CBSE Class 6 Mathematics Revision Worksheet Set 14

CHAPTER 7 - FRACTIONS

Q.1 What fraction is:
a) 6 hours of a day
b) 70 mm of a km
c) 290 ml of a liter
Answer:
a) Total hours in a day = 24 hours
Fraction = \(\frac{6}{24} = \frac{1}{4}\)
b) Total mm in 1 km = \(1 \text{ km} = 1000 \text{ m} = 1000 \times 1000 \text{ mm} = 1000000 \text{ mm}\)
Fraction = \(\frac{70}{1000000} = \frac{7}{100000}\)
c) Total ml in 1 liter = 1000 ml
Fraction = \(\frac{290}{1000} = \frac{29}{100}\)
In simple words: Compare the given quantity to the total standard units by writing them as a fraction and simplifying.

Exam Tip: Always convert both quantities to the same unit before forming the fraction.

Q.2 Write in ascending order
a) \(\frac{3}{5}, \frac{4}{5}, \frac{2}{5}, \frac{9}{5}, \frac{1}{5}\)
b) \(\frac{8}{11}, \frac{8}{9}, \frac{8}{15}, \frac{8}{18}, \frac{8}{13}\)
Answer:
a) Since denominators are equal, compare numerators: 1 < 2 < 3 < 4 < 9
Ascending order: \(\frac{1}{5}, \frac{2}{5}, \frac{3}{5}, \frac{4}{5}, \frac{9}{5}\)
b) Since numerators are equal, fractions with larger denominators are smaller: 18 > 15 > 13 > 11 > 9
Ascending order: \(\frac{8}{18}, \frac{8}{15}, \frac{8}{13}, \frac{8}{11}, \frac{8}{9}\)
In simple words: When denominators are the same, sort by the top numbers. When numerators are the same, sort by the bottom numbers in reverse.

Exam Tip: For like denominators, smaller numerator means smaller fraction. For like numerators, larger denominator means smaller fraction.

Q.3 Add the following:
a) \(\frac{13}{5}, \frac{4}{5}, \frac{22}{5}\)
b) \(\frac{6}{7}, \frac{5}{7}, \frac{3}{7}\)
c) \(\frac{2}{7}, \frac{9}{11}\)
Answer:
a) \(\frac{13 + 4 + 22}{5} = \frac{39}{5}\)
b) \(\frac{6 + 5 + 3}{7} = \frac{14}{7} = 2\)
c) LCM of 7 and 11 = 77
\(\frac{2 \times 11}{7 \times 11} + \frac{9 \times 7}{11 \times 7} = \frac{22}{77} + \frac{63}{77} = \frac{85}{77}\)
In simple words: Add numerators directly if denominators are the same. If different, find the LCM first.

Exam Tip: Always reduce the final fraction to its simplest form if possible.

Q.4 Rohit and Rahul have study tables of the same size. Rohit used \(\frac{5}{6}\) of its space and Rahul used \(\frac{2}{5}\) of its space. Whose study table is more occupied and by what fraction?
Answer:
Rohit's table space used = \(\frac{5}{6}\)
Rahul's table space used = \(\frac{2}{5}\)
LCM of 6 and 5 = 30
Rohit = \(\frac{5 \times 5}{6 \times 5} = \frac{25}{30}\)
Rahul = \(\frac{2 \times 6}{5 \times 6} = \frac{12}{30}\)
Since \(\frac{25}{30} > \frac{12}{30}\), Rohit's table is more occupied.
Difference = \(\frac{25}{30} - \frac{12}{30} = \frac{13}{30}\)
In simple words: Convert both fractions to have the same bottom number to easily compare them and find the difference.

Exam Tip: Cross-multiplication (\(5 \times 5 = 25\) and \(6 \times 2 = 12\)) is a quick way to compare two fractions.

Q.5 a) Is \(\frac{4}{5}\) equal to \(\frac{5}{9}\)?
b) Is \(\frac{8}{60}\) equal to \(\frac{1}{15}\)?
Answer:
a) Cross multiply: \(4 \times 9 = 36\) and \(5 \times 5 = 25\). Since \(36 \neq 25\), they are not equal.
b) Simplify \(\frac{8}{60}\) by dividing numerator and denominator by 4: \(\frac{8 \div 4}{60 \div 4} = \frac{2}{15}\). Since \(\frac{2}{15} \neq \frac{1}{15}\), they are not equal.
In simple words: Fractions are equal if their cross-products are equal or if their simplified forms match.

Exam Tip: Use cross-multiplication to check equality of two fractions quickly.

Q.6 Write three equivalent fractions of:
a) \(\frac{2}{5}\)
b) \(\frac{1}{7}\)
c) \(\frac{5}{11}\)
d) \(\frac{6}{23}\)
Answer:
a) Multiply numerator and denominator by 2, 3, and 4: \(\frac{4}{10}, \frac{6}{15}, \frac{8}{20}\)
b) Multiply by 2, 3, and 4: \(\frac{2}{14}, \frac{3}{21}, \frac{4}{28}\)
c) Multiply by 2, 3, and 4: \(\frac{10}{22}, \frac{15}{33}, \frac{20}{44}\)
d) Multiply by 2, 3, and 4: \(\frac{12}{46}, \frac{18}{69}, \frac{24}{92}\)
In simple words: Multiply the top and bottom numbers by the same counting numbers (2, 3, 4).

Q.7 Mika was given \(\frac{2}{9}\) of a pack of toffees. What fraction of toffees was left in the pack?
Answer:
Total pack = 1
Fraction left = \(1 - \frac{2}{9} = \frac{9 - 2}{9} = \frac{7}{9}\)
In simple words: Subtract the taken fraction from a whole (1) to find the remaining part.

Exam Tip: Treat whole 1 as \(\frac{9}{9}\) when the denominator is 9.

Q.8 Compute:
a) \(\frac{23}{67} - \frac{21}{67}\)
b) \(\frac{11}{23} - \frac{9}{23} + \frac{13}{23}\)
Answer:
a) \(\frac{23 - 21}{67} = \frac{2}{67}\)
b) \(\frac{11 - 9 + 13}{23} = \frac{15}{23}\)
In simple words: Perform addition and subtraction on numerators directly while keeping the common denominator.

Exam Tip: Follow the order of operations (PEMDAS/BODMAS) from left to right.

Q.9 Nitin bought a piece of rope \(11\frac{1}{8}\) m in length. He uses \(3\frac{1}{4}\) m of it. What is the length of the piece left?
Answer:
Initial length = \(11\frac{1}{8} = \frac{89}{8}\) m
Length used = \(3\frac{1}{4} = \frac{13}{4} = \frac{26}{8}\) m
Length left = \(\frac{89}{8} - \frac{26}{8} = \frac{63}{8} = 7\frac{7}{8}\) m
In simple words: Convert mixed fractions to improper fractions, subtract them, and convert back.

Exam Tip: Always make denominators equal before subtracting mixed fractions.

