Selina Concise Solutions for ICSE Class 6 Mathematics Chapter 15 Decimal Fractions

Important Points

Decimal Fraction : A fraction, whose denominator is 10 or a higher power of 10 e.g. 100, 1,000, 10,000 etc. is known as decimal fraction.

Number of Decimal Places: The number of digits in the decimal part of a number is the number of decimal places in it.
When the given number has only decimal part in it. It is always written 0 before it as 0.7, 0.55 are written as 0.7, 0.55.

Conversion of a Fraction into a Decimal Fraction :

• 1. When the denominator is 10, 100, 1000, 10,000 etc. : Counting from right to left of the numerator of the given fraction, mark the decimal point after as many digits as the number of zeroes in it denominator
  e.g. \( \frac{2}{10} = 0.2 \), \( \frac{24}{1000} = 0.024 \), \( \frac{221}{100} = 2.21 \)
• 2. When the denominator is not, 10, 100, 1000, 10,000 etc.
  Multiply both, the numerator and denominator of the given fraction, by a suitable number to get the denominator 10 or a power of 10 and then proceed as above,
  e.g.
  \( \frac{1}{2} = \frac{1 \times 50}{2 \times 50} = \frac{50}{100} = 0.50 = 0.5 \),
  \( \frac{2}{25} = \frac{2 \times 4}{25 \times 4} = \frac{8}{100} = 0.08 \)
• 3. Conversion of a given Decimal Fraction into a Non-Decimal Fraction : Remove the decimal point and at the same time write 1 in the denominator, as many zeroes to the right of 1 as there are digits in the decimal part e.g.,
  \( 0.42 = \frac{42}{100} \), \( 0.031 = \frac{31}{1000} \),
  \( 3.79 = \frac{379}{100} = 3 \frac{79}{100} \)
  \( 10^2 = 10 \times 10 \), \( 10^3 = 10 \times 10 \times 10 = 1,000 \),
  \( 10^5 = 10 \times 10 \times 10 \times 10 \times 10 = 1,00,000 \)

Zero or zeores written at the right of a decimal number does not change its value, e.g. 3.4 is the same as 3.40, 3.400, 3.4000 etc.

Exercise 15(A)

Question 1. Write the number of decimal places in each of the following :
(i) 7.03
(ii) 0.509
(iii) 146.2
(iv) 0.0065
(v) 8.03207
Answer:
(i) 7.03, the decimal part is .03 which contains two digits. Number 7.03 has 2 decimal places.
(ii) 0.509, the decimal part is 0.509 which contains three digits. Number 0.509 has 3 decimal places
(iii) 146.2, the decimal part is .2 which contains one digits. Number 146.2 has 1 decimal places.
(iv) 0.0065, the decimal part is .0065 which contains four digits. Number 0.0065 has 4 decimal places
(v) 8.03207, the decimal part is .03207 which contains five digits. Number 8.03207 has 5 decimal places.
In simple words: To find the number of decimal places, just count how many digits are written after the dot (decimal point).

📝 Teacher's Note: Use a highlighter on the chalkboard to separate the whole number part from the decimal part to help students visualize the "decimal part" specifically.

🎯 Exam Tip: Do not count the digits before the decimal point; only count the ones to the right of it.

Question 2. Convert the given unlike decimal fractions into like decimal fractions:
(i) 1.36, 239.8 and 47.008
(ii) 507.0752, 8.52073 and 0.808
(iii) 459.22, 7.03093 and 0.200037
Answer:
(i) 1.36 = 1.360
239.8 = 239.800
47.008 = 47.008
(ii) 507.0752 = 507.07520
8.52073 = 8.52073
0.808 = 0.80800
(iii) 459.22 = 459.220000
7.03093 = 7.030930
0.200037 = 0.200037
In simple words: "Like" decimals have the same amount of digits after the dot. You can add extra zeros at the very end of a decimal number to make them match without changing the value.

📝 Teacher's Note: Explain that adding zeros at the end is like saying "no thousandths" or "no ten-thousandths" — the actual size of the number stays exactly the same.

🎯 Exam Tip: To convert to like decimals, find the number with the most decimal places and add zeros to the others until they all match that maximum count.

Question 3. Change each of following fractions to a decimal fraction :
(i) \( \frac{7}{10} \)
(ii) \( \frac{47}{10} \)
(iii) \( \frac{343}{100} \)
(iv) \( \frac{3}{10^3} \)
(v) \( \frac{7295}{10^5} \)
(vi) \( \frac{289}{10^6} \)
(vii) 95-hundredths
Answer:
(i) \( \frac{7}{10} = 0.7 \)
(ii) \( \frac{47}{10} = 4.7 \)
(iii) \( \frac{343}{100} = 3.43 \)
(iv) \( \frac{3}{10^3} = \frac{3}{10 \times 10 \times 10} = \frac{3}{1000} = 0.003 \)
(v) \( \frac{7295}{10^5} = \frac{7295}{10 \times 10 \times 10 \times 10 \times 10} = \frac{7295}{100000} = 0.07295 \)
(vi) \( \frac{289}{10^6} = \frac{289}{10 \times 10 \times 10 \times 10 \times 10 \times 10} = \frac{289}{1000000} = 0.000289 \)
(vii) 95-hundredths = \( \frac{95}{100} = 0.95 \)
In simple words: Look at the number of zeros in the bottom number. Move the decimal point that many places to the left in the top number.

📝 Teacher's Note: Practice "jumping" the decimal point from the invisible spot at the end of the whole number numerator.

🎯 Exam Tip: If you run out of digits when moving left, fill the empty spaces with zeros before the digits you have.

Question 4. Convert into a decimal fraction :
(i) \( \frac{3}{4} \)
(ii) \( \frac{3}{40} \)
(iii) \( \frac{1}{125} \)
(iv) \( \frac{7}{25} \)
Answer:
(i) \( \frac{3}{4} = \frac{3 \times 25}{4 \times 25} = \frac{75}{100} = 0.75 \)
(ii) \( \frac{3}{40} = \frac{3 \times 25}{40 \times 25} = \frac{75}{1000} = 0.075 \)
(iii) \( \frac{1}{125} = \frac{1 \times 8}{125 \times 8} = \frac{8}{1000} = 0.008 \)
(iv) \( \frac{7}{25} = \frac{7 \times 4}{25 \times 4} = \frac{28}{100} = 0.28 \)
In simple words: Multiply the top and bottom by the same number to make the bottom 10, 100, or 1000. This makes it easy to turn it into a decimal.

📝 Teacher's Note: Teach "friendly number pairs" for decimals: \( 2 \times 5 \), \( 4 \times 25 \), and \( 8 \times 125 \).

🎯 Exam Tip: Always multiply the numerator and denominator by the same factor to keep the fraction's value unchanged.

