CBSE Class 7 Mathematics Practice Worksheet Set 04

Question 1. (a) Write a pair of unequal negative integers whose sum is –10.
(b) Write one positive and one negative integer whose sum is –10
(c) Write a pair of positive integers whose difference is -10
Answer:
(a) A pair of unequal negative integers is \( -3 \) and \( -7 \).
Check: \( (-3) + (-7) = -10 \), and \( -3 \neq -7 \).
(b) One positive integer and one negative integer are \( 5 \) and \( -15 \).
Check: \( 5 + (-15) = -10 \).
(c) A pair of positive integers is \( 5 \) and \( 15 \).
Check: \( 5 - 15 = -10 \).
In simple words: Two negative numbers like -3 and -7 add up to -10. A positive 5 and negative -15 add to -10. Subtracting 15 from 5 gives -10.

Exam Tip: For part (c), make sure the smaller positive number comes first so that the difference turns out negative.

Question 2. Evaluate the following:
(a) – 4 + (16 – 7)
(b) - 322 - ( - 400)
(c) (-1)X2X(-3)x4X(-5)
(d) [(- 36) ÷ 12] ÷ (- 3)
Answer:
(a) Solve the brackets first:
\( 16 - 7 = 9 \)
Now add:
\( -4 + 9 = 5 \).
(b) Subtracting a negative number is the same as addition:
\( -322 - (-400) = -322 + 400 = 78 \).
(c) Count the negative factors. There are 3 negative numbers, so the final product is negative:
\( (-1) \times 2 \times (-3) \times 4 \times (-5) \)
\( = - (1 \times 2 \times 3 \times 4 \times 5) \)
\( = - (6 \times 20) = -120 \).
(d) Evaluate the square brackets first:
\( (-36) \div 12 = -3 \)
Now divide by \( -3 \):
\( (-3) \div (-3) = 1 \).
In simple words: Work out the numbers inside brackets first. Remember that two minus signs together turn into a plus.

Exam Tip: When multiplying several integers, an odd number of minus signs gives a negative result, while an even number gives a positive result.

Question 3. Evaluate the following using Distributive property of integers:
(a) 9999 X 999
(b) 35 X(- 45) +(- 45) X65
Answer:
(a) Rewrite 999 as \( (1000 - 1) \):
\( 9999 \times 999 = 9999 \times (1000 - 1) \)
Apply the distributive property \( a \times (b - c) = (a \times b) - (a \times c) \):
\( = (9999 \times 1000) - (9999 \times 1) \)
\( = 9999000 - 9999 = 9989001 \).
(b) Take out the common factor \( -45 \) using \( (a \times b) + (a \times c) = a \times (b + c) \):
\( 35 \times (-45) + (-45) \times 65 = (-45) \times (35 + 65) \)
\( = (-45) \times 100 = -4500 \).
In simple words: In (a), change 999 to 1000 - 1 to multiply easily. In (b), pull out the shared -45 and add 35 and 65 to get 100.

Exam Tip: Do not multiply directly when the question specifies using the distributive property, or you will lose method marks.

Question 4. \( 4\frac{1}{8} \) kg of flour s used to make 11 cakes.how much flour will be needed to make 8 cakes?
Answer: Change the mixed fraction to an improper fraction:
Total flour used for 11 cakes = \( 4\frac{1}{8}\text{ kg} = \frac{33}{8}\text{ kg} \).
Find the flour needed for 1 cake using the unitary method:
\( \text{Flour for 1 cake} = \frac{33}{8} \div 11 = \frac{33}{8} \times \frac{1}{11} = \frac{3}{8}\text{ kg} \).
Now calculate the flour needed for 8 cakes:
\( \text{Flour for 8 cakes} = 8 \times \frac{3}{8} = 3\text{ kg} \).
Hence, 3 kg of flour is needed to make 8 cakes.
In simple words: Find the flour for 1 cake first, which is 3/8 kg. Then multiply by 8 to get 3 kg.

Exam Tip: Convert mixed fractions into improper fractions before performing division or multiplication.