CHAPTER - 8 (DECIMALS)

Q.1 Write as decimals:
i) Three ones and 2-tenths
ii) Sixty and 3-tenths
iii) Nineteen and 9-tenths
iv) One hundred six and 5-tenths
Answer:
i) 3.2
ii) 60.3
iii) 19.9
iv) 106.5
In simple words: Write the whole number part, put a decimal point, and follow with the tenth digit.

Exam Tip: 'Tenths' means one digit after the decimal point (\(\frac{1}{10}\)).

Q.2 Write each of the following as decimals:
i) \(\frac{6}{10}\)
ii) \(300 + 50 + 3 + \frac{1}{10} + \frac{3}{100}\)
iii) \(6\frac{2}{5}\)
Answer:
i) 0.6
ii) 353.13
iii) \(6\frac{4}{10} = 6.4\)
In simple words: Combine whole number places and convert fraction parts into decimal places.

Exam Tip: Denominators of 10, 100, 1000 directly determine decimal place values.

Q.3 Express as fraction:
i) 61.2
ii) 6.4
iii) 0.05
iv) 2.4
v) 2.56
Answer:
i) \(\frac{612}{10} = \frac{306}{5}\)
ii) \(\frac{64}{10} = \frac{32}{5}\)
iii) \(\frac{5}{100} = \frac{1}{20}\)
iv) \(\frac{24}{10} = \frac{12}{5}\)
v) \(\frac{256}{100} = \frac{64}{25}\)
In simple words: Remove the decimal point and divide by 10, 100, etc., depending on decimal places, then simplify.

Exam Tip: One decimal place gives denominator 10; two decimal places give denominator 100.

Q.4 Express the following as kilometers (km) using decimals:
i) 2m
ii) 16m
iii) 530m
iv) 915m
v) 830m
Answer:
Since \(1 \text{ km} = 1000 \text{ m}\):
i) \(\frac{2}{1000} = 0.002 \text{ km}\)
ii) \(\frac{16}{1000} = 0.016 \text{ km}\)
iii) \(\frac{530}{1000} = 0.530 \text{ km}\)
iv) \(\frac{915}{1000} = 0.915 \text{ km}\)
v) \(\frac{830}{1000} = 0.830 \text{ km}\)
In simple words: Divide meters by 1000 to convert into kilometers.

Exam Tip: Always place three digits after the decimal point when converting meters to kilometers.

Q.5 Express as Rupees using decimals:
i) 120 paise
ii) 450 paise
iii) 7 paise
iv) 500 paise
v) 6 rupees 70 paise
Answer:
Since \(1 \text{ Rupee} = 100 \text{ paise}\):
i) \(\frac{120}{100} = \text{Rs. } 1.20\)
ii) \(\frac{450}{100} = \text{Rs. } 4.50\)
iii) \(\frac{7}{100} = \text{Rs. } 0.07\)
iv) \(\frac{500}{100} = \text{Rs. } 5.00\)
v) \(6 + \frac{70}{100} = \text{Rs. } 6.70\)
In simple words: Divide paise by 100 to convert into rupees.

Q.6 Express as kilogram (kg) using decimals:
i) 5g
ii) 305g
iii) 6325g
iv) 2 kg 65g
v) 7 kg 750g
Answer:
Since \(1 \text{ kg} = 1000 \text{ g}\):
i) \(\frac{5}{1000} = 0.005 \text{ kg}\)
ii) \(\frac{305}{1000} = 0.305 \text{ kg}\)
iii) \(\frac{6325}{1000} = 6.325 \text{ kg}\)
iv) \(2 + \frac{65}{1000} = 2.065 \text{ kg}\)
v) \(7 + \frac{750}{1000} = 7.750 \text{ kg}\)
In simple words: Divide grams by 1000 to convert into kilograms.

Exam Tip: Grams must occupy three decimal places in kilograms.

Q.7 Find the sum of:
i) 25.43, 6.735 and 39
ii) 0.3, 6.2 and 8.932
iii) 4.003, 2.65 and 7.1
Answer:
i) 25.430 + 6.735 + 39.000 = 71.165
ii) 0.300 + 6.200 + 8.932 = 15.432
iii) 4.003 + 2.650 + 7.100 = 13.753
In simple words: Align decimal points vertically by adding trailing zeros and add normally.

Exam Tip: Make all decimal numbers have the same number of decimal places before adding.

Q.8 Find the value of:
i) 8.735 - 6.27
ii) 99.009 - 19.9
iii) 1000 - 27.35
Answer:
i) 8.735 - 6.270 = 2.465
ii) 99.009 - 19.900 = 79.109
iii) 1000.00 - 27.35 = 972.65
In simple words: Line up the decimal points and subtract column by column.

Exam Tip: Fill empty decimal places with zeros to avoid borrowing errors.

Q.9 Simplify:
i) 3.2 X 2.65 X 1.05
ii) 639.5 X 0.05
iii) 888.88 X 80.08
Answer:
i) 3.2 X 2.65 = 8.48
8.48 X 1.05 = 8.904
ii) 639.5 X 0.05 = 31.975
iii) 888.88 X 80.08 = 71177.3044
In simple words: Multiply as whole numbers, then place the decimal point by counting total decimal places.

Exam Tip: The total number of decimal places in the product equals the sum of decimal places in the factors.

CHAPTER - 9 (DATA HANDLING)

Q.1 Marks scored by students (out of 50) in a General Knowledge test are as follows:
40, 43, 45, 39, 41, 50, 45, 44, 40, 43, 42, 40, 45, 50, 50, 49, 48, 40, 30, 42, 43, 50, 49, 47, 48, 42, 45, 41
(Draw frequency table)
Answer:

In simple words: Count how many times each mark occurs and record it in a table.

Exam Tip: Cross off numbers from the raw list as you count them to avoid missing any data point.

Q.2 Smita rolled a dice 35 times and recorded the observation as follows:
3, 6, 1, 2, 6, 3, 4, 5, 4, 4, 3, 1, 2, 1, 2, 6, 5, 2, 6, 5, 3, 1, 2, 3, 4, 6, 5, 4, 1, 6, 5, 5, 3, 2, 3
Make a frequency table using tally marks & answer the following:
i.) Which number appeared the maximum number of times?
ii.) Which number appeared the minimum number of times?
Answer:

i) Number 3 appeared the maximum number of times (7 times).
ii) Numbers 1 and 4 appeared the minimum number of times (5 times each).
In simple words: Organize data into tally marks, then find the highest and lowest counts.

Exam Tip: Total frequency must equal the total number of observations (35).

Q.3 Total number of students of a school in different years is given below:
Prepare a pictograph using symbol = 100 students.

Answer:
Scale: \(\text{Symbol} = 100 \text{ students}, \text{Half symbol} = 50 \text{ students}\)
2001: 4.5 symbols
2002: 6 symbols
2003: 8.5 symbols
2004: 10.5 symbols
2005: 12.5 symbols
2006: 18 symbols
In simple words: Represent numbers using picture symbols where each full symbol stands for 100.

Exam Tip: Always mention the scale clearly at the bottom or top of the pictograph.

