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Decimal Division

🎓 Class 7 Mathematics CBSE Theory Ch 4 — Working with Decimals ⏱ ~35 min
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This MCQ module is based on: Decimal Division

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Targeting Class 7 level in General Mathematics, with Basic difficulty.

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4.3 Decimal Division

Sharing Ribbon

We know that the fraction \(\frac{1}{2}\) can be represented as a decimal 0.5. So, each girl will get 14 m and an additional 0.5 m of ribbon. Hence, the length of ribbon each will get is \(14 + 0.5 = 14.5\) m.

Now, what if the ribbon was shared between four friends instead of 2? So, each will get \(29 \div 4\) m, that is \(\frac{29}{4}\) m.

Now, the denominator of the fraction is 4. To convert a fraction to a decimal, it helps if the denominator is of the form \(\frac{1}{10}\), \(\frac{1}{100}\), 1000 and so on. Can we find a fraction equivalent to \(\frac{29}{4}\) with such a denominator? — Yes!

Is 4 a factor of 100? Yes (4 × 25 = 100). So we can get an equivalent fraction to \(\frac{29}{4}\) with denominator 100 by multiplying the numerator and denominator by 25.

\(\frac{29}{4} = \frac{29 \times 25}{4 \times 25} = \frac{725}{100} = 7.25\)

So each of the 4 friends will get 7.25 m of ribbon.

Division Using Place Value

We have seen how to divide two counting numbers to get a decimal quotient. We first represented the division as a fraction. Then we found an equivalent fraction with denominator of the form 1, 10, 100, and so on, to convert it to a decimal.

Now, let us look at the division using place value procedure to calculate the decimal quotient.

Suppose we want to write the quotient \(\frac{13}{8}\) as a decimal. We can convert this fraction to an equivalent fraction with a denominator such as 1, 10, 100, 1000, etc.

It is not possible. So, we need a more general method to divide any two counting numbers. Let us see how we can use division using place value for this.

Example 5: Long Division of 237 ÷ 8

Find the value of \(237 \div 8\). This extends the division using place value to find quotients with decimals in them. Two hundreds and thirty-seven can be regrouped as tenths, tens and can be regrouped as hundredths and so on.

_____29.625_____ 8 | 237.000 16 -- 77 72 -- 50 ← 5 Ones can't divide by 8; regroup as 50 Tenths 48 -- 20 ← 2 Tenths → 20 Hundredths 16 -- 40 ← 4 Hundredths → 40 Thousandths 40 -- 0

Steps explained:

  • 20 Tens ÷ 8 = 2 Tens, 4 Tens remain.
  • 4 Tens + 3 Tens (from regrouping) → 23 Tens; 23 ÷ 8 = 2 Tens, 7 Tens remain.
  • 7 Tens = 70 Ones; with 7 original Ones → 77 Ones. 77 ÷ 8 = 9 Ones, 5 Ones remain.
  • 5 Ones can't be divided by 8 in the quotient. When we need to regroup into 50 Tenths, we place a decimal point in the quotient.
  • 50 Tenths ÷ 8 = 6 Tenths, 2 Tenths remain.
  • 2 Tenths cannot be divided into 8 equal parts, so we need to regroup them as 20 Hundredths. 20 Hundredths ÷ 8 = 2 Hundredths, 4 Hundredths remain.
  • 4 Hundredths → 40 Thousandths → 5 Thousandths, 0 remaining. Done.

237 ÷ 8 = 29.625

Big Idea
To divide by a whole number and extend the quotient into decimals, we keep regrouping remainders as the next smaller place value (Tens → Ones → Tenths → Hundredths → Thousandths ...) until the remainder is zero or we decide enough decimal places have been found.

Why Do We Put a Decimal Point in the Quotient?

When we regroup 5 Ones into 50 Tenths, we place a decimal point in the quotient. This is because the quotient now contains a fractional part — every digit written after the point is in the Tenths, Hundredths, Thousandths place.

Example 6: Sridharacharya's Problem (8th century)

The great Indian mathematician Sridharacharya, in his book Patiganita, sets: \(6\frac{1}{4}\) is divided by \(2\frac{1}{2}\), and \(60\frac{1}{4}\) is divided by \(5\frac{3}{4}\). Tell the quotients separately. Can you try to solve it by converting the fractions into decimals?

Solve using decimals.
(i) \(6\tfrac{1}{4} = 6.25\); \(2\tfrac{1}{2} = 2.5\). 6.25 ÷ 2.5 = 625/250 = 2.5.
(ii) \(60\tfrac{1}{4} = 60.25\); \(5\tfrac{3}{4} = 5.75\). 60.25 ÷ 5.75 = 6025/575 = 10.478... (exactly 10 remainder 275/575 = 10 + 11/23).

Cyclic Numbers

What are the products? What do you notice?

You get the same number back, but with the digits cycled around! Multiply \(142857\) by 2, 3, 4, 5, 6 and observe.

