CBSE Class 7 Mathematics Another Peek beyond the Point Assignment Set 01

Short Answer Questions

Question 1. Write each of the following fractions as a sum of tenths, hundredths and thousandths, and then as a decimal:
(i) \( \frac{324}{1000} \)
(ii) \( \frac{47}{100} \)
Answer:
Split the fraction according to the digits of the numerator.
(i) \( \frac{324}{1000} \) = \( \frac{300}{1000} \) + \( \frac{20}{1000} \) + \( \frac{4}{1000} \) = \( \frac{3}{10} \) + \( \frac{2}{100} \) + \( \frac{4}{1000} \) = 0.3 + 0.02 + 0.004 = 0.324.
(ii) \( \frac{47}{100} \) = \( \frac{40}{100} \) + \( \frac{7}{100} \) = \( \frac{4}{10} \) + \( \frac{7}{100} \) = 0.4 + 0.07 = 0.47.
In simple words: Each digit of the numerator goes into its own place after the point.

Teacher's Note:
1. Simplify 300/1000 to 3/10 and so on.
2. The number of decimal digits matches the zeroes in the denominator.

Question 2. Find the following products and write each answer as a decimal:
(i) 8 × 3 tenths
(ii) 6 × 7 hundredths
Answer:
Multiply the counting numbers and keep the unit the same.
(i) 8 × 3 tenths = 24 tenths = \( \frac{24}{10} \) = 2.4.
(ii) 6 × 7 hundredths = 42 hundredths = \( \frac{42}{100} \) = 0.42.
In simple words: Multiply the numbers and keep the tenths or hundredths as they are.

Teacher's Note:
1. Like 8 × 3 apples = 24 apples.
2. Then change the unit into a decimal.

Question 3. Multiply 3.46 by 10, by 100 and by 1000. Write the rule you used in one sentence.
Answer:
3.46 × 10 = 34.6, 3.46 × 100 = 346, 3.46 × 1000 = 3460.
Rule: to multiply a decimal by 1 followed by zeroes, move the decimal point to the right as many places as there are zeroes, adding zeroes at the end if needed.
In simple words: More zeroes in the multiplier, more places the point moves right.

Teacher's Note:
1. Notice the extra zero needed in 3460.
2. The digits 3, 4, 6 stay in order.

Question 4. Divide 52.7 by 10, by 100 and by 1000. What happens to the decimal point each time?
Answer:
52.7 ÷ 10 = 5.27, 52.7 ÷ 100 = 0.527, 52.7 ÷ 1000 = 0.0527.
Each time the point moves one more place to the left — as many places as there are zeroes in the divisor. Zeroes are written in front when there are not enough digits.
In simple words: Dividing by 10, 100, 1000 moves the point 1, 2, 3 places left.

Teacher's Note:
1. Note the zero placed in 0.0527.
2. Dividing makes the number smaller.

Question 5. It is given that 43 × 25 = 1075. Using this, write the value of 4.3 × 2.5 without multiplying again, and state the rule you used to place the decimal point.
Answer:
Without the points, 43 × 25 = 1075.
4.3 has 1 decimal digit and 2.5 has 1, so the product needs 1 + 1 = 2 decimal digits.
Putting the point two places from the right in 1075: 4.3 × 2.5 = 10.75.
Rule: decimal digits in the product = total decimal digits in the two numbers.
In simple words: Use the same digits and put the point after counting decimal places.

Teacher's Note:
1. Estimate 4 × 2.5 = 10 to check.
2. Count places from the right end.

Question 6. A shopkeeper sells one gel pen for ₹12.75. Find the amount to be paid for 4 such pens. Show the multiplication step and write the answer in rupees.
Answer:
Cost of 4 pens = 4 × 12.75.
4 × 1275 = 5100; 12.75 has 2 decimal digits, so the product is 51.00.
So the amount to be paid is ₹51.
In simple words: Four pens at ₹12.75 each cost ₹51.

Teacher's Note:
1. Multiply as whole numbers, then place the point.
2. ₹51.00 = ₹51.

Question 7. Express the following fractions as decimals by first making the denominator 10, 100 or 1000:
(i) \( \frac{3}{5} \)
(ii) \( \frac{7}{8} \)
Answer:
Multiply the numerator and the denominator by the same number.
(i) \( \frac{3}{5} \) = \( \frac{3 × 2}{5 × 2} \) = \( \frac{6}{10} \) = 0.6.
(ii) \( \frac{7}{8} \) = \( \frac{7 × 125}{8 × 125} \) = \( \frac{875}{1000} \) = 0.875.
In simple words: Turn the denominator into 10 or 1000, then write the decimal.

Teacher's Note:
1. 5 × 2 = 10 and 8 × 125 = 1000.
2. Equivalent fractions keep the value the same.

