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# New Set of Mixed Questions!!!

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Manager
Joined: 25 Jun 2012
Posts: 64
Location: India
WE: General Management (Energy and Utilities)
Re: New Set of Mixed Questions!!!  [#permalink]

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01 Apr 2013, 18:20
1
6. A team contributes total of $399 from its members. If each member contributed at least$10, and no one contributed $19, what is the greatest number of members the club could have? Going by the options given : Suppose each member paid exactly 10 dollars. A. 37 = 37*10 = 370, balance = 29 which we cant tell who paid what some one might have paid more then 19 alos B. 38 = 38*10 = 380 , balance = 19 which No one paid 19 dollars so remove C. 39 = 39*10 = 390 , balance 9 dollars which can be paid by any member D. 40 = exceeds 399 dollar E. 41 = exceeds 399 dollars So , taking 39 members as the answer I dont know whether my approach is right or not. Manager Status: Looking to improve Joined: 15 Jan 2013 Posts: 158 GMAT 1: 530 Q43 V20 GMAT 2: 560 Q42 V25 GMAT 3: 650 Q48 V31 Re: New Set of Mixed Questions!!! [#permalink] ### Show Tags 01 Apr 2013, 18:54 Q11) P(No Blue at 7) = 1 - P(Blue at 7) using slot method _ _ _ _ _ _ _ = 5/8*4/7*3/6*2/5*3/3*2/2*1/1 = 1/56 and there are 14 such combinations. P(Blue at 7) = 14/56 = 1/4. Answer: B _________________ KUDOS is a way to say Thank You Senior Manager Status: Final Lap Joined: 25 Oct 2012 Posts: 257 Concentration: General Management, Entrepreneurship GPA: 3.54 WE: Project Management (Retail Banking) Re: New Set of Mixed Questions!!! [#permalink] ### Show Tags 01 Apr 2013, 21:39 2 1. The distance from the Y-axis to point K is 1/3 of the distance from the X-axis to point K. If the coordinates of K are (-3, y), what is the distance between point K and X-axis? A. 1/2 B. 1 C. 3 D. 4.5 E. 9 Since (-3, y) are the coordinates of K, -3 is the distance from K to the Y-axis and y is the distance from the X-axis to K. Given that the distance from the Y-axis to point K is 1/3 of the distance from the X-axis to point K, Thus : 3 = 1/3 * y --> y = 9 Answer : E _________________ KUDOS is the good manner to help the entire community. "If you don't change your life, your life will change you" Senior Manager Status: Final Lap Joined: 25 Oct 2012 Posts: 257 Concentration: General Management, Entrepreneurship GPA: 3.54 WE: Project Management (Retail Banking) Re: New Set of Mixed Questions!!! [#permalink] ### Show Tags 01 Apr 2013, 21:48 1 2. What is the area of a region enclosed by |x/3|+|y/9|=10? A. 675 B. 1350 C. 2700 D. 5400 E. 10800 In quadrant 1, this is just the line 3x+y=90. Which is a line with X-int=30 and Y-int=90. So area enclosed with axes in Q1, is just 0.5*30*90 = 1350 (a right angled triangle) Now notice, if you replace x by -x the equation does not change Hence, the figure must be symmetric about the Y-axis and Q2, must also have a right angled triangle, of area 1350 with the axis Finally, if you replace y by -y, the equation is still the sameand hence, the figure must be symmetric about the X-axis So Q3 and Q4 have a mirror image of the above two triangles So total area = 1350*4 = 5400 Answer : D Attachments Rock.png [ 29.43 KiB | Viewed 7099 times ] _________________ KUDOS is the good manner to help the entire community. "If you don't change your life, your life will change you" Verbal Forum Moderator Joined: 10 Oct 2012 Posts: 614 Re: New Set of Mixed Questions!!! [#permalink] ### Show Tags 01 Apr 2013, 22:15 1 8. Usually Holly leaves home to school at 9:00, however today she left home 20 minutes later. In order to be at school on time she increased her usual speed by 20% and still was at school 15 minutes later than usual. What is her usual time from home to school? Let the usual time, speed and the distance between her house and school be t,v, and s. We know that s = vt. Again, s = 1.2v*(t-20/60+15/60) or s = 1.2v*(t-1/12) On comparison, t = 1.2(t-1/12) or t = 30 mins D. _________________ Senior Manager Status: Final Lap Joined: 25 Oct 2012 Posts: 257 Concentration: General Management, Entrepreneurship GPA: 3.54 WE: Project Management (Retail Banking) Re: New Set of Mixed Questions!!! [#permalink] ### Show Tags 01 Apr 2013, 22:16 2 2 3. Three workers, A, B, and C, can complete a certain task in 