a) Prism
(b) Pyramid
(c) Cylinder
(d) Cone
Answer: (d) Cone
In simple words: A cone has a circular base and comes to a single sharp point at the top.

Exam Tip: A cone has exactly 1 vertex (apex), while a cylinder has 0 vertices.

Question Q2. True or False.
a) The shadow of a 3-D object is 2-D.
b) The net of a solid is unique.
c) A cylinder has 2 faces.
d) The front, side & top view of a sphere are different.
e) The given net is of a closed cone.
Answer:
a) True. A shadow cast on a flat screen or surface is two-dimensional.
b) False. A solid can have several different valid nets (for instance, a cube has 11 nets).
c) False. A cylinder has 3 faces: two flat circular bases and one curved surface.
d) False. All views of a sphere from any direction look like identical circles.
e) False. The net shows only a circular sector (lateral surface) without the circular base, so it forms an open cone.
In simple words: Shadows are flat, solids can unfold in different ways, and spheres look circular from all sides.

Exam Tip: A net for a closed cone must include both the curved sector and the circular base disk.

Question Q3. Fill in the blanks:
a) The base of a pyramid is a _______ & its faces are _______ having a common vertex.
b) The two faces of a solid meet to form an _______.
c) A point where three or more faces meet is called a _______.
d) A solid that has congruent & parallel polygons as top & base joined by rectangular faces is known as________.
e) A pyramid having 4 equilateral triangles as its faces is called a _______.
Answer:
a) polygon, triangles
b) edge
c) vertex
d) prism
e) regular tetrahedron
In simple words: Pyramids have triangular sides that meet at a point, while prisms have matching top and bottom shapes.

Exam Tip: A triangular pyramid whose four faces are all congruent equilateral triangles is called a regular tetrahedron.

Construction

Question Q1. (I) The number of parallel lines that can be drawn from a point outside the line to it is:
(a) 0
(b) 1
(c) 2
(d) infinite
Answer: (b) 1
In simple words: Through any point outside a line, exactly one line can run parallel to it.

Exam Tip: This is Playfair's axiom, a fundamental property of parallel lines.

Question Q1. (II) A triangle which is impossible to construct has sides measuring:-
(a) 3cm,4cm,5cm
(b) 7cm,8cm,9cm
(c) 4cm,6cm,10cm
(d) 10cm,18cm,12cm
Answer: (c) 4cm,6cm,10cm
In simple words: The two shorter sides add up to 10, which is not strictly greater than 10.

Exam Tip: By triangle inequality, the sum of any two sides must be strictly greater than the third side (\( 4 + 6 = 10 \), not \( > 10 \)).

Question Q1. (III) A unique triangle cannot be constructed with which of the following sets of measurements?
(a) 3 sides
(b) 3 angles
(c) 2 angles & an included side
(d) the hypotenuse & a leg
Answer: (b) 3 angles
In simple words: Knowing only three angles fixes the shape but allows the triangle to be any size.

Exam Tip: AAA criterion produces similar triangles of different sizes, not a unique triangle.

Question Q2. True or False.
a) An acute angle & the length of hypotenuse is sufficient to construct a right triangle.
b) In a triangle the angle opposite the largest side is the smallest.
c) A ∆ABC whose dimensions are shown in the given figure can be uniquely constructed, i.e. SSA construction criterion is valid.
Answer:
a) True. Knowing one acute angle gives the other acute angle (\( 90^\circ - \text{angle} \)), and the hypotenuse fixes the scale uniquely.
b) False. The angle opposite the largest side is always the largest angle.
c) False. SSA is not a valid congruence condition; an angle, an adjacent side, and an opposite side can create two different triangles or none.
In simple words: The largest side always faces the largest angle, and SSA does not guarantee one unique triangle.

Exam Tip: SSA is known as the ambiguous case in geometry and cannot guarantee unique construction.

