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In 1950, Richard was 4 times as old as Robert. In 1955, Richard was 3

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Bunuel
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In 1950, Richard was 4 times as old as Robert. In 1955, Richard was 3 [#permalink] New post  01 Dec 2015, 04:28
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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
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Re: In 1950, Richard was 4 times as old as Robert. In 1955, Richard was 3 [#permalink] New post  06 Dec 2015, 17:56
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Abhishek009 wrote:
From the picture above we have 4r+5 = 3r + 15

Or, r = 10

Quote:
In which year was Richard twice as old as Robert?


Let the number of years be " y "

so, 40 + x = 2 ( 10 + x )
Or, 40 + x = 20 + 2x

Or, x = 20

So, After 20 years from 1950 ie 1970 Richard will be twice as old as Robert.


i like your method of solving this question - simple and clean.
one typo though.. Let the number of years be " y " but in your equations, it is x.

thanks again!
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In 1950, Richard was 4 times as old as Robert. In 1955, Richard was 3 [#permalink] New post  Updated on: 07 Dec 2015, 07:20
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From the picture above we have 4r+5 = 3r + 15

Or, r = 10

Quote:
In which year was Richard twice as old as Robert?


Let the number of years be "x"

so, 40 + x = 2 ( 10 + x )
Or, 40 + x = 20 + 2x

Or, x = 20

So, After 20 years from 1950 ie 1970 Richard will be twice as old as Robert.

PS : Edited , all thanks to whitehalo
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Originally posted by Abhishek009 on 01 Dec 2015, 11:35.
Last edited by Abhishek009 on 07 Dec 2015, 07:20, edited 1 time in total.
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Re: In 1950, Richard was 4 times as old as Robert. In 1955, Richard was 3 [#permalink] New post  02 Dec 2015, 05:02
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In 1950 - Rich = 4 * Rob

In 1955 i.e. 5 years later (Rich + 5) = 4 * (Rob + 5) -- here replace the value of Rich as 4*Rob

Age of Rich in 1950 = 40
Age of Rob in 1950 = 10

In YEAR - (Rich,Rob)
In 1950 - (40 ,10),
In 1955 - (45 ,15),
In 1960 - (50 ,20),
In 1965 - (55 ,25),
In 1970 - (60 ,30) - twice

Option C is the correct answer.
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Re: In 1950, Richard was 4 times as old as Robert. In 1955, Richard was 3 [#permalink] New post  07 Dec 2015, 07:21
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whitehalo wrote:
i like your method of solving this question - simple and clean.
one typo though.. Let the number of years be " y " but in your equations, it is x.

thanks again!


Edited , thanks for pointing out !!
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In 1950, Richard was 4 times as old as Robert. In 1955, Richard was 3 [#permalink] New post  01 Dec 2015, 06:03
In 1950,
Let Richard's age = Ri
Robert's age = Ro

Ri = 4*Ro -- 1

In 1955,
Ri + 5 = 3(Ro + 5)
=>Ri + 5 = 3Ro + 15
=>Ri = 3Ro + 10 -- 2

From equations 1 and 2 ,
4Ro = 3Ro + 10
=> Ro = 10
Ri = 40

Let the number of years needed from 1950 be x
40 + x = 2(10+x) = 20 + 2x
=> x = 20

Therefore , the year is 1970.
Answer C
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Re: In 1950, Richard was 4 times as old as Robert. In 1955, Richard was 3 [#permalink] New post  01 Dec 2015, 18:24
In 1950: Ri = 4 Ro -------------- eq1
In 1955: Ri +5 = 3(Ro+5)---------eq2

Thus in 1950, solving eq1 and eq2
Ro= 10 , Ri =40

Now for each year we can calculate:
1960: Ri = 50 , Ro =20
1965: Ri= 55, Ro=25
1970: Ri= 60, Ro = 30

Thus Ans:C
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Re: In 1950, Richard was 4 times as old as Robert. In 1955, [#permalink] New post  14 Jan 2021, 09:20
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Re: In 1950, Richard was 4 times as old as Robert. In 1955, [#permalink]
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