CBSE Class 7 Mathematics Practice Worksheet Set 16

Question 1. Simplify (- 50) + (-10) – (-90) + 8
Answer: Simplify the signs step by step:
\( (-50) + (-10) - (-90) + 8 \)
\( = -50 - 10 + 90 + 8 \)
Combine the negative numbers and positive numbers:
\( = -60 + 98 = 38 \).
In simple words: Add the negative numbers to get -60, add the positive numbers to get 98, then find 98 minus 60, which is 38.

Exam Tip: Remember that two minus signs together turn into a plus sign: \( -(-90) = +90 \).

Question 2. What is the probability of getting, when a dice is thrown?
i. a multiple of 3
ii. an odd number
iii. An even number
iv. A number 3 or 4
v. a number between 3 and 6
Answer: The possible outcomes on a standard die are 1, 2, 3, 4, 5, and 6. Total outcomes = 6.
i. Multiples of 3 are 3 and 6 (2 outcomes):
\( \text{Probability} = \frac{2}{6} = \frac{1}{3} \).
ii. Odd numbers are 1, 3, and 5 (3 outcomes):
\( \text{Probability} = \frac{3}{6} = \frac{1}{2} \).
iii. Even numbers are 2, 4, and 6 (3 outcomes):
\( \text{Probability} = \frac{3}{6} = \frac{1}{2} \).
iv. The numbers 3 or 4 give 2 outcomes:
\( \text{Probability} = \frac{2}{6} = \frac{1}{3} \).
v. Numbers between 3 and 6 are 4 and 5 (2 outcomes):
\( \text{Probability} = \frac{2}{6} = \frac{1}{3} \).
In simple words: Count how many numbers match what you want, then put that over 6 and reduce the fraction.

Exam Tip: "Between 3 and 6" excludes both 3 and 6, leaving only 4 and 5 as favorable outcomes.

Question 3. Which is greater? \( \frac{3}{4} \) of \( \frac{28}{9} \) or \( \frac{2}{3} \) of \( \frac{21}{8} \)
Answer: Calculate the value of each expression:
First value = \( \frac{3}{4} \times \frac{28}{9} = \frac{1 \times 7}{1 \times 3} = \frac{7}{3} = 2\frac{1}{3} \approx 2.33 \).
Second value = \( \frac{2}{3} \times \frac{21}{8} = \frac{1 \times 7}{1 \times 4} = \frac{7}{4} = 1\frac{3}{4} = 1.75 \).
Comparing the two values, \( \frac{7}{3} > \frac{7}{4} \).
Therefore, \( \frac{3}{4} \text{ of } \frac{28}{9} \) is greater.
In simple words: The first part simplifies to 7/3 and the second part simplifies to 7/4. Since 7/3 is bigger, the first fraction is greater.

Exam Tip: Cancel common factors between numerators and denominators before multiplying to make comparisons easier.

Question 4. Compare using < or >
\( \frac{7}{9} \ \square \ \frac{10}{12} \)
Answer: Cross-multiply to compare the two fractions:
\( 7 \times 12 = 84 \)
\( 10 \times 9 = 90 \)
Since \( 84 < 90 \), it follows that:
\( \frac{7}{9} < \frac{10}{12} \).
In simple words: Cross-multiplying gives 84 on the left and 90 on the right. Since 84 is smaller, write < in the box.

Exam Tip: Cross-multiplication is the fastest way to compare two fractions without finding a common denominator.

Question 5. Find the perimeter of a
i. Triangle HOE
ii. Rectangle OUSE
Whose perimeter is lesser?
Answer: From the given figure:
Sides of Triangle HOE are \( \text{HO} = \frac{4}{3}\text{ m} \), \( \text{HE} = 2\frac{1}{2}\text{ m} = \frac{5}{2}\text{ m} \), and base \( \text{OE} = 2\frac{1}{2}\text{ m} = \frac{5}{2}\text{ m} \).
i. Perimeter of Triangle HOE:
\( = \frac{4}{3} + \frac{5}{2} + \frac{5}{2} = \frac{4}{3} + 5 = \frac{4 + 15}{3} = \frac{19}{3}\text{ m} = 6\frac{1}{3}\text{ m} \).
ii. Perimeter of Rectangle OUSE:
Length = \( 2\frac{1}{2}\text{ m} = \frac{5}{2}\text{ m} \), Breadth = \( \frac{7}{3}\text{ m} \).
\( \text{Perimeter} = 2 \times (\text{Length} + \text{Breadth}) = 2 \times \left(\frac{5}{2} + \frac{7}{3}\right) \)
\( = 2 \times \left(\frac{15 + 14}{6}\right) = 2 \times \frac{29}{6} = \frac{29}{3}\text{ m} = 9\frac{2}{3}\text{ m} \).
Comparing the two perimeters: \( \frac{19}{3}\text{ m} < \frac{29}{3}\text{ m} \).
Therefore, Triangle HOE has a lesser perimeter.In simple words: The triangle perimeter is 19/3 m and the rectangle perimeter is 29/3 m. The triangle has the smaller perimeter.

Exam Tip: When both final perimeters share the same denominator, you can compare their numerators directly.

Question 6. Verify; a – (-b) = a + b for the following values of a and b
a = -20, b = 15
Answer: Given \( a = -20 \) and \( b = 15 \).
Left Hand Side (LHS):
\( \text{LHS} = a - (-b) = -20 - (-15) = -20 + 15 = -5 \).
Right Hand Side (RHS):
\( \text{RHS} = a + b = -20 + 15 = -5 \).
Since \( \text{LHS} = \text{RHS} = -5 \), the relation is verified.
In simple words: Both sides equal -5, so the rule holds true.

Exam Tip: Write out LHS and RHS separately and conclude with "LHS = RHS, hence verified".

Question 7. Write equations for the following statements and solve:
i. Eight times a number is 120.
ii. The sum of a number and 12 is 27.
iii. 8 less than three times a number.
iv. Twenty less than a number is 54.
v. Adding 1 to one third of m is 5.
Answer: Let the unknown number be \( x \).
i. Equation: \( 8x = 120 \)
Solution: \( x = \frac{120}{8} = 15 \).
ii. Equation: \( x + 12 = 27 \)
Solution: \( x = 27 - 12 = 15 \).
iii. Expression: \( 3x - 8 \).
iv. Equation: \( x - 20 = 54 \)
Solution: \( x = 54 + 20 = 74 \).
v. Equation: \( \frac{1}{3}m + 1 = 5 \)
Solution: \( \frac{1}{3}m = 4 \implies m = 4 \times 3 = 12 \).
In simple words: Turn each sentence into math symbols, then find the unknown value.

Exam Tip: Be careful with "less than": "20 less than a number" means \( x - 20 \), not \( 20 - x \).

Question 8. A car covers a distance of 638.55 km in 5.5 litre of petrol. Find the distance covered by it in one litre of petrol.
Answer: Total distance covered = 638.55 km.
Total petrol used = 5.5 litres.
Distance covered in 1 litre of petrol:
\( = \frac{638.55}{5.5} = \frac{6385.5}{55} = 116.1\text{ km} \).
Therefore, the car covers 116.1 km in one litre of petrol.
In simple words: Divide the total distance by 5.5 litres to get 116.1 km per litre.

Exam Tip: Move the decimal point one place to the right in both numbers before dividing by 55.

Question 9. Find the mode and median of the scores.
4,5,6,7,7,8,9,13,12,8,8,9,8,10,11
Answer: Total number of scores = 15.
Arrange the scores in ascending order:
4, 5, 6, 7, 7, 8, 8, 8, 8, 9, 9, 10, 11, 12, 13.
1. Mode:
The number 8 occurs 4 times, which is more frequent than any other score.
\( \text{Mode} = 8 \).
2. Median:
Since \( n = 15 \) is odd, the median is the \( \left(\frac{15 + 1}{2}\right)\text{th} = 8\text{th} \) term.
The 8th score is 8.
\( \text{Median} = 8 \).
In simple words: The number 8 appears the most times, and 8 also sits right in the middle of the sorted list.

