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12 Mar 2023, 06:16
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B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
45%
(medium)
Question Stats:
71% (03:06) correct
29%
(03:21)
wrong
based on 104
sessions
History
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Not Attempted Yet
Ian regularly empties his pockets of coins into a jar. One day, he counts the money and finds that he has $22.50 in nickels (worth 5 cents each), dimes (each worth 10 cents), and quarters (each worth 25 cents.) He has 3 times as many nickels as dimes, and 6 more quarters than dimes. How much money does he have in quarters?
A $11.25
B $12.00
C $15.50
D $19.75
E $20.50
A $11.25
B $12.00
C $15.50
D $19.75
E $20.50
10 Apr 2023, 05:07
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Bunuel
Solution:
- let n = no. of nickels,
- d = no. of dimes
- q = no. of quarters
- According to question, \(\frac{n}{20}+\frac{d}{10}+\frac{q}{4}=22.5\)
\(⇒\frac{n}{20}+\frac{d}{10}+\frac{q}{4}=\frac{225}{10}\)
\(⇒n+2d+5q=450......(i)\) - According to question, 3 times as many nickels as dimes or \(n=3d\)
- 6 more quarters than dimes or \(q=d+6\)
- Plugging \(n=3d\) and \(q=d+6\) in equation \(i\), we get \(3d+2d+5(d+6)=450\) or \(d=42\)
- So, \(q=d+6=42+6=48\)
- Amount in quater \(=\frac{q}{4}=\frac{48}{4}=$12\)
Hence the right answer is Option B
10 Apr 2023, 05:44
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Bunuel
If the instinct is to set up an algebraic equation and implement variables that don't appear in the question, pause for a second and ask if Plugging In The Answers (PITA) might be a faster, better route to answering the question. In this case, it probably is. I like trying B or D first.
B: $12 in quarters is 48 quarters. That means 42 dimes, worth $4.20. And 126 nickels, worth $6.30. Does $12+$4.20+$6.30 = $22.50? Yep; done!
Answer choice B.
It's worth pointing out why I like trying B or D. When trying B, there are three possible outcomes. 1-in-5 that we got lucky and picked the correct one. If we ended up with too much money on this question, we would be able to eliminate C, D, and E and would therefore also be done. If we ended up with too little money on this question, we would be able to eliminate A, leaving us with C, D, and E, at which point we would simply try D...if D were too big, it would be C; if D were too small, it would be E. No matter what, we never need to try more than two answer choices.
ThatDudeKnowsPITA
ThatDudeKnowsPluggingInTheAnswers
