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Baker's Dozen
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08 Mar 2012, 12:27
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 Solution: bakersdozen12878220.html#p10575022. If \(y=\frac{(3^53^2)^2}{(5^75^4)^{2}}\), 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 Solution: bakersdozen12878220.html#p10575033. 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 Solution: bakersdozen12878220.html#p10575044. 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 Solution: bakersdozen12878220.html#p10575055. 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 Solution: bakersdozen12878220.html#p10575076. 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. \(\frac{yz}{yz+xzxy}\) C. \(\frac{yz}{yz+xz+xy}\) D. \(\frac{xyz}{yz+xzxy}\) E. \(\frac{yz+xzxy}{yz}\) Solution: bakersdozen12878220.html#p10575087. 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 Solution: bakersdozen12878220.html#p10575098. 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 Solution: bakersdozen12878220.html#p10575129. If x and y are negative numbers, what is the value of \(\frac{\sqrt{x^2}}{x}\sqrt{y*y}\)?A. 1+y B. 1y C. 1y D. y1 E. xy Solution: bakersdozen12878220.html#p105751410. If x^2<81 and y^2<25, what is the largest prime number that can be equal to x2y?A. 7 B. 11 C. 13 D. 17 E. 19 Solution: bakersdozen12878220.html#p105751511. 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 Solution: bakersdozen12878240.html#p105751712. 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 Solution: bakersdozen12878240.html#p105751913. 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 Solution: bakersdozen12878240.html#p1057520
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Re: Baker's Dozen
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13 Mar 2012, 02:53
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.
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Re: Baker's Dozen
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08 Mar 2012, 17:01
1) Ans B
The 3 digits that will be '6' can be selected in 5c3 = 10 ways
The remaining two digits can be any digit from 0 to 9 other than 6 i.e 9 digits.
so favorable outcomes = 5c3 *9*9*1*1*1 = 810 total outcome = 10*10*10*10 *10 = 10^5
P(exactly three 6) = 810/10^5
2) Ans E
y = (3^53^2)^2/ (5^7 5^4)^2
Numerator (3^2(3^31))^2 = (3^2(26) )^2 = 3^4*13^2*2^2
Denominator (5^4(5^31))^2 = 1/(5^4(124))^2 = 1/5^8*4^2*31^2 = 1/5^8*2^4*31^2
Numerator/denominator = 3^4 *13^2 *2^2 *5^8 *2^4*31^2 = 3^4*13^2*31^2*2^6
From answer choices we see that 52^4 is not divisble by y
3) Ans B
Old sum/k = 55 old sum = 55 k
old sum +100 /k+1 = 60 old sum+100 = 60k+60
55k+100 = 60k+60
5k = 40 k= 8
4) Ans D
126 * sqrt k = a^2 where a is a +ve integer
7*2*3^2 *sqrt k = a^2
So we need one two and one 7 to make 'a' a perfect square if k = 196 , 7*2*3^2*sqrt 196 = 2^2*7^2*3^2
5) For selecting 8 marbles from 12 marbles, a maximim of 4 marble can remain in the jar . The scenarios for atleast 1 red marble and 1 blue marble to remain are:
1r1b 7c1*5c1 = 35 1r2b 7c1*5c2 = 70 2r1b 7c2*5c1 = 105 2r2b 7c2*5c2 = 210 3r1b 7c3*5c1 = 175 1r3b 7c1*5c3 = 70
Total 665 . Not sure if the logic is right..
6) Ans E
Work done by pump a,b,c in 1 hr = 1/x , 1/y, 1/z respectively
Working simultaneously, work done by them in 1 hr is:
1/x+1/y1/z
yz +xz xy/xyz job done in 1 hr
So whole job will be completed in xyz/yz+xzxy hrs
Since pump A work for x hrs, fraction of job completed by pump A is :
x/(xyz/yz+xzxy) = yz+xzxy/yz
8) Ans) D
let the 7 consecutive odd integers be:
x, x+2, x+4, x+6, x+8, x+10, x+12 (in ascending order)
Sum of the 5 largest integers is x+12 +x+10 +x+8 +x+6 +x+4 = 5x+40
So 5x +40 = 185 5x = 225 x = 45
Sum of the 5 smallest integers = x+x+2+x+4+x+6+x+8 = 5x+20
5x+20 = (5*45) +20 = 225+20 = 205
9) Ans c
sqrt x^2 = x = x (since X<0)
sqrt (y*y) = sqrt(y*y) sqrt y^2 = y
sqrt x^2/x  sqrt(y*y) = x/x  y = 1y
10) Ans c
x^2<81 9<x<9 y^2<25 5<y<5
For x 2y to be a maximim prime number, x should be maximim prime number and y minimum prime number
So x2y = 7  (2*3) =13
11) Ans B
nth term of GP sereis is an = a1*r^n1 where r =3 common ratio , a1 =1
a13 = 1*3^12
a15 = 1^3^14
a13+a15 = 3^12 +3^14 = 3^12(1+3^2) = 3^12*10
Similarly a12 = 3^11 a14 = 3^13
a12+14 = 3^11+3^13 = 3^11*10
(a13+a15)  (a12+a14) = 3^12*10  3^11*10
3^11*10(31) = 3^11 *20
12)Ans B
x = yq1+3 y>3 since divisor should be greater than remainder
y = zq1+8 z>8 since divisor should be greater than remainder
so least value of z is 9 and least value of y 8 since for 8/9 remainder is 8
least value of x is 3 since for 3/8 remainder is 3
Least value of x+y+z = 3+8+9 = 20
13) Ans D
x = (8!)^10  (8!)^6 / (8!)^5 (8!)^3
Numerator, taking 8!^6 common ,(8!)^6 ((8!)^41) = (8!)^6 [(8!)^2+1)(8!)^21)]
Denominator, taking 8!^3 common (8!)^3 [(8!)^21]
x = (8!)^6 [(8!)^2+1)(8!)^21)] / (8!)^3 [(8!)^21]
x = (8!)^3[(8!)^2+1)
x = (8!)^5 +(8!)^3
x/(8!)^3 39 = [(8!)^5 +(8!)^3/(8!)^3] 39
= [(8!)^2 +1]39
=(8!)^238
8!^2 is a big number and will have 2 trailing zeros..
So some big number ending in 00  38 will have digits 2,6 in units and tens palce respectively
So product is 2*6 =12




