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A purse contains 5-cent coins and 10-cent coins worth a total of $1.75

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A purse contains 5-cent coins and 10-cent coins worth a total of $1.75. If the 5-cent coins were replaced with 10-cent coins and the 10-cent coins were replaced with 5-cent coins, the coins would be worth a total of $2.15. How many coins are in the purse?

A. 26
B. 27
C. 28
D. 29
E. 30
[Reveal] Spoiler: OA

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New post 17 Sep 2017, 18:28
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Let the number of 5 cent coins be a and the number of 10 cent coins be b

we have been give 5a + 10b = 175 (total value in cents)
on reversing the numbers we get 5b + 10a = 215 (total value in cents)
add both the equations to get 15(a + b) = 390
therefore a + b = 26

Answer is (A)

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Re: A purse contains 5-cent coins and 10-cent coins worth a total of $1.75 [#permalink]

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New post 17 Sep 2017, 19:53
Let the total 5 cent coins be x, and the 10 cent coins be y

From the question stem, \(5x + 10y = 175\)
Value of y(by the following equation) : \(y = \frac{175 - 5x}{10}\)

Substituting this value of y in the second equation
\(10x + \frac{5(175 - 5x)}{10} = 215\)
\(100x + 875 - 25x = 2150\)
\(75x = 1275\)
\(x = 17\)

Substituting this value of x in equation\(5x+10y = 175\)
\(85 + 10y = 175\)
\(y = 9\)

Therefore the sum of coins in the p = 17 + 9 = 26(Option A)
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A purse contains 5-cent coins and 10-cent coins worth a total of $1.75 [#permalink]

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New post 17 Sep 2017, 21:26
Gnpth wrote:
A purse contains 5-cent coins and 10-cent coins worth a total of $1.75. If the 5-cent coins were replaced with 10-cent coins and the 10-cent coins were replaced with 5-cent coins, the coins would be worth a total of $2.15. How many coins are in the purse?

A. 26
B. 27
C. 28
D. 29
E. 30

I solved with two variables, and it works, but I have a question about my own method that applies to vishalbalwani 's method, too.

Let A = the number of 5-cent coins
Let B = the number of 10- cent coins

In one scenario, the coins, in their respective unknown quantities, total $1.75

In a second scenario, the 5-cent coins and 10-cent coins exchange quantities exactly, and they total $2.15

5A + 10B = 175 (P)
10A + 5B = 215 (Q)

Multiply (Q) by two, and subtract (P)

20 A + 10B = 430
__5A + 10B = 175
15A = 255
A = 17

There are 17 coins worth 5 cents.

Use (P) to find the number of 10-cent pieces

5(17) + 10B = 175
10B = 90
B = 9
There are 9 coins worth 19
0 cents.

A = 17, B = 9, total = 26 coins

Answer A

I have run this problem three ways, including vishalbalwani 's, and working from answer choices. This combination (number of coins), is the only one that works.

Question: Is the method sound? The variables work, but they seem inconsistent.

I use A as a quantity for 5-cent coins. The coefficients of the variables -- 5 and 10 -- are the values of the coins in cents.

But in the second equation, (Q) I do not think I have switched quantities, per the prompt. I think have switched values. In the second equation, (Q), the 5, a value, is in front of B -- which is supposed to be the quantity of 10-cent coins. vishalbalwani did the same.

What am I missing? We are supposed to be switching quantities. But switching values works.

I think pushpitkc avoids the whole problem by immediately defining y in terms of x from one equation.

Is my method sound?

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Re: A purse contains 5-cent coins and 10-cent coins worth a total of $1.75 [#permalink]

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New post 22 Sep 2017, 07:49
Gnpth wrote:
A purse contains 5-cent coins and 10-cent coins worth a total of $1.75. If the 5-cent coins were replaced with 10-cent coins and the 10-cent coins were replaced with 5-cent coins, the coins would be worth a total of $2.15. How many coins are in the purse?

A. 26
B. 27
C. 28
D. 29
E. 30


We can create two equations in which a = the initial number of 5-cent coins and b = the initial number of 10-cent coins. We can create the “money” equation by recalling that a 5-cent coin is $0.05 and a 10-cent coin is $0.10; thus:

0.05a + 0.1b = 1.75

And, after we reverse the coinage, we have:

0.1a + 0.05b = 2.15

Multiplying the first equation by -2, we have:

-0.1a - 0.2b = -3.50

Adding the two equations, we have:

-0.15b = -1.35

b = 9

Thus:

0.05a + (0.1)(9) = 1.75

0.05a = 0.85

a = 17

So, there is a total of 17 + 9 = 26 coins.

