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:

53% (04:55) correct
47% (06:08) wrong based on 192 sessions

HideShow timer Statistics

Please find below new set of PS problems:

1. A family consisting of one mother, one father, two daughters and a son is taking a road trip in a sedan. The sedan has two front seats and three back seats. If one of the parents must drive and the two daughters refuse to sit next to each other, how many possible seating arrangements are there? (A) 28 (B) 32 (C) 48 (D) 60 (E) 120

2. What is the probability that a 3-digit positive integer picked at random will have one or more "7" in its digits? (A) 271/900 (B) 27/100 (C) 7/25 (D) 1/9 (E) 1/10

3. A sphere is inscribed in a cube with an edge of 10. What is the shortest possible distance from one of the vertices of the cube to the surface of the sphere? (A 10( sqrt3- 1) (B) 5 (C) 10( sqrt2 - 1) (D) 5( sqrt3 - 1) (E) 5( sqrt2 - 1)

4. A contractor estimated that his 10-man crew could complete the construction in 110 days if there was no rain. (Assume the crew does not work on any rainy day and rain is the only factor that can deter the crew from working). However, on the 61-st day, after 5 days of rain, he hired 6 more people and finished the project early. If the job was done in 100 days, how many days after day 60 had rain? (A) 4 (B) 5 (C) 6 (D) 7 (E) 8

5. If s and t are positive integer such that s/t=64.12, which of the following could be the remainder when s is divided by t? (A) 2 (B) 4 (C) 8 (D) 20 (E) 45

6. A committee of 6 is chosen from 8 men and 5 women so as to contain at least 2 men and 3 women. How many different committees could be formed if two of the men refuse to serve together? (A) 3510 (B) 2620 (C) 1404 (D) 700 (E) 635

7. If x is positive, which of the following could be the correct ordering of 1/x,2x and x^2 ? I. x^2<2x<1/x II. x^2<1/x<2x III. 2x<x^2<1/x

(A) None (B) I only (C) III only (D) I and II only (E) I II and III

8. In the xy plane, Line k has a positive slope and x-intercept 4. If the area of the triangle formed by line k and the two axes is 12, What is the y-intercept of line K ? (A) 3 (B) 6 (C) -3 (D) -6 (E) -4

9. Of the applicants passes a certain test, 15 applied to both college X and Y. If 20 % of the applicants who applied college X and 25% of the applicants who applied college Y applied both college X and Y, how many applicants applied only college X or college Y? (A) 135 (B) 120 (C) 115 (D) 105 (E) 90

10. What is the lowest positive integer that is divisible by each of the integers 1 through 7, inclusive? (A) 420 (B) 840 (C) 1260 (D) 2520 (E) 5040

Please share your way of thinking, not only post the answers.

1. A family consisting of one mother, one father, two daughters and a son is taking a road trip in a sedan. The sedan has two front seats and three back seats. If one of the parents must drive and the two daughters refuse to sit next to each other, how many possible seating arrangements are there? (A) 28 (B) 32 (C) 48 (D) 60 (E) 120

The total possible position given that one of the parents must drive is: 2x4x3x2x1 = 48 (the first two represents the driver seat).

I want to find the number of positions where the two daughters sit together and subtract this from 48. There are 4 possibilities where they are sitting together and that is: D1, D2, _ D2, D1, _ _, D1, D2 _, D2, D1 For each of these 4 possibilities the other 3 members can be seated in 2x2x1 ways. So the number of positions where the two daughters sit together = 4x2x2x1 = 16

Total positions that meet the question requirements = 48-16 = 32 ANS = B

Looking back on this I'm sure there's more elegant methods

Bunuel wrote:

2. What is the probability that a 3-digit positive integer picked at random will have one or more "7" in its digits? (A) 271/900 (B) 27/100 (C) 7/25 (D) 1/9 (E) 1/10

P(at least one 7) = 1 - P(no 7's) Total 3 digit numbers are 9x10x10 = 900 (first digit cannot have a 0) P(no 7's) = (8x9x9)/900 = 648/900 P(at least one 7) = 1 - P(no 7's) = 1 - 648/900 = 252/900 = 7/25

