Erica has $460 in 5-and 10-dollar bills only. If she has fewer 10-than 5-dollar bills, what is the least possible number of 5-dollar bills she could have?
(A) 27
(B) 28
(C) 29
(D) 30
(E) 32
she has fewer 10
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- sanju09
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- ajith
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Say Erica has x 5 Dollar Bills and y 10 Dollar billssanju09 wrote:Erica has $460 in 5-and 10-dollar bills only. If she has fewer 10-than 5-dollar bills, what is the least possible number of 5-dollar bills she could have?
(A) 27
(B) 28
(C) 29
(D) 30
(E) 32
5x+10y =460
and
x>y
5x +10x >460
15x>460
x>31.66
The least integer value x could have is 32
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since the amount is $ 460 the number of $5 notes cannot be odd.
Possible answers 28, 30, 32
pluggin values
28 $5 but 32 $ 10
30 $ 5 31 $ 10
32 is the answer
or
If we try to divide the sum $ 460 into equal no. of $10 and $5,
then $450 can be divided into 30 notes each of $ 10 and $ 5
the remaining $10 (460-450) cannot be a $ 10 note
so it must be 2 $ 5 notes
minimum $5 among the choices = 30+2
Possible answers 28, 30, 32
pluggin values
28 $5 but 32 $ 10
30 $ 5 31 $ 10
32 is the answer
or
If we try to divide the sum $ 460 into equal no. of $10 and $5,
then $450 can be divided into 30 notes each of $ 10 and $ 5
the remaining $10 (460-450) cannot be a $ 10 note
so it must be 2 $ 5 notes
minimum $5 among the choices = 30+2
Another way to solve this problem is to look at the equation.
5x+10y = 460
where is the x is number of 5 dollar notes and y is the number of 10 dollar notes.
460 has the units digit of 0.
and 10y will always have a unit digit of 0.
5x then should also have the unit digit of 0.
Out of the options only 32 and 30 satisfy the condition.
Plug in the values
You'll find 32 as the answer.
5x+10y = 460
where is the x is number of 5 dollar notes and y is the number of 10 dollar notes.
460 has the units digit of 0.
and 10y will always have a unit digit of 0.
5x then should also have the unit digit of 0.
Out of the options only 32 and 30 satisfy the condition.
Plug in the values
You'll find 32 as the answer.
- cans
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let no. of 10 dollars = x and no. of 5 dollars = y
y>x
also 10x + 5y = 460
2x + y = 92
y should be even (as 2x is even and 92 is even)
let y=x+1 (y has to be greater than x)
3x+1=91; x is not an integer. Thus not possible
y=x+2;
3x+2=92 ->x=30;y=32
Thus least value of y = 32
y>x
also 10x + 5y = 460
2x + y = 92
y should be even (as 2x is even and 92 is even)
let y=x+1 (y has to be greater than x)
3x+1=91; x is not an integer. Thus not possible
y=x+2;
3x+2=92 ->x=30;y=32
Thus least value of y = 32
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- sanju09
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That's an excellent approach in which the properties of integers have been used to their best in deducing only 2 out of 5 choices as working, and then the GMAT's favorite plugging in course of action takes us to the right answer. Wonderful work! But, still Ajith's approach is the ideal one because ideally we need to deal in inequalities when solving word problems of the form where in "the least" or "the most" sort of values are to be answered.peelamedu wrote:Another way to solve this problem is to look at the equation.
5x+10y = 460
where is the x is number of 5 dollar notes and y is the number of 10 dollar notes.
460 has the units digit of 0.
and 10y will always have a unit digit of 0.
5x then should also have the unit digit of 0.
Out of the options only 32 and 30 satisfy the condition.
Plug in the values
You'll find 32 as the answer.
The mind is everything. What you think you become. -Lord Buddha
Sanjeev K Saxena
Quantitative Instructor
The Princeton Review - Manya Abroad
Lucknow-226001
www.manyagroup.com
Sanjeev K Saxena
Quantitative Instructor
The Princeton Review - Manya Abroad
Lucknow-226001
www.manyagroup.com
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I agree with "peelamedu's" and "ajith's" approach...
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