In a birthday party all children got 3 candy's but 1 child got 2 candy's then if all children got 2 candy's then 8 candy is remaining how many andy's in total

  22

Step-by-step explanation:

If there are n children, the number of candies in the first distribution is ...

  3n +1

If the same number of candies is distributed to the same n children, then the second distribution tells us ...

  2n +8 = 3n +1

  7 = n . . . . . . . . . subtract 2n+1. This is the number of children.

The number of candies is ...

  3(7) +1 = 22

There were 22 candies in total.

step by step resolutions

1) n+n+1+n+2=51

3n+3=51

3n=48

n=16

the three numbers are 16, 17 and 18.

2) x+15=4x

3x=15

x=5

radha is now 5 years old

3) the two digits are x and y.

y=3x

(10x+y)+(10y+x)=88

y=3x

11x+11y=88

y=3x

x+y=8

x+3x=8

x=2

the original number can be either 28 or 82.

4) n+(n+2)+(n+4)=48

3n+6=48

3n=42

n=14

the numbers are 14, 16 and 18.

5) y=x+3

10y+x+10x+y=143

y=x+3

11x+11y=143

y=x+3

x+y=13

x+x+3=13

2x=10

x=5

the original number can be 57 or 75.

hope this !

i dont know what you want me to answer for this but im guessing 2: 11?