Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized for You
we will pick new questions that match your level based on your Timer History
Track Your Progress
every week, we’ll send you an estimated GMAT score based on your performance
Practice Pays
we will pick new questions that match your level based on your Timer History
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
It appears that you are browsing the GMAT Club forum unregistered!
Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club
Registration gives you:
Tests
Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.
Applicant Stats
View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more
Books/Downloads
Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!
Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:
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
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:
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 ,
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. _________________
[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. 1-y C. -1-y D. y-1 E. x-y
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/x-sqrt(-y*|y|) = |x|/x-|y|= -1-y
[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. 1-y C. -1-y D. y-1 E. x-y
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/x-sqrt(-y*|y|) = |x|/x-|y|= -1-y
Hint: what is |y| if y<0? and what is -1-|y| then? _________________
-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
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 _________________
Fire the final bullet only when you are constantly hitting the Bull's eye, till then KEEP PRACTICING.
-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 -1-|y|=-1-(-y)=-1+y. _________________
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
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
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 -x/x-|y|=-1-(-y)=y-1 since y<0 ; x<0
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 _________________
Happy are those who dream dreams and are ready to pay the price to make them come true
I am still on all gmat forums. msg me if you want to ask me smth
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
-9<x<9 -5<y<5
-9<x<9 -10<2y<10 x-2y<9-(-10) x-2y<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)-(3^11-3^13)= 3^11(3-1)+3^13(3-1)= 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 _________________
Happy are those who dream dreams and are ready to pay the price to make them come true
I am still on all gmat forums. msg me if you want to ask me smth
-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 -1-|y|=-1-(-y)=-1+y.
isn't -1 -|y|=-1-y 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=-1-y please help bunuel
-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 -1-|y|=-1-(-y)=-1+y.
isn't -1 -|y|=-1-y 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=-1-y 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 _________________
Happy are those who dream dreams and are ready to pay the price to make them come true
I am still on all gmat forums. msg me if you want to ask me smth
The “3 golden nuggets” of MBA admission process With ten years of experience helping prospective students with MBA admissions and career progression, I will be writing this blog through...
You know what’s worse than getting a ding at one of your dreams schools . Yes its getting that horrid wait-listed email . This limbo is frustrating as hell . Somewhere...