CBSE Class 6 Mathematics Revision Worksheet Set 04

POINTS TO REMEMBER:
1. Types of Angles:
- Acute angle: \(0^\circ < \angle A < 90^\circ\)
- Right angle: \(\angle A = 90^\circ\)
- Obtuse angle: \(90^\circ < \angle A < 180^\circ\)
- Straight angle: \(\angle A = 180^\circ\)
- Reflex angle: \(180^\circ < \angle A < 360^\circ\)
- Complete angle: \(\angle A = 360^\circ\)

2. Types of Triangles:
- ON THE BASIS OF SIDES:
- Equilateral Triangle: \(\overline{AB} = \overline{BC} = \overline{CA}\)
- Isosceles Triangle: \(\overline{AB} = \overline{BC} \neq \overline{CA}\)
- Scalene Triangle: \(\overline{AB} \neq \overline{BC} \neq \overline{CA}\)
- ON THE BASIS OF ANGLES:
- Acute Angled Triangle: all angles measure \(< 90^\circ\)
- Obtuse Angled Triangle: only one angle is obtuse and other two are acute
- Right Angled Triangle: only one angle is right and other two are acute

3. Classification of Quadrilaterals:
- Parallelograms (Rectangle, Square, Rhombus)
- Kite
- Trapezium

4. Three dimensional shapes having length, breadth and height:

5. Rectangle of length \(l\) and breadth \(b\):
Perimeter = \(2(l + b)\)
Area = \(l \times b\)
Length = Area \(\div\) breadth ; Length = \(\frac{\text{Perimeter}}{2} - \text{Breadth}\)
Breadth = Area \(\div\) length ; Breadth = \(\frac{\text{Perimeter}}{2} - \text{Length}\)

6. Square of side \(s\):
Perimeter = \(4 \times s\)
Area = side \(\times\) side
Side = Perimeter \(\div\) 4

Question 1. What least value should be given to * so that the number: i) \(915^*26\) is divisible by 9. ii) \(63576^*2\) is divisible by 8. iii) \(653^*47\) is divisible by 11.
Answer: i) For a number to be divisible by 9, the sum of its digits must be divisible by 9.
Sum of digits = \(9 + 1 + 5 + * + 2 + 6 = 23 + *\)
The nearest multiple of 9 greater than 23 is 27.
So, \(23 + * = 27 \implies * = 4\).
ii) For a number to be divisible by 8, the number formed by its last three digits must be divisible by 8.
Last three digits form the number \(*26\).
Testing values from 0 to 9 for \(*\):
If \(* = 3\), \(326\) is not divisible by 8.
If \(* = 5\), \(526\) is not divisible by 8.
If \(* = 3\): \(326 \div 8 = 40\) remainder 6.
Let us check carefully: Last three digits are \(*26\). For \(* = 3\), \(326 \div 8 = 40\) R 6. For \(* = 7\), \(726 \div 8 = 90\) R 6. Let us re-verify: \(8 \times 90 = 720\), \(720 + 6 = 726\). Wait, \(732 \div 8 = 91.5\). Let's check \(* = 3\): \(326 \div 8 = 40.75\). Let's check all digits for \(*26\): \(026\) (no), \(126\) (no), \(226\) (no), \(326\) (no), \(426\) (no), \(526\) (no), \(626\) (no), \(726\) (no), \(826\) (no), \(926\) (no). Wait, let's check \(63576*2\) where last three digits are \(6*2\). Let's test values of \(*\):
If \(* = 0\), \(602 \div 8 = 75.25\)
If \(* = 1\), \(612 \div 8 = 76.5\)
If \(* = 2\), \(622 \div 8 = 77.75\)
If \(* = 3\), \(632 \div 8 = 79\) (divisible!). So \(* = 3\).
iii) For a number to be divisible by 11, the difference between the sum of digits at odd places (from the right) and the sum of digits at even places must be 0 or a multiple of 11.
Number is \(653*47\).
Sum of digits at odd places (from right) = \(7 + * + 5 = 12 + *\)
Sum of digits at even places = \(4 + 3 + 6 = 13\)
Difference = \((12 + *) - 13 = * - 1\)
For the difference to be 0, \(* - 1 = 0 \implies * = 1\).
In simple words: Apply divisibility rules for 9, 8, and 11 to find the missing digit.

Exam Tip: Memorize divisibility rules thoroughly, especially for 8 and 11 which are frequently tested.

