CBSE
Arushi and Devesh's Painting, Plus Real Class 10 Quadratic Word Problems
Arushi alone finishes the painting in 30 minutes and the pair together take 20, so Devesh's one-minute share is 1/20 - 1/30 = 1/60. Ten minutes of joint work completes half the painting, so Devesh needs 30 more minutes, which is 40 minutes from the start. Full working is below, followed by genuine quadratic word problems.
Here is something most pages will not tell you: the Arushi and Devesh painting question is not actually a quadratic equations problem. It is a time and work question, usually set in Class 8, and it solves with simple fractions. It gets searched alongside Class 10 Chapter 4 because painting, tank and journey setups also appear there, so this page does both jobs. The painting question is solved first, in full, and after it come five genuine quadratic word problems, area, speed, ages, taps and numbers, laid out the way CBSE wants working shown. I sit Cambridge exams myself, not CBSE, so everything here follows the NCERT method rather than personal exam memory.
The painting question is not actually quadratic
Search traffic sends a lot of people looking for Arushi and Devesh to quadratic equations pages, but the question itself is a time and work problem, the kind usually met in Class 8. There is no x^2 anywhere in it: you convert each painter to a per-minute work rate, subtract to isolate Devesh, and divide the leftover work by his rate. It is still solved in full above, because it is what you came for. The genuinely quadratic relatives of this question are the tap and pipe problems, like problem 5 on this page, where the two times are unknown and related, say x and x + 5, and the rate equation 1/x + 1/(x + 5) = 1/6 multiplies out into a quadratic. If your teacher set the painting question inside a quadratic worksheet, that is probably the connection they had in mind.
A four-move routine for every quadratic word problem
Move one: name the unknown, in writing, with units. Let the speed be x km/h. Move two: translate the story into one equation, keeping fractions as fractions. Move three: clear denominators and rearrange into ax^2 + bx + c = 0, and only then decide whether to factorise or use the formula; if you can spot two numbers multiplying to ac and adding to b, factorising is faster, otherwise x = (-b ± sqrt(b^2 - 4ac))/2a always works. Move four: solve, then test both roots against the story and reject the impossible one with a written reason. The four moves matter more than speed, because CBSE marking gives most of the credit to the setup and the rearrangement, not the final number. A perfect answer with no visible equation can earn less than a wrong answer with a correct setup.
Rejecting a root is a scoring line, not an apology
Almost every quadratic word problem produces one answer that fits the story and one that does not: a breadth of -12 m, a speed of -60 km/h, an age of -10 years. Writing both roots and circling the positive one is not enough; the marking scheme wants a reason, one short sentence like speed cannot be negative, so x = 50. This costs five seconds and is often a dedicated mark. The second habit worth building is substituting your accepted root back into the original story, not into your own equation, since your equation might be the thing that is wrong. For the bus problem above, checking means computing 300/50 = 6 hours and 300/60 = 5 hours and seeing the difference really is 1 hour. When the check works, say so in one line.
Worked questions, step by step
Arushi and Devesh are working on a painting together. Arushi on her own could finish the whole painting in 30 minutes, while the two of them working together can finish it in 20 minutes. After 10 minutes of painting together they argue, and Arushi walks off. How much longer does Devesh need to finish the painting alone?
- Given: Arushi alone takes 30 minutes, both together take 20 minutes, and they work jointly for the first 10 minutes. To find: Devesh's remaining time alone.
- In one minute the pair completes 1/20 of the painting and Arushi alone completes 1/30, so Devesh's one-minute share is 1/20 - 1/30.
- 1/20 - 1/30 = 3/60 - 2/60 = 1/60, so Devesh alone would need 60 minutes for a whole painting.
- In the first 10 minutes together they finish 10 × (1/20) = 1/2 of the painting, leaving 1/2 undone.
- Time for Devesh = work left divided by his rate = (1/2) ÷ (1/60) = 30 minutes.
- Answer: Devesh needs 30 more minutes, so the painting is completed 40 minutes after they first started.
Answer: 30 minutes more for Devesh (40 minutes from the start)
Where marks slip: Half the wrong answers to this question come from not saying which clock you are reading: 30 is the time after Arushi leaves and 40 is the total, so state in words which one the question asked for.
Try one yourself: Sana alone can weed a garden in 40 minutes, and Sana and Bilal together take 24 minutes. They work together for 12 minutes, then Sana leaves. How long does Bilal need to finish alone? (Answer: 30 minutes)
A rectangular community hall has a length 4 m greater than its breadth, and its floor area is 96 square metres. Find the length and breadth of the hall.
- Given: length = breadth + 4 and area = 96 m^2. To find: both dimensions. Let the breadth be x metres, so the length is (x + 4) metres.
- Formula: area of a rectangle = length × breadth, so x(x + 4) = 96.
- Expand and rearrange into standard form: x^2 + 4x - 96 = 0.
- Factorise: x^2 + 12x - 8x - 96 = 0 gives (x + 12)(x - 8) = 0, so x = -12 or x = 8.
- A breadth cannot be negative, so reject x = -12 and keep x = 8.
- Answer: breadth = 8 m and length = 8 + 4 = 12 m. Check: 8 × 12 = 96 m^2.
Answer: Breadth = 8 m, length = 12 m
Where marks slip: The mark scheme expects you to reject the negative root with a written reason such as breadth cannot be negative, and leaving both roots standing without comment usually drops a mark.
Try one yourself: A rectangular garden has its length 6 m more than its breadth and an area of 91 square metres. Find its dimensions. (Answer: 7 m by 13 m)
A bus covers 300 km at a steady speed. If it had travelled 10 km/h faster, the same trip would have taken 1 hour less. Find the actual speed of the bus.