Q.4 Draw a pictograph to represent the following information:

Answer:
Let a suitable scale be chosen, such as \(\text{Symbol} = 10 \text{ saplings}\).
2002 (70): 7 symbols
2003 (90): 9 symbols
2004 (120): 12 symbols
2005 (150): 15 symbols
2006 (180): 18 symbols
In simple words: Draw symbols corresponding to each year's total divided by the scale unit.

Exam Tip: Choose a scale that avoids fractions of symbols where possible.

Q.5 Read the above Bar Graph & answer the following:
i) What information does the above bar graph give?
ii) Find the total number of tickets to be sold in five cities.
iii) In which city the maximum numbers of tickets were sold?
iv) Find the difference of tickets sold in Bangalore & Kolkata.
Answer:
i) The bar graph shows the number of tickets sold in five different cities (Chennai, Mumbai, Kolkata, Delhi, Bangalore).
ii) Ticket counts: Chennai = 60, Mumbai = 100, Kolkata = 50, Delhi = 80, Bangalore = 70.
Total tickets = 60 + 100 + 50 + 80 + 70 = 360 tickets.
iii) Mumbai (100 tickets).
iv) Bangalore = 70, Kolkata = 50. Difference = 70 - 50 = 20 tickets.
In simple words: Read values directly from the bar heights corresponding to each city.

Exam Tip: Always multiply the bar height by the scale factor (1 unit = 20 tickets).

Q.6 The following table shows the number of students of different ages participating in 'environment protection programme' on the world environment day. Draw a bar graph using the given information:

Answer:
Steps to draw bar graph:
1. Take Age on the horizontal axis (X-axis) and Number of Students on the vertical axis (Y-axis).
2. Choose a suitable scale for the Y-axis (e.g., 1 unit = 50 students).
3. Draw bars of uniform width for each age with heights proportional to the student counts (Age 8: 550, Age 9: 350, Age 10: 400, Age 11: 375, Age 12: 425, Age 13: 575).
In simple words: Plot each age against its student count using rectangular bars of equal width.

Exam Tip: Maintain equal spacing between consecutive bars on the axis.

CHAPTER - 10 (MENSURATION)

Q.1 Find the perimeter of a rectangle whose length and breadth are 20m and 16.5m respectively.
Answer:
Length (\(l\)) = 20 m
Breadth (\(b\)) = 16.5 m
Perimeter = \(2(l + b) = 2(20 + 16.5) = 2(36.5) = 73 \text{ m}\)
In simple words: Add length and breadth, then multiply the sum by 2.

Exam Tip: Always include proper units in the final answer.

Q.2 Find the perimeter of a square whose side is 6.5cm.
Answer:
Side (\(s\)) = 6.5 cm
Perimeter = \(4 \times s = 4 \times 6.5 = 26 \text{ cm}\)
In simple words: Multiply the length of one side by 4.

Exam Tip: A square has four equal sides.

Q.3 Find the cost of fencing a park of length 200m and breadth 150m at the rate of Rs. 33 per meter.
Answer:
Length = 200 m, Breadth = 150 m
Perimeter = \(2(200 + 150) = 2(350) = 700 \text{ m}\)
Cost per meter = Rs. 33
Total Cost = \(700 \times 33 = \text{Rs. } 23100\)
In simple words: Find the total boundary length (perimeter) and multiply by the cost per meter.

Exam Tip: Fencing always takes place along the perimeter of a field.

Q.4 How many rectangles can be drawn with perimeter as 28cm, given that the sides are positive integers in cm?
Answer:
Perimeter = \(2(l + b) = 28\)
\(l + b = 14\)
Since \(l\) and \(b\) are positive integers, possible pairs of \((l, b)\) are:
(13, 1), (12, 2), (11, 3), (10, 4), (9, 5), (8, 6), (7, 7 - square, usually considered a special rectangle).
Total rectangles = 7.
In simple words: Find all pairs of positive integers that add up to half the perimeter (14).

Q.5 A marble tile measures 25cm by 20cm. How many tiles will be required to cover a floor of size 5m by 4m?
Answer:
Floor dimensions: Length = \(5 \text{ m} = 500 \text{ cm}\), Breadth = \(4 \text{ m} = 400 \text{ cm}\)
Area of floor = \(500 \times 400 = 200000 \text{ sq. cm}\)
Area of one tile = \(25 \times 20 = 500 \text{ sq. cm}\)
Number of tiles = \(\frac{\text{Area of floor}}{\text{Area of one tile}} = \frac{200000}{500} = 400 \text{ tiles}\)
In simple words: Divide the total floor area by the area of a single tile.

Exam Tip: Always convert all dimensions to the same unit (cm) before calculating areas.

Q.6 Find the perimeter of a rectangle whose area is 500 sq. cm. and breadth is 20cm.
Answer:
Area = \(l \times b = 500\)
\(l \times 20 = 500\)
\(l = \frac{500}{20} = 25 \text{ cm}\)
Perimeter = \(2(l + b) = 2(25 + 20) = 2(45) = 90 \text{ cm}\)
In simple words: Find the length by dividing area by breadth, then calculate the perimeter.

Q.7 The area of a square field is 225 sq. m. Find the length of its side.
Answer:
Area of square = \(s^2 = 225\)
\(s = \sqrt{225} = 15 \text{ m}\)
In simple words: Find the number that when multiplied by itself gives 225.

Exam Tip: Side of square = \(\sqrt{\text{Area}}\).

Q.8 The length and breadth of a rectangular park are in the ratio 4:3. Its area is 2352 sq. m. What is the length and breadth of the field?
Answer:
Let length \(l = 4x\) and breadth \(b = 3x\)
Area = \(l \times b = (4x) \times (3x) = 12x^2\)
\(12x^2 = 2352\)
\(x^2 = \frac{2352}{12} = 196\)
\(x = \sqrt{196} = 14\)
Length \(l = 4 \times 14 = 56 \text{ m}\)
Breadth \(b = 3 \times 14 = 42 \text{ m}\)
In simple words: Use variables with the given ratio, find the common multiplier, and calculate dimensions.

Exam Tip: When given in ratio, assign \(4x\) and \(3x\) to length and breadth.

Q.9 The sides of a triangle are in the ratio 3:4:5. If the perimeter of the triangle is 180cm, what are its sides?
Answer:
Let sides be \(3x, 4x, 5x\)
Perimeter = sum of all sides = \(3x + 4x + 5x = 12x\)
\(12x = 180\)
\(x = \frac{180}{12} = 15 \text{ cm}\)
Sides are:
First side = \(3 \times 15 = 45 \text{ cm}\)
Second side = \(4 \times 15 = 60 \text{ cm}\)
Third side = \(5 \times 15 = 75 \text{ cm}\)
In simple words: Set the sum of ratio parts equal to the perimeter to solve for the multiplier.

Exam Tip: Perimeter of a triangle is the sum of all its three sides.

CHAPTER - 11 (ALGEBRA)

Q.1 Write algebraic expression for the following:
i) 6 times y to which 2 is added.
ii) -C divided by 9.
iii) 11 subtracted from two times x.
iv) x multiplied by itself three times.
v) Three times a number s subtracted from 22.
Answer:
i) \(6y + 2\)
ii) \(\frac{-C}{9}\)
iii) \(2x - 11\)
iv) \(x^3\)
v) \(22 - 3s\)
In simple words: Translate the English statement directly into mathematical symbols and variables.

