CBSE Class 6 Mathematics Revision Worksheet Set 20

Ch. 6 : INTEGERS

I Fill in the blanks

Question 1. The integer 5 more than (-2) is ______________
Answer: 3
In simple words: Moving 5 units to the right of (-2) on the number line gives 3.

Exam Tip: "More than" means addition: \(-2 + 5 = 3\).

Question 2. The greatest negative integer is ________________
Answer: -1
In simple words: (-1) is the closest negative integer to zero, making it the largest among all negative numbers.

Exam Tip: As negative numbers increase in magnitude, their value decreases.

Question 3. All natural numbers are ________________ integers
Answer: positive
In simple words: Natural numbers start from 1, 2, 3... which are all positive numbers.

Exam Tip: Natural numbers do not include zero or negative numbers.

Question 4. 25 – (-2) = ________________
Answer: 27
In simple words: Subtracting a negative number is the same as adding its positive counterpart.

Exam Tip: Remember the rule: minus multiplied by minus gives plus (\(-\times- = +\)). So, \(25 - (-2) = 25 + 2 = 27\).

Question 5. (-35) + (-15) = ________________
Answer: -50
In simple words: When adding two negative numbers, add their absolute values and keep the negative sign.

Exam Tip: Combine debts: spending 35 and then spending 15 results in a total debt of 50.

Question 6. 46 + (-98) = ________________
Answer: -52
In simple words: Subtract the smaller absolute value from the larger one and give the result the sign of the larger number.

Exam Tip: \(98 - 46 = 52\), and since 98 has a minus sign, the answer is -52.

Question 7. The additive inverse of (-13) is ________________
Answer: 13
In simple words: The additive inverse is the number you add to a given number to get zero.

Exam Tip: Changing the sign of a number gives its additive inverse.

Question 8. The integer which is neither positive nor negative is ________________
Answer: 0
In simple words: Zero is neutral; it has no sign.

Exam Tip: Zero separates positive and negative integers on the number line.

Question 9. 0 is larger than every ________________ integer.
Answer: negative
In simple words: Any negative number is always less than zero.

Exam Tip: Zero lies to the right of all negative numbers on the number line.

Question 10. ________________ + (-14) = -9
Answer: 5
In simple words: Find what number added to (-14) gives (-9).

Exam Tip: Use transposition: \(x = -9 - (-14) = -9 + 14 = 5\).

II Answer the following :

Question 1. Represent as an integer with suitable sign.
a. Saving Rs. 2500. b. Going 450 m below the sea level.
Answer:
a. +Rs. 2500 (or 2500)
b. -450 m
In simple words: Savings represent a gain (positive), while going below sea level represents a depth (negative).

Exam Tip: Use positive signs for gains, heights above sea level, and credits. Use negative signs for losses, depths, and debits.

Question 2. Add using number line.
a. 5 more than (-3)
b. (-2) + (-6)
Answer:
a. Start at (-3) on the number line and move 5 units to the right. We reach 2.
b. Start at (-2) on the number line and move 6 units to the left. We reach (-8).
In simple words: Addition on a number line means moving right for positive numbers and left for negative numbers.

Exam Tip: Always state your starting point and the direction of movement clearly.

Question 3. Arrange in descending order .
(-45), (-86), (-17), (-95)
Answer: -17, -45, -86, -95
In simple words: Descending order means arranging numbers from largest to smallest. Among negative numbers, the one with the smaller numerical value is larger.

Exam Tip: Think of negative numbers as temperatures or debts; smaller debt means a larger value.

Question 4. Write six integers less than (-4).
Answer: -5, -6, -7, -8, -9, -10
In simple words: Numbers to the left of (-4) on the number line are smaller than (-4).

Exam Tip: Move further left on the number line to find smaller integers.

Question 5. Which number will be reach if we move 5 numbers to the left of (-6)?
Answer: -11
In simple words: Moving left means subtracting: \(-6 - 5 = -11\).

Exam Tip: Left movement corresponds to subtraction or adding a negative number.

