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Re: New Set of Mixed Questions!!!
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05 Apr 2013, 04:51
2. What is the area of a region enclosed by x/3+y/9=10?A. 675 B. 1350 C. 2700 D. 5400 E. 10800 Find the x and y intercepts. When y=0, then x=30 or x=30. When x=0, then y=90 or x=90. So, we have 4 points: (30, 0), (30, 0) (0, 90), (0, 90). When joining these points we get the rhombus: Attachment:
2.png [ 7.85 KiB  Viewed 31837 times ]
The area of a rhombus is \(\frac{d_1*d_2}{2}\) (where \(d_1\) and \(d_2\) are the lengths of the diagonals), thus the area of the enclosed figure is 60*180/2=5,400. Answer: D.
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Re: New Set of Mixed Questions!!!
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05 Apr 2013, 05:22
4. A draining pipe can empty a pool in 4 hours. On a rainy day, when the pool is full, the draining pipe is opened and the pool is emptied in 6 hours. If rain inflow into the pool is 3 liters per hour, what is the capacity of the pool?A. 9 liters B. 18 liters C. 27 liters D. 36 liters E. 45 liters Let the rate of the draining pipe be \(x\) liters per hour. Then the capacity of the tank will be \(C=time*rate=4x\); Now, when raining, the net outflow is x3 liters per hour, and we are told that at this new rate the pool is emptied in 6 hours. So, the capacity of the pool also equals to \(C=time*rate=6(x3)\); Thus we have: \(4x=6(x3)\) > \(x=9\) > \(C=4x=36\). Answer: D.
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Re: New Set of Mixed Questions!!!
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05 Apr 2013, 05:33
5. For a certain set of numbers, if x is in the set, then both x^2 and x^3 are also in the set. If the number 1/2 is in the set , which of the following must also be in the set ?
I. 1/64 II. 1/64 III. 1/2^(1/3)A. I only, B. II only, C. III only, D. I and II only E. I, II and III Since 1/2 is in the set, the so must be: x^2 = 1/4; x^3 = 1/8. Since 1/4 is in the set, the so must be: x^3 = 1/64; Since 1/8 is in the set, the so must be: x^2 = 1/64. The only number we cannot get is 1/2^(1/3). Answer: D.
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Re: New Set of Mixed Questions!!!
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Re: New Set of Mixed Questions!!!
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05 Apr 2013, 06:06
10. If n is a nonnegative integer and the remainder when 3^n is divided by 4 is a multiple of 3, then which of the following must be true?
I. n^2 divided by 4 yields the reminder of 1 II. (2)^n is less than 0 III. n is a prime numberA. I only B. II only C. III only D. I and II only E. II and III only 3^0=1 > the remainder when 1 is divided by 4 is 1; 3^1=3 > the remainder when 3 is divided by 4 is 3; 3^2=9 > the remainder when 9 is divided by 4 is 1; 3^3=27 > the remainder when 27 is divided by 4 is 3; ... We can see that in order the condition to hold true n must be odd. I. n^2 divided by 4 yields the reminder of 1 > odd^2 divided by 4 always yields the reminder of 1. So, this statement must be true. II. (2)^n is less than 0 > (2)^odd<0. So, this statement must be true. III. n is a prime number. Not necessarily true. Answer: D.
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Re: New Set of Mixed Questions!!!
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05 Apr 2013, 06:07
11. Certain bowl contains 5 red marbles and 3 blue marbles only. One by one, every marble is drawn at random and without replacement. What is the probability that the seventh marble drawn is NOT blue?A. 7/8 B. 3/4 C. 2/3 D. 5/8 E. 3/8 Basically we need to find the probability that the seventh marble drawn is red (so not blue). Now, the initial probability of drawing red marble is 5/8. Without knowing the other results, the probability of drawing red marble will not change for ANY successive draw: second, third, fourth, ..., seventh. Thus the probability that the seventh marble is red is 5/8. The same for blue marble: the probability of drawing blue marble is 3/8, the probability that for instance the 8th marble drawn is blue is still 3/8. There is simply no reason to believe WHY is any draw different from another (provided we don't know the other results). Answer: D.
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Re: New Set of Mixed Questions!!!
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07 Apr 2013, 05:51
Bunuel wrote: 5. For a certain set of numbers, if x is in the set, then both x^2 and x^3 are also in the set. If the number 1/2 is in the set , which of the following must also be in the set ?