Exam Tip: Pay close attention to words like 'subtracted from' which reverse the order.

Q.2 Write the next two numbers in each of the following number patterns:
i) 5, 11, 23, 47, -----, -----
ii) 243, 81, 27, -----, -----
iii) 10, 30, 90, 270, ----, ----
iv) 16, 23, 30, 37, -----, -----
v) 3, 5, 8, 12, -----, -----
Answer:
i) Rule: Multiply by 2 and add 1 (\(5 \times 2 + 1 = 11, 11 \times 2 + 1 = 23\)...). Next two: 95, 191.
ii) Rule: Divide by 3. Next two: 9, 3.
iii) Rule: Multiply by 3. Next two: 810, 2430.
iv) Rule: Add 7. Next two: 44, 51.
v) Rule: Add consecutive numbers starting from 2 (\(+2, +3, +4, +5, +6\)...). Next two: 17, 23.
In simple words: Find the hidden rule or pattern linking consecutive numbers.

Exam Tip: Check differences between terms or multipliers between terms to find the pattern.

Q.3 Which of the following are equations?
i) \(6x - 7 > 12\)
ii) \(2x + 3 = 9\)
iii) \(\frac{7a}{11} < 19\)
iv) \(\frac{-5b}{8} = 2\)
v) \(3a + 7 < 8\)
vi) \(\frac{m}{3 - 5} = 9\)
Answer:
Equations must contain an equals sign (\(=\)).
ii) \(2x + 3 = 9\) (Equation)
iv) \(\frac{-5b}{8} = 2\) (Equation)
vi) \(\frac{m}{3 - 5} = 9\) (Equation)
(Note: i, iii, and v are inequalities because they contain \(>\) or \(<\)).
In simple words: An equation is a mathematical statement with an equals sign.

Exam Tip: Expressions and inequalities are not equations.

Q.4 Solve the equations by trial & error method:
i) \(3x + 7 = 25\)
ii) \(7y = 84\)
iii) \(3m - 6 = 15\)
Answer:
i) \(3x + 7 = 25\)
Try \(x = 5\): \(3(5) + 7 = 22 \neq 25\)
Try \(x = 6\): \(3(6) + 7 = 25\) (Correct)
Solution: \(x = 6\)
ii) \(7y = 84\)
Try \(y = 10\): \(7(10) = 70 \neq 84\)
Try \(y = 12\): \(7(12) = 84\) (Correct)
Solution: \(y = 12\)
iii) \(3m - 6 = 15\)
Try \(m = 6\): \(3(6) - 6 = 12 \neq 15\)
Try \(m = 7\): \(3(7) - 6 = 15\) (Correct)
Solution: \(m = 7\)
In simple words: Substitute different values until both sides of the equation balance.

Exam Tip: Systematically test nearby integer values.

Q.5 If Vidushi's present age is x years:
i) What will be her age 5 years from now?
ii) What was her age 2 years ago?
iii) Her grandmother's age is 7 times her age. What is her grandmother's age?
Answer:
i) \((x + 5)\) years
ii) \((x - 2)\) years
iii) \(7x\) years
In simple words: Add for future ages, subtract for past ages, and multiply for 'times'.

Exam Tip: 'From now' means addition, 'ago' means subtraction.

CHAPTER - 12 (RATIO AND PROPORTION)

Q.1 Express as ratios:
i) 2km to 6km
ii) 8g to 120g
iii) 6hr to 1 day
iv) 5m to 650cm
v) Rs. 3 to 600p
Answer:
i) \(\frac{2}{6} = 1:3\)
ii) \(\frac{8}{120} = 1:15\)
iii) \(1 \text{ day} = 24 \text{ hr} \Rightarrow \frac{6}{24} = 1:4\)
iv) \(5 \text{ m} = 500 \text{ cm} \Rightarrow \frac{500}{650} = 10:13\)
v) \(\text{Rs. } 3 = 300\text{p} \Rightarrow \frac{300}{600} = 1:2\)
In simple words: Convert both quantities to the same unit and simplify the fraction.

Exam Tip: Ratios have no units because units cancel out.

Q.2 In a class there are 25 boys and 35 girls. Find the ratio of:
i) number of boys to the number of girls.
ii) number of girls to the total number of students.
Answer:
Boys = 25, Girls = 35
Total students = 25 + 35 = 60
i) Ratio of boys to girls = \(\frac{25}{35} = 5:7\)
ii) Ratio of girls to total students = \(\frac{35}{60} = 7:12\)
In simple words: Write the requested quantities as a fraction and reduce to lowest terms.

Exam Tip: Total students must be calculated before finding ratios involving total counts.

Q.3 Write two equivalent ratios of:
i) 3:5
ii) 11:8
iii) 6:4
iv) 2:7
Answer:
i) Multiply by 2 and 3: 6:10, 9:15
ii) Multiply by 2 and 3: 22:16, 33:24
iii) Multiply by 2 and 3: 12:8, 18:12 (or simplify to 3:2, then 6:4, 9:6)
iv) Multiply by 2 and 3: 4:14, 6:21
In simple words: Multiply both terms of the ratio by the same number.

Exam Tip: Equivalent ratios maintain the same proportional value.

Q.4 Divide 20 pens between Samiara and Raima in the ratio 3:2.
Answer:
Total parts = 3 + 2 = 5
Samiara's share = \(\frac{3}{5} \times 20 = 12 \text{ pens}\)
Raima's share = \(\frac{2}{5} \times 20 = 8 \text{ pens}\)
In simple words: Divide total pens by total ratio parts, then multiply by each person's share.

Exam Tip: Sum of individual shares must equal the total quantity (12 + 8 = 20).

Q.5 Determine if the following are in proportion:
i) 32, 48, 70, 210
ii) 25, 30, 40, 48
iii) 45, 60, 12, 15
Answer:
i) Product of extremes = \(32 \times 210 = 6720\). Product of means = \(48 \times 70 = 3360\). Not in proportion.
ii) Product of extremes = \(25 \times 48 = 1200\). Product of means = \(30 \times 40 = 1200\). In proportion.
iii) Product of extremes = \(45 \times 15 = 675\). Product of means = \(60 \times 12 = 720\). Not in proportion.
In simple words: Check if (First \(\times\) Fourth) equals (Second \(\times\) Third).

Exam Tip: Four numbers \(a, b, c, d\) are in proportion if \(a \times d = b \times c\).

Q.6 Find 'a' if:
i) a : 30 :: 7 : 15
ii) 12 : a :: 96 : 160
iii) 72 : 32 :: a : 144
Answer:
i) \(a \times 15 = 30 \times 7 \Rightarrow 15a = 210 \Rightarrow a = 14\)
ii) \(12 \times 160 = a \times 96 \Rightarrow 1920 = 96a \Rightarrow a = 20\)
iii) \(72 \times 144 = 32 \times a \Rightarrow 10368 = 32a \Rightarrow a = 324\)
In simple words: Use the property that product of extremes equals product of means.