Question 6. Which is greater ? (-23) or (-45)
Answer: -23 is greater.
In simple words: (-23) is closer to zero than (-45), making it greater.

Exam Tip: The number positioned further to the right on the number line is always greater.

Question 7. Find the sum of (-39) and (-63)
Answer:
\((-39) + (-63) = -39 - 63\)
\(= -102\)
In simple words: Add the two numbers together and attach a negative sign to the total.

Exam Tip: Sum of two negative integers is always negative.

Question 8. Subtract (-8) from 25.
Answer:
\(25 - (-8) = 25 + 8\)
\(= 33\)
In simple words: Subtracting negative 8 is the same as adding positive 8.

Exam Tip: Pay close attention to the order of subtraction: "subtract A from B" means \(B - A\).

Question 9. If we are at 5 on the number line, in which direction should we move to reach (-9)?
Answer: Left direction.
In simple words: Since (-9) is smaller than 5, we must move to the left.

Exam Tip: Moving towards a smaller number always requires moving left.

Question 10. Add :- (-460) + 125 + (-325) + 250
Answer:
Group positive and negative numbers:
\(= (-460 - 325) + (125 + 250)\)
\(= -785 + 375\)
\(= -410\)
In simple words: Combine all negative numbers together and all positive numbers together before finding the final difference.

Exam Tip: Grouping like-signed integers simplifies addition and minimizes calculation errors.

Question 11. Subtract : a) (-40) - (25) b) 105 - (-76)
Answer:
a) \((-40) - (25) = -40 - 25 = -65\)
b) \(105 - (-76) = 105 + 76 = 181\)
In simple words: Combine negative quantities for part (a), and change subtraction of a negative to addition for part (b).

Exam Tip: Double check sign conversions before final calculation.

Question 12. Find a) (-42) + 26 - (-15) b) (-25) - 14 - (-75)
Answer:
a) \(-42 + 26 + 15 = -42 + 41 = -1\)
b) \(-25 - 14 + 75 = -39 + 75 = 36\)
In simple words: Convert double negatives to additions, then combine positives and negatives step by step.

Exam Tip: Always perform arithmetic operations from left to right after clearing parentheses.

Ch. 9 : DATA HANDLING

I Fill in the blanks

Question 1. The graph represents data in the form of pictures is known as _____________.
Answer: pictograph
In simple words: A pictograph uses symbols or pictures to represent data quantities.

Exam Tip: Remember the key prefix "picto-" relates to pictures.

Question 2. A graph drawn using bars of uniform width is called _______________.
Answer: bar graph
In simple words: Bar graphs use rectangular bars whose heights or lengths represent values.

Exam Tip: Bar widths must always be uniform for accurate comparison.

Question 3. A collection of numbers gathered to give some information is called _______________.
Answer: data
In simple words: Data is a collection of facts and figures gathered for analysis.

Exam Tip: Data can be raw (unorganized) or grouped.

Question 2. The height (in cm) of 20 students of class VI are given below
125, 130, 135, 142, 125, 139, 150, 147,
142, 139, 125, 130, 125, 150, 147, 139,
152, 130, 152, 142.
Prepare a table using tally marks.
Answer:

In simple words: Group identical heights together and count them using tally marks (groups of five with a diagonal slash).

Exam Tip: Always cross off items from the raw list as you record them to avoid missing any numbers or double counting.

Question a. Find how many students are of the height 142 cm ?
Answer: 3 students
In simple words: Looking at our tally table, the frequency for height 142 cm is 3.

Exam Tip: Verify counts directly from your frequency table.

Question b. How many students are there of height less than or equal to 135 cm?
Answer: Students with height \(\le 135\) cm include heights 125 cm, 130 cm, and 135 cm.
Number of students = \(4 + 3 + 1 = 8\)
In simple words: Add up the frequencies of all heights that are 135 cm or shorter.

Exam Tip: "Less than or equal to" means include the boundary value (135 cm).