Question 5. Change the given decimals fractions to fractions in their lowest terms :
(i) 0.05
(ii) 3.95
(iii) 4.005
(iv) 0.876
(v) 50.06
(vi) 0.01075
(vii) 4.8806
Answer:
(i) \( 0.05 = \frac{5}{100} = \frac{1}{20} \)
(ii) \( 3.95 = \frac{395}{100} = \frac{79}{20} = 3 \frac{19}{20} \)
(iii) \( 4.005 = \frac{4005}{1000} = \frac{801}{200} = 4 \frac{1}{200} \)
(iv) \( 0.876 = \frac{876}{1000} = \frac{219}{250} \)
(v) \( 50.06 = \frac{5006}{100} = \frac{2503}{50} = 50 \frac{3}{50} \)
(vi) \( 0.01075 = \frac{1075}{100000} = \frac{43}{4000} \)
(vii) \( 4.8806 = \frac{48806}{10000} = \frac{24403}{5000} = 4 \frac{4403}{5000} \)
In simple words: Write the number without the decimal on top. Put 1 followed by as many zeros as there were decimal places on the bottom. Then, divide both top and bottom by their common factors until they can't be reduced anymore.

📝 Teacher's Note: Remind students that "lowest terms" means the numerator and denominator should share no common factors other than 1.

🎯 Exam Tip: If the numerator is larger than the denominator, convert it to a mixed fraction for the final answer.

Exercise 15(B)

Question 1. Add the following :
(i) 0.243, 2.47 and 3.009
(ii) 0.0736, 0.6095 and 0.9107
(iii) 1.01, 257 and 0.200
(iv) 18, 200.35, 11.72 and 2.3
(v) 0.586, 0.0586 and 0.00586
Answer:
(i)
   0.243
+ 2.470
+ 3.009
----------
   5.722

(ii)
   0.0736
+ 0.6095
+ 0.9107
----------
   1.5938

(iii)
     1.010
+ 257.000
+   0.200
----------
  258.210

(iv)
  18.00
+ 200.35
+ 11.72
+   2.30
----------
  232.37

(v)
  0.58600
+ 0.05860
+ 0.00586
----------
  0.65046
In simple words: Line up the decimal points first. Add zeros at the end to make all numbers the same length, then add just like regular addition.

📝 Teacher's Note: The most common mistake is misaligning the place values. Using grid paper can help students keep the decimal points in a straight vertical line.

🎯 Exam Tip: Always convert whole numbers (like 18 or 257) into decimals by adding ".00" to help with alignment.

Question 2. Find the value of :
(i) 6.8 - 2.64
(ii) 2 - 1.0304
(iii) 0.1 - 0.08
(iv) 0.83 - 0.342
Answer:
(i) 6.80 - 2.64 = 4.16
(ii) 2.0000 - 1.0304 = 0.9696
(iii) 0.10 - 0.08 = 0.02
(iv) 0.830 - 0.342 = 0.488
In simple words: Just like addition, line up the decimals. Add zeros to the top number so it has as many places as the bottom number, then subtract.

📝 Teacher's Note: Emphasize "borrowing" across multiple zeros, which often confuses students (e.g., in part ii).

🎯 Exam Tip: Never subtract without adding the placeholder zeros first, or you might incorrectly subtract "empty" space.

Question 3. Subtract :
(i) 0.43 from 0.97
(ii) 2.008 from 22.1058
(iii) 0.18 from 0.6
(iv) 1.002 from 17
(v) 83 from 92.05
Answer:
(i) 0.97 - 0.43 = 0.54
(ii) 22.1058 - 2.0080 = 20.0978
(iii) 0.60 - 0.18 = 0.42
(iv) 17.000 - 1.002 = 15.998
(v) 92.05 - 83.00 = 9.05
In simple words: Remember that "Subtract A from B" means B - A. Put the second number on top.

📝 Teacher's Note: Clarify the language "Subtract X from Y" to ensure students don't reverse the numbers.

🎯 Exam Tip: Double-check the wording. The number after "from" is the larger value that goes at the top.

Question 4. Simplify :
(i) 3.5 - 2.43 + 0.075
(ii) 7.84 + 0.3 - 4.016
(iii) 2.987 - 1.25 - 0.54
(iv) 52.9 - 231.666 + 204
(v) 8.57 - 6.4432 - 1.70 + 0.683
Answer:
(i) 3.5 - 2.43 + 0.075
= 3.500 + 0.075 - 2.43
= 3.575 - 2.430 = 1.145

(ii) 7.84 + 0.3 - 4.016
= 7.840 + 0.300 - 4.016
= 8.140 - 4.016 = 4.124

(iii) 2.987 - 1.25 - 0.54
= 2.987 - 1.790 = 1.197

(iv) 52.9 - 231.666 + 204
= 52.9 + 204.0 - 231.666
= 256.900 - 231.666 = 25.234

(v) 8.57 - 6.4432 - 1.70 + 0.683
= (8.5700 + 0.6830) - (6.4432 + 1.7000)
= 9.2530 - 8.1432 = 1.1098
In simple words: For multi-step math, group the "plus" numbers together and add them, then group the "minus" numbers together and add them. Finally, take the minus total away from the plus total.

📝 Teacher's Note: Introduce the BODMAS/PEMDAS rule here to show the correct order of operations.

🎯 Exam Tip: Re-arranging numbers so you add the positive values first often prevents silly subtraction mistakes.

Question 5. From the sum of 75.75 and 4.9 subtract 28.465.
Answer:
Step 1: Sum = 75.75 + 4.90 = 80.65
Step 2: 80.650 - 28.465 = 52.185
In simple words: First find the total of the two numbers mentioned, then take 28.465 away from that total.

📝 Teacher's Note: This is a two-step word problem. Encourage students to write down the result of the first step before starting the second.

🎯 Exam Tip: Watch for precision; if one number has three decimal places, ensure your first result also has a placeholder zero to match.

Question 6. Subtract the sum of 8.14 and 12.9 from 32.7.
Answer:
Step 1: Sum = 8.14 + 12.90 = 21.04
Step 2: 32.70 - 21.04 = 11.66
In simple words: Add 8.14 and 12.9 together first. Then take that total and subtract it from 32.7.

📝 Teacher's Note: Emphasize that the sum must be subtracted *from* 32.7, so 32.7 goes on top.

🎯 Exam Tip: Label your steps "Step 1: Addition" and "Step 2: Subtraction" to make your logic clear to the examiner.

Question 7. Subtract the sum of 34.27 and 159.8 from the sum of 20.937 and 200.6.
Answer:
Step 1: Sum 1 = 34.27 + 159.80 = 194.07
Step 2: Sum 2 = 20.937 + 200.600 = 221.537
Step 3: 221.537 - 194.070 = 27.467
In simple words: Find two separate totals, then find the difference between them.

📝 Teacher's Note: This requires careful bookkeeping. Advise students to organize their working columns clearly.