10, 5 and x hours re ,spectively. A starts working alone and 2 hours later B joins. After another 2 hours joins C. After that A, B, and C together complete the task in 15 minutes. What is the value of x? A. 1 B. 1.25 C. 2 D. 2.5 E. 4 A alone can do the work in 10 hours B alone can do the work in 5 hours C alone can do the work in x hours Hence, in 1hour A alone will do W/10 and B alone W/5 and C alone W/x From the figure below, W = W/5 + 3W/5 + W/40 + W/20 + W/4x -----> x = 2 Answer : C Attachments Rock2.png [ 10.04 KiB | Viewed 7090 times ] _________________ KUDOS is the good manner to help the entire community. "If you don't change your life, your life will change you" Senior Manager Status: Final Lap Joined: 25 Oct 2012 Posts: 257 Concentration: General Management, Entrepreneurship GPA: 3.54 WE: Project Management (Retail Banking) Re: New Set of Mixed Questions!!! [#permalink] ### Show Tags 01 Apr 2013, 23:08 2 1 4. A draining pipe can empty a pool in 4 hours. On a rainy day, when the pool is full, the draining pipe is opened and the pool is emptied in 6 hours. If rain inflow into the pool is 3 liters per hour, what is the capacity of the pool? A. 9 liters B. 18 liters C. 27 liters D. 36 liters E. 45 liters To empty X liters, 4 hours are needed To empty X + (3*6) liters, 6 hours are needed Hence, $$X/4 = (X+18)/6$$ --> $$X=36$$ _________________ KUDOS is the good manner to help the entire community. "If you don't change your life, your life will change you" Director Joined: 25 Apr 2012 Posts: 698 Location: India GPA: 3.21 WE: Business Development (Other) Re: New Set of Mixed Questions!!! [#permalink] ### Show Tags 01 Apr 2013, 23:49 1 10. If n is a non-negative integer and the remainder when 3^n is divided by 4 is a multiple of 3, then which of the following must be true? I. n^2 divided by 4 yields the reminder of 1 II. (-2)^n is less than 0 III. n is a prime number A. I only B. II only C. III only D. I and II only E. II and III only Sol: From Q stem, we have n can be 0 or positive integer Possible values of n = 0,1,2...... When 3^n/4 =3I where I is any Integer Consider n= 1 so we have 3/4 and remainder is 3 which is of the form 3I Consider n= 2 so we have 9/4 and remainder is 1 which is not of the form 3I Consider n =3 , so we have 27/4 and remainder is 3 which is of the form 3I Therefore n= 1,3,5...for Odd nos we have that when 3^n/4 will give remainder of the form 3I Now St 1: n^2/4 yields remainder 1 for n= 1, remainder is 1 for n= 2, reminder is zero n=3, remainder is 1 We can conclude that for all odd values of n i.e.1,3, 5 and so on remainder will be 1 and hence st 1 will be true St 2: (-2)^n is less than 0 for any odd value of n the expression will always be less than 0 . Ex for n = 1, we have (-2)^1 ----> -2<0 For n= 3, we have (-2)^3------> -8<0 Hence st2 is also true St 3: n is a prime number Now if n =3,5,7 then yes n is a prime no But if n=15, n is not a prime no Hence Ans should be Option D _________________ “If you can't fly then run, if you can't run then walk, if you can't walk then crawl, but whatever you do you have to keep moving forward.” VP Status: Far, far away! Joined: 02 Sep 2012 Posts: 1086 Location: Italy Concentration: Finance, Entrepreneurship GPA: 3.8 Re: New Set of Mixed Questions!!! [#permalink] ### Show Tags 02 Apr 2013, 01:46 BONUS QUESTION: 11. Certain bowl contains 5 red marbles and 3 blue marbles only. One by one, every marble is drawn at random and without replacement. What is the probability that the seventh marble drawn is NOT blue? We have 8 positions to fill: R,R,R,R,R,B,B,B My method is: the probability that a given position has a red marble is 5/8, the probability that a given position has a blue marble is 3/8. So the probability that the 7th position has a NOT blue marlble ( a red one ) is 5/8. IMO D. _________________ It is beyond a doubt that all our knowledge that begins with experience. Kant , Critique of Pure Reason Tips and tricks: Inequalities , Mixture | Review: MGMAT workshop Strategy: SmartGMAT v1.0 | Questions: Verbal challenge SC I-II- CR New SC set out !! , My Quant Rules for Posting in the Verbal Forum - Rules for Posting in the Quant Forum[/size][/color][/b] Intern Joined: 15 Jan 2013 Posts: 26 Concentration: Finance, Operations GPA: 4 Re: New Set of Mixed Questions!!! [#permalink] ### Show Tags 02 Apr 2013, 04:50 1 1. The distance