Question 2. The H.C.F and L.C.M of two numbers are 13 and 1989 respectively. If one of the numbers is 117, determine the other.
Answer: We know that for any two numbers \(a\) and \(b\):
\(\text{HCF} \times \text{LCM} = \text{Product of the two numbers}\)
Let the other number be \(x\).
\(13 \times 1989 = 117 \times x\)
\(x = \frac{13 \times 1989}{117}\)
Since \(117 = 13 \times 9\):
\(x = \frac{1989}{9} = 221\)
In simple words: The product of two numbers always equals the product of their HCF and LCM.

Exam Tip: Always simplify fractions before performing long division to save time.

Question 3. What is the smallest number which when divided by 24, 36 and 54 gives a remainder of 5 each time?
Answer: First, find the LCM of 24, 36, and 54.
Prime factorization:
\(24 = 2^3 \times 3\)
\(36 = 2^2 \times 3^2\)
\(54 = 2 \times 3^3\)
\(\text{LCM} = 2^3 \times 3^3 = 8 \times 27 = 216\)
The smallest number that leaves a remainder of 5 when divided by 24, 36, and 54 is:
\(\text{LCM} + 5 = 216 + 5 = 221\)
In simple words: Find the LCM of the divisors and add the common remainder to it.

Exam Tip: "Smallest number" in such contexts almost always requires finding the LCM.

Question 4. Find the greatest number of 6 digits exactly divisible by 24, 15 and 36.
Answer: First, find the LCM of 24, 15, and 36.
Prime factorization:
\(24 = 2^3 \times 3\)
\(15 = 3 \times 5\)
\(36 = 2^2 \times 3^2\)
\(\text{LCM} = 2^3 \times 3^2 \times 5 = 8 \times 9 \times 5 = 360\)
The greatest 6-digit number is 999999.
Divide 999999 by 360 to find the remainder:
\(999999 \div 360 = 2777\) with a remainder of 279.
Subtract the remainder from 999999:
\(999999 - 279 = 999720\)
In simple words: Find the LCM of the divisors, divide the largest 6-digit number by this LCM, and subtract the remainder.

Exam Tip: For greatest number divisible by multiple numbers, always subtract the remainder from the target number.

Question 5. The length, breadth and height of a room are 8m 25 cm, 6m 75 cm, 4m 50 cm respectively. Determine the longest rod which can measure the three dimensions of the room exactly.
Answer: Convert all dimensions into centimeters:
Length = \(8\text{ m } 25\text{ cm} = 825\text{ cm}\)
Breadth = \(6\text{ m } 75\text{ cm} = 675\text{ cm}\)
Height = \(4\text{ m } 50\text{ cm} = 450\text{ cm}\)
The length of the longest rod is the HCF of 825, 675, and 450.
Find HCF using prime factorization or division method:
\(825 = 3 \times 5^2 \times 11\)
\(675 = 3^3 \times 5^2\)
\(450 = 2 \times 3^2 \times 5^2\)
\(\text{HCF} = 3^1 \times 5^2 = 3 \times 25 = 75\text{ cm}\)
Length of the rod = \(75\text{ cm}\) (or 0.75 m).
In simple words: "Longest rod to measure exactly" means we need to find the HCF of the given dimensions.

Exam Tip: Always convert all units to the smallest unit (cm) before calculating HCF or LCM.

Question 6. Anjali scores 100 marks in Mathematics and \(x\) marks in Science. What is her total score in Science and Maths.
Answer: Marks in Mathematics = 100
Marks in Science = \(x\)
Total score = \(100 + x\)
In simple words: Add the marks of both subjects together to get the total algebraic expression.

Exam Tip: In algebraic word problems, simply translate the words into mathematical operations.

Question 7. Rohit covers a total distance of \((7x + 3y)\) Km in his journey. If he covers \((3x - 8y)\) Km by train, how many Kilometers does he cover by plane?
Answer: Total distance = \(7x + 3y\)
Distance covered by train = \(3x - 8y\)
Distance covered by plane = Total distance - Distance covered by train
\(= (7x + 3y) - (3x - 8y)\)
\(= 7x + 3y - 3x + 8y\)
\(= (7x - 3x) + (3y + 8y)\)
\(= 4x + 11y \text{ Km}\)
In simple words: Subtract the train distance from the total distance by carefully changing signs inside the parentheses.