Exam Tip: Always sort the numbers from smallest to largest before finding the median.

Question 10. A box of 600 electric bulbs contains 12 defective bulbs. One bulb is taken out at random from this box. What is the probability that it is a non-defective bulb?
Answer: Total number of bulbs = 600.
Number of defective bulbs = 12.
Number of non-defective bulbs = \( 600 - 12 = 588 \).
Probability of drawing a non-defective bulb:
\( \text{Probability} = \frac{588}{600} = \frac{147}{150} = \frac{49}{50} = 0.98 \).
In simple words: There are 588 good bulbs out of 600. Dividing 588 by 600 simplifies to 49/50 or 0.98.

Exam Tip: You can also find this by subtracting the probability of a defective bulb from 1: \( 1 - \frac{12}{600} = \frac{588}{600} \).

Question 11. Find the area of a rectangle whose length is 36.8cm and breadth is 11.5cm.
Answer: Length (\( l \)) = 36.8 cm.
Breadth (\( b \)) = 11.5 cm.
\( \text{Area of rectangle} = l \times b \)
\( = 36.8 \times 11.5 = 423.2\text{ cm}^2 \).
Hence, the area is 423.2 cm².
In simple words: Multiply length by breadth to get an area of 423.2 square centimetres.

Exam Tip: Always include the proper unit of area (\( \text{cm}^2 \)) in your final statement.

Question 12. Construct 2 equations starting with x = 3
Answer: Starting with \( x = 3 \):
1. Multiply both sides by 2 and add 5:
\( 2x + 5 = 2(3) + 5 \implies 2x + 5 = 11 \).
2. Multiply both sides by 4 and subtract 2:
\( 4x - 2 = 4(3) - 2 \implies 4x - 2 = 10 \).
Two equations starting with \( x = 3 \) are \( 2x + 5 = 11 \) and \( 4x - 2 = 10 \).
In simple words: Apply the same math operations to both sides of x = 3 to create new valid equations.

Exam Tip: Keep operations simple, such as multiplying by a small number and then adding or subtracting a constant.

Question 13. Find
i. 158.1 × 100
ii. 7.9 × 100
iii. 82.7 × 1000
iv. 536.4 ÷ 1000
Answer:
i. \( 158.1 \times 100 = 15810 \).
ii. \( 7.9 \times 100 = 790 \).
iii. \( 82.7 \times 1000 = 82700 \).
iv. \( 536.4 \div 1000 = 0.5364 \).
In simple words: Move the decimal point right when multiplying by 10, 100, or 1000, and move it left when dividing.

Exam Tip: Count the number of zeros in powers of 10 to know exactly how many places to shift the decimal point.

Question 14. Solve the equation: 20 = 2 + 3 [y + 2]
Answer: Subtract 2 from both sides:
\( 20 - 2 = 3[y + 2] \)
\( 18 = 3[y + 2] \)
Divide both sides by 3:
\( \frac{18}{3} = y + 2 \)
\( 6 = y + 2 \)

\( \implies y = 6 - 2 = 4 \).
Therefore, \( y = 4 \).
In simple words: Subtract 2 to get 18, divide by 3 to get 6, and subtract 2 to find y = 4.

Exam Tip: Dividing by 3 first avoids expanding the brackets, making calculations simpler.

Question 15. Solve the following.
i. The sum of three consecutive numbers is 21.Find the numbers.
ii. Radhika’s father is three times as old as Radhika.If the sum of their ages is 56 years , what are their ages?
iii. The length of a rectangle is twice its breadth. If its perimeter is 18cm, find its length and breadth.
Answer:
i. Let the three consecutive numbers be \( x, x+1, x+2 \):
\( x + (x + 1) + (x + 2) = 21 \)
\( 3x + 3 = 21 \implies 3x = 18 \implies x = 6 \).
The numbers are 6, 7, and 8.
ii. Let Radhika's age be \( x \) years. Her father's age = \( 3x \) years.
\( x + 3x = 56 \implies 4x = 56 \implies x = 14 \).
Radhika is 14 years old and her father is \( 3 \times 14 = 42 \) years old.
iii. Let breadth = \( b \). Length = \( 2b \).
\( \text{Perimeter} = 2(2b + b) = 6b = 18 \implies b = 3\text{ cm} \).
Breadth = 3 cm, Length = \( 2 \times 3 = 6\text{ cm} \).
In simple words: The consecutive numbers are 6, 7, 8. Radhika is 14 and her dad is 42. The rectangle has width 3 cm and length 6 cm.

Exam Tip: Set up clear variable representations at the start of word problems to earn full method marks.

Question 16. A bag contains 3 red and 2 blue marbles. A marble is drawn at random. What is the probability of drawing a blue marble.
Answer: Total number of marbles = \( 3 + 2 = 5 \).
Number of blue marbles = 2.
\( \text{Probability of blue marble} = \frac{\text{Number of blue marbles}}{\text{Total marbles}} = \frac{2}{5} \).
In simple words: 2 out of the 5 marbles are blue, so the probability is 2/5.

Exam Tip: Add all marbles together to get the total possible outcomes for the denominator.

Question 17. Evaluate:
i. [(-45) ÷ 9] ÷ 5
ii. [8 + (-20)] ÷ [(-3) + 5]
Answer:
i. Evaluate the bracket first:
\( [(-45) \div 9] = -5 \).
Now divide by 5:
\( -5 \div 5 = -1 \).
ii. Evaluate each bracket separately:
First bracket = \( 8 + (-20) = -12 \).
Second bracket = \( (-3) + 5 = 2 \).
Now divide:
\( -12 \div 2 = -6 \).
In simple words: Work out the inside of the brackets first, then complete the division.

Exam Tip: A negative number divided by a positive number always produces a negative result.

Question 18. Express 48 cm in meter and kilometer.
Answer: Since \( 1\text{ m} = 100\text{ cm} \) and \( 1\text{ km} = 1000\text{ m} = 100,000\text{ cm} \):
1. In metres:
\( 48\text{ cm} = \frac{48}{100}\text{ m} = 0.48\text{ m} \).
2. In kilometres:
\( 48\text{ cm} = \frac{48}{100000}\text{ km} = 0.00048\text{ km} \).
In simple words: Divide by 100 to get 0.48 metres, and divide by 100,000 to get 0.00048 kilometres.

Exam Tip: Check that your decimal in kilometres has four zeros before 48 since \( 10^5 \) has five zeros.

Question 19. Thickness of 12 sheets of paper is 2.4 mm. Find the thickness of 1 sheet of paper.
Answer: Total thickness of 12 sheets = 2.4 mm.
Thickness of 1 sheet of paper:
\( = \frac{2.4}{12} = 0.2\text{ mm} \).
Therefore, one sheet is 0.2 mm thick.
In simple words: Divide 2.4 mm by 12 to find that a single sheet is 0.2 mm thick.

Exam Tip: Notice that \( 24 \div 12 = 2 \), so \( 2.4 \div 12 = 0.2 \).

Question 20. Two different states of India’s experts of garments in the years 2000 to 2004 are given in the following table. Draw a double bar graph.

Answer: Draw the double bar graph by taking years along the horizontal axis and garment exports (in crores of Rs.) along the vertical axis with a scale of 1 unit = 2 crores.In simple words: Draw side-by-side bars for each year to compare garment exports between Kerala and Karnataka.

Exam Tip: Include a clear legend showing which color represents Kerala and which represents Karnataka.