Question 8. A ribbon of length 7.5 m is cut into 5 pieces of equal length. Find the length of each piece and check your answer by multiplication.
Answer:
Length of each piece = 7.5 ÷ 5 = \( \frac{7.5}{5} \) = \( \frac{75}{10 × 5} \) = \( \frac{15}{10} \) = 1.5.
So each piece is 1.5 m long.
Check: 5 × 1.5 = 7.5 m, the whole ribbon. ✓
In simple words: 7.5 m shared into 5 equal parts gives 1.5 m each.

Teacher's Note:
1. Equal pieces means divide.
2. Multiply back to check.

Question 9. Find 27.6 × 4.5 by first removing the decimal points and multiplying as counting numbers. Explain clearly how you decided where to put the decimal point in the product.
Answer:
Without the points: 276 × 45 = 12420.
27.6 has 1 decimal digit and 4.5 has 1, so the product needs 1 + 1 = 2 decimal digits (tenths × tenths = hundredths, since 10 × 10 = 100).
Counting two places from the right in 12420 gives 124.20.
So 27.6 × 4.5 = 124.2.
In simple words: Multiply 276 × 45, then put the point two places from the right.

Teacher's Note:
1. Place the point first, then drop the end zero.
2. Estimate: 28 × 4.5 ≈ 126.

Question 10. Without actually multiplying, decide which of these products are less than 1, and give the reason in each case:
(i) 0.6 × 0.5
(ii) 6 × 0.5
(iii) 0.06 × 0.5
Answer:
(i) 0.6 × 0.5 is less than 1: both numbers are between 0 and 1, so the product is smaller than both of them.
(ii) 6 × 0.5 is not less than 1: multiplying by 0.5 means taking half of 6, which is 3.
(iii) 0.06 × 0.5 is less than 1: both numbers are again between 0 and 1, so the product is even smaller.
In simple words: Two numbers below 1 always give a product below 1.

Teacher's Note:
1. Compare each factor with 1.
2. Half of 6 is clearly more than 1.

Question 11. Given that 324 × 18 = 5832, find:
(i) 3.24 × 1.8
(ii) 0.324 × 0.018
Answer:
The digits of the product are 5832; only the point changes.
(i) 3.24 has 2 decimal digits and 1.8 has 1, total 3: 3.24 × 1.8 = 5.832.
(ii) 0.324 has 3 and 0.018 has 3, total 6: 0.324 × 0.018 = 0.005832.
In simple words: Count the decimal places and put the point in 5832.

Teacher's Note:
1. Add zeroes in front when there are not enough digits.
2. Estimate (i): 3 × 2 = 6.

Question 12. Find 4.56 ÷ 1.2 by first making the divisor a counting number. Write the two steps you used and verify the quotient by multiplying.
Answer:
Step 1: Multiply the dividend and the divisor by 10: \( \frac{4.56}{1.2} \) = \( \frac{4.56 × 10}{1.2 × 10} \) = \( \frac{45.6}{12} \).
Step 2: Long division: 45.6 ÷ 12 = 3.8.
Check: 1.2 × 3.8 = 4.56. ✓
In simple words: Make the divisor 12, then 45.6 ÷ 12 = 3.8.

Teacher's Note:
1. Multiply both numbers by the same power of 10.
2. Always verify by multiplying back.

Question 13. Apples are sold at ₹84.50 per kilogram. Find the cost of 2.4 kg of apples. Also state how many digits the product must have after the decimal point and why.
Answer:
Cost = 84.50 × 2.4.
Without the points: 8450 × 24 = 202800.
84.50 has 2 decimal digits and 2.4 has 1, so the product must have 2 + 1 = 3 decimal digits: 202.800.
So 2.4 kg of apples cost ₹202.80.
In simple words: Multiply rate by weight and place the point after 3 digits from the right.

Teacher's Note:
1. Keep the zero of 84.50 while counting places.
2. Money is written with two decimal digits.

Question 14. A trader has 16.8 kg of rice. He packs it equally into 12 bags.
(i) Find the mass of rice in one bag.
(ii) Find the mass of rice in 5 such bags.
Answer:
(i) One bag = 16.8 ÷ 12 = \( \frac{168}{10 × 12} \) = \( \frac{14}{10} \) = 1.4, so one bag holds 1.4 kg.
(ii) 5 bags = 5 × 1.4 = 7.0 = 7 kg.
Check: 12 × 1.4 = 16.8 kg, the whole rice. ✓
In simple words: Share 16.8 kg into 12 bags, then multiply one bag by 5.

Teacher's Note:
1. Divide first, then multiply.
2. Check against the total.