Hi Bunuel, For the above question, can you please let me know where I am going wrong. I'm able to solve till here Since 1/2 is in the set, the so must be: x^2 = 1/4; x^3 = 1/8. [b]Doubt 1 However, what I failed to understand is why set should continue beyond 1/8 as we don't know the number of elements in the set, and since this is a must be true question, the below solution seems redundant to me. Bunuel wrote: Since 1/4 is in the set, the so must be: x^3 = 1/64;
Since 1/8 is in the set, the so must be: x^2 = 1/64.
The only number we cannot get is 1/2^(1/3).
Doubt 2I am hopeful that I lack some understanding the above solution will most probably be correct. In that case, I want to understand what will be maximum no of elements in this set. Will this be an Infinite set. Doubt 3Will the answer behaved differently had this question been "Could be true" Kindly elaborate. Thanks H



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Re: New Set of Mixed Questions!!!
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07 Apr 2013, 13:37
imhimanshu wrote: Doubt 1
However, what I failed to understand is why set should continue beyond 1/8 as we don't know the number of elements in the set, and since this is a must be true question, the below solution seems redundant to me. ` The logic is, if 1/2 is in the set, then 1/8 MUST be in the set. Knowing that 1/8 MUST be in the set, by the rule of formation of the set, (1/8)ˆ3 and (1/8)ˆ2 MUST also be in the set.. and so on. imhimanshu wrote: Doubt 2 I am hopeful that I lack some understanding the above solution will most probably be correct. In that case, I want to understand what will be maximum no of elements in this set. Will this be an Infinite set.
Given that 1/2 (different from 0) is part of the set, then Yes, this is an infinite set! imhimanshu wrote: Doubt 3 Will the answer behaved differently had this question been "Could be true" [/b] "could be true" problems try to play on POSSIBILITIES when a lack of some sort of restriction allows the items to be true, somehow. If the question was posed as "could be true", then ANY real number could be a part of the set. Think of it this way... Could a real number X be a part of the set? Yes, if cbrt(x) is part of the set (hypothesis), then x COULD be part of the set. KUDOS +1 IF THIS HELPED YOU



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Re: New Set of Mixed Questions!!!
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19 Apr 2013, 12:50
Bunuel wrote: 2. What is the area of a region enclosed by x/3+y/9=10?A. 675 B. 1350 C. 2700 D. 5400 E. 10800 Find the x and y intercepts. When y=0, then x=30 or x=30. When x=0, then y=90 or x=90. So, we have 4 points: (30, 0), (30, 0) (0, 90), (0, 90). When joining these points we get the rhombus: Attachment: 2.png The area of a rhombus is \(\frac{d_1*d_2}{2}\) (where \(d_1\) and \(d_2\) are the lengths of the diagonals), thus the area of the enclosed figure is 60*180/2=5,400. Answer: D. Bunuel what about the following points that also meets the given conditions of x/3 +  y/9 = 10 but come with a different area. (x,y) = (3,81) (x,y) = (3,81) (x,y) = (3,81) (x,y) = (3,81) In this case Area of the rectangle will be = 972 or you can have (x,y) = (6,72) (x,y) = (6,72) (x,y) = (6,72) (x,y) = (6,72) In this case the area of the rectangle will be = 1728. Can you please clarify
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Re: New Set of Mixed Questions!!!
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19 Apr 2013, 13:55
Bunuel wrote: 4. A draining pipe can empty a pool in 4 hours. On a rainy day, when the pool is full, the draining pipe is opened and the pool is emptied in 6 hours. If rain inflow into the pool is 3 liters per hour, what is the capacity of the pool? A. 9 liters B. 18 liters C. 27 liters D. 36 liters E. 45 liters
Let the rate of the draining pipe be \(x\) liters per hour. Then the capacity of the tank will be \(C=time*rate=4x\);
Now, when raining, the net outflow is x3 liters per hour, and we are told that at this new rate the pool is emptied in 6 hours. So, the capacity of the pool also equals to \(C=time*rate=6(x3)\);
Thus we have: \(4x=6(x3)\) > \(x=9\) > \(C=4x=36\).
Answer: D. Bunuel, Here is how I solved the problem. I must admit though I got it incorrect the first time I tried and when the answer choices didn't showed my answer I gave it another shot. Let x be the capacity of the pool. The rate at which the pipe drains the pool is x/4. Now on the rainy day the pull is already full and its raining at the rate of 3 lit/hr. The total inflow of rain in 6 hrs is 6*3 = 18 lit. So the pipe must drain x+18 lit in 6 hrs. The new rate is (x+18)/6 and we are given that these two rates are equal. Hence x/4 = (x+18)/6 > solving for x gives 36lits Is this approach correct. I must say this didn't strike the first time I tried.
