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Manager  Joined: 26 Feb 2013
Posts: 51
Concentration: Strategy, General Management
GMAT 1: 660 Q50 V30 WE: Consulting (Telecommunications)

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4. Two machines, A and B, each working at a constant rate, can complete a certain task working together in 6 days. In how many days, working alone, can machine A complete the task?

(1) The average time A and B can complete the task working alone is 12.5 days.
(2) It would take machine A 5 more days to complete the task alone than it would take for machine B to complete the task

Time taken by A alone is a and by B is b.

Stmt 1: (a+b)/2= 12.5 => a+b=25. together they complete the task in 6 days => (1/a)+(1/b)=1/6 =>(ab)/(a+b) =6
a+b=25 => ab = 150. So (a,b) can be (10,15) or (15,10). insufficient.

Stmt 2: b=a-5.=> (1/a)+(1/a-5) = 1/6. on simplification we get a quadratic a = a^2-17a+30=0 solving , a=15 or 2.
a cant be 2 ,as a=2=> b=-3 which on this earth is not possible. Hence a=15. sufficient.
Manager  Joined: 11 Jun 2010
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Q1. Let 3 integers be (x-1), x and (x+1)

St1: At least 1 integer is positive.
Case 1 (1 Integers positive): 1 -- 2 -- 3 (Product = 6)
Case 2 (2 Integers positive): 0 -- 1 -- 2 (Product = 0)
Case 3 (All 3 Integers positive): -1 -- 0 -- 1 (Product = 0)
Hence Not Sufficient

St2: sum of 3 integers < 6
x - 1 + x + x + 1 < 6
=> x < 2
if integers are 0, 1, 2 (product = 0)
if integers are -6, -2, -1 (product = -12)
Insufficient

Together: atleast 1 integer is negative and sum < 6
Case1: if 2 integers >0, 0, 1, 2 (product = 0)
Case1: if 1 integer is positive, -1, 0, 1 (product = 0)
both case will include a 0 hence the product will be 0
Together sufficient.
Ans C

Originally posted by srcc25anu on 10 Apr 2013, 13:57.
Last edited by srcc25anu on 10 Apr 2013, 15:15, edited 1 time in total.
Manager  Joined: 26 Feb 2013
Posts: 51
Concentration: Strategy, General Management
GMAT 1: 660 Q50 V30 WE: Consulting (Telecommunications)

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6. The ratio of the number of employees of three companies X, Y and Z is 3:4:8, respectively. Is the average age of all employees in these companies less than 40 years?

(1) The total age of all the employees in these companies is 600
(2) The average age of employees in X, Y, and Z, is 40, 20, and 50, respectively.

Let the number of employees in X,Y,Z be 3p,4p,8p.

Stmt 1: Total age of all the employees = 600. Avg age of all the employees = 600/(15p) = 40/p. Hence avg age is<= 40. It will be equal to 40 when p=1 and for integer values of p its less than 40. Hence insufficient.

stmt 2: total age of X= 120p, of Y = 80p and of Z= 400p. total age of x,y and Z= 600P. Avg age = 600p/(15p) =40 . Hence its not less than 40. Hence sufficient.
Manager  Joined: 11 Jun 2010
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Q2: remainder (x/y) = ?

St 1: Y = 2 digit rpime.
No information about X hence insufficient

St2: x = qY + 9
remainder when x / y = 9
hence sufficient

And B
Manager  Joined: 26 Feb 2013
Posts: 51
Concentration: Strategy, General Management
GMAT 1: 660 Q50 V30 WE: Consulting (Telecommunications)

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7. Was the average (arithmetic mean) temperature in city A in March less than the average (arithmetic mean) temperature in city B in March?

(1) The median temperature in City A in March was less than the median temperature in city B
(2) The ratio of the average temperatures in A and B in March was 3 to 4, respectively

Stmt 1 is clearly insufficient.
From statement 2 since ratio of avg temp A to B is less than 1.(3/4) Avg temp of A is less than avg temp of B. Sufficient.
Manager  Joined: 26 Feb 2013
Posts: 51
Concentration: Strategy, General Management
GMAT 1: 660 Q50 V30 WE: Consulting (Telecommunications)

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9. If x is an integer, is x^2>2x?

(1) x is a prime number.
(2) x^2 is a multiple of 9.

Stmt 1: x can be 2 when X^2=2X. insufficient.

Stmt 2 : x^2 is a multiple of 9. for any multiple of X, X^2 will always be greater than 2X. Sufficient
Stmt 2 : X^2 can be 0, 9,36 etc..in sufficient.

combining we get X as 3. sufficient.

made the obvious mistake of not considering X=0.

Originally posted by mdbharadwaj on 10 Apr 2013, 14:03.
Last edited by mdbharadwaj on 10 Apr 2013, 14:51, edited 1 time in total.
Manager  Joined: 11 Jun 2010
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Q3.
St1: ABC is isoceles traingle.
We dont know which 2 sides are equal or nor do we know any of the angles measures.
Hence insufficient.

