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The next set of medium/hard DS questions. I'll post OA's with detailed explanations after some discussion. Please, post your solutions along with the answers.

1. What is the product of three consecutive integers?

(1) At least one of the integers is positive (2) The sum of the integers is less than 6

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

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

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.

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

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. What is the product of three consecutive integers?

(1) At least one of the integers is positive.

We can have three cases: (i) All three integers are positive. In this case the product will obviously be positive. (ii) Two of the integers are positive: {0, 1, 2}. In this case the product will be zero. (iii) Only one of the integers is positive: {-1, 0, 1}. In this case the product will be zero.

Not sufficient.

(2) The sum of the integers is less than 6. Clearly insufficient, consider {-1, 0, 1} and {-3, -2, -1}.

(1)+(2) The second statement implies that we cannot have case (i) from (1), since the least sum of three positive consecutive integers is 1+2+3=6. Thus we have either case (ii) or case (iii). Therefore the product of the integers is zero. Sufficient.

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

If \(x\) and \(y\) are positive integers, there exist unique integers \(q\) and \(r\), called the quotient and remainder, respectively, such that \(y =divisor*quotient+remainder= xq + r\) and \(0\leq{r}<x\).

(1) y is a two-digit prime number. Clearly insufficient since we know nothinf about x.

(2) x=qy+9, for some positive integer q. It's tempting to say that this statement is sufficient and \(r=9\), since given equation is very similar to \(y =divisor*quotient+remainder= xq + r\) . But we don't know whether \(y>9\): remainder must be less than divisor.

For example: If \(x=10\) and \(y=1\) then \(10=1*1+9\), then the remainder upon division 10 by 1 is zero. If \(x=11\) and \(y=2\) then \(11=1*2+9\), then the remainder upon division 11 by 2 is one. Not sufficient.

(1)+(2) From (2) we have that \(x=qy+9\) and from (1) that y is more than 9 (since it's a two-digit number), so we have direct formula of remainder, as given above. Sufficient.

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

Sol:

We know SD >/ 0 and hence Set A is Positive does not tell us about y in statement B

So St 1 is alone not sufficent

St 2: y =3

Set A : SD \sqrt{10X^2/5}

Set B : SD is \sqrt{10x^2/6}

Since x is an integer we have SD of A as x \sqrt{2} and SD of B as x\sqrt{5/3}

Clearly SD of A is greater than that of B and hence ans should be B

Notice that x can be 0 for (2), so this statement is NOT sufficient.

3. The length of the median BD in triangle ABC is 12 centimeters, what is the length of side AC?

(1) ABC is an isosceles triangle. Clearly insufficient.

(2) AC^2 = AB^2 + BC^2. This statement implies that ABC is a right triangle and AC is its hypotenuse. Important property: median from right angle is half of the hypotenuse, hence BD=12=AC/2, from which we have that AC=24. Sufficient.

Answer: B.

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.

(1) No number in set A is less than the average (arithmetic mean) of set A.

Since no number is less than the average, then no number is more than the average, which implies that the list contains identical elements: A={x, x, x, ...}. From this it follows that (the average)=(the median). But we don't know the value of x, thus this statement is NOT sufficient.

(2) The average (arithmetic mean) of set A is equal to the range of set A.

Not sufficient: if A={0, 0, 0, 0}, then (the median)=0, but if A={1, 2, 2, 3}, then (the median)=2.

(1)+(2) From (1) we have that the list contains identical elements. The range of all such sets is 0. Therefore, from (2) we have that (the average)=(the range)=0 and since from (1) we also know that (the average)=(the median), then (the median)=0. Sufficient.

Answer: C..

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.

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.

(1) The average time A and B can complete the task working alone is 12.5 days. This statement implies that A+B=2*12.5=25. Now, since we don't know which machine works faster then even if we substitute B with 25-A (1/A + 1/(25-A) = 1/6) we must get two different answers for A and B: A<B and A>B. 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, so we have that 1/A+1/(A-5)=1/6. From this we can find that A=2 (not a valid solution since in this case B will be negative) or A=15. Sufficient.

Answer: B.

Hi Bunuel!

Could you please tell me where I made a mistake in my solution to statement 1?

statement 1:

(1) A+B=25

--> from the question stem we have that (1/A)+(1/B)=1/6 --> [(A+B)/AB]=1/6 ---> [25/AB]=1/6 --> 150=AB --> B=150/A (2)

insert 2 into 1: A+[150/A]=25 --> A²-25A-150=0 --> (A-30)(A+5)=0 --> A=30 or (-5), both answers obviously can't be right, since A+B=25 and no value can be negative here. But where exactly was my approach wrong?

Thanks a lot for your help!

Max

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 \(B=10\).

Q8) Whether the probably of drawing 2 red marbles from a jar of 10 marbles is > 3/5

From Stmt 1, the probability that both marbles selected will be blue is < 1/10 and since there are only red and blue marbles in the jar

The probability of selecting both marbles red = 1 - LT 1/10 ==> GT 9/10 - Sufficient

From Stmt 2, the number of red marbles in the jar is at least 60 % is >= 6 of 10 marbles in the jar

The probability of selecting both marbles red >= 6/10 * 5/9 >= 0.333 which can be less than 3/5 when there are 6 red marbles and more than 3/5 when there are 9 red marbles - Not Sufficient

This implies all the numbers in the set are equal to the average otherwise the Avg of the set A will be more if any one number is more than the Avg. Hence the median of set A in which all the numbers are same as the Avg. is none other than the Avg. value. - Sufficient. From Stmt 2, Avg. of set A is = Amax - Amin of Set A.

This only implies the lower and upper limit of Set A and they could be even or odd number of values in the set. This will result in various values for median within Amin and Amin. - Not Sufficient.

Answer is A _________________

KUDOS is a way to say Thank You

Last edited by nt2010 on 10 Apr 2013, 08:34, edited 1 time in total.

Q6) Is Avg. age of employees of three companies X, Y and Z < 40 years?

Using the unknown multiplier approach, the number of employees in companies X, Y, and Z are 3x, 4x, and 8x respectively

From Stmt 1, the total age of all the employees in 3 companies is 600 - Not Sufficient

If x = 1, the total employees in companies X, Y and Z is 3 + 4 + 8 = 15, so the avg. age of employees is 600 / 15 = 40 years If x = 2, the total employees in companies X, Y and Z is 6 + 8 + 16 = 30, so the avg. age of employees is 600 / 30 = 20 years

From Stmt 2, avg age of X is 40, avg age of Y is 20, and avg age of z is 50 - Sufficient

If x = 1, the total employees in companies X, Y and Z is 3 + 4 + 8 = 15, so the avg. age of employees is (40*3+20*4+50*8) / 15 = 40 years If x = 2, the total employees in companies X, Y and Z is 6 + 8 + 16 = 30, so the avg. age of employees is (40*6+20*8+50*16) / 30 = 40 years

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

(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.

My answer E

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.

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

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

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

My answer C

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.

Back to hometown after a short trip to New Delhi for my visa appointment. Whoever tells you that the toughest part gets over once you get an admit is...