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Probability
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22 Sep 2024, 04:18
22 Sep 2024, 04:18
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Question Stats:
33% (01:43) correct
66% (02:55) wrong based on 6 sessions
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If x and y are both integers chosen at random between 1 and 100, inclusive, what is the approximate probability that x/y is an integer?
(A) 1%
(B) 3%
(C) 5%
(D) 7%
(E) 9%
(A) 1%
(B) 3%
(C) 5%
(D) 7%
(E) 9%
Re: Probability
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22 Sep 2024, 05:05
22 Sep 2024, 05:05
Expert Reply
This is a problem solving question. As such. you must post it under the correct sub section.
NOT under quantitative comparison.
I move it
NOT under quantitative comparison.
I move it
Re: Probability
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22 Sep 2024, 05:07
22 Sep 2024, 05:07
Expert Reply
Total # of x/y possible is 100*100=10,000 (since both x and y can tale 100 values each).
Now, if:
y=1 then all 100 possible values for x will give integer value for x/y;
y=2 then all 50 even values for x will give integer value for x/y;
y=3 then all 33 multiples of 3 for x will give integer value for x/y;
y=4 then all 25 multiples of 4 for x will give integer value for x/y;
y=5 then all 20 multiples of 5 for x will give integer value for x/y;
y=6 then all 16 multiples of 6 for x will give integer value for x/y;
y=7 then all 14 multiples of 7 for x will give integer value for x/y;
y=8 then all 12 multiples of 8 for x will give integer value for x/y;
You can continue with same logic for other values of y...
...
If y is from 51 to 100 then only one value for x (namely the value when x=y) will give integer value for x/y.
Next, if you sum all those numbers: 100+50+33+25+20+16+14+12+...+50=~500.
P=~500/10,000=5%.
Or: (100+50+33+25+20+16+14+12)+50=320, values of y from 9 to 50 (42 values), inclusive would add at least 42*2=84 more numbers, so total will be at least 320+84=404, which is closer to 5% then to 3%.
Answer: C.
P.S. There is zero chances of getting something like this on actual exam, not because this question is too hard but because of tedious math you have to do (though in the middle you can apply some approximation), so I wouldn't worry about this problem at all.
I would skip such questions and rather I would take a look at out
GRE PREMIUM Quant Question Banks - Topic-Wise 2700 Questions
https://gre.myprepclub.com/forum/gre-pr ... 34207.html
They are specifically for the GRE
Now, if:
y=1 then all 100 possible values for x will give integer value for x/y;
y=2 then all 50 even values for x will give integer value for x/y;
y=3 then all 33 multiples of 3 for x will give integer value for x/y;
y=4 then all 25 multiples of 4 for x will give integer value for x/y;
y=5 then all 20 multiples of 5 for x will give integer value for x/y;
y=6 then all 16 multiples of 6 for x will give integer value for x/y;
y=7 then all 14 multiples of 7 for x will give integer value for x/y;
y=8 then all 12 multiples of 8 for x will give integer value for x/y;
You can continue with same logic for other values of y...
...
If y is from 51 to 100 then only one value for x (namely the value when x=y) will give integer value for x/y.
Next, if you sum all those numbers: 100+50+33+25+20+16+14+12+...+50=~500.
P=~500/10,000=5%.
Or: (100+50+33+25+20+16+14+12)+50=320, values of y from 9 to 50 (42 values), inclusive would add at least 42*2=84 more numbers, so total will be at least 320+84=404, which is closer to 5% then to 3%.
Answer: C.
P.S. There is zero chances of getting something like this on actual exam, not because this question is too hard but because of tedious math you have to do (though in the middle you can apply some approximation), so I wouldn't worry about this problem at all.
I would skip such questions and rather I would take a look at out
GRE PREMIUM Quant Question Banks - Topic-Wise 2700 Questions
https://gre.myprepclub.com/forum/gre-pr ... 34207.html
They are specifically for the GRE
