MathsCBSE Class 10 · Pair of Linear Equations in Two Variables

Word Problems on Pair of Linear Equations, Class 10

Updated 14 October 2026 · Fact-checked

Word problems on a pair of linear equations describe a real situation with two unknowns and two conditions. You define the unknowns, turn each condition into a linear equation, and solve by substitution or elimination. The board asks these on ages, digits, fractions, speed, money, fares and geometry, as 3-mark, 5-mark and case-based questions.

Understand Word Problems on Linear Equations

A word problem gives you a situation in sentences instead of equations. If the situation has two unknown quantities and gives two separate conditions linking them, you can write each condition as a linear equation and solve the pair. Your real job is translation: turning English into algebra.

Why does it work? Each condition is a statement that is true for the actual values. So the actual values satisfy both equations at once, which means they are the solution of the pair. If you translate correctly, the algebra must give the true answer.

This topic ties the whole chapter together. You use substitution or elimination to solve. The ratio conditions tell you whether a situation is even possible: if the two conditions give parallel lines, no answer exists, and if they give the same line, the second condition adds no new information.

The method holds only when each condition is linear in the unknowns: the unknowns appear only to the power 1 and are never multiplied together. In a fraction problem, $x$ and $y$ are the numerator and the denominator, treated as two separate unknowns. A condition like $\frac{x + 1}{y + 4} = \frac{1}{2}$ becomes $2 (x + 1) = y + 4$ once you multiply both sides by the denominators, and that is linear in $x$ and $y$ . You must also have two independent conditions; one condition alone allows infinitely many answers.

Finally, answers must make sense. Ages, prices, speeds and lengths must be positive. Digits must be whole numbers from 0 to 9, and the tens digit cannot be 0. Always check your answer against the original words, not just your equations.

Word Problems on Linear Equations: formulas to remember

Two-digit number: $N = 10 x + y, reversed = 10 y + x$; x is the tens digit, y the units digit. Note N − reversed = 9(x − y).
Age shift: $t years ago : x - t, y - t; t years later : x + t, y + t$; Shift every person's age by the same t.
Fraction: $fraction = \frac{x}{y}, \frac{x + p}{y + q} = \frac{m}{n} \Rightarrow n (x + p) = m (y + q)$; Cross-multiply to get a linear equation.
Speed, distance, time: $distance = speed \times time$; Same direction: relative speed u − v. Opposite directions: u + v.
Boat in a stream: $downstream = u + v, upstream = u - v$; u is speed in still water, v speed of the stream.
Fixed plus variable charge: $Cost = f + r \times n$; f is the fixed charge, r the rate per unit, n the number of units.
Simple interest for one year: $SI = \frac{P \times R}{100}$; Used in investment word problems.
Conditions for number of solutions: $\frac{a _{1}}{a _{2}} \neq = \frac{b _{1}}{b _{2}} : unique; \frac{a _{1}}{a _{2}} = \frac{b _{1}}{b _{2}} \neq = \frac{c _{1}}{c _{2}} : none; \frac{a _{1}}{a _{2}} = \frac{b _{1}}{b _{2}} = \frac{c _{1}}{c _{2}} : infinitely many$; In a word problem: unique means one definite answer; none means the two conditions contradict each other; infinitely many means the second condition only repeats the first. In 'find k' questions, set the right ratios equal (or unequal) and solve for k.

How to solve Word Problems on Linear Equations questions

Use the same four stages every time: define, translate, solve, check. Write each equation next to the sentence it came from.

1Read the problem twice and identify the two unknowns being asked for.
2Write 'Let ... be x and ... be y' with units, for example 'let the cost of one pen be Rs x'.
3Translate the first condition into an equation and number it (1). Simplify to the form ax + by = c.
4Translate the second condition into equation (2) the same way.
5Solve by elimination or substitution, whichever is quicker for these coefficients.
6Check both values in the original words (not just the equations) and reject impossible values.
7Write the final answer in a sentence with units, answering exactly what was asked.

