solve the problem? (math)?

Question

if you distribute 1 sweet to each child, rest will be 1 sweet. if you distribute 2 to each, 1 child can't get any sweet. how many sweets and clildren are there?

Answer 1

if c is nr of childs , then nr of sweets is c+1

each child 2 sweets , then one child no sweets at all thus
2*(c-1) = c+1 => c =3;

3childs ,4 sweets

Answer 2

let x be the number of children
y be the number of sweets
-> x * 1 = y - 1
x * 2 = y + 2 (1 child can't get any so if you add 2 more, it will be enough)
-> x = 3
y = 4
-> 3 children and 4 sweets

Answer 3

There will be 3 children & 4 sweets

Answer 4

There is 3 children and 4 sweets.

Answer 5

Let number of sweets be S
Let number of children be C

1 x C + 1 = S ... (1)

2 x (C -1) = S ... (2)

Equate (1) = (2), solve for C,

C + 1 = 2C - 2

C = 3

Substitute C = 3 into (1)

S = 1 x (3) + 1

S = 4

Therefore there C=3 Children and S=4 Sweets.

Answer 6

Number of sweets = S
Number of children = C

First statement
C = S - 1 (number of sweets is one greater than children)
2 x (C -1) = S
Solve equations to get 3 children and S=4 sweets

Answer 7

3 children and 4 sweets.

Answer 8

2 children, 2 sweets

Answer 9

there are two children n two sweets!!! is u give one child 1 sweet the other child gets 1 too but is u give one child 2 sweets the other have none!!

Answer 10

there are 3 sweets & 2 children

Answer 11

3 kids. 4 sweets