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Re: Baker's Dozen
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09 Mar 2012, 11:28
In regards to question 13, how do you know that (8!)^2 will have two 0's at the end?



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Re: Baker's Dozen
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09 Mar 2012, 22:41
BN1989 wrote: In regards to question 13, how do you know that (8!)^2 will have two 0's at the end? You figure it out by knowing that there is a 5 as a factor in the final number. example 8! = 8*7*6*5...*1 Now if u know how many fives are there you will have that number of zeros in the end because at least that many number of 2 will there for sure ... Now IF u want to figure out how may fives are there in (which is equivalent to finding how many zeros in the end of the number ) then simply keep dividing the factorial by 5 and adding it to the result until 5^x exceeds the factorial, i know i have used very confusing language but a example will simplify things for sure how many zeros at the end of 312! answer : 312!/5 + 312!/5^2 + 312!/5^3 = 62 + 12 + 2 (now note i stopped at 5^3, because 5^4 =25*25=625 which is more that 312 ) hence : 62 + 12 + 2 =76 number of 5s in the factorial and same number of zeros in the end because these many number of 2s will obviously be there in factorial and will form 10s . IF you are still confused I would suggest you to download Bunuel's math's notes and even if you are not confused i would still suggest you to do the same ,



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Re: Baker's Dozen
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09 Mar 2012, 23:01
2)e 3)b 4)d 6)e 7)d 8)b 9)c 10)c 11)b 12) b 13)d
Bunuel could you please provide us/me with the answer keys so that I can cross check thanks once again for the questions and a amazing post .



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Re: Baker's Dozen
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09 Mar 2012, 23:22



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Re: Baker's Dozen
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09 Mar 2012, 23:38
Bunuel wrote: utkarshlavania wrote: 2)e 3)b 4)d 6)e 7)d 8)b 9)c 10)c 11)b 12) b 13)d
Bunuel could you please provide us/me with the answer keys so that I can cross check thanks once again for the questions and a amazing post . I'll provide OA's with detailed solutions after some discussion in a couple of days. If you need OA's asap pm me and I'll send you them. Bunuel i have messaged you, looking forward thanks once again ...



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Re: Baker's Dozen
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10 Mar 2012, 00:24
[quote2]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. 1y C. 1y D. y1 E. xy
Correction/suggestion/advice/confirmation  needed Answer to 9th question If x and y are negative numbers, what is the value of \(\frac{\sqrt{x^2}}{x}\sqrt{y*y}\)?[/b] Sqrtx^2/xsqrt(y*y) = x/xy= 1y



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Re: Baker's Dozen
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10 Mar 2012, 00:41



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Re: Baker's Dozen
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10 Mar 2012, 01:38
y = +y and y no matter what, comes out as positive hence Sqrt (y*y ) comes out as possitive y and sqrtx^2 =x= positive x , of course from your tone i get that i'm going wrong some where but