1. A family consisting of one mother, one father, two daughters and a son is taking a road trip in a sedan. The sedan has two front seats and three back seats. If one of the parents must drive and the two daughters refuse to sit next to each other, how many possible seating arrangements are there? (A) 28 (B) 32 (C) 48 (D) 60 (E) 120

As most of the combination problems this one can be solved in more than 1 way:

Sisters sit separately: 1. one of them is on the front seat (2 ways). Others (including second sister) can be arranged in: 2 (drivers seat)*3! (arrangements of three on the back seat)=12 ways. Total for this case: 2*12=24 Or 2. both by the window on the back seat (2 ways). Others can be arranged in: 2 (drivers seat)*2 (front seat)*1(one left to sit between the sisters on the back seat)=4 ways. Total for this case=8. Total=24+8=32.

Another way: Total number of arrangements-arrangements with sisters sitting together=2*4*3!-2*2(sisters together)*2*2*1(arrangement of others)=48-16=32

Answer: B.

2. What is the probability that a 3-digit positive integer picked at random will have one or more "7" in its digits? (A) 271/900 (B) 27/100 (C) 7/25 (D) 1/9 (E) 1/10

Total 3 digit numbers 900, 3 digit number with no 7 =8*9*9=648, P(at least one 7)=1-P(no 7)=1-648/900=252/900=7/25

Answer: C.

3. A sphere is inscribed in a cube with an edge of 10. What is the shortest possible distance from one of the vertices of the cube to the surface of the sphere? (A) \(10(\sqrt{3}- 1)\) (B) \(5\) (C) \(10(\sqrt{2} - 1)\) (D) \(5(\sqrt{3} - 1)\) (E) \(5(\sqrt{2} - 1)\)

Shortest distance=(diagonal of cube-diameter of sphere)/2= \(\frac{10*\sqrt{3}-10}{2}=5(\sqrt{3}-1)\)

Answer: D.

4. A contractor estimated that his 10-man crew could complete the construction in 110 days if there was no rain. (Assume the crew does not work on any rainy day and rain is the only factor that can deter the crew from working). However, on the 61-st day, after 5 days of rain, he hired 6 more people and finished the project early. If the job was done in 100 days, how many days after day 60 had rain? (A) 4 (B) 5 (C) 6 (D) 7 (E) 8

This one was solved incorrectly: Days to finish the job for 10 people 110 days. On the 61-st day, after 5 days of rain --> 5 days was rain, 55 days they worked, thus completed 1/2 of the job, 1/2 is left (55 days of work for 10 people). Then 6 more people was hired --> speed of construction increased by 1.6, days needed to finish 55/1.6=34.375, BUT after they were hired job was done in 100-60=40 days --> so 5 days rained. They needed MORE than 34 days to finish the job, so if it rained for 6 days they wouldn't be able to finish the job in 100(40) days.

Answer: B.

5. If s and t are positive integer such that s/t=64.12, which of the following could be the remainder when s is divided by t? (A) 2 (B) 4 (C) 8 (D) 20 (E) 45

\(s\) divided by \(t\) yields the remainder of \(r\) can always be expressed as: \(\frac{s}{t}=q+\frac{r}{t}\) (which is the same as \(s=qt+r\)), where \(q\) is the quotient and \(r\) is the remainder.

Given that \(\frac{s}{t}=64.12=64\frac{12}{100}=64\frac{3}{25}=64+\frac{3}{25}\), so according to the above \(\frac{r}{t}=\frac{3}{25}\), which means that \(r\) must be a multiple of 3. Only option E offers answer which is a multiple of 3

Answer: E.

6. A committee of 6 is chosen from 8 men and 5 women so as to contain at least 2 men and 3 women. How many different committees could be formed if two of the men refuse to serve together? (A) 3510 (B) 2620 (C) 1404 (D) 700 (E) 635

Committee can have either: 2 men and 4 women OR 3 men and 3 women (to meet the condition of at least 2 men and 3 women).

Ways to chose 6 members committee without restriction (two men refuse to server together): \(C^2_8*C^4_5+C^3_8*C^3_5 = 700\)

Ways to chose 6 members committee with two particular men serve together: \(C^2_2*C^4_5+C2^_2*C^1_6*C^3_5=5+60=65\)

700-65 = 635

Answer: E.