Question c. How many students are of the height more than 147 cm?
Answer: Students with height \(> 147\) cm include heights 150 cm and 152 cm.
Number of students = \(2 + 2 = 4\)
In simple words: Add the frequencies of all heights strictly greater than 147 cm.

Exam Tip: "More than" means strictly greater (exclude 147 cm itself).

Question III. Observe the following bar graph and answer the following questions :
(Bar graph showing shirt sales: Mon: 45, Tue: 32, Wed: 40, Thu: 16, Fri: 50, Sat: 22)
a. What information does the above bar graph give ?
Answer: The bar graph gives the number of shirts sold on different days of a week.
In simple words: The graph shows shirt sales across various days.

Exam Tip: Read the title and axis labels to state the graph's purpose accurately.

Question b. How many shirts were sold on Friday?
Answer: 50 shirts
In simple words: The bar for Friday reaches the level of 50 on the vertical axis.

Exam Tip: Trace horizontally from the top of the Friday bar to the Y-axis.

Question c. On which day the minimum no. of shirts sold?
Answer: Thursday (16 shirts)
In simple words: Thursday has the shortest bar among all days.

Exam Tip: The shortest bar represents the minimum value.

Question d. What is the scale chosen on the vertical line representing no. of shirts ?
Answer: 1 unit = 5 shirts (or 1 cm = 5 units)
In simple words: The markings on the vertical axis increase by intervals of 5.

Exam Tip: Look at the spacing between consecutive numbers on the Y-axis.

Question e. How many more shirts sold on Wednesday than Tuesday ?
Answer: Shirts sold on Wednesday = 40
Shirts sold on Tuesday = 32
Difference = \(40 - 32 = 8\) shirts
In simple words: Subtract Tuesday's sales from Wednesday's sales.

Exam Tip: "How many more" always indicates subtraction between two values.

Question IV. Following table shows the monthly expenditure of a family on various items. Draw a bar graph.
(House Rent: 900, Food: 500, Education: 700, Electricity: 200, Transport: 400, Clothes: 800)
Answer:
Steps to draw the bar graph:
1. Draw horizontal X-axis (Items) and vertical Y-axis (Expenditure in Rs.).
2. Choose a suitable scale on the Y-axis: e.g., 1 unit = 100 Rs. (so 100, 200, 300, ..., 1000).
3. Draw bars of uniform width for each item corresponding to its expenditure value.
In simple words: Represent each expenditure item as a vertical bar whose height is proportional to the money spent.

Exam Tip: Ensure equal spacing between bars and uniform bar width.

Question V. The following are the number of bulbs purchased for a house during the first months of a year. Represent the details by a pictograph using one symbol to represent 10 bulbs :
Months: January (30), February (35), March (20), April (26)
Answer:
Given scale: 1 symbol (\(\odot\)) = 10 bulbs.
Calculations for symbols:
- January (30 bulbs) = \(30 \div 10 = 3\) symbols
- February (35 bulbs) = \(35 \div 10 = 3.5\) symbols (3 full symbols and 1 half symbol)
- March (20 bulbs) = \(20 \div 10 = 2\) symbols
- April (26 bulbs) = \(26 \div 10 = 2.6\) symbols (2 full symbols and 1 fraction symbol or approximate representation)

In simple words: Divide each month's total by 10 to find out how many bulb symbols to draw.

Exam Tip: Always mention the key/scale clearly below or alongside the pictograph.

Ch. 11 : Algebra

I Write the expression for the following :-

Question a. sum of 5 and x divided by 8
Answer: \(\frac{5 + x}{8}\)
In simple words: Add 5 and x first, then divide the result by 8.

Exam Tip: Use parentheses or fraction bars to indicate order of operations.

Question b. Product of 7 and y is added to twice ‘x’
Answer: \(7y + 2x\)
In simple words: Multiply 7 by y, then add 2 times x to it.

Exam Tip: "Product of 7 and y" is written as \(7y\), and "twice x" is \(2x\).

