# HELP!!! (方程)complete There have two questions

1)The sum of three numbers A,B and C is 405.A is four times B,and C is greater than A by 45.Find these three numbers.

2)100 buns were distributed to 100 monks.If each elderly monk got 3 buns,and every three young monks had to share 1 bun,how many elderly monks and young monks were three ?

### 2 個解答

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1)

A+B+C =405

A=4B

C=A+45

4B+B+(A+45)=405

4B+B+(4B+45)=405

5B+4B+45=405

9B=405-45

9B=360

B=40

A=4B=4*40=160

C=A+45=160+45=205

Final check

A+B+C=160+40+205=405 (GOOD!)

2)

Let x be the no. of elderly monks.

So, the no. of younger monks is (100-x).

100buns were distributed with elderly monks got 3buns per head and 3 young monks had to share 1 bun.

100 = 3x + (100-x)/3 * 1

100 = 3x + (100-x)/3

100*3 = 3x * 3 + (100-x)/3*3

300 = 9x + 100 - x

200 = 8x

x=200/8=25

So, there are 25 elderly monks and (100-25)=75young monks.

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• 1)

A+B+C=405

A=4B

C=A+45

then:

4B+B+4B+45=405

9B=360

B=40

A=160

C=205

2)Let there are a elderly monks and b young monks, x be get the numbers of buns

then

a=3x

3b=x

3x+x=100

x=25

so a=75

b=25

because every elderly monk can get three buns, so 75/3=25 elderly monks

every three young monks share one bun, so 25*3=75 young monks

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