the sum of the age of wendy and her mother is 72. In how many years time will their age be 90?

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# the sum of the age of wendy and her mother is 72. In how many years time will their age be 90?

The sum of the age of wendy and her mother is 72. In how many years time will their age be 90?

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### 1.9 Practice - Age Problems - Saylor Academy

1.9 Practice - Age Problems 1. A boy is 10 years ... The sum of Clyde and Wendy’s age is 64. In four years, Wendy ... In eight years the sum of their ages will be 72.

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Ages of Three Children ... "The product of their ages is 72. The sum of their ages is the ... twins could be born in different years, but still be the same age ...

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Wendy Marvell (ウェンディ ... She was one of the orphans chosen to be a Dragon Slayer four hundred years prior to X777, serving as her ... During her time in ...

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... when she was 11 years old. At the time ... At 14 years of age, Wendy was ... Davis' daughters each released letters in defense of their mother ...

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... she has written seven books and started her own fashion line in addition to her daily television show, “The Wendy ... their time off. I believe in ... age is 84 ...

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### the sum of the age of wendy and her mother is 72. In how many years time will their age be 90?

time=(90-72/2=18/2=9 yeers

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### Jack is 2 2 times as old as Lacy. 7 7 years ago the sum of their ages was 13 13. How old are they now?

Jack is 2 2 times as old as Lacy. 7 7 years ago the sum of their ages was 13 13. How old are they now? OK, there seems to be a bit of repitition in your question, so I'm going to reword it as follows. Jack is 2 times as old as Lacy. 7 years ago the sum of their ages was 13. How old are they n

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### What will be their present agees?

Every problem has the answer built in. No-one is asking you to make up a process. Use the data given and simple equations, manipulate the equations, and the answer pops out. The sum of the ages of A and his father is 100. When A is as old as his father is now, he will be five times as old as his so

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### Age-related Algebra problem

Let the current ages be S (Selina) and N (Nancy). 10 years ago: S-10+N-10=20, so S+N=40. Therefore, N=40-S, so neither current age can exceed 40, and assuming ages are whole years, the minimum is 11 and maximum 29. Selina is 6 times as old as Nancy was x years ago. That means S is a multiple

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### a father and his sons ages are in the ratio 10:3.his son is 15years old now.how many years time will the fathers age be in the ratio 2:1

a father and his sons ages are in the ratio 10:3.his son is 15years old now.how many years time will the fathers age be in the ratio 2:1 Let F be the father's current age Let S be the son's current age   (and S = 15) Then, F:S = 10:3    i.e. 3F = 10S Since

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### Word Problem where you have to figure out Goerge's wife's age in 1941.

In 1941 George's children were both between the ages of 10 and 20. The sum of the cube of one child's age and the square of the other child's age gives the year in which George's wife was born. How old was his wife in 1941? Start by looking for the possible cubed age: 10^3 = 1000 11^3 = 1331 12^3 =

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### Father is aged three times more than his son ronit. after 8years,he would be two and half times of ronit age.after further 8 years,how many times would be of ronit's age?

Let F=age of father and S=age of son. F=3S. In 8 years' time they will be F+8 and S+8 years old and F+8=5/2(S+8), because 2 1/2=5/2, so, multiplying through by 2, 2F+16=5S+40; 2F-5S=24; substituting for F, 6S-5S=24; S=24 and F=3S=72. In 16 years' time, son will be 40 and father wil

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### Find the age of each?

Let the ages be represented by the first letter of their names. C=R/2; J=C+11; (R+4)/2=(J+4)/2+C-9. From these we have R=2C. Substitute for R and C in the third equation: (2C+4)/2=(C+15)/2+C-9. This equation contains only one variable so should be solvable. Multiply through by 2: 2C+4=C+15+

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### Earl is 6 years older than Simin. In 2 years the sum of their ages will be 20. How old is Simim?

6+x+2+x+2=20 10+2x=20 2x=20-10 2x=10 x=5 (in 2 years time  Earl is 13;Simin70 before 2 years time(EARL 11 years ;SIMIN 5 years)

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### Monica is 7 years older than cheena.Aroma is 2 times younger than monica.if the sum of their ages is 58,what is the age of each child?

Let the ages be M, C, A for Monica, Cheena, Aroma. M=C+7, A=M/2. M+C+A=58. C+7+C+(C+7)/2=58, 4C+14+C+7=116 5C+21=116, 5C=116-21=95, C=95/5=19. Cheena is 19. M=19+7=26, Monica is 26 and Aroma is 26/2=13.

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