7. If x is positive, which of the following could be the correct ordering of 1/x,2x and x^2 ? I. x^2<2x<1/x II. x^2<1/x<2x III. 2x<x^2<1/x

(A) None (B) I only (C) III only (D) I and II only (E) I II and III

First note that we are asked "which of the following COULD be the correct ordering" not MUST be. Basically we should determine relationship between \(x\), \(\frac{1}{x}\) and \(x^2\) in three areas: \(0<1<2<\).

\(x>2\)

\(1<x<2\)

\(0<x<1\)

When \(x>2\) --> \(x^2\) is the greatest and no option is offering this, so we know that x<2. If \(1<x<2\) --> \(2x\) is greatest then comes \(x^2\) and no option is offering this.

So, we are left with \(0<x<1\): In this case \(x^2\) is least value, so we are left with:

I. \(x^2<2x<\frac{1}{x}\) --> can \(2x<\frac{1}{x}\)? Can \(\frac{2x^2-1}{x}<0\), the expression \(2x^2-1\) can be negative or positive for \(0<x<1\). (You can check it either algebraically or by picking numbers)

II. \(x^2<\frac{1}{x}<2x\) --> can \(\frac{1}{x}<2x\)? The same here \(\frac{2x^2-1}{x}>0\), the expression \(2x^2-1\) can be negative or positive for \(0<x<1\). (You can check it either algebraically or by picking numbers)

Answer: D.

8. In the xy plane, Line k has a positive slope and x-intercept 4. If the area of the triangle formed by line k and the two axes is 12, What is the y-intercept of line K ? (A) 3 (B) 6 (C) -3 (D) -6 (E) -4

9. Of the applicants passes a certain test, 15 applied to both college X and Y. If 20 % of the applicants who applied college X and 25% of the applicants who applied college Y applied both college X and Y, how many applicants applied only college X or college Y? (A) 135 (B) 120 (C) 115 (D) 105 (E) 90

20%X=X&Y=15 --> X=75 --> Only X=75-15=60 25%Y=X&Y=15 --> Y=60 --> Only Y=60-15=45 Only X or Y=60+45=105

Answer: D.

10. What is the lowest positive integer that is divisible by each of the integers 1 through 7, inclusive. (A) 420 (B) 840 (C) 1260 (D) 2520 (E) 5040

The integer should be divisible by: 2, 3, 4(=2^2), 5, 6(=2*3), and 7. LCM=2^2*3*5*7=420

Easiest one so far Bunuel... your questions are killers (to me anyway haha)

You are right not every question is 700+... Though I try to post toughest problems from my collection.

BTW 7 is not correct, try again. You have good speed and almost every answer from you is correct. Check DS set too if you like such "killers"
_________________

3. A sphere is inscribed in a cube with an edge of 10. What is the shortest possible distance from one of the vertices of the cube to the surface of the sphere? (A 10( sqrt3- 1) (B) 5 (C) 10( sqrt2 - 1) (D) 5( sqrt3 - 1) (E) 5( sqrt2 - 1)

The shortest distance from a vertice to the cube would follow the line that is the diagonal to the cube. Shortest distance = (Diagonal of cube - Diameter of cube) / 2 We divide by 2 otherwise we get the distance from the cube to the vertex on each side of the diagonal. (Sorry don't have graphics software to draw it out).

Diagonal^2 = (diagonal of base)^2 + (height)^2 = \((10*\sqrt{2})^2 + 10^2 = 300\) Diagonal = \sqrt{300} Diameter = same as height of cube = 10 Shortest Distance = (sqrt300-10)/2 = 5(sqrt 3 - 1)

6. A committee of 6 is chosen from 8 men and 5 women so as to contain at least 2 men and 3 women. How many different committees could be formed if two of the men refuse to serve together? (A) 3510 (B) 2620 (C) 1404 (D) 700 (E) 635

The committee can be formed in two ways: 1) 2 men and 4 women 2) 3 men and 3 women The answer is the sum of these.