Question 21. In a class test, containing 20 questions, 6 marks are awarded for every correct and [-2] for every Incorrect answer and 0 for answer not attempted.
i. Ramesh gets 9 correct answers and 11 incorrect answers. Find his score.
ii. Syed scored 36 marks though he got 8 correct answers. How many questions had he attempted incorrectly.
Answer:
i. Marks for 9 correct answers = \( 9 \times 6 = 54 \).
Marks deducted for 11 incorrect answers = \( 11 \times (-2) = -22 \).
Total score for Ramesh = \( 54 + (-22) = 32 \).
ii. Marks for 8 correct answers = \( 8 \times 6 = 48 \).
Total score = 36.
Marks lost due to incorrect answers = \( 36 - 48 = -12 \).
Number of incorrect questions = \( \frac{-12}{-2} = 6 \).
Therefore, Syed answered 6 questions incorrectly.
In simple words: Ramesh scored 32 marks. Syed lost 12 marks on wrong answers, which means he got 6 questions wrong.

Exam Tip: Find the difference between actual score and marks from correct answers to determine incorrect attempts.

Question 22. 2.5 X 0.001 = ____________
Answer: 0.0025
Multiply 25 by 1 to get 25. Count the total decimal places: 1 place in 2.5 and 3 places in 0.001 gives 4 decimal places in total:
\( 2.5 \times 0.001 = 0.0025 \).
In simple words: Shift the decimal point three spots to the left to get 0.0025.

Exam Tip: Add leading zeros as placeholders when placing the decimal point across 4 places.

Question 23. Solve the following equations.
i. 3x – 14 = 16
ii. 2 [f – 2] + 3 [4f – 1] = 7
iii. 4 [2x + 1] = 36
iv. 3 [2x – 1] + 5 = 14
v. 2y – [7 – 5y] – 21 = 0
vi. 3[x - \( \frac{1}{3} \)] = 2 [2x – 6]
Answer:
i. \( 3x = 16 + 14 = 30 \implies x = 10 \).
ii. \( 2f - 4 + 12f - 3 = 7 \implies 14f - 7 = 7 \implies 14f = 14 \implies f = 1 \).
iii. \( 2x + 1 = \frac{36}{4} = 9 \implies 2x = 8 \implies x = 4 \).
iv. \( 6x - 3 + 5 = 14 \implies 6x + 2 = 14 \implies 6x = 12 \implies x = 2 \).
v. \( 2y - 7 + 5y - 21 = 0 \implies 7y - 28 = 0 \implies 7y = 28 \implies y = 4 \).
vi. \( 3x - 1 = 4x - 12 \implies 4x - 3x = 12 - 1 \implies x = 11 \).
In simple words: Open brackets, collect like terms, and isolate the variable in each equation.

Exam Tip: In part (v), watch the sign when expanding \( -[7 - 5y] \): it becomes \( -7 + 5y \).

Question 24. _______________ X -12 = 132
Answer: -11
Let the blank be \( x \):
\( x = \frac{132}{-12} = -11 \).
In simple words: Divide 132 by -12 to find that the missing number is -11.

Exam Tip: A positive number divided by a negative number gives a negative quotient.

Question 25. Find the mean of first 5 prime numbers.
Answer: The first 5 prime numbers are 2, 3, 5, 7, and 11.
Sum of numbers = \( 2 + 3 + 5 + 7 + 11 = 28 \).
\( \text{Mean} = \frac{28}{5} = 5.6 \).
In simple words: Add the first five prime numbers to get 28. Divide by 5 to get a mean of 5.6.

Exam Tip: Remember that 1 is neither prime nor composite; the smallest prime number is 2.

Question 26. A certain freezing process requires that room temperature be lowered from 35°C at the rate of 5°C every hour. What will be the room temperature in 10 hours after the process begins?
Answer: Initial room temperature = 35°C.
Rate of cooling = 5°C per hour.
Total drop in temperature in 10 hours = \( 10 \times 5^\circ\text{C} = 50^\circ\text{C} \).
Final temperature = \( 35^\circ\text{C} - 50^\circ\text{C} = -15^\circ\text{C} \).
Therefore, the room temperature will be -15°C.
In simple words: The temperature drops by 50 degrees over 10 hours, moving from 35 down to -15 degrees.

Exam Tip: Always subtract the total drop from the initial temperature: \( 35 - 50 = -15 \).

Question 27. 78 ÷ ___________ = - 78
Answer: -1
Dividing 78 by \( -1 \) yields \( -78 \).
In simple words: Dividing any number by -1 changes its sign to negative.

Exam Tip: Recall the integer rule: \( a \div (-1) = -a \).

Question 28. Find the product using suitable property.
- 725 X [22] + [-78] X 725
Answer: Rewrite the expression by taking out common factor 725:
\( = 725 \times (-22) + 725 \times (-78) \)
Apply the distributive property \( a \times b + a \times c = a \times (b + c) \):
\( = 725 \times [(-22) + (-78)] \)
\( = 725 \times [-100] = -72500 \).
In simple words: Factor out 725 to add -22 and -78, which makes -100. Multiplying 725 by -100 gives -72500.

Exam Tip: Pulling out common factors using the distributive property makes big multiplications quick and simple.

Question 29. \( 4\frac{3}{7} \div 4\frac{3}{7} \) = _______________
Answer: 1
Any non-zero number divided by itself equals 1.
In simple words: Any number divided by itself gives 1.

Exam Tip: Avoid converting into improper fractions here since any identical fraction division simplifies directly to 1.

Question 30. Write down a pair of integers whose
i. Sum is 0
ii. Sum is -8
Answer:
i. A pair of integers whose sum is 0: \( -5 \) and \( 5 \) (since \( -5 + 5 = 0 \)).
ii. A pair of integers whose sum is -8: \( -5 \) and \( -3 \) (since \( -5 + (-3) = -8 \)).
In simple words: Opposite numbers like -5 and 5 add to 0. Two negative numbers like -5 and -3 add to -8.

Exam Tip: Any integer and its additive inverse add up to 0.

Question 31. The number of hours spend by two students; Devi and Deepthi on different activities on a working day is given below:
Represent it on a double bar graph.

Answer: Draw a double bar graph representing the activities on the horizontal axis and hours on the vertical axis with a scale of 1 unit = 2 hours.In simple words: Draw pairs of bars for each activity to compare how Devi and Deepthi spend their day.

Exam Tip: Draw bars with equal widths and leave equal gaps between consecutive activity groups.

Question 32. Express 7cm in meter and kilometer.
Answer:
1. In metres: \( 7\text{ cm} = \frac{7}{100}\text{ m} = 0.07\text{ m} \).
2. In kilometres: \( 7\text{ cm} = \frac{7}{100000}\text{ km} = 0.00007\text{ km} \).
In simple words: Divide by 100 to get 0.07 m, and divide by 100,000 to get 0.00007 km.

Exam Tip: Make sure 0.00007 km has 4 zeros between the decimal point and 7.

Question 33. Find the product using suitable property
[-25] X 103
Answer: Express 103 as \( 100 + 3 \) to apply the distributive property:
\( (-25) \times 103 = (-25) \times (100 + 3) \)
\( = [(-25) \times 100] + [(-25) \times 3] \)
\( = -2500 + (-75) = -2575 \).
In simple words: Split 103 into 100 and 3. Multiply both by -25 and add to get -2575.

Exam Tip: Breaking numbers into multiples of 100 avoids manual long multiplication.

Question 34. How much less is 16.05 kg than 39 kg?
Answer: Subtract 16.05 kg from 39 kg:
\( 39.00 - 16.05 = 22.95\text{ kg} \).
Therefore, 16.05 kg is 22.95 kg less than 39 kg.
In simple words: Take 16.05 away from 39.00 to find the difference of 22.95 kg.