Word Problems on Linear Equations in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Ages

Imagine your grandfather says at Diwali dinner that the sum of his age and your father's age is 110 years. He adds that ten years ago, he was twice as old as your father was then.

Find the present ages of your grandfather and your father. In how many years will the grandfather's age be 1.5 times the father's age?

Try it first, then show the solutionSolution

Let grandfather's age be x years and father's age be y years.
x + y = 110 ... (1)
Ten years ago: x − 10 = 2(y − 10), so x − 2y = −10 ... (2)
(1) − (2): 3y = 120, so y = 40.
x = 110 − 40 = 70.
Check: 60 = 2 × 30. Correct.
Let t years: 70 + t = 1.5(40 + t), so 70 + t = 60 + 1.5t.
0.5t = 10, so t = 20.

Answer: Grandfather is 70 years, father is 40 years; after 20 years the ratio becomes 1.5.

Memory hook: Family ages: shift everyone by the same years.

Money (fixed plus variable charge)

Suppose a mobile data plan charges a fixed monthly fee plus a rate per GB used. One month you use 10 GB and pay Rs 250. Another month you use 16 GB and pay Rs 340.

Find the fixed fee and the rate per GB, and the bill for 22 GB. A second plan charges a fixed fee of Rs 160 plus Rs 12 per GB. For how many GB do the two plans cost the same, and which plan is cheaper for 30 GB?

Try it first, then show the solutionSolution

Let fixed fee be Rs f and rate Rs r per GB.
f + 10r = 250 ... (1)
f + 16r = 340 ... (2)
(2) − (1): 6r = 90, so r = 15.
f = 250 − 150 = 100.
22 GB: 100 + 22 × 15 = 100 + 330 = 430.
Equal cost: $100 + 15 n = 160 + 12 n$ , so $3 n = 60$ and $n = 20$ GB (both cost Rs 400).
30 GB: first plan $100 + 450 = 550$ ; second plan $160 + 360 = 520$ . The second plan is cheaper by Rs 30.

Answer: Fixed fee Rs 100, rate Rs 15 per GB; 22 GB costs Rs 430. The plans cost the same (Rs 400) at 20 GB; for 30 GB the second plan is cheaper (Rs 520 against Rs 550).

Memory hook: Bill = fixed fee + rate × units: subtract two bills to kill the fixed fee, then set two plans equal to find the break-even point.

Tricks to remember Word Problems on Linear Equations

Short memory aids. Say each one aloud once, then use it on a question the same day.

Define, Translate, Solve, Check
Remember DTSC: Do The Sums Carefully.
Use it when: Follow these four stages in every word problem so you never skip the check.
Reverse a number, nine times the gap
A two-digit number minus its reverse is 9 times the difference of its digits.
Use it when: 'Digits reversed when 27 is added' instantly gives y − x = 3.
Same way subtract, face to face add
Same direction: speeds subtract. Towards each other: speeds add.
Use it when: Speed and boat problems.
Ages move together
Time passes for everyone equally.
Use it when: Add or subtract t from every person's age, never just one.

Word Problems on Linear Equations: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Cost of pens and notebooks · 3 marks · Short answerBoard level

The cost of 4 pens and 3 notebooks is Rs 190, and the cost of 3 pens and 5 notebooks is Rs 225. Find the cost of one pen and one notebook.

Try it first, then show the solutionSolution

Let one pen cost Rs x and one notebook Rs y. (½)
4x + 3y = 190 ... (1); 3x + 5y = 225 ... (2) (½)
(1) × 3: 12x + 9y = 570; (2) × 4: 12x + 20y = 900. (½)
Subtract: 11y = 330, so y = 30. (½)
From (1): 4x = 190 − 90 = 100, so x = 25. Check in (2): 75 + 150 = 225. (1)

Answer: One pen costs Rs 25 and one notebook costs Rs 30.

Example 2: Two-digit number · 3 marks · Short answerBoard level

The sum of the digits of a two-digit number is 11. The number obtained by reversing the digits exceeds the original number by 27. Find the number.