Exam Tip: Be very careful with minus signs when subtracting polynomial expressions.

Question 8. Add \(-7xy\), \(5xy\), \(-2xy\), \(9xy\). Write the coefficient of "x" in the sum.
Answer: Sum = \(-7xy + 5xy - 2xy + 9xy\)
Combine like terms:
\(= (-7 + 5 - 2 + 9)xy\)
\(= (-4 + 9)xy = 5xy\)
Coefficient of \(x\) in \(5xy\) is \(5y\).
In simple words: Add all the like terms together, then identify what multiplies \(x\).

Exam Tip: The coefficient of a variable in a term is everything else multiplied by that variable.

Question 9. Solve the following equations:
i) \(3x + 2 = 5x + 7\)
ii) \(4x + 3 = 11\)
iii) \(5x + 3 = 28\)
iv) \(7 + 3y = -14\)
v) \(y + \frac{3}{2} = 6\)
Answer:
i) \(3x + 2 = 5x + 7\)
\(2 - 7 = 5x - 3x\)
\(-5 = 2x\)
\(x = \frac{-5}{2} = -2.5\)

ii) \(4x + 3 = 11\)
\(4x = 11 - 3\)
\(4x = 8\)
\(x = \frac{8}{4} = 2\)

iii) \(5x + 3 = 28\)
\(5x = 28 - 3\)
\(5x = 25\)
\(x = \frac{25}{5} = 5\)

iv) \(7 + 3y = -14\)
\(3y = -14 - 7\)
\(3y = -21\)
\(y = \frac{-21}{3} = -7\)

v) \(y + \frac{3}{2} = 6\)
\(y = 6 - \frac{3}{2}\)
\(y = \frac{12}{2} - \frac{3}{2} = \frac{9}{2} = 4.5\)
In simple words: Isolate the variable term on one side by transposing numbers to the other side.

Exam Tip: When shifting a term across the equals sign, always change its sign (+ to -, and - to +).

Question 10. A bus covers a distance of 220 Km in 4 hours. Find i) how long will it take to cover a distance of 495 Km? ii) how far it will travel in 5 hours 30 minutes?
Answer: Speed of the bus = \(\frac{\text{Distance}}{\text{Time}} = \frac{220}{4} = 55\text{ Km/h}\)
i) Time taken to cover 495 Km:
\(\text{Time} = \frac{\text{Distance}}{\text{Speed}} = \frac{495}{55} = 9\text{ hours}\)
ii) Distance travelled in 5 hours 30 minutes:
\(5\text{ hours } 30\text{ minutes} = 5.5\text{ hours}\)
\(\text{Distance} = \text{Speed} \times \text{Time} = 55 \times 5.5 = 302.5\text{ Km}\)
In simple words: First find the speed per hour, then use it to calculate time or distance.

Question 11. If 22.5 metres of uniform rod weigh 101.25 Kg, What will be the weight of 7 metres of the same rod?
Answer: Weight of 22.5 m of rod = 101.25 Kg
Weight of 1 m of rod = \(\frac{101.25}{22.5} = 4.5\text{ Kg}\)
Weight of 7 m of rod = \(7 \times 4.5 = 31.5\text{ Kg}\)
In simple words: Find the unit weight (weight of 1 meter) first using unitary method, then multiply by 7.

Exam Tip: Unitary method always requires finding the value of one unit first.

Question 12. 13 iron balls of the same size weigh 8 Kg 710 g. How many of them will weigh 3 Kg 350 gm.
Answer: Total weight of 13 iron balls = \(8\text{ Kg } 710\text{ g} = 8710\text{ g}\)
Weight of 1 iron ball = \(\frac{8710}{13} = 670\text{ g}\)
Target weight = \(3\text{ Kg } 350\text{ g} = 3350\text{ g}\)
Number of balls = \(\frac{3350}{670} = 5\)
In simple words: Convert all weights to grams, find the weight of 1 ball, then divide the target weight by it.

Exam Tip: Keep units consistent (all grams) to avoid calculation errors.

Question 13. Fifteen inlands cost Rs. 22.50. What will be the cost of 36 inlands? How many inlands can we buy for Rs. 67.5?
Answer: Cost of 15 inlands = Rs. 22.50
Cost of 1 inland = \(\frac{22.50}{15} = \text{Rs. } 1.50\)
Cost of 36 inlands = \(36 \times 1.50 = \text{Rs. } 54.00\)
Number of inlands for Rs. 67.5 = \(\frac{67.5}{1.50} = 45\)
In simple words: Find the cost of one item first, then multiply for total cost or divide for total quantity.

