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I'm posting the next set of medium/hard PS questions. I'll post OA's with detailed explanations after some discussion. Please, post your solutions along with the answers. Good luck!

1. A password on Mr. Wallace's briefcase consists of 5 digits. What is the probability that the password contains exactly three digit 6?

A. 860/90,000 B. 810/100,000 C. 858/100,000 D. 860/100,000 E. 1530/100,000

3. For the past k days the average (arithmetic mean) cupcakes per day that Liv baked was 55. Today Bibi joined and together with Liv they baked 100 cupcakes, which raises the average to 60 cupcakes per day. What is the value of k? A. 6 B. 8 C. 9 D. 10 E. 12

5. There are 7 red and 5 blue marbles in a jar. In how many ways 8 marbles can be selected from the jar so that at least one red marble and at least one blue marble to remain in the jar? A. 460 B. 490 C. 493 D. 455 E. 445

6. A pool has two water pumps A and B and one drain C. Pump A alone can fill the whole pool in x hours, and pump B alone can fill the whole pool in y hours. The drain can empty the whole pool in z hours, where z>x. With pumps A and B both running and the drain C unstopped till the pool is filled, which of the following represents the amount of water in terms of the fraction of the pool which pump A pumped into the pool? A. \(\frac{yz}{x+y+z}\)

7. Metropolis Corporation has 4 shareholders: Fritz, Luis, Alfred and Werner. Number of shares that Fritz owns is 2/3 rd of number of the shares of the other three shareholders, number of the shares that Luis owns is 3/7 th of number of the shares of the other three shareholders and number of the shares that Alfred owns is 4/11 th of number of the shares of the other three shareholders. If dividends of $3,600,000 were distributed among the 4 shareholders, how much of this amount did Werner receive? A. $60,000 B. $90,000 C. $100,000 D. $120,000 E. $180,000

8. A set A consists of 7 consecutive odd integers. If the sum of 5 largest integers of set A is -185 what is the sum of the 5 smallest integers of set A? A. -165 B. -175 C. -195 D. -205 E. -215

11. In an infinite sequence 1, 3, 9, 27, ... each term after the first is three times the previous term. What is the difference between the sum of 13th and 15th terms and the sum of 12th and 14th terms of the sequence? A. 10*3^11 B. 20*3^11 C. 10*3^12 D. 40*3^11 E. 20*3^12

12. x, y and z are positive integers such that when x is divided by y the remainder is 3 and when y is divided by z the remainder is 8. What is the smallest possible value of x+y+z? A. 12 B. 20 C. 24 D. 29 E. 33

13. If \(x=\frac{(8!)^{10}-(8!)^6}{(8!)^{5}-(8!)^3}\), what is the product of the tens and the units digits of \(\frac{x}{(8!)^3}-39\)? A. 0 B. 6 C. 7 D. 12 E. 14

12. x, y and z are positive integers such that when x is divided by y the remainder is 3 and when y is divided by z the remainder is 8. What is the smallest possible value of x+y+z? A. 12 B. 20 C. 24 D. 29 E. 33

Given \(x=qy+3\), where \(q\) is a quotient, an integer \(\geq0\). Which means that the least value of \(x\) is when \(q=0\), in that case \(x=3\). This basically means that \(x\) is less than \(y\). For example 3 divided by 4 yields remainder of 3.

Thus we have that: \(x\) is divided by \(y\) the remainder is 3 --> minimum value of \(x\) is 3; \(y\) is divided by \(z\) the remainder is 8 --> minimum value of \(y\) is 8 and minimum value of \(z\) is one more than 8, so 9 (8 divided by 9 yields the remainder of 8);

So, the smallest possible value of \(x+y+z\) is 3+8+9=20.

13. If \(x=\frac{(8!)^{10}-(8!)^6}{(8!)^{5}-(8!)^3}\), what is the product of the tens and the units digits of \(\frac{x}{(8!)^3}-39\)? A. 0 B. 6 C. 7 D. 12 E. 14

Now, since \(8!\) has 2 and 5 as its multiples, then it will have 0 as the units digit, so \((8!)^2\) will have two zeros in the end, which means that \((8!)^2-38\) will have 00-38=62 as the last digits: 6*2=12.

