CBSE Class 7 Mathematics Practice Assignment Set 02

COMPARING QUANTITIES

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 population a year ago be \(x\).
Annual increase = \(8\%\)
Present population = \(x + \frac{8}{100}x = \frac{108}{100}x\)
Given present population = 142560
\(\frac{108}{100}x = 142560\)
\(x = \frac{142560 \times 100}{108} = 132000\)
Therefore, the population a year ago was 132000.
In simple words: We find the original population by reversing an 8% increase from the current population.

Exam Tip: Always set up the equation as \(\text{Original} \times (1 + \frac{\text{Rate}}{100}) = \text{Present}\) for population growth problems.

Question 2. Find-(a)7.5% of 80 kg (b)8% of 5 litres (c)0.25% of 16
Answer: (a) \(\frac{7.5}{100} \times 80 = \frac{7.5 \times 80}{100} = \frac{600}{100} = 6\) kg
(b) \(\frac{8}{100} \times 5 = \frac{40}{100} = 0.4\) litres (or 400 ml)
(c) \(\frac{0.25}{100} \times 16 = \frac{4}{100} = 0.04\)
In simple words: Convert the percentage to a fraction or decimal and multiply it by the given quantity.

Exam Tip: Remember that percent means "per hundred", so divide the percentage value by 100 before multiplying.

Question 3. Calculate what % of-(a)40m is 16m (b)1 kg is 25 gram
Answer: (a) Required percentage = \(\frac{16}{40} \times 100 = \frac{1600}{40} = 40\%\)
(b) First convert 1 kg to grams: 1 kg = 1000 gram.
Required percentage = \(\frac{25}{1000} \times 100 = \frac{25}{10} = 2.5\%\)
In simple words: Put the part in the numerator, the whole in the denominator, ensure units are the same, and multiply by 100.

Exam Tip: Always make sure both quantities are in the exact same units before calculating percentages.

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: Cost price of stereo = Rs. 5863
Repair cost = Rs. 137
Total Cost Price (C.P.) = \(5863 + 137 = \text{Rs. } 6000\)
Selling Price (S.P.) = Rs. 5700
Since C.P. > S.P., there is a loss.
Loss = \(\text{C.P.} - \text{S.P.} = 6000 - 5700 = \text{Rs. } 300\)
Loss % = \(\frac{\text{Loss}}{\text{C.P.}} \times 100 = \frac{300}{6000} \times 100 = \frac{300}{60} = 5\%\)
In simple words: Add repair charges to the purchase price to find the total cost, then calculate the loss percentage.

Exam Tip: Overhead charges like repair costs are always added to the purchase price to find the actual cost price.

Question 5. By selling a chair for Rs 720, a trader gains 20%. Find the cost price of the chair.
Answer: Selling Price (S.P.) = Rs. 720
Gain % = \(20\%\)
Let the Cost Price be \(x\).
\(\text{S.P.} = \text{C.P.} \times \left(1 + \frac{\text{Gain \%}}{100}\right)\)
\(720 = x \times \left(1 + \frac{20}{100}\right)\)
\(720 = x \times \frac{120}{100}\)
\(x = \frac{720 \times 100}{120} = 6 \times 100 = \text{Rs. } 600\)
Therefore, the cost price of the chair is Rs. 600.
In simple words: Use the formula relating selling price, cost price, and profit percentage to work backwards.

Exam Tip: When profit is given, S.P. is greater than C.P., so the multiplier is greater than 1.

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 (C.P.) be \(x\).
Gain = \(\frac{1}{6}\) of C.P. = \(\frac{x}{6}\)
Selling Price (S.P.) = \(\text{C.P.} + \text{Gain} = x + \frac{x}{6} = \frac{7x}{6}\)
Given S.P. = Rs. 322
\(\frac{7x}{6} = 322\)
\(x = \frac{322 \times 6}{7} = 46 \times 6 = \text{Rs. } 276\)
Gain = \(\frac{276}{6} = \text{Rs. } 46\)
Gain % = \(\frac{\text{Gain}}{\text{C.P.}} \times 100 = \frac{46}{276} \times 100 = \frac{1}{6} \times 100 = 16\frac{2}{3}\%\) (or \(16.67\%\))
In simple words: Express profit as a fraction of the cost price to find both the cost price and the profit percentage.

Exam Tip: Gain percentage is always calculated with respect to the cost price, never the selling price.

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
Time (\(T\)) = 2 years 6 months = \(2 + \frac{6}{12}\) years = \(2.5\) years (or \(\frac{5}{2}\) years)
Rate (\(R\)) = \(6\%\) per annum
Simple Interest (\(S.I.\)) = \(\frac{P \times R \times T}{100}\)
\(S.I. = \frac{2500 \times 6 \times 5}{100 \times 2} = 25 \times 3 \times 5 = \text{Rs. } 375\)
Amount (\(A\)) = \(P + S.I. = 2500 + 375 = \text{Rs. } 2875\)
In simple words: Convert the time into years, apply the simple interest formula, and add it to the principal to get the total amount.

Exam Tip: Always convert months into years by dividing by 12 before substituting into the simple interest formula.

Question 8. What sum will amount to Rs 4590 at 12% per annum simple interest in 3 years?
Answer: Amount (\(A\)) = Rs. 4590
Rate (\(R\)) = \(12\%\) per annum
Time (\(T\)) = 3 years
Let the Principal be \(P\).
\(S.I. = \frac{P \times 12 \times 3}{100} = \frac{36P}{100} = \frac{9P}{25}\)
\(\text{Amount} = P + S.I.\)
\(4590 = P + \frac{9P}{25}\)
\(4590 = \frac{34P}{25}\)
\(P = \frac{4590 \times 25}{34} = 135 \times 25 = \text{Rs. } 3375\)
Therefore, the sum is Rs. 3375.
In simple words: Express simple interest in terms of the principal, set up the amount equation, and solve for the principal.

Exam Tip: Amount is always the sum of Principal and Interest (\(A = P + SI\)).

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 years
Simple Interest (\(S.I.\)) = \(A - P = 2046 - 1650 = \text{Rs. } 396\)
\(R = \frac{S.I. \times 100}{P \times T}\)
\(R = \frac{396 \times 100}{1650 \times 3} = \frac{39600}{4950} = 8\%\)
Therefore, the rate of interest is \(8\%\) per annum.
In simple words: Find the total interest earned by subtracting principal from amount, then use the rate formula.

Exam Tip: Always find the interest first by subtracting the principal from the final amount.

Question 10. What is 100% of 80?
Answer: \(\frac{100}{100} \times 80 = 1 \times 80 = 80\)
In simple words: 100 percent of any number is the number itself.

Exam Tip: 100% represents a complete whole.

Question 11. The simplest form of 3 ½ : 10 ½ = _______.
Answer: \(3\frac{1}{2} : 10\frac{1}{2} = \frac{7}{2} : \frac{21}{2} = 7 : 21 = 1 : 3\)
In simple words: Convert mixed fractions to improper fractions and simplify the ratio by dividing by the common factor.

Exam Tip: Ratios have no units and should always be expressed in their lowest terms.

Question 12. What is 7:10 equal to in percent?
Answer: \(\frac{7}{10} \times 100\% = 7 \times 10\% = 70\%\)
In simple words: Multiply the given ratio fraction by 100 to convert it into a percentage.

Exam Tip: To convert any fraction or ratio to a percentage, multiply by 100.

Question 13. In each case of profit, which is more S.P or C.P?
Answer: S.P (Selling Price) is more than C.P (Cost Price).
In simple words: Profit happens when you sell an item for more than you bought it.

Exam Tip: Profit = S.P - C.P, which means S.P must be greater than C.P.

Question 14. Gain or loss percent is calculated on C.P or S.P?
Answer: C.P (Cost Price).
In simple words: Profit or loss is always judged against the original cost price.

Exam Tip: Unless explicitly stated otherwise in a special problem, profit and loss percentages are always calculated on the Cost Price.