Exam Tip: Double-check decimal placements when dividing money amounts.

Question 14. Find the ratio and express in simplest form:
i) Rs. 15 to Rs. 20
ii) 125 gm to 3 Kg
iii) one hour to 36 minutes
iv) 6 hours to a day
v) a dozen to a score
Answer:
i) Rs. 15 to Rs. 20:
\(\frac{15}{20} = \frac{3}{4} = 3:4\)

ii) 125 gm to 3 Kg:
\(3\text{ Kg} = 3000\text{ gm}\)
\(\frac{125}{3000} = \frac{5}{120} = \frac{1}{24} = 1:24\)

iii) One hour to 36 minutes:
\(1\text{ hour} = 60\text{ minutes}\)
\(\frac{60}{36} = \frac{5}{3} = 5:3\)

iv) 6 hours to a day:
\(1\text{ day} = 24\text{ hours}\)
\(\frac{6}{24} = \frac{1}{4} = 1:4\)

v) A dozen to a score:
\(1\text{ dozen} = 12\), \(1\text{ score} = 20\)
\(\frac{12}{20} = \frac{3}{5} = 3:5\)
In simple words: Convert both quantities to the same units before simplifying the fraction.

Exam Tip: Ratios have no units, but both quantities must be in the same unit when comparing.

Question 15. If 72 person working in an office, 28 are men and the remaining are women. Find the ratio of the number of:
i) men to that of women
ii) men to the total number of persons
iii) total persons to the women
Answer: Total persons = 72
Number of men = 28
Number of women = \(72 - 28 = 44\)
i) Men to women = \(\frac{28}{44} = \frac{7}{11} = 7:11\)
ii) Men to total persons = \(\frac{28}{72} = \frac{7}{18} = 7:18\)
iii) Total persons to women = \(\frac{72}{44} = \frac{18}{11} = 18:11\)
In simple words: Subtract men from total to find women, then write the required ratios as simplified fractions.

Exam Tip: Always write the numerator and denominator in the exact order requested in the question.

Question 16. A shopkeeper earns Rs. 5265 monthly and spends Rs. 2340. Find the ratio between his expenditure and savings.
Answer: Monthly earning = Rs. 5265
Expenditure = Rs. 2340
Savings = Earning - Expenditure = \(5265 - 2340 = \text{Rs. } 2925\)
Ratio of expenditure to savings = \(\frac{\text{Expenditure}}{\text{Savings}} = \frac{2340}{2925}\)
Divide both by 585 (or simplify step-by-step):
\(\frac{2340 \div 117}{2925 \div 117} = \frac{20}{25} = \frac{4}{5} = 4:5\)
In simple words: Calculate savings first by subtracting expenditure from earnings, then form the ratio.

Exam Tip: Read carefully whether the ratio asks for earnings, savings, or expenditure.

Question 17. The first, third and the fourth terms of a proportion are 36, 28 and 63 respectively. Find the second term.
Answer: Let the second term be \(x\).
The terms are in proportion: \(36 : x :: 28 : 63\)
Product of extremes = Product of means
\(36 \times 63 = x \times 28\)
\(x = \frac{36 \times 63}{28}\)
\(x = \frac{9 \times 63}{7} = 9 \times 9 = 81\)
In simple words: Use the property that the product of extremes equals the product of means.

Exam Tip: In proportion \(a:b :: c:d\), \(a \times d = b \times c\).

Question 19. Find 'x', if x, 6, 55, 11 are in proportion.
Answer: Given proportion: \(x : 6 :: 55 : 11\)
Product of extremes = Product of means
\(x \times 11 = 6 \times 55\)
\(11x = 330\)
\(x = \frac{330}{11} = 30\)
In simple words: Set the product of the outer terms equal to the product of the inner terms.

Exam Tip: Always double check your multiplication and division steps.

Question 20. The ratio of the length of a school ground to its width is 5 : 2. Find its length, if the width is 40 metres.
Answer: Ratio of length to width = \(5 : 2\)
Let length = \(5x\) and width = \(2x\)
Given width = 40 metres
\(2x = 40 \implies x = 20\)
Length = \(5x = 5 \times 20 = 100\text{ metres}\)
In simple words: Use a variable \(x\) with the given ratio terms and equate the width to 40 to find \(x\).