St2: AC^2 = AB^2 + BC^2
This implies Tr.ABC is a RIGHT ANGLED TRAINGLE at Point B
But we dont know the ratio of length of remaining 2 sides hence INSUFFICIENT

Together: we know Angle B = 90 degree and ABC is isoceles traingle (thus AB = AC) With this information. we can calculate the length of side AC.
each of traingle ABD and BCD are 45-45-90 triangles with sides in the ratio 1:1:root2
since 1 corresponds to 12, AC = 2 * 12 = 24

Ans C
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GMAT 1: 530 Q43 V20 GMAT 2: 560 Q42 V25 GMAT 3: 650 Q48 V31 ### Show Tags

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Q2) If x and y are both positive integers and x>y, what the remainder when x is divided by y?

From Stmt 1, y is 11, 13, 17, 19 and x = y + k where k is positive - Not Sufficient

x/y can end up with any remainder such as 0, 1, 2,..

From Stmt 2, x=qy+9, for some positive integer q - Not Sufficient

x/y end up with o remainder iff q = y = 1, 3 and other remainders for other values of q and y.

Considering Stmt 1 and Stmt 2 - Sufficient
y = 11, 13, 17, 19, and q = 1, 2, 3,... the remainder will be always 9

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Manager  Joined: 11 Jun 2010
Posts: 70

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Q4:

A+B in 1 day = 1/6 work
A in 1 day = 1/a work or A takes 'a' days to finish the work alone
B in 1 day does 1/6 - 1/a work = (a-6) / 6a
B takes '6a/(a-6)' days to finish the job working alone.

ST1: Average (a, 6a/(a-6)) = 12.5
a^2 - 25a + 150 = 0
a = 10 or a = 15
Insufficient

St2: A takes 'a' days to finish working alone
B takes 'a-5' days to finish working alone
average (a, a-5) = 12.5
a = 15
therefore A can finish the job in 15 days working alone and B can finish it in 10 days working alone.
Hence sufficient

Ans B
Manager  Joined: 26 Feb 2013
Posts: 51
Concentration: Strategy, General Management
GMAT 1: 660 Q50 V30 WE: Consulting (Telecommunications)

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10. What is the value of the median of set A?

(1) No number in set A is less than the average (arithmetic mean) of set A.
(2) The average (arithmetic mean) of set A is equal to the range of set A.

Stmt 1: Since no number is less than average all the numbers must be equal to average. all the numbers in the set are equal. But we dont know the number. hence insufficient.

Stmt 2: Set A (1,2,3) range and avg =2. median is 2 as well. But in set (1,4,4) avg & range = 3 and median =4 . Clearly insufficient.

combining , the set can be of all 0s or 1 or 2 or 2.5 or any other number. Hence insufficient.
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4. Two machines, A and B, each working at a constant rate, can complete a certain task working together in 6 days. In how many days, working alone, can machine A complete the task?

$$(\frac{1}{A}+\frac{1}{B})*6=1$$

(1) The average time A and B can complete the task working alone is 12.5 days.
A+B=25, from statement above we get that $$(\frac{B+A}{AB})*6=1$$ so $$6(A+B)=AB$$ , $$6*25=AB$$, in this case many combination could work (6,25) (3,50) ...
Not Sufficient

(2) It would take machine A 5 more days to complete the task alone than it would take for machine B to complete the task
A-B=5, $$(\frac{1}{A}+\frac{1}{A-5})*6=1$$ so $$(2A-5)*6=A(A-5)$$ and we find that $$A^2-17A+30=0$$ A=15 or A=2, but since A-B=5 A cannot be 2 (because otherwise the time B would be negative). From 2 => A=15 Sufficient
IMO B
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Manager  Joined: 11 Jun 2010
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Q5.
St1: SD of A > 0 implies X <> 0. No information about Y in set B hence insufficient

St2: y = 3
average of Set A = 3, y = 3 implies y = mean of set A Adding a term = mean of the set will lower the sd of new set.
if sd of A = 0 or all terms = 3, then sd of A will be equal to sd of B
but if sd of A is positive (i.e. x <>0), then sd of A will be different from sd of B
not sufficient

Together:if sd of A > 0 and y = 3
sd of A > sd of B
hence sufficient to answer the question

Ans C
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Q6:
ratio of employees x:Y:Z = 3:4:8

St1: Sum of all ages in 3 companies = 600
we dont know anything about the distribution of ages or how many employees are there
if # employees in X Y and Z are 3, 4 and 8 then 600 / 15 = 40 (Not LEss than 40)
But if # employees in X Y and Z are 6, 8 and 16 then 600 / 30 = 4 years (LEss than 40)
hence INSUFFICIENT

St2: avg age of employee ratio = 40 : 20: 50
let emp in X be 3x
let emp in y be 4x
let emp in z be 8x
Total emp = 15x

Weighted average Age = 40 * 3x + 20 * 4x + 50 * 8x / 15x
=> x = 40 thus average age of all employees in 3 companies is NOT lESS tHAN 40

hence B
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Q1) What is the product of three consecutive integers?