Exam Tip: Write 39 as 39.00 to align decimals correctly before subtracting.

Question 35. Construct 2 equations starting with x = -5
Answer: Starting with \( x = -5 \):
1. Multiply both sides by 2:
\( 2x = 2(-5) \implies 2x = -10 \).
2. Multiply both sides by 3 and add 5:
\( 3x + 5 = 3(-5) + 5 \implies 3x + 5 = -10 \).
Two equations are \( 2x = -10 \) and \( 3x + 5 = -10 \).
In simple words: Do the same math step to both sides of x = -5 to build two new equations.

Exam Tip: You can apply any simple operation (multiplication, addition) to both sides to form equations.

Question 36. Find the perimeter of a triangle with sides \( 7\frac{1}{6} \)cm, \( 3\frac{2}{3} \)cm and \( 5\frac{1}{6} \)cm
Answer: Convert mixed fractions to improper fractions:
\( 7\frac{1}{6} = \frac{43}{6}\text{ cm} \)
\( 3\frac{2}{3} = \frac{11}{3} = \frac{22}{6}\text{ cm} \)
\( 5\frac{1}{6} = \frac{31}{6}\text{ cm} \)
\( \text{Perimeter} = \frac{43}{6} + \frac{22}{6} + \frac{31}{6} = \frac{43 + 22 + 31}{6} = \frac{96}{6} = 16\text{ cm} \).
Hence, the perimeter is 16 cm.
In simple words: Write all fractions with denominator 6, add them to get 96/6, which equals 16 cm.

Exam Tip: Make denominators equal before adding fractions.

Question 37. The quotient in 0.07 ÷ 7 = ________________
Answer: 0.01
Divide 0.07 by 7:
\( \frac{0.07}{7} = 0.01 \).
In simple words: 7 divided by 7 is 1, so 0.07 divided by 7 is 0.01.

Exam Tip: Keep the decimal place in the same position as in the dividend.

Question 38. Find the mean of the scores
8,6,10,12,1,3,4,4
Find the range of the data.
Answer: Total number of scores = 8.
1. Mean:
Sum = \( 8 + 6 + 10 + 12 + 1 + 3 + 4 + 4 = 48 \).
\( \text{Mean} = \frac{48}{8} = 6 \).
2. Range:
Highest score = 12, Lowest score = 1.
\( \text{Range} = 12 - 1 = 11 \).
In simple words: Add all 8 numbers to get 48, then divide by 8 to get a mean of 6. The range is 12 - 1 = 11.

Exam Tip: Range is always the highest value minus the lowest value.

Question 39. When a die is thrown, what is the probability of getting?
i. 7
ii. 5
Answer: Possible outcomes when throwing a die are 1, 2, 3, 4, 5, 6 (total = 6).
i. The number 7 cannot appear on a standard die (0 outcomes):
\( \text{Probability} = \frac{0}{6} = 0 \).
ii. The number 5 appears once (1 outcome):
\( \text{Probability} = \frac{1}{6} \).
In simple words: A die cannot roll a 7, so probability is 0. A 5 appears once, so probability is 1/6.

Exam Tip: The probability of an impossible event is always 0.

Question 40. The height [in cm] of 6 girls in a group are given below
150,141,137,147,138,147
i. What is the height of tallest girl?
ii. What is the range of the data?
iii. What is the mean height of the girls?
Answer:
i. The height of the tallest girl is 150 cm.
ii. Shortest height = 137 cm.
\( \text{Range} = 150 - 137 = 13\text{ cm} \).
iii. Mean height:
Sum = \( 150 + 141 + 137 + 147 + 138 + 147 = 860\text{ cm} \).
\( \text{Mean} = \frac{860}{6} = 143.33\text{ cm} \).
In simple words: The tallest is 150 cm. The range is 13 cm. The mean height is 143.33 cm.

Exam Tip: Include units (cm) with your answers for height, range, and mean.

Question 41. An elevator descends into a mineshaft at the rate of 6 meters per minute. If it begins to descends from 20 m above the ground.
i. What will be its position after 45 minutes?
ii. How long will it take to reach -400 m?
Answer: Starting position = \( +20\text{ m} \). Descent rate = \( -6\text{ m/min} \).
i. Distance descended in 45 minutes = \( 45 \times 6 = 270\text{ m} \).
Position after 45 minutes = \( 20 - 270 = -250\text{ m} \) (250 m below ground).
ii. Total distance to descend from +20 m to -400 m = \( 20 - (-400) = 420\text{ m} \).
Time taken = \( \frac{420}{6} = 70\text{ minutes} \) (1 hour 10 minutes).
In simple words: After 45 minutes it is at -250 m. To reach -400 m, it travels 420 m, which takes 70 minutes.

Exam Tip: Remember to add the 20 m above ground to the 400 m below ground to get the total descent distance.

Question 42. Find the product using suitable property
− 56 X 102
Answer: Write 102 as \( 100 + 2 \) and use the distributive property:
\( (-56) \times 102 = (-56) \times (100 + 2) \)
\( = [(-56) \times 100] + [(-56) \times 2] \)
\( = -5600 + (-112) = -5712 \).
In simple words: Break 102 into 100 + 2. Multiply each part by -56 to get -5712.

Exam Tip: Always state the distributive property used to secure full marks.

Question 43. Construct 4 equations starting with x = 3
Answer: Starting with \( x = 3 \):
1. \( 2x = 6 \)
2. \( 3x + 1 = 10 \)
3. \( 5x - 5 = 10 \)
4. \( \frac{x}{3} = 1 \).
In simple words: Multiply, add, or divide both sides of x = 3 to create four equations.

Exam Tip: Use different operations on each side to create distinct linear equations.

Question 44. Kathy plants 5 saplings in a row in his garden. The distance between two adjacent saplings is \( \frac{5}{6} \)m.
Find the distance between the first and last sapling.
Answer: With 5 saplings in a row, there are \( 5 - 1 = 4 \) gaps between them.
Distance of each gap = \( \frac{5}{6}\text{ m} \).
Total distance = \( 4 \times \frac{5}{6} = \frac{20}{6} = \frac{10}{3}\text{ m} = 3\frac{1}{3}\text{ m} \).
Hence, the distance between the first and last sapling is \( 3\frac{1}{3}\text{ m} \).
In simple words: 5 plants have 4 spaces between them. 4 times 5/6 m gives 10/3 m or 3 1/3 m.

Exam Tip: \( n \) items placed in a row always create \( n - 1 \) gaps.

Question 45. Find
i. 76.5 ÷ 0.15
ii. 1.07 X 0.02
Answer:
i. \( \frac{76.5}{0.15} = \frac{7650}{15} = 510 \).
ii. \( 1.07 \times 0.02 = 0.0214 \).
In simple words: In (i), shift the decimal two spots to divide 7650 by 15, giving 510. In (ii), multiply to get 0.0214.

Exam Tip: Count all decimal places in factors to place the decimal point correctly in the product.

Question 46. Find the following using suitable properties.
i. 625 X [-25] X [-16] X [-4]
ii. 348 X [-37] + 348 X [-63]
Answer:
i. Group terms conveniently using the associative property:
\( = 625 \times [(-25) \times (-4)] \times (-16) \)
\( = 625 \times [100] \times (-16) \)
\( = 62500 \times (-16) = -1,000,000 \).
ii. Use the distributive property:
\( = 348 \times [(-37) + (-63)] \)
\( = 348 \times [-100] = -34,800 \).
In simple words: In (i), -25 times -4 gives 100, leading to -1,000,000. In (ii), combine -37 and -63 to get -100, giving -34,800.

Exam Tip: Grouping numbers that multiply to 100 or 1000 makes long calculations much easier.

Question 47. The additive inverse of –n is _________________.
Answer: n
The additive inverse of a number is the value that adds to it to yield 0: \( (-n) + n = 0 \).
In simple words: Changing the sign of -n gives n.