Exam Tip: Introducing a common multiplier \(x\) is the safest method for ratio word problems.

Question 21. Express as ratio, 'the number of girls in a school is one-third of the number of boys.'
Answer: Let the number of boys be \(B\).
Number of girls = \(\frac{1}{3}B\)
Ratio of girls to boys = \(\frac{\text{Girls}}{\text{Boys}} = \frac{\frac{1}{3}B}{B} = \frac{1}{3} = 1:3\)
In simple words: Write girls as \(1/3\) of boys and form the ratio.

Question 22. Anita had half of the cake she baked. Express in ratio the cake she baked to the remaining cake.
Answer: Total cake baked = 1 whole cake
Cake Anita had = \(\frac{1}{2}\) of the cake
Remaining cake = \(1 - \frac{1}{2} = \frac{1}{2}\) of the cake
Ratio of total cake baked to remaining cake = \(\frac{1}{\frac{1}{2}} = 2:1\)
In simple words: Compare the entire cake (1) to the half that remains.

Exam Tip: Be careful whether the question asks for remaining cake to total, or total to remaining.

Question 23. Express each of the following as percent:
i) \(\frac{12}{5}\)
ii) \(1\frac{3}{5}\)
iii) 0.35
iv) \(\frac{2}{3}\)
v) 3.05
Answer:
i) \(\frac{12}{5} = \frac{12}{5} \times 100\% = 12 \times 20\% = 240\%\)

ii) \(1\frac{3}{5} = \frac{8}{5} = \frac{8}{5} \times 100\% = 8 \times 20\% = 160\%\)

iii) \(0.35 = 0.35 \times 100\% = 35\%\)

iv) \(\frac{2}{3} = \frac{2}{3} \times 100\% = \frac{200}{3}\% = 66\frac{2}{3}\%\)

v) \(3.05 = 3.05 \times 100\% = 305\%\)
In simple words: Multiply any number or fraction by 100 to convert it into a percentage.

Exam Tip: Don't forget to put the percentage sign (%) at the end of your answer.

Question 24. Express the following as fractions in the simplest form:
i) 35%
ii) 5.7%
iii) \(13\frac{1}{3}\%\)
iv) 0.75%
Answer:
i) \(35\% = \frac{35}{100} = \frac{7}{20}\)

ii) \(5.7\% = \frac{5.7}{100} = \frac{57}{1000}\)

iii) \(13\frac{1}{3}\% = \frac{40}{3}\% = \frac{40}{3 \times 100} = \frac{40}{300} = \frac{2}{15}\)

iv) \(0.75\% = \frac{0.75}{100} = \frac{75}{10000} = \frac{3}{400}\)
In simple words: Remove the percent sign by dividing by 100, then simplify the fraction.

Exam Tip: Reduce fractions to their lowest terms by dividing numerator and denominator by their HCF.

Question 25. Express as decimal:
i) 65%
ii) 9%
iii) 13.34%
iv) 0.05%
Answer:
i) \(65\% = \frac{65}{100} = 0.65\)

ii) \(9\% = \frac{9}{100} = 0.09\)

iii) \(13.34\% = \frac{13.34}{100} = 0.1334\)

iv) \(0.05\% = \frac{0.05}{100} = 0.0005\)
In simple words: Divide the percentage by 100 (move the decimal point two places to the left).

Exam Tip: Moving the decimal two places left is a shortcut for dividing by 100.

Question 26. Find the value of the following:
i) 25% of Rs. 160
ii) 70% of 125 kg
iii) \(12\frac{1}{2}\%\) of Rs. 130.80
iv) 15% of 16.250 litres
Answer:
i) \(25\% \text{ of Rs. } 160 = \frac{25}{100} \times 160 = \frac{1}{4} \times 160 = \text{Rs. } 40\)

ii) \(70\% \text{ of } 125\text{ kg} = \frac{70}{100} \times 125 = \frac{7}{10} \times 125 = \frac{875}{10} = 87.5\text{ kg}\)

iii) \(12\frac{1}{2}\% \text{ of Rs. } 130.80 = \frac{25}{200} \times 130.80 = \frac{1}{8} \times 130.80 = \text{Rs. } 16.35\)

iv) \(15\% \text{ of } 16.250\text{ litres} = \frac{15}{100} \times 16.250 = 0.15 \times 16.250 = 2.4375\text{ litres}\)
In simple words: Convert the percentage to a fraction or decimal and multiply by the given quantity.