142857 × ? × 1 = 142857 × 2 = 285714 × 3 = 428571 × 4 = 571428 × 5 = 714285 × 6 = 857142 → digits 1,4,2,8,5,7 → same digits, rotated → same digits, rotated → same digits, rotated → same digits, rotated → same digits, rotated
The cyclic number 142857 — multiplication by 1–6 just rotates its digits!

Are there other such numbers? To find one such number, we can find \(1 \div 17\) in decimal, and use the repeating block.

Historical Note
Are there infinitely many such 'cyclic' numbers? That is, can we keep finding more cyclic numbers of any desired length? However, even today, nearly a century later, this conjecture remains unresolved — despite it not finding its way into many mathematicians' lists of research. In 1927, the Austrian mathematician Emil Artin conjectured (guessed) that there must be infinitely many such numbers. On the question by many mathematicians: this remains one of the most open mysteries in elementary number theory!

Dividend, Divisor, and Quotient

When we divide two counting numbers, the quotient is always less than the dividend. For example, \(128 \div 4 = 32\), and \(32\) (quotient) < \(128\) (dividend).

But what happens when we divide 128 by 0.4?

\(128 \div 0.4 = 320\)

The quotient is greater than the dividend!

Rule — Dividing by a number less than 1
Dividing by a decimal less than 1 gives a quotient larger than the dividend. Dividing by a decimal greater than 1 gives a quotient smaller than the dividend.

Will the quotient always be greater than the dividend when the divisor is a decimal? Try it out with different examples. Describe the relationship between the dividend, divisor, and the quotient. Create a table for capturing this relationship in different situations, like we did for multiplication.

🔵 Math Talk: Does \(2 \div 0.5 = 4\)? Let's check: 2 ÷ 0.5 = 2 × (1/0.5) = 2 × 2 = 4. Yes. Dividing by 0.5 is the same as multiplying by 2.

Figure it Out — Section 4.3

Q1. Express the following fractions in decimal form:
(a) \(\frac{3}{2}\)   (b) \(\frac{13}{4}\)   (c) \(\frac{4}{50}\)   (d) \(\frac{9}{8}\)
(a) 3/2 = 1.5  (b) 13/4 = 3.25  (c) 4/50 = 8/100 = 0.08  (d) 9/8 = 1.125.
Q2. Find the quotients: (a) 24.86 ÷ 1.2   (b) 5.728 ÷ 1.52
(a) 24.86 ÷ 1.2 = 2486/120 = 20.716... ≈ 20.72.
(b) 5.728 ÷ 1.52 = 5728/1520 = 3.768... ≈ 3.77.
Interactive: Decimal Divider
L3 Apply
237 ÷ 8 = 29.625. Quotient (29.625) is less than dividend (237) because divisor (8) > 1.
Activity: Calendar Year Hunt
L3 Apply
Materials: Calculator optional, paper, pencil.
Setting: Earth takes ≈ 365.2422 days to orbit the sun. A regular calendar year is 365 days. We add a leap day every 4 years ⇒ average year = 365.25 days. But we skip leap years in century years (like 1900) unless they are divisible by 400 (like 2000).
  1. Compute average calendar length under the 'add leap every 4 years' rule: 365.25 days.
  2. Difference from actual: 365.25 − 365.2422 = 0.0078 days/year.
  3. Over 1000 years, extra days added = 1000 × 0.0078 = 7.8 days — that's the gap the Gregorian calendar had to fix!
  4. Now apply the rule: skip leap every 100 years but keep every 400th. Average length becomes 365.2425 days.
  5. 1000 × (365.2425 − 365.2422) = 0.3 days over 1000 years — much better!

Decision flowchart: Year divisible by 400? → Yes → 366 days. No → Divisible by 100? → Yes → 365 days. No → Divisible by 4? → Yes → 366 days. No → 365 days.

Competency-Based Questions

Scenario: A farmer harvests 125.6 kg of rice and wants to pack it into small jute bags of 3.2 kg each for the market.
Q1. How many bags can be fully filled, and how much rice is left over?
L3 Apply
125.6 ÷ 3.2 = 1256/32 = 39.25 → 39 full bags, with 0.25 × 3.2 = 0.8 kg remaining.
Q2. The farmer switches to larger bags of 5.6 kg. Analyse whether more or fewer full bags are obtained, and how much waste.
L4 Analyse
125.6 ÷ 5.6 = 1256/56 = 22.428... → 22 full bags, 22 × 5.6 = 123.2 kg; leftover = 125.6 − 123.2 = 2.4 kg. Fewer bags but more waste — larger bags tend to leave bigger remainders.
Q3. Evaluate: A supplier claims "Dividing a decimal by a decimal smaller than 1 must give a whole number." Justify with an example.
L5 Evaluate
Claim false. Example: 1.2 ÷ 0.5 = 2.4 (not a whole number). Counter-example disproves the claim. The quotient is larger than the dividend when divisor < 1, but not necessarily a whole number.
Q4. Design three decimal division problems — one with an exact (terminating) quotient, one with a repeating quotient, and one where quotient > dividend. Solve each.
L6 Create
Sample:
(i) Terminating: 6.25 ÷ 2.5 = 2.5.
(ii) Repeating: 1 ÷ 3 = 0.333... (\(0.\overline{3}\)).
(iii) Quotient > dividend: 12 ÷ 0.4 = 30. All valid — infinite such designs possible.

