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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

Can any one please explain: why it is (9*9*1*1*1* 5C3) in Question no. 1 (password problem)? Why not : 9*9*5P3 ??

That's a tough 700+ problem.

We are interested 666XX numbers. Now, two this can be arranged in 10 ways; 666XX; 66X6X; 6X66X; X666X;

66XX6; 6X6X6; X66X6;

6XX66; XX666;

X6X66.

So, basically with \(C^3_5=10\) we are choosing which 3 places out of 5 will be occupied by X's. Or we could do the other way around: \(C^2_510\) choosing which 2 places out of 5 will be occupied by 6's.

Next, each X can take any value from 0 to 9, inclusive but 6, so total of 9 values. Notice that the first digit can be zero here since the 5-digit password unlike 5-digit number can start with 0.

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.

Hey Bunuel,

Sorry to trouble you. If you could, I would appreciate if you could clarify more about 5C3. So this is how I approached the problem:

We have three - 6's. With two slots that can be filled by 9 numbers.

666 99 . Now the question is how many ways can we rearrange these codes of three 6's and 2 non 6's? Well, have 5 elements 3 identical (because we have 3 identical 6's and two non-6's) Hence we have 5!/(3!2!) ways of rearranging the numbers. Is this why we have 5C3 in the problem?

But the two non-6's can be identical or different. For example 66611 66622 66633 66644 66655 ... 66699... etc all satisfy the condition of only three 6's. The number of ways to rearrange the letters {aaabb} is 5!/(3!2!). Moreover, we can also have the two non-6's be different, 66612 66654 66624, etc . the number of ways to rearrange the letters {aaabc} is 5!/(3!).

What do we do in this situation? Do we look at this problem as (#6)(#6)(#6)(not #6)(not #6) therefore 5!/(3!2!). Three repeating elements of #6 and two repeating elements of not #6.

OR

Do we look at the problem as (#6)(#6)(#6)(x)(y), where x and y two numbers that are not 6, that may or may not be equal, therefore 5!/3! (3! for the 3 repeating 6's but we cannot say anything about whether x and y are equal to each other)

Bunuel, as always, thank you so much!! And again, sorry to trouble you.

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.

Hey Bunuel,

Sorry to trouble you. If you could, I would appreciate if you could clarify more about 5C3. So this is how I approached the problem:

We have three - 6's. With two slots that can be filled by 9 numbers.

666 99 . Now the question is how many ways can we rearrange these codes of three 6's and 2 non 6's? Well, have 5 elements 3 identical (because we have 3 identical 6's and two non-6's) Hence we have 5!/(3!2!) ways of rearranging the numbers. Is this why we have 5C3 in the problem?

But the two non-6's can be identical or different. For example 66611 66622 66633 66644 66655 ... 66699... etc all satisfy the condition of only three 6's. The number of ways to rearrange the letters {aaabb} is 5!/(3!2!). Moreover, we can also have the two non-6's be different, 66612 66654 66624, etc . the number of ways to rearrange the letters {aaabc} is 5!/(3!).

What do we do in this situation? Do we look at this problem as (#6)(#6)(#6)(not #6)(not #6) therefore 5!/(3!2!). Three repeating elements of #6 and two repeating elements of not #6.

OR

Do we look at the problem as (#6)(#6)(#6)(x)(y), where x and y two numbers that are not 6, that may or may not be equal, therefore 5!/3! (3! for the 3 repeating 6's but we cannot say anything about whether x and y are equal to each other)

Bunuel, as always, thank you so much!! And again, sorry to trouble you.

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!

10. If x^2<81 and y^2<25, what is the largest prime number that can be equal to x-2y? A. 7 B. 11 C. 13 D. 17 E. 19

my solution: x^2<81 ==> Mod(x)<9 ==> -9 < x< 9----------------------------------# y^2<25 ==> Mod(y)<5 ==> -5 < y<5 so, -10<2y<10 and (-)-10<(-)2y<(-)10 {multiplying - to all. after multiplication, sign reverses.} so, it will be -10<-2y<10 ------------------------------------------@ adding @ & # -19<x-2y<19 largest prime no is 17. ANS : D

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

my solution : Difference = [(a13 + a15)- (a12 + a14)] : here series is in GP with common ratio r= 3 and first term a=1 = [ar^12 +ar^14 - ar^11- ar^13] = [ar^11 (r+r^3-1-r^2] = ar^11 ( 3+27-1-9) = 1.3^11 .20 = 20*3^11 Ans : B 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

My solution : x = y@ + 3 y= Z# +8 min value of z is 1 (given x, y and z are positive integers ) and Min value of # is 0 so, y = 0.1+8 =8 & z=1 min value of @ is 0 x=0+3 =3 X+Y+Z = 3+8+1 =12 .Ans :A

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

then x = (8!)^(6-3) [(8!)^2+1] x / (8!)^3 = [(8!)^2+1] we have to calculate \(\frac{x}{(8!)^3}-39\)?[/b] so, \(\frac{x}{(8!)^3}-39\)?[/b] = [(8!)^2+1] -39 = (8!)^2-38

now we have top find the product of the tens and the units digits : 8! consist one 5 so, its unit digit must be zero. on squaring the 8!, there will be two zero at the end i.e one at unit digit and tens digit.

so, xxxxxxx00-38 = yyyyyy62

Ans : 12 (D)
_________________

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Attitude determine everything. all the best and God bless you.