Exam Tip: Isolate 'a' by dividing both sides by its coefficient.

Q.7 A cricket set contains 3 balls and 2 bats. Can 24 balls and 8 bats be made into complete sets?
Answer:
Sets possible with 24 balls = \(\frac{24}{3} = 8 \text{ sets}\)
Sets possible with 8 bats = \(\frac{8}{2} = 4 \text{ sets}\)
Since the number of sets formed by bats (4) is less than balls (8), only 4 complete sets can be made.
Therefore, 24 balls and 8 bats cannot be made into complete uniform sets.
In simple words: Check how many sets can be made from each item; the limiting component decides the total sets.

Exam Tip: Complete sets require matching proportions of all items.

Q.8 6 packets of pens cost Rs. 120. How many packets can be bought in Rs. 360?
Answer:
Cost of 6 packets = Rs. 120
Cost of 1 packet = \(\frac{120}{6} = \text{Rs. } 20\)
Number of packets in Rs. 360 = \(\frac{360}{20} = 18 \text{ packets}\)
In simple words: Find the cost of one item first using division, then divide total money by unit cost.

Exam Tip: This is a standard unitary method problem.

Q.9 A worker is paid Rs. 162.50 for 5 days. What should be paid to him for 29 days?
Answer:
Pay for 5 days = Rs. 162.50
Pay for 1 day = \(\frac{162.50}{5} = \text{Rs. } 32.50\)
Pay for 29 days = \(32.50 \times 29 = \text{Rs. } 942.50\)
In simple words: Calculate daily wage first, then multiply by 29.

Exam Tip: Always calculate the unit value (1 day) before finding the required multiple.

CHAPTER 13 & 14 (SYMMETRY & PRACTICAL GEOMETRY)

1. Which of the digits from 0 to 9 have lines of symmetry? Draw them.
Answer:
Digits with lines of symmetry:
- 0 (vertical and horizontal lines of symmetry)
- 1 (vertical line of symmetry)
- 3 (horizontal line of symmetry)
- 8 (vertical and horizontal lines of symmetry)
In simple words: Digits 0, 1, 3, and 8 can be folded symmetrically along specific lines.

Exam Tip: Test folding each digit horizontally and vertically.

2. Write four letters of the English alphabet that have both vertical and horizontal lines of symmetry.
Answer:
H, I, O, X
In simple words: These letters look identical when reflected across both vertical and horizontal mirrors.

3. Draw a regular polygon that has five lines of symmetry.
Answer:
A regular pentagon has five lines of symmetry.
Construction steps:
1. Draw a regular pentagon with 5 equal sides and 5 equal angles.
2. Draw lines passing through each vertex and the midpoint of the opposite side. There will be 5 such lines of symmetry.
In simple words: A regular pentagon has 5 lines of symmetry passing through its vertices.

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

4. Draw a kite and draw its line of symmetry.
Answer:
Construction steps:
1. Draw a kite shape with two pairs of equal adjacent sides.
2. Draw the main diagonal connecting the vertices between the unequal sides. This diagonal acts as the single line of symmetry.
In simple words: A kite has 1 line of symmetry along its main vertical diagonal.

Exam Tip: A standard kite has exactly one line of symmetry.

5. Write 'True' or 'False' for the following statements.
a. A circle has infinite number of lines of symmetry.
b. A scalene triangle has one line of symmetry.
c. A square has two lines of symmetry.
d. A kite has one line of symmetry.
e. A rhombus has four lines of symmetry.
Answer:
a. True
b. False (A scalene triangle has 0 lines of symmetry)
c. False (A square has 4 lines of symmetry)
d. True
e. False (A rhombus has 2 lines of symmetry)
In simple words: Evaluate each geometric shape's actual mirror symmetry lines.

Exam Tip: Remember that a square has 4 lines of symmetry, while a rhombus has 2.

6. Draw a semicircle of radius 5.8cm.
Answer:
Steps of construction:
1. Draw a straight line segment acting as the diameter.
2. Mark the center point and set the compass radius to 5.8 cm.
3. Draw an arc of 180 degrees (a semicircle) above or below the diameter line.
In simple words: Use a compass opened to 5.8 cm to draw half of a circle.

Exam Tip: Ensure the compass is measured correctly against a ruler.

7. Using a protractor draw an angle of \(75^\circ\). Draw the angle bisector of this angle.
Answer:
Steps of construction:
1. Draw a ray AB.
2. Place protractor on point A and mark \(75^\circ\), then draw ray AC.
3. With center A and any convenient radius, draw an arc cutting AB and AC at points P and Q.
4. With centers P and Q and radius greater than half of PQ, draw two intersecting arcs inside the angle.
5. Join the intersection point to vertex A to form the angle bisector (\(37.5^\circ\)).
In simple words: Draw a 75 degree angle and use compass arcs to divide it into two equal halves.

Exam Tip: Angle bisector divides the angle into two equal parts.

8. Draw any line segment AB and take a point P outside it. Construct a perpendicular on AB passing through P.
Answer:
Steps of construction:
1. Draw a line segment AB and take an external point P.
2. With P as center and a suitable radius, draw an arc intersecting AB at two points, say X and Y.
3. With X and Y as centers and radius greater than half of XY, draw two arcs on the other side of AB intersecting at Q.
4. Join P and Q. PQ is perpendicular to AB.
In simple words: Drop a perpendicular line from an outside point to a given line segment using a compass.

Exam Tip: Keep compass width unchanged while drawing intersecting arcs from points X and Y.

9. Draw a line segment AB = 9cm. Using compass find a point C on it such that AC = BC.
Answer:
Steps of construction:
1. Draw line segment AB = 9 cm.
2. With A as center and radius greater than half of AB (more than 4.5 cm), draw arcs above and below AB.
3. With B as center and same radius, cut the previous arcs at points X and Y.
4. Join XY. The point where XY intersects AB is point C.
In simple words: Construct the perpendicular bisector of the line segment to find its exact midpoint C.

Exam Tip: Midpoint divides the segment into two equal halves of 4.5 cm each.

10. Draw a line AB and a point C on it. Draw a line CD perpendicular to AB.
Answer:
Steps of construction:
1. Draw a line AB and mark a point C on it.
2. With C as center and any convenient radius, draw an arc intersecting AB at points P and Q.
3. With P and Q as centers and radius greater than PC, draw two arcs intersecting each other above C at point D.
4. Join C and D. CD is perpendicular to AB.
In simple words: Erect a perpendicular at a specific point on a line using compass arcs.

1. Show \(1, \frac{1}{3}, \frac{1}{2}, \frac{1}{6}, \frac{0}{6}\) on the number line.
Answer:
Convert fractions to have a common denominator (6):
\(1 = \frac{6}{6}\)
\(\frac{1}{3} = \frac{2}{6}\)
\(\frac{1}{2} = \frac{3}{6}\)
\(\frac{1}{6} = \frac{1}{6}\)
\(\frac{0}{6} = 0\)
Number line points from left to right: \(0 (\frac{0}{6}), \frac{1}{6}, \frac{2}{6} (\frac{1}{3}), \frac{3}{6} (\frac{1}{2}), \frac{6}{6} (1)\).
In simple words: Convert all fractions to denominator 6 and plot them from 0 to 1 on a number line.