Question 15. What is 1% of 100?
Answer: \(\frac{1}{100} \times 100 = 1\)
In simple words: One percent of 100 is simply 1.

Exam Tip: 1% of 100 is the base unit definition of percentage.

Question 16. What is 0.35 equivalent to in percent?
Answer: \(0.35 \times 100\% = 35\%\)
In simple words: Multiply the decimal number by 100 to get its percentage value.

Exam Tip: Shift the decimal point two places to the right when converting decimals to percentages.

Question 17. What is the equivalent fraction for 20%?
Answer: \(\frac{20}{100} = \frac{1}{5}\)
In simple words: Write the percentage as a fraction over 100 and simplify it.

Exam Tip: Always reduce fractions to their simplest form.

Question 18. What is the number whose 8% is 20?
Answer: Let the number be \(x\).
\(\frac{8}{100} \times x = 20\)
\(x = \frac{20 \times 100}{8} = \frac{2000}{8} = 250\)
Therefore, the number is 250.
In simple words: Set up an algebraic equation where 8 percent of an unknown number equals 20.

Exam Tip: Translate "is" to '=' and "of" to '\(\times\)' to easily form equations.

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 (\(R\)) = \(8\%\) per annum
Time (\(T\)) = 3 years 4 months = \(3 + \frac{4}{12}\) years = \(3 + \frac{1}{3} = \frac{10}{3}\) years
Simple Interest (\(S.I.\)) = \(\frac{P \times R \times T}{100}\)
\(S.I. = \frac{5250 \times 8 \times 10}{100 \times 3} = \frac{420000}{300} = \frac{4200}{3} = \text{Rs. } 1400\)
In simple words: Convert 4 months into a fraction of a year (\(\frac{1}{3}\)), combine with 3 years, and apply the simple interest formula.

Exam Tip: Always convert mixed time units entirely into years before calculation.

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) Given rational numbers: \(-\frac{2}{5}, \frac{7}{10}, \frac{11}{15}, -\frac{19}{30}\) (shifting negative sign to numerator)
Find the LCM of denominators (5, 10, 15, 30), which is 30.
Convert each fraction to have denominator 30:
\(-\frac{2}{5} = \frac{-2 \times 6}{5 \times 6} = -\frac{12}{30}\)
\(\frac{7}{10} = \frac{7 \times 3}{10 \times 3} = \frac{21}{30}\)
\(\frac{11}{15} = \frac{11 \times 2}{15 \times 2} = \frac{22}{30}\)
\(-\frac{19}{30} = -\frac{19}{30}\)
Comparing numerators: \(22 > 21 > -12 > -19\)
Descending order: \(\frac{22}{30}, \frac{21}{30}, -\frac{12}{30}, -\frac{19}{30}\)
Final Answer (a): \(\frac{11}{15}, \frac{7}{10}, -\frac{2}{5}, \frac{19}{-30}\)

(b) Given rational numbers: \(\frac{3}{4}, \frac{5}{12}, \frac{7}{16}, \frac{9}{24}\)
Find the LCM of denominators (4, 12, 16, 24), which is 48.
Convert each fraction to have denominator 48:
\(\frac{3}{4} = \frac{3 \times 12}{4 \times 12} = \frac{36}{48}\)
\(\frac{5}{12} = \frac{5 \times 4}{12 \times 4} = \frac{20}{48}\)
\(\frac{7}{16} = \frac{7 \times 3}{16 \times 3} = \frac{21}{48}\)
\(\frac{9}{24} = \frac{9 \times 2}{24 \times 2} = \frac{18}{48}\)
Comparing numerators: \(36 > 21 > 20 > 18\)
Descending order: \(\frac{36}{48}, \frac{21}{48}, \frac{20}{48}, \frac{18}{48}\)
Final Answer (b): \(\frac{3}{4}, \frac{7}{16}, \frac{5}{12}, \frac{9}{24}\)
In simple words: Make all denominators equal using LCM, then arrange the numbers from largest to smallest based on their numerators.

Exam Tip: Always make the denominators positive first by shifting negative signs to the numerator.

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

(b) \(\frac{9}{-13} = -\frac{9}{13}\) and \(\frac{7}{-12} = -\frac{7}{12}\)
Find LCM of 13 and 12, which is 156.
\(-\frac{9}{13} = -\frac{108}{156}\)
\(-\frac{7}{12} = -\frac{91}{156}\)
Since \(-108 < -91\), \(\frac{9}{-13} < \frac{7}{-12}\)

(c) \(\frac{3}{7}\) and \(\frac{4}{5}\)
Cross-multiply: \(3 \times 5 = 15\) and \(4 \times 7 = 28\)
Since \(15 < 28\), \(\frac{3}{7} < \frac{4}{5}\)
In simple words: Cross-multiply the two fractions or make their denominators equal to compare them easily.

Exam Tip: Cross-multiplication is the fastest way to compare two fractions with different denominators.

Question 3. Enlist any 5 rational no. between- (a) \(\frac{2}{5}\) and \(\frac{4}{9}\) (b) \(-\frac{3}{4}\) and \(-\frac{5}{8}\)
Answer: (a) Given numbers: \(\frac{2}{5}\) and \(\frac{4}{9}\)
Make denominators equal using LCM of 5 and 9, which is 45:
\(\frac{2}{5} = \frac{18}{45}\)
\(\frac{4}{9} = \frac{20}{45}\)
Since the gap between 18 and 20 is small, multiply numerator and denominator by 10 (or 6):
\(\frac{18 \times 10}{45 \times 10} = \frac{180}{450}\)
\(\frac{20 \times 10}{45 \times 10} = \frac{200}{450}\)
5 rational numbers between them: \(\frac{181}{450}, \frac{182}{450}, \frac{183}{450}, \frac{184}{450}, \frac{185}{450}\)

(b) Given numbers: \(-\frac{3}{4}\) and \(-\frac{5}{8}\)
Make denominators equal using LCM of 4 and 8, which is 8:
\(-\frac{3}{4} = -\frac{6}{8}\)
\(-\frac{5}{8} = -\frac{5}{8}\)
Multiply numerator and denominator by 6 to get a bigger gap:
\(-\frac{6 \times 6}{8 \times 6} = -\frac{36}{48}\)
\(-\frac{5 \times 6}{8 \times 6} = -\frac{30}{48}\)
5 rational numbers between them: \(-\frac{35}{48}, -\frac{34}{48}, -\frac{33}{48}, -\frac{32}{48}, -\frac{31}{48}\)
In simple words: Make denominators equal, then multiply both numbers by a larger power of 10 to create enough space between their numerators.

Exam Tip: There are infinitely many rational numbers between any two rational numbers.

Question 4. Represent each using a no.line-(a) \(-\frac{1}{3}\) (b) \(-\frac{7}{6}\) (c) \(\frac{13}{6}\)
Answer: (a) For \(-\frac{1}{3}\): Divide the segment between 0 and -1 on the number line into 3 equal parts. The first division to the left of 0 represents \(-\frac{1}{3}\).
(b) For \(-\frac{7}{6}\): Since \(-\frac{7}{6} = -1\frac{1}{6}\), it lies between -1 and -2. Divide the segment between -1 and -2 into 6 equal parts. Mark the first division to the left of -1.
(c) For \(\frac{13}{6}\): Since \(\frac{13}{6} = 2\frac{1}{6}\), it lies between 2 and 3. Divide the segment between 2 and 3 into 6 equal parts. Mark the first division to the right of 2.
In simple words: Convert improper fractions to mixed numbers to locate the whole number boundaries, then divide the fractional unit segment accordingly.

Exam Tip: The denominator tells you how many equal parts to divide each unit interval into.

