CBSE Class 7 Mathematics Practice Worksheet Set 07

Fill in the blanks

Question 1. (-3) X 1 = 1 X ( )
Answer: -3
According to the commutative property of multiplication, changing the order of two numbers does not alter their product: \( a \times b = b \times a \). Therefore, \( (-3) \times 1 = 1 \times (-3) \).
In simple words: You can multiply numbers in any order. The answer stays the same, so the blank is -3.

Exam Tip: Always remember the rule \( a \times b = b \times a \) to fill in missing numbers quickly.

Question 2. Predecessor of zero is ______
Answer: -1
The predecessor of an integer is found by subtracting 1 from it. Here, \( 0 - 1 = -1 \).
In simple words: The number that comes right before 0 on the number line is -1.

Exam Tip: Do not confuse predecessor (subtract 1) with successor (add 1).

Question 3. 40.07 X 100 = ______
Answer: 4007
When multiplying a decimal number by 100, shift the decimal point two places to the right: \( 40.07 \times 100 = 4007 \).
In simple words: Move the dot two steps to the right to get 4007.

Exam Tip: Count the number of zeros in the multiplier to know how many places to move the decimal point.

Solve the following

Question 1. Give the supplement of 116°
Answer: Two angles are supplementary if their sum is \( 180^\circ \).
Supplement of \( 116^\circ = 180^\circ - 116^\circ = 64^\circ \).
In simple words: Subtract 116 from 180 to find the supplement, which is 64 degrees.

Exam Tip: Supplementary angles add up to \( 180^\circ \), while complementary angles add up to \( 90^\circ \).

Question 2. Sum of complemetary angles is always equal to ______
Answer: 90°
By definition, two angles are called complementary when the sum of their measures equals \( 90^\circ \).
In simple words: Complementary angles always add up to 90 degrees.

Exam Tip: Keep the rule clear: C for Complementary is \( 90^\circ \), S for Supplementary is \( 180^\circ \).

Question 3. Change 25kg 34gm in to decimal.
Answer: We know that \( 1\text{ kg} = 1000\text{ g} \).
Therefore, \( 1\text{ g} = \frac{1}{1000}\text{ kg} = 0.001\text{ kg} \).
Now convert 34 grams into kilograms:
\( 34\text{ g} = \frac{34}{1000}\text{ kg} = 0.034\text{ kg} \).
Total weight = \( 25\text{ kg} + 0.034\text{ kg} = 25.034\text{ kg} \).
In simple words: 34 grams is 0.034 kg, so the full weight is 25.034 kg.

Exam Tip: Remember to write 34 grams as 0.034 kg with three decimal digits because 1 kg has 1000 g.

Question 4. Find an angle which is equal to its supplement.
Answer: Let the required angle be \( x \).
Its supplement is also \( x \).
Since supplementary angles sum to \( 180^\circ \):
\( x + x = 180^\circ \)
\( 2x = 180^\circ \)

\( \implies x = \frac{180^\circ}{2} = 90^\circ \).
Thus, the angle equal to its supplement is \( 90^\circ \).
In simple words: An angle of 90 degrees is equal to its own supplement.

Exam Tip: Set up the equation \( x + x = 180^\circ \) to show your complete step-by-step working.

Question 5. Divide 13.6 by 4
Answer: Divide 13.6 by 4:
\( \frac{13.6}{4} = 3.4 \).
Working:
13 divided by 4 gives 3 with a remainder of 1.
Place the decimal point in the quotient and bring down 6 to get 16.
16 divided by 4 gives 4.
So, \( 13.6 \div 4 = 3.4 \).
In simple words: 13.6 split into 4 equal parts gives 3.4.

Exam Tip: Align the decimal point in the quotient directly above the decimal point in the dividend.

Question 6. The weight of 30 packets of butter is 900.30kg. Find the weight of 1 packet.
Answer: Total weight of 30 packets = \( 900.30\text{ kg} \).
Weight of 1 packet = \( \frac{900.30}{30}\text{ kg} \).
Divide by 30:
\( \frac{900.30}{30} = \frac{90.03}{3} = 30.01\text{ kg} \).
Hence, each packet weighs \( 30.01\text{ kg} \).
In simple words: Divide the total weight by 30 to find that one packet weighs 30.01 kg.

Exam Tip: Cancel the ending zeros or simplify by dividing by 10 first to make the division easier.