1. A password on Mr. Wallace's briefcase consists of 5 digits. What is the probability that the password contains exactly three digit 6?

A. 860/90,000 B. 810/100,000 C. 858/100,000 D. 860/100,000 E. 1530/100,000

Total # of 5 digit codes is 10^5, notice that it's not 9*10^4, since in a code we can have zero as the first digit.

# of passwords with three digit 6 is \(9*9*C^3_5=810\): each out of two other digits (not 6) has 9 choices, thus we have 9*9 and \(C^3_5\) is ways to choose which 3 digits will be 6's out of 5 digits we have.

\(P=\frac{favorable}{total}=\frac{810}{10^5}\)

Answer: B.

Did this way- prob of choosing 6 will be 1/10 on each of he three times (and these choices can be in \(5C3\)=10 ways) and the rest two digits can be any thing from set 0-9 (except 6) so in total we have 9 choices (i..e a prob of 9/10 for the rest of the 2 positions). So total prob will be\(10*(1/10*1/10*1/10*9/10*9/10)= 810/10^5\)(B)
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9. If x and y are negative numbers, what is the value of \(\frac{\sqrt{x^2}}{x}-\sqrt{-y*|y|}\)? A. 1+y B. 1-y C. -1-y D. y-1 E. x-y

Note that \(\sqrt{a^2}=|a|\). Next, since \(x<0\) and \(y<0\) then \(|x|=-x\) and \(|y|=-y\).

So, \(\frac{\sqrt{x^2}}{x}-\sqrt{-y*|y|}=\frac{|x|}{x}-\sqrt{(-y)*(-y)}=\frac{-x}{x}-\sqrt{y^2}=-1-|y|=-1+y\)

Answer: D.

Hi Bunuel,

I didn't quite understand this ..

X and Y are odd then how can |x| = -x it has to be +ive.

x and y are negative numbers, not odd numbers.

If \(x\geq{0}\) then \(|x|=x\); If \(x<{0}\) then \(|x|=-x\). So if \(x\) is negative then \(|x|=-x=-negative=positive\). For example, if \(x=-2\) then \(|x|=|-2|=2=-x\).

Got 9 correct with @ 2.5 mins/Q ... how about others ? (2 Q left without going to deep as killed more than 4 mins on those Q4,7- this proves that not to waste time on something you don't get idea in 1.5-2 mins time frame and missed 2 narrowly- Q9, 10: 10 I din't read 2y but only y so...)..will post in details some explanations, esp. those which are little different from already posted by the very own legend Mr B!!
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13. If \(x=\frac{(8!)^{10}-(8!)^6}{(8!)^{5}-(8!)^3}\), what is the product of the tens and the units digits of \(\frac{x}{(8!)^3}-39\)? A. 0 B. 6 C. 7 D. 12 E. 14

Now, since \(8!\) has 2 and 5 as its multiples, then it will have 0 as the units digit, so \((8!)^2\) will have two zeros in the end, which means that \((8!)^2-38\) will have 00-38=62 as the last digits: 6*2=12.

Answer: D.

Although there is hardly anything left after your's explanations but still I'll post my approach, which might help few people out here (esp. on the # trailing 0's as you people have kept in the Math book)--

first I simplified the whole x value as it looks so clumsy/heavy: given \(x = (8!)^6 [(8!)^4 - 1]/ (8!)^3[(8!)^2 - 1] = (8!)^3[(8!)^2 + 1][(8!)^2 - 1]/[(8!)^2 - 1] so, (x/ (8!)^3) - 39 = [(8!)^2 + 1] -39 = (8!)^2 -38\) Now as per the Q we need to know unit & tenth digit of the above number- as we know factorials of numbers > 4 have trailing zero's so here we can determine how many trailing 0's in 8!. Since \(8 < 5^2\) so that means we have only 1 trailing 0 in here (alternately you can have manual calc in this case as well because it's small number). So \((8!)^2\) has 2 trailing 0's. Now it does not matter what hundredth position is becasue we need to subtract only 38- this would result in 6 in tenth pos and 2 in unit pos. so product is = 12...