1) 2 men and 4 women = (8C2 - 1) x 5C4 = 27 x 5 = 135 Subtract 1 since there is one combo of men that are not allowed. 2) 3 men and 3 women = (8C3 - 6) x 5C3 = (56-6) x 10 = 500 Subtract 6 since there are 6 groups of men that can include those the two that refuse to work together. Adding these together we get 135+500 = 635 ANS = E

4. A contractor estimated that his 10-man crew could complete the construction in 110 days if there was no rain. (Assume the crew does not work on any rainy day and rain is the only factor that can deter the crew from working). However, on the 61-st day, after 5 days of rain, he hired 6 more people and finished the project early. If the job was done in 100 days, how many days after day 60 had rain? (A) 4 (B) 5 (C) 6 (D) 7 (E) 8

This one was solved incorrectly: Days to finish the job for 10 people 110 days. On the 61-st day, after 5 days of rain --> 5 days was rain, 55 days they worked, thus completed 1/2 of the job, 1/2 is left (55 days of work for 10 people). Then 6 more people was hired --> speed of construction increased by 1.6, days needed to finish 55/1.6=34.375, BUT after they were hired job was done in 100-60=40 days --> so 5 days rained. They needed MORE than 34 days to finish the job, so if it rained for 6 days they wouldn't be able to finish the job in 100(40) days.

Answer: B.

I solved in a more easier way I think:

1) 10 man 110 days --> need for 1100 man.days 2) 55 days with 10 men --> 550 man.days 3) 40 days with 16 men --> 640 man.days

--> total man.days equals 1190 vs need for 1100 --> days of rain equals 90/16 max --> 5.625 --> rounded to 5

5. If s and t are positive integer such that s/t=64.12, which of the following could be the remainder when s is divided by t? (A) 2 (B) 4 (C) 8 (D) 20 (E) 45

s/t = 64.12 = 6412/100 => 6412 div 100 Remainder = 12 = 3206/50 => 3206 div 50 Remainder = 6 = 1603/25 => 1603 div 25 Remainder = 3 No more common factors. I don't see how the remainder could be anything but 3,6,12,24,48. What am I doing wrong here?

7. If x is positive, which of the following could be the correct ordering of 1/x,2x and x^2 ? I. x^2<2x<1/x II. x^2<1/x<2x III. 2x<x^2<1/x

(A) None (B) I only (C) III only (D) I and II only (E) I II and III

Two situations we need to consider: 0 < x < 1, or x > 1 (since squaring a number for the first makes the number smaller for the former and larger for the latter).

Situation 1: let X = 1/3 x^2 = 1/9 2x = 2/3 1/x = 3 Correct ordering: x^2 < 2x < 1/x Option I is the only possibility. You could choose B based on this.

Check Situation 2: Let X = 3 x^2 = 9 2x = 6 1/x = 1/3 Correct ordering 1/x < 2x < x^2 Not in any of the possibilities.

9. Of the applicants passes a certain test, 15 applied to both college X and Y. If 20 % of the applicants who applied college X and 25% of the applicants who applied college Y applied both college X and Y, how many applicants applied only college X or college Y? (A) 135 (B) 120 (C) 115 (D) 105 (E) 90

10. What is the lowest positive integer that is divisible by each of the integers 1 through 7, inclusive? (A) 420 (B) 840 (C) 1260 (D) 2520 (E) 5040

Q9) 20% of total applied at X = 15 100% of total applied at X = 75 Only applied at X = 60

25% of total applied at Y = 15 100% of total applied at Y = 60 Only applied at Y = 45

Only applied at X + Only applied at Y = 60 + 45 = 105 ANS = D

Q10) 1 x 2 x 3 x 4 x 5 x 6 x 7 = 2^4 x 3^2 x 5 x 7 Option A = 420 = 2^2 x 3 x 5 x 7 (factored this out) Option B = 2 x 420 = 2^3 x 3 x 5 x 7 (just add a 2 to A) Option C = 3 x 420 = 2^2 x 3^2 x 5 x 7 (just add a 3 to A) Option D = 2 x 1260 = 2^3 x 3^2 x 5 x 7 (just add a 2 to C) Option E = 2 x 2520 = 2^4 x 3^2 x 5 x 7 (just add a 2 to D)

Actually after doing up to A you can quickly just figure out how many more 2's and 3's and determine it's E.