Question c. (-6) is multiplied by ‘n’ and the result is added to 8.
Answer: \(-6n + 8\)
In simple words: Multiply (-6) by n and then add 8.

Exam Tip: \((-6) \times n\) is written compactly as \(-6n\).

Question d. ‘z’ is subtracted from (-12)
Answer: \(-12 - z\)
In simple words: Subtract z from (-12).

Exam Tip: "Subtracted from" reverses the order of terms: A subtracted from B is \(B - A\).

Question e. 5 more than (-x) gives 14.
Answer: \(-x + 5 = 14\)
In simple words: Adding 5 to (-x) results in 14.

Exam Tip: "Gives" translates to the equals sign \((=)\).

II Fill in the blanks.

Question 1. The solution of the equation x + 6 = (-20)
Answer: -26
In simple words: Solving for x gives \(x = -20 - 6 = -26\).

Exam Tip: Transpose +6 to the other side as -6.

Question 2. The variables in the expression 3x + y are ___________
Answer: x and y
In simple words: Letters used to represent unknown quantities in an expression are variables.

Exam Tip: Numbers like 3 are constants, while letters are variables.

Question 3. The expression for the perimeter of a regular pentagon whose side is ‘p’ cm is ______________
Answer: 5p
In simple words: A regular pentagon has 5 equal sides, so its perimeter is 5 times the side length.

Exam Tip: Perimeter of any regular polygon = number of sides \(\times\) side length.

Question 4. Identify the equation in the following
2m < 30, 7 – x = 5, (t-7) > 5
Answer: 7 - x = 5
In simple words: An equation must contain an equality sign \((=)\).

Exam Tip: Inequalities use \(<\) or \(>\) symbols, whereas equations use \(=\).

Question 5. Express the diameter (d) of a circle in terms of its radius (r).
Answer: \(d = 2r\)
In simple words: The diameter is always twice the length of the radius.

Exam Tip: Standard circle relation: \(d = 2r\) or \(r = \frac{d}{2}\).

Question III. Complete the following table and find the solution of the equation 3 + 4x = 23
x: 0, 1, 2, 3, 4, 5
Answer:

Solution of the equation is \(x = 5\).
In simple words: Substitute each value of x into the expression \(3 + 4x\) until you get 23.

Exam Tip: The value of x that yields 23 is the solution of the equation.

Question IV. Pick out the solution and show that other values do not satisfy the equation.
4m + (-5) = 15 (0, 7, 3, 5)
Answer:
Let us test each value in \(4m - 5\):
- For \(m = 0\): \(4(0) - 5 = -5 \neq 15\) (Not satisfied)
- For \(m = 7\): \(4(7) - 5 = 28 - 5 = 23 \neq 15\) (Not satisfied)
- For \(m = 3\): \(4(3) - 5 = 12 - 5 = 7 \neq 15\) (Not satisfied)
- For \(m = 5\): \(4(5) - 5 = 20 - 5 = 15 = 15\) (Satisfied)
Therefore, the solution is \(m = 5\).
In simple words: Test each given option one by one by substituting it into the equation.

Exam Tip: Show substitution for all values to earn full credit.

Question V. Let Raju’s age be ‘x’ years.
a. His father is 2 years more than 3 times his age. What is the fathers age?
b. His sister is 3 years younger than him what is her age?
Answer:
a. Father's age = \(3x + 2\) years
b. Sister's age = \(x - 3\) years
In simple words: Translate the word statements into algebraic expressions step by step.

Exam Tip: "3 times" means \(3x\), "2 years more than" means adding 2, and "younger" means subtracting.

Question VI. Form expressions using x, 4 and 5. Every expression must have ‘x’ in it. Use only two number operations.
Answer:
Possible expressions:
1. \(4x + 5\)
2. \(5x - 4\)
3. \(\frac{x}{4} + 5\)
4. \(4(x + 5)\)
In simple words: Combine x, 4, and 5 using operations like addition, subtraction, multiplication, or division.