P.S.- just for refrence # trailing 0's in factorial N is determined by- \(N/5^1+N/5^2+....+N/5^k (where 5^k < N)\)for each division we take integer part (quotient) and then add them up. this should give # trailing 0's
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5. There are 7 red and 5 blue marbles in a jar. In how many ways 8 marbles can be selected from the jar so that at least one red marble and at least one blue marble to remain in the jar? A. 460 B. 490 C. 493 D. 455 E. 445

Total ways to select 8 marbles out of 7+5=12 is \(C^8_{12}\); Ways to select 8 marbles so that zero red marbles is left in the jar is \(C^7_7*C^1_5\); Ways to select 8 marbles so that zero blue marbles is left in the jar is \(C^5_5*C^3_7\);

Hence ways to select 8 marbles so that at least one red marble and at least one blue marble to remain the jar is \(C^8_{12}-(C^7_7*C^1_5+C^5_5*C^3_7)=495-(5+35)=455\).

Answer: D.

I tried feagure out this way first but thought to calculate little differently-

Since there are a toatl of 12 Marbels and after picking the required # marbels we are left with 4 marbels in the Jar- so our task is determine out of these 4 in how many ways we can pick R, B marbels so that there exists at least 1 marbel of each type- (R-1,B-3), (R-2,B-2),(R-3,B-1)

6. A pool has two water pumps A and B and one drain C. Pump A alone can fill the whole pool in x hours, and pump B alone can fill the whole pool in y hours. The drain can empty the whole pool in z hours, where z>x. With pumps A and B both running and the drain C unstopped till the pool is filled, which of the following represents the fraction of the pool which pump A filled? A. \(\frac{yz}{x+y+z}\)

B. \(\frac{yz}{yz+xz-xy}\)

C. \(\frac{yz}{yz+xz+xy}\)

D. \(\frac{xyz}{yz+xz-xy}\)

E. \(\frac{yz+xz-xy}{yz}\)

With pumps A and B both running and the drain unstopped the pool will be filled in a rate \(\frac{1}{x}+\frac{1}{y}-\frac{1}{z}=\frac{yz+xz-zy}{xyz}\) pool/hour. So, the pool will be filled in \(\frac{xyz}{yz+xz-xy}\) hours (time is reciprocal of rate).

In \(\frac{xyz}{yz+xz-xy}\) hours pump A will do \(\frac{1}{x}*\frac{xyz}{yz+xz-xy}=\frac{yz}{yz+xz-xy}\) part of the job.

Answer: B.

hi , i feel the ans to this Q is wrong. as we have to tell what fraction of the pool was filled up by pump A. it will not depend on the drain C as the filling was done only by pump A and B. ans shud be y/(x+y). the given ans would be correct if we were to find "the amount of water pumped by A in that duration"
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6. A pool has two water pumps A and B and one drain C. Pump A alone can fill the whole pool in x hours, and pump B alone can fill the whole pool in y hours. The drain can empty the whole pool in z hours, where z>x. With pumps A and B both running and the drain C unstopped till the pool is filled, which of the following represents the fraction of the pool which pump A filled? A. \(\frac{yz}{x+y+z}\)

B. \(\frac{yz}{yz+xz-xy}\)

C. \(\frac{yz}{yz+xz+xy}\)

D. \(\frac{xyz}{yz+xz-xy}\)

E. \(\frac{yz+xz-xy}{yz}\)

With pumps A and B both running and the drain unstopped the pool will be filled in a rate \(\frac{1}{x}+\frac{1}{y}-\frac{1}{z}=\frac{yz+xz-zy}{xyz}\) pool/hour. So, the pool will be filled in \(\frac{xyz}{yz+xz-xy}\) hours (time is reciprocal of rate).

In \(\frac{xyz}{yz+xz-xy}\) hours pump A will do \(\frac{1}{x}*\frac{xyz}{yz+xz-xy}=\frac{yz}{yz+xz-xy}\) part of the job.

Answer: B.

hi , i feel the ans to this Q is wrong. as we have to tell what fraction of the pool was filled up by pump A. it will not depend on the drain C as the filling was done only by pump A and B. ans shud be y/(x+y). the given ans would be correct if we were to find "the amount of water pumped by A in that duration"

Actually that was the intended meaning of the question. I edited it so to avoid ambiguity: "which of the following represents the amount of water in terms of the fraction of the pool which pump A pumped into the pool?"