Exam Tip: The additive inverse of any value \( -a \) is always \( +a \).

Question 48. The sum of 3 times a number and 11 is 32. Find the number?
Answer: Let the number be \( x \).
\( 3x + 11 = 32 \)
\( 3x = 32 - 11 = 21 \)

\( \implies x = \frac{21}{3} = 7 \).
Therefore, the number is 7.
In simple words: 3 times the number plus 11 equals 32. Subtract 11 and divide by 3 to get 7.

Exam Tip: Set up the equation directly from the sentence description and isolate \( x \).

Question 49. Write 305.142 in expanded form.
Answer: Write each digit with its place value:
\( 305.142 = (3 \times 100) + (0 \times 10) + (5 \times 1) + \left(1 \times \frac{1}{10}\right) + \left(4 \times \frac{1}{100}\right) + \left(2 \times \frac{1}{1000}\right) \)
\( = 300 + 5 + \frac{1}{10} + \frac{4}{100} + \frac{2}{1000} \).
In simple words: Expand each digit according to its place value: 300 + 5 + 1/10 + 4/100 + 2/1000.

Exam Tip: Digits after the decimal point represent fractional place values: tenths, hundredths, thousandths.

Question 50. Solve 54 = 14 + 5 [t – 17]
Answer: Subtract 14 from both sides:
\( 54 - 14 = 5[t - 17] \)
\( 40 = 5[t - 17] \)
Divide both sides by 5:
\( 8 = t - 17 \)

\( \implies t = 8 + 17 = 25 \).
Hence, \( t = 25 \).
In simple words: Subtract 14 to get 40, divide by 5 to get 8, then add 17 to get t = 25.

Exam Tip: Dividing both sides by 5 before expanding brackets makes solving simpler.

Question 51. Write three integers between – 8 and 15?
Answer: Any three integers strictly between -8 and 15 can be chosen, such as \( -5, 0, 10 \).
In simple words: -5, 0, and 10 all lie between -8 and 15 on the number line.

Exam Tip: Make sure your chosen numbers do not include the boundary values -8 and 15.

Question 52. Subtract the sum of – 1050 and 813 from −23.
Answer: First, find the sum of -1050 and 813:
\( \text{Sum} = -1050 + 813 = -237 \).
Now, subtract this sum from -23:
\( -23 - (-237) = -23 + 237 = 214 \).
Therefore, the result is 214.
In simple words: Add -1050 and 813 to get -237. Then calculate -23 minus -237, which gives 214.

Exam Tip: "Subtract A from B" always means \( B - A \).

Question 53. Study the following bar graph and answer the questions given below.
a. The maximum rainfall received is _____________ mm on _____________.
b. On which days the rainfall was same?
c. The least rainfall received is _______________ mm on ______________.Answer: Reading the bar graph:
a. The maximum rainfall received is 12 mm on Friday.
b. The rainfall was the same on Monday and Thursday (5 mm each), as well as on Tuesday and Saturday (6 mm each).
c. The least rainfall received is 2 mm on Wednesday.
In simple words: Friday had the most rain at 12 mm, Wednesday had the least at 2 mm, and Monday and Thursday had equal rain.

Exam Tip: Check bar heights carefully against the vertical axis scale to read values accurately.

Question 54. Write a pair of integers whose sum is -7.
Answer: A pair of integers whose sum is -7 is \( -3 \) and \( -4 \) (since \( -3 + (-4) = -7 \)).
In simple words: -3 and -4 add together to make -7.

Exam Tip: Multiple correct pairs exist, such as (-10, 3) or (-1, -6).

Question 55. Find the product using suitable property
750 X [-45] + [-750] X 55
Answer: Rewrite the expression by factoring out 750:
\( = 750 \times (-45) + 750 \times (-55) \)
Use the distributive property:
\( = 750 \times [(-45) + (-55)] \)
\( = 750 \times [-100] = -75,000 \).
In simple words: Take 750 out as a common factor. Add -45 and -55 to get -100, then multiply by 750 to get -75,000.

Exam Tip: Rewrite \( [-750] \times 55 \) as \( 750 \times (-55) \) to extract 750 cleanly.

Question 56. \( 2\frac{2}{3} + 3\frac{1}{2} \) = ________________
Answer: \( 6\frac{1}{6} \)
Convert to improper fractions:
\( \frac{8}{3} + \frac{7}{2} = \frac{16 + 21}{6} = \frac{37}{6} = 6\frac{1}{6} \).
In simple words: Make denominators 6 and add 16/6 and 21/6 to get 37/6 or 6 1/6.

Exam Tip: Find the LCM of 3 and 2, which is 6, to add fractions with unlike denominators.

Question 57. [-24] ÷ ______________ = 8
Answer: -3
Divide -24 by 8:
\( \frac{-24}{8} = -3 \).
In simple words: -24 divided by -3 equals positive 8.

Exam Tip: Dividing a negative by a negative gives a positive result.

Question 58. The most common representative value of a group of data is __________________
Answer: Arithmetic Mean (or Mean)
The arithmetic mean is the most widely used representative value for a given set of data.
In simple words: The mean or average is the most common representative value of a data group.

Exam Tip: The three central measures are mean, median, and mode, with mean being the most common.

Question 59. The solution of the equation 2q + 6 = 0 is __________.
Answer: -3
\( 2q = -6 \implies q = \frac{-6}{2} = -3 \).
In simple words: Move 6 across to get -6, then divide by 2 to get q = -3.

Exam Tip: Transposing +6 across the equals sign changes its sign to -6.

Question 60. Identify the property used in the following.
a. [-5] + [-8] = [-8] + [-5]
b. 75 + 0 = 75
c. [13 + 7] + [-9] = 13 + [7 + (-9)]
Answer:
a. Commutative property of addition.
b. Additive identity property.
c. Associative property of addition.
In simple words: (a) shows commutative property, (b) shows additive identity, and (c) shows associative property.

Exam Tip: Commutative is about order of terms, associative is about grouping brackets, and identity involves adding 0.

Question 61. Reciprocal of a mixed fraction is always ______________.
Answer: a proper fraction
Every mixed fraction is greater than 1, so its reciprocal is always less than 1, which forms a proper fraction.
In simple words: A mixed fraction is bigger than 1, so flipping it upside down always gives a proper fraction.

Exam Tip: Since a mixed fraction has numerator > denominator, its reciprocal has numerator < denominator.

Question 62. Mean of the first five natural numbers is _____________.
Answer: 3
The first five natural numbers are 1, 2, 3, 4, and 5.
\( \text{Mean} = \frac{1 + 2 + 3 + 4 + 5}{5} = \frac{15}{5} = 3 \).
In simple words: Add 1 through 5 to get 15, then divide by 5 to get 3.

Exam Tip: Natural numbers begin at 1, unlike whole numbers which start at 0.

Question 63. 6 cm = ____________ km
Answer: 0.00006
Since \( 1\text{ km} = 100,000\text{ cm} \):
\( 6\text{ cm} = \frac{6}{100000}\text{ km} = 0.00006\text{ km} \).
In simple words: Divide 6 by 100,000 to get 0.00006 km.

Exam Tip: Shift the decimal point five places left to convert from centimetres to kilometres.

Question 64. If you multiply [-1] 12 times, the sign of the product will be _________________.
Answer: positive
Multiplying -1 an even number of times gives a positive result: \( (-1)^{12} = +1 \).
In simple words: 12 is an even number, so multiplying -1 twelve times gives a positive sign.

Exam Tip: Negative numbers raised to an even power become positive.

Question 65. [-10] ÷ 0 = ____________.
Answer: undefined (or not defined)
Division of any real number by zero is not defined in mathematics.
In simple words: You cannot divide any number by zero, so the result is undefined.