🎯 Exam Tip: Keep your decimal points aligned throughout all three steps to ensure no place value shift occurs.

Question 8. From the sum of 2.43 and 4.349 subtract the sum of 0.8 and 3.15.
Answer:
Step 1: Sum 1 = 2.430 + 4.349 = 6.779
Step 2: Sum 2 = 0.80 + 3.15 = 3.95
Step 3: 6.779 - 3.950 = 2.829
In simple words: Calculate the first pair's sum, calculate the second pair's sum, and subtract the second sum from the first.

📝 Teacher's Note: This is a great exercise for identifying which "sum" is the minuend and which is the subtrahend.

🎯 Exam Tip: Read carefully: "From [A] subtract [B]" means A - B.

Question 9. By how much does the sum of 18.0495 and 34.9644 exceed the sum of 7.6752 and 24.876 ?
Answer:
Step 1: Sum 1 = 18.0495 + 34.9644 = 53.0139
Step 2: Sum 2 = 7.6752 + 24.8760 = 32.5512
Step 3: Difference = 53.0139 - 32.5512 = 20.4627
In simple words: "Exceed" means one number is bigger than the other. Find out how much bigger by subtracting the second sum from the first sum.

📝 Teacher's Note: Vocabulary check: "Exceed" implies subtraction of the smaller value from the larger.

🎯 Exam Tip: When dealing with 4 decimal places, work slowly to avoid simple calculation errors in carrying or borrowing.

Question 10. What least number must be added to 89.376 to get 1000?
Answer:
To find the number to be added, subtract 89.376 from 1000.
1000.000 - 89.376 = 910.624
\( \implies \) The number to add to receive 1000 is 910.624.
In simple words: To find the missing amount to reach a goal, subtract what you already have from the goal total.

📝 Teacher's Note: This is an inverse operation problem. Help students relate it to simpler numbers like "What is added to 2 to get 10?"

🎯 Exam Tip: Be very careful when subtracting from numbers ending in multiple zeros like 1000; the borrowing cascades through every place value.

Exercise 15(C)

Question 1. Multiply :
(i) 5.6 and 8
(ii) 38.46 and 9
(iii) 0.943 and 62
(iv) 0.0453 and 35
(v) 7.5 and 2.5
(vi) 4.23 and 0.8
(vii) 83.54 and 0.07
(viii) 0.636 and 1.83
(ix) 6.4564 and 1000
(x) 0.076 and 100
Answer:
(i) 5.6 x 8 = 44.8
(ii) 38.46 x 9 = 346.14
(iii) 0.943 x 62: Since \( 943 \times 62 = 58466 \), and 0.943 has 3 decimal places, \( 0.943 \times 62 = 58.466 \).
(iv) 0.0453 x 35: Since \( 453 \times 35 = 15855 \), and 0.0453 has 4 decimal places, \( 0.0453 \times 35 = 1.5855 \).
(v) 7.5 x 2.5: Since \( 75 \times 25 = 1875 \), and both have 1 decimal place (total 2), \( 7.5 \times 2.5 = 18.75 \).
(vi) 4.23 x 0.8: Since \( 423 \times 8 = 3384 \), and there are total 3 decimal places, \( 4.23 \times 0.8 = 3.384 \).
(vii) 83.54 x 0.07: Since \( 8354 \times 7 = 58478 \), and total 4 decimal places, \( 83.54 \times 0.07 = 5.8478 \).
(viii) 0.636 x 1.83: Since \( 636 \times 183 = 116388 \), and total 5 decimal places, \( 0.636 \times 1.83 = 1.16388 \).
(ix) 6.4564 x 1000 = 6456.4
(x) 0.076 x 100 = 7.6
In simple words: Multiply as if the dots aren't there. Then count how many digits are after the dots in both original numbers combined. Put that many digits after the dot in your final answer.

📝 Teacher's Note: Teach students that multiplying by 10, 100, or 1000 just moves the decimal to the right by 1, 2, or 3 spots — no complex multiplication needed!

🎯 Exam Tip: Always double-check your total decimal count. For example, if you multiply \( 0.2 \times 0.3 \), the answer must have two decimal places (\( 0.06 \)).

Question 2. Evaluate :
(i) 0.0008 x 26
(ii) 0.038 x 95
(iii) 1.2 x 2.4 x 3.6
(iv) 0.9 x 1.8 x 0.27
(v) 1.5 x 1.5 x 1.5
(vi) 0.025 x 0.025
(vii) 0.2 x 0.002 x 0.001
Answer:
(i) 0.0008 x 26: Since \( 8 \times 26 = 208 \), and we need 4 decimal places, \( 0.0008 \times 26 = 0.0208 \).
(ii) 0.038 x 95: Since \( 38 \times 95 = 3610 \), and total 3 decimal places, \( 0.038 \times 95 = 3.610 = 3.61 \).
(iii) 1.2 x 2.4 x 3.6: Since \( 12 \times 24 \times 36 = 10368 \), and total 3 decimal places, \( 1.2 \times 2.4 \times 3.6 = 10.368 \).
(iv) 0.9 x 1.8 x 0.27: Since \( 9 \times 18 \times 27 = 4374 \), and total 4 decimal places, \( 0.9 \times 1.8 \times 0.27 = 0.4374 \).
(v) 1.5 x 1.5 x 1.5: Since \( 15 \times 15 \times 15 = 3375 \), and total 3 decimal places, \( 1.5 \times 1.5 \times 1.5 = 3.375 \).
(vi) 0.025 x 0.025: Since \( 25 \times 25 = 625 \), and total 6 decimal places, \( 0.025 \times 0.025 = 0.000625 \).
(vii) 0.2 x 0.002 x 0.001: Since \( 2 \times 2 \times 1 = 4 \), and total 7 decimal places, \( 0.2 \times 0.002 \times 0.001 = 0.0000004 \).
In simple words: For multiplying three numbers, just do the first two, get an answer, then multiply that by the third number. Keep counting the total decimal spots.

📝 Teacher's Note: In part (vii), emphasize that leading zeros are placeholders and essential for correct value.

🎯 Exam Tip: Count decimal places carefully in triple multiplications. It's easy to miss one!

Question 3. Multiply each of the following numbers by 10, 100 and 1000 :
(i) 3.9
(ii) 2.89
(iii) 0.0829
(iv) 40.3
(v) 0.3725
Answer:
(i) 3.9 x 10 = 39; 3.9 x 100 = 390; 3.9 x 1000 = 3900
(ii) 2.89 x 10 = 28.9; 2.89 x 100 = 289; 2.89 x 1000 = 2890
(iii) 0.0829 x 10 = 0.829; 0.0829 x 100 = 8.29; 0.0829 x 1000 = 82.9
(iv) 40.3 x 10 = 403; 40.3 x 100 = 4030; 40.3 x 1000 = 40300
(v) 0.3725 x 10 = 3.725; 0.3725 x 100 = 37.25; 0.3725 x 1000 = 372.5
In simple words: When you multiply by 10, 100, or 1000, the dot jumps to the right. 10 has one zero (1 jump), 100 has two zeros (2 jumps), and 1000 has three zeros (3 jumps).

