Five years ago Jim was three times as old as Raoul was and Monica was

Question

Five years ago Jim was three times as old as Raoul was and Monica was six years older than Raoul was. If all three are still living in five years, which of the following must be true about their ages five years from now?
 
I. Monica is older than Jim.
II. Raoul is six years younger than Monica
III. The combined ages of Jim and Raoul are more than Monica's age.

A. I only
B. II only
C. I and II
D. I and III
E. II and III

BrentGMATPrepNow · Accepted Answer

Let R = Raoul's PRESENT age
So, R - 5 = Raoul's age 5 YEARS AGO

Five years ago .... Monica was six years older than Raoul was.  
So, (R - 5) + 6 = Monica's age 5 YEARS AGO
In other words, R + 1 = Monica's age 5 YEARS AGO

Five years ago Jim was three times as old as Raoul was  
So, 3(R - 5) = Jim's age 5 YEARS AGO

IMPORTANT: In order for us to know the information about Raoul's age 5 years ago, it must be the case that Raoul's PRESENT age is greater than 5. Otherwise, Raoul wouldn't have been alive 5 years ago

To find the ages 5 years in the FUTURE, we must take these ages for 5 years ago and add 10 years.

So, (R - 5) + 10 = Raoul's age 5 YEARS IN THE FUTURE
R + 1 + 10 = Monica's age 5 YEARS IN THE FUTURE
3(R - 5) + 10 = Jim's age 5 YEARS IN THE FUTURE

SIMPLIFY to get: 
 R + 5 = Raoul's age 5 YEARS IN THE FUTURE
R + 11 = Monica's age 5 YEARS IN THE FUTURE
3R - 5 = Jim's age 5 YEARS IN THE FUTURE

Now, let's examine the statements:

I. Monica is older than Jim.  
MUST it be the case that  R + 11  is greater than  3R - 5 ?
No. 
If R = 10, then  R + 11 = 21  and  3R - 5 = 25 
So, if R = 10, Monica is NOT older than Jim (5 years from now)
So, statement 1 need not be true.

We can ELIMINATE answer choices A, C and D

IMPORTANT: Notice that the remaining answer choices (B and E) both say that statement II is correct. 
So, we need not check statement II, since it MUST be correct.

III. The combined ages of Jim and Raoul are more than Monica's age.

Is it true that  (3R - 5)  +  (R + 5)  >  (R + 11) ?
Let's simplify to get: 4R > R + 11
Subtract R from both sides to get: 3R > 11
Divide both sides by 3 to get: R > 11/3
MUST this be TRUE?
Yes. It must be true, because we earlier concluded that   it must be the case that R is greater than 5  
So statement III must be true.

Answer: E

rohit8865 · Answer

Case 1:- 
Let present age of raoul =6yrs
then 5 yrs back his age =1 yrs
then jim =3*Roaul= 3 yrs...
but monica's age was then =1+6=7 yrs

Case 2:- 
Let present age of raoul =25yrs
then 5 yrs back his age =20 yrs
then jim =3*Roaul= 60yrs...
but monica's age was then =20+6=26yrs

(1) From case1 --> monica is older than jim
From case2-----> monica is younger than jim
Not true

(2)true for both cases

(3) true for case 2
False for case 1
not true

Ans B

gmatexam439 · Answer

Let the present ages of Jim, Raoul and Monica be J, R and M respectively.

5 years ago, we will have the following equations as per the question:
(J-5)=3(R-5) which can be simplified to => J=3R-10 .................... 
and, M=R+6 ...................................................................................

As per   ,the minimum value of R should be minimum 4 for J to be alive.
Case 1: When R=4, J=2 and M=10
Case 2: When R=20, J=40 and M=26

Therefore, the answer should be II ---> B
  
Hi Bunuel, please correct me if I am wrong in my assumption regarding the question.

rohit8865 · Answer

Hope bunuel too respond....

but for now the highlighted part is not correct....
R should be minimum of 5 Yrs ...

gmatexam439 · Answer

I am sorry my silly mistake  ... but I think it should 6 and not 5 because if R=5, J will be 0 five years ago which is not as per our assumption .... therefore if R>=6, then 
case1: When R=6, J=8 and M=12
case2: as per above post;

Therefore answer should be E and not C.