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Re: New Set of Mixed Questions!!!
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20 Apr 2013, 05:15
pikachu wrote: Bunuel wrote: 2. What is the area of a region enclosed by x/3+y/9=10?A. 675 B. 1350 C. 2700 D. 5400 E. 10800 Find the x and y intercepts. When y=0, then x=30 or x=30. When x=0, then y=90 or x=90. So, we have 4 points: (30, 0), (30, 0) (0, 90), (0, 90). When joining these points we get the rhombus: Attachment: 2.png The area of a rhombus is \(\frac{d_1*d_2}{2}\) (where \(d_1\) and \(d_2\) are the lengths of the diagonals), thus the area of the enclosed figure is 60*180/2=5,400. Answer: D. Bunuel what about the following points that also meets the given conditions of x/3 +  y/9 = 10 but come with a different area. (x,y) = (3,81) (x,y) = (3,81) (x,y) = (3,81) (x,y) = (3,81) In this case Area of the rectangle will be = 972 or you can have (x,y) = (6,72) (x,y) = (6,72) (x,y) = (6,72) (x,y) = (6,72) In this case the area of the rectangle will be = 1728. Can you please clarify Below is the graph of x/3+y/9=10: The points you mention are on it and not making some other figure.
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Re: New Set of Mixed Questions!!!
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23 May 2013, 13:26
SaraLotfy wrote: Bunuel wrote: 5. For a certain set of numbers, if x is in the set, then both x^2 and x^3 are also in the set. If the number 1/2 is in the set , which of the following must also be in the set ?
I. 1/64 II. 1/64 III. 1/2^(1/3)
A. I only, B. II only, C. III only, D. I and II only E. I, II and III Wouldn't x^2 = +1/4 since the power is even? or should I treat this as X^2? Since 1/2 is in the set, the so must be: x^2 = 1/4; x^3 = 1/8.
Since 1/4 is in the set, the so must be: x^3 = 1/64;
Since 1/8 is in the set, the so must be: x^2 = 1/64.
The only number we cannot get is 1/2^(1/3).
Answer: D. \(x^2=(\frac{1}{2})^2=\frac{1}{4}\) (its x^2 not (x)^2). Hope it's clear.
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Re: New Set of Mixed Questions!!!
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01 Jan 2014, 04:06
Bunuel wrote: 11. Certain bowl contains 5 red marbles and 3 blue marbles only. One by one, every marble is drawn at random and without replacement. What is the probability that the seventh marble drawn is NOT blue? A. 7/8 B. 3/4 C. 2/3 D. 5/8 E. 3/8
Basically we need to find the probability that the seventh marble drawn is red (so not blue).
Now, the initial probability of drawing red marble is 5/8. Without knowing the other results, the probability of drawing red marble will not change for ANY successive draw: second, third, fourth, ..., seventh. Thus the probability that the seventh marble is red is 5/8.
The same for blue marble: the probability of drawing blue marble is 3/8, the probability that for instance the 8th marble drawn is blue is still 3/8. There is simply no reason to believe WHY is any draw different from another (provided we don't know the other results).
Answer: D. Hi Bunuel, Why is the probability not changing? If we keep withdrawing marbles withoout replacing then following things are changing: 1. total no. of marble (hence the denominator in probability) 2. No. of Red marbles left. 3. No. of blue marbles left.



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Re: New Set of Mixed Questions!!!
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01 Jan 2014, 05:44
Nilabh_s wrote: Bunuel wrote: 11. Certain bowl contains 5 red marbles and 3 blue marbles only. One by one, every marble is drawn at random and without replacement. What is the probability that the seventh marble drawn is NOT blue? A. 7/8 B. 3/4 C. 2/3 D. 5/8 E. 3/8
Basically we need to find the probability that the seventh marble drawn is red (so not blue).
Now, the initial probability of drawing red marble is 5/8. Without knowing the other results, the probability of drawing red marble will not change for ANY successive draw: second, third, fourth, ..., seventh. Thus the probability that the seventh marble is red is 5/8.
The same for blue marble: the probability of drawing blue marble is 3/8, the probability that for instance the 8th marble drawn is blue is still 3/8. There is simply no reason to believe WHY is any draw different from another (provided we don't know the other results).