5. If s and t are positive integer such that s/t=64.12, which of the following could be the remainder when s is divided by t? (A) 2 (B) 4 (C) 8 (D) 20 (E) 45

s/t = 64.12

we know -> s,t are +ve integers, remainder r is a +ve integer (from options).

s/t = 64.12

so from this we can infer :-

r= 12% of t

r = (12/100) x t

t = (r x 100)/12

t needs to be an integer, which is only satisfied by option (E) 45.
_________________

9. Of the applicants passes a certain test, 15 applied to both college X and Y. If 20 % of the applicants who applied college X and 25% of the applicants who applied college Y applied both college X and Y, how many applicants applied only college X or college Y? (A) 135 (B) 120 (C) 115 (D) 105 (E) 90

10. What is the lowest positive integer that is divisible by each of the integers 1 through 7, inclusive? (A) 420 (B) 840 (C) 1260 (D) 2520 (E) 5040

Q9) 20% of total applied at X = 15 100% of total applied at X = 75 Only applied at X = 60

25% of total applied at Y = 15 100% of total applied at Y = 60 Only applied at Y = 45

Only applied at X + Only applied at Y = 60 + 45 = 105 ANS = D

Q10) 1 x 2 x 3 x 4 x 5 x 6 x 7 = 2^4 x 3^2 x 5 x 7 Option A = 420 = 2^2 x 3 x 5 x 7 (factored this out) Option B = 2 x 420 = 2^3 x 3 x 5 x 7 (just add a 2 to A) Option C = 3 x 420 = 2^2 x 3^2 x 5 x 7 (just add a 3 to A) Option D = 2 x 1260 = 2^3 x 3^2 x 5 x 7 (just add a 2 to C) Option E = 2 x 2520 = 2^4 x 3^2 x 5 x 7 (just add a 2 to D)

Actually after doing up to A you can quickly just figure out how many more 2's and 3's and determine it's E.

for q10 we need to find the lowest number - so should be 420 its divisible by all the integers from 1-7 inclusive

4. A contractor estimated that his 10-man crew could complete the construction in 110 days if there was no rain. (Assume the crew does not work on any rainy day and rain is the only factor that can deter the crew from working). However, on the 61-st day, after 5 days of rain, he hired 6 more people and finished the project early. If the job was done in 100 days, how many days after day 60 had rain? (A) 4 (B) 5 (C) 6 (D) 7 (E) 8

Nobody attempted 4, let me give a try. total work = 110*10 = 1100 man days now, from day 1 to day 55, 10 men worked = 550 man days of work was done. from day 61 to day 100, 16 men worked = 640 man days of work was done.

so total work done should be 1190, the 90 days offset is due to rain on few days between 61 to 100th day...so the number of rainy days should be 90/16=5.625 ~ 6. C.

10. What is the lowest positive integer that is divisible by each of the integers 1 through 7, inclusive? (A) 420 (B) 840 (C) 1260 (D) 2520 (E) 5040

Q10) 1 x 2 x 3 x 4 x 5 x 6 x 7 = 2^4 x 3^2 x 5 x 7 Option A = 420 = 2^2 x 3 x 5 x 7 (factored this out) Option B = 2 x 420 = 2^3 x 3 x 5 x 7 (just add a 2 to A) Option C = 3 x 420 = 2^2 x 3^2 x 5 x 7 (just add a 3 to A) Option D = 2 x 1260 = 2^3 x 3^2 x 5 x 7 (just add a 2 to C) Option E = 2 x 2520 = 2^4 x 3^2 x 5 x 7 (just add a 2 to D)

Actually after doing up to A you can quickly just figure out how many more 2's and 3's and determine it's E.

What does the ques ask...which of the following is divisible by EACH of the integers from 1 thru 7...so I interpret it as which is the option that is divisible by 1 and 2 and 3...It does not ask : each of the following is divisible by the product of each of the integers from 1 thru 7 ! or am I missing something badly !!

Your answer as well as Asterixmatrix's are correct according to the wording of the question. I just assumed it would be harder, since Bunuel posts tough questions haha. Let's see what the OA says.

