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Garcia purchased two kinds of premium coffee from Harbucks Colorado Co

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Garcia purchased two kinds of premium coffee from Harbucks Colorado Co  [#permalink]

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New post 24 Sep 2017, 23:04
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Question Stats:

56% (02:44) correct 44% (02:37) wrong based on 91 sessions

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Garcia purchased two kinds of premium coffee from Harbucks Colorado Coffee Emporium. If premium coffee is only sold in one pound bags and each pound of premium mild coffee costs $2.10 and each pound of premium hearty costs $2.30, how many pounds of premium mild did Garcia buy?

(1) The total value of the coffee Garcia purchased was $17.80
(2) If Garcia had purchased 1 more pound of each type of coffee he would have spent $22.20

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Garcia purchased two kinds of premium coffee from Harbucks Colorado Co  [#permalink]

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New post 25 Sep 2017, 02:00
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If the quantity of premium mild coffee is x and the quantity of premium hearty is y
From the question stem, the cost of the coffee is 2.1x + 2.3y

1. The total value of the coffee Garcia purchased was $17.80
2.1x + 2.3y = 17.8
y=5 is the only possibility, where x = 3(Sufficient)

2. If Garcia had purchased 1 more pound of each type of coffee he would have spent $22.20
2.1(x+1) + 2.3(y+1) = 22.20
2.1x + 2.3y = 17.8(on rewriting the equation) (Sufficient)(Option D)
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Garcia purchased two kinds of premium coffee from Harbucks Colorado Co  [#permalink]

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New post 26 Sep 2017, 00:05
pushpitkc wrote:
If the quantity of premium mild coffee is x and the quantity of premium hearty is y
From the question stem, the cost of the coffee is 2.1x + 2.3y

1. The total value of the coffee Garcia purchased was $17.80
2.1x + 2.3y = 17.8
y=6 is the only possibility, where x = 0 (Sufficient)

2. If Garcia had purchased 1 more pound of each type of coffee he would have spent $22.20
2.1(x+1) + 2.3(y+1) = 22.20
2.1x + 2.3y = 17.8(on rewriting the equation) (Sufficient)(Option D)


1) 2.1x + 2.3y = 17.8 .. if y=6 and x=0 >>> 0+ 2.3 x 6 = 13.8 is not equal to 17.8. However, x=3 and y=5 can solely solve the problem>>>> 2.1 x 3 (6.3) + 2.3 x 5 ( 11.5) = 17.8

2) the same thing
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Garcia purchased two kinds of premium coffee from Harbucks Colorado Co  [#permalink]

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New post 26 Sep 2017, 08:46
Thanks for bringing that up tjkim88
Made the necessary change!

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Re: Garcia purchased two kinds of premium coffee from Harbucks Colorado Co  [#permalink]

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New post 08 Oct 2017, 23:39
Can anyone please suggest a simpler way to solve this question? We surely can do trial and error. It will take ages! 178 is not divisible by either 21 or 23. The LCM of 21 and 23 is 483. How do we get 5 and 3?
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Re: Garcia purchased two kinds of premium coffee from Harbucks Colorado Co  [#permalink]

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New post 09 Oct 2017, 00:23
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tejas0999 wrote:
Can anyone please suggest a simpler way to solve this question? We surely can do trial and error. It will take ages! 178 is not divisible by either 21 or 23. The LCM of 21 and 23 is 483. How do we get 5 and 3?




1) total cost is 17.80. focusing on tenth decimal 0.8. 0.1(2.1)a and 0.3(2.3)a should make 0.8. therefore, we can easily to figure out either 0.1X5 + 0.3X1 or 0.3X5 + 0.1X3 (0.1(2.1) X2+0.3(2.3) X2 is too small)
1. 2.1X5 + 2.3X1 = 10.5+2.3= 12.8 (x)
2. 2.1X3 + 2.3X5 = 6.3+11.5= 17.8 (o)
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Re: Garcia purchased two kinds of premium coffee from Harbucks Colorado Co  [#permalink]

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New post 09 Oct 2017, 01:38
Bunuel wrote:
Garcia purchased two kinds of premium coffee from Harbucks Colorado Coffee Emporium. If premium coffee is only sold in one pound bags and each pound of premium mild coffee costs $2.10 and each pound of premium hearty costs $2.30, how many pounds of premium mild did Garcia buy?

(1) The total value of the coffee Garcia purchased was $17.80
(2) If Garcia had purchased 1 more pound of each type of coffee he would have spent $22.20



Yes, it is difficult to test numbers but there are ways to make it less calculation intensive..

Since cost of ONE pound of both is given, so A,B and C are out...

The equation we get is ...
2.1x+2.3y=17.8...........2.1x=17.8-2.3y
As 2.1 is a MULTIPLE of 3, let's make other part multiple of 3....
17.8 is 1+7+8=16 so would leave a Remainder of 1...
But 2.3 will leave a Remainder of 2 when div by 3....
So y will be 3a+2.....
a is 0, y is 2.......17.8-2.3*2=17.8-4.6=13.2, not a multiple of 2.1
a is 1, y is 5....... 17.8-2.3*5=17.8-11.5=6.3=2.1*3..
y as 8 will go beyond 17.8..
So only value y is 5..
Sufficient

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Re: Garcia purchased two kinds of premium coffee from Harbucks Colorado Co  [#permalink]

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New post 10 Dec 2017, 13:42
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Hi All,

We're told that coffee is only sold in one pound bags, each pound of mild coffee costs $2.10 and each pound of hearty coffee costs $2.30. We're asked for the number of pounds of mild coffee that Garcia bought.

1) The total value of the coffee Garcia purchased was $17.80

Since each type of coffee costs a little more than $2, this $17.80 total means that there were likely 8-9 total pounds of coffee purchased. This severely limits the number of possible 'combinations' of coffee, so we can use a bit of 'brute force' math to get to the solution.

Multiples of $2.10: $2.10, $4.20, $6.30, $8.40, $10.50, $12.60, $14.70, $16.80

Multiples of $2.30: $2.30, $4.60, $6.90, $9.20, $11.50, $13.80, $16.10

So, how many different ways can you get to $17.80 using one value from each row (hint: you can 'count down' from $16.80 and try to find a multiple of $2.30 that will increase the sum to $17.80)?

$6.30 and $11.50
This is the only option that fits the given information - and the answer to the question is 3 pounds.
Fact 1 is SUFFICIENT

2) If Garcia had purchased 1 more pound of each type of coffee he would have spent $22.20

Purchasing one more pound of each type would have increased the total by $2.10+$2.30 = $4.40. The total in Fact 2 is $22.20, which is exactly $4.40 more than the $17.80 we were dealing with in Fact 1. Since Fact 1 had just one solution, spending $4.40 more can only lead to $22.20 from that one solution.
Fact 2 is SUFFICIENT

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Re: Garcia purchased two kinds of premium coffee from Harbucks Colorado Co &nbs [#permalink] 10 Dec 2017, 13:42
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