Exam Tip: Do not confuse \( 0 \div a = 0 \) with \( a \div 0 \), which is undefined.

Question 66. 5.63 X 1000 = _____________.
Answer: 5630
Move the decimal point 3 places to the right: \( 5.63 \times 1000 = 5630 \).
In simple words: Shift the decimal point three spots right to get 5630.

Exam Tip: Add a trailing zero as a placeholder when multiplying 5.63 by 1000.

Question 67. The solution of the equation 5m + 7 = 17 is ______________.
Answer: 2
\( 5m = 17 - 7 = 10 \implies m = \frac{10}{5} = 2 \).
In simple words: Subtract 7 from 17 to get 10, then divide by 5 to get m = 2.

Exam Tip: Isolate the variable term first before dividing by its coefficient.

Question 68. In a class of 40 students 1/5 of the total number of students like to study English. 2/5 of the total number of students like to study Maths and the remaining students like to study Science?
a. How many students like to study Maths?
b. How many students like to study English?
c. What fraction of the total number of students like to study Science?
Answer: Total students = 40.
a. Students who like Maths = \( \frac{2}{5} \times 40 = 16 \).
b. Students who like English = \( \frac{1}{5} \times 40 = 8 \).
c. Fraction who like Science = \( 1 - \left(\frac{1}{5} + \frac{2}{5}\right) = 1 - \frac{3}{5} = \frac{2}{5} \).
In simple words: 16 students like Maths, 8 like English, and 2/5 of the class like Science.

Exam Tip: Subtract the sum of known fractions from 1 to find the remaining fraction.

Question 69. Solve the following equation
a. \( \frac{y}{2} - 3 = 8 \)
b. 4p – 20 = 8
Answer:
a. \( \frac{y}{2} = 8 + 3 = 11 \implies y = 11 \times 2 = 22 \).
b. \( 4p = 8 + 20 = 28 \implies p = \frac{28}{4} = 7 \).
In simple words: In (a), add 3 and multiply by 2 to get y = 22. In (b), add 20 and divide by 4 to get p = 7.

Exam Tip: Transpose the constant term first, then multiply or divide by the coefficient.

Question 70. Find the value of \( 6\frac{2}{3} \div 13\frac{1}{3} \)
Answer: Convert mixed fractions to improper fractions:
\( 6\frac{2}{3} = \frac{20}{3} \)
\( 13\frac{1}{3} = \frac{40}{3} \)
\( \frac{20}{3} \div \frac{40}{3} = \frac{20}{3} \times \frac{3}{40} = \frac{20}{40} = \frac{1}{2} \).
In simple words: Convert to 20/3 and 40/3. Dividing gives 20/40, which simplifies to 1/2.

Exam Tip: Multiply by the reciprocal of the second fraction to complete division.

Question 71. Verify the following
17 X [ 6 + ( - 2 ) ] = [ 17 X 6 ] + [ 17 x ( - 2 ) ]
Answer:
LHS = \( 17 \times [6 + (-2)] = 17 \times 4 = 68 \).
RHS = \( [17 \times 6] + [17 \times (-2)] = 102 + (-34) = 68 \).
Since \( \text{LHS} = \text{RHS} = 68 \), the statement is verified.
In simple words: Both sides simplify to 68, proving the distributive property works.

Exam Tip: Evaluate both sides separately to verify equality clearly.

Question 72. A certain freezing process requires that room temperature be lowered from 35°C at the rate of 4°C every hour. What will be the room temperature 10 hours after the process begins.
Answer: Initial temperature = 35°C.
Rate of cooling = 4°C per hour.
Total decrease in 10 hours = \( 10 \times 4^\circ\text{C} = 40^\circ\text{C} \).
Final temperature = \( 35^\circ\text{C} - 40^\circ\text{C} = -5^\circ\text{C} \).
Therefore, the room temperature after 10 hours will be -5°C.
In simple words: The temperature drops by 40 degrees in 10 hours, moving from 35 down to -5 degrees.

Exam Tip: Subtract total temperature drop from starting temperature: \( 35 - 40 = -5 \).

Question 73. Find the mean , median and mode of the data : 7,9,8,11,8,12,8,9,9
Answer: Total values (\( n \)) = 9.
Sorted values: 7, 8, 8, 8, 9, 9, 9, 11, 12.
1. Mean:
Sum = \( 7 + 8 + 8 + 8 + 9 + 9 + 9 + 11 + 12 = 81 \).
\( \text{Mean} = \frac{81}{9} = 9 \).
2. Median:
Median is the 5th term = 9.
3. Mode:
Both 8 and 9 occur 3 times each.
\( \text{Mode} = 8 \text{ and } 9 \) (bimodal).
In simple words: The mean is 9, the median is 9, and the modes are 8 and 9.

Exam Tip: A dataset can have more than one mode if two numbers share the highest frequency.

Question 74. In a test +5 marks are given for every correct answers and [-2] marks are given for every incorrect answers. Rayan answered all the questions and scored 30 marks though he got 10 correct answers.
How many questions has he attempted incorrectly?
Answer: Marks for 10 correct answers = \( 10 \times 5 = 50 \).
Total score = 30.
Marks lost due to incorrect answers = \( 30 - 50 = -20 \).
Marks awarded for each incorrect answer = -2.
Number of incorrect questions = \( \frac{-20}{-2} = 10 \).
Hence, Rayan attempted 10 questions incorrectly.
In simple words: Rayan earned 50 marks from correct answers. He ended up with 30, meaning he lost 20 marks on 10 wrong answers.

Exam Tip: Divide the lost marks by -2 to get the number of incorrect responses.

Question 75. Verify 25 X [ 7 + ( - 3 ) ] = { 25 X 7 ] + [ 25 X ( - 3 ) ]
Answer:
LHS = \( 25 \times [7 + (-3)] = 25 \times 4 = 100 \).
RHS = \( [25 \times 7] + [25 \times (-3)] = 175 + (-75) = 100 \).
Since \( \text{LHS} = \text{RHS} = 100 \), the equation is verified.
In simple words: Both sides work out to 100, verifying the distributive property.

Exam Tip: Clearly show calculations for both sides separately to demonstrate equality.

Question 76. Write the property used in the following.
i. 25 + [-25] = 0
ii. 13 + [ - 12 ] + [ - 7 ] = 13 + [ ( - 12 ) + ( - 7 ) ]
Answer:
i. Additive inverse property.
ii. Associative property of addition.
In simple words: (i) shows the additive inverse property and (ii) shows the associative property.

Exam Tip: Adding a number to its opposite to get 0 is the additive inverse property.

Question 77. Put correct sign <,> or = in the box
47 + [-24] – 12 [ ] [-47] – [-24] + 12
Answer:
Evaluate Left Side:
\( 47 + (-24) - 12 = 47 - 24 - 12 = 47 - 36 = 11 \).
Evaluate Right Side:
\( -47 - (-24) + 12 = -47 + 24 + 12 = -47 + 36 = -11 \).
Since \( 11 > -11 \), the correct sign is >.
\( 47 + [-24] - 12 > [-47] - [-24] + 12 \).
In simple words: The left side is 11 and the right side is -11. Since 11 is greater, use >.

Exam Tip: Simplify each side to a single integer before comparing.

Question 78. Write two pairs of integers [a,b] such that a ÷ b = [-5]
Answer: Two pairs of integers \( [a, b] \) are:
1. \( [-25, 5] \), since \( -25 \div 5 = -5 \).
2. \( [10, -2] \), since \( 10 \div (-2) = -5 \).
In simple words: Choose pairs like [-25, 5] and [10, -2] where division gives -5.

Exam Tip: Make sure exactly one integer in each pair is negative so the quotient is -5.

