Chapter 1.5 — Exercise 1.3 — Decimal Representation

Question 1 — Terminating Decimals to p/q

A terminating decimal ends after a finite number of digits. Write the digits (ignoring the decimal point) as the numerator, and use 10 raised to the number of decimal places as the denominator, then simplify.

• 0.57 → 57/100 (already in lowest terms).
• 0.176 → 176/1000, which simplifies to 22/125 after dividing both by 8.
• 1.00001 → 100001/100000 (5 decimal places give a denominator of 100000).
• 25.125 → 25125/1000, which simplifies to 201/8 after dividing both by 125.

Question 2 — Repeating Decimals to p/q

Repeating (recurring) decimals need a different rule, and it depends on whether the repeating block starts right after the decimal point or only after some non-repeating digits.

• 0.9̄ (9 repeating forever) → 9/9 = 1. This is the classic proof that a string of recurring 9s equals the next whole number.
• 0.5̄7̄ (both digits repeat) → 57/99, which simplifies to 19/33.
• 0.7̄2̄9̄ (all three digits repeat) → (729 − 7)/990 = 722/990, which simplifies to 361/495. Subtracting the 7 accounts for the single non-repeating digit before the block restarts its cycle in the derivation.
• 12.2̄8̄ (2 is fixed, 8 repeats) → the decimal part 0.2̄8̄ = (28 − 2)/90 = 26/90 = 13/45, so the full value is 12 + 13/45 = 553/45.

Question 3 — Algebraic Substitution

Find (x + y) ÷ (x − y) for two given pairs of values, computing each sum and difference with cross-multiplication first.

• x = 5/2, y = −3/4: x + y = 10/4 − 3/4 = 7/4; x − y = 10/4 + 3/4 = 13/4. So (7/4) ÷ (13/4) = 7/13.
• x = 1/4, y = 3/2: x + y = 1/4 + 6/4 = 7/4; x − y = 1/4 − 6/4 = −5/4. So (7/4) ÷ (−5/4) = −7/5.

Questions 4 to 10 — Word Problems

• Q4 — Sum ÷ product: the sum of −13/5 and 12/7 is −31/35; the product of −13/7 and −1/2 is 13/14. Dividing the sum by the product: (−31/35) ÷ (13/14) = −62/65.
• Q5 — Finding an unknown number: if 2/5 of a number exceeds 1/7 of the same number by 36, then 2x/5 − x/7 = 36, which simplifies to 9x/35 = 36, giving x = 140.
• Q6 — Rope problem: two pieces measuring 2⅗ m and 3 3/10 m are cut from an 11 m rope. Together they measure 13/5 + 33/10 = 59/10 m, leaving 11 − 59/10 = 5 1/10 m of rope.
• Q7 — Cost per metre: 7⅔ m of cloth costs ₹12¾, so the rate per metre is (51/4) ÷ (23/3) = 153/92 = ₹1 61/92.
• Q8 — Area of a park: a park 18⅗ m long (93/5 m) and 8⅔ m broad (26/3 m) has area 93/5 × 26/3 = 806/5 = 161⅕ sq. m.
• Q9 — Finding the divisor: if −33/16 divided by an unknown number gives −11/4, the number is (−33/16) ÷ (−11/4) = 3/4.
• Q10 — Cloth per trouser: 64 m of cloth stitches 36 identical trousers, so each needs 64/36 = 16/9 = 1 7/9 m.

Question 11 — A Repeating Decimal, Start to Finish

Write 0.363636… in the form p/q and find p + q. Since "36" is the two-digit block that repeats immediately after the decimal point, this is a pure recurring decimal: 0.3̄6̄ = 36/99, which simplifies to 4/11 after dividing both by 9. Here p = 4 and q = 11, so p + q = 15.

Terminating vs. Recurring — Telling Them Apart in Advance

Before converting any decimal, it helps to know which rule applies without trial and error. A fraction p/q in lowest terms gives a terminating decimal exactly when the denominator q has no prime factors other than 2 and 5 (for example, 1/8 = 0.125, since 8 = 2³). If q has any other prime factor — 3, 7, 11, and so on — the decimal will recur instead. This is why 1/125 (125 = 5³) terminates at 0.008, while 1/33 (33 = 3 × 11) recurs as 0.0303…. Checking the denominator's prime factors first tells you immediately which of the two conversion methods in Question 2 to reach for.

Where the Recurring-Decimal Rule Actually Comes From

The 9s-in-the-denominator rule isn't an arbitrary shortcut — it comes directly from algebra, and seeing the derivation once makes it far easier to trust and remember. Take x = 0.5̄7̄ (57 repeating). Multiplying by 100 shifts the decimal point past one full repeating block: 100x = 57.5̄7̄. Subtracting the original equation cancels the repeating part entirely: 100x − x = 57.5̄7̄ − 0.5̄7̄, so 99x = 57, giving x = 57/99 = 19/33 — matching the rule exactly, with the 99 coming from 100 − 1 (two 9s for a two-digit repeating block). The same subtraction trick, using the right power of 10 to align the repeating blocks, is what justifies every case in Question 2, including the mixed-recurring cases where a non-repeating digit or two sits before the block begins — the only difference is an extra subtraction step to strip those leading digits out first. It's a good exercise to redo the 12.2̄8̄ conversion from Question 2 using this 100x − 10x style approach yourself, purely as a check that the shortcut formula and the algebra agree — 100x accounts for shifting past the non-repeating "2" and the first repeat of "8", while 10x shifts past just the "2", so subtracting the two cancels every repeating digit and leaves a clean integer equation to solve for x.

Three Recurring Slip-Ups

• Missing the 0s in a mixed recurring denominator. If any digits before the repeating block don't repeat, the denominator needs a 0 for each of those digits in addition to the 9s — leaving them out gives a wrong fraction that doesn't simplify cleanly.
• Computing with mixed numbers directly. Always convert 2⅗ or 8⅔ to an improper fraction (13/5, 26/3) before multiplying or dividing — mixed-number arithmetic without this step is a frequent source of errors.
• Inverting the wrong fraction when dividing. Dividing by a fraction means multiplying by its reciprocal — double-check which of the two fractions gets flipped.

Terminating and Recurring, Reframed in Class 9

Recognising which decimals terminate and which recur is exactly the distinction Class 9 builds on when it separates rational numbers (terminating or recurring decimals) from irrational numbers (non-terminating, non-recurring decimals) in Real Numbers. The word-problem techniques here also carry forward into Linear Equations in One Variable. For the properties this exercise assumes you already know, revisit Exercise 1.1 and the Introduction to Rational Numbers.