Exam Tip: Like denominators make it easy to mark exact positions on a number line.

2. Convert (i) \(\frac{25}{6}\) into mixed fraction and (ii) \(3\frac{4}{5}\) into improper fraction.
Answer:
i) \(25 \div 6\): Quotient = 4, Remainder = 1. Mixed fraction = \(4\frac{1}{6}\)
ii) \(3\frac{4}{5} = \frac{3 \times 5 + 4}{5} = \frac{19}{5}\)
In simple words: Divide numerator by denominator for mixed fractions; multiply whole number by denominator and add numerator for improper fractions.

Exam Tip: Denominator always remains unchanged during conversions.

3. Are \(\frac{3}{4}, \frac{5}{6}, \frac{7}{8}\) equivalent fractions? Give reason.
Answer:
No, they are not equivalent fractions.
Reason: Converting them to like denominators (LCM of 4, 6, 8 is 24):
\(\frac{3}{4} = \frac{18}{24}\)
\(\frac{5}{6} = \frac{20}{24}\)
\(\frac{7}{8} = \frac{21}{24}\)
Since \(\frac{18}{24}, \frac{20}{24}, \frac{21}{24}\) are not equal, the fractions are not equivalent.
In simple words: Fractions are not equivalent because their simplified values or common-denominator forms are different.

Exam Tip: Equivalent fractions must represent the exact same numerical value.

4. Show this pictorically (i) \(\frac{1}{3} + \frac{2}{3}\) (ii) \(\frac{4}{5} - \frac{1}{5}\)
Answer:
i) Draw a shape divided into 3 equal parts. Shade 1 part, then shade 2 more parts. Total shaded parts = 3 out of 3 (= 1 whole).
ii) Draw a shape divided into 5 equal parts. Shade 4 parts, then cross out 1 shaded part. Remaining shaded parts = 3 out of 5 (\(\frac{3}{5}\)).
In simple words: Use shaded regions of figures to show combining or removing fractional parts.

Exam Tip: Pictorial representation helps visualize fraction addition and subtraction.

5. Convert \(\frac{48}{112}\) into the simplest form.
Answer:
HCF of 48 and 112 is 16.
\(\frac{48 \div 16}{112 \div 16} = \frac{3}{7}\)
In simple words: Divide both numerator and denominator by their HCF.

Exam Tip: Successive division by common prime factors also works if HCF is not immediately obvious.

6. Write the first five equivalent fraction of \(\frac{4}{5}\).
Answer:
Multiply numerator and denominator by 2, 3, 4, 5, and 6:
\(\frac{8}{10}, \frac{12}{15}, \frac{16}{20}, \frac{20}{25}, \frac{24}{30}\)
In simple words: Multiply top and bottom by 2, 3, 4, 5, and 6 sequentially.

Exam Tip: Always show multiplication steps clearly.

7. Correct the given figure, shaded portion is \(\frac{1}{2}\) of the given figure.
Answer:
(Note: Source contains a triangle with a shaded corner or part).
To correct the figure so that the shaded portion represents \(\frac{1}{2}\), the total figure must be divided into 2 equal parts, and exactly 1 equal half must be shaded.
In simple words: Ensure exactly half of the total equal parts of the geometric figure are shaded.

Exam Tip: Fractions always require equal parts.

8. Give an improper fraction:
(i) Whose denominator is 4 less than the numerator?
(ii) Whose numerator & denominator add upto 10?
(iii) Difference between numerator and denominator is 12.
Answer:
i) Example: \(\frac{7}{3}\) (Numerator 7, denominator 3 = 7 - 4)
ii) Example: \(\frac{7}{3}\) (7 + 3 = 10, and \(7 > 3\))
iii) Example: \(\frac{15}{3}\) (15 - 3 = 12, and \(15 > 3\))
In simple words: Create fractions where the top number is greater than the bottom number satisfying the given conditions.

Exam Tip: An improper fraction always has a numerator greater than or equal to its denominator.

9. Arrange in the ascending order:
(i) \(\frac{3}{2}, \frac{3}{6}, \frac{3}{11}, \frac{3}{5}, \frac{3}{8}\)
(ii) \(\frac{8}{11}, \frac{10}{11}, \frac{3}{11}, \frac{5}{11}, \frac{9}{11}\)
(iii) \(\frac{5}{4}, \frac{1}{2}, \frac{5}{6}, \frac{3}{8}, \frac{1}{3}\)
Answer:
i) Same numerators, larger denominator means smaller fraction:
Ascending order: \(\frac{3}{11}, \frac{3}{8}, \frac{3}{6}, \frac{3}{5}, \frac{3}{2}\)
ii) Same denominators, sort by numerators:
Ascending order: \(\frac{3}{11}, \frac{5}{11}, \frac{8}{11}, \frac{9}{11}, \frac{10}{11}\)
iii) Convert to common denominator (LCM of 4, 2, 6, 8, 3 = 24):
\(\frac{30}{24}, \frac{12}{24}, \frac{20}{24}, \frac{9}{24}, \frac{8}{24}\)
Ascending order: \(\frac{8}{24} (\frac{1}{3}), \frac{9}{24} (\frac{3}{8}), \frac{12}{24} (\frac{1}{2}), \frac{20}{24} (\frac{5}{6}), \frac{30}{24} (\frac{5}{4})\)
In simple words: Compare fractions using like numerators, like denominators, or by converting to a common denominator.

Exam Tip: LCM method is safest when both numerators and denominators are different.

10. Write in the place value chart (i) 308.758 (ii) 25.99
Answer:
i) 308.758: Hundreds = 3, Tens = 0, Ones = 8, Tenths = 7, Hundredths = 5, Thousandths = 8.
ii) 25.99: Tens = 2, Ones = 5, Tenths = 9, Hundredths = 9.
In simple words: Place each digit into its respective whole number and decimal place-value column.

Exam Tip: Decimal places start immediately to the right of the decimal point as tenths, hundredths, thousandths.

11. Write each of the following as decimals:
(i) \(\frac{5}{2}\)
(ii) \(\frac{8}{5}\)
(iii) \(200 + 60 + \frac{1}{10} + \frac{3}{100}\)
(iv) \(3\frac{3}{5}\)
Answer:
i) \(\frac{5}{2} = 2.5\)
ii) \(\frac{8}{5} = 1.6\)
iii) \(260.13\)
iv) \(3\frac{3}{5} = 3.6\)
In simple words: Convert fractions to tenths/hundredths or divide to write in decimal form.

Exam Tip: Making the denominator 10, 100, or 1000 makes decimal conversion easy.

12. Write as fraction and then reduce it to the lowest form (i) 3.8 (ii) 31.4
Answer:
i) \(3.8 = \frac{38}{10} = \frac{19}{5}\)
ii) \(31.4 = \frac{314}{10} = \frac{157}{5}\)
In simple words: Remove decimal point, put over 10, and simplify by dividing by 2.