Question 15. A car covers 189 km using 12.5 litres of petrol. Find the distance covered by the car per litre of petrol. Show how you made the divisor a counting number.
Answer:
Distance per litre = 189 ÷ 12.5.
Multiply both by 10: \( \frac{189}{12.5} \) = \( \frac{189 × 10}{12.5 × 10} \) = \( \frac{1890}{125} \).
Long division: 1890 ÷ 125 = 15.12.
So the car runs 15.12 km per litre. Check: 12.5 × 15.12 = 189 km. ✓
In simple words: 189 km ÷ 12.5 litres = 15.12 km per litre.

Teacher's Note:
1. Make the divisor whole by × 10.
2. Distance per litre = distance ÷ litres.

Question 16. Arrange these four numbers in increasing order and explain your order without doing the full working: 36.4 × 0.8, 36.4, 36.4 ÷ 0.8, 36.4 × 1.5.
Answer:
Increasing order: 36.4 × 0.8, 36.4, 36.4 ÷ 0.8, 36.4 × 1.5.
Multiplying by 0.8 (less than 1) makes the number smaller: 36.4 × 0.8 = 29.12.
Dividing by 0.8 makes the number larger: 36.4 ÷ 0.8 = 45.5.
Multiplying by 1.5 makes it one and a half times as big: 36.4 × 1.5 = 54.6.
So 29.12 < 36.4 < 45.5 < 54.6.
In simple words: × 0.8 shrinks, ÷ 0.8 grows a little, × 1.5 grows more.

Teacher's Note:
1. Compare multipliers and divisors with 1.
2. Values confirm the order.

Question 17. Divide 10 by 3 using long division up to three decimal places. Write the remainder obtained at each step and explain why this division can never come to an end.
Answer:
10 ones ÷ 3 = 3 ones, remainder 1 one.
1 one = 10 tenths; 10 tenths ÷ 3 = 3 tenths, remainder 1 tenth.
1 tenth = 10 hundredths; ÷ 3 gives 3 hundredths, remainder 1 hundredth; the same happens at thousandths (remainder 1).
So 10 ÷ 3 = 3.333 …. The remainder is always 1, so the same step repeats for ever and the division never ends.
In simple words: The remainder 1 keeps coming back, so the 3s never stop.

Teacher's Note:
1. Write each remainder clearly.
2. A repeating remainder gives repeating digits.

Question 18. Fill in the missing numbers and justify each answer:
(i) 25 ÷ ____ = 0.025
(ii) 25 ÷ ____ = 250
Answer:
(i) The missing number is 1000. 25 becomes 0.025, so the point moves 3 places left, which is what dividing by 1000 does: 25 ÷ 1000 = 0.025.
(ii) The missing number is 0.1. 25 ÷ 0.1 = \( \frac{25 × 10}{0.1 × 10} \) = \( \frac{250}{1} \) = 250. Dividing by 0.1 asks how many tenths are in 25, so the answer is ten times 25.
In simple words: Smaller answer: divide by 1000; ten times bigger answer: divide by 0.1.

Teacher's Note:
1. Look at how the point has moved.
2. Dividing by 0.1 works like × 10.

Question 19. Given that 912 ÷ 24 = 38, write the value of each of the following and explain why the digits stay the same while the decimal point shifts:
(i) 91.2 ÷ 2.4
(ii) 9.12 ÷ 24
(iii) 912 ÷ 0.24
Answer:
(i) 91.2 ÷ 2.4 = \( \frac{912}{24} \) = 38.
(ii) 9.12 ÷ 24 = 0.38.
(iii) 912 ÷ 0.24 = \( \frac{91200}{24} \) = 3800.
In each case the divisor is made a counting number by multiplying both numbers by the same power of 10, which does not change the quotient. So the digits 38 always appear; only the number of tens or tenths changes, which shifts the point.
In simple words: Same digits 38, but the point moves with the powers of 10.

Teacher's Note:
1. Compare each dividend with 912 after scaling.
2. Estimate to check the size.

Question 20. The Earth takes 365.2422 days to go once around the Sun, but a calendar year is taken as 365 days. Find the total number of extra days left over after 50 such calendar years, and explain in one or two lines why leap years are needed.
Answer:
Extra time in one year = 365.2422 − 365 = 0.2422 day.
Extra time in 50 years = 0.2422 × 50. 2422 × 50 = 121100, with 4 decimal digits: 12.1100 = 12.11 days.
So the calendar falls about 12.11 days behind in 50 years. Unless a leap day is added every fourth year, the months would slowly slip away from the seasons.
In simple words: About a quarter day is lost each year, so an extra day every 4 years keeps the calendar in step.

Teacher's Note:
1. Find the extra part first.
2. 0.2422 is close to a quarter, hence 'every 4 years'.