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# New DS set!!!

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New DS set!!!  [#permalink]

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10 Apr 2013, 08:10
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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

Solution: http://gmatclub.com/forum/new-ds-set-15 ... l#p1211902

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

(1) y is a two-digit prime number
(2) x=qy+9, for some positive integer q

Solution: http://gmatclub.com/forum/new-ds-set-15 ... l#p1211903

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
(2) AC^2 = AB^2 + BC^2

Solution: http://gmatclub.com/forum/new-ds-set-15 ... l#p1211904

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

Solution: http://gmatclub.com/forum/new-ds-set-15 ... l#p1211906

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

Solution: http://gmatclub.com/forum/new-ds-set-15 ... l#p1211907

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.

Solution: http://gmatclub.com/forum/new-ds-set-15 ... l#p1211908

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

(1) The median temperature in degrees Celsius in City A in March was less than the median temperature in degrees Celsius in city B.

(2) The ratio of the average temperatures in degrees Celsius in A and B in March was 3 to 4, respectively.

Solution: http://gmatclub.com/forum/new-ds-set-15 ... l#p1211909

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

Solution: http://gmatclub.com/forum/new-ds-set-15 ... l#p1211910

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

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

Solution: http://gmatclub.com/forum/new-ds-set-15 ... l#p1211911

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.

Solution: http://gmatclub.com/forum/new-ds-set-15 ... l#p1211912

Kudos points for each correct solution!!!
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14 Apr 2013, 09:02
9
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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. We know that the standard deviation of any set is more than or equal to zero. The standard deviation of a set is zero only when the set consists of identical elements. So, this statement implies that set A does NOT consists of identical elements or that x does not equal to zero. Still this statement is not sufficient to answer the question.

(2) y=3. The mean of set A is 3. Now, if $$x\neq{0}$$ for example if x=1, then the standard deviation of B would be smaller that the standard deviation A, since the elements of B would be less widespread than the element of set A. But if x=0, then A={3, 3, 3, 3, 3} and B={3, 3, 3, 3, 3, 3}, so both will have the standard deviation of zero. Bot sufficient.

(1)+(2) Since from (1) $$x\neq{0}$$, then adding a new element which equals to the mean will shrink the standard deviation, thus SD(A)>SD(B). Sufficient.

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Re: New DS set!!!  [#permalink]

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11 Apr 2013, 21:52
4
2
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.

##### General Discussion
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Re: New DS set!!!  [#permalink]

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14 Apr 2013, 09:03
12
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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?

The question: is P(R and R)=R/10*(R-1)/9>3/5?
Is R(R-1)>54?
Is R>7? (By number plugging) So, the question asks whether the number of red marbles is more than 7 (8, 9, or 10).

(1) The probability that both marbles selected will be blue is less than 1/10. This implies that B/10*(B-1)/9<1/10. So, we have that B(B-1)<9, thus B<4, so the number of red marbles in the jar is 7, 8, 9, or 10. Not sufficient.

(2) At least 60% of the marbles in the jar are red. This implies that the number of red marbles is 6 or more. Not sufficient.

(1)+(2) From above we have that R>6. Not sufficient.

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New DS set!!!  [#permalink]

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14 Apr 2013, 09:03
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Was the average (arithmetic mean) temperature in degrees Celsius in city A in March less than the average (arithmetic mean) temperature in degrees Celsius in city B in March?

(1) The median temperature in degrees Celsius in City A in March was less than the median temperature in degrees Celsius in city B. Clearly insufficient.

(2) The ratio of the average temperatures in degrees Celsius in A and B in March was 3 to 4, respectively. Temperatures can be negative, thus this statement is also not sufficient. Consider $$T(A)=3$$ and $$T(B)=4$$ AND $$T(A)=-3$$ and $$T(B)=-4$$.

(1)+(2) We have no additional useful info. Not sufficient.

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14 Apr 2013, 09:00
7
29
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.

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14 Apr 2013, 09:01
6
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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.

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Re: New DS set!!!  [#permalink]

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14 Apr 2013, 08:59
5
7
SOLUTION:

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.

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14 Apr 2013, 09:01
5
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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?

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.

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Re: New DS set!!!  [#permalink]

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12 Apr 2013, 03:48
3
1
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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Re: New DS set!!!  [#permalink]

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16 Apr 2013, 03:33
3
Dipankar6435 wrote:
Bunuel wrote:
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.

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.
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Re: New DS set!!!  [#permalink]

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Updated on: 10 Apr 2013, 09:57
2
9. If x is an integer, is x^2>2x?
$$x^2-2x>0$$
$$x(x-2)>0$$
The question asks is x in one of those intervals?
$$x<0 , x>2$$

(1) x is a prime number.
Sufficient All prime numbers are greater than 2 so $$x\geq{2}$$ so we are in the right interval.
What a stupid mistake!Not suff, x can be 2
(2) x^2 is a multiple of 9.
Not sufficient. If x=0, x^2 =0 is a multiple of 9 but 0 isn't in the intervals, if x=3 x^2=9 is a multiple of nine and 3 is in our intervals.