📝 Teacher's Note: Use a "bouncing ball" analogy to show how the decimal point moves over the digits.

🎯 Exam Tip: If the decimal point jumps past all the numbers, just keep adding zeros at the end.

Question 4. Evaluate :
(i) 8.64 \( \div \) 8
(ii) 0.0072 \( \div \) 6
(iii) 20.64 \( \div \) 16
(iv) 1.602 \( \div \) 15
(v) 13.08 \( \div \) 4
(vi) 3.204 \( \div \) 9
(vii) 3.024 \( \div \) 12
(viii) 5.15 \( \div \) 5
(ix) 3 \( \div \) 5
Answer:
(i) 8.64 \( \div \) 8 = 1.08
(ii) 0.0072 \( \div \) 6 = 0.0012
(iii) 20.64 \( \div \) 16 = 1.29
(iv) 1.602 \( \div \) 15 = 0.1068
(v) 13.08 \( \div \) 4 = 3.27
(vi) 3.204 \( \div \) 9 = 0.356
(vii) 3.024 \( \div \) 12 = 0.252
(viii) 5.15 \( \div \) 5 = 1.03
(ix) 3 \( \div \) 5 = 0.6
In simple words: Divide like normal numbers, but keep the decimal point in the same column as it is in the number you are dividing.

📝 Teacher's Note: Remind students to place the decimal point in the quotient directly above the decimal point in the dividend before they start dividing.

🎯 Exam Tip: If the divisor cannot go into the first few digits, don't forget to put zeros in the quotient to hold those places.

Question 5. Divide each of the following numbers by 10, 100 and 1000 :
(i) 49.79
(ii) 0.923
(iii) 0.0704
Answer:
(i) 49.79 \( \div \) 10 = 4.979; 49.79 \( \div \) 100 = 0.4979; 49.79 \( \div \) 1000 = 0.04979
(ii) 0.923 \( \div \) 10 = 0.0923; 0.923 \( \div \) 100 = 0.00923; 0.923 \( \div \) 1000 = 0.000923
(iii) 0.0704 \( \div \) 10 = 0.00704; 0.0704 \( \div \) 100 = 0.000704; 0.0704 \( \div \) 1000 = 0.0000704
In simple words: Dividing by 10, 100, or 1000 makes the number smaller, so the dot jumps to the left. The more zeros, the more jumps!

📝 Teacher's Note: Compare this to multiplication where the decimal point moves right. This contrast helps clarify the concept of place value shifting.

🎯 Exam Tip: Always add a '0' before the decimal point (e.g., 0.4979) to make your answer professional and clear.

Question 6. Evaluate :
(i) 9.4 \( \div \) 0.47
(ii) 6.3 \( \div \) 0.09
(iii) 2.88 \( \div \) 1.2
(iv) 8.64 \( \div \) 1.6
(v) 37.188 \( \div \) 3.6
(vi) 16.5 \( \div \) 0.15
(vii) 3.2 \( \div \) 0.005
(viii) 3.24 \( \div \) 0.0016
Answer:
(i) \( 9.4 \div 0.47 = \frac{94 \times 100}{47 \times 10} = 2 \times 10 = 20 \)
(ii) \( 6.3 \div 0.09 = \frac{63 \times 100}{9 \times 10} = 7 \times 10 = 70 \)
(iii) \( 2.88 \div 1.2 = \frac{288 \times 10}{12 \times 100} = \frac{288}{120} = 2.4 \)
(iv) \( 8.64 \div 1.6 = \frac{8.64 \times 10}{1.6 \times 10} = \frac{86.4}{16} = 5.4 \)
(v) \( 37.188 \div 3.6 = \frac{371.88}{36} = 10.33 \)
(vi) \( 16.5 \div 0.15 = \frac{165 \times 100}{15 \times 10} = 11 \times 10 = 110 \)
(vii) \( 3.2 \div 0.005 = \frac{3.2 \times 1000}{0.005 \times 1000} = \frac{3200}{5} = 640 \)
(viii) \( 3.24 \div 0.0016 = \frac{324 \times 10000}{100 \times 16} = \frac{32400}{16} = 2025 \)
In simple words: When dividing by a decimal, move the dots in both numbers to the right until the number you are dividing by is a whole number. Then divide as usual.

📝 Teacher's Note: This "equivalent fraction" method (multiplying both by 10, 100, etc.) is the safest way to handle division by decimals without getting confused.

🎯 Exam Tip: Ensure you move the decimal point the *exact same* number of times in both numbers.

Question 7. Fill in the blanks with 10, 100, 1000, or 10000 etc.:
(i) 7.85 x .............. = 78.5
(ii) 0.442 x .......... = 442
(iii) 0.0924 x ............ = 9.24
(iv) 0.00187 x ............. = 18.7
(v) 2.6 x ............. = 2600
(vi) 0.08 x ............ = 80
(vii) 96.7 \( \div \) .............. = 0.967
(viii) 5.2 \( \div \) ................ = 0.52
(ix) 33.15 \( \div \) .................... = 0.03315
(x) 0.7 \( \div \) ............ = 0.007
(xi) 0.00672 x ......... = 67.2
Answer:
(i) 7.85 x 10 = 78.5
(ii) 0.442 x 1000 = 442
(iii) 0.0924 x 100 = 9.24
(iv) 0.00187 x 10000 = 18.7
(v) 2.6 x 1000 = 2600
(vi) 0.08 x 1000 = 80
(vii) 96.7 \( \div \) 100 = 0.967
(viii) 5.2 \( \div \) 10 = 0.52
(ix) 33.15 \( \div \) 1000 = 0.03315
(x) 0.7 \( \div \) 100 = 0.007
(xi) 0.00672 x 10000 = 67.2
In simple words: Count how many places the dot moved. If it moved right, it's multiplication. If left, it's division. The number of jumps tells you how many zeros you need.

📝 Teacher's Note: This exercise reinforces mental math and the conceptual understanding of how decimal points interact with powers of 10.

🎯 Exam Tip: Count the "steps" the decimal point took to decide if you need 10 (1 step), 100 (2 steps), or 1000 (3 steps).

