If 5 dollars and 35 crowns is equivalent to 7 pounds, and 4

7p−5d35=d+p147p−5d35=d+p14 --> 105d = 63p, doesn't exactly equal to 5/3 though.

To solve this problem, we need to determine the equivalent of 1 pound to dollars and 1 crown to dollars. Let d = value of 1 dollar, c = value of 1 crown and p = value of 1 pound. We are given that:

5d + 35c = 7p

4d + 4p = 56c

We can simplify the second equation as d + p = 14c. Let’s isolate d in each equation:

5d = 7p - 35c → [1]

d = -p + 14c → [2]

If we multiply equation [2] by 7 and add that to equation [1], we have:

12d = 63c

c = 12d/63 = 4d/21

Similarly, if we multiply equation [1] by 2 and equation [2] by 5 and add those together, we have:

15d = 9p

p = 15d/9 = 5d/3

Thus, 1 pound and 28 crowns is equivalent to:

5d/3 + 28 x 4d/21 = 5d/3 + 16d/3 = 21d/3 = 7d or 7 dollars

Answer: A

\(?\,\,\,:\,\,\,p + 28c = f\left( d \right)\)

\(\left( {\rm{I}} \right)\,\,5d + 35c = 7p\)

\(4d + 4p = 56c\,\,\,\, \Rightarrow \,\,\,\,\left( {{\rm{II}}} \right)\,\,d + p = 14c\)


\(\left\{ \matrix{\\
2 \cdot \left( {\rm{I}} \right)\,\,\,\,10d + 70c = 14p\,\,\, \hfill \cr \\
5 \cdot \,\left( {{\rm{II}}} \right)\,\,\,5d + 5p = 70c \hfill \cr} \right.\,\,\,\mathop \Rightarrow \limits^{\left( + \right)} \,\,\,\,\,\,15d = 9p\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,p = {5 \over 3}d\,\,\,\,\left( {{\rm{III}}} \right)\)


\(\left\{ \matrix{\\
\left( {\rm{I}} \right)\,\,\,\,5d + 35c = 7p\,\,\, \hfill \cr \\
7 \cdot \,\left( {{\rm{II}}} \right)\,\,\,7d + 7p = 98c \hfill \cr} \right.\,\,\,\mathop \Rightarrow \limits^{\left( + \right)} \,\,\,\,\,\,12d = 63p\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,c = {4 \over {21}}d\,\,\,\,\left( {{\rm{IV}}} \right)\)


\(?\,\,:\,\,\,p + 28c\,\,\,\mathop = \limits^{{\text{III}}\,,\,\,{\text{IV}}} \,\,\,\,\frac{5}{3}d + 28\left( {\frac{4}{{21}}d} \right) = \frac{5}{3}d + \frac{{16}}{3}d = 7d\,\,\,\,\, \Rightarrow \,\,\,\,\boxed{\,\,? = \$ 7\,\,}\)



This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.

Here's the long version of the answer -
7p - 35c = 5d
4p - 56c = -4d
solve this (this is the only step that takes time)
we get 252c = 48d
or 126c = 24d
or 42 c = 8d (hold onto this)

4d + 4p = 56c
1d + 1p = 14c
so 1p = 14c - 1d
so 1p + 28c = 42c - 1d
using 42c = 8d

Answer = 1p + 28c = 7d

It takes around 2.5-3 minutes to solve

Consider a dollar to be = D.
Crown = C.
Pound = P.
We have :
5*D + 35*C = 7*P. (1)
4*D + 4*P = 56*C or D + P = 14*C. (2)
Multiplying (1) with 2 we have :
10*D + 70*C = 14*P.
Multiplying (2) with 5 we have :
5*D +5*P = 70*C.
Equating the 70*C in the two we have :
5*D + 5*P = 14*P - 10*D.
5*D = 3*P.
Similarly equating the 1 and 2 we have :
4P = 35C.
Hence \(28C = \frac{16P}{5}\). Simialrly one P .
= \(\frac{21P}{5}\)
= Since 5D = 3P.
= 7 Dollars.


Asked: If 5 dollars and 35 crowns is equivalent to 7 pounds, and 4 dollars and 4 pounds is equivalent to 56 crowns, 1 pound and 28 crowns is equivalent to how many dollars?

Let D denote dollars, C denote crowns and P denote pounds

5D + 35C = 7P
D + 7C = 1.4P

4D + 4P = 56C
D + P = 14C

2.8P - 2D = D + P
3D = 1.8P
P = 3D/1.8 = 30D/18 = 10D/6 = 5D/3

D + 7C = 1.4*5D/3 = 7D/3
7C = 4D/3
C = 4D/21

P + 28C = 5D/3 + 28*4D/21 = 5D/3 + 16D/3 = 21D/3 = 7D = $7

IMO A

Let X = number of crowns/pound
Y = crowns/dollar

therefore number of dollars/pound = X/Y

Since you're trying to find the number of dollars, rearrange the two given statements to equal dollars:

5 = 7*X/Y - 35/Y

4 = 56/Y - 4*X/Y

Use the first statement to solve for Y. Multiply both sides by Y: 5*Y = 7*x -35> therefore Y=(7*X-35)/5

Multiply the second statement both sides by Y: 4*Y = 56-4*X

Substitute Y=(7*X-35)/5: 4(7*X-35)/5 = 56 - 4*X

Solve for X= 105/12 = 35/4.

Substitute this value of X to solve for Y = (7*(35/4) - 35)/5 = 21/4

The question asks for how many dollars equals 1 pound and 28 crowns. Algebraically, this equals :

1*(X/Y) + 28/Y = (35/21) + 28*(4/21)

= 7(5 + 16)/21 = 7

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