Try it first, then show the solutionSolution

Let tens digit be x, units digit y; number = 10x + y. (½)
x + y = 11 ... (1) (½)
(10y + x) − (10x + y) = 27, so 9y − 9x = 27, y − x = 3 ... (2) (1)
Add (1) and (2): 2y = 14, y = 7; x = 4. (½)
Number = 47. Check: 74 − 47 = 27, 4 + 7 = 11. (½)

Answer: The number is 47.

Example 3: Boat upstream and downstream · 5 marks · Long answerBoard level

A boat goes 36 km downstream in 3 hours and 24 km upstream in 4 hours. Find the speed of the boat in still water and the speed of the stream. How long will the boat take to go 30 km downstream and come back to its starting point?

Try it first, then show the solutionSolution

Let the speed of the boat in still water be u km/h and of the stream v km/h. Downstream speed = u + v, upstream speed = u − v. (½)
Downstream: 36 km in 3 h, so u + v = 36 ÷ 3 = 12 ... (1) (1)
Upstream: 24 km in 4 h, so u − v = 24 ÷ 4 = 6 ... (2) (1)
Add (1) and (2): 2u = 18, so u = 9. (1)
From (1): v = 12 − 9 = 3. Check: 3 × 12 = 36 and 4 × 6 = 24. (½)
Round trip: downstream 30 ÷ 12 = 2.5 h, upstream 30 ÷ 6 = 5 h; total = 7.5 h. (1)

Answer: Boat in still water 9 km/h, stream 3 km/h; the 30 km round trip takes 7 hours 30 minutes.

Example 4: Speed and time: the product cancels · 5 marks · Long answerHarder board level

A train covers a fixed distance at a uniform speed. If it were 6 km/h faster, it would take 4 hours less. If it were 6 km/h slower, it would take 6 hours more. Find its speed, the time taken and the distance. At what speed would it cover the distance in 18 hours?

Try it first, then show the solutionSolution

Let the speed be $x$ km/h and the time $y$ hours, so the distance is $x y$ km. (½)
Faster: $(x + 6) (y - 4) = x y$ gives $x y - 4 x + 6 y - 24 = x y$ , so $- 4 x + 6 y = 24$ , i.e. $2 x - 3 y = - 12$ ... (1). The $x y$ terms cancel, so the equation is linear. (1)
Slower: $(x - 6) (y + 6) = x y$ gives $6 x - 6 y - 36 = 0$ , i.e. $x - y = 6$ ... (2). (1)
From (2), $x = y + 6$ . Substitute in (1): $2 y + 12 - 3 y = - 12$ , so $y = 24$ and $x = 30$ . (1)
Distance $= 30 \times 24 = 720$ km. Check: $36 \times 20 = 720$ and $24 \times 30 = 720$ . Speed for 18 hours $= 720 \div 18 = 40$ km/h. (1½)

Answer: Speed 30 km/h, time 24 hours, distance 720 km; it would need 40 km/h to finish in 18 hours.

Example 5: Case study: school canteen · 4 marks · Case-basedHOTS / competency

Suppose a school canteen sells samosas and juice. On Monday, 50 samosas and 30 juices earned Rs 1350. On Tuesday, 40 samosas and 45 juices earned Rs 1500.
(i) Write the pair of equations taking samosa price Rs x and juice price Rs y. (1 mark)
(ii) Is a unique pair of prices possible? Justify using ratios. (1 mark)
(iii) Find the prices, and Wednesday's earnings if 60 samosas and 20 juices are sold. (2 marks)

Try it first, then show the solutionSolution

(i) 50x + 30y = 1350, i.e. 5x + 3y = 135; 40x + 45y = 1500, i.e. 8x + 9y = 300. (1)
(ii) 5/8 ≠ 3/9, so lines intersect: unique solution. (1)
(iii) First equation × 3: 15x + 9y = 405. Subtract 8x + 9y = 300: 7x = 105, x = 15; 3y = 135 − 75 = 60, y = 20. (1)
Wednesday: 60 × 15 + 20 × 20 = 900 + 400 = Rs 1300. Check Monday: 750 + 600 = 1350. (1)

Answer: Samosa Rs 15, juice Rs 20; Wednesday earnings Rs 1300.

