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by ddm » Mon Sep 08, 2008 1:45 pm
In 1950, Richard was 4 times as old as Robert. In 1955, Richard was 3 times as old as Robert. In which year was Richard twice as old as Robert?

(A) 1960

(B) 1965

(C) 1970

(D) 1975

(E) 1980

Your Answer : A
Actual Answer : C
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Re: age

by sudhir3127 » Mon Sep 08, 2008 9:26 pm
ddm wrote:In 1950, Richard was 4 times as old as Robert. In 1955, Richard was 3 times as old as Robert. In which year was Richard twice as old as Robert?

(A) 1960

(B) 1965

(C) 1970

(D) 1975

(E) 1980

Your Answer : A
Actual Answer : C
i wud do this Question by plugging in

1950 : Ro (X) = Ri ( 4x)

assume robert to be 10 yrs old

then richard will be 40

1955
Ro = 15, Ri = 45 ............... 3 times

1960
Ro= 20 Ri = 50

1965
Ro = 25 Ri = 55

1970

Ro = 30 Ri = 60 ...........................2 times

hence its 1970..

it takes not more than 30 secs to do it this way .. though u can solve it algebraically a well
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Re: age

by Stuart@KaplanGMAT » Mon Sep 08, 2008 10:53 pm
sudhir3127 wrote:
ddm wrote:In 1950, Richard was 4 times as old as Robert. In 1955, Richard was 3 times as old as Robert. In which year was Richard twice as old as Robert?

(A) 1960

(B) 1965

(C) 1970

(D) 1975

(E) 1980

Your Answer : A
Actual Answer : C
i wud do this Question by plugging in

1950 : Ro (X) = Ri ( 4x)

assume robert to be 10 yrs old

then richard will be 40

1955
Ro = 15, Ri = 45 ............... 3 times

1960
Ro= 20 Ri = 50

1965
Ro = 25 Ri = 55

1970

Ro = 30 Ri = 60 ...........................2 times

hence its 1970..

it takes not more than 30 secs to do it this way .. though u can solve it algebraically a well
This method works, but you got lucky that it only took 30 seconds.

You started by assuming starting ages of 10 and 40. These are actually the ONLY numbers that work on this question, since that's the only pair that give you 15 and 45 (x and 3x) 5 years later. If you had picked any numbers other than 10 and 40, you'd have had to start again.

So, while picking numbers is a great strategy, you did get a bit lucky :)

We could also solve algebraically:

If we let Ri be Richard's starting age and Ro be Robert's, we know that:

Ri = 4Ro
Ri + 5 = 3 (Ro + 5)

Subbing in, we get:

4Ro + 5 = 3Ro + 15
Ro = 10
Ri = 40

Then to actually solve, we let x represent the number of years since 1950:

Ri + x = 2(Ro + x)
40 + x = 2(10) + 2x
20 = x

So, we started in 1950, 20 years later it's 1970: choose (c).

As you see, picking numbers is much quicker, but there's no guarantee that you'll hit the right numbers on the first shot.
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