Question 79. Find the area of a rectangular field whose length and breadth is 15.7 m and 11.8 m respectively.
Answer: Length = 15.7 m, Breadth = 11.8 m.
\( \text{Area} = \text{Length} \times \text{Breadth} \)
\( = 15.7 \times 11.8 = 185.26\text{ m}^2 \).
Hence, the area is 185.26 m².
In simple words: Multiply 15.7 by 11.8 to get an area of 185.26 square metres.

Exam Tip: Count two decimal places in total for the product.

Question 80. Write 12.034 in expanded form.
Answer:
\( 12.034 = (1 \times 10) + (2 \times 1) + \left(0 \times \frac{1}{10}\right) + \left(3 \times \frac{1}{100}\right) + \left(4 \times \frac{1}{1000}\right) \)
\( = 10 + 2 + \frac{3}{100} + \frac{4}{1000} \).
In simple words: Write out each digit with its place value: 10 + 2 + 3/100 + 4/1000.

Exam Tip: Note that the tenths place is 0, so the first fraction is hundredths.

Question 81. The number of children in the families of 20 students in class VII are –
2,3,3,1,4,5,3,3,2,1,1,4,3,2,2,4,3,5,6,5
Organize this data in a tabular form.
Answer: Count the frequency of each number of children:

In simple words: Organize the counts into a table showing how many families have 1, 2, 3, 4, 5, or 6 children.

Exam Tip: Always sum the frequencies at the bottom of the table to verify they match the total (20).

Question 82. [6 + (- 12)] ÷ [(- 18) + 12]. Find the value.
Answer: First bracket = \( 6 + (-12) = -6 \).
Second bracket = \( (-18) + 12 = -6 \).
Divide:
\( (-6) \div (-6) = 1 \).
In simple words: Both brackets equal -6. Dividing -6 by -6 gives 1.

Exam Tip: Any negative number divided by itself equals positive 1.

Question 83. Write a negative integer and a positive integer whose sum is -7.
Answer: A negative integer and a positive integer are \( -10 \) and \( 3 \) (since \( -10 + 3 = -7 \)).
In simple words: -10 plus 3 gives -7.

Exam Tip: Pick a negative number larger in magnitude than the desired negative sum.

Question 84. Provide the number in the box such that \( \frac{3}{4} \times \square = \frac{15}{40} \)
ii. The simplest form of the number obtained in \( \square \) is _____________.
Answer:
1. Let the missing fraction be \( \frac{a}{b} \):
\( \frac{a}{b} = \frac{15}{40} \div \frac{3}{4} = \frac{15}{40} \times \frac{4}{3} = \frac{5}{10} \).
Therefore, the box contains \( \frac{5}{10} \).
ii. The simplest form of \( \frac{5}{10} \) is \( \frac{1}{2} \).
In simple words: The box holds 5/10. Simplifying 5/10 gives 1/2.

Exam Tip: Divide 15/40 by 3/4 by multiplying by the reciprocal 4/3.

Question 85. Use the sign <,> or = in the box
49 + [-34] – 15 [ ] 26 + [-42] – [- 26]
Answer:
Left Side: \( 49 - 34 - 15 = 49 - 49 = 0 \).
Right Side: \( 26 - 42 + 26 = 52 - 42 = 10 \).
Since \( 0 < 10 \), the correct sign is <.
\( 49 + [-34] - 15 < 26 + [-42] - [-26] \).
In simple words: The left side is 0 and the right side is 10, so 0 is less than 10.

Exam Tip: Notice that \( -[-26] = +26 \) on the right-hand side.

Question 86. Solve 2 [x + 5] = 18
Answer: Divide both sides by 2:
\( x + 5 = \frac{18}{2} = 9 \)

\( \implies x = 9 - 5 = 4 \).
In simple words: Divide 18 by 2 to get 9, then subtract 5 to get x = 4.

Exam Tip: Dividing by 2 first is quicker than expanding \( 2x + 10 = 18 \).

Question 87. Find:
i. 7.9 ÷ 1000
ii. 0.9 X 100
Answer:
i. \( 7.9 \div 1000 = 0.0079 \).
ii. \( 0.9 \times 100 = 90 \).
In simple words: Dividing by 1000 moves the dot left to 0.0079. Multiplying by 100 moves it right to 90.

Exam Tip: Check decimal shifts: 3 places left for dividing by 1000, 2 places right for multiplying by 100.

Question 88. Solve \( \frac{12x}{5} = 6 \)
Answer: Multiply both sides by 5:
\( 12x = 6 \times 5 = 30 \)

\( \implies x = \frac{30}{12} = \frac{5}{2} = 2.5 \).
In simple words: Multiply 6 by 5 to get 30, then divide by 12 to find x = 5/2 or 2.5.

Exam Tip: Reduce 30/12 by dividing numerator and denominator by 6.

Question 89. The ages of 10 teachers of a school are:
32,41,28,54,35,26,23,33,38,40
i. What is the mean age of these ages?
ii. What is the range of the ages of the teachers?
Answer:
i. Mean age:
Sum = \( 32 + 41 + 28 + 54 + 35 + 26 + 23 + 33 + 38 + 40 = 350 \).
\( \text{Mean age} = \frac{350}{10} = 35\text{ years} \).
ii. Range:
Highest age = 54, Lowest age = 23.
\( \text{Range} = 54 - 23 = 31\text{ years} \).
In simple words: The average age is 35 years, and the age range is 54 - 23 = 31 years.

Exam Tip: Sum all 10 ages carefully before dividing by 10.

Question 90. Solve: 4 [x – 3] = 24
Answer: Divide both sides by 4:
\( x - 3 = \frac{24}{4} = 6 \)

\( \implies x = 6 + 3 = 9 \).
In simple words: Divide 24 by 4 to get 6, then add 3 to get x = 9.

Exam Tip: Add 3 to both sides to solve for \( x \) after dividing by 4.

Question 91. What will be the sign of the product if we multiply?
i. 8 negative integers and 3 positive integers.
ii. 5 negative integers and 4 positive integers.
Answer: The sign of the product is determined by the count of negative factors:
i. 8 is an even number, so the product of 8 negative integers is positive. Positive times positive is positive. Sign = positive (+).
ii. 5 is an odd number, so the product of 5 negative integers is negative. Negative times positive is negative. Sign = negative (-).
In simple words: An even number of negative factors gives a positive product; an odd number gives a negative product.

Exam Tip: Only the count of negative integers matters; positive integers never change the sign of the product.

Question 92. Construct two equations starting with x = 4
Answer: Starting with \( x = 4 \):
1. Multiply by 3: \( 3x = 12 \).
2. Multiply by 2 and add 3: \( 2x + 3 = 2(4) + 3 \implies 2x + 3 = 11 \).
Two equations are \( 3x = 12 \) and \( 2x + 3 = 11 \).
In simple words: Apply matching steps to both sides of x = 4 to make two new equations.

Exam Tip: Keep operations simple so you can easily verify the equation holds true for \( x = 4 \).

Question 93. The _______________ of a set of observations is the observation that occurs most often.
Answer: mode
By definition, the mode is the data value that appears with the highest frequency.
In simple words: Mode is the value that shows up most often in a dataset.

Exam Tip: Mode = most frequent, Median = middle value, Mean = average.

Question 94. The probability of getting a head when a coin is tossed is ______________.
Answer: \( \frac{1}{2} \) (or 0.5)
A coin has 2 equally likely outcomes (Head or Tail). Favorable outcome = 1.
\( \text{Probability} = \frac{1}{2} \).
In simple words: A coin has 2 sides, so the chance of landing on a head is 1 out of 2.

Exam Tip: Express probability as a simplified fraction \( \frac{1}{2} \).