Exam Tip: Ensure every expression contains variable x and uses at least two distinct operations.

Question VII. Identify the operations involved in the following expressions ;-
a) 2y + 17
b) \(\frac{y}{8} - 6\)
c) -7m + 3
Answer:
a) Multiplication (between 2 and y) and Addition (adding 17)
b) Division (dividing y by 8) and Subtraction (subtracting 6)
c) Multiplication (between -7 and m) and Addition (adding 3)
In simple words: List the mathematical operations used to build each algebraic expression.

Exam Tip: Look closely at coefficients and constant terms to identify operations.

Question VIII. Leela bought some toffees. She gives some toffees to her friends and family members. Still 9 toffees remain. If the no of toffees she gives away is ‘q’, how many toffees she bought?
Answer:
Total toffees bought = \(q + 9\)
In simple words: Add the toffees given away to the toffees remaining to find the total bought.

Exam Tip: Total = Given away + Remaining.

Question IX. Radha has some pencils with her she distributed the pencils equally to three of her children. Still 5 pencils were remaining. If the no. of pencils each child gets is taken as ‘y’, then how many pencils were there in total?
Answer:
Total pencils = \(3y + 5\)
In simple words: Multiply the number of children (3) by pencils each child gets (y) and add the 5 remaining pencils.

Exam Tip: Distributed equally among 3 children gives \(3y\), plus remaining 5 gives \(3y + 5\).

Ch. 12 : Ratio and proportion

Question 1. Find the ratio of the following
a. 15 minutes to an hour
b. 50 gm to 1.5 kg
c. 40 cm to 2.5 m
d. 15 paise to five rupees
e. 45 seconds to 2 minutes
f. 10 metres to 25 cm
g. 250 gm to a kilogram
Answer:
a. 15 minutes to 60 minutes (\(1\text{ hour} = 60\text{ minutes}\)):
\(\frac{15}{60} = \frac{1}{4} = 1:4\)
b. 50 gm to 1500 gm (\(1.5\text{ kg} = 1500\text{ gm}\)):
\(\frac{50}{1500} = \frac{1}{30} = 1:30\)
c. 40 cm to 250 cm (\(2.5\text{ m} = 250\text{ cm}\)):
\(\frac{40}{250} = \frac{4}{25} = 4:25\)
d. 15 paise to 500 paise (\(\text{Rs. } 5 = 500\text{ paise}\)):
\(\frac{15}{500} = \frac{3}{100} = 3:100\)
e. 45 seconds to 120 seconds (\(2\text{ minutes} = 120\text{ seconds}\)):
\(\frac{45}{120} = \frac{3}{8} = 3:8\)
f. 1000 cm to 25 cm (\(10\text{ m} = 1000\text{ cm}\)):
\(\frac{1000}{25} = \frac{40}{1} = 40:1\)
g. 250 gm to 1000 gm (\(1\text{ kg} = 1000\text{ gm}\)):
\(\frac{250}{1000} = \frac{1}{4} = 1:4\)
In simple words: Always convert both quantities to the same unit before writing their ratio in simplest form.

Exam Tip: Ratios have no units; always convert to identical units first.

Question 2. In a box containing 80 bulbs, 15 were found to be defective. Find the ratio of defective to good bulbs.
Answer:
Total bulbs = 80
Defective bulbs = 15
Good bulbs = \(80 - 15 = 65\)
Ratio of defective to good bulbs = \(\frac{15}{65} = \frac{3}{13} = 3:13\)
In simple words: Find the number of good bulbs by subtracting defective ones from the total, then simplify the ratio.

Exam Tip: Read carefully: ratio of defective to good means \(\frac{\text{Defective}}{\text{Good}}\).

Question 3. The ratio of the length and breadth of a football ground is 3 : 2. Find the length if the breadth is 28 metres.
Answer:
Let length = \(3x\) and breadth = \(2x\).
Given breadth = 28 m
\(2x = 28 \implies x = 14\)
Length = \(3x = 3 \times 14 = 42\text{ metres}\)
In simple words: Use a common variable x for the ratio parts, solve for x, and multiply to find the length.

