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Director  P
Joined: 20 Jul 2017
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2 unit of a% alcohol is mixed with 3 units of b% alcohol to  [#permalink]

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7 00:00

Difficulty:   95% (hard)

Question Stats: 28% (02:04) correct 72% (02:36) wrong based on 57 sessions

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2 unit of a% alcohol is mixed with 3 units of b% alcohol to give 40% alcohol. If a > b, how many integer values can a take? (a & b are integers)

A. 13
B. 17
C. 19
D. 20
E. None of these

Posted from my mobile device

Originally posted by Dillesh4096 on 15 Aug 2019, 06:42.
Last edited by Dillesh4096 on 20 Aug 2019, 23:53, edited 1 time in total.
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Re: 2 unit of a% alcohol is mixed with 3 units of b% alcohol to  [#permalink]

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1
This is a very difficult question, but let's break down the question into small steps to find the solution

If a>b and the resulting mixture has to be 40%, that means that a>40%, otherwise a and b would never average 40%. Check this out:

b.......40%........a

I hope you see more clearly now that a has to be more than 40%. If it was 39%, b would be lower, and it would be impossible to average 40%

However, there is another method to come to the conclusion that x>40%, by applying the mixture formula (well known among GMATers)

(a-40)/(40-b)=3/2

a=(3/2)b-20

now if:

a=30 -> b=33.333 (b>a, so a cannot be 30)
a=70 -> b=60 (a>b, so a can be 70)
a=40 -> b=40 (breakeven)

so for a to be higher than b, a needs to be higher than 40

(Note this calculation would be impossible to do in the GMAT, so you need to apply the logic to find that a has to be higher than 40%)

Now that we know that a>40% but a<=100% (the liquid cannot have more alcohol than the whole liquid itself ) We can assume that there are 60 integers that x can take (why is not 60 among the answers?)

because there is one tricky thing we should be aware of:

Since we are mixing 2 units of a with 3 units of b, the average will always be 2 units away from b and 3 units away from a.

So if b = 58, a = 43
if b = 56, a = 46
and so on...

This means than from those 60 integers we were thinking of at the beginnning, we can only take the ones that are divisible by 3.

To calculate the number we use the formula to count numbers = ((an-a1)/d)+1, where:

an is the last number of the range
a1 the first number of the range
d the distance between numbers.

In our case a1=42 (first number divisible by 3 after 40) and an is 99 (last digit disible by 3) and d=3.

Therefore the number of integers that fulfill this conditions is ((99-42)/3)+1 =20

OPTION D
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Re: 2 unit of a% alcohol is mixed with 3 units of b% alcohol to  [#permalink]

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Dillesh4096 wrote:

2 unit of a% alcohol is mixed with 3 units of b% alcohol to give 40% alcohol. If a > b, how many integer values can a take?

A. 13
B. 17
C. 19
D. 20
E. None of these

SOLUTION:

Vol of Final Solution (FS): 5 units. Total qty of Alcohol in FS = 2 units.
Qty of alcohol obtained from the a% solution = (a/100)*2
Qty of alcohol obtained from the b% solution = (b/100)*3
Total qty of alcohol in FS = 2a/100 + 3b/100 = 2
2a +3b = 200....> a = 100 - (3/2)b
Now let us calculate the minimum value that 'a' can take considering that a>b. If we assume that 'a' is equal to 'b' then a=40. Therefore, 'a' must be greater than 40 (i.e. 41, 42, 43....). Thus, the minimum value of 'a' could be 41 and the maximum could be 99.

Now, I want to address a crucial point regarding the wording of the question. It states that 'a' must have integer values but nothing is said regarding 'b' so it would be logical to assume that 'b' can take fractional values. However, since there is an interpretative confusion, I am presenting both solutions:

(A) ASSUMING 'b' CAN TAKE FRACTIONAL VALUES:
a = 100 - (3/2)b. For 'a' to be an integer, the 2nd term of the RHS of this equation has to be an integer so 'b' can be 2/3, 4/3, 6/3.... and consequently 'a' will be 99, 98, 97.....41.
The number of values 'a' can take = (99 - 41) + 1 = 59. ANS: E

(B) ASSUMING 'b' CAN TAKE ONLY INTEGER VALUES:
a = 100 -(3/2)b. 'b' has to be a multiple of 2 (i.e. 2, 4, 6 and so on) which means 'a' will be 97, 94, 91...... This is an Arithmetic Series in which, from a maximum of 97, the values decline at the rate of 3 until it reaches a value which is just above 40. If we assume that the min value is 40, the total number of values (n)=
40 (nth term) = 97 (1st term) + (n-1)(-3)....> 40 = 100 - 3n....> n = 20. Since 'a' has to be more than 40, the next highest number, 43, must be the min value of 'a'. So the total number of values that 'a' can take is 19. ANS: C

NOTE: I would be obliged if Bunuel, VeritasKarishma, chetan2u or any of the other experts could take a look at my solution and point out any mistakes I might have made.
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Re: 2 unit of a% alcohol is mixed with 3 units of b% alcohol to  [#permalink]

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Hey, guys! Since none of my above answers match the OA, this problem has been bugging me. Could ANYONE throw some light on it - or maybe Dillesh4096 could provide an Official Solution. Could you do that please?
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Re: 2 unit of a% alcohol is mixed with 3 units of b% alcohol to  [#permalink]

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effatara wrote:
Hey, guys! Since none of my above answers match the OA, this problem has been bugging me. Could ANYONE throw some light on it - or maybe Dillesh4096 could provide an Official Solution. Could you do that please?

Hey effatara, I hope this helps!