Question 8. Evaluate :
(i) 9.32 - 28.54 \( \div \) 10
(ii) 0.234 x 10 + 62.8
(iii) 3.06 x 100 - 889.4 \( \div \) 100
(iv) 2.86 x 7.5 + 45.4 \( \div \) 0.2
(v) 97.82 x 0.03 - 0.54 \( \div \) 0.3
Answer:
(i) 9.32 - 28.54 \( \div \) 10 = 9.32 - 2.854 = 6.466
(ii) 0.234 x 10 + 62.8 = 2.34 + 62.8 = 65.14 (Using BODMAS)
(iii) 3.06 x 100 - 889.4 \( \div \) 100 = 306 - 8.894 = 297.106 (Using BODMAS)
(iv) 2.86 x 7.5 + 45.4 \( \div \) 0.2 = 21.45 + 227 = 248.45 (Using BODMAS)
(v) 97.82 x 0.03 - 0.54 \( \div \) 0.3 = 2.9346 - 1.8 = 1.1346
In simple words: Remember the order: always do Division and Multiplication first, then Addition and Subtraction later!

📝 Teacher's Note: Students often just calculate from left to right. Force the habit of circling the multiplication or division parts before starting.

🎯 Exam Tip: Underline the operations that must be done first so you don't accidentally add or subtract too early.

Exercise 15(D)

Question 1. Express in paise :
(i) Rs. 8.40
(ii) Rs. 0.97
(iii) Rs. 0.09
(iv) Rs. 62.35
Answer:
(i) Rs. 8.40 = 8.40 x 100 paise = 840 Paise [1Rs. = 100 Paise]
(ii) Rs. 0.97 = 0.97 x 100 paise = 97 paise
(iii) Rs. 0.09 = 0.09 x 100 Paise = 9 paise
(iv) Rs. 62.35 = 62.35 x 100 Paise = 6235 Paise.
In simple words: Since 1 Rupee is 100 paise, just multiply the amount by 100 to change it to paise.

📝 Teacher's Note: Relate this to jumping the decimal point two places to the right because you are multiplying by 100.

🎯 Exam Tip: Always state your conversion unit like "[1Rs. = 100 Paise]" to show your method clearly.

Question 2. Express in rupees :
(i) 55 P
(ii) 8 P
(iii) 695 P
(iv) 3279 P
Answer:
(i) 55P = \( \frac{55}{100} \) = Rs. 0.55
(ii) 8P = \( \frac{8}{100} \) = Rs. 0.08
(iii) 695P = \( \frac{695}{100} \) = Rs. 6.95
(iv) 3279P = \( \frac{3279}{100} \) = Rs. 32.79
In simple words: To turn small coins (paise) into whole Rupees, divide by 100. This moves the decimal dot two jumps to the left.

📝 Teacher's Note: Common mistake is writing 8 paise as Rs. 0.8 instead of Rs. 0.08. Emphasize that hundredths require two decimal places.

🎯 Exam Tip: Don't forget the "Rs." symbol in your final answer.

Question 3. Express in centimetre (cm) :
(i) 6 m
(ii) 8.54 m
(iii) 3.08 m
(iv) 0.87 m
(v) 0.03 m
(vi) 25.04 m
Answer:
(i) 6 x 100 = 600 cm
(ii) 8.54 x 100 = 854 cm
(iii) 3.08 x 100 = 308 cm
(iv) 0.87 x 100 = 87 cm
(v) 0.03 x 100 = 3 cm
(vi) 25.04 x 100 = 2504 cm
In simple words: There are 100 centimetres in 1 metre. So, multiply by 100 to convert metres to centimetres.

📝 Teacher's Note: Bring a meter ruler to class to visually demonstrate that it takes 100 small cm marks to make one large meter.

🎯 Exam Tip: Multiplication by 100 moves the decimal two places to the right.

Question 4. Express in metre (m) :
(i) 250 cm
(ii) 2328 cm
(iii) 86 cm
(iv) 4 cm
(v) 107 cm
Answer:
(i) \( \frac{250}{100} \) = 2.50 m
(ii) \( \frac{2328}{100} \) = 23.28 m
(iii) \( \frac{86}{100} \) = 0.86 m
(iv) \( \frac{4}{100} \) = 0.04 m
(v) \( \frac{107}{100} \) = 1.07 m
In simple words: Divide by 100 to turn centimetres into metres. Just move the dot two places to the left!

📝 Teacher's Note: Help students realize that metres are the "base" unit and cm is smaller, requiring division.

🎯 Exam Tip: Always double-check that you haven't forgotten the 'm' unit label.

Question 5. Express in gramme (gm) :
(i) 6 kg
(ii) 5.543 kg
(iii) 0.078 kg
(iv) 3.62 kg
(v) 4.5 kg
Answer:
(i) 6 x 1000 = 6000 gm
(ii) 5.543 x 1000 = 5543 gm
(iii) 0.078 kg = 0.078 x 1000 g = 78 g (1 kg = 1000 g)
(iv) 3.62 x 1000 = 3620 gm
(v) 4.5 x 1000 = 4500 gm
In simple words: 1 Kilogram is 1000 grams. Multiply by 1000 (move the dot 3 places right) to turn kg into gm.

📝 Teacher's Note: Explain that "kilo" means 1000, which makes it easy to remember the conversion factor.

🎯 Exam Tip: Be careful with part (iii); 0.078 becomes 78, not 7.8 or 780.

Question 6. Express in kilogramme (kg) :
(i) 7000 gm
(ii) 6839 gm
(iii) 445 gm
(iv) 8 gm
(v) 93 gm
(vi) 13545 gm
Answer:
(i) \( \frac{7000}{1000} \) = 7 kg
(ii) \( \frac{6839}{1000} \) = 6.839 kg
(iii) \( \frac{445}{1000} \) = 0.445 kg
(iv) \( \frac{8}{1000} \) = 0.008 kg
(v) \( \frac{93}{1000} \) = 0.093 kg
(vi) \( \frac{13545}{1000} \) = 13.545 kg
In simple words: To change grams into kilograms, divide by 1000. Move the dot three jumps to the left!

📝 Teacher's Note: Again, watch for placeholders. 8 gm becomes 0.008 kg, because there are no hundreds or tens of grams.

🎯 Exam Tip: Count exactly three places left. If you land on a blank spot, put a zero there.

Question 7. Add (giving answer in rupees) :
(i) Rs. 5.37 and Rs. 12
(ii) Rs. 24.03 and 532 paise
(iii) 73 paise and Rs. 2.08
(iv) 8 paise and Rs. 15.36
Answer:
(i) Rs. 5.37 + Rs. 12.00 = Rs. 17.37
(ii) Rs. 24.03 + \( \frac{532}{100} \) = Rs. 24.03 + 5.32 = Rs. 29.35
(iii) \( \frac{73}{100} \) + Rs. 2.08 = Rs. 0.73 + 2.08 = Rs. 2.81
(iv) \( \frac{8}{100} \) + Rs. 15.36 = Rs. 0.08 + 15.36 = Rs. 15.44
In simple words: You can't add different things like apples and oranges. First, change the paise into Rupees, then add the Rupee amounts together.

📝 Teacher's Note: Emphasize conversion as the very first step before any arithmetic begins.

🎯 Exam Tip: Use the same unit (Rupees) for the whole addition to avoid confusion.