Example 6: Fraction problem · 3 marks · Short answerBoard level

The denominator of a fraction is 4 more than twice its numerator. When both the numerator and the denominator are decreased by 6, the new denominator is 12 times the new numerator. Find the fraction.

Try it first, then show the solutionSolution

Let the fraction be x/y, where x is the numerator and y the denominator. (½)
First condition: y = 2x + 4 ... (1) (½)
Second condition: y − 6 = 12(x − 6), so 12x − y = 66 ... (2) (½)
Substitute (1) in (2): 12x − (2x + 4) = 66, so 10x = 70, x = 7. (1)
y = 2(7) + 4 = 18. Check: (7 − 6)/(18 − 6) = 1/12, and 18 = 2 × 7 + 4. (½)

Answer: The fraction is 7/18.

Example 7: A 2-mark notes problem · 2 marks · Very short answerBoard level

A purse holds only Rs 10 and Rs 20 notes. There are 25 notes in all, worth Rs 380. How many notes of each kind are there?

Try it first, then show the solutionSolution

Let there be $x$ notes of Rs 10 and $y$ of Rs 20: $x + y = 25$ and $10 x + 20 y = 380$ , i.e. $x + 2 y = 38$ . (½ + ½)
Subtract: $y = 13$ , so $x = 12$ . Check: $120 + 260 = 380$ . (1)

Answer: 12 notes of Rs 10 and 13 notes of Rs 20.

Example 8: Rectangle: changes in area · 5 marks · Long answerHarder board level

If the length of a rectangle is reduced by 5 units and its breadth is increased by 3 units, the area decreases by 9 square units. If the length is increased by 3 units and the breadth by 2 units, the area increases by 67 square units. Find the length and breadth of the rectangle.

Try it first, then show the solutionSolution

Let the length be x units and the breadth y units, so the area is xy square units. (½)
(x − 5)(y + 3) = xy − 9 gives xy + 3x − 5y − 15 = xy − 9, so 3x − 5y = 6 ... (1) (1)
(x + 3)(y + 2) = xy + 67 gives xy + 2x + 3y + 6 = xy + 67, so 2x + 3y = 61 ... (2) (1)
(1) × 3: 9x − 15y = 18; (2) × 5: 10x + 15y = 305. (½)
Add: 19x = 323, so x = 17. (1)
From (1): 51 − 5y = 6, so y = 9. Check: area 153; 12 × 12 = 144 = 153 − 9 and 20 × 11 = 220 = 153 + 67. (1)

Answer: Length 17 units and breadth 9 units (area 153 square units).

Example 9: Finding k in a shopping context · 3 marks · Short answerHarder board level

One record says 2 kg of apples and 3 kg of grapes cost Rs 390. A second record says 4 kg of apples and $k$ kg of grapes cost Rs 700 at the same prices. (i) For which value of $k$ do the two records contradict each other? (ii) If in fact $k = 5$ , find the price per kg of each fruit.

Try it first, then show the solutionSolution

Pair: $2 a + 3 g = 390$ and $4 a + k g = 700$ . Contradiction means no solution: $\frac{2}{4} = \frac{3}{k} \neq = \frac{390}{700}$ . (½)
$\frac{3}{k} = \frac{1}{2}$ gives $k = 6$ , and $\frac{390}{700} = \frac{39}{70} \neq = \frac{1}{2}$ , so the records contradict when $k = 6$ . (1)
With $k = 5$ : double the first, $4 a + 6 g = 780$ ; subtract $4 a + 5 g = 700$ : $g = 80$ . Then $2 a = 390 - 240 = 150$ , so $a = 75$ . (1)
Check: $4 (75) + 5 (80) = 300 + 400 = 700$ . (½)

Answer: (i) $k = 6$ . (ii) Apples Rs 75 per kg, grapes Rs 80 per kg.

Where students lose marks in Word Problems on Linear Equations

The slips examiners see most often, and what to write instead.

