New DS set!!!

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


Assuming A and B takes x and y days respectively to finish the work alone.
In 6 days, work done by A = 6/x, work done by B = 6/y, total work done by A and B together = 6/x + 6/y
6/x + 6/y = 1 --> 6x + 6y = xy …. (eq1)

1) (x+y)/2 = 12.5 --> x+y = 25
Substituting this in eq1, 6*25 = x(25-x) --> x^2 -25x + 150 = 0 --> (x-10)(x-15) = 0 --> x = 10 or 15
Not sufficient.

2) y = x-5
Substituting this in eq1, 6x + 6x – 30 = x(x-5) --> x^2 -17x + 30 = 0 --> (x-2)(x-15) = 0 --> x = 2 or 15
If x = 2, y become -3 which is impossible. --> x = 15.
Sufficient.

Answer is B.

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


1) Standard deviation can be positive or zero (in case of x=0). Here the information only says that x is not zero and all elements have different values. But, without knowing y, we cannot conclude. Not sufficient.
2) This says mean of both the sets is 3, no. of elements are 5 and 6 for A and B respectively. If x is not equal to 0, elements in set A will be more spread than those in set B. If x = 0, Standard deviation of both the sets will be zero. Not sufficient.

Together: x is not equal to 0 --> elements in set A will be more spread than those in set B --> standard deviation of set A is more than that of set B. Sufficient.

Answer is C.

8. Two marbles are drawn from a jar with 10 marbles. If all marbles are either red of blue, is the probability that both marbles selected will be red greater than 3/5?
(1) The probability that both marbles selected will be blue is less than 1/10
(2) At least 60% of the marbles in the jar are red


1) If there are 4 blue marbles, probability of drawing 2 blue marbles = 4/10 * 3/9 = 2/15 > 1/10
If there are 3 blue marbles, probability of drawing 2 blue marbles = 3/10 * 2/9 = 1/15 < 1/10
So, there are maximum 3 blue marbles and at least 7 red marbles in the jar.
If there are 7 red marbles, probability of drawing 2 red marbles = 7/10 * 6/9 = 7/15 < 3/5
If there are 8 red marbles, probability of drawing 2 red marbles = 8/10 * 7/9 = 28/45 > 3/5
Not sufficient.

2) As seen from calculations against option 1, with 6 or 7 red marbles, probability of drawing 2 red marbles will be less than 3/5, and 8 or more red marbles, probability of drawing 2 red marbles will be more than 3/5.
Not sufficient.

Together: There is no added information found from option 1 and 2 together. Number of red marbles can be 7 or 8 or 9 10 which may result in probability of drawing 2 red marbles to be less than or more than 3/5.
Not sufficient.

Answer is E.

9. If x is an integer, is x^2>2x?
(1) x is a prime number.
(2) x^2 is a multiple of 9.


1) If x = 2, x^2 = 2x. For any other integer, x^2 > 2x. Not sufficient.
2) As x^2 is a multiple of 9, x is not 2. So, x^2 > 2x. Sufficient.

Together: As x is prime number, x cannot be 0. As x^2 is multiple of 9, x can be 3, 6, 9, 12, …..
So, x^2 > 2x as x must be > 2. Sufficient.

Answer is C.

10. What is the value of the media 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.


1) This means all elements in the set have the same value and that is same to mean and median as well. But, we do not know the number. Not sufficient.
2) There are many sets for which average and range will be the same. Examples are: [10,10,30,30] where median is 20, or [2,2,3,3,5] where median is 3. Not sufficient.

Together: This is possible only when all elements have the same value as 0 --> average = range = 0 --> median = 0

Answer is C.

Note: I believe the question should be read as “What is the value of the median of set A?”

1. (1) Insufficient. It could be {0,1,2} and the product is 0, or it could be {1,2,3} and their product is 6.
(2) Insufficient. It could be {0,1,2} and the product is 0, or it could be {-3,-2,-1} and the product is -6

(1)+(2) Sufficient. Suppose the least number is n, so the second is (n+1) and the third is (n+2). Their sum must be less than 6: n+(n+1)+(n+2)<6 or 3n+3<6 or n<1. By first statement at least one of the integers must be positive, it means that the largest is positive: n+2>0 or n>-2. So n=-1 or n=0. Therefore, there are two possible sets {-1, 0, 1} or {0, 1, 2}. The product anyway is 0.