Exam Tip: Letting parts be \(3x\) and \(2x\) is the standard method for ratio problems.

Question 4. Sumita gets pocket money of Rs. 500 per month. Out of which she saves Rs. 40 per month. Find the ratio of her savings to the amount she gets.
Answer:
Savings = Rs. 40
Total amount = Rs. 500
Ratio = \(\frac{40}{500} = \frac{4}{50} = \frac{2}{25} = 2:25\)
In simple words: Put savings in the numerator and total pocket money in the denominator, then simplify.

Exam Tip: Both quantities are already in the same unit (Rupees).

Question 5. Sumit buys pencils at the rate of Rs. 36 per dozen and pens at the rate of Rs. 72 per dozen. Find the ratio of cost of a pencil to cost of a pen.
Answer:
Cost of 1 dozen pencils = Rs. 36 \(\implies\) Cost of 1 pencil = \(\frac{36}{12} = \text{Rs. } 3\)
Cost of 1 dozen pens = Rs. 72 \(\implies\) Cost of 1 pen = \(\frac{72}{12} = \text{Rs. } 6\)
Ratio of cost of a pencil to a pen = \(\frac{3}{6} = \frac{1}{2} = 1:2\)
In simple words: Find the cost of 1 pencil and 1 pen by dividing the dozen cost by 12, then find their ratio.

Exam Tip: 1 dozen equals 12 items.

Question 6. Divide Rs. 200 between Rita and Gita in the ratio 2 : 3
Answer:
Total parts = \(2 + 3 = 5\)
Rita's share = \(\frac{2}{5} \times 200 = 2 \times 40 = \text{Rs. } 80\)
Gita's share = \(\frac{3}{5} \times 200 = 3 \times 40 = \text{Rs. } 120\)
In simple words: Add the ratio parts to find the total parts, then multiply each person's fraction by the total amount.

Exam Tip: Check your answer by adding the shares: \(80 + 120 = 200\).

Question 7. Determine if the following ratios from a proportion
a. 20 cm : 1 m and 3.5 \(\ell\) : 17.5 \(\ell\)
b. 2 kg : 80 kg and 25 g : 625 g
c. 200 ml : 2.5 \(\ell\) and 4 : Rs. 50
d. 440 m : 2 km and 55 cm : 3m
Answer:
a. \(20\text{ cm} : 100\text{ cm} = 1:5\); \(3.5 : 17.5 = 1:5\). Yes, they form a proportion.
b. \(2 : 80 = 1:40\); \(25 : 625 = 1:25\). No, they do not form a proportion.
c. \(200\text{ ml} : 2500\text{ ml} = 2:25\); Ratio 4 : 50 = \(2:25\). Yes, they form a proportion.
d. \(440\text{ m} : 2000\text{ m} = 11:50\); \(55\text{ cm} : 300\text{ cm} = 11:60\). No, they do not form a proportion.
In simple words: Convert quantities to the same units and check if both ratios simplify to the exact same fraction.

Exam Tip: Two ratios \(a:b\) and \(c:d\) are in proportion if \(a \times d = b \times c\).

Question 8. A worker is paid Rs. 560 for 5 days. What should be paid to the worker for 28 days.
Answer:
Pay for 5 days = Rs. 560
Pay for 1 day = \(\frac{560}{5} = \text{Rs. } 112\)
Pay for 28 days = \(112 \times 28 = \text{Rs. } 3136\)
In simple words: Find the pay for 1 day using division, then multiply by 28 for the total days.

Exam Tip: This is a standard unitary method problem.

Question 9. A family of 4 members consumes 6 kg of sugar in a month. What will be the monthly consumption of sugar if the number of family members becomes 6.
Answer:
Consumption for 4 members = 6 kg
Consumption for 1 member = \(\frac{6}{4} = 1.5\text{ kg}\)
Consumption for 6 members = \(1.5 \times 6 = 9\text{ kg}\)
In simple words: Calculate sugar needed per person first, then multiply by 6.