Question 8. Subtract :
(i) Rs. 35.74 from Rs. 63.22
(ii) 286 paise from Rs. 7.02
(iii) Rs. 0.55 from 121 paise
Answer:
(i) Rs. 63.22 - Rs. 35.74 = Rs. 27.48
(ii) Rs. 7.02 - 286 paise = Rs. 7.02 - \( \frac{286}{100} \) = Rs. 7.02 - Rs. 2.86 = Rs. 4.16
(iii) 121 paise - Rs. 0.55 = \( \frac{121}{100} \) - Rs. 0.55 = Rs. 1.21 - Rs. 0.55 = Rs. 0.66
In simple words: Just like addition, convert to the same units first! Then subtract the "from" number on top.

📝 Teacher's Note: The word "from" is critical. It indicates the minuend.

🎯 Exam Tip: Ensure that your final answer includes the "Rs." symbol clearly.

Question 9. Add (giving answer in metre) :
(i) 2.4 m and 1.78 m
(ii) 848 cm and 2.9 m
(iii) 0.93 m and 64 cm
Answer:
(i) 2.40 m + 1.78 m = 4.18 m
(ii) \( \frac{848}{100} \) m + 2.9 m = 8.48 m + 2.90 m = 11.38 m
(iii) 0.93 m + \( \frac{64}{100} \) m = 0.93 m + 0.64 m = 1.57 m
In simple words: Convert everything to metres first by dividing centimetres by 100, then line up the dots and add.

📝 Teacher's Note: Remind students that adding 2.4 and 1.78 requires making them "like" decimals by adding a zero to 2.4.

🎯 Exam Tip: Check your unit conversions first; 848 cm is 8.48 m, not 84.8 m.

Question 10. Subtract (giving answer in metre) :
(i) 5.03 m from 19.6 m
(ii) 428 cm from 1033 m
(iii) 0.84 m from 122 cm
Answer:
(i) 19.60 m - 5.03 m = 14.57 m
(ii) 1033 m - \( \frac{428}{100} \) m = 1033.00 m - 4.28 m = 1028.72 m
(iii) \( \frac{122}{100} \) m - 0.84 m = 1.22 m - 0.84 m = 0.38 m
In simple words: "Subtract X from Y" means Y - X. Change any cm into metres first by dividing by 100.

📝 Teacher's Note: The scale of (ii) is quite large; help students keep their subtraction alignment correct over four digits.

🎯 Exam Tip: Always convert all numbers to the unit specified in the question before starting the calculation.

Question 11. Add (giving answer in kg) :
(i) 2.06 kg and 57.864 kg
(ii) 778 gm and 1.939 kg
(iii) 0.065 kg and 4023 gm
Answer:
(i) 2.060 kg + 57.864 kg = 59.924 kg
(ii) \( \frac{778}{1000} \) kg + 1.939 kg = 0.778 kg + 1.939 kg = 2.717 kg
(iii) 0.065 kg + \( \frac{4023}{1000} \) kg = 0.065 kg + 4.023 kg = 4.088 kg
In simple words: Turn all grams into kilograms by dividing by 1000. Then add them up just like regular decimals.

📝 Teacher's Note: Grams are thousandths of a kilogram, so dividing by 1000 is key.

🎯 Exam Tip: Line up those decimal points! In part (i), add a zero to 2.06 so it has 3 decimal places to match 57.864.

Question 12. Subtract (giving answer in kg) :
(i) 9.462 kg from 15.6 kg
(ii) 4317 gm from 23 kg
(iii) 0.798 kg from 4169 gm
Answer:
(i) 15.600 kg - 9.462 kg = 6.138 kg
(ii) 23.000 kg - \( \frac{4317}{1000} \) kg = 23.000 kg - 4.317 kg = 18.683 kg
(iii) \( \frac{4169}{1000} \) kg - 0.798 kg = 4.169 kg - 0.798 kg = 3.371 kg
In simple words: Move the "from" number to the top. Change gm to kg by dividing by 1000.

📝 Teacher's Note: This requires students to be very comfortable with placeholder zeros, especially when subtracting from a whole number like 23.

🎯 Exam Tip: If the question asks for a specific unit (like kg), never leave your final answer in grams.

Exercise 15(E)

Question 1. The cost of a fountain pen is Rs. 13.25. Find the cost of 8 such pens.
Answer:
Cost of 1 fountain Pen = Rs. 13.25
Cost of 8 fountain Pens = 13.25 x 8 = Rs. 106.00 = Rs. 106
In simple words: If you know the price of one, multiply it by the number of things you want to buy.

📝 Teacher's Note: This is a simple repeated addition problem framed as multiplication.

🎯 Exam Tip: Show the multiplication calculation clearly to get full marks for the method.

Question 2. The cost of 25 identical articles is Rs. 218.25. Find the cost of one article.
Answer:
Cost of 25 articles = Rs. 218.25
Cost of 1 article = \( \frac{218.25}{25} \) = Rs. 8.73
In simple words: To find the price of just one thing, take the big total price and divide it by how many things there are.

📝 Teacher's Note: Relate this to sharing fairly. If you share the total cost among all articles, each gets the single price.

🎯 Exam Tip: Division of decimals by a whole number can be tricky; double-check the placement of your decimal point in the quotient.

Question 3. The length of an iron rod is 10.32 m. The rod is divided into 4 pieces of equal lengths. Find the length of each piece.
Answer:
The length of iron rod = 10.32 m
Dividing in 4 equal parts = \( \frac{10.32}{4} \) = 2.58 m
In simple words: When you cut something into equal pieces, you use division. Divide the total length by the number of pieces.

📝 Teacher's Note: A visual aid like a string being cut can help explain this concept.

🎯 Exam Tip: Don't forget to include the unit 'm' in your final length.

Question 4. What will be the total length of cloth required to make 5 shirts, if 2.15 m of cloth is needed for each shirt ?
Answer:
Cloth required for each shirt = 2.15 m
Cloth required for 5 shirts = 2.15 x 5 m = 10.75 m
In simple words: Take the cloth for one shirt and multiply it by 5.

📝 Teacher's Note: Remind students to count two decimal places in their final product since 2.15 has two decimal places.

🎯 Exam Tip: Read word problems carefully to identify whether multiplication or division is required.

Question 5. Find the distance walked by a boy in \( 1 \frac{1}{2} \) hours, if he walks at 2.150 km every hour.
Answer:
Distance covered in one hour = 2.150 km
Distance covered in \( 1 \frac{1}{2} \) hour = 2.150 x \( 1 \frac{1}{2} \)
= 2.150 x 1.5 = 3.225 km
In simple words: Multiply the speed (km per hour) by the time (hours) to find the distance.

📝 Teacher's Note: Show students that \( 1 \frac{1}{2} \) is the same as 1.5 to make the decimal multiplication easier.

🎯 Exam Tip: Always convert mixed numbers to decimals first when working with other decimal values.

