Helen, like jumping cars.
On her first jump, she jumps over 5 cars and 2 vans, which is a total of 14m. On her second jump she jumps over 5 cars and 4 vans, which is a total of 19m.

What is the width of a car and the width of a van?

I used algebra to solve this problem, which I think is right (?)

5c + 2v=14 and 5c + 4v= 19

By rearranging the first to get 5c (5c = 14-2v) as the subject I rewrote the equation as

14-2v + 4v =19 (then solve)

14 +2v = 19

2v = 19-14

2v = 5

V = 2.5

C = 1.8

Think this is correct, but happy to be corrected....

Need more of these types of question to practise. Can anyone tell me where to find and what this would be called?

I would not use algebra, but I got the same answer as you.

I would just minus 19m from 14m to get 5m = 2 vans (the only difference between the 2 amounts)

5 divided by 2 vans =2.5m

so if 4 vans = 10m in the second amount, and equal 5m in the first amount, then the cars are 9m divided 5 = 1.8m.

If you look for maths word problem books, you might find more of these.

Yes, as Sookie says, its spotting what the difference it. The algebra is quite complicated (albeit that its technically correct).

The method (as taught to my DC by her tutor) is to write the 2 sums above each other :

5 cars + 2 vans = 14m
5 cars + 4 vans = 19m

Its easy to see then that 2 vans must be 5m.

Then its quite straightforward to say vans must be 2.5m and so cars must be 1.8m.

P.S. They're called Simultaneous equations.

Hi all.... that’s brill. Hadn’t looked at it that way.

That exam does lend itself to being compared, but are you able to do the same with:

4t + A = 265
T + A = 115

What is T and A?

These are the only examples I have been able to find so far. Trying to get my head wrapped around an easier way (other than using algebra - only way I know) to teach little one... more suggestions pls where not so easily comparable

That was meant to say ‘example’

For questions like these use common sense and forget Algebra.
The difference between the two totals is 3T.
The difference between the two totals is 150. So T must = 50, that is 150 divided by 3.
T + A = 115 so 50 + A = 115.
115 - 50 = 65 so A must = 65.

As Sookie says, this is spot the difference.
Where are you getting these questions from and what exams are you preparing for?

_________________
mad?

Great... the first from the recent grammar sample questions. Making sure if little one is called back next week, this is something spited and dealt with without issue.

Many thanks. That common sense approach is exactly what i’ll use. Just looking for more problems online to solve...