Answer: A
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A purse contains 5-cent coins and 10-cent coins worth a total of $1.75 [#permalink]

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New post 22 Sep 2017, 08:38
ScottTargetTestPrep wrote:
Gnpth wrote:
A purse contains 5-cent coins and 10-cent coins worth a total of $1.75. If the 5-cent coins were replaced with 10-cent coins and the 10-cent coins were replaced with 5-cent coins, the coins would be worth a total of $2.15. How many coins are in the purse?

A. 26
B. 27
C. 28
D. 29
E. 30

We can create two equations in which a = the initial number of 5-cent coins and b = the initial number of 10-cent coins. We can create the “money” equation by recalling that a 5-cent coin is $0.05 and a 10-cent coin is $0.10; thus:

0.05a + 0.1b = 1.75

And, after we reverse the coinage, we have:

0.1a + 0.05b = 2.15

Multiplying the first equation by -2, we have:

-0.1a - 0.2b = -3.50

Adding the two equations, we have:

-0.15b = -1.35

b = 9

Thus:

0.05a + (0.1)(9) = 1.75

0.05a = 0.85

a = 17

So, there is a total of 17 + 9 = 26 coins.

Answer: A

ScottTargetTestPrep , where you reverse the coinage, what do variables \(a\) and \(b\) stand for? (I had the same question about my own method, above.)

Perhaps the variables stand simply for the total number of coins, and absent any total number of coins, we cannot write one variable in terms of another.

Conceptually, however, that reasoning does not make a lot of sense to me.

We have two unknown quantities in two scenarios. In the first equation, the quantity b is defined as the number of $0.10 coins (and qty "a" initially is # of $.05 coins)

If "b" then switches to the number of $0.05 coins (vice versa for "a"), how are those two a's and b's the same, such that we can use them as an identity in solving two equations?

Sorry if this question is sophomoric; usually I can find my way out of a muddle. Not this time.

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Re: A purse contains 5-cent coins and 10-cent coins worth a total of $1.75 [#permalink]

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Gnpth wrote:
A purse contains 5-cent coins and 10-cent coins worth a total of $1.75. If the 5-cent coins were replaced with 10-cent coins and the 10-cent coins were replaced with 5-cent coins, the coins would be worth a total of $2.15. How many coins are in the purse?

A. 26
B. 27
C. 28
D. 29
E. 30



Hi....

A logical way...

When you interchange, the total amount increases, so 5-cents are more.
How much ? Say x, so 5x=215-175=40... X=8...

So initial amount is 5(y+8)+10y=175.....15y+40=175.....15y=135....y=9
Total coins = y+8+y=9+8+9=26

A
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New post 01 Oct 2017, 12:26
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genxer123 wrote:
ScottTargetTestPrep wrote:
Gnpth wrote:
A purse contains 5-cent coins and 10-cent coins worth a total of $1.75. If the 5-cent coins were replaced with 10-cent coins and the 10-cent coins were replaced with 5-cent coins, the coins would be worth a total of $2.15. How many coins are in the purse?

A. 26
B. 27
C. 28
D. 29
E. 30

We can create two equations in which a = the initial number of 5-cent coins and b = the initial number of 10-cent coins. We can create the “money” equation by recalling that a 5-cent coin is $0.05 and a 10-cent coin is $0.10; thus:

0.05a + 0.1b = 1.75

And, after we reverse the coinage, we have:

0.1a + 0.05b = 2.15

Multiplying the first equation by -2, we have:

-0.1a - 0.2b = -3.50

Adding the two equations, we have:

-0.15b = -1.35

b = 9

Thus:

0.05a + (0.1)(9) = 1.75

0.05a = 0.85

a = 17

So, there is a total of 17 + 9 = 26 coins.

Answer: A

ScottTargetTestPrep , where you reverse the coinage, what do variables \(a\) and \(b\) stand for? (I had the same question about my own method, above.)

Perhaps the variables stand simply for the total number of coins, and absent any total number of coins, we cannot write one variable in terms of another.

Conceptually, however, that reasoning does not make a lot of sense to me.

We have two unknown quantities in two scenarios. In the first equation, the quantity b is defined as the number of $0.10 coins (and qty "a" initially is # of $.05 coins)

If "b" then switches to the number of $0.05 coins (vice versa for "a"), how are those two a's and b's the same, such that we can use them as an identity in solving two equations?

Sorry if this question is sophomoric; usually I can find my way out of a muddle. Not this time.


The variables a and b stand for the initial number of 5- and 10-cent coins, before and after reversing the coinage. The only thing that changes when the coinage is reversed is that the number of 5-cent coins is now b and the number of 10-cent coins is now a.
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Re: A purse contains 5-cent coins and 10-cent coins worth a total of $1.75   [#permalink] 01 Oct 2017, 12:26
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