CBSE
The Fraction That Becomes 9/11: Class 10 Linear Equations Word Problems
Call the fraction x/y. Adding 2 to both parts gives 11x - 9y = -4, and adding 3 to both parts gives 6x - 5y = -3. Eliminating y gives x = 7, then y = 9, so the fraction is 7/9. Check: 9/11 and 10/12 = 5/6, so both conditions hold.
A fraction becomes 9/11 if 2 is added to both the numerator and the denominator: this is one of the most-searched lines from Class 10 Chapter 3, Pair of Linear Equations in Two Variables. The trick is to stop thinking of it as a fraction problem at all. Call the numerator x and the denominator y, cross multiply each condition, and you get two straight linear equations you can solve by elimination. The full solution is below, followed by four more word problems of the kinds that turn up again and again: two-digit numbers, ages, taxi fares with a fixed charge, and hostel charges with a daily rate.
Turning a sentence into a pair of equations
Every Chapter 3 word problem hides two unknowns and two facts about them. The whole game is naming the unknowns precisely and translating each fact separately. For fractions, call the numerator x and the denominator y, never the fraction f, because the conditions act on the parts. For two-digit numbers, the number itself is 10x + y where x is the tens digit. For fare and hostel questions, the unknowns are a fixed charge and a per-unit rate. Then take the sentences one at a time: becomes 9/11 when 2 is added to both is exactly (x + 2)/(y + 2) = 9/11, nothing more. Cross multiply, collect terms on one side, and you have a clean linear equation. Do the same with the second sentence, and the setup is finished before any solving starts.
Elimination or substitution?
Both methods are accepted and both earn full marks, so the choice is purely practical. Substitution shines when one equation already gives a variable alone, like y - x = 3 from the digit problem, where y = x + 3 drops straight into the other equation. Elimination is better when both equations are in the form ax + by = c with awkward coefficients, which is exactly the shape the fraction problem produces: 11x - 9y = -4 and 6x - 5y = -3. There, multiplying to match the y coefficients at 45 kills y in one subtraction. A useful habit is to glance at both equations before choosing and ask which variable is cheapest to remove. Whichever route you take, keep the multiplication step visible, because that line carries a method mark of its own.
The verification line examiners look for
A pair of linear equations gives you two numbers, and the fastest way to be certain of them is to push both back through the original words of the question, not through your own equations. For the fraction problem, that means actually computing (7 + 2)/(9 + 2) and (7 + 3)/(9 + 3) and confirming 9/11 and 5/6 appear. This catches the most common failure in the whole chapter, a single sign flipped during rearrangement, which produces tidy-looking but wrong values that satisfy your corrupted equation perfectly. It also earns the closing statement mark that CBSE schemes routinely attach to word problems. One sentence is enough: both conditions are satisfied, so the fraction is 7/9. If a check fails, do not start over; recheck the rearrangement lines first, because that is where the slip nearly always lives.
Worked questions, step by step
There is a fraction with this property: add 2 to both its numerator and its denominator and it becomes 9/11, but add 3 to both instead and it becomes 5/6. Find the fraction.
- Given: (x + 2)/(y + 2) = 9/11 and (x + 3)/(y + 3) = 5/6, where the fraction is x/y. To find: x and y.
- Cross multiply the first condition: 11(x + 2) = 9(y + 2), which gives 11x - 9y = -4.
- Cross multiply the second condition: 6(x + 3) = 5(y + 3), which gives 6x - 5y = -3.
- Eliminate y: multiply the first equation by 5 and the second by 9 to get 55x - 45y = -20 and 54x - 45y = -27.
- Subtract: (55x - 54x) = -20 - (-27), so x = 7.
- Substitute into 6x - 5y = -3: 42 - 5y = -3, so 5y = 45 and y = 9.
- Check both conditions: (7 + 2)/(9 + 2) = 9/11 and (7 + 3)/(9 + 3) = 10/12 = 5/6. Both hold.
- Answer: the fraction is 7/9.
Answer: The fraction is 7/9
Where marks slip: Nearly every dropped mark here is a sign slip while rearranging 11x + 22 = 9y + 18 into 11x - 9y = -4, so move the terms one at a time and then verify both original conditions at the end.
Try one yourself: A fraction becomes 3/4 when 2 is added to both numerator and denominator, and 2/3 when 1 is subtracted from both. Find the fraction. (Answer: 7/10)
The digits of a two-digit number add up to 11. If the digits are swapped, the new number is 27 more than the original one. What is the original number?
- Given: digit sum 11, and reversing the digits raises the number by 27. To find: the number. Let the tens digit be x and the units digit be y, so the number is 10x + y.
- First equation from the digit sum: x + y = 11.
- Second equation from reversing: (10y + x) - (10x + y) = 27, which simplifies to 9(y - x) = 27, so y - x = 3.
- Add the equations x + y = 11 and y - x = 3: 2y = 14, so y = 7 and then x = 4.
- Check: the number 47 reversed is 74, and 74 - 47 = 27, with digits summing to 11.
- Answer: the original number is 47.
Answer: The number is 47
Where marks slip: Writing the two-digit number as xy or x + y instead of 10x + y is the error examiners see most; the place-value form 10x + y is the whole point of the question.