The correct answer is C.

2. (1) Insufficient. For x=11 and y=10 the remainder when x is divided by y is 1, for x=12 and y=2 the remainder when x is divied by y is 2.
(2) Insufficient. The main point here is the remainder must be less than the divisor. We don't know is y>9 or y<=9. For y=1 the remainder when x is divided by y is 0, for y=2, x=2q+9 the remainder is 1 when x is divided by 2.

(1)+(2) Sufficient. If y is a two-digit number the when qy is divided by y there is remainder 0, and when 9 is divided by y the remainder is 9, since 9<y. So, x when divided by y gives the remainder 9.

The correct answer C

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Hi Bunuel,

Doubt in this Q. Pls help & suggest where am I going Wrong in this Question ......

Q .. 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

Now, it gives the ratio of number of employees i.e.; 3:4:8 & then we can take this as real number of employees like 3 employees in Company X, 4 in Y & 8 in company Z ...therefore, we have 3+4+8=15 , 15 employees in total....& average age of employees in company X must be something, lets say its x for Company X, so the total age for company X will be 3x, similarly, for Company Y the average age is y therefore, total age for company Y will be 4y & for Company Z, the total age will be 8z. & the average of the total ages will be 3x+4y+8z/15 ...... & as the question asks, Is the average age of all employees in these companies less than 40 years ?? i.e.; 3x+4y+8z/15 <40 or 3x+4y+8z<600...... so we have to find out if 3x+4y+8z<600 ..... rephrasing complete.

now statement (1) The total age of all the employees in these companies is 600.
that means 3x+4y+8z=600 but we have to check if 3x+4y+8z<600. so it is clearly NO.

statement (2) The average age of employees in X, Y, and Z, is 40, 20, and 50, respectively
therefore, total age = 3*40+4*20+8*50 = 600. as we have to check if 3x+4y+8z<600. so it is also a clear NO. & hence , answer is D.

Now please let me know where am I going wrong. Your help will be appreciated. Thanks !!!!

Given that the ratio of the number of employees of three companies X, Y and Z is 3:4:8, respectively, so the number of employees could be:
3, 4, 8;
3*2=6, 4*2=8, 8*2=16;
3*3=9, 4*3=12, 8*3=24;
3*4=12, 4*4=16, 8*4=32;
...

Notice that the multiple is the same in each case. Thus the ratio of the number of employees is 3x:4x:8x, for some positive multiple x, not 3x:4y:8z.

Hope it's clear.

Hi Bunnel
Could you elaborate how this is true - median from right angle is half of the hypotenuse??
Thanks

Sure.

Imagine a right triangle inscribed in a circle. We know that if a right triangle is inscribed in a circle, then its hypotenuse must be the diameter of the circle, hence half of the hypotenuse is radius. The line segment from the third vertex to the center is on the one hand radius of the circle=half of the hypotenuse and on the other hand as it's connecting the vertex with the midpoint of the hypotenuse it's median too.



Hope it's clear.

Hi Bunuel.. How to be sure that A={0,0,0,0} and A={1,2,2,3} are the only sets possible from Statement2. Is there any quick method to find this. infact i was unable to find A={1,2,2,3}.

A={0, 0, 0, 0} and A={1, 2, 2, 3} are NOT the only sets possible. For example A={0, 0, 0} and A={1, 2, 3}. You can find these sets by trial and error.

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?

Given that 1/A+1/B=1/6, where A is the time needed for machine A to complete the task working alone and B is the time needed for machine B to complete the task working alone.

Can anyone please explain how 1/A + 1/B = 1/6

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?

A is the time needed for machine A to complete the task working alone, thus the rate of A is 1/A job/day.
B is the time needed for machine B to complete the task working alone, thus the rate of A is 1/B job/day.

Their combined rate is 1/A+1/B, which given to be equal to 1/6.



Hope this helps.

Bununel, what if Set A only contains one factor "1"?

The range of one element set is 0. If set A={1}, then it's range (0) does not equal to its mean (1). Thus this example contradicts the second statement and therefore is not valid.

Hope it's clear.

Bunuel..

in question..A+b=6..
but in statement 1..its a+b=25?
are these both not contradict with eachother?

Seems like m missing something :/