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Re: Baker's Dozen
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10 Mar 2012, 04:20
Ans 1 Total number of ways = 10 x 10 x 10 x 10 x 10 = 100,000 total number ways with exactly three six = 5C3 * 9*9 = 810 correct Answer B p.s IMO one more zero is required in the denominator. Ans 2 resolving the numerator/denominator, we get 3^4 * 5^8 * 2^2 *13^2 * 2^2 *62^2 so A to D is divisible by the above Correct Answer E Ans 3 This one is little simple the equation is 55*k +100 = 60(k + 1) thus 5k = 40 k = 8 Correct Answer B Ans 4 3*3*7*2*\sqrt{k} is the square of a positive integer( suppose X^2) so we need 7 and 2 and put them in square root as 14 * 14= 196 Correct Answer D
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Re: Baker's Dozen
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Re: Baker's Dozen
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10 Mar 2012, 06:42
Quote: 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 x ;x+2;x+2*2;x+2*3....x+2*6 since the set has consecutive odd integers, then mean =median mean of 5 largest integers=185/5= 37 median of 5 largest integers is the 3d of the largest or the 5th of the whole set A= x+2*4 x+8=37 ;x=45 median of the first 5 small integers =45+2*2=41 the sum of the first small odd integers=number of integers*mean (or median)=5*median=5*(41)=205 answ is
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Re: Baker's Dozen
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Updated on: 10 Mar 2012, 08:08
Quote: 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
x=((8!)^6*((8!)^41))/((8!)^3*((8!)^21))=(8!)^3*((8!)^2+1)
((8!)^3*((8!)^2+1) /(8!)^3)) 39=(8!)^2+1)39 note, that 8! has 5 and 2. so 8! ends with 0. then (8!)^2 will end by two 0s. ***0038=62 6*2=12 answ is
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*\sqrt{k}=3*7*3*2*sqrt{k}=3^2*7*2 so we need one more 7*2 (14) sqrootk=14 k=196 answ is
what an irony. all the 3 questions are heh
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. 1y C. 1y D. y1 E. xy x/xy=1(y)=y1 since y<0 ; x<0 is the answ
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Originally posted by LalaB on 10 Mar 2012, 06:54.
Last edited by LalaB on 10 Mar 2012, 08:08, edited 1 time in total.



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Re: Baker's Dozen
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10 Mar 2012, 07:03
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 x/y the remainder is 3 . then the smallest x is 3 y/z the remainder is 8. then the smallest y is 8 and z is 9 3+8+9=20 answ is B
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Re: Baker's Dozen
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10 Mar 2012, 07:17
10. If x^2<81 and y^2<25, what is the largest prime number that can be equal to x2y?A. 7 B. 11 C. 13 D. 17 E. 19 9<x<9 5<y<5 9<x<9 10<2y<10 x2y<9(10) x2y<19 the answer is 17. D again the mystical letter for today heh 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 1, 3, 9, 27 ok please pay attention to the fact that the term of sequence has 3 one less than the number of the term,i.e. the 3d term is 9, but the number has two 3s (one less than the term). the 4th term is 3*3*3 ,which has three 3s. ok, then we have the following the q. asks us to find out  (13th term+15th term) (12th term +14th term) (3^12+3^14)(3^113^13)= 3^11(31)+3^13(31)= 2*3^11(1+3^2)= 20*3^11 the answer is B p.s. also note, that answer choices B and D are leaders of the day heh
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Re: Baker's Dozen
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10 Mar 2012, 08:03
Bunuel wrote: utkarshlavania wrote: y = +y and y no matter what, comes out as positive hence Sqrt (y*y ) comes out as possitive y and sqrtx^2 =x= positive x , of course from your tone i get that i'm going wrong some where but What I meant is that if y<0 then y=y and 1y=1(y)=1+y. isn't 1 y=1y for the same reason x/x became 1 because numerator x was positive and denominator x was negative . sqrt(negative y *y) so negative y =postive y and y = positive hence when y comes out it comes as positive y and because of the negative sign 1  y=1y please help bunuel



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Re: Baker's Dozen
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10 Mar 2012, 08:12
utkarshlavania wrote: Bunuel wrote: utkarshlavania wrote: y = +y and y no matter what, comes out as positive hence Sqrt (y*y ) comes out as possitive y and sqrtx^2 =x= positive x , of course from your tone i get that i'm going wrong some where but What I meant is that if y<0 then y=y and 1y=1(y)=1+y. isn't 1 y=1y for the same reason x/x became 1 because numerator x was positive and denominator x was negative . sqrt(negative y *y) so negative y =postive y and y = positive hence when y comes out it comes as positive y and because of the negative sign 1  y=1y please help bunuel sqroot of (negative number ^2) = (negative number ^2)^1/2=negative number for example sqroot (5*5)=sqroot (5^2)=((5)^2)^1/2=5
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Re: Baker's Dozen
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10 Mar 2012, 10:55
LalaB wrote: ( sqroot of (negative number ^2) = (negative number ^2)^1/2=negative number for example sqroot (5*5)=sqroot (5^2)=((5)^2)^1/2=5[/quote] agreed but here isn't the concept little different, as it is already mentioned that y is negative hence y= positive y




Re: Baker's Dozen &nbs
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