New DS set!!!

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
Not sufficient.ie with x=1
A={1,2, 3, 4, 5} B={1, 2, 3, 4, 5, y} All depends on y.
(2) y=3
Not sufficient.ie if x=0
A={3,3,3,3,3} B={3,3,3,3,3,3} STD of B is = STD of A
if x=1
A={1,2, 3, 4, 5} B={1, 2, 3, 4, 5, 3} STD of B is < STD of A

(1)+(2) From 1 we know that \(x\neq{0}\) and from 2 that y=3. Sufficient. ie:x=1 : A={1,2, 3, 4, 5} B={1, 2, 3, 4, 5, 3} STD of B is < STD of A
x=1000 : A={-1997,-997, 3, 1003, 2003} B={-1997,-997, 3, 1003, 2003, 3} STD of B is < STD of A
A and B share 4 elements in common that are different from 3, but because B has one more 3 than X it will have a STD lesser than A

QUESTION 1
(1) We might have a product equal to 0: 0.1.2 = 0 but we might also have a product equal to 6: 1.2.3 = 6
INSUFFICIENT

(2) We might have a product equal to -6: -3.(-2).(-1) = -6 but we might also have a product equal to 0: 0.1.2 = 0
INSUFFICIENT

(1)+(2) If the integers are k, k-1, k-2, and we know (stmt2) that: k+ k-1 + k-2 < 6 => k < 3, we can only have:
{0,1,2}or {-1,0,1}, and both have a product of ZERO
SUFFICIENT

ANS: C

QUESTION 2
(1) Clearly INSUFFICIENT. Take x = 13, y = 11, for which the division gives remainder 2. Take x=12, y=11, for which the division gives remainder 1.

(2) x = q.y + 9. We'll know that the remainder of the division is 9, IF y > 9. Take the case x=15,y=2, q=3 (which gives a remainder 1). Now, take the
case x= 29, y=10, q= 9 (which gives a remainder 9). INSUFFICIENT

(1) + (2): Given, from stmt1, that y is larger than 9, then x=q.y+9 will always give remainder 9. SUFFICIENT

ANS: C

QUESTION 3
(1) Clearly INSUFFICIENT. If ABC is isosceles, the median BD is also the height of the triangle relative to AC.
We can increase the value of AC, as much as we want while also keeping BD equal to 12 cm

(2) AC^2 = AB^2 + BC^2 means B is a right angle (reverse property of the Pythagorean theorem). Which means, that AC is
a diameter to the circle to which ABC is inscribed. This means that the the median BD is also equal to AD and to CD, all
equal to the radius of the circle. AC = 24 cm SUFFICIENT

ANS: B

QUESTION 8
Let r + b = 10, where r and b are the number of red marbles and blue marbles, respectively.
The probability that both will be red is: C(r,2)/C(10,2) = r.(r-1)/90.
For it to be greater than 3/5, we need to have: r.(r-1)/90 > 3/5 => r.(r-1) > 54 => r can be 9 or 8.

(1) Probability = C(b,2)/C(10,2) = b.(b-1)/(10.9) < 1/10. Therefore, b.(b-1) < 9. Which means, b can be 2 (r = 8) or 3 (r=7). In
the first, the answer to the stem question is YES, and in the second NO. INSUFFICIENT

(2) r > 6. If r = 7, the answer to the stem is NO. If r=8, the answer to the stem is YES. INSUFFICIENT

(1) + (2), No new information: r can be 7 or r can be 8. In the first case, the answer is NO. In the second, YES. INSUFFICIENT


ANS: E

See Img for my solutions.These are very good question...

I just wanna say: thanks Bunuel !

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.

From \(A+\frac{150}{A}=25\) --> \(A^2-25A+150=0\) (not A^2 - 25A - 150 = 0) --> \((A-10)(A-15)=0\) --> \(A=10\) and \(B=15\) OR \(A=15\) and \(B=10\).

Hope it's clear.

{0, 1, 2} is the only set possible --> the product is 0.

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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

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.