Question 7. Express 7kg650gm in kg.
Answer: We know that \( 1\text{ kg} = 1000\text{ g} \).
Convert 650 g into kg:
\( 650\text{ g} = \frac{650}{1000}\text{ kg} = 0.65\text{ kg} \).
Total weight = \( 7\text{ kg} + 0.65\text{ kg} = 7.65\text{ kg} \) (or \( 7.650\text{ kg} \)).
In simple words: 650 grams is 0.65 kg, so adding 7 kg gives 7.65 kg.

Exam Tip: Writing either 7.65 kg or 7.650 kg is correct since trailing zeros after a decimal do not change the value.

Question 8. The difference in the measure of two complementary angles is 12°. Find the measure.
Answer: Let the two complementary angles be \( x \) and \( y \) where \( x > y \).
Since the angles are complementary:
\( x + y = 90^\circ \) ... (1)
Their difference is given as:
\( x - y = 12^\circ \) ... (2)
Add equation (1) and equation (2):
\( 2x = 102^\circ \)

\( \implies x = \frac{102^\circ}{2} = 51^\circ \).
Substitute \( x = 51^\circ \) into equation (1):
\( 51^\circ + y = 90^\circ \)

\( \implies y = 90^\circ - 51^\circ = 39^\circ \).
Thus, the two angles are \( 51^\circ \) and \( 39^\circ \).
In simple words: The two angles are 51 degrees and 39 degrees. They add up to 90 and differ by 12.

Exam Tip: Always verify your final angles by checking both conditions: \( 51 + 39 = 90 \) and \( 51 - 39 = 12 \).

Question 9. Two angles of a linear pair are in the ratio 3:6. Find two angles.
Answer: Angles forming a linear pair add up to \( 180^\circ \).
Let the angles be \( 3x \) and \( 6x \).
\( 3x + 6x = 180^\circ \)
\( 9x = 180^\circ \)

\( \implies x = \frac{180^\circ}{9} = 20^\circ \).
Now calculate each angle:
First angle = \( 3x = 3 \times 20^\circ = 60^\circ \).
Second angle = \( 6x = 6 \times 20^\circ = 120^\circ \).
Hence, the two angles are \( 60^\circ \) and \( 120^\circ \).
In simple words: The angles add up to 180 degrees. The two angles are 60 degrees and 120 degrees.

Exam Tip: Notice that the ratio 3:6 simplifies to 1:2, which makes solving even faster.

Question 10. Solve (a) 7 – (-4)
(b) 140 – (-50)
Answer:
(a) Subtracting a negative number is equivalent to adding its opposite:
\( 7 - (-4) = 7 + 4 = 11 \).
(b) Similarly, simplify the double negative:
\( 140 - (-50) = 140 + 50 = 190 \).
In simple words: Two negative signs together become a plus, giving 11 for (a) and 190 for (b).

Exam Tip: Remember the sign rule: \( -(-a) = +a \).

Question 11. (a) Find the cost of 25.4m of cloth, if cost of 1m cloth is Rs 50.50
(b) Divide 14.49 by 7
Answer:
(a) Cost of 1 m of cloth = Rs 50.50.
Cost of 25.4 m of cloth = \( 25.4 \times 50.50 \).
Multiply the numbers:
\( 25.4 \times 50.50 = 1282.70 \).
So, the cost of the cloth is Rs 1282.70.
(b) Divide 14.49 by 7:
\( \frac{14.49}{7} = 2.07 \).
In simple words: For (a), multiply to get Rs 1282.70. For (b), divide to get 2.07.

Exam Tip: In division of decimals, remember to write 0 in the tenths place when 4 is smaller than 7 in 14.49.

Question 12. Represent 0.75 as fraction.
Answer: Write the decimal as a fraction with denominator 100:
\( 0.75 = \frac{75}{100} \).
Divide both numerator and denominator by their greatest common factor, 25:
\( \frac{75 \div 25}{100 \div 25} = \frac{3}{4} \).
Thus, 0.75 as a fraction in simplest form is \( \frac{3}{4} \).
In simple words: 0.75 is 75 over 100, which reduces to 3/4.

Exam Tip: Always reduce the fraction to its lowest terms to gain full marks.

Question 13. Find the sum of (-4) + 3 +(-5)
Answer: Group the negative integers first:
\( (-4) + (-5) = -9 \).
Now add the positive integer:
\( -9 + 3 = -6 \).
Therefore, the sum is \( -6 \).
In simple words: -4 and -5 make -9. Adding 3 brings it up to -6.

Exam Tip: Group integers with the same sign together before combining opposites.