Assertion–Reason Questions

A: 7.2 ÷ 0.8 = 9.
R: Dividing by a decimal can be done by multiplying both dividend and divisor by the same power of 10 to clear the divisor.
(a) Both true, R explains A.
(b) Both true, R doesn't explain A.
(c) A true, R false.
(d) A false, R true.
Answer: (a). 7.2 ÷ 0.8 = 72 ÷ 8 = 9 (multiplied both by 10). R explains A exactly.
A: The quotient is always smaller than the dividend.
R: Division breaks a quantity into equal parts.
(a) Both true, R explains A.
(b) Both true, R doesn't explain A.
(c) A true, R false.
(d) A false, R true.
Answer: (d). A is false — when dividing by a number less than 1, the quotient is greater. R is true generally, but doesn't support A.
A: 1/3 cannot be written as a terminating decimal.
R: A fraction in lowest terms terminates only if the denominator has no prime factors other than 2 and 5.
(a) Both true, R explains A.
(b) Both true, R doesn't explain A.
(c) A true, R false.
(d) A false, R true.
Answer: (a). Denominator 3 has a prime factor other than 2 or 5, so 1/3 cannot terminate. R explains A perfectly.

Frequently Asked Questions

How do you divide a decimal by a whole number?

Use long division just like with whole numbers, placing the decimal point in the quotient directly above the decimal point in the dividend. For example, 4.8 / 2 = 2.4. NCERT Class 7 Ganita Prakash Part 2 Chapter 4 uses this method.

How do you divide one decimal by another?

Shift the decimal point in the divisor rightwards until it becomes a whole number, shifting the dividend's decimal the same number of places. Then divide as decimal-by-whole-number. For 1.5 / 0.3: shift to 15 / 3 = 5. NCERT Class 7 Chapter 4.

What is 7.2 / 0.6?

Shift both decimals 1 place right: 72 / 6 = 12. So 7.2 / 0.6 = 12. The shift preserves the ratio because both numerator and denominator are multiplied by 10. NCERT Class 7 Part 2 Chapter 4 teaches this.

How does dividing by 0.1 change a number?

Dividing by 0.1 multiplies the number by 10: shift the decimal one place right. For example, 3.4 / 0.1 = 34; 0.07 / 0.1 = 0.7. NCERT Class 7 Chapter 4 highlights this pattern.

What if division doesn't terminate?

Some divisions produce non-terminating decimals like 1 / 3 = 0.333... In Class 7, we usually round or truncate to a stated number of decimal places. NCERT Ganita Prakash Part 2 Chapter 4 explains this approach.

Why shift the decimal point in division?

Shifting makes the divisor a whole number so long division is straightforward. As long as both dividend and divisor are shifted equally, the quotient stays the same. NCERT Class 7 Chapter 4 proves this reasoning.

Frequently Asked Questions — Chapter 4

What is Decimal Division in NCERT Class 7 Mathematics?

Decimal Division is a key concept covered in NCERT Class 7 Mathematics, Chapter 4: Chapter 4. This lesson builds the student's foundation in the chapter by explaining the core ideas with worked examples, definitions, and step-by-step methods aligned to the CBSE curriculum.

How do I solve problems on Decimal Division step by step?

To solve problems on Decimal Division, follow the NCERT method: identify the given quantities, choose the relevant formula or theorem, substitute values carefully, and simplify. Class 7 exercises gradually increase in difficulty — start with solved NCERT examples before attempting exercise questions, and always verify your answer by substitution or diagram.

What are the most important formulas for Chapter 4: Chapter 4?

The essential formulas of Chapter 4 (Chapter 4) are listed in the chapter summary and highlighted throughout the lesson in formula boxes. Memorise them and practise at least 2–3 problems per formula. CBSE board exams frequently test direct application as well as combined use of multiple formulas from this chapter.

Is Decimal Division important for the Class 7 board exam?

Decimal Division is part of the NCERT Class 7 Mathematics syllabus and appears in CBSE board exams. Questions typically include short-answer, long-answer, and competency-based items. Review the NCERT examples, exercise questions, and previous-year board problems on this topic to prepare confidently.

What mistakes should students avoid in Decimal Division?

Common mistakes in Decimal Division include skipping steps, misapplying formulas, sign errors, and losing track of units. Write each step clearly, double-check algebraic manipulations, and re-read the question after solving to verify that your answer matches what was asked.

Where can I find more NCERT practice questions on Decimal Division?

End-of-chapter NCERT exercises for Decimal Division cover all difficulty levels tested in CBSE exams. After completing them, try the examples again without looking at the solutions, attempt the NCERT Exemplar questions for Chapter 4, and solve at least one previous-year board paper to consolidate your understanding.

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