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Nezavi [6.7K]
3 years ago
6

Please help, doing age word problems. Tysm if you do, really appreciated :D

Mathematics
2 answers:
topjm [15]3 years ago
5 0

Answer:

D (12, 21); B (11, 17); C (18,28); D (42); C (8).

Step-by-step explanation:

First:

Let I represent Isabel and M represent Marie.

We know that currently, Isabel is 9 years older than Marie, or:

I=9+M.

In three years, Isabel will be six years less than twice of Marie's age. In other words:

(I+3)= 2(M+3)-6

Now solve. Substitute I.

(9+M+3)=2M+6-6

12+M=2M

M=12; I=21. Marie is 12 while Isabel is 21.

Second:

Let I represent Isabel and M represent Marie.

Isabel is 6 years older than Marie; in other words: I=6+M

In 4 years, Isabel will be 9 years less than twice Marie's age. Or:

(I+4)=2(M+4)-9

Solve. Substitute I.

(6+M+4)=2(M+4)-9

10+M=2M+8-9

10+M=2M-1

11=M; I=17; Marie is 11 while Isabel is 17.

Third:

Let I represent Isabel and M represent Marie.

Isabel is 10 years older than Marie, or: I=10+M

In 2 years, twice Isabel's age is three times Marie's age. Or:

2(I+2)=3(M+2)

Solve. Substitute for I.

2(10+M+2)=3(M+2)

24+2M=3M+6

18=M; I=28. Marie is 18 while Isabel is 28.

Fourth:

Let M represent Mary and A represent Ann.

Mary is 3 time as old as Ann. Or: M=3A

7 years ago Mary was 5 times as old as Ann. In other words:

(M-7)=5(A-7)

Solve for this system. Substitute M.

(3A-7)=5A-35

-2A=-28

A=14; M=42; Mary is 42.

Fifth:

Let T represent Tammy and L represent Laurel.

We know that Tammy is 42 while Laurel is 9. In other words:

T=42 and L=9.

We need to find in how many years will 3 times Laurel's age be 1 more than Tammy's age. In other words, let's let y represent the amount of years. Thus:

3(L+y)=(T+y)+1

We already know L and T:

3(9+y)=(42+y)+1

27+3y=43+y

2y=16

y=8

In 8 years, when Laurel is 17 and Tammy is 50. (3 times 17 is 51, one more than 50).

Digiron [165]3 years ago
4 0

Step-by-step explanation:

  • Problem 1 :

Since we have the answers no need to write an equation and solve it

  • 4 and 13 :
  • substract 3 from each : 1 , 10 so Isabel is really 9 years older than Marie
  • add 3 : 7 and 16
  • 16+6 = 21 = 7*3 so that's right

the answer is A

  • Problem 2 :
  • 11 and 17
  • 17 -11 = 6 so that's right
  • add 4 to each : 17+4 = 21   11+4= 15
  • 21+9 = 30= 15*2 so that's right

b is the answer

  • Problem 3 :
  • 18 and 28
  • 28-18= 10 so it's a good choice
  • add 2 :  30 and 20
  • 30*2= 60 and 20*3 = 60
  • so that's true !

C is the answer

  • Problem 4 :
  • here i have a method
  • let x be the age of Marie and y the age of Ann
  • Mary is 3 times as old as Ann so : x = 3y
  • 7 years ago Mary was 5 times as old as An so x-7 = 5(y-7)
  • solve the equations and you'll get y= 14
  • so x = 3*y = 3*14= 42

So the answer is D

  • Problem 5 :
  • 42 + 8 = 50 and 8+9= 17
  • 17*3 = 51
  • 51-50 = 1
  • so the answer is 8 years

The answer is c

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