Exam Tip: More family members require more sugar (direct proportion).

Question 10. A car travels 165 km in 3 hrs.
a. How long will it take to travel 440 km?
b. How far will it travel in 7 hrs.
Answer:
Speed = \(\frac{165}{3} = 55\text{ km/hr}\)
a. Time to travel 440 km = \(\frac{440}{55} = 8\text{ hours}\)
b. Distance in 7 hours = \(55 \times 7 = 385\text{ km}\)
In simple words: Find the speed per hour first, then use it to find time or distance.

Exam Tip: Remember formulas: \(\text{Speed} = \frac{\text{Distance}}{\text{Time}}\), \(\text{Time} = \frac{\text{Distance}}{\text{Speed}}\), \(\text{Distance} = \text{Speed} \times \text{Time}\).

Question 11. 15 boys can type 270 pages a day how many pages 36 boys can type a day.
Answer:
Pages typed by 15 boys = 270
Pages typed by 1 boy in a day = \(\frac{270}{15} = 18\text{ pages}\)
Pages typed by 36 boys in a day = \(18 \times 36 = 648\text{ pages}\)
In simple words: Find out how many pages 1 boy types in a day, then multiply by 36.

Exam Tip: Unitary method applies directly here.

Question 12. The weight of 15 bags of rice is 112.5 kg
a. What is the weight of 10 such bags
b. How many bags will weigh 750 kg.
Answer:
Weight of 15 bags = 112.5 kg
Weight of 1 bag = \(\frac{112.5}{15} = 7.5\text{ kg}\)
a. Weight of 10 bags = \(7.5 \times 10 = 75\text{ kg}\)
b. Number of bags for 750 kg = \(\frac{750}{7.5} = 100\text{ bags}\)
In simple words: Find the weight of 1 bag, then multiply or divide to find the answers for parts (a) and (b).

Exam Tip: Double check decimal divisions when finding the weight of a single unit.

Ch. 14 : Practical Geometry

Question 1. Draw any circle and mark points x, y and z such that ‘x’ is on the circle, y in the interior of the circle and ‘z’ is in the exterior of the circle.
Answer:
Construction Steps:
1. Open the compass to any convenient radius, place the pointer at O, and draw a circle.
2. Mark point x on the boundary (circumference) of the circle.
3. Mark point y anywhere inside the region enclosed by the circle.
4. Mark point z anywhere outside the circle.
In simple words: Draw a circle and place x on the edge, y inside, and z outside.

Exam Tip: Label all points clearly with capital or lowercase letters as requested.

Question 2. Draw any line segment AB, construct PQ without measuring AB such that the length of PQ is twice that of AB.
Answer:
Construction Steps:
1. Draw a line segment \(\overline{AB}\) of any length using a ruler.
2. Draw a line ℓ and mark a point P on it.
3. Open compasses equal in length to \(\overline{AB}\).
4. With the pointer at P, cut an arc on line ℓ at point X.
5. With the pointer at X, cut another arc further along line ℓ at point Q.
6. Thus, \(\overline{PQ} = 2 \times \overline{AB}\).
In simple words: Copy the length of AB twice consecutively using a compass to get PQ.

Exam Tip: Do not use a ruler to measure; use compasses to transfer exact lengths.

Question 3. Draw any line segment XY. Mark any point M on it. Through M draw a perpendicular to XY. (use ruler and compasses)
Answer:
Construction Steps:
1. Draw line segment \(\overline{XY}\) and mark point M on it.
2. With M as center and a convenient radius, draw an arc intersecting \(\overline{XY}\) at two points A and B.
3. With A and B as centers and radius greater than AM, draw two arcs intersecting each other at point C.
4. Join M and C. Then \(\overline{MC} \perp \overline{XY}\).
In simple words: Construct a perpendicular at a point on a line using arcs from that point and intersecting arcs above/below.