Question 6. 83 note-books are sold at Rs. 15.25 each. Find the total money (in rupees) obtained by selling these note-books.
Answer:
Sale price of 1 note-book = Rs. 15.25
Sale of 83 books = Rs. 15.25 x 83 = Rs. 1265.75
In simple words: Multiply the price of one notebook by the number of notebooks sold.

📝 Teacher's Note: This is a standard long multiplication of decimals. Review the digit-by-digit process with the class.

🎯 Exam Tip: Be very tidy with your column multiplication to avoid adding the wrong digits.

Question 7. If length of one bed-cover is 2.1 m, find the total length of 17 bed-covers.
Answer:
Length of one bed-cover = 2.1 m
Length of 17 bed-covers = 17 x 2.1 = 35.7 m
In simple words: To find the total, multiply the length of one cover by the total number of covers.

📝 Teacher's Note: Explain that "length of one" is the unit rate.

🎯 Exam Tip: One decimal place in 2.1 means your final answer should have one decimal place.

Question 8. A piece of rope is 10 m 67 cm long. Another rope is 16 m 32 cm long. By how much is the second rope longer than the first one ?
Answer:
Length of one rope = 10 m 67 cm = 10.67 m
Length of another rope = 16 m 32 cm = 16.32 m
Difference in length = 16.32 m - 10.67 m = 5.65 m
= 5 m 65 cm.
In simple words: Convert the metres and centimetres into a decimal (like 16.32) and subtract the smaller rope's length from the bigger one.

📝 Teacher's Note: This is a practical example of why converting to decimal form makes subtraction easier.

🎯 Exam Tip: If the question provides mixed units (m and cm), it's best to provide the final answer in mixed units too (5 m 65 cm).

Question 9. 12 cakes of soap together weigh 5 kg and 604 gm. Find the weight of
(i) One cake in both kg and gramme
(ii) 5 cakes in kg.
Answer:
Weight of 12 cakes of soap = 5 kg and 604 gm = 5.604 kg
(i) Weight of 1 cake = \( \frac{5.604}{12} \) = 0.467 kg
Weight in gm = 0.467 x 1000 = 467 gm
(ii) Weight of five cakes = 0.467 x 5 = 2.335 kg.
In simple words: First, find the weight of one single soap by dividing the total weight by 12. Then, to find the weight of 5 soaps, multiply the weight of one soap by 5.

📝 Teacher's Note: Remind students to convert grams into the decimal part of the kilogram (604 gm = .604 kg) before dividing.

🎯 Exam Tip: Always double-check your division: 5.604 divided by 12 gives 0.467.

Question 10. Three strings of lengths 50 m 75 cm; 68 m 58 cm and 121 m 3 cm, respectively, are joined together to get a single string of greatest length, And the length of the single string obtained. If this single string is then divided into 12 equal pieces ; find the length of each piece.
Answer:
1st string 50 m 75 cm = 50.75 m
2nd string 68 m 68 m 58 cm = 68.58 m
3rd string 121 m 3 cm = 121.03 m
On joining three total length = 50.75 + 68.58 + 121.03 = 240.36 m
Dividing 12 parts = \( \frac{240.36}{12} \) = 20.03 m.
In simple words: Add the three string lengths together to get one big string. Then divide that big total by 12 to see how long each piece is.

📝 Teacher's Note: Careful with 3 cm; it is 0.03 m, not 0.3 m.

🎯 Exam Tip: Step 1 is addition of unlike decimals; Step 2 is decimal division. Show both clearly.

Revision Exercise

Question 1. Write the following decimal numbers in ascending order of value
(i) 5.054, 5.250, 5.245 and 5.0543
(ii) 62.443, 62.434, 62.344 and 62.444
Answer:
(i) 5.054, 5.250, 5.245 and 5.0543
Writing them in like decimals : 5.0540, 5.2500, 5.2450, 5.0543
Now arranging in ascending order : 5.0540, 5.0543, 5.2450, 5.2500
\( \implies \) 5.054 < 5.0543 < 5.245 < 5.250

(ii) 62.443, 62.434, 62.344 and 62.444
Writing them in like decimals : 62.443, 62.434, 62.344, 62.444
Now writing in ascending order : 62.344, 62.434, 62.443, 62.444
or 62.344 < 62.434 < 62.443 < 62.444
In simple words: To see which decimal is bigger, first make them all have the same number of digits after the dot by adding zeros. Then compare them like they are whole numbers.

📝 Teacher's Note: Comparing decimals digit-by-digit from left to right is the best technique to teach here.

🎯 Exam Tip: Ascending means "small to big". Use the '<' symbol to show the relationship correctly.

Question 2. What number added to 0.805 gives 1 ?
Answer:
The required number will be formed by subtracting 0.805 from 1
Required number = 1 - 0.805 = 1.000 - 0.805 = 0.195
In simple words: To find the missing part of 1, subtract the decimal you have from 1.

📝 Teacher's Note: Review place value; remind students that 1 is the same as 1.000.

🎯 Exam Tip: Don't forget to align the decimal of 1.000 with 0.805 before subtracting.

Question 3. What must be subtracted from 3 to get 2.462 ?
Answer:
The required number can be formed by subtracting 2.462 from 3
Required number = 3 - 2.462 = 3.000 - 2.462 = 0.538
In simple words: To find out what was taken away from 3, subtract the result you ended up with from the 3.

📝 Teacher's Note: This is a missing subtrahend problem. Use a number line to show the gap.

🎯 Exam Tip: Precision in borrowing is vital here.

Question 4. By how much should 83.407 be decreased to get 27.78 ?
Answer:
The required number can be formed by subtracting 27.78 from 83.407
Required number = 83.407 - 27.78 = 83.407 - 27.780 = 55.627
In simple words: Subtract the target number from the starting number to see how much of a decrease is needed.

📝 Teacher's Note: "Decreased" is another term for subtraction.

🎯 Exam Tip: Make the decimals "like" before subtracting by adding a 0 to 27.78.

Question 5. Two articles weigh 32.674 kg and 40.038 kg respectively. Find :
(i) the total weight of both the articles.
(ii) the difference in the weights of both the articles.
Answer:
Weight of first article = 32.674 kg
Weight of second article = 40.038 kg
(i) Total weight of both the articles = (32.674 + 40.038) kg = 72.712 kg
(ii) Difference between the weights of the articles = (40.038 - 32.674) kg = 7.364 kg
In simple words: Part one is adding the weights together. Part two is subtracting the smaller weight from the bigger weight.

📝 Teacher's Note: This problem tests basic arithmetic on decimals in a real-world scenario.

🎯 Exam Tip: Keep the "kg" unit in your final answers for both parts.