1. A family consisting of one mother, one father, two daughters and a son is taking a road trip in a sedan. The sedan has two front seats and three back seats. If one of the parents must drive and the two daughters refuse to sit next to each other, how many possible seating arrangements are there? (B) 32 - Total ways to seat them with one parent driving: 2 options for driver * 4 options for other seats = 2*4*3*2*1 = 48. Assuming the 2 sisters do sit together, front seat we have 2 options for driver and 2 options for passenger (the other parent + son) = 4 ways. Back seats: sisters sit together so either they sit in seats 1 and 2 or seats 2 and 3 = 2 ways. Also sisters themselves can sit in 2 ways so total = 2*2 = 4. So total ways sisters can sit together = 4*4 =16. So ways they wont sit together = 48-16 = 32 ways.

2. What is the probability that a 3-digit positive integer picked at random will have one or more "7" in its digits? (C) 7/25 - Total #'s between 100 and 999 inclusive = 9*10*10 = 900. # of numbers with no digit 7 = 8*9*9 = 648. So #'s with a 7 = 1 - 648/900 = 252/900 = 7/25.

3. A sphere is inscribed in a cube with an edge of 10. What is the shortest possible distance from one of the vertices of the cube to the surface of the sphere? (D) 5( sqrt3 - 1) - Length of the diagonal of cube = ((10^2+10^2) + 10^2)^(1/2) = 10 sqrt3. Diagonal includes diameter of circle (10) + 2*length from vertex to sphere. Length of vertex to sphere = (10*sqrt3- 10)/2 = 5( sqrt3 - 1)

4. A contractor estimated that his 10-man crew could complete the construction in 110 days if there was no rain. (Assume the crew does not work on any rainy day and rain is the only factor that can deter the crew from working). However, on the 61-st day, after 5 days of rain, he hired 6 more people and finished the project early. If the job was done in 100 days, how many days after day 60 had rain? (C) 6 - rains for 5 days from day 56-60. So 10 guys worked for 55 days and accomplished half of the work. If 6 more guys are added to the job then the rate is 16/1100. (since one man's rate is 1/1100). Half the job left means 550/1100 is left. Therefore 550/16 = 34.375 days of more work. Since there were 40 days between day 60 and job completion, it must've rained for 40-34.375 = 5.625 or ~6 days. (I'm not sure if this is correct)

5. If s and t are positive integer such that s/t=64.12, which of the following could be the remainder when s is divided by t? (E) 45 - 64.12 = 6412/100 or 1603/25. 1603/25 gives a remainder of 3, 3206/50 gives remainder of 6 and so on ..pattern = factors of 3. so to get remainder of 45, we multiply everything by 15: 1603*15/(25*15) = 24045/375.

6. A committee of 6 is chosen from 8 men and 5 women so as to contain at least 2 men and 3 women. How many different committees could be formed if two of the men refuse to serve together? (E) 635 - 2 options: 2 men and 4 women -(i), or 3 men and 3 women -(ii). Starting with women: for (i): 5C4=5 and for (ii): 5C3 = 10. Men for (i): 2 further options if 2 men - a) neither of the problematic men chosen = 6C2 = 15 or b) one of them is chose - 2C1*6C1 = 12 so 12+15= 27 ways if (i) (ii): same 2 options - 6C3 + 2C1*6C2 = 20 + 30 = 50. Combining men and women - (i): 5*27 = 135 and (ii): 10*50 = 500 so total ways = 635.

7. If x is positive, which of the following could be the correct ordering of 1/x,2x and x^2 ? I. x^2<2x<1/x II. x^2<1/x<2x III. 2x<x^2<1/x (D) I and II only - if x = 1/3, then I is correct, if x = 4/5, then II is correct. I couldnt find a way to get III to work.

8. In the xy plane, Line k has a positive slope and x-intercept 4. If the area of the triangle formed by line k and the two axes is 12, What is the y-intercept of line K ? (D) -6 - Area of bxh of triangle = 24. Given base is 4, and drawing the line, easy to see that y intercept has to be -6.

9. Of the applicants passes a certain test, 15 applied to both college X and Y. If 20 % of the applicants who applied college X and 25% of the applicants who applied college Y applied both college X and Y, how many applicants applied only college X or college Y? (D) 105 - Let x be # who applied to only X and y who applied to only Y. Then 0.2(x+15) = 15 and 0.25(y+15) = 15. So x = 60 and y = 45 so x+y = 105.