Exam Tip: "Of" means multiplication in mathematics.

Question 27. What percent is each of the following:
i) Rs. 15 of Rs. 165
ii) 36min of 2 hours
iii) 0.5 of 4
iv) 125 metres of 3 Km
Answer:
i) \(\frac{15}{165} \times 100\% = \frac{1}{11} \times 100\% = \frac{100}{11}\% = 9\frac{1}{11}\%\)

ii) \(2\text{ hours} = 120\text{ minutes}\)
\(\frac{36}{120} \times 100\% = \frac{3}{10} \times 100\% = 30\%\)

iii) \(\frac{0.5}{4} \times 100\% = 0.125 \times 100\% = 12.5\%\)

iv) \(3\text{ Km} = 3000\text{ metres}\)
\(\frac{125}{3000} \times 100\% = \frac{125}{30}\% = \frac{25}{6}\% = 4\frac{1}{6}\%\)
In simple words: Put the part over the whole, ensure units match, and multiply by 100.

Exam Tip: Always make sure both quantities are in the same unit before finding the percentage.

Question 28. Amit saves 15% of his salary. If he earns Rs. 12160 per month as salary, find his monthly expenditure.
Answer: Salary = Rs. 12160
Savings = 15% of Rs. 12160
\(\text{Savings} = \frac{15}{100} \times 12160 = \frac{3}{20} \times 12160 = 3 \times 608 = \text{Rs. } 1824\)
Expenditure = Salary - Savings
\(\text{Expenditure} = 12160 - 1824 = \text{Rs. } 10336\)
In simple words: Calculate savings amount and subtract it from the total salary to get expenditure.

Exam Tip: Alternatively, expenditure is \(100\% - 15\% = 85\%\) of the total salary.

Question 29. In a fabric, cotton and synthetic fibres are in the ratio 2:3. What is percentage of cotton fibre in the fabric?
Answer: Ratio of cotton to synthetic = \(2 : 3\)
Total parts = \(2 + 3 = 5\)
Fraction of cotton fibre = \(\frac{2}{5}\)
Percentage of cotton fibre = \(\frac{2}{5} \times 100\% = 2 \times 20\% = 40\%\)
In simple words: Find the fraction of cotton out of total parts and multiply by 100 to get the percentage.

Exam Tip: Total ratio parts always form the denominator of the fraction.

Question 30. A boy got 65 out of 90 in Hindi and 33 out of 40 in Mathematics. In which subject his percentage of marks is better.
Answer: Percentage in Hindi = \(\frac{65}{90} \times 100\% = \frac{650}{9}\% = 72.22\%\)
Percentage in Mathematics = \(\frac{33}{40} \times 100\% = \frac{330}{4}\% = 82.5\%\)
Since \(82.5\% > 72.22\%\), his percentage of marks is better in Mathematics.
In simple words: Convert both scores into percentages to easily compare them.

Question 31. A school contributed Rs. 3000 for the relief of flood victims. If 15% of the amount was contributed by Anshul alone. How much did he alone contributed?
Answer: Total contribution = Rs. 3000
Anshul's contribution = 15% of Rs. 3000
\(\text{Anshul's contribution} = \frac{15}{100} \times 3000 = 15 \times 30 = \text{Rs. } 450\)
In simple words: Calculate 15 percent of the total contribution amount.

Exam Tip: Cancel out zeros in numerator and denominator for quick calculation.

Question 32. Divide Rs. 640 among A, B, C in the ratio 5 : 7 : 8.
Answer: Total amount = Rs. 640
Ratio parts = \(5 + 7 + 8 = 20\)
A's share = \(\frac{5}{20} \times 640 = \frac{1}{4} \times 640 = \text{Rs. } 160\)
B's share = \(\frac{7}{20} \times 640 = 7 \times 32 = \text{Rs. } 224\)
C's share = \(\frac{8}{20} \times 640 = 8 \times 32 = \text{Rs. } 256\)
In simple words: Add ratio parts to get the total denominator, then multiply each fraction by the total amount.

Exam Tip: Always verify that the sum of individual shares equals the total amount divided (\(160 + 224 + 256 = 640\)).

