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

My answer B

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.

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.

My answer B

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.

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

My answer is B

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.

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

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

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
_________________

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
sufficient to answer

hence B

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

Answer B

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

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