Answer: D. Hi Bunuel, Why is the probability not changing? If we keep withdrawing marbles withoout replacing then following things are changing: 1. total no. of marble (hence the denominator in probability) 2. No. of Red marbles left. 3. No. of blue marbles left. Consider this: there are 8 cards, 5 spades, 3 hearts. What is the probability that the first card you pick will be spade? Clearly 5/8. Now next question, suppose I throw away three cards at random, not telling you which cards I thrown away. What is the probability NOW that you pick a spade out of 5 remaining cards? I've just reduced the sample from which you pick, but does the probability changed? WHY should it change? Another question, the same cards 5 spades and 3 hearts: I say that you can pick one card but only out of randomly selected 5 cards. What is the probability that the card you pick will be spade? There is no difference in ALL above cases and the probability will remain 5/8. If it changed, increased (or decreased), it would mean that the probability of picking heart on the other hand decreased (or increased). Why would it? Similar questions to practice: aboxcontains3yellowballsand5blackballsonebyone90272.htmlabagcontains3whiteballs3blackballs2redballs100023.htmleachoffourdifferentlockshasamatchingkeythekeys101553.htmlif40peoplegetthechancetopickacardfromacanister97015.htmlabagcontains3whiteballs3blackballs2redballs100023.htmlHope this helps.
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Re: New Set of Mixed Questions!!!
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01 Jan 2014, 11:40
11. Certain bowl contains 5 red marbles and 3 blue marbles only. One by one, every marble is drawn at random and without replacement. What is the probability that the seventh marble drawn is NOT blue?
A. 7/8 B. 3/4 C. 2/3 D. 5/8 E. 3/8
answer:
We fix the seventh ball to be red. Since all balls are same it doesnt matter which of the red balls.
So the rest 7 balls can be arranged in 7!/(4!*3!)= 35 ways becoz there are 4 identical red balls and 3 identical blue balls.
Without any restriction 8 balls can be arranged in 8!/(5!*3!) = 56 ways.
Therefore probability is 35/56 = 5/8.



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Re: New Set of Mixed Questions!!!
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01 Jan 2014, 19:21
Bunuel wrote: Nilabh_s wrote: Bunuel wrote: .Consider this: there are 8 cards, 5 spades, 3 hearts. What is the probability that the first card you pick will be spade? Clearly 5/8.
Now next question, suppose I throw away three cards at random, not telling you which cards I thrown away. What is the probability NOW that you pick a spade out of 5 remaining cards? I've just reduced the sample from which you pick, but does the probability changed? WHY should it change?
Another question, the same cards 5 spades and 3 hearts: I say that you can pick one card but only out of randomly selected 5 cards. What is the probability that the card you pick will be spade?
There is no difference in ALL above cases and the probability will remain 5/8. If it changed, increased (or decreased), it would mean that the probability of picking heart on the other hand decreased (or increased). Why would it?
Hope this helps. For the statement " NOW that you pick a spade out of 5 remaining cards" If you throw 3 cards then how are we sure than 5 spades are still remaining. Following three cards may have been thrown: 1. All 3 spades. 2. All three hearts. 3. 2 spade & 1 heart & viceversa. Hence how can we be sure that 5 spades are till remaining?



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Re: New Set of Mixed Questions!!!
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02 Jan 2014, 05:51
Nilabh_s wrote: Bunuel wrote: Bunuel wrote: .Consider this: there are 8 cards, 5 spades, 3 hearts. What is the probability that the first card you pick will be spade? Clearly 5/8.
Now next question, suppose I throw away three cards at random, not telling you which cards I thrown away. What is the probability NOW that you pick a spade out of 5 remaining cards? I've just reduced the sample from which you pick, but does the probability changed? WHY should it change?
Another question, the same cards 5 spades and 3 hearts: I say that you can pick one card but only out of randomly selected 5 cards. What is the probability that the card you pick will be spade?
There is no difference in ALL above cases and the probability will remain 5/8. If it changed, increased (or decreased), it would mean that the probability of picking heart on the other hand decreased (or increased). Why would it?
Hope this helps. For the statement " NOW that you pick a spade out of 5 remaining cards" If you throw 3 cards then how are we sure than 5 spades are still remaining. Following three cards may have been thrown: 1. All 3 spades. 2. All three hearts. 3. 2 spade & 1 heart & viceversa. Hence how can we be sure that 5 spades are till remaining? Yes, but we don't know which cards are thrown and this won't affect the probability. Please follow the links in my previous post for better understanding of such problems.
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Re: New Set of Mixed Questions!!! &nbs
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02 Jan 2014, 05:51



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