Mistake: Shifting only one person's age
Students forget time passes for everyone.
Wrong working
Five years ago: x − 5 = 3y
Right working
x − 5 = 3(y − 5)
Do this: Subtract t from every age, then apply the ratio.
Usually costs: Usually the whole equation mark and the answer.
Mistake: Writing the two-digit number as x + y
Confusing the number with its digit sum.
Wrong working
Number = xy or x + y
Right working
Number = 10x + y
Do this: Write 10 × tens digit + units digit.
Usually costs: All marks after the set-up.
Mistake: Wrong relative speed
Mixing up same and opposite directions.
Wrong working
Towards each other: 2(u − v) = 180
Right working
Towards each other: 2(u + v) = 180
Do this: Facing: add speeds; chasing: subtract.
Usually costs: 1-2 marks and a wrong answer.
Mistake: No final statement or units
Stopping at x = 15, y = 20.
Wrong working
x = 15, y = 20
Right working
A samosa costs Rs 15 and a juice Rs 20.
Do this: Answer in words with units.
Usually costs: Often ½ mark.

Exam tips

Write a clear 'Let...' line; examiners award a mark for correct variables and equations even if later arithmetic slips.
Number your equations (1) and (2) and show which operation you perform.
Check your answer against the original wording; the check also catches translation errors.
In case studies, keep the same variables in every part since later parts reuse the solution.
Reject impossible answers such as negative ages or digits above 9, and say why.