We can infer from the stem that if $$a>b$$ and $$40$$ is the weighted average,

$$b<40<a$$

Set up the weighted average equation

---a%---------------------b%---
---|-------------40%------|-----
---2------------------------3---

$$\frac{2}{3}=\frac{40-b}{a-40}$$

$$2a-80=120-3b$$

$$2a=200-3b$$

$$a=100-\frac{3}{2}b$$

So for $$a$$ to be an integer, $$b$$ must be even

But wait, what if $$b$$ is something like $$60$$? We get $$a=10$$

But we know from the stem that $$a>b$$

So to find the number of values $$b$$ can take, we must put $$a=100-\frac{3}{2}b$$ in $$a>b$$

$$100-\frac{3}{2}b>b$$

$$\frac{5}{2}b<100$$

$$b<40$$

So for $$a$$ to be an integer and $$a>b$$, $$b$$ is less than $$40$$ and $$b$$ is even and so the maximum value that $$b$$ can take is $$38$$

So $$b$$ can be $$0,2,4,6....38$$ which is $$20$$ values

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Re: 2 unit of a% alcohol is mixed with 3 units of b% alcohol to  [#permalink]

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effatara wrote:
Dillesh4096 wrote:

2 unit of a% alcohol is mixed with 3 units of b% alcohol to give 40% alcohol. If a > b, how many integer values can a take?

A. 13
B. 17
C. 19
D. 20
E. None of these

SOLUTION:

Vol of Final Solution (FS): 5 units. Total qty of Alcohol in FS = 2 units.
Qty of alcohol obtained from the a% solution = (a/100)*2
Qty of alcohol obtained from the b% solution = (b/100)*3
Total qty of alcohol in FS = 2a/100 + 3b/100 = 2
2a +3b = 200....> a = 100 - (3/2)b
Now let us calculate the minimum value that 'a' can take considering that a>b. If we assume that 'a' is equal to 'b' then a=40. Therefore, 'a' must be greater than 40 (i.e. 41, 42, 43....). Thus, the minimum value of 'a' could be 41 and the maximum could be 99.

Now, I want to address a crucial point regarding the wording of the question. It states that 'a' must have integer values but nothing is said regarding 'b' so it would be logical to assume that 'b' can take fractional values. However, since there is an interpretative confusion, I am presenting both solutions:

(A) ASSUMING 'b' CAN TAKE FRACTIONAL VALUES:
a = 100 - (3/2)b. For 'a' to be an integer, the 2nd term of the RHS of this equation has to be an integer so 'b' can be 2/3, 4/3, 6/3.... and consequently 'a' will be 99, 98, 97.....41.
The number of values 'a' can take = (99 - 41) + 1 = 59. ANS: E

(B) ASSUMING 'b' CAN TAKE ONLY INTEGER VALUES:
a = 100 -(3/2)b. 'b' has to be a multiple of 2 (i.e. 2, 4, 6 and so on) which means 'a' will be 97, 94, 91...... This is an Arithmetic Series in which, from a maximum of 97, the values decline at the rate of 3 until it reaches a value which is just above 40. If we assume that the min value is 40, the total number of values (n)=
40 (nth term) = 97 (1st term) + (n-1)(-3)....> 40 = 100 - 3n....> n = 20. Since 'a' has to be more than 40, the next highest number, 43, must be the min value of 'a'. So the total number of values that 'a' can take is 19. ANS: C

NOTE: I would be obliged if Bunuel, VeritasKarishma, chetan2u or any of the other experts could take a look at my solution and point out any mistakes I might have made.

Hey effatara,

b can take 0 also right? then a would be 100.
--> Possible values of a are {100, 97, 94, . . . . . . . 43}
--> Number of values possibles = (100-43)/3 + 1 = 19 + 1 = 20

Hope its clear!!

Also, agree with you that b can take fractional values. missed to indicate in the question. Shall be edited.
Thanks
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Re: 2 unit of a% alcohol is mixed with 3 units of b% alcohol to  [#permalink]

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Dillesh4096 wrote:
2 unit of a% alcohol is mixed with 3 units of b% alcohol to give 40% alcohol. If a > b, how many integer values can a take? (a & b are integers)

A. 13
B. 17
C. 19
D. 20
E. None of these

Posted from my mobile device

Given:
1. 2 unit of a% alcohol is mixed with 3 units of b% alcohol to give 40% alcohol.
2. a > b
3. a & b are integers

Asked: How many integer values can a take?

2a + 3b = 40*5 = 200
Since 2a and 200 are even => 3b = even => b=even
b= 0; a = 100
b = 2; a = 97
b =4 ; a = 94
b= 38; a = 43
b = 40; a = 40 Not feasible wince a>b

b will take values from a= 0 to l = 38 d=2 => 38/2 + 1 = 20 values and a also will take corresponding 20 values

IMO D
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Re: 2 unit of a% alcohol is mixed with 3 units of b% alcohol to  [#permalink]

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Thanks, guys. It's clear now.
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Re: 2 unit of a% alcohol is mixed with 3 units of b% alcohol to  [#permalink]

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b a

40

3 2

Now,

3/2= a-40/40-b

2a+3b=200

a b

100 0

Now for the next co efficients

a= 100-coefficient of b = 97
b=0+ co efficient of a = 2

Thus, second values are

97 2

it can extend until anyone of them become negative.

But, here a>b

To make sure when to stop,

let us assume a=b

100+(n-1) -3=0+(n-1) 2

It gives n=21

Thus, we have to stop at n=20

Thus, a or b can have 20 values Re: 2 unit of a% alcohol is mixed with 3 units of b% alcohol to   [#permalink] 21 Aug 2019, 08:10
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