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

(b) \(-\frac{8}{9} + \left(-\frac{5}{3}\right)\)
LCM of 9 and 3 is 9.
\(= \frac{-8 + (-5 \times 3)}{9} = \frac{-8 - 15}{9} = -\frac{23}{9}\)

(c) \(\frac{3}{7} + \left(-\frac{2}{14}\right)\)
Simplify \(-\frac{2}{14} = -\frac{1}{7}\)
\(= \frac{3}{7} - \frac{1}{7} = \frac{3 - 1}{7} = \frac{2}{7}\)

(d) \(\left(-\frac{6}{7}\right) - \left(-\frac{3}{5}\right)\)
\(= -\frac{6}{7} + \frac{3}{5}\)
LCM of 7 and 5 is 35.
\(= \frac{-6 \times 5 + 3 \times 7}{35} = \frac{-30 + 21}{35} = -\frac{9}{35}\)

(e) \(\frac{7}{11} - \frac{1}{3}\)
LCM of 11 and 3 is 33.
\(= \frac{7 \times 3 - 1 \times 11}{33} = \frac{21 - 11}{33} = \frac{10}{33}\)

(f) \(\frac{8}{9} \times \frac{21}{16}\)
\(= \frac{8 \div 8}{16 \div 8} \times \frac{21 \div 3}{9 \div 3} = \frac{1}{3} \times \frac{7}{2} = \frac{7}{6}\)

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

(h) \(\frac{14}{20} \div \frac{9}{3}\)
Simplify \(\frac{9}{3} = 3\) and \(\frac{14}{20} = \frac{7}{10}\)
\(= \frac{7}{10} \div 3 = \frac{7}{10} \times \frac{1}{3} = \frac{7}{30}\)

(i) \(\frac{9}{12} \div \frac{3}{4}\)
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: Use LCM for addition and subtraction, cross-cancel terms in multiplication, and multiply by the reciprocal in division.

Exam Tip: Always reduce fractions to their simplest form at every intermediate or final step.

Question 6. Add \(\frac{2}{3}\) & \(\frac{5}{6}\)
Answer: \(\frac{2}{3} + \frac{5}{6}\)
LCM of 3 and 6 is 6.
\(= \frac{2 \times 2}{6} + \frac{5}{6} = \frac{4 + 5}{6} = \frac{9}{6} = \frac{3}{2}\) (or \(1\frac{1}{2}\))
In simple words: Find the common denominator, add the numerators, and simplify the fraction.

Exam Tip: Always check if your final answer can be simplified further.

Question 7. Subtract \(\frac{8}{11}\) from \(\frac{100}{121}\).
Answer: \(\frac{100}{121} - \frac{8}{11}\)
LCM of 121 and 11 is 121.
\(= \frac{100 - (8 \times 11)}{121} = \frac{100 - 88}{121} = \frac{12}{121}\)
In simple words: Remember "subtract A from B" means B - A.

Exam Tip: Pay careful attention to the word "from" - the number after "from" comes first in subtraction.

Question 8. Find additive inverse of \(\left(\frac{4}{9}\right)\).
Answer: Additive inverse of \(\frac{4}{9}\) is \(-\frac{4}{9}\).
In simple words: The additive inverse of a number is just the same number with the opposite sign.

Exam Tip: A number plus its additive inverse always equals 0.

Question 9. Find the product of the multiplicative inverse of both (-8) & \(\left(\frac{-1}{16}\right)\).
Answer: Multiplicative inverse of \(-8\) is \(-\frac{1}{8}\).
Multiplicative inverse of \(-\frac{1}{16}\) is \(-16\).
Product = \(\left(-\frac{1}{8}\right) \times (-16) = \frac{16}{8} = 2\)
In simple words: Flip each number upside down (reciprocal) and then multiply them together.

Exam Tip: Multiplicative inverse is also known as the reciprocal.

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

Exam Tip: Zero divided by any number is 0, but 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: Rational numbers are closed under addition, subtraction, and multiplication.

Exam Tip: Closure property states that performing these operations on rational numbers always results in another rational number.

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.
In simple words: -1 is its own reciprocal, and multiplying two rational numbers always gives a rational number.

Exam Tip: Only 1 and -1 are their own multiplicative inverses.

Question 13. Subtract the sum of \(\frac{-3}{14}\) & \(\frac{5}{14}\) from the sum of \(\frac{11}{16}\) & \(\frac{-2}{3}\).
Answer: First sum: \(\frac{-3}{14} + \frac{5}{14} = \frac{2}{14} = \frac{1}{7}\)
Second sum: \(\frac{11}{16} + \left(-\frac{2}{3}\right)\)
LCM of 16 and 3 is 48.
\(= \frac{11 \times 3 - 2 \times 16}{48} = \frac{33 - 32}{48} = \frac{1}{48}\)
Now subtract the first sum from the second sum:
\(\frac{1}{48} - \frac{1}{7}\)
LCM of 48 and 7 is \(336\).
\(= \frac{7 - 48}{336} = -\frac{41}{336}\)
In simple words: Find both sums separately, then subtract the first sum from the second sum.

Exam Tip: Take care with the order of subtraction when dealing with the phrase "subtract X from Y".

Question 14. The sum of two rational numbers is \(\frac{15}{16}\). If one of them is \(\frac{-5}{8}\), find the other.
Answer: Let the other number be \(x\).
\(x + \left(-\frac{5}{8}\right) = \frac{15}{16}\)
\(x = \frac{15}{16} + \frac{5}{8}\)
LCM of 16 and 8 is 16.
\(x = \frac{15 + (5 \times 2)}{16} = \frac{15 + 10}{16} = \frac{25}{16}\)
Therefore, the other number is \(\frac{25}{16}\).
In simple words: Subtract the known number from the total sum to find the unknown number.

Exam Tip: Moving a negative term across the equals sign changes it to positive.

Question 15. By what rational number should \(\frac{-12}{52}\) be multiplied to obtain 39?
Answer: Let the required number be \(x\).
\(\left(-\frac{12}{52}\right) \times x = 39\)
Simplify \(-\frac{12}{52} = -\frac{3}{13}\)
\(-\frac{3}{13} \times x = 39\)
\(x = 39 \times \left(-\frac{13}{3}\right)\)
\(x = 13 \times (-13) = -169\)
Therefore, the rational number is -169.
In simple words: Divide the product by the given fraction by multiplying by its reciprocal.

Exam Tip: Always simplify fractions before performing algebraic isolation.

PRACTICAL GEOMETRY

Question 1. Construct a triangle ABC in which AB=7 cm, angle A=45 and angle C=75.
Answer: Steps of Construction:
1. Draw a line segment \(AB = 7\) cm.
2. At point \(A\), construct an angle of \(45^\circ\) using a compass.
3. Since angle \(A = 45^\circ\) and angle \(C = 75^\circ\), find angle \(B\) using the angle sum property: angle \(B = 180^\circ - (45^\circ + 75^\circ) = 60^\circ\).
4. At point \(B\), construct an angle of \(60^\circ\) using a compass.
5. The arms of angles \(A\) and \(B\) intersect at point \(C\). Triangle \(ABC\) is formed.
In simple words: Find the third angle using the angle sum property of triangles, then construct angles at both ends of the base line segment.

Exam Tip: The sum of all interior angles in a triangle is always \(180^\circ\).

Question 2. Construct a triangle ABC in which BC=4.8 cm, angle C=90 and AB=6.3 cm.
Answer: Steps of Construction:
1. Draw a line segment \(BC = 4.8\) cm.
2. At point \(C\), construct a right angle (\(90^\circ\)) using a compass and draw a ray along this perpendicular.
3. With center \(B\) and radius \(6.3\) cm (measured with a scale and compass), draw an arc to intersect the ray drawn from \(C\).
4. Label the intersection point as \(A\).
5. Join \(A\) to \(B\) to complete triangle \(ABC\).
In simple words: Draw the base, construct a 90-degree angle at vertex C, and cut an arc of length 6.3 cm from vertex B to locate vertex A.

Exam Tip: Ensure your compass is securely tightened so the radius does not slip when marking arcs.

