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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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Wendy is 8 years old. The product of Wendy's and Andy's age is


... many years time will the sum of their age be 90? ... time that Sean's age and his mother's age ... her age the sum of their age is 44 years hope Wendy plans ...
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Solving Linear Equations - Age Problems


In the change column because we don’t know the time ... The sum of Clyde and Wendy’s age is 64. ... age is double Daniel’s age. Eight years ago the sum of their ...
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math age problem - search results - jiskha.com


What would be their age 3 years from now if the sum of ... times her age in 3 years' time minus ... years ago mother's age was square of her son. 10 years ...
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Math Forum - Ask Dr. Math


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 | Fairy Tail Wiki | Fandom powered by Wikia


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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Wendy Davis (politician) - Wikipedia


... 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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Wendy Williams says people think she 'was ... - Daily Mail Online


She is famed for her larger than life personality. And Wendy ... as 20 years after her ... Chloe Lattanzi shares adorable throwback photo of her cuddling up to mother ...
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Wendy Williams on why you should always open (and pay) your ...


... 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 now? Let Jack's age be X. Let Lacy's age be Y. Using the information given, we have Jack is 2 times as old as Lacy: this means X = 2Y 7 years ago the sum of their ages was 13: Since Jack is now X years old, then 7 years ago his age would be (X - 7), So ... 7 years ago the sum of their ages was 13: This means (X - 7) + (Y - 7) = 13, or, X + Y - 14 = 13, i.e. X + Y = 27 Our two equations then are, X = 2Y X + Y = 27 Substituting for X = 2Y into the 2nd equation, 2Y + Y = 27 3Y = 27  Y = 9 X = 2Y = 18 This give the ages as: Jack is 18, Lacy is 9  
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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 son B is now. B will be eight years older than A is now, when A is as old as his father is now. Start with the first statement. 1)   A + F = 100 The solution revolves around A being as old as his father is now, but x years from now 2)   F = A + x Substitute the value of F from equation 2 into equation 1 3)   A + (A + x) = 100 At that time in the future, A will be 5 times as old as his son is now 4)   A + x = 5B At that future date, the son will be 8 years older than A is right now 5)   B + x = A + 8 Solving that equation for B gives us the son's age right now 6)   B = A - x + 8 That takes care of the preliminaries. Substitute the value of B from equation 6 into equation 4 7) A + x = 5 (A - x + 8) Multiply through on the right side 8) A + x = 5A - 5x + 40 Add 5x to both sides, subtract A from both sides 9) 6x = 4A + 40 Divide both sides by 6 10) x = 2/3 A + 6 2/3 Substitute the value of x from equation 10 into equation 3 11) A + A + (2/3 A + 6 2/3) = 100 Combine the A terms, then subtract 6 2/3 from both sides 12) 2 2/3 A = 93 1/3 Multiply both sides by 3 13) 8A = 280 Divide both sides by 8 14) A = 35 We now know that A is 35 years old.   From equation 1, we find the age of A's father A + F = 100;    35 + F = 100;   F = 100 - 35;    F = 65 Putting that into equation 2, we find how far we are looking into the future F = A + x;    65 = 35 + x;    30 = x Substituting the appropriate values into equation 6, we find B's age B = A - x + 8;    B = 35 - 30 + 8;     B = 13   Check: 30 years in the future, B will be B + 30 = ??;    13 + 30 = 43 That should be 8 years older than A is now A + 8 = ??;      35 + 8 = 43 Done
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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 of 6, limiting the possibilities to just 3 pairs of ages (S,N)=(12,28), (18,22), (24,16). Two of these put Selina younger than Nancy, and one older than Nancy. 10 years ago the ages would have been (S-10,N-10)=(2,18), (8,12), (14,6), which add up to 20. The (S,N) pairs tell us for each pair what age Nancy was x years ago: (2), (3), (4), when Selina is now 12, 18 or 24 and Nancy is now 28, 22 or 16. From this we can deduce x: 26, 19 or 12, and we can work out (S-x,N-x): (-14,2), (-1,3), (12,4). The only sensible solution is (12,4) making (S,N)=(24,16), so Selina is 24 and Nancy is 16 years old. 12 years ago Selina was 12, half the age she is now. Another way of solving the problem is to use algebra and substitution: x years ago Nancy was N-x years old and S=6(N-x) and S=2(S-x). S=2S-2x, making x=S/2. N=40-S, so S=6(40-S-x)=240-6S-6x=240-6S-3S=240-9S, and 10S=240, making S=24 and N=16. The question actually contains too much information, because it is not necessary to know that Selina was "twice younger" when she was 6 times Nancy's age. S=24 and N=16 is the only logical answer.
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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 S = 15, then F = 50 Now add on T years (from now), this gives us, (F+T):(S+T) = 2:1     i.e. F + T = 2(S + T) 50 + T = 30 + 2T 20 = T So 20 years from now, the father's and son's ages will be 70 and 35, which is in the ratio of 2:1
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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 = 1728 13^3 = 2197 We can stop there. Obviously the child's age that is cubed can't be 13; 2197 is nearly 200 years in the future. Now, look for the age that is squared. Start at the top, 20. 20^2 = 400 Add 400 to 10^3: we get 1400, too far in the past. Add 400 to 11^3: we get 1731, also too far in the past. Add 400 to 12^3: we get 2128, at least a hundred years in the future. We have narrowed the cubed age to 12. Now we need a squared age that gives a reasonable sum for the mother's birth year. 10^2 = 100: add to 12^3: we get 1828, too far in the past. 11^2 = 121: add to 12^3: we get 1849, still too far in the past. 12^2 = 144: add to 12^3: we get 1872, too far back. 13^2 = 169: add to 12^3: we get 1897; age in 1941 = 44; possible. 14^2 = 196: add to 12^3: we get 1924; age in 1941 = 17; impossible. George's wife was born in 1897, and she was 44 in 1941. One child was 12, the other was 13.
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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 will be 88, so 88/40=11/5=2 1/11. 
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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+2C-18, C=7. So we can calculate R=2*C=14; and J=C+11=18.  Check: Carlos is half as old as Ramon, C is half of 14; Ramon and Jun will be 18 and 22 in 4 years' time. Half of Jun's age then (11) plus Carlos' age now (7) is 18. Half of Ramon's age in 4 years will be 9 which is 9 less than this sum. So the facts check out and Carlos is 7, Ramon is 14 and Jun is 18.  
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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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