Question 95. A batsman scored the following number of runs in six innings – 36, 35, 50, 46, 60, 55.
Calculate the mean runs scored by him in an inning.
Answer: Total innings = 6.
Sum of runs = \( 36 + 35 + 50 + 46 + 60 + 55 = 282 \).
\( \text{Mean runs} = \frac{282}{6} = 47 \).
Therefore, the batsman scored an average of 47 runs per inning.
In simple words: Add all six scores to get 282. Dividing 282 by 6 gives a mean of 47 runs.

Exam Tip: Mean = (Sum of all observations) / (Number of observations).

Question 96. Solve: 9 + 3 [x – 1] = 27
Answer: Subtract 9 from both sides:
\( 3[x - 1] = 27 - 9 = 18 \)
Divide both sides by 3:
\( x - 1 = \frac{18}{3} = 6 \)

\( \implies x = 6 + 1 = 7 \).
In simple words: Subtract 9 to get 18, divide by 3 to get 6, then add 1 to get x = 7.

Exam Tip: Subtract the outside constant before dealing with the bracket.

Question 97. Draw a number line and represent \( \frac{-7}{4} \) and \( \frac{3}{4} \)
Answer: Divide each unit gap into 4 equal segments.
\( \frac{-7}{4} = -1\frac{3}{4} \) lies between \( -1 \) and \( -2 \) (3 units left of -1).
\( \frac{3}{4} \) lies between 0 and 1 (3 units right of 0).In simple words: Split each unit into 4 parts. Count 3 steps right from 0 for 3/4, and 7 steps left from 0 for -7/4.

Exam Tip: Since both fractions share denominator 4, use 4 equal sub-parts per integer unit.

Question 98. Construct two equations starting with x = -1
Answer: Starting with \( x = -1 \):
1. Multiply by 2: \( 2x = -2 \).
2. Multiply by 5 and add 3: \( 5x + 3 = 5(-1) + 3 \implies 5x + 3 = -2 \).
Two equations are \( 2x = -2 \) and \( 5x + 3 = -2 \).
In simple words: Perform the same operations on both sides of x = -1 to construct two equations.

Exam Tip: Check that \( x = -1 \) satisfies both constructed equations.

Question 99. The greatest negative integer is ____________
Answer: -1
On the number line, \( -1 \) lies furthest to the right among all negative integers.
In simple words: -1 is the largest negative whole number.

Exam Tip: Negative numbers increase in value as they get closer to 0, making -1 the greatest.

Question 100. Find the following
i. \( 3\frac{5}{8} - 2\frac{3}{4} \)
ii. 30.94 ÷ 7
Answer:
i. Convert to improper fractions with common denominator 8:
\( 3\frac{5}{8} = \frac{29}{8} \)
\( 2\frac{3}{4} = \frac{11}{4} = \frac{22}{8} \)
\( \frac{29}{8} - \frac{22}{8} = \frac{7}{8} \).
ii. Divide 30.94 by 7:
\( \frac{30.94}{7} = 4.42 \).
In simple words: In (i), subtract 22/8 from 29/8 to get 7/8. In (ii), divide to get 4.42.

Exam Tip: Change 11/4 into 22/8 to make denominators matching before subtracting.

Question 101. \( \frac{5}{6} \) of 18 = _______________
Answer: 15
\( \frac{5}{6} \times 18 = 5 \times 3 = 15 \).
In simple words: Divide 18 by 6 to get 3, then multiply by 5 to get 15.

Exam Tip: "Of" always translates to multiplication in fraction operations.

Question 102. Find the product using suitable properties
[-48] X 26 + [-48] X [-36]
Answer: Factor out -48 using the distributive property:
\( = (-48) \times [26 + (-36)] \)
\( = (-48) \times [-10] = 480 \).
In simple words: Pull out -48. Add 26 and -36 to get -10. Multiplying -48 by -10 gives 480.

Exam Tip: Two negative numbers multiplied together always give a positive product.

Question 103. 5 cm = ______________ km
Answer: 0.00005
\( 5\text{ cm} = \frac{5}{100000}\text{ km} = 0.00005\text{ km} \).
In simple words: Divide 5 by 100,000 to get 0.00005 km.

Exam Tip: 1 km = 1000 m = 100,000 cm.

Question 104. Find the area of a rectangle whose length is 11.5 m and breadth is 3.3 m.
Answer: Length = 11.5 m, Breadth = 3.3 m.
\( \text{Area} = 11.5 \times 3.3 = 37.95\text{ m}^2 \).
In simple words: Multiply 11.5 by 3.3 to get an area of 37.95 square metres.

Exam Tip: The product has 2 decimal places because each factor has 1 decimal place.

Question 105. The difference between 28 kg and 42.6 kg is _______________.
Answer: 14.6 kg
\( 42.6 - 28.0 = 14.6\text{ kg} \).
In simple words: Subtract 28 from 42.6 to find a difference of 14.6 kg.

Exam Tip: Add a decimal point and zero to 28 (28.0) to align columns correctly.

Question 106. Sale of English and Hindi books in the years 1995,1996, 1997 and 1998 are given below.

i. Draw a double bar graph choosing and appropriate scale.
ii. In which year was the difference in the sale of the two language books least?
iii. In which year the difference in the sale of two language books more?
Answer:
i. Draw the double bar graph using a vertical scale of 1 unit = 100 books.ii. Differences in book sales:
- 1995: \( 500 - 350 = 150 \)
- 1996: \( 530 - 400 = 130 \)
- 1997: \( 600 - 450 = 150 \)
- 1998: \( 650 - 620 = 30 \).
The difference was least in the year 1998 (30 books).
iii. The difference was greatest in the years 1995 and 1997 (150 books each).
In simple words: The sales gap between Hindi and English books was smallest in 1998 and largest in 1995 and 1997.

Exam Tip: Calculate the difference for each year explicitly to support your conclusions.

Question 107. \( \frac{1}{2} \) of 32 is ___________
Answer: 16
\( \frac{1}{2} \times 32 = 16 \).
In simple words: Half of 32 is 16.

Exam Tip: Half of any even number is obtained by dividing it by 2.

Question 108. 61.29 ÷ 1000
Answer: 0.06129
Move the decimal point 3 places to the left:
\( 61.29 \div 1000 = 0.06129 \).
In simple words: Shift the dot three places left to get 0.06129.

Exam Tip: Place a leading zero before 6 when moving the decimal 3 spots left.

Question 109. Write a pair of negative integers whose difference is 10.
Answer: A pair of negative integers is \( -5 \) and \( -15 \).
Difference = \( -5 - (-15) = -5 + 15 = 10 \).
In simple words: -5 minus -15 equals -5 + 15, which gives positive 10.

Exam Tip: For the difference to be positive 10, the first negative number must be greater than the second.

Question 110. In a class test containing 10 questions, 5 marks are awarded for every correct answer [-2] marks are awarded for every incorrect answer and 0 for questions not attempted.
i. Lena gets four correct and six incorrect answers. What is her score?
ii. Alan gets five incorrect answers and five incorrect answers. What is his score?
Answer:
i. Marks for 4 correct answers = \( 4 \times 5 = 20 \).
Marks for 6 incorrect answers = \( 6 \times (-2) = -12 \).
Lena's total score = \( 20 + (-12) = 8 \).
ii. Interpreting the statement as 5 correct and 5 incorrect answers:
Marks for 5 correct answers = \( 5 \times 5 = 25 \).
Marks for 5 incorrect answers = \( 5 \times (-2) = -10 \).
Alan's score = \( 25 + (-10) = 15 \).
(Note: If all 10 were attempted incorrectly, the score would be \( 10 \times (-2) = -20 \)).
In simple words: Lena scored 8 marks. Alan with 5 correct and 5 wrong answers scored 15 marks.

Exam Tip: Multiply correct answers by +5 and incorrect answers by -2, then add both totals together.