Practice questions on Word Problems on Linear Equations

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 3 marks · Short answerHarder board level
A two-digit number is 3 more than 4 times the sum of its digits. If 18 is added to the number, its digits are reversed. Find the number. Then show that the sum of the number and its reversal is 11 times the sum of the digits.
Hint
Write the number as $10 x + y$ . The last part works for every two-digit number; show why.
Answer
$10 x + y = 4 (x + y) + 3$ gives $2 x - y = 1$ . $10 x + y + 18 = 10 y + x$ gives $x - y = - 2$ . Subtract: $x = 3$ , so $y = 5$ . The number is 35 (check: $4 \times 8 + 3 = 35$ and $35 + 18 = 53$ ). Sum $= 35 + 53 = 88 = 11 \times 8$ ; in general $(10 x + y) + (10 y + x) = 11 (x + y)$ .
Q2 · 3 marks · Short answerBoard level
In a cyclic quadrilateral ABCD, angle A = (2x + 4)°, angle B = (y + 3)°, angle C = (2y + 10)° and angle D = (4x − 5)°. Find all four angles.
Hint
Opposite angles of a cyclic quadrilateral add up to 180°: use A + C and B + D.
Answer
A + C = 180: 2x + 2y + 14 = 180, so x + y = 83 ... (1). B + D = 180: 4x + y − 2 = 180, so 4x + y = 182 ... (2). (2) − (1): 3x = 99, x = 33; y = 50. Angles: A = 70°, B = 53°, C = 110°, D = 127°. Check: 70 + 110 = 180, 53 + 127 = 180.
Q3 · 5 marks · Long answerBoard level
Two places A and B are 100 km apart. Two cars start at the same time from A and B. Moving in the same direction they meet in 5 hours; moving towards each other they meet in 1 hour. Find their speeds.
Hint
Same direction: 5(u − v) = 100. Opposite: u + v = 100.
Answer
Let the faster car (from A) have speed $u$ km/h and the other $v$ km/h. (½)
Same direction: the faster car gains 100 km in 5 hours, so $5 (u - v) = 100$ , i.e. $u - v = 20$ ... (1). (1)
Towards each other: together they cover 100 km in 1 hour, so $u + v = 100$ ... (2). (1)
Add (1) and (2): $2 u = 120$ , so $u = 60$ ; then $v = 40$ . (1)
Check: in 5 hours, $300 - 200 = 100$ km gained; in 1 hour, $60 + 40 = 100$ km. (1)
The speeds are 60 km/h and 40 km/h. (½)
Q4 · 5 marks · Long answerHarder board level
A man invests a total of Rs 22000 in two schemes paying 5% and 8% simple interest per annum. His total interest for one year is Rs 1460. Find the amount in each scheme. If he had swapped the two amounts between the schemes, by how much would his yearly interest have changed?
Hint
x + y = 22000 and 5x/100 + 8y/100 = 1460, i.e. 5x + 8y = 146000.
Answer
Multiply the first equation by 5: 5x + 5y = 110000. Subtract from 5x + 8y = 146000: 3y = 36000, y = 12000, x = 10000. So Rs 10000 at 5% and Rs 12000 at 8% (check: 500 + 960 = 1460). Swapped: 5% of 12000 = 600 and 8% of 10000 = 800, total Rs 1400. The interest would fall by Rs 60.
Q5 · 3 marks · Short answerHOTS / competency
The cost of 2 notebooks and 3 pens is Rs 40. A shopkeeper's second bill claims that 4 notebooks and k pens cost Rs 80. For which value of k is the price of each item not uniquely determined? Is there any value of k for which the two bills contradict each other? Justify using ratios.
Hint
Write 2x + 3y = 40 and 4x + ky = 80, then compare a1/a2, b1/b2 and c1/c2.
Answer
a1/a2 = 2/4 = 1/2, b1/b2 = 3/k, c1/c2 = 40/80 = 1/2. Not unique when 3/k = 1/2, i.e. k = 6: then all three ratios are equal, the second bill is just double the first, and there are infinitely many solutions. Contradictory bills need a1/a2 = b1/b2 ≠ c1/c2, but b1/b2 = 1/2 forces k = 6, which makes c1/c2 equal too. So no value of k makes the bills contradict; for every k ≠ 6 the prices are unique.
Q6 · 5 marks · Long answerHarder board level
Suppose a hostel charges a fixed monthly amount plus a fixed cost for each day a student takes food in the mess. A student who takes food for 20 days in a month pays Rs 3800, and one who takes food for 26 days pays Rs 4400. Find the fixed charge and the cost of food per day. Next month the fixed charge rises by 10% and the food cost rises by Rs 10 per day. What will a student who takes food for 25 days pay?
Hint
Let the fixed charge be Rs f and food cost Rs r per day: f + 20r = 3800 and f + 26r = 4400.
Answer
Subtract: 6r = 600, r = 100; f = 3800 − 2000 = 1800. Check: 1800 + 2600 = 4400. New fixed charge = 1800 + 180 = Rs 1980; new food cost = Rs 110 per day. Bill for 25 days = 1980 + 25 × 110 = 1980 + 2750 = Rs 4730.
Q7 · 4 marks · Case-basedHOTS / competency
Case study: Suppose your school runs a two-day fair. Adult tickets cost Rs x and child tickets Rs y. On Day 1, 30 adult and 50 child tickets were sold for Rs 3300. On Day 2, 40 adult and 20 child tickets were sold for Rs 3000.