From Stmt 1, one of the three consecutive numbers is > 0 - Not Sufficient

n, n+1, n+2 are 3 consecutive numbers and either n > 0 or n+1 > 0 or n+2 > 0. n*(n+1)*(n+2) will have more than 1 value

From Stmt 2, sum of the 3 integers is less than 6
- Not Sufficient

n + n+1 + n+2 < 6 results in n < 1, so n can be 0, and any negative numbers. n*(n+1)*(n+2) will have more than 1 value

Combining Stmt 1 and Stmt 2 - Not Sufficient

n can be 0, -1 and n*(n+1)*(n+2) will have more than 1 value

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Manager  Joined: 11 Jun 2010
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Q7:
Is avg temp in A < avg temp in B?

St1: median temp in A < median temp in B
case 1: Temp in A for 3 days = 1 - 1 - 10 (mean = 4, median = 1)
Temp in B for 3 days = 2 - 2 - 2 (mean = 2, median = 2)
MEAN of A > MEAN of B

Case 2:Temp in A for 3 days = 1 - 1 - 2 (mean = 4/3, median = 1)
Temp in B for 3 days = 2 - 2 - 2 (mean = 2, median = 2)
MEAN of A < MEAN of B
hence insufficient

St2: Ratio of avg March temp in A : Ratio of avg March temp in B = 3:4
clearly implies Avg Temp of A for March < Avg temp of B for MArch
Sufficient

Manager  Joined: 26 Feb 2013
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5. Set A={3-2x, 3-x, 3, 3+x, 3+2x}, where x is an integer. Is the standard deviation of set A more than the standard deviation of set B={3-2x, 3-x, 3, 3+x, 3+2x, y}

(1) The standard deviation of set A is positive
(2) y=3

SD of A = 2*x^2. and of B = (10*x^2 +(3-y)^2)/6. we need to check if SD(A)-SD(B)>0

Stmt 1: from this we know that x^2>=1. But not the value of Y. SD(A)-SD(B)= .34*x^2-((3-y)^2)/6. We dont know the values of X or Y. hence cant decide.

Stmt 2: Y =3. hence SD(B)=1.66*x^2. Hence SD(A)-SD(B)=.34*x^2>=0. 0 for x=0. Hence insufficient.

Combining We know X^2>=1 and Y=3 hence SD(A)>SD(B). Sufficient.
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Q8:
Total marbles = 10
All marbles either RED or BLUE
is Prob (R,R) > 3/5 or 0.6?

St1: Prob (B,B) < 0.1
case A: if 2 B and 8 R marbles,
P(b,b) = 1/45 < 0.1
P(r,r) = 7/45 > 0.6

case B: if 3 B and 7 R marbles,
P(b,b) = 1/15 < 0.1
P(r,r) = 7/15 < 0.6

Not sufficient

St2: Red marbles >=6
as per explanation in St1: if R = 7 and B = 3, P(r,r) < 0.6 and if R = 8 and B = 2, P(r,r) > 0.6
hence insufficeint

Taking both statements together still insufficient (different answers for 7 Red versus 8 red marbles)
Ans E
Manager  Joined: 11 Jun 2010
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Q9:
Is x^2 > 2x or x(x-2)>0
we need to know whether x > 2 or X < 0

St1: x = prime number
if x = 2, above condition is not satisfied
for any other prime number (3,5,7 ... ) the above condition IS satisfied
hence insufficient

ST2: x^2 = mult of 9
for x = 0, x^2 = mult. of 9 (as 0 is a multiple of any integer). X does not satify the main question though (whether x > 2 or X < 0)
for x = 3, x^2 = mult of 9 and IT DOES SATISFY the above condition (whether x > 2 or X < 0 )
hence insufficient

Together, sufficient only value that satifies this is 3.
hence C
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Q10.
what is median of A?

St1: No number is less than average IMPLIES ALL the TERMS IN SET A ARE EQUAL but A can be: {0,0,0} or {1,1,1} or {2,2,2}
hence we cannot say what the median will be as it could be 0, 1 or 2 for the above 3 sets
INSUFFICIENT

St2: average = range of set A
case A: set = {1,2,3}
both mean and range = 2 and median = 2
case B: set = {0,0,0}
both mean and range = 0 and median is also 0 in this case
hence insufficient

together only set that will satify is when all the elements of set A = 0
Only in that case all elements are 0, average = 0, range = 0 and median = 0
Sufficient

Ans C
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Q3) The length of the median BD in triangle ABC is 12 centimeters, what is the length of side AC?

From Stmt 1, ABC is a isosceles triangle
- Not Sufficient
AB = BC or BC = AC or AB = AC, so AC length cannot be determined from median BD = 12 cm.

From Stmt 2, AC^2 = AB^2 + BC^2 - Sufficient
<B = 90 and AC is the hypotenuse, so BD is the median to the hypotenuse. Using right angle isosceles triangle properties BD = AD = CD
AC = AD + CD = 2*BD = 24 cm.

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