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21 Aug 2017, 11:04
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
15%
(low)
Question Stats:
83% (01:45) correct
17%
(01:31)
wrong
based on 73
sessions
History
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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
A. 5
B. 7
C. 11
D. 13
E. 17
21 Aug 2017, 13:59
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Mahmud6
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.
septwibowo
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22 Aug 2017, 09:51
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Mahmud6
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.