(i) Write the pair of linear equations in simplest form. (1 mark)
(ii) An organiser's notebook says that on Day 3, 60 adult and 100 child tickets were sold for Rs 6000 at the same prices. Using ratios with the Day 1 equation, decide whether the Day 1 and Day 3 records can both be correct. (1 mark)
(iii) Find the price of each ticket, and the collection from 25 adult and 40 child tickets. (2 marks)
Hint
Divide each equation by its common factor first. For (ii), compare a1/a2, b1/b2 and c1/c2.
Answer
(i) 3x + 5y = 330 and 2x + y = 150. (ii) Day 3: 60x + 100y = 6000. Ratios with Day 1: 30/60 = 50/100 = 1/2 but 3300/6000 = 11/20, so a1/a2 = b1/b2 ≠ c1/c2: parallel lines, no solution, so both records cannot be correct. (iii) From 2x + y = 150, y = 150 − 2x. Then 3x + 750 − 10x = 330, so 7x = 420, x = 60, y = 30. Check Day 1: 1800 + 1500 = 3300. Collection = 25 × 60 + 40 × 30 = 1500 + 1200 = Rs 2700.
Q8 · 1 mark · MCQBoard level
MCQ: In a two-digit number, the tens digit is $x$ and the units digit is $y$ . The number formed by reversing the digits is 9 less than the original number. Which equation is correct?
(a) $x - y = 1$
(b) $y - x = 1$
(c) $x + y = 9$
(d) $x - y = 9$
Hint
Original $10 x + y$ , reversed $10 y + x$ .
Answer
(a). $(10 x + y) - (10 y + x) = 9$ gives $9 x - 9 y = 9$ , so $x - y = 1$ .
Q9 · 1 mark · MCQBoard level
MCQ: A taxi charges a fixed amount plus a fixed rate per km. An 8 km ride costs Rs 200 and a 12 km ride costs Rs 280. The rate per km is
(a) Rs 25
(b) Rs 20
(c) Rs 40
(d) Rs 35
Hint
Subtract the two bills.
Answer
(b) Rs 20. The extra 4 km cost Rs 80, so the rate is Rs 20 per km (and the fixed charge is Rs 40).
Q10 · 1 mark · Assertion-reasonHarder board level
Assertion (A): If 2 adult and 3 child tickets cost Rs 400, and 4 adult and 6 child tickets cost Rs 750 at the same prices, then the ticket prices cannot be found.
Reason (R): A pair with $\frac{a _{1}}{a _{2}} = \frac{b _{1}}{b _{2}} \neq = \frac{c _{1}}{c _{2}}$ has no solution.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Hint
Compare the ratios of the coefficients and of the totals.
Answer
(a). $\frac{2}{4} = \frac{3}{6} = \frac{1}{2}$ but $\frac{400}{750} = \frac{8}{15} \neq = \frac{1}{2}$ . The lines are parallel, so no prices fit both bills and they cannot be found. R is true and is exactly the reason.
Q11 · 4 marks · Case-basedHOTS / competency
Case study: Suppose a stationery shop keeps this record of sales at fixed prices.
Monday: 5 notebooks and 8 pens, Rs 410.
Tuesday: 3 notebooks and 4 pens, Rs 230.
Wednesday: 10 notebooks and 16 pens, Rs 800.
(i) Write the equations for Monday and Tuesday, taking a notebook as Rs $n$ and a pen as Rs $p$ . (1 mark)
(ii) Find $n$ and $p$ . (1 mark)
(iii) Without using your answer to (ii), use ratios to show that the Wednesday entry cannot be right if Monday's is. Then find by how much Wednesday's total was under-recorded. (2 marks)
Hint
Wednesday sold exactly twice Monday's items.
Answer
(i) $5 n + 8 p = 410$ and $3 n + 4 p = 230$ . (1)
(ii) Double Tuesday: $6 n + 8 p = 460$ ; subtract Monday: $n = 50$ . Then $4 p = 230 - 150 = 80$ , so $p = 20$ . Check: $250 + 160 = 410$ . (1)
(iii) Monday and Wednesday: $\frac{5}{10} = \frac{8}{16} = \frac{1}{2}$ but $\frac{410}{800} \neq = \frac{1}{2}$ , so the two lines are parallel and cannot both be true. The correct Wednesday total is $10 (50) + 16 (20) = 820$ , so it was under-recorded by Rs 20. (2)

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Two-digit number with tens digit x, units y?Show answer
10x + y
2.Upstream speed if boat speed u, stream v?Show answer
u − v
3.Relative speed of cars moving towards each other?Show answer
Sum of speeds
4.Fare with fixed charge f and rate r for n km?Show answer
f + rn
5.Number minus its reverse equals?Show answer
9(x − y)

Word Problems on Linear Equations: frequently asked questions

How do I know which method to use in a word problem?

Either method is accepted unless the question names one. Use substitution when a variable has coefficient 1, and elimination otherwise. Both earn full marks if correct.

Do I need to check my answer?

A check is not always marked, but it catches errors in translation. Checking against the original sentences takes under a minute.

Is cross-multiplication allowed in the 2026-27 syllabus?

No. The cross-multiplication method is not in the 2026-27 syllabus, so solve using substitution or elimination.

What types of word problems come in the board exam?

Typical contexts are ages, digits, fractions, speed, money, fares and geometry. They appear as short answers, long answers and case studies.