Gerald and James each had a few one dollar bills. After James gave 16

Question

Competition Mode Question

Gerald and James each had a few one dollar bills. After James gave 16 one dollar bills to Gerald, the ratio of the number of bills that James and Gerald respectively had was 14/11. Gerald now returned 8 one dollar bills to James, thereby increasing the ratio of the number of bills that James and Gerald respectively had to 16/9. What was the ratio of the number of bills that James and Gerald respectively had originally?

A. 18/7
B. 46/29
C. 3/2
D. 74/51
E. 2/3

A. 18/7
B. 46/29
C. 3/2
D. 74/51
E. 2/3

sambitspm · Answer

See the attachment
IMO A

nick1816 · Answer

\frac{J-16}{G+16 }= \frac{14}{11}

\frac{J-8}{G+8} = \frac{16}{9}

J/G must be greater than 16/9 or 1.78

We don't have to solve the question algebraically. 
Option A > 2 
Option B= 1.6ish
As the options in GMAT are sorted, C, D and E must be less than 1.6. (Don't need to check them either)

Eliminate B,C, D and E.

A

unraveled · Answer

Gerald and James each had a few one dollar bills. After James gave 16 one dollar bills to Gerald, the ratio of the number of bills that James and Gerald respectively had was 14/11. Gerald now returned 8 one dollar bills to James, thereby increasing the ratio of the number of bills that James and Gerald respectively had to 16/9. What was the ratio of the number of bills that James and Gerald respectively had originally?

A. 18/7
B. 46/29
C. 3/2
D. 74/51
E. 2/3

A. 18/7
B. 46/29
C. 3/2
D. 74/51
E. 2/3
 \frac{J}{G} = ? 
Note that the ratio is increasing so E is out for  \frac{2}{3}  < 1.

\frac{J - 16}{G + 16} = \frac{14}{11}  i.e. 11J - 14G = 25*16
 \frac{J - 8}{G + 8} = \frac{16}{9}  i.e. 9J - 16G = 200

Solving the 2 eqn. 
J + G = 100

Hence the sum of the numerator and denominators or their multiples must be a factor of 100.
Only option A and C are suitable.

But since  \frac{16}{9} > \frac{3}{2} > \frac{14}{11} , A it is ( \frac{18}{7} > \frac{16}{9} > \frac{14}{11} ).

Alternatively,

Option B, D and E can be eliminated since B and D are ~ \frac{3}{2} . So, only A and C are the candidates. Equations are not needed since we can see that  \frac{16}{9} > \frac{3}{2} . Only  \frac{18}{7} > \frac{16}{9}

Answer A.

HoneyLemon · Answer

A , IMO .

As per question 
 (J-16) / (G+16) = 14/11 ----------------eq-1
(J-8) / (G+8) = 16/9 ----------------eq-2

Solving 2 linear equation we can find that J = 72 and G = 28

so J / G = 72 / 28 = 18 / 7 ... So option A is correct .

Saasingh · Answer

IMO A

Gerald --- James 
Originally gerald has G bills, James has J bills.

According to question

G+16 ---------- J-16 yields ratio 14:11 which is roughly 1.36

and G+16-(8) = (G+8) -------- J-16+(8)= (J-8) yields ratio 16:9 which is roughly 1.77

We see ratio has increased as Gerald's bills are reduced. (Also mentioned in the question)

Therefore, if we were to transfer 8 more bills from gerald to james (which by the way, leads to original ratio needed) the ratio must increase more.

G +8 -(8) = G -------- J-8+(8)= J

Ratio will be more than 16/9. Therefore more than 1.77.

Clearly 18/7 which is greater than 2 satisfies.

ALSO,

Its a good quality question IMO where one can fall in traps of lengthy solutions or wrong guesstimation.

When I saw the options I was sure it would be E thinking its the odd one out.

Please correct me if my approach is wrong.

Posted from my mobile device

exc4libur · Answer

j-16/g+16=14/11,
11j-11.16=14g+16.14,
11j-14g=16(11+14),
11j-14g=16(25)

j-8/g+8=16/9,
9j-8.9=16g+16.8,
9j-16g=8(9+16),
 *2,
18j-32g=16(25)

18j-32g=16(25)
11j-14g=16(25)
7j-18g=0,
j/g=18/7

Ans (A)

MayankSingh · Answer

IMO A

J-16/ G+16 = 14/11
J-8 / G+8 = 16/9

Easy Approach:

J-16/ G+16 = 14/11 = 28/22 = 42/33 = 56/44 
 J/G  = (28+16)/(22-16) Or, (42+16)/(33-16) Or, (56+16)/(44-16)
J/G = 44/6 Or 58/17 Or 72/28
J/G = 72/28 = 18/7 (matches A)

Ans- A. 18/7

Alternate: 
(J-16)/ (G+16) + 1= 14/11+1
J+G/G+16 = 25/11 -------(I)

(J-8)/ (G+8) + 1= 16/9+1
J+G/G+8 = 25/9 .......(II)

Dividing I by II
G+8/G+16 = 9/11
G=28
Putting G in I , J=72
so , J/G= 72/28 = 18/7 
Ans A.

lacktutor · Answer

James gave 16 one dollar bills to Gerald: 
 \frac{J-16}{ G+16}= \frac{14}{11}

Now, Gerald returned 8 one dollar bills to James:

\frac{14x +8}{11x- 8} = \frac{16}{9}

--->  126x +72 = 176x - 128

50x = 200

x = 4

J -16 = 14x= 56 
 J = 72

G +16 = 11x = 44 
 G = 28

\frac{J}{G }= \frac{72}{28} = \frac{18}{7}

Answer (A).

bumpbot · Answer

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Gerald and James each had a few one dollar bills. After James gave 16

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