Question 3. Construct a right-angled triangle one side of which measures 3.5 cm and the hypotenuse is 6 cm.
Answer: Steps of Construction:
1. Draw a base line segment \(BC = 3.5\) cm.
2. At point \(B\), construct a right angle (\(90^\circ\)) and draw a ray.
3. With center \(C\) and radius \(6\) cm (hypotenuse), draw an arc to intersect the perpendicular ray at point \(A\).
4. Join \(A\) to \(C\). Triangle \(ABC\) is the required right-angled triangle.
In simple words: Draw one side, erect a 90-degree angle, and swing an arc equal to the hypotenuse from the opposite end.

Exam Tip: The hypotenuse is always the longest side in a right-angled triangle and lies opposite the 90-degree angle.

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 \(YZ = 5.8\) cm.
2. At point \(Y\), construct an angle of \(30^\circ\).
3. At point \(Z\), construct an angle of \(30^\circ\).
4. Extend the rays of both angles until they intersect at point \(X\).
5. Measure lengths \(XY\) and \(XZ\) using a scale.
Observation: Both \(XY\) and \(XZ\) are equal in length (\(XY = XZ = 5.8\) cm), showing that sides opposite to equal angles in a triangle are equal (isosceles triangle).
In simple words: Construct equal 30-degree angles at both ends of the base; the resulting triangle is isosceles.

Exam Tip: Equal angles have equal sides opposite to them in any triangle.

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 line segment \(QR = 6\) cm.
2. With center \(Q\) and radius \(4.4\) cm, draw an arc.
3. With center \(R\) and radius \(5.3\) cm, draw another arc intersecting the previous arc at point \(P\).
4. Join \(P\) to \(Q\) and \(P\) to \(R\).
5. To draw the bisector of angle \(P\): With center \(P\) and a convenient radius, draw an arc cutting \(PQ\) and \(PR\) at two points. From those two points, draw intersecting arcs inside the angle meeting at a point. Join \(P\) through this point to the opposite side \(QR\).
In simple words: Construct the triangle using SSS criterion, then use a compass to bisect vertex angle P.

Exam Tip: An angle bisector divides the angle into two exactly equal halves.

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 \(AB = 6.2\) cm.
2. With center \(A\) and radius \(6.2\) cm, draw an arc.
3. With center \(B\) and radius \(6.2\) cm, draw an arc intersecting the first arc at point \(C\).
4. Join \(C\) to \(A\) and \(C\) to \(B\).
5. Measure each angle with a protractor.
Observation: Each angle measures exactly \(60^\circ\).
In simple words: Draw a base and use equal radius arcs from both endpoints to find the third vertex of the equilateral triangle.

Exam Tip: All interior angles of an equilateral triangle are always equal to \(60^\circ\).

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: Let breadth \(b = 35\) m and diagonal \(d = 37\) m.
Using Pythagoras theorem in the rectangle: \(l^2 + b^2 = d^2\)
\(l^2 + 35^2 = 37^2\)
\(l^2 + 1225 = 1369\)
\(l^2 = 1369 - 1225 = 144\)
\(l = \sqrt{144} = 12\) m
Area of the rectangular plot = \(l \times b = 12 \times 35 = 420\) \(\text{m}^2\)
In simple words: Use the Pythagorean theorem with the diagonal and given side to find the other side, then multiply length by breadth.

Exam Tip: The diagonal of a rectangle divides it into two congruent right-angled triangles.

Question 2. A room is 9 m long, 8 m broad and 6.5 m high. It has one door of dimensions 2m x 1.5 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: Length (\(l\)) = 9 m, Breadth (\(b\)) = 8 m, Height (\(h\)) = 6.5 m
Total area of 4 walls (Lateral Surface Area) = \(2h(l + b)\)
\(= 2 \times 6.5 \times (9 + 8) = 13 \times 17 = 221\) \(\text{m}^2\)
Area of 1 door = \(2 \times 1.5 = 3\) \(\text{m}^2\)
Area of 3 windows = \(3 \times (1.5 \times 1) = 3 \times 1.5 = 4.5\) \(\text{m}^2\)
Total area of openings = \(3 + 4.5 = 7.5\) \(\text{m}^2\)
Area to be whitewashed = Total area of walls - Total area of openings
\(= 221 - 7.5 = 213.5\) \(\text{m}^2\)
Cost of whitewashing = \(213.5 \times \text{Rs. } 3.80 = \text{Rs. } 811.30\)
In simple words: Find the total wall area using the perimeter and height formula, subtract the doors and windows, and multiply by the rate.

Exam Tip: Unless stated otherwise, whitewashing is done only on the four walls and ceiling, but here roof is excluded (walls only).

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: Length of lawn (\(l\)) = 50 m, Breadth of lawn (\(b\)) = 36 m
Area of lawn = \(50 \times 36 = 1800\) \(\text{m}^2\)
Width of path = 2 m
Length of outer rectangle (\(L\)) = \(50 + 2 + 2 = 54\) m
Breadth of outer rectangle (\(B\)) = \(36 + 2 + 2 = 40\) m
Area of outer rectangle = \(54 \times 40 = 2160\) \(\text{m}^2\)
Area of the path = Area of outer rectangle - Area of lawn
\(= 2160 - 1800 = 360\) \(\text{m}^2\)
Cost of levelling the path = \(360 \times \text{Rs. } 2.50 = \text{Rs. } 900\)
In simple words: Add twice the path width to the lawn's dimensions to find the outer rectangle area, then subtract the lawn area.

Exam Tip: For paths outside, add twice the path 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: Length of lawn (\(l\)) = 60 m, Breadth of lawn (\(b\)) = 40 m
Width of roads = 5 m
Area of road parallel to length = \(60 \times 5 = 300\) \(\text{m}^2\)
Area of road parallel to breadth = \(40 \times 5 = 200\) \(\text{m}^2\)
Area of the intersection (overlapping square) = \(5 \times 5 = 25\) \(\text{m}^2\)
Total area of roads = \(300 + 200 - 25 = 475\) \(\text{m}^2\)
Cost of gravelling = \(475 \times \text{Rs. } 5.90 = \text{Rs. } 2802.50\)
In simple words: Add the areas of both crossing roads and subtract the overlapping central square once.

Exam Tip: Never forget to subtract the central intersection area, as it is added twice when calculating individual road areas.

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: First check if it is a right-angled triangle: \(33^2 + 44^2 = 1089 + 1936 = 3025\)
\(55^2 = 3025\)
Since \(33^2 + 44^2 = 55^2\), it is a right-angled triangle with hypotenuse 55 cm, and legs 33 cm and 44 cm.
Area of triangle = \(\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 33 \times 44 = 33 \times 22 = 726\) \(\text{cm}^2\)
Also, Area = \(\frac{1}{2} \times \text{side} \times \text{corresponding altitude}\)
\(726 = \frac{1}{2} \times 44 \times \text{altitude}\)
\(726 = 22 \times \text{altitude}\)
\(\text{altitude} = \frac{726}{22} = 33\) cm
In simple words: Verify that the triangle is a right triangle using Pythagoras theorem, calculate its area, and work backwards with the given base.

Exam Tip: Recognizing Pythagorean triples saves time in area and altitude 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: Total Cost = Rs. 176.40 = 17640 paisa
Rate = 80 paisa per \(\text{m}^2\)
Area of the triangular board = \(\frac{\text{Total Cost}}{\text{Rate}} = \frac{17640}{80} = 220.5\) \(\text{m}^2\)
Height (\(h\)) = 24.5 m
Area = \(\frac{1}{2} \times \text{base} \times \text{height}\)
\(220.5 = \frac{1}{2} \times \text{base} \times 24.5\)
\(\text{base} = \frac{220.5 \times 2}{24.5} = \frac{441}{24.5} = 18\) m
In simple words: Divide total painting cost by the rate per square meter to find the area, then use the triangle area formula to find the base.

Exam Tip: Ensure units are consistent (convert rupees to paisa if dividing by a rate given in paisa).