Question 14. Write 2 integers just succeeding -7
Answer: The successor of an integer is obtained by adding 1.
First integer succeeding \( -7 = -7 + 1 = -6 \).
Second integer succeeding \( -6 = -6 + 1 = -5 \).
Thus, the two integers just succeeding \( -7 \) are \( -6 \) and \( -5 \).
In simple words: The next two numbers after -7 when counting up are -6 and -5.

Exam Tip: On a number line, succeeding integers lie to the right of the given number.

Question 15. Sum of two integers is -30.If one of them is 15,Find the other integer.
Answer: Let the other integer be \( x \).
According to the problem:
\( x + 15 = -30 \)

\( \implies x = -30 - 15 \)

\( \implies x = -45 \).
Hence, the other integer is \( -45 \).
In simple words: Subtract 15 from -30 to find that the other number is -45.

Exam Tip: When moving a positive number across the equal sign, it becomes negative: \( -30 - 15 = -45 \).

Question 16. Simplify using distributive property -37 × 98
Answer: Express 98 as \( (100 - 2) \):
\( -37 \times 98 = -37 \times (100 - 2) \).
Apply the distributive property \( a \times (b - c) = (a \times b) - (a \times c) \):
\( = (-37 \times 100) - (-37 \times 2) \)
\( = -3700 - (-74) \)
\( = -3700 + 74 \)
\( = -3626 \).
Thus, the value is \( -3626 \).
In simple words: Break 98 into 100 minus 2, multiply each by -37, and combine to get -3626.

Exam Tip: Writing 98 as \( (100 - 2) \) makes large multiplications quick and easy using the distributive law.

Question 17. A cooling machine requires that room temperature to be lowered from 50°c at the rate of 5°c every hour. What will be the room temperature after 8hours after process begins?
Answer: Initial room temperature = \( 50^\circ\text{C} \).
Rate of cooling = \( 5^\circ\text{C} \) per hour.
Total decrease in temperature in 8 hours = \( 8 \times 5^\circ\text{C} = 40^\circ\text{C} \).
Room temperature after 8 hours = \( 50^\circ\text{C} - 40^\circ\text{C} = 10^\circ\text{C} \).
Therefore, the room temperature will be \( 10^\circ\text{C} \).
In simple words: The temperature drops by 40 degrees in 8 hours, leaving the room at 10 degrees.

Exam Tip: Subtract the total temperature decrease from the starting temperature.

Question 18. Evaluate { (-16) + 12} ÷ {(-4) + 2}
Answer: First evaluate the expressions inside each bracket:
First bracket: \( (-16) + 12 = -4 \).
Second bracket: \( (-4) + 2 = -2 \).
Now perform the division:
\( (-4) \div (-2) = 2 \).
Therefore, the result is 2.
In simple words: The first bracket gives -4 and the second gives -2. Dividing -4 by -2 gives 2.

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

Question 19. Subtract 3 1/2 from 5 1/4
Answer: Convert both mixed numbers into improper fractions:
\( 5\frac{1}{4} = \frac{5 \times 4 + 1}{4} = \frac{21}{4} \).
\( 3\frac{1}{2} = \frac{3 \times 2 + 1}{2} = \frac{7}{2} \).
Make the denominators common by multiplying \( \frac{7}{2} \) by \( \frac{2}{2} \):
\( \frac{7}{2} = \frac{14}{4} \).
Now subtract:
\( \frac{21}{4} - \frac{14}{4} = \frac{21 - 14}{4} = \frac{7}{4} = 1\frac{3}{4} \).
Hence, the answer is \( 1\frac{3}{4} \) (or \( \frac{7}{4} \)).
In simple words: Turn the mixed fractions into 21/4 and 14/4, then subtract to get 7/4 or 1 3/4.

Exam Tip: Always convert mixed numbers to improper fractions with a common denominator before subtracting.

Question 20. A bus covers a distance of 60 3/4 km in an hour. Find the distance covered by bus in 2 2/3 hours.
Answer: Distance covered in 1 hour = \( 60\frac{3}{4}\text{ km} = \frac{243}{4}\text{ km} \).
Time taken = \( 2\frac{2}{3}\text{ hours} = \frac{8}{3}\text{ hours} \).
Distance covered = \( \text{Speed} \times \text{Time} \):
\( \text{Distance} = \frac{243}{4} \times \frac{8}{3} \).
Simplify the multiplication:
Divide 243 by 3: \( \frac{243}{3} = 81 \).
Divide 8 by 4: \( \frac{8}{4} = 2 \).
\( \text{Distance} = 81 \times 2 = 162\text{ km} \).
Therefore, the bus covers a distance of 162 km.
In simple words: Multiply 243/4 by 8/3 to find the total distance of 162 km.