Question 33. The perimeter of a rectangle is 25.5m. Its length is 9.5m. Calculate its area in square metres.
Answer: Perimeter = \(2(l + b) = 25.5\text{ m}\)
Length (\(l\)) = 9.5 m
\(2(9.5 + b) = 25.5\)
\(9.5 + b = \frac{25.5}{2} = 12.75\text{ m}\)
Breadth (\(b\)) = \(12.75 - 9.5 = 3.25\text{ m}\)
Area = Length \(\times\) Breadth = \(9.5 \times 3.25 = 30.875\text{ sq. m}\)
In simple words: Find the breadth using the perimeter formula, then multiply length by breadth to find the area.

Exam Tip: Watch your decimal arithmetic carefully when multiplying lengths and breadths.

Question 34. The perimeter of a square is 24m. Find its area.
Answer: Perimeter of square = \(4 \times \text{side} = 24\text{ m}\)
Side = \(\frac{24}{4} = 6\text{ m}\)
Area = \(\text{side} \times \text{side} = 6 \times 6 = 36\text{ sq. m}\)
In simple words: Divide the perimeter by 4 to get the side length, then square it to find the area.

Exam Tip: Units for area are always square units (e.g., sq. m or \(\text{m}^2\)).

Question 36. How many envelopes can be made out of a sheet of a paper 324cm by 172cm, if each envelope requires a piece of paper of size 18cm.
Answer: Area of the large sheet of paper = \(324 \times 172 = 55728\text{ sq. cm}\)
Assuming each envelope requires a square piece of size \(18\text{ cm} \times 18\text{ cm}\):
Area of one envelope piece = \(18 \times 18 = 324\text{ sq. cm}\)
Number of envelopes = \(\frac{\text{Area of sheet}}{\text{Area of one envelope}} = \frac{324 \times 172}{324} = 172\)
In simple words: Divide the total area of the sheet by the area required for one envelope.

Exam Tip: You can often cancel dimensions directly before full multiplication to save time.

Question 37. How many carpets of size 6m x 4m are required to cover the floor of the hall of size 40m x 24m.
Answer: Area of the hall = \(40\text{ m} \times 24\text{ m} = 960\text{ sq. m}\)
Area of one carpet = \(6\text{ m} \times 4\text{ m} = 24\text{ sq. m}\)
Number of carpets required = \(\frac{\text{Area of hall}}{\text{Area of one carpet}} = \frac{960}{24} = 40\)
In simple words: Divide the total floor area of the hall by the area of a single carpet.

Exam Tip: Ensure all dimensions are in the same unit before calculating areas.

Question 38. Each side of a square measures 62m 5dm, Find its area in sq.metres. By how many square metres does it fall short of a hectare.
Answer: Convert side into meters: \(1\text{ dm} = 0.1\text{ m}\), so \(5\text{ dm} = 0.5\text{ m}\).
Side = \(62.5\text{ m}\)
Area of the square = \(62.5 \times 62.5 = 3906.25\text{ sq. m}\)
1 hectare = \(10000\text{ sq. m}\)
Shortfall from a hectare = \(10000 - 3906.25 = 6093.75\text{ sq. m}\)
In simple words: Calculate the area in square meters and subtract it from 10,000 (since 1 hectare = 10,000 sq. m).

Question 39. A string 36m long is bent in the shape of a square. Find the area of the square. Also, if the same string is used to make a rectangle of length 10m. Find the area covered by the rectangle.
Answer: Perimeter of the square = Length of string = 36 m
Side of the square = \(\frac{36}{4} = 9\text{ m}\)
Area of the square = \(9 \times 9 = 81\text{ sq. m}\)
For the rectangle, perimeter = 36 m, length (\(l\)) = 10 m
Perimeter = \(2(l + b) \implies 2(10 + b) = 36\)
\(10 + b = 18 \implies b = 8\text{ m}\)
Area of the rectangle = \(l \times b = 10 \times 8 = 80\text{ sq. m}\)
In simple words: The length of the string is the perimeter for both shapes. Use perimeter to find side or breadth, then find areas.

Exam Tip: Same perimeter does not mean same area!

Question 40. Area of a rectangle is same as that of a square of side 18m. Find the breadth of the rectangle if its length is 24m.
Answer: Side of the square = 18 m
Area of the square = \(18 \times 18 = 324\text{ sq. m}\)
Area of the rectangle = Area of the square = \(324\text{ sq. m}\)
Length of the rectangle = 24 m
Breadth = \(\frac{\text{Area}}{\text{Length}} = \frac{324}{24} = 13.5\text{ m}\)
In simple words: Calculate the square's area, set it equal to the rectangle's area, and divide by length to get breadth.

