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Mona has $2.05 in quarters and dimes. How many quarters does she have? 17 Oct 2020, 11:59
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D
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55% (hard)

Question Stats:

58% (02:03) correct 42% (01:36) wrong based on 99 sessions

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Mona has $2.05 in quarters and dimes. How many quarters does she have? (A dime is worth 10 cents. A quarter is worth 25 cents.)

(1) She has more quarters than dimes.
(2) She has a total of ten coins.
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D
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Mona has $2.05 in quarters and dimes. How many quarters does she have? Updated on: 18 Oct 2020, 08:41
Originally posted by stne on 18 Oct 2020, 08:35.
Last edited by stne on 18 Oct 2020, 08:41, edited 2 times in total.
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Bunuel
Mona has $2.05 in quarters and dimes. How many quarters does she have? (A dime is worth 10 cents. A quarter is worth 25 cents.)

(1) She has more quarters than dimes.
(2) She has a total of ten coins.
Show more

Lets convert to cents and solve, she has a total of 205 cents
so \(10y+25x = 205\) where \(y\) is no. of cents and \(x\) is no. quarters.

By testing various combinations we see the following pattern emerges:
10(3)+25 (7)=205
10(8)+25(5)=205
10(13)+25(3)=205
We can see the pattern so for the further cases obviously dimes > quarters.



(1) She has more quarters than dimes.
Only one case fulfills this criteria
10(3)+25 (7)=205
We have quarters = 7
SUFF.

(2) She has a total of ten coins.
Again we can have only the following cases
10(9)+25(1)
10(8)+25(2)
10(7)+25(3)
10(6)+25(4)
10(5)+25(5)
10(4)+25(6)
10(3)+25(7)=205 This is the only case that works
10(2)+25(8)
10(1)+25(9)

SUFF.

IMO D
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Mona has $2.05 in quarters and dimes. How many quarters does she have? 18 Oct 2020, 08:37
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Given: Mona has $2.05 in quarters and dimes. (A dime is worth 10 cents. A quarter is worth 25 cents.)
To find: How many quarters does she have?

Solution: Let X = Quarters and Y = Dime
25X+10Y = 205 (2.05$ = 205 cents)

(1) She has more quarters than dimes.
X>Y
205 is not divisible by either 25 or 10

case 1: let there are 8 quarters = 200 cents , but 5 remaining cents are not divisible by 10 (Wrong)

case 2: let there are 7 quarters = 175 cents , but 30 remaining cents are divisible by 3 (10s) dimes (Keep this) (X>Y)

case 3: let there are 6 quarters = 150 cents , but 55 remaining cents are not divisible by 10 (Wrong)

case 4: let there are 5 quarters = 125 cents , but 80 remaining cents are divisible by 8 (10s) dimes (Keep this) (X<Y)

Only case 2 holds true.
(sufficient)

(2) She has a total of ten coins.
case 1: let there are 8 quarters = 200 cents , but 5 remaining cents are not divisible by 10 (Wrong)

case 2: let there are 7 quarters = 175 cents , but 30 remaining cents are divisible by 3 (10s) dimes (Keep this) (7+3 =10)

case 3: let there are 6 quarters = 150 cents , but 55 remaining cents are not divisible by 10 (Wrong)

case 4: let there are 5 quarters = 125 cents , but 80 remaining cents are divisible by 8 (10s) dimes (Keep this) (5+8 =13)
going further will increase no. of coins, only 2nd case hold true
(Sufficient)

Answer is D
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Mona has $2.05 in quarters and dimes. How many quarters does she have? 19 Oct 2020, 02:33
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Forget the conventional way to solve DS questions.

We will solve this DS question using the variable approach.

DS question with 2 variables and 1 Equation: Let the original condition in a DS question contain 2 variables and 1 Equation. Now, 2 variables and 1 Equation would generally require 1 more equation for us to be able to solve for the value of the variable.

We know that each condition would usually give us an equation, and Since we need 1 more equation to match the numbers of variables and equations in the original condition, the logical answer is D.

To master the Variable Approach, visit https://www.mathrevolution.com and check our lessons and proven techniques to score high in DS questions.

Let’s apply the 3 steps suggested previously. [Watch lessons on our website to master these 3 steps]

Step 1 of the Variable Approach: Modifying and rechecking the original condition and the question.

Let us assign the variables: Quarters (q) and Dimes(d)

We have to find the numebr of 'quarters(q)'. - where 1 Dime = 10 cents and 1 Quarter = 25 cents

=> 25q + 10b = $2.05 = 205 [1 dollar = 100 cents and therefore 2.05 * 100 = 205 cents]

=> For q = 7 and d = 3: 25 (7) + 10(3) = 175 + 30 = 205
=> For q = 5 and d = 8: 25 (5) + 10(8) = 125 + 80 = 205

Second and the third step of Variable Approach: From the original condition, we have 2 variables (q and b) and 1 Equation (25q + 10b = 205).To match the number of variables with the number of equations, we need 1 more equation. Since conditions (1) and (2) will provide 1 equation each, D would most likely be the answer.

Let’s take look at each condition separately.

Condition(1) tells us that she has more quarters than dimes.

=> Possible only for q = 7 and d = 3: 25 (7) + 10(3) = 175 + 30 = 205

Since the answer is unique , Condition(1) is alone sufficient by CMT 2.


Condition(2) tells us that she has a total of ten coins.

=> For ten coins we can have possibel combinations (q = 5 and d = 5, q =1 and d = 9, and so on.)

=> Only possible combination of q = 7 and d = 3: 25 (7) + 10(3) = 175 + 30 = will give 205

Since the answer is unique , Condition(2) is alone sufficient by CMT 2.


Each condition alone is sufficient.

So, D is the correct answer.

Answer: D


SAVE TIME: By Variable Approach, when you know that value of Con(1) = Con(2), then 'D' is the correct answer.
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