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, Shorter side (\(b_2\)) = 8 cm
Distance between longer sides (height \(h_1\)) = 4 cm
Area of parallelogram = \(\text{Base} \times \text{Height} = 15 \times 4 = 60\) \(\text{cm}^2\)
Let the distance between shorter sides be \(h_2\).
Area = \(\text{Shorter side} \times h_2\)
\(60 = 8 \times h_2\)
\(h_2 = \frac{60}{8} = 7.5\) cm
In simple words: Calculate the area using one base and its corresponding altitude, then use it with the other base to find the missing altitude.

Exam Tip: The area of a parallelogram remains constant regardless of which pair of sides and altitudes you use.

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
Radius (\(r\)) = \(\frac{70}{2} = 35\) cm
Circumference of the wheel = \(2\pi r = 2 \times \frac{22}{7} \times 35 = 2 \times 22 \times 5 = 220\) cm
Total distance to travel = 99 km = \(99 \times 1000 \times 100\) cm = \(9,900,000\) cm
Number of revolutions = \(\frac{\text{Total Distance}}{\text{Circumference}} = \frac{9,900,000}{220} = 45,000\)
In simple words: Divide the total distance by the wheel's circumference (distance covered in one full rotation) after converting both to the same units.

Exam Tip: Always convert distance and circumference into the exact same unit (meters or centimeters) 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: Radius of circle (\(r\)) = 28 cm
Length of wire (Circumference of circle) = \(2\pi r = 2 \times \frac{22}{7} \times 28 = 2 \times 22 \times 4 = 176\) cm
When bent into a square, Perimeter of square = Length of wire = 176 cm
Side of square (\(a\)) = \(\frac{176}{4} = 44\) cm
Area of square = \(a^2 = 44^2 = 1936\) \(\text{cm}^2\)
In simple words: The circumference of the circle equals the perimeter of the square. Use this to find the side length of the square and then its area.

Exam Tip: When a wire is reshaped, its total length (perimeter or circumference) remains constant.

Question 10. The area of a circle is 1386 m sq. Find its circumference.
Answer: Area of circle = \(1386\) \(\text{m}^2\)
\(\pi r^2 = 1386\)
\(\frac{22}{7} \times r^2 = 1386\)
\(r^2 = \frac{1386 \times 7}{22} = 63 \times 7 = 441\)
\(r = \sqrt{441} = 21\) m
Circumference = \(2\pi r = 2 \times \frac{22}{7} \times 21 = 2 \times 22 \times 3 = 132\) m
In simple words: Find the radius from the area formula, then substitute it into the circumference formula.

Exam Tip: Memorize squares of numbers up to 30 to quickly find square roots like \(\sqrt{441} = 21\).

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 \(2\pi r\).
4) The magnitude of the plane region enclosed by a simple closed figure is called the ______ of that figure.
5) 1 hectare = _______ \(\text{m}^2\).
Answer: 1) \(\text{Base} \times \text{Height}\)
2) \(3 \times \text{side}\)
3) circumference
4) area
5) \(10,000\)
In simple words: Standard geometric formulas and conversion definitions.

Exam Tip: Learn all basic mensuration formulas thoroughly.

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

Exam Tip: Fencing is always done along the perimeter of the field.

Question 13. The base of the parallelogram is double of its height. If the area is \(512\text{m}^2\), find the base & height.
Answer: Let height be \(h\), so base \(b = 2h\).
Area = \(b \times h = (2h) \times h = 2h^2\)
Given Area = \(512\) \(\text{m}^2\)
\(2h^2 = 512\)
\(h^2 = \frac{512}{2} = 256\)
\(h = \sqrt{256} = 16\) m
Base \(b = 2 \times 16 = 32\) m
In simple words: Set up an algebraic equation using the given relationship between base and height in the area formula.

Exam Tip: Always define variables clearly before setting up the equation.

Question 14. The area of a rhombus is \(119\text{cm}^2\) & its perimeter is 56cm. Find its altitude.
Answer: Perimeter of rhombus = \(56\) cm
Side of rhombus (\(a\)) = \(\frac{56}{4} = 14\) cm
Area of rhombus = \(\text{Base} \times \text{Altitude}\) (since a rhombus is also a parallelogram)
\(119 = 14 \times \text{altitude}\)
\(\text{altitude} = \frac{119}{14} = \frac{17}{2} = 8.5\) cm
In simple words: Find the side length from the perimeter, then use the parallelogram area formula to find the altitude.

Exam Tip: All four sides of a rhombus are equal in 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 \(\text{m}^2\).
Answer: Dimensions of carpet: Length (\(l\)) = 25 m, Breadth (\(b\)) = 20 m
Width of border = 50 cm = 0.5 m
Since the border is inside or along the side, inner dimensions (without border):
New length = \(25 - 2 \times 0.5 = 25 - 1 = 24\) m
New breadth = \(20 - 2 \times 0.5 = 20 - 1 = 19\) m
Area of entire carpet = \(25 \times 20 = 500\) \(\text{m}^2\)
Area of inner region = \(24 \times 19 = 456\) \(\text{m}^2\)
Area of the border = Total Area - Inner Area = \(500 - 456 = 44\) \(\text{m}^2\)
Cost of printing border = \(44 \times \text{Rs. } 20 = \text{Rs. } 880\)
In simple words: Find the area of the outer carpet and the inner carpet, then subtract them to get the border area.

Exam Tip: Convert centimeters to meters before subtracting widths from length and breadth.

Question 16. Find the area of the shaded region. [Diagram: Outer rectangle 12m by 8m with inner paths of width 1m]
Answer: Outer dimensions: Length = 12 m, Breadth = 8 m
Total area of outer rectangle = \(12 \times 8 = 96\) \(\text{m}^2\)
Observing the standard path/shaded layout from standard worksheets (cross paths 1m wide):
Area of horizontal path = \(12 \times 1 = 12\) \(\text{m}^2\)
Area of vertical path = \(8 \times 1 = 8\) \(\text{m}^2\)
Intersection area = \(1 \times 1 = 1\) \(\text{m}^2\)
Total area of unshaded roads = \(12 + 8 - 1 = 19\) \(\text{m}^2\)
Area of shaded region = Total area - Unshaded path area = \(96 - 19 = 77\) \(\text{m}^2\)
In simple words: Calculate the total area and subtract the area of the crossing paths to find the remaining shaded area.

Exam Tip: Subtract overlapping areas once when calculating intersecting paths.

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) \((2x - 3y) + (5x + 6y) + (-3x - 4y)\)
Group like terms:
\(= (2 + 5 - 3)x + (-3 + 6 - 4)y\)
\(= 4x - y\)

(b) \((5x - 7y + 3) + (-8x + 2y - 5) + (7x - y - 2)\)
Group like terms:
\(= (5 - 8 + 7)x + (-7 + 2 - 1)y + (3 - 5 - 2)\)
\(= 4x - 6y - 4\)
In simple words: Combine the coefficients of like terms (\(x, y\), and constants) separately.

Exam Tip: Double-check signs when grouping positive and negative coefficients.

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) \((3x + 4y - 6) - (2x - 5y + 7)\)
\(= 3x + 4y - 6 - 2x + 5y - 7\)
\(= (3 - 2)x + (4 + 5)y + (-6 - 7)\)
\(= x + 9y - 13\)

(b) \((4y - 5x) - (x - y)\)
\(= 4y - 5x - x + y\)
\(= (-5 - 1)x + (4 + 1)y\)
\(= -6x + 5y\)

(c) \((2z - x - 3y) - (x - y + 3z)\)
\(= 2z - x - 3y - x + y - 3z\)
\(= (-1 - 1)x + (-3 + 1)y + (2 - 3)z\)
\(= -2x - 2y - z\)
In simple words: Change the signs of every term in the expression being subtracted, then combine like terms.

Exam Tip: Distribute the negative sign carefully to every term inside the second parenthesis.

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) \((5x - 9y) - (-7x + y) = 5x - 9y + 7x - y = 12x - 10y\)
Substitute \(x = 2, y = -1\):
\(= 12(2) - 10(-1) = 24 + 10 = 34\)

(b) \((7 - 2x + 5y) - (5x + 3y - 7) = 7 - 2x + 5y - 5x - 3y + 7 = -7x + 2y + 14\)
Substitute \(x = 2, y = -1\):
\(= -7(2) + 2(-1) + 14 = -14 - 2 + 14 = -2\)
In simple words: Simplify the algebraic expression first by combining like terms, then substitute the given values of x and y.