Question 13 number of 6's can be arranged among themselves in 1 way. Number of ways we can select three position from 5= 5C3. For the remaining two position we can choose any digit but 6. So number of possible patterns=(5C3)*9*9 Hence required probability=(5C3)*9*9/10^5 I hope those who have doubt can understand this explanation

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.

Could some one explain this to me how does one arrive at 2/2+3 = 2/5 and so on?

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.

Could some one explain this to me how does one arrive at 2/2+3 = 2/5 and so on?

Can any one please explain: why it is (9*9*1*1*1* 5C3) in Question no. 1 (password problem)? Why not : 9*9*5P3 ??

That's a tough 700+ problem.

We are interested 666XX numbers. Now, two this can be arranged in 10 ways; 666XX; 66X6X; 6X66X; X666X;

66XX6; 6X6X6; X66X6;

6XX66; XX666;

X6X66.

So, basically with \(C^3_5=10\) we are choosing which 3 places out of 5 will be occupied by X's. Or we could do the other way around: \(C^2_510\) choosing which 2 places out of 5 will be occupied by 6's.

Next, each X can take any value from 0 to 9, inclusive but 6, so total of 9 values. Notice that the first digit can be zero here since the 5-digit password unlike 5-digit number can start with 0.

Total \(9*9*C^3_5=810\).

Hope it's clear.

Hi Bunuel..can you tell me if I am thinking correct.

1/10 * 1/10 * 1/10 * 9/10 * 9/10 this gives me 3 chances of getting 6 and rest 2 for getting any other number. now the above scenario can be arranged in 5C3 ways so

1/10 * 1/10 * 1/10 * 9/10 * 9/10 * 5C3..

I hope I am correct in this way..?

Edit: made it more clear..
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Life is very similar to a boxing ring. Defeat is not final when you fall down… It is final when you refuse to get up and fight back!

Hi, Can someone please provide the working links for the solutions of all "Baker's dozen" questions? The questions are interesting, but the answers are not collated and the link below every question does not lead to solution for that question. It's painful to go through all the four pages but still not get answers for few of the questions. Also, for the shareholder's question, someone asked for an explanation of how 2/5 shares, 3/10 shares and 4/15 shares were calculated, the post says that there is an explanation in the given link, but unfortunately the link does not lead to how the fractions were derived. Please help me with this. Thanks!

Hi, Can someone please provide the working links for the solutions of all "Baker's dozen" questions? The questions are interesting, but the answers are not collated and the link below every question does not lead to solution for that question. It's painful to go through all the four pages but still not get answers for few of the questions. Also, for the shareholder's question, someone asked for an explanation of how 2/5 shares, 3/10 shares and 4/15 shares were calculated, the post says that there is an explanation in the given link, but unfortunately the link does not lead to how the fractions were derived. Please help me with this. Thanks!

Switch view mode of the topic from "Best Reply" to "Oldest" and the links from the initial post will lead you to the posts with solutions. The question about the shareholders is answered here: baker-s-dozen-128782-40.html#p1059585

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.

Hi Bunnel,

I didn't understand how you got 5C3, can you please explain more this...
_________________

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.

Hi Bunnel,

I didn't understand how you got 5C3, can you please explain more this...

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

In 7 such numbers the middle three would remain same. So we are basically replacing 6th and 7th by 1st and 2nd. Since these are consecutive odd numbers => 7th and 2nd differ by 10..... similarly 6th and 1st by 10...total difference = 20 so our answer should be = sum - 20 = -185-20 = -205
_________________

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.

Answer: B.

Hey Bunuel,

Thank you for the explanation. I am trying to understand why we can impose the condition that \(z>8\) be greater than its remainded, but that we don't impose the same condition on \(y>3\)

4. What is the smallest positive integer \(k\) such that \(126*\sqrt{k}\) is the square of a positive integer? A. 14 B. 36 C. 144 D. 196 E. 441

\(126=2*3^2*7\), so in order \(126*\sqrt{k}\) to be a square of an integer \(\sqrt{k}\) must complete the powers of 2 and 7 to even number, so the least value of \(\sqrt{k}\) must equal to 2*7=14, which makes the leas value of \(k\) equal to 14^2=196.Answer: D.

I did a silly mistake here....after coming to the digit as 14. saw the option and marked A..!!! but didnt think that its sqrt of K...

