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

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
\(\frac{J}{G} = ?\)
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.

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 .

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

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),
[9j-16g=8(25)]*2,
18j-32g=16(25)

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

Ans (A)

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.

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).

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