Question 5. Kanta finished her homework in \( \frac{7}{12} \) hour and Rakhi finished the same homework in \( \frac{5}{8} \) hour. Who was faster and by how much?
Answer: Time taken by Kanta = \( \frac{7}{12} \) hour.
Time taken by Rakhi = \( \frac{5}{8} \) hour.
Find a common denominator by taking the LCM of 12 and 8, which is 24:
\( \text{Kanta's time} = \frac{7 \times 2}{12 \times 2} = \frac{14}{24}\text{ hour} \).
\( \text{Rakhi's time} = \frac{5 \times 3}{8 \times 3} = \frac{15}{24}\text{ hour} \).
Comparing the two fractions:
\( \frac{14}{24} < \frac{15}{24} \).
Since Kanta took less time, Kanta was faster.
Find the difference in time:
\( \frac{15}{24} - \frac{14}{24} = \frac{1}{24}\text{ hour} \).
Thus, Kanta was faster by \( \frac{1}{24} \) hour.
In simple words: Kanta took 14/24 of an hour and Rakhi took 15/24. Kanta finished earlier, so she was faster by 1/24 of an hour.

Exam Tip: Remember that in a race or task, the person who takes less time is the faster one.

Question 6. Solve the following:
(a) 441.709 ÷ 18.14
(b) 9.487+ 3.58
(c) 12.23 X6.3
(d) (i) 0.76 ÷ 1000 (ii) 2.4 X100 (iii) 7.36 ÷ 100 (iv) 0.004 X 10000
Answer:
(a) Move the decimal point two places to the right in both numbers to make the divisor a whole number:
\( \frac{441.709}{18.14} = \frac{44170.9}{1814} = 24.35 \).
(b) Line up the decimal points and add:
\( 9.487 + 3.580 = 13.067 \).
(c) Multiply without decimals: \( 1223 \times 63 = 77049 \).
Count total decimal places (\( 2 + 1 = 3 \)) and place the point:
\( 12.23 \times 6.3 = 77.049 \).
(d)
(i) Shift the decimal point 3 places to the left:
\( 0.76 \div 1000 = 0.00076 \).
(ii) Shift the decimal point 2 places to the right:
\( 2.4 \times 100 = 240 \).
(iii) Shift the decimal point 2 places to the left:
\( 7.36 \div 100 = 0.0736 \).
(iv) Shift the decimal point 4 places to the right:
\( 0.004 \times 10000 = 40 \).
In simple words: When dividing by 10, 100, or 1000, move the dot left. When multiplying, move the dot right.

Exam Tip: When adding decimals like in part (b), add a trailing zero to 3.58 to make it 3.580 so all columns align neatly.

Question 7. The sides of a triangle are \( 2\frac{1}{2} \) cm, \( 3\frac{3}{4} \) cm and \( 3\frac{1}{5} \) cm. Convert the lengths of the sides in decimal and then find the perimeter. Wrie your answer in metre also.
Answer: Convert each side length into decimal form:
First side = \( 2\frac{1}{2}\text{ cm} = 2.5\text{ cm} \).
Second side = \( 3\frac{3}{4}\text{ cm} = 3.75\text{ cm} \).
Third side = \( 3\frac{1}{5}\text{ cm} = 3.2\text{ cm} \).
Calculate the perimeter by adding all three sides:
\( \text{Perimeter in cm} = 2.5 + 3.75 + 3.2 = 9.45\text{ cm} \).
Now convert the perimeter into metres (\( 1\text{ m} = 100\text{ cm} \)):
\( \text{Perimeter in metres} = \frac{9.45}{100} = 0.0945\text{ m} \).
Therefore, the perimeter is 9.45 cm or 0.0945 m.
In simple words: Turn the fractions into 2.5, 3.75, and 3.2. Add them up to get 9.45 cm, then divide by 100 to get 0.0945 m.

Exam Tip: Be sure to write the final answer in both units (cm and m) as requested by the question.

Question 8. How many seconds are there in \( \frac{2}{3} \) of an hour?
Answer: First, find the total seconds in 1 hour:
\( 1\text{ hour} = 60\text{ minutes} = 60 \times 60\text{ seconds} = 3600\text{ seconds} \).
Now calculate \( \frac{2}{3} \) of 3600 seconds:
\( \frac{2}{3} \times 3600 = 2 \times 1200 = 2400\text{ seconds} \).
Therefore, there are 2400 seconds in \( \frac{2}{3} \) of an hour.
In simple words: One hour has 3600 seconds. Two-thirds of 3600 gives 2400 seconds.

Exam Tip: Convert directly using \( 1\text{ hour} = 3600\text{ seconds} \) to keep the calculation quick and avoid multi-step errors.