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

Answer B

Number of codes with exactly 3 digits of 6 is 5C3*9*9= 810. Total number of possible codes \(10^5.\) Requested probability \(\frac{810}{10^5}\)

2. If \(y=\frac{(3^5-3^2)^2}{(5^7-5^4)^{-2}}\), then y is NOT divisible by which of the following? A. 6^4 B. 62^2 C. 65^2 D. 15^4 E. 52^4

Answer E \(y=\frac{(3^5-3^2)^2}{(5^7-5^4)^{-2}}=(3^5-3^2)^2(5^7-5^4)^2=3^4*5^8(3^2-1)^2*(5^3-1)^2=3^4*5^8*8^2*124^2\). The number is not divisible by 13, which is a factor of 52.

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

Answer B 55k+100 = 60(k+1), solve for k, and get k = 8.

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

Answer D The total number of possibilities to chose 8 marbles is 12C8 = \(\frac{12*11*10*9}{2*3*4}=\frac{990}{2}=495.\). Count the number of possibilities when no red marble is left, so 7 red and 1 blue were selected, which is 5 possibilities. Count the number of possibilities that no blue marble is left, so 3 red and 5 blue were selected, which is 7C3=\(\frac{7*6*5}{2*3}=35\). Therefore, the required number is 495 - 5 - 35 = 455.

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. \(\frac{yz}{x+y+z}\)

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

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

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

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

Answer B Total time it takes to fill the whole pool is \(T=\frac{1}{\frac{1}{x}+\frac{1}{y}-\frac{1}{z}}=\frac{xyz}{yz+xz-xy}\). Rate of pump A is 1/x, and T/x gives exactly the expression in B.

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

Answer D 2/3 + 1 =5/3, 3/7+1=10/7, 4/11+1=15/11, therefore consider the total amount a multiple of 30, so 30N. From the given information we can deduce that Fritz owns 12N, Luis 9N, and Alfred 8N. So, there is N left for Werner, which is 3,600,000/30 = 120,000.

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

Answer D We can deduce that the fifth number in the sequence is -185/5 = -37. Then, the third number is -37 - 2 -2 = -41. The required sum is 5*(-41) = -205.

10. If x^2<81 and y^2<25, what is the largest prime number that can be equal to x-2y? A. 7 B. 11 C. 13 D. 17 E. 19

Answer D We can deduce that \(-9<x<9\) and \(-5<y<5\) or \(-10<-2y<10\), and altogether \(-19<x-2y<19\). Therefore, the largest prime number which fulfills the above requirement is 17.

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

Answer B Let's denote by \(a_1=1, a_2=3,...\) the terms of the given sequence. \(a_{13}+a_{15}-(a_{12}+a_{14})=a_{13}+9a_{13}-(a_{12}+9a_{12})=10a_{13}-10a_{12}=30a_{12}-10a_{12}=20a_{12}=20*3^{11}\)

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

Answer B We can write \(x = ay+3\), where \(a\) is a non-negative integer. We can deduce that \(x\geq3\) and \(y>3\). Also, \(y=bz+8\), for some non-negative integer \(b\). It follows from here that \(y\geq8\) and \(z>8\). The minimum sum for \(x+y+z\) is obtained for \(x=3, y=8, z=9\).

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

Answer D \(x=\frac{(8!)^{10}-(8!)^6}{(8!)^{5}-(8!)^3}=\frac{(8!)^6((8!)^4-1)}{(8!)^3((8!)^2-1)}\). Since \((8!)^4-1=((8!)^2+1)((8!)^2-1)\), \(\frac{x}{(8!)^3}=(8!)^2+1\), which is an integer which ends in 1 and has more than 2 zeros before the 1. *001-39 ends in 62. The product of the last two digits is 6*2=12.

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.

Bunuel, i am confused with the answer...

i have followed the method...

{sq-rt[(-x)sq]/(-x) } - {sq-rt[-(-y)*mod(-y)]}

{sq-rt(x2)/(-x)} - sq-rt(y2)

[x/(-x)] - y

-1-y

since we had used the negative values before, the final answer would have to be -(y+1) which is C. Can you please explain me why we have again substituted y with -y just before the final step to give -1+y.

Thanks
_________________

Regards, Harsha

Note: Give me kudos if my approach is right , else help me understand where i am missing.. I want to bell the GMAT Cat

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.

Bunuel, i am confused with the answer...

i have followed the method...

{sq-rt[(-x)sq]/(-x) } - {sq-rt[-(-y)*mod(-y)]}

{sq-rt(x2)/(-x)} - sq-rt(y2)

[x/(-x)] - y

-1-y

since we had used the negative values before, the final answer would have to be -(y+1) which is C. Can you please explain me why we have again substituted y with -y just before the final step to give -1+y.

Thanks

\(\sqrt{y^2}=|y|\) and \(|y|=-y,\) because \(y<0.\) So you should have \(-(-y)=y.\)
_________________

PhD in Applied Mathematics Love GMAT Quant questions and running.

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