13. Write (i) 108.56 (ii) 5.008 in words.
Answer:
i) One hundred eight point five six (or One hundred eight and fifty-six hundredths)
ii) Five point zero zero eight (or Five and eight thousandths)
In simple words: Read the whole number part, say 'point', and read decimal digits individually.

14. Fill in the blanks using <, >, =:
(i) 0.099 ____ 0.19
(ii) 2.542 ____ 2.501
(iii) 1.5 ____ 1.50
Answer:
i) 0.099 < 0.19
ii) 2.542 > 2.501
iii) 1.5 = 1.50
In simple words: Compare decimals place by place from left to right.

15. Express: (i) 75 paise as rupees (ii) 60 cm as metre (iii) 11cm 5mm as cm (iv) 85 km 7m as km (v) 62kg 70g as kg
Answer:
i) \(\frac{75}{100} = \text{Rs. } 0.75\)
ii) \(\frac{60}{100} = 0.60 \text{ m}\)
iii) \(11 + \frac{5}{10} = 11.5 \text{ cm}\)
iv) \(85 + \frac{7}{1000} = 85.007 \text{ km}\)
v) \(62 + \frac{70}{1000} = 62.070 \text{ kg}\)
In simple words: Divide smaller units by 10, 100, or 1000 to convert to larger standard units.

17. By how much is the difference of 208 & 802.28 less than their sum?
Answer:
Sum = \(802.28 + 208 = 1010.28\)
Difference = \(802.28 - 208 = 594.28\)
Required difference = \(1010.28 - 594.28 = 416.00\)
In simple words: Find sum and difference separately, then subtract the difference from the sum.

18. What should be added to the difference of 3.14 & 1.673 to get 11?
Answer:
Difference = \(3.140 - 1.673 = 1.467\)
Let \(x\) be added: \(1.467 + x = 11\)
\(x = 11.000 - 1.467 = 9.533\)
In simple words: Subtract the difference from 11 to find the missing number.

19. Sameer spent \(\frac{2}{4}\) of his pocket money on a movie and \(\frac{2}{5}\) on a new pen. What fraction of his pocket money did he spend?
Answer:
Fraction spent = \(\frac{2}{4} + \frac{2}{5} = \frac{1}{2} + \frac{2}{5}\)
LCM of 2 and 5 = 10
\(= \frac{5}{10} + \frac{4}{10} = \frac{9}{10}\)
In simple words: Add the two fractions representing money spent after finding a common denominator.

20. The horticulture department made a table to show the number of trees of different types planted inside the garden and public buildings of a city. Prepare a pictograph to represent the above data.

Answer:
Scale: Let \(\text{Symbol} = 5 \text{ trees}\) (Half symbol = 2 or 3 trees approximately, or choose scale 2 trees per symbol).
Using scale: \(\text{Symbol} = 2 \text{ trees}\):
Neem (22): 11 symbols
Coconut (35): 17.5 symbols
Banyan (20): 10 symbols
Mango (42): 21 symbols
Christmas (27): 13.5 symbols
In simple words: Draw picture symbols representing tree counts according to the chosen scale.

21. The following table shows the marks obtained by Aziz in Annual Examination in different subjects. Prepare a bar graph to represent the above information.

Answer:
Steps to draw bar graph:
1. Plot subjects on the X-axis and marks on the Y-axis.
2. Choose a scale on Y-axis: \(1 \text{ unit} = 10 \text{ marks}\).
3. Draw vertical bars of equal width for Hindi (85), English (72), Maths (98), Science (89), and SST (70).
In simple words: Represent subject marks using proportional vertical rectangular bars.

22. A rectangle is 25m long and 16m broad. Find the side of a square whose area is equal to the area of the rectangle.
Answer:
Area of rectangle = \(25 \times 16 = 400 \text{ sq. m}\)
Area of square = 400 sq. m
Side of square = \(\sqrt{400} = 20 \text{ m}\)
In simple words: Calculate the rectangular area, then take its square root to find the side of the square.

23. Find the area of a square of side 36cm in terms of square metre.
Answer:
Side = \(36 \text{ cm} = 0.36 \text{ m}\)
Area = \(s^2 = 0.36 \times 0.36 = 0.1296 \text{ sq. m}\)
In simple words: Convert side to meters first, then square it to get area in square meters.

24. On a squared paper, sketch the following:
(i) A hexagon with exactly two lines of symmetry.
(ii) A quadrilateral with a horizontal line of symmetry but no vertical line of symmetry.
Answer:
i) Sketch an elongated hexagon (like a stretched regular hexagon) with 2 lines of symmetry.
ii) Sketch a kite or isosceles trapezoid oriented horizontally so that it has a horizontal line of symmetry but no vertical one.
In simple words: Draw shapes on grid paper matching the specified mirror-line criteria.

25. Draw any segment PQ. Take a point X not on it. Through X, draw a perpendicular to PQ.
Answer:
Steps of construction:
1. Draw line segment PQ.
2. Take an external point X.
3. Using compass, draw arcs from X to intersect PQ at two points, then construct perpendicular dropping through X.
In simple words: Drop a perpendicular line from an external point to a segment.

26. Draw an angle of \(135^\circ\) and divide it into four equal parts using compass.
Answer:
Steps of construction:
1. Draw an angle of \(135^\circ\) using a protractor.
2. Bisect the \(135^\circ\) angle to get two \(67.5^\circ\) angles.
3. Bisect each \(67.5^\circ\) angle to get four equal parts of \(33.75^\circ\) each.
In simple words: Draw a 135 degree angle and use angle bisection repeatedly to divide it into 4 equal parts.

27. Construct with ruler and compass, angles of the following measures - (i) \(30^\circ\) (ii) \(90^\circ\) (iii) \(75^\circ\) (iv) \(45^\circ\) (v) \(105^\circ\)
Answer:
i) \(30^\circ\): Bisect a \(60^\circ\) angle.
ii) \(90^\circ\): Construct perpendicular at the endpoint or midpoint using standard 60-120 arc intersections.
iii) \(75^\circ\): Bisect the angle between \(60^\circ\) and \(90^\circ\).
iv) \(45^\circ\): Bisect a \(90^\circ\) angle.
v) \(105^\circ\): Bisect the angle between \(90^\circ\) and \(120^\circ\).
In simple words: Use standard angle construction and bisection rules with a compass.

28. Draw angle bisector of \(150^\circ\).
Answer:
Steps of construction:
1. Construct \(150^\circ\) (which is \(90^\circ + 60^\circ\) or supplementary to \(30^\circ\)).
2. Using compass bisection on the arms of the \(150^\circ\) angle, draw the angle bisector to get two \(75^\circ\) angles.
In simple words: Construct 150 degrees and bisect it into two 75-degree angles.

29. Make a copy of \(80^\circ\) using ruler and compass.
Answer:
Steps of construction:
1. Draw initial angle \(80^\circ\) with protractor.
2. Use compass to measure arc distance between arms and transfer it to a new ray to replicate the exact angle.
In simple words: Use a compass to copy the span and width of the 80-degree angle onto a new line.