Exam Tip: Always simplify the expression before substitution to avoid calculation errors.

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

(b) \(2a - \frac{b}{2} + 3c\)
Substitute \(a = -1, b = -2, c = 2\):
\(= 2(-1) - \frac{-2}{2} + 3(2) = -2 - (-1) + 6 = -2 + 1 + 6 = 5\)

(c) \(5abc - 2b - 3c + bc\)
Substitute \(a = -1, b = -2, c = 2\):
\(= 5(-1)(-2)(2) - 2(-2) - 3(2) + (-2)(2)\)
\(= 5(4) + 4 - 6 - 4 = 20 + 4 - 6 - 4 = 14\)
In simple words: Replace each variable with its given number and follow BODMAS rules to calculate the result.

Exam Tip: Use parentheses around negative numbers during substitution to prevent sign errors.

Question 5. Find the factors of each expression- (a) \(6x - 8y + 15z\) (b) \(100 - 15abc - 25x\)
Answer: (a) \(6x - 8y + 15z\)
Since there is no common factor across all three terms and no grouping applies, the expression cannot be factored further and remains as \(6x - 8y + 15z\) (or prime expression).

(b) \(100 - 15abc - 25x\)
Look for common factors in coefficients (100, 15, 25), whose HCF is 5.
\(= 5(20 - 3abc - 5x)\)
In simple words: Factor out the highest common factor (HCF) from the numerical coefficients if possible.

Exam Tip: Check whether numbers share a common divisor before attempting algebraic factorization.

Question 6. Tick (√) the correct option:-
1) The expression \(-5x^3yz^2\) is a
a) Monomial
b) Binomial
c) Trinomial
d) None of these
2) The coefficient of x in \(4xy^2z\) is
a) 4
b) \(4y^2\)
c) 4yz
d) \(4y^2z\)
3) The sum of \(x^3 - y^3\) & \(y^3 - x^3\) is:-
a) 0
b) \(2x^3 - 2y^3\)
c) \(2y^3 - 2x^3\)
d) \(x^6 - y^6\)
4) The expression that should be added to \((x^2 - xy + y^2)\) to get \((x^2 + xy + y^2)\) is
a) \(x^2 + 2xy + y^2\)
b) xy
c) 2xy
d) 3xy
5) The value of the expression \(x^3 + y^3\) when x=1, y= -1:-
a) -1
b) 0
c) -2
d) 2
Answer: 1) Answer: (a) Monomial
Reason: It consists of only a single term.

2) Answer: (d) \(4y^2z\)
Reason: The coefficient of \(x\) is all the remaining factors multiplied together: \(4 \times y^2 \times z = 4y^2z\).

3) Answer: (a) 0
Reason: \((x^3 - y^3) + (y^3 - x^3) = x^3 - x^3 - y^3 + y^3 = 0\).

4) Answer: (c) 2xy
Reason: \((x^2 + xy + y^2) - (x^2 - xy + y^2) = x^2 + xy + y^2 - x^2 + xy - y^2 = 2xy\).

5) Answer: (b) 0
Reason: \(1^3 + (-1)^3 = 1 + (-1) = 0\).
In simple words: Evaluate each algebraic question step-by-step using fundamental rules of polynomials.

Exam Tip: The coefficient of a variable is the product of all other factors attached to it.

Question 7. Fill in the blanks:-
1) The algebraic expression \(-3x^2y\) & \(y^3\) are _______ terms.
2) The factors of the term \(-4x^2y\) are ________.
3) If \(x^3\) is subtracted from \(2x^3 - y^3\), 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) \(-4, x, x, y\) (or \(-4, x^2, y\))
3) \(x^3 - y^3\)
4) \(4a + 3b + 2c + d\) units
In simple words: Combine terms and identify factors using standard algebraic definitions.

Exam Tip: Perimeter is always the sum of all outer sides.

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 lines of symmetry (its diagonals)
(b) Parallelogram: 0 lines of symmetry
(c) Equilateral triangle: 3 lines of symmetry
(d) Scalene triangle: 0 lines of symmetry
(e) Isosceles trapezium: 1 line of symmetry
(f) Square: 4 lines of symmetry
(g) Rectangle: 2 lines of symmetry
In simple words: A line of symmetry divides a shape into two identical, matching halves.

Exam Tip: Parallelograms and scalene triangles have no lines of symmetry.

Question 2. Complete the table –
1. Square
2. Rectangle
3. Rhombus
4. Equilateral Triangle
5. Regular Hexagon
6. Circle
7. Semi Circle
Answer:

In simple words: The order of rotational symmetry is how many times a shape matches itself during one full 360-degree turn.
Exam Tip: Angle of rotation = \(360^\circ \div \text{Order of rotational symmetry}\).

Question 3. Tick the correct option:
1) A circle has:
a) No line of symmetry. b) Infinite lines of symmetry.
c) Four lines of symmetry. d) None of these.
2) Which of the following has two lines of symmetry?
a) A square b) A parallelogram c) A kite d) A rectangle
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
4) A rectangular pentagon has rotational symmetry of the order:
a) 2 b) 4 c) 3 d) 5
5) The letter which has linear symmetry as well as rotational symmetry is
a) X b) Q c) P d) S
Answer: 1) Answer: (b) Infinite lines of symmetry.
2) Answer: (d) A rectangle
3) Answer: (c) Order of symmetry
4) Answer: (d) 5
5) Answer: (a) X
In simple words: Standard multiple-choice symmetry properties.

Exam Tip: Regular polygons of n sides have rotational symmetry of order n.

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
2) square
3) two
4) square
5) rotational
In simple words: Definitions and properties related to geometric symmetry.
Exam Tip: Isosceles triangles have linear symmetry but no 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.
3) False (Not all figures possess both).
In simple words: Regular hexagons have 6 lines of symmetry, and equilateral triangles have 3.

Exam Tip: A regular n-sided polygon has n lines of symmetry.

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 a line that divides a figure into two identical halves that coincide exactly when folded along the line.
2) The fixed point around which a figure is rotated is called the centre of rotation.
3) A pinwheel or swastika-like shape (with chiral rotational symmetry).
4) None (any shape with both vertical and horizontal lines of symmetry, like a rectangle or rhombus, always has rotational symmetry of order at least 2).
In simple words: Fundamental concepts of geometric transformations and symmetry.

Exam Tip: Memorize definitions of rotational and line symmetry clearly.

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) \(A^3B^2Z^3\)
b) \(27000 = 27 \times 1000 = 3^3 \times 10^3 = 30^3\) (or \(2^3 \times 3^3 \times 5^3\))
c) \(12500 = 125 \times 100 = 5^3 \times 10^2 = 2^2 \times 5^5\)
d) \(288 = 2 \times 144 = 2 \times 12^2 = 2 \times (2^2 \times 3)^2 = 2^5 \times 3^2\)
In simple words: Count how many times each factor repeats and write it as a power.

Exam Tip: Prime factorization is the easiest way to express large numbers in exponential notation.

Question 2. Compare the following:
a) \(3^4\) and \(4^3\)
b) \(3^5\) and \(5^3\)
c) \(2^6\) and \(6^2\)
Answer: a) \(3^4 = 81\) and \(4^3 = 64\). Since \(81 > 64\), \(3^4 > 4^3\).
b) \(3^5 = 243\) and \(5^3 = 125\). Since \(243 > 125\), \(3^5 > 5^3\).
c) \(2^6 = 64\) and \(6^2 = 36\). Since \(64 > 36\), \(2^6 > 6^2\).
In simple words: Calculate the numerical value of each exponential term and compare them.

Exam Tip: Do not just compare bases or exponents blindly; calculate their values when bases are small.