AARONRAMSEY · Answer

Assume ages of Joe,Roula and monica five years back = J ,R,M respectivily

given that J = 3R and M=R+6 (this is 5 years back)
so 5 years from now the ages will be (we need to add 10 to each age)
J = 3R+10
M= R+16
and R = R+10

I. Monica is older than Jim.
Not true. because R should be at least 10 years old then J >M

II. Raoul is six years younger than Monica
True. From the equation R-M = 16-10 =6

III. The combined ages of Jim and Raoul are more than Monica's age.
True. J+M = 4R+20
M = R+16 
therefore J+M > M

Hence answer is E

Statement 1 :

Mo2men · Answer

Let CURRENT ages of Jim=j, Raoul=r, and Monica=m

Five Years ago: Jim=j-5, Raoul=r-5, and Monica=m-5

Set up equations: 
j-5=3(r-5)...........j+10=3r
m-5=r-5+6........m=r+6

Because we have only 2 equations in 3 variables, there is open cases. So we need to be careful to choose numbers that cover multiple cases.

Plug in numbers:
 case 1:  r=7, j=11, m=13, After 5 Yeas: r=12, j=16, m=18

case 2:  r=10, j=20, m=16, After 5 Years: r=15, j=25, m=21

Checking Numeral I as it is most frequent.

From case 1: m>j

From case 2: m<j

Not always true...........Eliminate A,C & D

To save time, check Numeral III not II. Because if you do II and get correct then you will move to III. But we do III first, you will eliminate one choice in one step.

From case 1: r+j>m

From case 2: r+j>m

Therefore, Eliminate choice B

Answer: E

ScottTargetTestPrep · Answer

We can let Jim’s age today = J, Raoul’s age today = R, and Monica’s age today = M. Let’s set up their ages 5 years ago: Jim was (J - 5), Rauol was (R - 5), and Monica was (M - 5). Since five years ago Jim was three times as old as Raoul was: (J - 5) = 3(R - 5) J - 5 = 3R - 15 J = 3R - 10 Since five years ago Monica was six years older than Raoul was: (M - 5) = (R - 5) + 6 M - 5 = R + 1 M = R + 6 Notice that Raoul is the youngest of the three people, and Raoul must be more than 5 years old since only then can we talk about their ages 5 years ago. Let’s now test each Roman numeral: I. Monica is older than Jim. We can represent Monica's age in 5 years as M + 5, or R + 6 + 5 = R + 11. We can represent Jim’s age in 5 years as J + 5, or 3R - 10 + 5 = 3R - 5. Is M + 5 > J + 5 ? Is R + 11 > 3R - 5 ? Is 16 > 2R ? Is 8 > R ? Is R < 8? We know that R > 5, however, we can’t determine whether R < 8. Thus, we cannot determine whether Monica is older than Jim. II. Raoul is six years younger than Monica Since M = R + 6, in 5 years Raoul will still be 6 years younger than Monica. Roman numeral II is true. III. The combined ages of Jim and Raoul are more than Monica's age. We already see that in 5 years, Monica’s age will be R + 11, Jim’s age will be 3R - 5, and Raoul’s age will be R + 5. We can create the following inequality: Is 3R - 5 + R + 5 > R + 11 ? Is 4R > R + 11 ? Is 3R > 11 ? Is R > 11/3 ? We know that R > 5, so R > 11/3. Thus Roman numeral III is true. Answer: E

energetics · Answer

I tried this way and got E. Don't know if it's good to do it this way though, please tell me if I made a mistake in my logic or setup!

1) Five years ago Jim was three times as old as Raoul
\(J=3x\) and\(R=x\)
2) Monica was six years older than Raoul was
\(M = x+6\)

If all three are still living in five years, which of the following must be true about their ages five years from now?
So the question is asking, what is true about their age NOW+5, given the limitation that x≥5 because Raoul can't be <0(we know x is a positive number because it's age). Since I set up the relationship to x 5 years in the past I have to add +10 years to get their age NOW+5. But, you can just calculate NOW by adding +5 and get the same result because all the relationships should hold regardless.

I. Monica is older than Jim.
\(M > J ?\)
\(x+6+5>3(x+5)?\)
\(5+11 > 3(5+5)?\)
\(16 > 30?\)
NO

II. Raoul is six years younger than Monica
\(R = M-6?\)
\(x+5 = x+6+5?\)
\(5+5 = 5+6+5\)
\(R=10 M=16\)
YES

III. The combined ages of Jim and Raoul are more than Monica's age.
\(J+R>M?\)
\(3(x+5) + x+5 > x+5+6?\)
\(3x+15+x+5 > x+11?\)
\(15+15+10 > 5+11?\)
\(40 > 16?\)
YES

So, answer is E. II and III
Time 3:38

Dooperman · Answer

Question becomes easier if you assume the ages 5 year ago for J, R, M. 
Ages 5 years ago 
Raoul's age = R; J = 3R; M= R+6

Ages 5 years from now 
Raoul's = R+10; J= 3R+10; M= R+16.

III) J+R = 4R + 20 which will always be greater than R + 16

I) M vs J. R+16 vs 3R+10. If R=2 then Yes. If R=4, then No.

bumpbot · Answer

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Five years ago Jim was three times as old as Raoul was and Monica was

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