Exam Tip: Cross-cancel common factors between numerators and denominators to simplify fractions easily.

Question 21. Product of two fractions is 3/4. If one of the fraction is 7 1/2, then find the other.
Answer: Let the required fraction be \( x \).
One fraction = \( 7\frac{1}{2} = \frac{15}{2} \).
Given that their product is \( \frac{3}{4} \):
\( x \times \frac{15}{2} = \frac{3}{4} \)

\( \implies x = \frac{3}{4} \div \frac{15}{2} = \frac{3}{4} \times \frac{2}{15} \).
Simplify by cancelling:
\( x = \frac{3 \times 2}{4 \times 15} = \frac{6}{60} = \frac{1}{10} \).
Hence, the other fraction is \( \frac{1}{10} \).
In simple words: Divide 3/4 by 15/2, which flips to 3/4 times 2/15, giving 1/10.

Exam Tip: To divide by a fraction, multiply by its reciprocal.

Question 22. What are like fractions? Give example.
Answer: Fractions that have the same denominator are known as like fractions.
Example: \( \frac{2}{9}, \frac{5}{9}, \frac{7}{9} \) are like fractions because all of them have the common denominator 9.
In simple words: Like fractions are fractions that share the exact same bottom number, like 2/9 and 5/9.

Exam Tip: Clearly state that the denominators must be identical and provide at least two examples.

Question 23. Express 415cm in metres.
Answer: We know that \( 100\text{ cm} = 1\text{ m} \).
Therefore, \( 1\text{ cm} = \frac{1}{100}\text{ m} \).
\( 415\text{ cm} = \frac{415}{100}\text{ m} = 4.15\text{ m} \).
In simple words: Divide 415 by 100 to get 4.15 metres.

Exam Tip: When converting from a smaller unit (cm) to a larger unit (m), always divide by 100.

Question 24. How much is 37.005 more than 23.5
Answer: To find the difference, subtract 23.5 from 37.005:
\( 37.005 - 23.500 = 13.505 \).
Therefore, 37.005 is 13.505 more than 23.5.
In simple words: Take 23.5 away from 37.005 to find that it is 13.505 more.

Exam Tip: Add placeholder zeros so that both decimal numbers have the same number of decimal places before subtracting.

Question 25. Find area of a square whose side is 6.25m.
Answer: Side of the square (\( s \)) = \( 6.25\text{ m} \).
\( \text{Area of square} = \text{side} \times \text{side} \).
\( \text{Area} = 6.25 \times 6.25 = 39.0625\text{ m}^2 \).
Hence, the area of the square is \( 39.0625\text{ m}^2 \).
In simple words: Multiply 6.25 by itself to get an area of 39.0625 square metres.

Exam Tip: Write the unit as \( \text{m}^2 \) or sq m, as area is always measured in square units.

Question 26. Divide 0.00942 by 0.314
Answer: To divide, make the divisor a whole number by multiplying both numbers by 1000:
\( \frac{0.00942}{0.314} = \frac{0.00942 \times 1000}{0.314 \times 1000} = \frac{9.42}{314} \).
Now divide 9.42 by 314:
\( \frac{9.42}{314} = 0.03 \).
Therefore, \( 0.00942 \div 0.314 = 0.03 \).
In simple words: Move the decimal three places to divide 9.42 by 314, giving 0.03.

Exam Tip: Eliminate the decimal from the divisor first before doing standard long division.

Question 27. Check whether x =2 is the solution of 2x+4=15 or not.
Answer: Given equation: \( 2x + 4 = 15 \).
Substitute \( x = 2 \) into the Left Hand Side (LHS):
\( \text{LHS} = 2(2) + 4 = 4 + 4 = 8 \).
The Right Hand Side (RHS) is 15.
Since \( \text{LHS} \neq \text{RHS} \) (\( 8 \neq 15 \)), \( x = 2 \) is not the solution of the equation.
In simple words: Plugging in 2 gives 8, which is not 15, so x = 2 is not the correct solution.

Exam Tip: Clearly show that LHS does not equal RHS to justify your conclusion.

Question 28. Give first steps you will use to separate the variable and then solve the equation.
(a) 4x = 256
(b) 3t+12 = 0
(c) 2x/3 + 5/2 = 4
(d) 8x – 3 = 9 + 4x
Answer:
(a) \( 4x = 256 \)
First step: Divide both sides by 4.
Solution: \( x = \frac{256}{4} = 64 \).