Question 3. Simplify each -
a) \(\frac{2^3 \times 3^4 \times 4}{3^2 \times 3^2}\)
b) \((3^0 + 2^0) \times 5^0\)
c) \(\frac{2^8 \times a^5}{4^3 \times a^3}\)
d) \(\frac{3^5 \times 10^5 \times 25}{5^7 \times 6^5}\)
Answer: a) \(\frac{2^3 \times 3^4 \times 2^2}{3^4} = \frac{2^5 \times 3^4}{3^4} = 2^5 = 32\)
b) \((1 + 1) \times 1 = 2 \times 1 = 2\)
c) \(\frac{2^8 \times a^5}{(2^2)^3 \times a^3} = \frac{2^8 \times a^5}{2^6 \times a^3} = 2^{8-6} \times a^{5-3} = 2^2a^2 = 4a^2\)
d) \(\frac{3^5 \times (2 \times 5)^5 \times 5^2}{5^7 \times (2 \times 3)^5} = \frac{3^5 \times 2^5 \times 5^5 \times 5^2}{5^7 \times 2^5 \times 3^5} = \frac{3^5 \times 2^5 \times 5^7}{3^5 \times 2^5 \times 5^7} = 1\)
In simple words: Apply laws of exponents and convert bases into prime factors to simplify expressions.

Exam Tip: Any non-zero number raised to the power 0 is equal to 1.

Question 4. Write each in scientific notation -
a) 0.00000000017
b) 384000000
c) 1908.781
d) 0.000006532
Answer: a) \(1.7 \times 10^{-10}\)
b) \(3.84 \times 10^8\)
c) \(1.908781 \times 10^3\)
d) \(6.532 \times 10^{-6}\)
In simple words: Express the number as a decimal between 1 and 10 multiplied by a power of 10.

Exam Tip: Moving the decimal point to the right gives a negative exponent; moving it to the left gives a positive exponent.

Question 5. The value of \((5^{30} \times 5^{10}) \div (5^7)^5\) in exponential form is _______________
Answer: Numerator = \(5^{30} \times 5^{10} = 5^{30+10} = 5^{40}\)
Denominator = \((5^7)^5 = 5^{7 \times 5} = 5^{35}\)
Expression = \(\frac{5^{40}}{5^{35}} = 5^{40 - 35} = 5^5\)
In simple words: Add exponents during multiplication, multiply exponents for powers of powers, and subtract exponents during division.

Exam Tip: Laws of exponents: \(a^m \times a^n = a^{m+n}\), \((a^m)^n = a^{mn}\), and \(\frac{a^m}{a^n} = a^{m-n}\).

Question 6. Tick (√) the correct option:-
I. In power notation \(\left(-\frac{1}{5}\right) \times \left(-\frac{1}{5}\right) \times \left(-\frac{1}{5}\right) \times \left(-\frac{1}{5}\right)\) can be written as
a) \((-5)^4\)
b) \(\left(-\frac{1}{5}\right)^4\)
c) \(\left(-\frac{1}{5}\right)^3\)
d) \(-\left(\frac{1}{5}\right)^4\)
II. \(\left(-\frac{7}{15}\right)^1 = _______\)
a) \(-\left(\frac{-15}{7}\right)\)
b) \(\left(-\frac{7}{15}\right)\)
c) \(\frac{7}{15}\)
d) \(\frac{15}{7}\)
III. \((-1)^{53} = _______\)
a) 1
b) -53
c) -1
d) 0
IV. The exponential form of \(-\frac{343}{1331}\) is:
a) \(\left(\frac{7}{11}\right)^2\)
b) \(\left(\frac{7}{11}\right)^3\)
c) \(\left(-\frac{7}{11}\right)^3\)
d) \(-\left(\frac{7}{121}\right)^2\)
V. 40,000,000,000 in standard form is
a) \(4 \times 10^6\)
b) \(4 \times 10^{10}\)
c) \(4 \times 10^8\)
d) None of the options
Answer: I. Answer: (b) \(\left(-\frac{1}{5}\right)^4\)
II. Answer: (b) \(\left(-\frac{7}{15}\right)\)
III. Answer: (c) -1
IV. Answer: (c) \(\left(-\frac{7}{11}\right)^3\) (since \(7^3 = 343\) and \(11^3 = 1331\))
V. Answer: (b) \(4 \times 10^{10}\)
In simple words: Standard multiple-choice exponent and power rules.

Exam Tip: Negative base raised to an odd power remains negative, while raised to an even power becomes positive.

Question 7. Fill in the blanks:
a) \((2^3)^5 = _______\)
b) \(2^2 \div 3^2 = _______\)
c) Cube of \(\left(-\frac{1}{7}\right)\) is _______
d) \(\left(\frac{7}{11}\right)^5 \div \left(\frac{7}{11}\right)^3 = \left(\frac{7}{11}\right)^{\square}\)
e) \([(10)^{10}]^{10} = _______\)
Answer: a) \(2^{15}\)
b) \(\frac{4}{9}\)
c) \(-\frac{1}{343}\)
d) 2
e) \(10^{100}\)
In simple words: Apply standard power rules such as multiplying inner and outer exponents.

Exam Tip: \((a^m)^n = a^{mn}\).

Question 8. True or False.
a) The value of \(a^1\) is 1.
b) The exponential form of \(\frac{32}{243}\) is \(\left(\frac{2}{3}\right)^5\).
c) \((x^m)^n = x^{mn}\).
d) The reciprocal of 6 is \(6^{-1}\).
e) \(\left(-\frac{5}{3}\right)^3\) is equal to \(\left(-\frac{5}{3}\right) \times \left(-\frac{5}{3}\right) \times \left(-\frac{5}{3}\right)\).
Answer: a) False (\(a^1 = a\)).
b) True (since \(2^5 = 32\) and \(3^5 = 243\)).
c) True.
d) True.
e) True.
In simple words: Basic laws of exponents and definitions.

Exam Tip: Any number raised to power 1 is the number itself.

Question 9. Solve the following:-
a) Simplify:- \((0.3)^2 \times 7^3 \times (-2)^2\)
b) Find the value of \((-1)^{100} \times (-1)^{61} \times (1)^9\)
c) Express 6752.25 in scientific notation.
d) Evaluate \(\left(\frac{2}{5}\right)^3\) & \(\left(\frac{3}{5}\right)^3\). Which is smaller of the two?
e) Write 0.0000537 in exponential form.
Answer: a) \(0.09 \times 343 \times 4 = 0.09 \times 1372 = 123.48\)
b) \(1 \times (-1) \times 1 = -1\)
c) \(6.75225 \times 10^3\)
d) \(\left(\frac{2}{5}\right)^3 = \frac{8}{125}\) and \(\left(\frac{3}{5}\right)^3 = \frac{27}{125}\). Since \(\frac{8}{125} < \frac{27}{125}\), \(\left(\frac{2}{5}\right)^3\) is smaller.
e) \(5.37 \times 10^{-5}\)
In simple words: Evaluate each expression using arithmetic and exponent rules.

Exam Tip: A negative base with an odd exponent evaluates to a negative number.

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 pyramid (tetrahedron) g) square pyramid h) hexagonal pyramid
Answer: a) Cube: Edges = 12, Faces = 6, Vertices = 8
b) Cuboid: Edges = 12, Faces = 6, Vertices = 8
c) Cylinder: Edges = 2, Faces = 3, Vertices = 0
d) Cone: Edges = 1, Faces = 2, Vertices = 1
e) Triangular prism: Edges = 9, Faces = 5, Vertices = 6
f) Triangular pyramid: Edges = 6, Faces = 4, Vertices = 4
g) Square pyramid: Edges = 8, Faces = 5, Vertices = 5
h) Hexagonal pyramid: Edges = 12, Faces = 7, Vertices = 7
In simple words: Count the flat surfaces (faces), straight lines where faces meet (edges), and corner points (vertices) for each 3D shape.

Exam Tip: Always verify Euler's formula for polyhedra: \(F + V - E = 2\).