from the Y-axis to point K is 1/3 of the distance from the X-axis to point K. If the coordinates of K are (-3, y), what is the distance between point K and X-axis? A. 1/2 B. 1 C. 3 D. 4.5 E. 9. Distance of a point(x, y) from the x-axis is given by the y-coordinate of the point i.e. |y| and the distance from the y-axis is given by the x-coordinate of the point i.e. |x| So, distance of the point k(-3 , y) from x-axis = y, from y-axis = 3 As per question $$3 = 1/3(y)$$ => $$y = 3*3 = 9$$ So distance of point k from x-axis = y = 9 (option E) Intern Joined: 15 Jan 2013 Posts: 26 Concentration: Finance, Operations GPA: 4 Re: New Set of Mixed Questions!!! [#permalink] ### Show Tags 02 Apr 2013, 04:57 1 2. What is the area of a region enclosed by |x/3|+|y/9|=10? A. 675 B. 1350 C. 2700 D. 5400 E. 10800 Putting x= o we get y = +90, -90 Putting y= 0 we get x = +30, -30 Required area will be set of four Pythagorean triangles with the two perpendicular sides as 30 and 90 So required area = $$4 * (1/2 * 30 * 90) = 2 * 30 * 90 = 5400$$(option D) Intern Joined: 15 Jan 2013 Posts: 26 Concentration: Finance, Operations GPA: 4 Re: New Set of Mixed Questions!!! [#permalink] ### Show Tags 02 Apr 2013, 05:07 1 3. Three workers, A, B, and C, can complete a certain task in 10, 5 and x hours respectively. A starts working alone and 2 hours later B joins. After another 2 hours joins C. After that A, B, and C together complete the task in 15 minutes. What is the value of x? A. 1 B. 1.25 C. 2 D. 2.5 E. 4 A alone can complete the task in 10 hours, in 1 hour he can do 1/10 of the task B alone can complete the task in 5 hours, in 1 hour he can do 1/5 of the task C alone can complete the task in x hours, in 1 hour he can do 1/x of the task A and B combined can do (1/10 + 1/5) = 3/10 of the task in 1 hour hour A, B and C combined can do (1/10 + 1/5 + 1/x) = (3/10 + 1/x) of the task in one hour As per question, A starts the work and works for 2 hours so he completes 1/10 * 2 of the task i.e. 1/5 of the task Then A and B work for next two hours i.e they complete 3/10 *2 = 3/5 of the task Remaining task = 1 - (1/5 + 3/5) = 1/5 This remaining task is done by A, B and C combined in 15 mins. So 1/5 = (1/10 + 1/5 + 1/x)* 15/60, that gives x =2 (option c) Intern Joined: 15 Jan 2013 Posts: 26 Concentration: Finance, Operations GPA: 4 Re: New Set of Mixed Questions!!! [#permalink] ### Show Tags 02 Apr 2013, 05:13 1 4. A draining pipe can empty a pool in 4 hours. On a rainy day, when the pool is full, the draining pipe is opened and the pool is emptied in 6 hours. If rain inflow into the pool is 3 liters per hour, what is the capacity of the pool? A. 9 liters B. 18 liters C. 27 liters D. 36 liters E. 45 iters Let the draining rate of the pipe be x liters per hour So the volume of the pool = 4x litres As per question, 4x + 3*6 = 6x => x = 9 So the capacity of the tank = 4*9 = 36 liters (option D) Intern Joined: 15 Jan 2013 Posts: 26 Concentration: Finance, Operations GPA: 4 Re: New Set of Mixed Questions!!! [#permalink] ### Show Tags 02 Apr 2013, 05:17 5. For a certain set of numbers, if x is in the set, then both -x^2 and -x^3 are also in the set. If the number 1/2 is in the set , which of the following must also be in the set ? I. -1/64 II. 1/64 III. 1/2^(1/3) A. I only, B. II only, C. II only, D. I and II only E. I, II and III Intern Joined: 15 Jan 2013 Posts: 26 Concentration: Finance, Operations GPA: 4 Re: New Set of Mixed Questions!!! [#permalink] ### Show Tags 02 Apr 2013, 05:20 1 5. For a certain set of numbers, if x is in the set, then both -x^2 and -x^3 are also in the set. If the number 1/2 is in the set , which of the following must also be in the set ? I. -1/64 II. 1/64 III. 1/2^(1/3) A. I only, B. II only, C. II only, D. I and II only E. I, II and III Since 1/2 is in the set, so -1/4 and -1/8 are also in the set Since -1/4 is in the set, so $$- (-1/4)^3$$ i.e. 1/64 is also in the set Since -1/8 is in the set, so $$-(-1/8)^2$$ i.es -1/64 is also in the set So option D. I and II only Intern Joined: 15 Jan 2013 Posts: 26 Concentration: Finance, Operations GPA: 4 Re: New Set of Mixed Questions!!! [#permalink] ### Show Tags 02 Apr 2013, 05:36 1 6. A team contributes total of$399 from its members. If each member contributed at least $10, and no one contributed$19, what is the greatest number of members the club could have?

