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Please solve this! It is about percentages and algebra!

Please solve this! It is about percentages and algebra!-example-1
User Geos
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1 Answer

5 votes

Answer:

a) my age is 36, sister's age is 24 b) my age is 48, the sister's age is 36

c) my age is 96, sister's age is 84 Extension: 21.7percents

Explanation:

Suppose now I am x , my sister is y (x years, y years)

Six years ago I was x-6, my sister was y-6, I was 7 times older than my sister

x-6= (y-6)*7

x-6=7y-42

x-7y=-36

Four years ago I was x-4, my sister was y-4. I was 4 times older.

x-4= (y-4)*4

x-4=4y-16

x-4y=-12

We have the system of equations

x-7y=-36

x-4y=-12

4y-x=12

x-7y+4y-x=-36+12

-3y=-24

y=8

x-32=-12

x=20

The difference between us is 20-8=12. It is constant measure.

So if my sister is m, I am m+12

a) (m+12)*2/3=m

2/3m+8=m

1/3m=8

m=24- the age of my sister

24+12=36- my age

b) 3/4*(m+12)= m

3/4m+9=m

m=36- her age

36+12=48- my age

c) 7/8(m+12)= m

1/8m= 84/8

m=84- her age

84+12=96- my age

Extension

6/1/7=42- the age of Ian when Jamie will be six

10-6=4- years (after four years Jamie will be 10)

42+4=46- the age of Ian after 4 years

46-100

10-x

x= 100/4.6= 21.7percents is an approximate percentage which Jamie's age is from Ian

User Shalamus
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