TszYing 發問於 科學及數學數學 · 1 十年 前

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

Complete to answer me.............

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

評分
  • Hunter
    Lv 6
    1 十年 前
    最佳解答

    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.

  • 1 十年 前

    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

    hope this can help you to solve your problem

    資料來源: myself
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