Question Q3. Fill in the blanks:-
a) The _______angle of a triangle is equal to the sum of the interior opposite angles.
b) In a right triangle, the square of the _______ is equal to the ________ of the squares of the other two sides.
Answer:
a) exterior
b) hypotenuse, sum
In simple words: An outside angle equals the sum of two opposite inside angles, and Pythagoras rule relates the squared sides.

Exam Tip: Memorize the Pythagoras theorem statement word for word: \( (\text{Hypotenuse})^2 = (\text{Base})^2 + (\text{Perpendicular})^2 \).

Question Q4. Draw any line l. Draw another line m parallel to l at a distance of 6.6cm from it.
Answer: Steps of Construction:
1. Draw a straight line \( l \) horizontally and choose any point P on it.
2. At point P, construct a line perpendicular to line \( l \) using a compass.
3. With P as centre and radius \( 6.6\text{ cm} \), cut an arc on the perpendicular line at point Q.
4. At point Q, construct a line \( m \) perpendicular to segment PQ.
Line \( m \) is the required line parallel to line \( l \) at a distance of \( 6.6\text{ cm} \). In simple words: Draw a straight line, make a 90-degree line upwards measuring 6.6 cm, and draw a new perpendicular line across the top.

Exam Tip: Since two lines perpendicular to the same transversal are parallel, constructing right angles at both P and Q guarantees parallelism.

Question Q5. Construct a ∆ABC in which:-
a) BC = 5.5cm, AB = 5.7cm, AC = 6cm
b) AB =6cm, AC = 6.5cm & ∠BAC = 60°.
c) BC = 6.5cm, ∠C = 45º & ∠B= 60º.
Answer:
(a) SSS Criterion:
1. Draw a line segment \( \text{BC} = 5.5\text{ cm} \).
2. With centre B and compass set to \( 5.7\text{ cm} \), draw an arc.
3. With centre C and compass set to \( 6\text{ cm} \), draw an arc intersecting the previous arc at A.
4. Join AB and AC. \( \Delta \text{ABC} \) is constructed.

(b) SAS Criterion:
1. Draw a line segment \( \text{AB} = 6\text{ cm} \).
2. At point A, construct ray AX making an angle of \( 60^\circ \) with AB.
3. With centre A and radius \( 6.5\text{ cm} \), cut an arc on ray AX at point C.
4. Join BC. \( \Delta \text{ABC} \) is constructed.

(c) ASA Criterion:
1. Draw a line segment \( \text{BC} = 6.5\text{ cm} \).
2. At point B, draw ray BY making an angle of \( 60^\circ \) with BC.
3. At point C, draw ray CZ making an angle of \( 45^\circ \) with CB.
4. Let rays BY and CZ intersect at point A. \( \Delta \text{ABC} \) is constructed.
In simple words: Use compass arcs for three sides, draw angle rays between two given sides, or build angles at both base ends to meet at the top.

Exam Tip: Clearly label the criteria used: SSS for part (a), SAS for part (b), and ASA for part (c).

Question Q6. Construct a right angled triangle having hypotenuse 7cm & one of its acute angles measuring 45º.
Answer: In a right-angled triangle, if one acute angle is \( 45^\circ \), the other acute angle is:
\( 180^\circ - (90^\circ + 45^\circ) = 45^\circ \).
Since both acute angles are \( 45^\circ \), this is an isosceles right-angled triangle.

Steps of Construction:
1. Draw a line segment \( \text{AB} = 7\text{ cm} \) representing the hypotenuse.
2. At point A, construct ray AX making an angle of \( 45^\circ \) with AB.
3. At point B, construct ray BY making an angle of \( 45^\circ \) with BA on the same side of AB.
4. The two rays AX and BY intersect at point C.
5. The angle at C will naturally be \( 180^\circ - (45^\circ + 45^\circ) = 90^\circ \).
\( \Delta \text{ABC} \) is the required right-angled triangle. In simple words: Draw a 7 cm base, build 45-degree angles on both corners, and let the rays cross at the 90-degree corner at the top.

Exam Tip: Mentioning that both base angles are \( 45^\circ \) justifies using the ASA construction method on the hypotenuse.