- Given: distance = 300 km, and raising the speed by 10 km/h cuts the time by 1 hour. To find: the speed. Let the speed be x km/h, so time = 300/x hours.
- Set up the time difference: 300/x - 300/(x + 10) = 1.
- Multiply every term by x(x + 10): 300(x + 10) - 300x = x(x + 10), which simplifies to 3000 = x^2 + 10x.
- Standard form: x^2 + 10x - 3000 = 0.
- By the quadratic formula, x = (-10 ± sqrt(100 + 12000))/2 = (-10 ± 110)/2, giving x = 50 or x = -60.
- Speed cannot be negative, so x = 50. Check: 300/50 = 6 hours and 300/60 = 5 hours, a difference of exactly 1 hour.
- Answer: the bus travels at 50 km/h.
Answer: Speed = 50 km/h
Where marks slip: Most marks vanish in the clearing-fractions line, so multiply every term by x(x + 10) in one visible step before simplifying, and keep the plus-minus until you reject the negative root.
Try one yourself: A van covers 200 km at a uniform speed. Going 10 km/h faster would save 1 hour. Find its speed. (Answer: 40 km/h)
The product of Aisha's age 4 years ago and her age 6 years from now is 56. How old is Aisha now?
- Given: (age 4 years ago) × (age 6 years from now) = 56. To find: her present age. Let the present age be x years.
- Translate: (x - 4)(x + 6) = 56.
- Expand: x^2 + 2x - 24 = 56, so x^2 + 2x - 80 = 0.
- Factorise: (x + 10)(x - 8) = 0, giving x = -10 or x = 8.
- An age cannot be negative, so x = 8. Check: 4 years ago she was 4, in 6 years she will be 14, and 4 × 14 = 56.
- Answer: Aisha is 8 years old.
Answer: Present age = 8 years
Where marks slip: Expanding (x - 4)(x + 6) with a middle term of -2x is the recurring slip; the middle term is 6x - 4x = +2x, so slow down on exactly that line.
Try one yourself: The product of Rohan's age 3 years ago and his age 5 years from now is 105. Find his present age. (Answer: 10 years)
Two taps running together can fill a water tank in 6 hours. On its own, the smaller tap takes 5 hours longer than the larger tap to fill the same tank. How long does each tap take alone?
- Given: together they take 6 hours, and the smaller tap's time is 5 hours more than the larger tap's. To find: each tap's time. Let the larger tap take x hours, so the smaller takes (x + 5) hours.
- Rates add in a filling problem: 1/x + 1/(x + 5) = 1/6.
- Multiply every term by 6x(x + 5): 6(x + 5) + 6x = x(x + 5).
- Simplify: 12x + 30 = x^2 + 5x, so x^2 - 7x - 30 = 0.
- Factorise: (x - 10)(x + 3) = 0, so x = 10 or x = -3; a time cannot be negative, so x = 10.
- Answer: the larger tap takes 10 hours and the smaller takes 15 hours. Check: 1/10 + 1/15 = 3/30 + 2/30 = 5/30 = 1/6.
Answer: Larger tap = 10 hours, smaller tap = 15 hours
Where marks slip: This is the quadratic cousin of the painting question, and the standard error is writing the smaller tap as x - 5; the smaller, slower tap takes more time, so it must be x + 5.
Try one yourself: Two pipes fill a tank together in 4 hours, and the narrower pipe alone needs 6 hours more than the wider one. Find each pipe's time. (Answer: 6 hours and 12 hours)
The squares of two consecutive positive odd numbers add up to 130. Find the two numbers.
- Given: two consecutive odd numbers whose squares sum to 130. To find: the numbers. Let them be x and x + 2.
- Translate: x^2 + (x + 2)^2 = 130.
- Expand: x^2 + x^2 + 4x + 4 = 130, so 2x^2 + 4x - 126 = 0.
- Divide the whole equation by 2 to keep the numbers small: x^2 + 2x - 63 = 0.
- Factorise: (x + 9)(x - 7) = 0, and since the numbers are positive, x = 7.
- Answer: the numbers are 7 and 9. Check: 49 + 81 = 130.
Answer: The numbers are 7 and 9
Where marks slip: Dividing the whole equation by 2 before factorising is worth doing every time, because most sign errors on this question happen while factorising 2x^2 + 4x - 126 directly.
Try one yourself: The squares of two consecutive positive even numbers add up to 164. Find them. (Answer: 8 and 10)
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Solve my question →Quiz me on this topicQuestions students ask
Is the Arushi and Devesh painting question really a quadratic problem?
No. It is a time and work question solved with fractions: rates of 1/30 and 1/20 per minute, with no squared term anywhere. It usually gets filed under quadratics because tap-and-pipe questions, which look similar, genuinely do produce quadratics when the two times are unknown. The full non-quadratic solution is the first problem on this page.
Is the answer to the painting question 30 minutes or 40 minutes?
Both numbers are correct answers to different questions. Devesh needs 30 minutes of solo work after Arushi leaves. Since the pair had already painted together for 10 minutes, the painting is finished 40 minutes after the start. Read the final sentence of the question carefully and state in words which duration your number represents.
Should I factorise or use the quadratic formula?
Factorise when two numbers multiplying to ac and adding to b jump out, as in x^2 + 2x - 80 = (x + 10)(x - 8); it is faster and less error-prone. When nothing jumps out within about twenty seconds, switch to x = (-b ± sqrt(b^2 - 4ac))/2a, which always works. Both methods receive full credit in CBSE marking.
Why do quadratic word problems give two answers?
The algebra cannot see the story, so it returns every x satisfying the equation, including physically impossible ones like negative speeds or ages. One root usually fits the situation and one does not. You are expected to reject the invalid root with a written reason, and that rejection line frequently carries its own mark in the scheme.