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Hadeeqa has 20 coins consisting of quarters and dimes. If she has a to

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New post 21 Aug 2017, 11:04
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Hadeeqa has 20 coins consisting of quarters and dimes. If she has a total of $3.05, how many dimes does she have?

A. 5
B. 7
C. 11
D. 13
E. 17
[Reveal] Spoiler: OA

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Hasan Mahmud

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Hadeeqa has 20 coins consisting of quarters and dimes. If she has a to [#permalink]

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New post 21 Aug 2017, 13:59
Mahmud6 wrote:
Hadeeqa has 20 coins consisting of quarters and dimes. If she has a total of $3.05, how many dimes does she have?

A. 5
B. 7
C. 11
D. 13
E. 17

Let x = # of dimes
Let y = # of quarters

x + y = 20
y = 20 - x

.10x + .25y = 3.05

10x + 25y = 305

10(x) + 25(20 - x) = 305

10x + 500 - 25x = 305

195 = 15x

\(\frac{195}{15}\) = 13 = x

There are 13 dimes.

Answer D

Check:
13 * .10 = $1.30
7 * .25 = $1.75

$1.30 + $1.75 = $3.05
------
Alternatively, use answer choices. Example with Answer A.

1) Multiply the number (5) by .10:

(5)(.10) = .50

2) Subtract from 3.05
3.05 - .50 = 2.55

3) 2.55 is NOT divisible by .25. Reject and move on

Keep going until you find the value which, after subtraction by the dimes' value, lets you divide by .25. That's D, where 1.75/.25 = 7

Then add # dimes + # quarters to be sure you have 20 coins.

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Re: Hadeeqa has 20 coins consisting of quarters and dimes. If she has a to [#permalink]

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New post 22 Aug 2017, 09:51
Mahmud6 wrote:
Hadeeqa has 20 coins consisting of quarters and dimes. If she has a total of $3.05, how many dimes does she have?

A. 5
B. 7
C. 11
D. 13
E. 17


I use principal from mixture problems.

Let X = quarters and Y = dimes.

\(X + Y = 20\) ...(equation 1)
\(0.25X + 0.1Y = 3.05\)
multiply both sides with 100 : \(25X + 10Y = 305\) ...(equation 2)

We want to know how manys dimes (Y), so we change our equation to Y.
- From eq.1 : \(X = 20-Y\).
- Plug into eq.2 : \(25(20-Y) + 10Y = 305\) --> \(500 - 25Y + 10Y = 305\) --> \(-15Y = -195.\) Y is 13.

D.
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Re: Hadeeqa has 20 coins consisting of quarters and dimes. If she has a to [#permalink]

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New post 22 Aug 2017, 23:42
Mahmud6 wrote:
Hadeeqa has 20 coins consisting of quarters and dimes. If she has a total of $3.05, how many dimes does she have?

A. 5
B. 7
C. 11
D. 13
E. 17


lets solve backwards

if 17 dimes then 170 + 25*3 = 245 not equal to 305
if 13 dimes then 130 + 25*7 = 130 + 175 = 305 answer

D
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Re: Hadeeqa has 20 coins consisting of quarters and dimes. If she has a to [#permalink]

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New post 24 Aug 2017, 16:01
Mahmud6 wrote:
Hadeeqa has 20 coins consisting of quarters and dimes. If she has a total of $3.05, how many dimes does she have?

A. 5
B. 7
C. 11
D. 13
E. 17


We will create a “number of coins” equation and a “money” equation. Let’s let d = the number of dimes and q = the number of quarters. We create the “number of coins” equation as:

q + d = 20

q = 20 - d

The “money” equation uses the facts that a quarter is 25 cents, a dime is 10 cents, and $3.05 is 305 cents:

25q + 10d = 305

5q + 2d = 61

We substitute (20 - d) for q and get:

5(20 - d) + 2d = 61

100 - 5d + 2d = 61

-3d = -39

d = 13

Answer: D
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Re: Hadeeqa has 20 coins consisting of quarters and dimes. If she has a to   [#permalink] 24 Aug 2017, 16:01
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