Hi Bunuel, Please help me in understanding the below text. I think I am lacking this concept.I always try to solve by taking LCM in order to get rid of the fractions. However, I found your approach short n simple. If you could shed some light on this, would be really helpful

Fritz owns \(\frac{2}{2+3}=\frac{2}{5}\)th of all shares

Thanks H

Bunuel wrote:

7. Metropolis Corporation has 4 shareholders: Fritz, Luis, Alfred and Werner. Number of shares that Fritz owns is 2/3 rd of number of the shares of the other three shareholders, number of the shares that Luis owns is 3/7 th of number of the shares of the other three shareholders and number of the shares that Alfred owns is 4/11 th of number of the shares of the other three shareholders. If dividends of $3,600,000 were distributed among the 4 shareholders, how much of this amount did Werner receive? A. $60,000 B. $90,000 C. $100,000 D. $120,000 E. $180,000

Fritz owns is \(\frac{2}{3}\)rd of the shares of the other three shareholders --> Fritz owns \(\frac{2}{2+3}=\frac{2}{5}\)th of all shares; Luis owns is \(\frac{3}{7}\)th of the shares of the other three shareholders --> Luis owns \(\frac{3}{3+7}=\frac{3}{10}\)th of all shares; Alfred owns is \(\frac{4}{11}\)th of the shares of the other three shareholders --> Alfred owns \(\frac{4}{4+11}=\frac{4}{15}\)th of all shares;

Together those three own \(\frac{2}{5}+\frac{3}{10}+\frac{4}{15}=\frac{29}{30}\)th of all shares, which means that Werner owns \(1-\frac{29}{30}=\frac{1}{30}\). Hence from $3,600,000 Werner gets \($3,600,000*\frac{1}{30}=$120,000\).

Hi Bunuel, Please help me in understanding the below text. I think I am lacking this concept.I always try to solve by taking LCM in order to get rid of the fractions. However, I found your approach short n simple. If you could shed some light on this, would be really helpful

Fritz owns \(\frac{2}{2+3}=\frac{2}{5}\)th of all shares

Thanks H

Bunuel wrote:

7. Metropolis Corporation has 4 shareholders: Fritz, Luis, Alfred and Werner. Number of shares that Fritz owns is 2/3 rd of number of the shares of the other three shareholders, number of the shares that Luis owns is 3/7 th of number of the shares of the other three shareholders and number of the shares that Alfred owns is 4/11 th of number of the shares of the other three shareholders. If dividends of $3,600,000 were distributed among the 4 shareholders, how much of this amount did Werner receive? A. $60,000 B. $90,000 C. $100,000 D. $120,000 E. $180,000

Fritz owns is \(\frac{2}{3}\)rd of the shares of the other three shareholders --> Fritz owns \(\frac{2}{2+3}=\frac{2}{5}\)th of all shares; Luis owns is \(\frac{3}{7}\)th of the shares of the other three shareholders --> Luis owns \(\frac{3}{3+7}=\frac{3}{10}\)th of all shares; Alfred owns is \(\frac{4}{11}\)th of the shares of the other three shareholders --> Alfred owns \(\frac{4}{4+11}=\frac{4}{15}\)th of all shares;

Together those three own \(\frac{2}{5}+\frac{3}{10}+\frac{4}{15}=\frac{29}{30}\)th of all shares, which means that Werner owns \(1-\frac{29}{30}=\frac{1}{30}\). Hence from $3,600,000 Werner gets \($3,600,000*\frac{1}{30}=$120,000\).

Answer: D.

It's quite simple: A has $2 and B has 3$ --> A has 2/3rd of B's amount and also 2/(2+3)=2/5th of total amount of $5.

Now, if you analyze each option you'll see that only \(52^4=2^4*13^4\) is not a factor of \(y\), since the power of 13 in it is higher than the power of 13 in \(y\).

Answer: E.

Pls correct the Typo 2^4 to 2^8
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The proof of understanding is the ability to explain it.