Try one yourself: A two-digit number has digits adding to 9, and the number formed by reversing its digits is 45 less than the original. Find the original number. (Answer: 72)
A father is currently four times as old as his daughter. Six years from now, he will be three times as old as she will be then. Find their present ages.
- Given: father = 4 × daughter now, and in 6 years father = 3 × daughter. To find: both ages. Let the daughter be x years old, so the father is 4x years old.
- In 6 years: 4x + 6 = 3(x + 6).
- Expand and solve: 4x + 6 = 3x + 18, so x = 12.
- Then the father is 4 × 12 = 48 years old.
- Check: in 6 years they will be 54 and 18, and 54 = 3 × 18.
- Answer: the daughter is 12 years old and the father is 48 years old.
Answer: Daughter = 12 years, father = 48 years
Where marks slip: The near-universal slip is writing 4x + 6 = 3x + 6, forgetting that the daughter also ages 6 years; the +6 must sit inside the bracket on the right-hand side.
Try one yourself: A mother is five times as old as her son. In 4 years she will be four times as old. Find their present ages. (Answer: son 12 years, mother 60 years)
A taxi service charges a fixed booking fee plus a set rate for every kilometre. A 10 km journey costs Rs 205 and a 15 km journey costs Rs 280. Find the fixed fee and the per-km rate, and then the cost of a 25 km journey.
- Given: 10 km costs Rs 205 and 15 km costs Rs 280. To find: the fixed fee, the rate, and the 25 km fare. Let the fixed fee be Rs f and the rate be Rs r per km.
- Equations: f + 10r = 205 and f + 15r = 280.
- Subtract the first from the second: 5r = 75, so r = 15.
- Substitute back: f + 10 × 15 = 205, so f = 205 - 150 = 55.
- Fare for 25 km = 55 + 25 × 15 = 55 + 375 = Rs 430.
- Answer: fixed fee Rs 55, rate Rs 15 per km, and a 25 km journey costs Rs 430.
Answer: Fixed fee = Rs 55, rate = Rs 15 per km, 25 km fare = Rs 430
Where marks slip: Subtracting the equations the wrong way round gives r = -15 and everything collapses from there, so subtract the smaller-distance equation from the larger one and sanity-check that both unknowns come out positive.
Try one yourself: A cab company charges a fixed fee plus a per-km rate; 8 km costs Rs 156 and 12 km costs Rs 220. Find the fixed fee and the rate. (Answer: Rs 28 fixed and Rs 16 per km)
A hostel bills each student a fixed monthly charge plus a daily food charge. A student who took food for 22 days paid Rs 4,250 in total, while another who took food for 28 days paid Rs 5,150. Find the fixed charge and the cost of food per day.
- Given: 22 days costs Rs 4,250 and 28 days costs Rs 5,150. To find: the fixed monthly charge and the per-day food cost. Let the fixed charge be Rs f and the food cost Rs d per day.
- Equations: f + 22d = 4250 and f + 28d = 5150.
- Subtract: 6d = 900, so d = 150.
- Substitute: f + 22 × 150 = 4250, so f = 4250 - 3300 = 950.
- Check with the second student: 950 + 28 × 150 = 950 + 4200 = Rs 5,150, which matches.
- Answer: the fixed charge is Rs 950 per month and food costs Rs 150 per day.
Answer: Fixed charge = Rs 950, food cost = Rs 150 per day
Where marks slip: This question is almost always worth a verification mark, so after finding f and d, plug them into the equation you did not use for substitution and state that the totals match.
Try one yourself: A hostel charges a fixed amount plus a per-day mess fee; 20 days costs Rs 3,900 and 26 days costs Rs 4,860. Find both charges. (Answer: fixed Rs 700, mess Rs 160 per day)
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Solve my question →Quiz me on this topicQuestions students ask
What fraction becomes 9/11 when 2 is added to both the numerator and denominator?
The fraction is 7/9. Adding 2 to both parts gives 9/11 directly, and adding 3 to both gives 10/12, which simplifies to 5/6, matching the second condition in the question. The two conditions convert to the linear equations 11x - 9y = -4 and 6x - 5y = -3, which solve to x = 7 and y = 9.
Which is better for this chapter, substitution or elimination?
Both earn full marks, so choose by shape. If a variable is already almost alone, like y = x + 3, substitute. If both equations look like ax + by = c with clumsy coefficients, eliminate by matching one coefficient and subtracting. The fraction problem is a textbook elimination case, since multiplying by 5 and 9 matches the y terms at 45y.
Do I need to verify my answer at the end?
Yes, and against the original wording rather than your own equations. Substituting 7 and 9 back into the story confirms both conditions and catches sign slips made during rearrangement, which are the most common error in this chapter. CBSE schemes also routinely reserve a mark for the final statement, so the checking sentence pays for itself.
Can I solve these word problems graphically?
In principle yes, since every pair of linear equations is two lines whose intersection is the answer, but in practice it is slow and imprecise for values like 7 and 9 unless the question demands a graph. Use algebra unless the paper explicitly asks for a graphical solution, and expect graph questions to be flagged clearly with grid space provided.