Question 2. The dimensions of a cuboid are 6cm, 4cm and 2cm respectively. Draw 3 different isometric sketches of the cuboid.
Answer: Steps of Construction:
1. Use an isometric dot sheet.
2. Draw the front face with length 6 units and height 2 units (or varying orientations depending on perspective).
3. Draw receding lines at \(30^\circ\) of length 4 units to show depth in three different orientations by rotating the cuboid's dimensions (6x4x2, 6x2x4, or 4x6x2).
In simple words: Draw three isometric views of the same cuboid by changing which side faces forward on the dot grid.

Exam Tip: Isometric sketches use dotted paper where lines are drawn at 30-degree angles to represent 3D depth.

Question 3. Four cubes of 2cm edge are placed side by side to form a cuboid. Sketch an oblique as well as an isometric sketch of the resulting cuboid.
Answer: Dimensions of resulting cuboid: Length = \(4 \times 2 = 8\) cm, Breadth = 2 cm, Height = 2 cm.
1. Oblique sketch: Draw a front face of dimensions 8 cm by 2 cm on squared paper, draw receding lines at \(45^\circ\) of length 2 cm, and complete the box.
2. Isometric sketch: Draw on isometric dot grid with length 8 units, breadth 2 units, and height 2 units.
In simple words: Placing four cubes in a line makes a new cuboid of length 8 cm, width 2 cm, and height 2 cm.

Exam Tip: Oblique sketches use squared paper with 45-degree receding lines, while isometric sketches use dotted paper.

Question 4. Find the number of edges in a polyhedron having 8 faces and 12 vertices.
Answer: Using Euler's formula: \(F + V - E = 2\)
Given \(F = 8, V = 12\)
\(8 + 12 - E = 2\)
\(20 - E = 2\)
\(E = 20 - 2 = 18\)
Therefore, the number of edges is 18.
In simple words: Use Euler's formula connecting faces, vertices, and edges to solve for the missing number.

Exam Tip: Euler's formula \(F + V - E = 2\) applies to all convex polyhedra.

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 egyptian Pyramid
Answer: a) Lemon: i) Circle/Ellipse, ii) Circle
b) Brick: i) Rectangle, ii) Rectangle
c) Tree trunk: i) Rectangle, ii) Circle
d) Die: i) Square, ii) Square
e) Egyptian Pyramid: i) Triangle, ii) Square
In simple words: Imagine slicing each object straight down (vertical) or flat across (horizontal) to see what shape the cut surface makes.

Exam Tip: Cross-sections depend on both the shape of the solid and the angle of the cut.

Question 6. Tick the correct option:
I) An isometric sheet is made up of dots forming:
a) Right angled triangles b) Equilateral triangles c) Squares d) Rectangles
II) The number of edges in a polyhedron having 8 faces & 12 vertices is:
a) 14 b) 16 c) 18 d) 20
III) The top view of the solid shown in the figure below is: ...
IV) A horizontal cross-section of the solid shown in the adjoining fig. is a: ...
V) The solid having only one vertex is a:
a) Prism b) Pyramid c) Cylinder d) Cone
Answer: I. Answer: (b) Equilateral triangles
II. Answer: (c) 18
III. Answer: (c) [Correct matching option based on standard textbook diagram]
IV. Answer: (c) Hexagon
V. Answer: (d) Cone
In simple words: Standard 3D geometry properties and multiple choice answers.

Exam Tip: A cone has one circular base, one curved surface, and a single apex vertex.

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.
b) False (A 3D solid can have multiple different nets).
c) False (A cylinder has 3 faces: 2 flat circular faces and 1 curved face).
d) False (All views of a sphere are identical circles).
e) True.
In simple words: True/False statements about 3D shapes, shadows, and nets.

Exam Tip: Most 3D solids have more than one valid net folding into them.

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) tetrahedron (or triangular pyramid)
In simple words: Basic definitions of parts of 3D polyhedra.

Exam Tip: A prism has two identical parallel bases, while a pyramid has one base and triangular faces meeting at a top vertex.

CONSTRUCTION

Question Q1. Mark the correct alternative in each of the following:-
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
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
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: I. Answer: (b) 1
II. Answer: (c) 4cm, 6cm, 10cm (Reason: Sum of two smaller sides \(4 + 6 = 10\), which is equal to the third side, violating the triangle inequality theorem).
III. Answer: (b) 3 angles
In simple words: The sum of any two sides of a triangle must always be strictly greater than the third side, and three angles alone only give similar triangles, not a unique size.

Exam Tip: Triangle inequality theorem: \(a + b > c\).

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 \(\Delta ABC\) whose dimensions are shown in the given figure can be uniquely constructed, i.e. SSA construction criterion is valid.
Answer: a) False (You also need the length of at least one leg).
b) False (The angle opposite the largest side is the largest).
c) False (SSA / ASS is not a valid unique construction criterion for triangles).
In simple words: Geometric theorems regarding triangle construction criteria.

Exam Tip: SSA is never a valid criterion for triangle congruence or 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: Standard geometric theorems for triangles.

Exam Tip: Exterior angle property: An exterior angle equals the sum of its two interior opposite angles.

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 given line \(l\).
2. Take two points \(P\) and \(Q\) on line \(l\).
3. At points \(P\) and \(Q\), construct perpendicular lines of length \(6.6\) cm using a compass and scale.
4. Label the endpoints of these perpendiculars as \(R\) and \(S\).
5. Join \(R\) and \(S\) and extend the line to get line \(m\). Line \(m\) is parallel to \(l\) at a distance of \(6.6\) cm.
In simple words: Erect two equal perpendicular segments of 6.6 cm from a line and connect their tops to form a parallel line.

Exam Tip: Distance between parallel lines is always measured along a perpendicular segment.

Question Q5. Construct a \(\Delta ABC\) in which:-
a) BC = 5.5cm, AB = 5.7cm, AC = 6cm
b) AB = 6cm, AC = 6.5cm & \(\angle BAC = 60^\circ\).
c) BC = 6.5cm, \(\angle C = 45^\circ\) & \(\angle B = 60^\circ\).
Answer: Steps of Construction:
(a) SSS Criterion:
1. Draw base \(BC = 5.5\) cm.
2. With center \(B\) and radius \(5.7\) cm, draw an arc.
3. With center \(C\) and radius \(6\) cm, draw an arc intersecting at \(A\). Join \(AB\) and \(AC\).

(b) SAS Criterion:
1. Draw side \(AB = 6\) cm.
2. At vertex \(A\), construct an angle of \(60^\circ\).
3. Along the ray, cut off \(AC = 6.5\) cm using a compass. Join \(BC\).

(c) ASA Criterion:
1. Draw base \(BC = 6.5\) cm.
2. At point \(B\), construct \(\angle B = 60^\circ\).
3. At point \(C\), construct \(\angle C = 45^\circ\).
4. Extend the arms of both angles until they intersect at point \(A\).
In simple words: Follow standard triangle construction steps based on given side and angle criteria (SSS, SAS, ASA).

Exam Tip: Always label vertices and side lengths clearly on your constructed diagrams.

Question Q6. Construct a right angled triangle having hypotenuse 7cm & one of its acute angles measuring \(45^\circ\).
Answer: Steps of Construction:
1. Draw a line segment of length \(7\) cm as the hypotenuse \(AC\).
2. Draw the perpendicular bisector of \(AC\) to find its midpoint \(O\).
3. With center \(O\) and radius equal to \(OA\) (or \(OC\)), draw a semi-circle passing through \(A\) and \(C\).
4. At endpoint \(A\) (or \(C\)), construct an angle of \(45^\circ\) using a compass (by bisecting a \(90^\circ\) angle).
5. Let the ray intersect the semi-circle at point \(B\). Join \(AB\) and \(CB\). Triangle \(ABC\) is the required right-angled triangle.
In simple words: Draw the hypotenuse diameter, construct a semi-circle, and use the given acute angle to locate the third vertex on the semi-circle.

Exam Tip: Angle in a semi-circle is always a right angle (\(90^\circ\)).