Now, if you analyze each option you'll see that only \(52^4=2^4*13^4\) is not a factor of \(y\), since the power of 13 in it is higher than the power of 13 in \(y\).

1. A password on Mr. Wallace's briefcase consists of 5 digits. What is the probability that the password contains exactly three digit 6?

A. 860/90,000 B. 810/100,000 C. 858/100,000 D. 860/100,000 E. 1530/100,000

Total # of 5 digit codes is 10^5, notice that it's not 9*10^4, since in a code we can have zero as the first digit.

# of passwords with three digit 6 is \(9*9*C^3_5=810\): each out of two other digits (not 6) has 9 choices, thus we have 9*9 and \(C^3_5\) is ways to choose which 3 digits will be 6's out of 5 digits we have.

\(P=\frac{favorable}{total}=\frac{810}{10^5}\)

Answer: B.

Hello Bunnel....

Can you kindly clarify this question and elaborate on the solution!!

1. A password on Mr. Wallace's briefcase consists of 5 digits. What is the probability that the password contains exactly three digit 6?

A. 860/90,000 B. 810/100,000 C. 858/100,000 D. 860/100,000 E. 1530/100,000

Total # of 5 digit codes is 10^5, notice that it's not 9*10^4, since in a code we can have zero as the first digit.

# of passwords with three digit 6 is \(9*9*C^3_5=810\): each out of two other digits (not 6) has 9 choices, thus we have 9*9 and \(C^3_5\) is ways to choose which 3 digits will be 6's out of 5 digits we have.

\(P=\frac{favorable}{total}=\frac{810}{10^5}\)

Answer: B.

Hello Bunnel....

Can you kindly clarify this question and elaborate on the solution!!

Thanks in advance.

Can you please specify which part didn't you understand? Thanks.
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6. A pool has two water pumps A and B and one drain C. Pump A alone can fill the whole pool in x hours, and pump B alone can fill the whole pool in y hours. The drain can empty the whole pool in z hours, where z>x. With pumps A and B both running and the drain C unstopped till the pool is filled, which of the following represents the amount of water in terms of the fraction of the pool which pump A pumped into the pool? A. \(\frac{yz}{x+y+z}\)

Reading the discussion on this question made me realize how I end up retracting on the suggested strategies again and again. Since we are talking about Quant, you would expect to have a definite set of rules but no sir, you don't.

I repeat this quite often: When you have a question using variables and the answer is in terms of those variables, your life is easier than if you get a question using numbers. The reason - you can assign your own 'cool' values to the variables for which the relations hold. Then just go on and check the option which gives you the right answer. I retract it with the following - But if the number of variables is high, say 3 or 4 or more, it might be too cumbersome to plug in values for each variable and keep in mind what stands for what. This could lead to a lot of confusion and errors. I retract it again - But if you can give some very convenient values to the variables, go ahead and plug it in.

This question has variables and the answer is in terms of the variables so plugging in values is a good option. But the number of variables is 3 which could make it cumbersome. But, you can give the variables such values that you get your answer quickly.

I need to find the rate of A. There are no constraints on the values x, y, and z can take except z > x (drain C empties slower than pipe A fille)

Let's say, x = 4, y = 8, z = 8 What did I do here? I made the rate of B same as the rate of C. This means, whenever both of them are working together, drain C cancels out the work of pipe B. So the entire pool will be filled by pipe A and the amount of water pumped in by A will be the entire pool. Hence, if y = z, pipe A fills the entire pool i.e. the amount of water in terms of fraction of the pool pumped by A is 1. In the options, put y = z and see which option gives you 1. Only options (B) and (E) do.

Now let's say, x = 8, y = 4, z = 8.00001 ( z should be greater than x but let's assume it is infinitesimally greater than x such that we can approximate it to 8 only) Rate of work of C is half the rate of work of B. Rate of work of C is same as rate of work of A. All the work done by pipe A is removed by drain C. So if pipe B fills the pool, drain C empties half of it in that time (this is the water pumped by pipe A) Put x = z in the options B and E. You get y/z = 4/8 = 1/2 in option (B).

Answer (B)

Even if you end up feeling that this method is complicated, try and wrap your head around it. It might give you some ideas of logical solutions in some other questions.
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