10. What is the lowest positive integer that is divisible by each of the integers 1 through 7, inclusive? (A) 420 - 420 is divisible by all #'s from 1 thru 7.

2. What is the probability that a 3-digit positive integer picked at random will have one or more "7" in its digits? (A) 271/900 (B) 27/100 (C) 7/25 (D) 1/9 (E) 1/10

Total 3 digit numbers 900, 3 digit number with no 7 =8*9*9=648, P(at least one 7)=1-P(no 7)=1-648/900=252/900=7/25

Answer: C.

Dear Bunuel!

Could you please explain hod did you get: 3 digit number with no 7=8*9*9- where it coms from?

Thank you in advance

3 digit number with no 7, I mean without 7 = 8*9*9 = 648:

First digit can take 8 values from 1 to 9 excluding 7 (1xx, 2xx, ... 9xx, but not 7xx); Second and third digits can take 9 values from 0 to 9 excluding 7 (eg. for second digit: x0x, x1x, ... x9x but not x7x).

2. What is the probability that a 3-digit positive integer picked at random will have one or more "7" in its digits? (A) 271/900 (B) 27/100 (C) 7/25 (D) 1/9 (E) 1/10

Total 3 digit numbers 900, 3 digit number with no 7 =8*9*9=648, P(at least one 7)=1-P(no 7)=1-648/900=252/900=7/25

Answer: C.

Dear Bunuel!

Could you please explain hod did you get: 3 digit number with no 7=8*9*9- where it coms from?

Thank you in advance

3 digit number with no 7, I mean without 7 = 8*9*9 = 648:

First digit can take 8 values from 1 to 9 excluding 7 (1xx, 2xx, ... 9xx, but not 7xx); Second and third digits can take 9 values from 0 to 9 excluding 7 (eg. for second digit: x0x, x1x, ... x9x but not x7x).

Hope it's clear.

Thank you very much.This will help me iin solving such problems.I used to do it manually

for the time and work question in this problem set can you pls explain this step in a bit more detail:

"Then 6 more people was hired --> speed of construction increased by 1.6, days needed to finish 55/1.6=34.375"

How did you figure out the speed increased by 1.6?

For such problems I tend to reduce the question to how much work did the workers do in 1 day and then to how much work did each worker do in one day and then multiply that by 6 to get what 6 workers would have done in a day; add that to what 10 workers would have done in a day; take reciprocal of the fraction to see how much time 16 workers would take for that work---

your method seems much better.... can you explain that step of figuring out the increased speed of 1.6.... thanks.

Time, rate and job in work problems are in the same relationship as time, speed (rate) and distance.

We know that 10 people need 55 days to complete the job --> let the combined rate of 10 people be x --> \(Time*Rate=x*55=Job \ done\); Now, if combined rate of 10 people is \(x\), then combined rate of 16 man will be \(1.6x\) (1.6 times more than x as 16 is 1.6 times more than 10), so \(new \ time*new \ rate=same \ job\) --> \(t_2*1.6x=x*55\) --> \(t_2=\frac{55}{1.6}\approx{34.8}\) (as rate increased 1.6 times then time needed to do the same job will decrease 1.6 times).

Why is it that the triangle will be in 4th quadrant? 8. In the xy plane, Line k has a positive slope and x-intercept 4. If the area of the triangle formed by line k and the two axes is 12, What is the y-intercept of line K ? (A) 3 (B) 6 (C) -3 (D) -6 (E) -4

Easiest one so far Bunuel... your questions are killers (to me anyway haha) If K has a positive slope it will create a triangle in the bottom right quadrant (4th) of the xy plane. This means that the Y intercept will be negative.

Area = 1/2 * base * height 12 = (1/2) * 4 * height height = 6 Y Intercept = -6

AND = D

I Couldnt get the triangle being in 4th quadrant part!Bunuel,pls help!????

After days of waiting, sharing the tension with other applicants in forums, coming up with different theories about invites patterns, and, overall, refreshing my inbox every five minutes to...

I was totally freaking out. Apparently, most of the HBS invites were already sent and I didn’t get one. However, there are still some to come out on...

In early 2012, when I was working as a biomedical researcher at the National Institutes of Health , I decided that I wanted to get an MBA and make the...