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ParvinderGoyat
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muskaan2204
My aim is also round 2 but also I should be able to give gmat by no November
muskaan2204, same here
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x=25a , y=25b with gcd(a,b)=1
Now 25(a+b)=350 yields a+b=14

Now (a,b) can be 1,13
3,11
5,9

So how many pairs?

So C I think
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a+b=14
E(14)=6
Number of unordered pairs=E(14)/2=6/2=3
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skgang_1
a+b=14
E(14)=6
Number of unordered pairs=E(14)/2=6/2=3
What’s E here?
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25a+25b=250, a+b=14, with gcd(a,b)=1, (a,b)=(1,13), (3,11), (5,9) c)
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Problem Solving Butler: October 2024
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it does not really matter how apart they are, just multiply the combined speed per time = (25+65)*1,5
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oswin123
it does not really matter how apart they are, just multiply the combined speed per time = (25+65)*1,5
Yes it’s just a distractor.
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Bunuel
Data Sufficiency Butler: October 2024
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for the first qu., Is it solvable?

solveable*

we have (A and B)-(neither A or B)= 100, A+B=500, but i don’t remember the exact equation for overlapping
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Let’s say both=a, neither=n
Only B=b, only C=c
Then a+b+c+n=500...(m)
Then
From 1)

(a+b)+(a+c)=2a+b+c=400

Not possible to find a
( plugging 1 with m gives 2)
From 2) n-a=100
Not possible find a
( plugging 2 with m gives 1)

So actually 1 is equivalent to 2. So can’t be determined .
Answer E
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Matthyrou
we have (A and B)-(neither A or B)= 100, A+B=500, but i don’t remember the exact equation for overlapping
Yes its E. Because Total is equal to = A+B - A∩B + Neither put values you will see we dont get anything
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A pilot required to fly a certain number of hours each day’s flew of the day’s required hours plus 2 more in the morning. In the afternoon, she flew 2/5 of the day’ remaining required hours. In the evening, she flew 4 remaining hours. Which of the following is closest to the number of hours she flew that day?
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Problem Solving Butler: October 2024
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Critical Reasoning Butler: October 2024
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