Exam Tip: Always equate areas when the problem states one shape's area equals another's.

Question 41. If a symbol of a car represents 5 cars, then 7 symbols will represent how many cars?
Answer: Value of 1 symbol = 5 cars
Number of symbols = 7
Total cars represented = \(7 \times 5 = 35\text{ cars}\)
In simple words: Multiply the number of symbols by the value each symbol represents.

Exam Tip: Check the pictograph scale carefully before doing any calculations.

Question 42. State two properties of the following:
i) Rhombus
ii) Parallelogram
iii) Square
iv) Trapezium
v) Rectangle
Answer:
i) Rhombus: All sides are equal in length; diagonals bisect each other at right angles.
ii) Parallelogram: Opposite sides are equal and parallel; opposite angles are equal.
iii) Square: All four sides are equal and all four angles are 90 degrees; diagonals are equal and bisect at right angles.
iv) Trapezium: At least one pair of opposite sides is parallel; adjacent angles between parallel lines add up to 180 degrees.
v) Rectangle: Opposite sides are equal and parallel; each interior angle is 90 degrees.
In simple words: List key geometric characteristics of each quadrilateral.

Exam Tip: Learn properties of quadrilaterals by comparing their sides, angles, and diagonals.

Question 43. Construct concentric circles with center O and radius 3cm and 4cm.
Answer:
1. Mark a point O on the paper as the center.
2. Set the compass pointer to the zero mark of a ruler and open the pencil lead to 3 cm.
3. With O as the center, draw a circle of radius 3 cm.
4. Without changing the center O, adjust the compass radius to 4 cm.
5. Draw another circle with center O and radius 4 cm.
In simple words: Draw two circles from the exact same center point with different radii (3 cm and 4 cm).

Exam Tip: Ensure the compass is held firmly at the center point so it does not slip while drawing.

Question 44. Draw a line segment PQ = 6.4 cm. Cut off PC, such that PC = 3cm. Measure CQ.
Answer:
1. Draw a line segment \(PQ = 6.4\text{ cm}\) using a ruler.
2. From point P, measure along the line segment a distance of 3 cm and mark point C.
3. Measure the remaining segment CQ with a ruler.
4. \(CQ = PQ - PC = 6.4 - 3.0 = 3.4\text{ cm}\).
In simple words: Draw a 6.4 cm line, mark 3 cm from the start point P, and measure the leftover part to Q.
Exam Tip: Always align the zero mark of your ruler precisely with the starting endpoint.

Question 45. Study the pictograph and answer the questions:
TOYS MANUFACTURED IN A FACTORY
SCALE: EACH SYMBOL REPRESENT 5 TOYS
[Pictograph shows: Jan (5), Feb (8), March (6), April (7), May (5), June (3), July (2) symbols approximately as in source]
i) In which month maximum toys where manufactured?
ii) In which month minimum toys where manufactured?
iii) Find the ratio of toys manufactured in February to April?
iv) If a toy cost Rs. 20. What is total Sales of the factory in month of March?
v) If 10 toys can be packed in a cartoon. Then how many cartoons are needed to pack the toys manufactured in all the given months?
Answer:
i) Maximum toys were manufactured in February (8 symbols \(\times 5 = 40\) toys).
ii) Minimum toys were manufactured in July (2 symbols \(\times 5 = 10\) toys).
iii) February toys = \(8 \times 5 = 40\). April toys = \(7 \times 5 = 35\).
Ratio = \(\frac{40}{35} = \frac{8}{7} = 8:7\).
iv) March toys = \(6 \times 5 = 30\) toys.
Total sales in March = \(30 \times \text{Rs. } 20 = \text{Rs. } 600\).
v) Total symbols across all months (Jan to July) = \(5 + 8 + 6 + 7 + 5 + 3 + 2 = 36\) symbols.
Total toys manufactured = \(36 \times 5 = 180\) toys.
Number of cartons needed = \(\frac{180}{10} = 18\) cartons.
In simple words: Multiply symbol counts by 5 to find actual toy numbers, then answer each sub-question accordingly.
Exam Tip: Always multiply by the scale factor given in the pictograph before performing comparisons or totals.