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8 years ago, the ratio of A’s age to B’s age was 4:5. At present the

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jackfr2
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8 years ago, the ratio of A’s age to B’s age was 4:5. At present the [#permalink] New post   29 Feb 2020, 14:36
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8 years ago, the ratio of A’s age to B’s age was 4:5. At present the ratio of B’s age to C’s age is 4:5. At present, the difference between A’s age and C ‘s age is 20 years, then what is the sum of the ages of A, B and C?


A. 138 years
B. 122 years
C. 164 years
D. 148 years
E. 152 years
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D
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GMATinsight
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Re: 8 years ago, the ratio of A’s age to B’s age was 4:5. At present the [#permalink] New post   29 Feb 2020, 23:29
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jackfr2 wrote:
8 years ago, the ratio of A’s age to B’s age was 4:5. At present the ratio of B’s age to C’s age is 4:5. At present, the difference between A’s age and C ‘s age is 20 years, then what is the sum of the ages of A, B and C?


A. 138 years
B. 122 years
C. 164 years
D. 148 years
E. 152 years


Current age of A = A
Current age of B = B = 4x
Current age of C = C = 5x


8 years ago, the ratio of A’s age to B’s age = (A-8)/(B-8) = 4:5

i.e. A's Age < B's Age

B:C = 4:5
i.e. if B = 4x then C = 5x

i.e. B's Age < C's Age

i.e. A < B < C

ALso, Difference of current ages of A and C = C-A = 20

i.e. A = C-20 = 5x-20

Now, using (A-8)/(B-8) = 4:5
i.e. 5A - 40 = 4B - 32
i.e. 5 (5x-20) - 4*(4x) = 40-32 = 8

i.e. 25x - 16x = 8+100 = 108
i.e. 9x = 108
i.e. x = 12

sum of the ages of A, B and C = A+B+C = (5x-20) + 4x + 5x = 14x - 20 = 14*12 - 20 = 148

Answer: Option D
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Re: 8 years ago, the ratio of A’s age to B’s age was 4:5. At present the [#permalink] New post   01 Mar 2020, 01:25
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need to notice the factor of age ∆ amongst a,b,c here
i.e c>b>a
so
a-8/b-8 = 4/5
b/c=4/5
c-a=20
solve in terms of a
5a-40=4b-32
5a-16/5*c= 8
25a-16c=40
25a-320-16a=40
a=40
then
200-4b=8
b=48
and
c = 48*5/4 ; 60
a+b+c = 40+48+60 ; 148
IMO D


jackfr2 wrote:
8 years ago, the ratio of A’s age to B’s age was 4:5. At present the ratio of B’s age to C’s age is 4:5. At present, the difference between A’s age and C ‘s age is 20 years, then what is the sum of the ages of A, B and C?


A. 138 years
B. 122 years
C. 164 years
D. 148 years
E. 152 years
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AbhaGanu
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Re: 8 years ago, the ratio of A’s age to B’s age was 4:5. At present the [#permalink] New post   19 May 2021, 07:08
Archit3110 how did you get 5a-40=4b-32?
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Re: 8 years ago, the ratio of A’s age to B’s age was 4:5. At present the [#permalink] New post   19 May 2021, 07:59
AbhaGanu wrote:
Archit3110 how did you get 5a-40=4b-32?


AbhaGanu
I have just simplified the expression of the ratio...
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AbhaGanu
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Re: 8 years ago, the ratio of A’s age to B’s age was 4:5. At present the [#permalink] New post   19 May 2021, 08:18
Can you please elaborate? I'm unable to get it that way. Archit3110

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Re: 8 years ago, the ratio of A’s age to B’s age was 4:5. At present the [#permalink] New post   19 May 2021, 08:35
AbhaGanu
question says 8 years ago, the ratio of A’s age to B’s age was 4:5
we can write it as : a-8/b-8 = 4/5 simplify this by cross multiplication we get 5a-40=4b-32 ---(1)

Later question says : At present the ratio of B’s age to C’s age is 4:5 i.e b/c=4/5 ---(2)

and At present, the difference between A’s age and C ‘s age is 20 years i.e c-a=20---(3)

Target what is the sum of the ages of A, B and C?

using 1,2 & 3 solve to get values of a,b,c
you can solve in terms of a
5a-40=4b-32
5a-16/5*c= 8
25a-16c=40
25a-320-16a=40
a=40
then
200-4b=8
b=48
and
c = 48*5/4 ; 60
a+b+c = 40+48+60 ; 148
IMO D

hope this helps :)


AbhaGanu wrote:
Can you please elaborate? I'm unable to get it that way. Archit3110

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Re: 8 years ago, the ratio of A’s age to B’s age was 4:5. At present the [#permalink] New post   22 May 2021, 04:40
Hi Archit3110 ! Thank you for the explanation. But I still haven't understood how has 5a-16/5*c= 8 arrived?

GMATinsight How have you got i.e. 5 (5x-20) - 4*(4x) = 40-32 = 8?
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Re: 8 years ago, the ratio of A’s age to B’s age was 4:5. At present the [#permalink] New post   22 May 2021, 04:49
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GMATinsight wrote:
jackfr2 wrote:
8 years ago, the ratio of A’s age to B’s age was 4:5. At present the ratio of B’s age to C’s age is 4:5. At present, the difference between A’s age and C ‘s age is 20 years, then what is the sum of the ages of A, B and C?


A. 138 years
B. 122 years
C. 164 years
D. 148 years
E. 152 years


Current age of A = A
Current age of B = B = 4x
Current age of C = C = 5x


8 years ago, the ratio of A’s age to B’s age = (A-8)/(B-8) = 4:5

i.e. A's Age < B's Age

B:C = 4:5
i.e. if B = 4x then C = 5x

i.e. B's Age < C's Age

i.e. A < B < C

ALso, Difference of current ages of A and C = C-A = 20

i.e. A = C-20 = 5x-20

Now, using (A-8)/(B-8) = 4:5
i.e. 5A - 40 = 4B - 32
i.e. 5 (5x-20) - 4*(4x) = 40-32 = 8

i.e. 25x - 16x = 8+100 = 108
i.e. 9x = 108
i.e. x = 12

sum of the ages of A, B and C = A+B+C = (5x-20) + 4x + 5x = 14x - 20 = 14*12 - 20 = 148

Answer: Option D



AbhaGanu

Check the highlighted parts

I just substituted teh values of A and B in the equation 5A - 40 = 4B - 32 i.e. 5A - 4B = 40 - 32

I hope this help! :)
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Re: 8 years ago, the ratio of A’s age to B’s age was 4:5. At present the [#permalink] New post   23 May 2021, 12:33
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Current Age of B=4x
Current age of C=5x
Current Age of A=C-20 = 5x-20
Sum of their ages = 4x+5x+5x-20 = 14x-20=?
Let's find x then,
Ages 8 years ago = A/B=4/5 = 5x-20-8/4x-8=4/5
After Solving x=12
14x-20 = 14(12)-20 = 168-20=148 (D)
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Re: 8 years ago, the ratio of A’s age to B’s age was 4:5. At present the [#permalink]
23 May 2021, 12:33
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