Question 6. By how much does the sum of 34.07 and 15.239 exceed the sum of 16.40 and 27.08?
Answer:
Sum of 34.07 and 15.239 = 34.070 + 15.239 = 49.309
Sum of 16.40 and 27.08 = 16.40 + 27.08 = 43.48
Difference between their sums = 49.309 - 43.48 = 49.309 - 43.480 = 5.829
In simple words: First, add the first pair. Second, add the second pair. Third, subtract the smaller total from the bigger total.

📝 Teacher's Note: This is a common multi-step logic problem. Advise students to label each sum clearly.

🎯 Exam Tip: Don't rush; check each sum individually before doing the final subtraction.

Question 7. The cost of 1 kg of fruit is Rs. 27.50. What is the cost of 3.6 kg of fruit ?
Answer:
Cost of 1 kg fruit = Rs. 27.50
Cost of 3.6 kg fruit = Rs. 27.50 x 3.6 = Rs. 99.00
In simple words: Multiply the price of 1 kilogram by how many kilograms you want to buy.

📝 Teacher's Note: Multiplication of decimals is required here. 27.50 x 3.6 is 99.00.

🎯 Exam Tip: Make sure to put the decimal in the correct place after multiplying \( 275 \times 36 \).

Question 8. Evaluate :
(i) 0.8 x 0.8 x 0.8
(ii) 0.8 \( \div \) 0.8 x 0.8
(iii) 0.8 x 0.8 \( \div \) 0.8
(iv) 0.8 \( \div \) 0.8 of 0.8
(v) 0.8 of 0.8 \( \div \) 0.8
Answer:
(i) 0.8 x 0.8 x 0.8 = 0.512
(ii) 0.8 \( \div \) 0.8 x 0.8 = 1 x 0.8 = 0.8
(iii) 0.8 x 0.8 \( \div \) 0.8 = 0.64 \( \div \) 0.8 = 0.8
(iv) 0.8 \( \div \) 0.8 of 0.8 = 0.8 \( \div \) 0.64 = \( \frac{80}{64} \) = 1.25
(v) 0.8 of 0.8 \( \div \) 0.8 = 0.64 \( \div \) 0.8 = 0.8
In simple words: Use BODMAS. "Of" is like multiplication and must be done before division.

📝 Teacher's Note: The "of" part in part (iv) is crucial. "0.8 of 0.8" means "0.8 \( \times \) 0.8" and it takes priority.

🎯 Exam Tip: Be careful with BODMAS; multiplication and division have equal priority left-to-right, but "Of" comes before both.

Question 9. Evaluate :
(i) 3.5 x (4.2 + 2.6)
(ii) 3.5 x 4.2 + 3.5 x 2.6
Are (i) and (ii) equal ?
Answer:
(i) 3.5 x (4.2 + 2.6) = 3.5 x (6.8) = 23.8
(ii) 3.5 x 4.2 + 3.5 x 2.6 = 14.7 + 9.1 = 23.8
Yes results of (i) and (ii) are equal.
In simple words: This shows that multiplying a number by a sum is the same as multiplying it by each part of the sum separately and then adding them.

📝 Teacher's Note: This is an introduction to the Distributive Property of multiplication over addition.

🎯 Exam Tip: Brackets indicate that the addition must be done first in part (i).

Question 10. Evaluate :
(i) (3.87 - 2.09) x 2.4
(ii) 3.87 x 2.4 - 2.09 x 2.4
Are (i) and (ii) equal ?
Answer:
(i) (3.87 - 2.09) x 2.4 = 1.78 x 2.4 = 4.272
(ii) 3.87 x 2.4 - 2.09 x 2.4 = 9.288 - 5.016 = 4.272
Yes, results of (i) and (ii) are equal.
In simple words: Just like with adding, the distributive property also works for subtraction!

📝 Teacher's Note: Similar to the previous question, this reinforces algebraic properties using decimal numbers.

🎯 Exam Tip: Multiplying 3.87 by 2.4 requires counting three decimal places in the final result.

Question 11. A 4.85 m long pole is divided into 5 equal parts. Find the length of each part.
Answer:
Length of pole = 4.85 m
It is divided into 5 equal parts Length of each part = 4.85 \( \div \) 5 m = 0.97 m
Hence length of each part = 0.97 m
In simple words: Divide the total length by the number of parts to find how long each piece is.

📝 Teacher's Note: Reinforce that division is the operation used for sharing or cutting into equal portions.

🎯 Exam Tip: Division by 5 is simple: \( 4.85 \div 5 = 0.97 \).

Question 12. A car can run 16.8 km consuming one litre of petrol. How many kilometres will it run on 3.7 litres of petrol ?
Answer:
A car can go in one litre = 16.8 km
It will go in 3.7 litres of petrol = 16.8 x 3.7 km = 62.16 km
In simple words: If you know how far you go on one litre, multiply that by the total litres you have.

📝 Teacher's Note: This is a rate problem. Distance = Fuel Efficiency \( \times \) Fuel Volume.

🎯 Exam Tip: Total decimals in 16.8 x 3.7 is two. Make sure 62.16 has exactly two.

Question 13. A certain amount of money is distributed among 28 persons. If each person gets Rs. 62.45 and Rs. 5.78 is left, find the original amount of money.
Answer:
Number of persons = 28
Each person gets = Rs. 62.45
Total amount distributed to 28 persons = Rs. 62.45 x 28 = Rs. 1748.60
Amount left undistributed = Rs. 5.78
Total amount = Rs. 1748.60 + 5.78 = Rs. 1754.38
In simple words: Multiply the money each person got by the number of people. Then add the small amount that was left over to find the starting total.

📝 Teacher's Note: This teaches the concept of "Total = (Quotient \( \times \) Divisor) + Remainder".

🎯 Exam Tip: Always multiply first, then add the remainder.

Question 14. Complete the following table :

Answer:
The table is completed by calculating Amount = Cost per kg \( \times \) Quantity for each item, and then adding all amounts.
A: 17.40 x 2.5 = Rs. 43.50
B: 42.25 x 1.6 = Rs. 67.60
C: 28.50 x 3.2 = Rs. 91.20
Total = 43.50 + 67.60 + 91.20 = Rs. 202.30
In simple words: This is like a grocery bill. Multiply price by weight for each item, then add them all up to get the final total.

📝 Teacher's Note: This is a highly practical problem. Encourage students to check their multiplications carefully as one error affects the final total.

🎯 Exam Tip: Fill each row clearly and double-check your final column addition.

Question 15. The difference between two numbers is 47.364. If the smaller number is 31.855 ; find the bigger one.
Answer:
Difference of two number = 47.364
Smaller number = 31.855
Bigger number = 47.364 + 31.855 = 79.219
In simple words: If you know the gap (difference) between two things and you know the smaller one, just add the gap to the small one to find the big one.

📝 Teacher's Note: Use the formula: Bigger Number = Smaller Number + Difference.

🎯 Exam Tip: Make sure you add, don't subtract! If you subtract, you'll get a number even smaller than 31.855.