Help with maths question please

[CalicoAnnie]

Is the only way to solve this question a process of going through all the options?
For example - 1 reading class of three, plus 6 groups of 5 for maths = 33, 2 reading groups of three plus 5 groups of 5 for maths = 31 and so on. Doing it this way we worked out the answer was 29 but it seemed a long process.

I would say trial and error is the most efficient method.

There are three unknown variables and only two equations that you could set up. Therefore you can't eliminate uncertainty from a set of liner equations.

Trial and error shouldn't be particularly laborious here because the forced solutions and limit of 7 help a lot.

Groups of 5:
5 (add six 3s - too small)
10 add 5x3 = 25
15 add 4x3 = 27
20 add 3 x3=29
25 (add two 3s - too big)

@Stevenage689

I would say trial and error is the most efficient method.

There are three unknown variables and only two equations that you could set up. Therefore you can't eliminate uncertainty from a set of liner equations.

Trial and error shouldn't be particularly laborious here because the forced solutions and limit of 7 help a lot.

Groups of 5:
5 (add six 3s - too small)
10 add 5x3 = 25
15 add 4x3 = 27
20 add 3 x3=29
25 (add two 3s - too big)

Appreciated thank you, I like the way you have set your answer out. I wasn’t sure if I was missing some easier way of working it out.

At primary trial and error is taught as a technique. It seems bizarre as adults would not think of approaching things with trial and error technique usually, because it takes so long.

But they are taught to do the logical combinations in order, not just flit about randomly hoping they hit on the answer.

So the PP is correct.

If your child is the top end of primary school, I guess algebra may be expected as a possible option.

(Isn't 21 a possible answer as well? i.e. all children doing reading)

@RedskyThisNight

If your child is the top end of primary school, I guess algebra may be expected as a possible option.

(Isn't 21 a possible answer as well? i.e. all children doing reading)

I would say that "there were groups of 5 for writing" rules this out, though it is semantics really, which shouldn't be part of a maths question at school level.

Algebra is ineffective here for the reasons I explained above. Whenever the number of variables is higher than the number of equations, you can't solve by algebra alone. You would still end up needing trial and error.

In summary, it's a crappy question.

Algebra is ineffective here for the reasons I explained above. Whenever the number of variables is higher than the number of equations, you can't solve by algebra alone. You would still end up needing trial and error.

on the basis that the question offers up 4 possible answers, you only have to see which ones meet your end criteria. Which, agreed, is still trial and error but much quicker and easier than trial and error from scratch.

I agree with you about the question not being clear and this being poor. I'd hope either answer was allowable if the reasoning was explained.

Just some convoluted musings ..

Adding the number of reading groups and the number of writing groups give 7 groups in total.

This means if the number of reading groups is even, the number of writing groups is odd and vice versa.

So the total number of students (The number of reading groups x 3 + the number of writing groups x 5) is always even x odd + odd x odd, that is, even + odd, that is, odd.

So the total number of students must be odd.

Now assume there is at least one reading group and at least one writing group.

To get the highest possible total number of students, let as many groups as possible be writing groups as writing groups contain the most students. -> 6 writing groups, 1 reading group.

This gives 6 x 5 + 1x3 = 33 students in total.

To get the lowest possible total number of students, let as many groups as possible be reading groups as reading groups contain the fewest students -> 6 reading groups, 1 writing group.

This gives 6 x 3 + 1x5 = 23 students in total.

So the actual number of students is an odd number which is larger than or equal to 23 and smaller than or equal to 33.

(I realise this is probably too complicated.)

If t is total number of children, r is number of reading groups and w is number of writing groups then:
3r + 5w = t so 3r = t -5w
and

r + w = 7 so 3r + 3w = 21
so putting substituting for 3r in the second equation just above,
t- 5w + 3w = 21 so t = 21 + 2w
So as w>0, the answer must be an odd number greater than 21 so the answer is 29.

The only possible answer is 29

3 groups of 3 = 9
4 groups of 5 = 20

No other combination works

The total must be odd and narrows it down to 21 or 29

You therefore work out the variations for both groups numbers of 3 and 5
1 x3 + 6 x5 =33
2X3 + 5X5 =31
3x3 + 4x5 = 29 *
4X3 + 3x5 = 27
5x3 + 2x5 = 25
6X3 + 1X5 = 23
The other way was to realise the pattern reduced by 2 each time and not need to continue passed the 29 total