A. 37
B. 38
C. 39
D. 40
E. 41

Since we need to maximize the number of members we need to minimize each member's contribution which is gives as 10
So if we take no. of members as 39, their contribution = 39 * 10 = 390, remaining 9 can be shared among other members.
So maximum number of members = 39 (option C)
Intern
Joined: 15 Jan 2013
Posts: 26
Concentration: Finance, Operations
GPA: 4
Re: New Set of Mixed Questions!!!  [#permalink]

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02 Apr 2013, 05:42
1
7. Mary spent 64 percent of her salary on food (including meat) and 16% of her salary on meat. What percent of the salary spent on food were not spent on meat?

A. 16%
B. 25%
C. 32%
D. 48%
E. 75%

Let Marry has 100 rupees with her
So amount spent on food = 64
amount spent on meat = 16
amount spent on food other then meat = 64 - 16 = 48
48 out of 64 is spent on food other than meat, % = 48/60 * 100 = 75% (option E)
Intern
Joined: 15 Jan 2013
Posts: 26
Concentration: Finance, Operations
GPA: 4
Re: New Set of Mixed Questions!!!  [#permalink]

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02 Apr 2013, 05:47
1
8. Usually Holly leaves home to school at 9:00, however today she left home 20 minutes later. In order to be at school on time she increased her usual speed by 20% and still was at school 15 minutes later than usual. What is her usual time from home to school?

A. 15 minutes
B. 20 minutes
C. 25 minutes
D. 30 minutes
E. 210 minutes

Holly increases her speed by 20% i.e by 1/5th of the original speed
Since, speed * time = distance (constant in this case)
Since speed increased by 1/5, for distance to be constant time must decrease by 1/6
Net decrease in time = 20 - 25 mins = 5 mins
So 1/6th of usual time = 5 mins
Usuall Time = 6 * 5 = 30 mins (option D)
Intern
Joined: 15 Jan 2013
Posts: 26
Concentration: Finance, Operations
GPA: 4
Re: New Set of Mixed Questions!!!  [#permalink]

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02 Apr 2013, 05:52
1
9. If x and y are integers and x + y = -12, which of the following must be true?

A. Both x and y are negative
B. xy > 0
C. If y < 0, then x > 0
D. If y > 0, then x < 0
E. x - y > 0

x + y = -12

A. Both x and y are negative; not necessary as x = 1 and y = -13 satisfies the equation
B. xy > 0; not necessary as x = 1 and y = -13 satisfies the equation
C. If y < 0, then x > 0; not necessary as y = -2 and x = -10 satisfies the equation
D. If y > 0, then x < 0; must be true
E. x - y > 0; not necessary as y = -2 and x = -10 satisfies the equation

Correct answer D. If y > 0, then x < 0
Intern
Joined: 15 Jan 2013
Posts: 26
Concentration: Finance, Operations
GPA: 4
Re: New Set of Mixed Questions!!!  [#permalink]

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02 Apr 2013, 05:58
1
10. If n is a non-negative integer and the remainder when 3^n is divided by 4 is a multiple of 3, then which of the following must be true?

I. n^2 divided by 4 yields the reminder of 1
II. (-2)^n is less than 0
III. n is a prime number

A. I only
B. II only
C. III only
D. I and II only
E. II and III only

n being a non-negative integer and when 3^n is divided by 4, it leaves a remainder of 3
So n must be 1, 3, 5, 7, ...........odd integers
I. n^2 divided by 4 yields the reminder of 1; True
take any possible value of n say n = 3 so n^2 ie. 3^2 =9 when divided by 4 leaves a remainder of 1
II.(-2)^n is less than 0; True as n is odd
III. n is a prime number; False as n = 9 also satisfies the condition

So correct option is D. I and II only
Re: New Set of Mixed Questions!!! &nbs [#permalink] 02 Apr 2013, 05:58

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# New Set of Mixed Questions!!!

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