(b) \( 3t + 12 = 0 \)
First step: Subtract 12 from both sides.
Solution: \( 3t = -12 \). Next, divide both sides by 3: \( t = \frac{-12}{3} = -4 \).

(c) \( \frac{2x}{3} + \frac{5}{2} = 4 \)
First step: Subtract \( \frac{5}{2} \) from both sides.
Solution: \( \frac{2x}{3} = 4 - \frac{5}{2} = \frac{8 - 5}{2} = \frac{3}{2} \).
Multiply both sides by 3: \( 2x = \frac{9}{2} \).
Divide both sides by 2: \( x = \frac{9}{4} = 2.25 \).

(d) \( 8x - 3 = 9 + 4x \)
First step: Subtract \( 4x \) from both sides.
Solution: \( 8x - 4x - 3 = 9 \implies 4x - 3 = 9 \).
Add 3 to both sides: \( 4x = 12 \).
Divide both sides by 4: \( x = 3 \).
In simple words: State the first step to isolate the letter in each equation, then solve to find the variable.

Exam Tip: Be sure to write down the name of the "first step" clearly, as the question specifically asks for it.

Question 29. Mukesh subtract thrice the number of notebooks he has from50. He finds the result to be 8.
Answer: Let the number of notebooks Mukesh has be \( x \).
Thrice the number of notebooks = \( 3x \).
According to the statement:
\( 50 - 3x = 8 \)
Rearrange the terms:
\( 50 - 8 = 3x \)
\( 42 = 3x \)

\( \implies x = \frac{42}{3} = 14 \).
Therefore, Mukesh has 14 notebooks.
In simple words: 50 minus 3 times the notebooks equals 8. Solving gives 14 notebooks.

Exam Tip: "Subtract from 50" means you must write 50 first: \( 50 - 3x = 8 \), not \( 3x - 50 = 8 \).

Question 30. The angles of a triangle are in the ratio 1:2:3, Find the angles.
Answer: Let the angles of the triangle be \( 1x \), \( 2x \), and \( 3x \).
The sum of angles in any triangle is \( 180^\circ \):
\( 1x + 2x + 3x = 180^\circ \)
\( 6x = 180^\circ \)

\( \implies x = \frac{180^\circ}{6} = 30^\circ \).
Now find the measure of each angle:
First angle = \( 1x = 1 \times 30^\circ = 30^\circ \).
Second angle = \( 2x = 2 \times 30^\circ = 60^\circ \).
Third angle = \( 3x = 3 \times 30^\circ = 90^\circ \).
Hence, the angles of the triangle are \( 30^\circ \), \( 60^\circ \), and \( 90^\circ \).
In simple words: Add the ratio parts to get 6. Divide 180 by 6 to find that the angles are 30, 60, and 90 degrees.

Exam Tip: Verify that the three calculated angles sum to \( 180^\circ \): \( 30 + 60 + 90 = 180 \).

Question 31. A man goes 7m east and then 24m to south. Find his distance from the starting point.
Answer: The path taken by the man forms a right-angled triangle where the East and South directions are perpendicular to each other.
Base (East) = \( 7\text{ m} \).
Perpendicular (South) = \( 24\text{ m} \).
The distance from the starting point is the hypotenuse (\( d \)).
By Pythagoras theorem:
\( d^2 = 7^2 + 24^2 \)
\( d^2 = 49 + 576 = 625 \)

\( \implies d = \sqrt{625} = 25\text{ m} \).
Therefore, the man is 25 m away from the starting point.
In simple words: Use the Pythagoras theorem: 7 squared plus 24 squared is 625, and the square root of 625 is 25 metres.

Exam Tip: Remember common Pythagorean triples like (7, 24, 25) to solve distance problems rapidly.

Question 32. The square of diagonal of a square is 1250 cm , find the length of the sides the square.
Answer: Let the side of the square be \( a \).
In a square, the diagonal (\( d \)) forms a right-angled triangle with two adjacent sides of length \( a \).
By Pythagoras theorem:
\( d^2 = a^2 + a^2 = 2a^2 \).
Given that the square of the diagonal is 1250:
\( 2a^2 = 1250 \)

\( \implies a^2 = \frac{1250}{2} = 625 \)

\( \implies a = \sqrt{625} = 25\text{ cm} \).
Hence, the length of each side of the square is 25 cm.
In simple words: The square of the diagonal is twice the side squared. Half of 1250 is 625, so the side is 25 cm.

Exam Tip: Use the direct formula \( \text{Diagonal}^2 = 2 \times \text{Side}^2 \) to find the side of a square.