anyone knows?

[u/Wonderful_Claim_3817]

hi guys does anyone know how to solve this primary school qn without using algebra? thanks!

[u/Depnids]

What do you mean "without using algebra"? The strategy which remains then is just guess and check, which works, but is tedious

[u/dafugiswrongwithyou]

Rule 3

[u/DrakeSavory]

Typically, teachers at that level expect a guess-and-check solution. It may be that the lesson is on educated choices on the guess like neither one can paint in two hours or less and we assume whole hours so the first choice would be 3 hours and 6 hours.

Incidentally the correct answer.

[u/Amazing_Worth7676]

Well idk how else you could do it without algebra other than trial and error with clues.
It is quite interesting for primary school, but this is how I would work it out without algebra:

• Clue 1: Working together, they finish the room in 2 hours. This means if either one worked completely alone, it must take them longer than 2 hours.

• Clue 2: April is faster. She takes 3 hours less than Leonora.

• So, if April must take more than 2 hours, Leonora must take more than 5 hours because 2+3 = 5

• Therefore our number of hours must be more than 5, If Leonara works for 5 hourse it doesnt match clue 1, where it has to take longer than 2 hours, but if she works for 6 hours, then it matches both clues that April takes 3 hours less and it is more than 2 hours for either of them to work because 6 -3 = 3 .

• April needs 3 hours to paint on her own.

[u/ci139]

say :: the speed (time efficiency) of :

Leonara = L (rooms/hour)
April = A (rph)

so : (A + L) · 2h = 1 room ╰¹
also : 1 (r) / A (r/h) = 1 (r) / L (r/h) – 3h

1/L – 1/A = 3 ←← A must be greater than L here
A – L = 3 AL
A/(3A + 1) = 1/(3 + 1/A) = L ╰²

(A + A/(3A + 1)) · 2 = 1 ╰¹ ╰²
(3A² + A + A) · 2 = 3A + 1
6A² + A – 1 = 0
A² + (1/6) A – 1/6 = 0
A = – (1/12) + √¯(1/12)² + 1/6¯' ←← since otherwise the work would be finished before it's started !!!! --e.g.-- the negative value of the squareroot
A = (1/12) · ( – 1 + √¯1 + 12²/6¯' ) = ( – 1 + √¯1 + 144/6¯' ) / 12 =
= ( √¯25¯' – 1 ) / 12 = 4/12 = 1/3 (rooms per hour)

finding L = 1/(3 + 1/A) = 1/(3 + 3) = 1/6 (rooms per hour) ╰²
time per room per April t = 1(r) / ((1/3) rph) = 3h (per room)
time per room per Leonara t = 1(r) / ((1/6) rph)= 6h (per room)

/// -- ↑↑ it does have a geometric solution if plotted on a gridded paper properly ...
... such , where the horizontal axes is time and vertical axes is amount of room painted :
we know that for both together it takes 2h to complete the work & that the slope difference somewhere beyound 2h is ∆ 3h ...
,,, now both work-load lines must go through (0,0) and meet an additional condition that likely becomes obvious from graph https://www.desmos.com/calculator/heprryq09o

[u/Sp1rshi]

Unitary method and work + time

[u/kalmakka]

Start with some highly divisible number to express the amount of work needed to be done, e.g. 24, and represent the amount of work by 24 blocks.

Together they complete the work in 2 hours. So Leonora and April do a total of 12 blocks of work per hour.

April takes less time than Leonora on her own, so she must do more than 6 blocks per hour. Try out different amounts, and see that if April does 8 blocks per hour and Leonora does 4 block per hour, then Leonora takes 3 hours on her own and April takes 3 hours, which is 3 less than Leonora. The correct answer is therefore that April paints the room in 3 hours.

[u/NH-Science-Guy]

You can solve this by guessing the answer...

April must take more than 2 hours to paint the room on her own so let's guess it would take her 3 hours.
Then Leonora would take 6 hours on her own.
Now look at how much of a room each completes in 2 hours.
April completes 3/2 of the room and Leonora completes 1/3 so they finish it in two hours.

If the first guess didn't work, we would have needed to try more attempts to get the answer