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souvonik2k
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x years ago, David's age was twice his son's age and 12 Feb 2018, 07:58
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75% (hard)

Question Stats:

60% (03:28) correct 40% (03:14) wrong based on 90 sessions

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x years ago, David's age was twice his son's age and 5x years ago, David's age was thrice his son's age. If the difference of their present ages is 24 years, then find the sum of their present ages (in years).
A) 65
B) 78
C) 88
D) 91
E) Cannot be determined
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B
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x years ago, David's age was twice his son's age and 12 Feb 2018, 08:15
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souvonik2k
x years ago, David's age was twice his son's age and 5x years ago, David's age was thrice his son's age. If the difference of their present ages is 24 years, then find the sum of their present ages (in years).
A) 65
B) 78
C) 88
D) 91
E) Cannot be determined
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Let son's current age be \(y\)

Son's age \(x\) years ago\(= y-x\) & David's age \(x\) years ago\(=2(y-x)=2y-2x\)

Hence David's current age \(= 2y-2x+x=2y-x\) -----(1)

Son's age \(5x\) years ago \(= y-5x\) & David's age \(5x\) years ago \(= 3(y-5x)=3y-15x\)

So from this equation David's current age \(= 3y-15x+5x=3y-10x\) -------(2), equating (1) & (2)

\(3y-10x=2y-x => y=9x\)

Hence David's current age \(= 2y-x=2*9x-x=17x\)

Difference of their current ages \(= 17x-9x=24 =>x=3\)

Hence current age of Son \(=9*3=27\) & David\(=17*3=51\)

So Sum of current ages \(= 27+51=78\)

Option B
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x years ago, David's age was twice his son's age and 12 Feb 2018, 08:36
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Let David's age be D and his son's age be S

From x years ago, D-x = 2(S-x) = 2S - 2x -> D = 2S - x

From 5x years ago, D-5x = 3(S-5x)
Using D = 2S - x the equation can be rewritten as 2S-6x = 3S-15x -> S = 9x
D = 2S - x = 18x - x = 17x -> D = 17x

Difference in ages is 24
D-S = 24 -> 8x = 24 -> x = 3

Therefore, the sum of their ages is 9x+17x = 26x = 78(Option B)
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x years ago, David's age was twice his son's age and 12 Feb 2018, 18:40
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x years ago, David's age was twice his son's age and 5x years ago, David's age was thrice his son's age. If the difference of their present ages is 24 years, then find the sum of their present ages (in years).
A) 65
B) 78
C) 88
D) 91
E) Cannot be determined
Show more
Even if you can't "see" the whole method, start with something that IS easily translated. How the equations work together is likely to become clear.
Let D = David's (father) age
Let S = son's age

The difference of their present ages is 24 years:
\(D - S = 24\)
\(D = S + 24\)

\(x\) years ago, D was twice his son's age:
\(D - x = 2(S - x)\)
\(D - x = 2S - 2x\)
Substitute \(D = (S + 24)\)
\((S + 24) - x = 2S - 2x\)
\(24 = S - x\) ······(P)

\(5x\) years ago, D was thrice his son's age:
\(D - 5x = 3(S - 5x)\)
\(D - 5x = 3S - 15x\)
Substitute \(D = (S + 24)\)
\((S + 24) - 5x = 3S - 15x\)
\(24 = 2S - 10x\) ·····(Q)

Now we have a system of equations:
\(24 = S - x\) ··········(P)
\(24 = 2S - 10x\) ·····(Q)

We need to eliminate one variable. Eliminate S (easier).
Multiply (P) by 2, then subtract (Q) from (P)

48 = 2S - 2x
-24 = -2S -(-)10x
24 = 8x
x = 3

Plug x = 3 into either original equation (here, P)
\(24 = S - x\)
\(24 = S - 3\)
\(27 = S\). Son is 27 years old. Dad is \((S+24) = (27 + 24) = 51\) years old

Sum their ages: \(27 + 51 = 78\)

Answer B
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x years ago, David's age was twice his son's age and 12 Feb 2018, 23:10
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souvonik2k
x years ago, David's age was twice his son's age and 5x years ago, David's age was thrice his son's age. If the difference of their present ages is 24 years, then find the sum of their present ages (in years).
A) 65
B) 78
C) 88
D) 91
E) Cannot be determined
Show more

It's always better to use fewer variables. Say son's current age is S so David's current age is S + 24.
We need to find 2S + 24.

"x years ago, David's age was twice his son's age"

(S + 24 - x) = 2(S - x)
x = S - 24

5x years ago, David's age was thrice his son's age

(S + 24 - 5x) = 3(S - 5x)
10x = 2S - 24 (substitute x from above)
10(S - 24) = 2S - 24
8S = 216
S = 27

2S + 24 = 78

Answer (B)
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x years ago, David's age was twice his son's age and 08 Jun 2020, 08:34
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