# Thread: Primary 5 problem sum. needs help to solve.

1. ## Primary 5 problem sum. needs help to solve.

Mrs lim has some sweets.If she gives 2 to each of her children. she will have 4 sweets left. If she gives 4 sweets to each of them, she will be sort
of 2 sweets.
How many children does she have?

2. Originally Posted by mytan1963
Mrs lim has some sweets.If she gives 2 to each of her children. she will have 4 sweets left. If she gives 4 sweets to each of them, she will be sort
of 2 sweets.
How many children does she have?
what do you mean she will be (i suppose you meant to write) short of two sweets? meaning the last child would only get 2 sweets if she tries to give everyone 4?

If so, let x be the number of children. Then we know she has 2x + 4 sweets. as she has 4 more than double the number of her children, by the second sentence. however, she has 2 less than 4 times the number of her children, so she also has 4x - 2 sweets. since this describes the same amount, we want

2x + 4 = 4x - 2

now find x

3. ## Word problem

Hello mytan1963
Originally Posted by mytan1963
Mrs lim has some sweets.If she gives 2 to each of her children. she will have 4 sweets left. If she gives 4 sweets to each of them, she will be sort
of 2 sweets.
How many children does she have?
If she has $x$ children and she gives them $2$ sweets each, that's $2x$ sweets. And if she has $4$ sweets left, she must have had $2x+4$ sweets to begin with.

Or, if she gives them $4$ sweets each, that's $4x$ sweets. But that is $2$ sweets more than she has - so she must have $4x-2$ sweets.

We now say that these two numbers of sweets must be the same. So:

$2x+4 = 4x-2$

$\Rightarrow 6 = 2x$

$\Rightarrow x =3$

So she has $3$ children (and $10$ sweets).

4. Thanks vey much for your help!

5. Originally Posted by mytan1963
Mrs lim has some sweets.If she gives 2 to each of her children. she will have 4 sweets left. If she gives 4 sweets to each of them, she will be sort of 2 sweets. How many children does she have?
For a primary-school student, think of the exercise as follows:

However many students she has, if she gives four sweets to each, she'll be short two. So, if she had two more sweets, she should have a multiple of four, and also would have enough for all of her students. That is, she would have a number of sweets equal to:

4, 8, 12, 16, 20, 24, 28, 32, 36, 40, 44, 48, 52,...

This means that she actually had a number of sweets equal to 2 less than the values in the above list, or:

2, 6, 10, 14, 18, 22, 26, 30, 34, 38, 42, 46, 50,...

If she gives two sweets to each, she'll have four left over. If she had two more students, she could have given away all of the sweets. So the number of sweets is some multiple of 2:

2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26,...

Since the number of sweets hasn't changed, the number has to be some value which is in both lists:

2, 6, 10, 14, 18, 22, 26,...

If she has only two sweets, she obviously cannot give four to anybody, so this can't work.

If she has six sweets, then she can give four to only one student, leaving two for a second student. If she has six sweets and two students, then she can give two to each student, but then she'll have only two left over, and she's supposed to have four left over. So this number won't work.

If she has ten sweets, then she can give four to two students, leaving only two for the third student. If she has ten sweets and three students, she can give two to each student, and she'll have four left over. This number works.