Exam Tip: Keep compass width unchanged while drawing the intersecting arcs from A and B.

Question 4. Draw a line segment of length 10 cm and construct its perpendicular bisector.
Answer:
Construction Steps:
1. Draw a line segment \(\overline{AB} = 10\text{ cm}\).
2. With A as center and radius greater than half of AB (\(> 5\text{ cm}\)), draw arcs above and below AB.
3. With B as center and the same radius, draw arcs cutting the previous arcs at C and D.
4. Join C and D. \(\overline{CD}\) is the perpendicular bisector of \(\overline{AB}\).
In simple words: Open compass more than half way, draw intersecting arcs from both endpoints, and join their intersection points.

Exam Tip: Ensure compass radius is strictly greater than 5 cm for accurate arc intersections.

Question 5. Draw a line segment of length 12 cm. using compasses divide it into four equal parts verify by actual measurement.
Answer:
Construction Steps:
1. Draw \(\overline{AB} = 12\text{ cm}\).
2. Construct the perpendicular bisector of \(\overline{AB}\) to divide it into two equal halves of 6 cm each (let M be the midpoint).
3. Similarly, construct perpendicular bisectors for segment \(\overline{AM}\) and segment \(\overline{MB}\) to divide them further.
4. This divides the 12 cm segment into four equal parts of 3 cm each.
5. Verification: Measuring with a ruler confirms each of the four parts is exactly 3 cm.
In simple words: Bisect the 12 cm line, then bisect each half again to get 4 equal parts of 3 cm each.

Exam Tip: Successive bisection of a line segment divides it into \(2^n\) equal parts (here \(2^2 = 4\)).

Question 6. Draw an angle of measure \(140^\circ\) and construct its bisector.
Answer:
Construction Steps:
1. Draw a ray OA.
2. Using a protractor, measure and draw an angle \(\angle AOB = 140^\circ\).
3. With O as center and a convenient radius, draw an arc cutting OA and OB at points P and Q.
4. With P and Q as centers, draw arcs inside the angle with radius greater than half of PQ.
5. Let these arcs intersect at point R. Join OR and extend it.
6. Ray OR is the required angle bisector dividing \(140^\circ\) into two \(70^\circ\) angles.
In simple words: Draw the 140 degree angle with a protractor, then use a compass to bisect it into two 70 degree angles.

Exam Tip: Half of 140 degrees is 70 degrees.

Question 7. Draw an angle measure \(150^\circ\) and divide it into four equal parts.
Answer:
Construction Steps:
1. Construct or draw an angle of \(150^\circ\).
2. Draw its angle bisector to divide it into two angles of \(75^\circ\) each.
3. Further bisect each \(75^\circ\) angle to divide them into four equal parts of \(37.5^\circ\) each.
In simple words: Draw a 150 degree angle, bisect it, and then bisect both halves again.

Exam Tip: Successive angle bisection divides an angle into \(2^n\) equal parts.

Question 8. Construct with ruler and compasses, angles of following measures.
a. \(60^\circ\), b. \(30^\circ\) c. \(90^\circ\) d. \(120^\circ\) e. \(45^\circ\)
Answer:
- **\(60^\circ\):** Draw a ray, draw an arc from the vertex cutting the ray, and with the same radius cut an arc on the first arc. Join vertex and intersection.
- **\(30^\circ\):** Construct a \(60^\circ\) angle and bisect it.
- **\(90^\circ\):** Construct a \(60^\circ\) and a \(120^\circ\) angle at the same vertex and bisect the angle between them.
- **\(120^\circ\):** Draw a ray, draw an arc, and cut two successive arcs along the circumference with the same radius.
- **\(45^\circ\):** Construct a \(90^\circ\) angle and bisect it.
In simple words: Use standard compass constructions for 60, 30, 90, 120, and 45 degree angles without relying on a protractor.

Exam Tip: Master the 60 degree arc construction as it is the foundation for 30, 90, 120, and 45 degree angles.