30. Divide Rs. 3500 among A, B and C in the ratio 3:2:2.
Answer:
Total parts = 3 + 2 + 2 = 7
A's share = \(\frac{3}{7} \times 3500 = \text{Rs. } 1500\)
B's share = \(\frac{2}{7} \times 3500 = \text{Rs. } 1000\)
C's share = \(\frac{2}{7} \times 3500 = \text{Rs. } 1000\)
In simple words: Divide total amount by sum of ratio parts, then multiply by individual ratio numbers.

31. If a dozen pens cost Rs. 15. How many pens can be bought for Rs. 43.75?
Answer:
1 dozen = 12 pens cost Rs. 15
Cost of 1 pen = \(\frac{15}{12} = \text{Rs. } 1.25\)
Number of pens for Rs. 43.75 = \(\frac{43.75}{1.25} = 35 \text{ pens}\)
In simple words: Find cost of one pen, then divide total budget by unit cost.

32. Find 'a' if \(12 : a :: 96 : 160\).
Answer:
Product of extremes = Product of means
\(12 \times 160 = a \times 96\)
\(1920 = 96a\)
\(a = \frac{1920}{96} = 20\)
In simple words: Multiply outer terms and inner terms to solve for the unknown variable a.

33. Find the ratio of 5 minutes to 40 seconds.
Answer:
Convert minutes to seconds: \(5 \text{ minutes} = 5 \times 60 = 300 \text{ seconds}\)
Ratio = \(\frac{300}{40} = \frac{30}{4} = 15:2\)
In simple words: Convert both time values to seconds before finding their ratio.

34. An engineer earns Rs. 60,000 per month. His monthly expenditure is Rs. 45,000. Find the ratio of -
(i) his expenditure to his saving
(ii) his income to his expenditure
(iii) his income to his savings
Answer:
Income = Rs. 60,000, Expenditure = Rs. 45,000
Savings = Income - Expenditure = \(60000 - 45000 = \text{Rs. } 15000\)
i) Expenditure to saving = \(\frac{45000}{15000} = 3:1\)
ii) Income to expenditure = \(\frac{60000}{45000} = 4:3\)
iii) Income to savings = \(\frac{60000}{15000} = 4:1\)
In simple words: Calculate savings first, then write the required ratios as simplified fractions.

37. Mother wants to divide Rs. 380 between her daughter Divyangi & Shubhangi in the ratio of their ages. If the age of Divyangi is 9 yrs. and age of Shubhangi is 10 yrs. Find how much Divyangi and Shubhangi will get?
Answer:
Ratio of ages = 9:10
Total parts = 9 + 10 = 19
Divyangi's share = \(\frac{9}{19} \times 380 = 9 \times 20 = \text{Rs. } 180\)
Shubhangi's share = \(\frac{10}{19} \times 380 = 10 \times 20 = \text{Rs. } 200\)
In simple words: Divide the amount in the ratio of their ages (9:10).

38. Present age of father is 42 years and that of son is 14 years. Find the ratio of:
(i) Present age of father to the present age of son.
(ii) Age of father to the age of son when father was 30 yrs. old.
(iii) Age of father to the age of son, when son was 12 yrs old.
Answer:
i) Present ratio = \(\frac{42}{14} = 3:1\)
ii) Difference in ages = \(42 - 14 = 28 \text{ years}\). When father was 30, son's age = \(30 - 28 = 2 \text{ years}\). Ratio = \(\frac{30}{2} = 15:1\)
iii) When son was 12 years old, father's age = \(12 + 28 = 40 \text{ years}\). Ratio = \(\frac{40}{12} = 10:3\)
In simple words: Age difference remains constant; calculate father's and son's ages for each given condition.

39. How many tiles with dimensions 12cm and 5cm will be needed to fit a region whose length and breadth are 144cm and 100 cm respectively?
Answer:
Area of region = \(144 \times 100 = 14400 \text{ sq. cm}\)
Area of one tile = \(12 \times 5 = 60 \text{ sq. cm}\)
Number of tiles = \(\frac{14400}{60} = 240 \text{ tiles}\)
In simple words: Divide total region area by single tile area.

40. The length and breadth of a rectangular park are 615m and 55m respectively. Find the cost of fencing the park at the rate of Rs. 9.25 per metre.
Answer:
Perimeter = \(2(615 + 55) = 2(670) = 1340 \text{ m}\)
Cost per meter = Rs. 9.25
Total Cost = \(1340 \times 9.25 = \text{Rs. } 12395\)
In simple words: Find perimeter and multiply by cost per meter.

41. Find the side of a square, if its area is \(324 \text{ m}^2\).
Answer:
Area = \(s^2 = 324\)
Side \(s = \sqrt{324} = 18 \text{ m}\)
In simple words: Take the square root of the area to find the side length.

42. Find the rule which gives the number of match sticks required to make a pattern of letter Z.
Answer:
Letter Z requires 3 match sticks for 1 Z.
Rule = \(3n\) (where \(n\) is the number of Z shapes).
In simple words: Multiply the number of shapes by 3.

43. State with reason which of the following are equations:
(i) \((x - 5) > 8\)
(ii) \(7 = (11 \times 5) - (12 \times 4)\)
(iii) \(20 = 5Y\)
(iv) \(7 - x = 5\)
Answer:
Equations (contain equals sign):
ii) \(7 = (11 \times 5) - (12 \times 4)\)
iii) \(20 = 5Y\)
iv) \(7 - x = 5\)
(Reason: They have an equality sign between LHS and RHS. Note: (i) is an inequality).
In simple words: Statements with an equals sign are equations.

44. Choose the value of 'x' from the value given in the bracket.
(i) \(x + 5 = 14 \quad (19, 9, -5, 0)\)
(iii) \(6x = 48 \quad (42, 6, 8, 7)\)
Answer:
i) For \(x + 5 = 14\), testing \(x = 9\) gives \(9 + 5 = 14\). Correct value = 9.
iii) For \(6x = 48\), testing \(x = 8\) gives \(6(8) = 48\). Correct value = 8.
In simple words: Substitute bracket values into the equation to find which one satisfies it.

45. Give expression for the following cases -
(i) 'n' multiplied by 2 and 1 subtracted from the product.
(ii) The property of associativity of addition of numbers by using variables a, b and c.
Answer:
i) \(2n - 1\)
ii) \((a + b) + c = a + (b + c)\)
In simple words: Write algebraic statements corresponding to the description.

46. Change the following statement using expression into statement in ordinary language. 'Jagya is z years old. His uncle is 4z years old and his aunt is (4z - 3) years old.'
Answer:
Uncle's age is 4 times Jagya's age, and Aunt's age is 3 years less than 4 times Jagya's age.
In simple words: Translate algebraic expressions back into ordinary descriptive sentences.

47. What is the other name for line of symmetry of circle?
Answer:
Diameter.
In simple words: Any line of symmetry of a circle passes through its center and is called a diameter.

48. Complete the following table.

In simple words: List the exact number of mirror-symmetry lines for each standard geometric shape.

Exam Tip: Memorize symmetry counts for standard polygons.