C
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Originally posted by Zarrolou on 10 Apr 2013, 09:12.
Last edited by Zarrolou on 10 Apr 2013, 09:57, edited 1 time in total.
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Re: New DS set!!!  [#permalink]

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Updated on: 11 Apr 2013, 20:00
2
Bunuel wrote:
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 -Can not determine the values
(2) The sum of the integers is less than 6- Can not determine the values

(1)+(2)-> at least one positive and <6 so can't be 1,2,3 -> so 0 must be part of the sequence .

IMO: C

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

(1) y is a two-digit prime number - Can not determine the values
(2) x=qy+9, for some positive integer q - Can not determine the values
1+2=>y>9 , as y is two digit ...so remainder 9

IMO: C

Bunuel wrote:
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 No idea about which sides are same
(2) AC^2 = AB^2 + BC^2 D is mid point of AC , SO AD= DC=BD = 12 => AC

IMO:B

Bunuel wrote:
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.say A can complete in x , then b in = 12.5*2 - x
hence $$\frac{1}{x}+\frac{1}{25-x}=\frac{1}{6}=> x= 15 or 10$$

(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
$$\frac{1}{x}+\frac{1}{x+5}=\frac{1}{6}=> x= 10$$
So A 15 days
[/color]

IMO: B

Bunuel wrote:
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 positiveNot sufficient
(2) y=3So we added 3 to the existing series , where 3 was mean=> SD will decrease

IMO: B

Bunuel wrote:
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 600Can 't do anything
(2) The average age of employees in X, Y, and Z, is 40, 20, and 50, respectively.can determine
$$40*3x + 20*4x + 50 *8x = total age , now dividing by 15x we can get avg age$$

IMO: B

Bunuel wrote:
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 BNo idea about avg
(2) The ratio of the average temperatures in A and B in March was 3 to 4, respectivelyit means Avg(A) < Avg(B)

IMO: B

Bunuel wrote:
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
say there are b blue and 10-b red=> \frac{(bC2)}{(10C2) < 0.1=> b(b-1) < 9 Not sufficient}
(2) At least 60% of the marbles in the jar are red
So Red GE 6 and Blue LE 4 => for 6 red p(selecting 2 red out of 6) = 0.33 < \frac{3}{5} but if all are red P(red) = 1 > \frac{3}{5}
so insufficient

Now adding 1+2 => blue can be 4 ,3,2,1,0 both blue should be less than 0.1 => so b can't be 4 it can only 3,2
So red balls can be 7 or 8-> no definite ans

IMO: E

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

(1) x is a prime number.not sufficient can be 2 or 3..
(2) x^2 is a multiple of 9. =>X is a multiple of 3 when we do squaring we multiply with a multiple of 3 and whem we do 2X we multiply with 2
so x^2>2x is always true except x= 0

adding 1 +2 => x can't be zero
IMO: C

Bunuel wrote:
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. so all the numbers must be same =>median is any number of the set, sufficient to determine
(2) The average (arithmetic mean) of set A is equal to the range of set A.not sufficient

IMO: A

Please let me know how many i got correct

Originally posted by focus2k13 on 10 Apr 2013, 10:08.
Last edited by focus2k13 on 11 Apr 2013, 20:00, edited 1 time in total.
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Re: New DS set!!!  [#permalink]

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11 Apr 2013, 10:07
2
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
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Re: New DS set!!!  [#permalink]

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11 Apr 2013, 10:08
2
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
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Re: New DS set!!!  [#permalink]

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11 Apr 2013, 10:08
2
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
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Re: New DS set!!!  [#permalink]

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11 Apr 2013, 16:37
2
See Img for my solutions.These are very good question...
Attachments

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Re: New DS set!!!  [#permalink]

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11 Apr 2013, 21:45
2
4
2. If x and y are both positive integers and x>y, what the remainder when x is divided by y?
(1) y is a two-digit prime number
(2) x=qy+9, for some positive integer q

1) Information on y is only given. No information on x. Not sufficient.
2) x/y = q + 9/y. For y > 9, remainder will be 9 when x is divided by y. But remainder will be different for other y’s, Not sufficient.

Together: y>9 --> remainder will be 9 when x is divided by y. Sufficient.

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Re: New DS set!!!  [#permalink]

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11 Apr 2013, 21:47
2
2
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
(2) AC^2 = AB^2 + BC^2

1) There is no information on angles and which two sides are equal. Not sufficient.

2) It says that angle at B is right angle and AC is hypotenuse. If any right-angled triangle is inscribed in a circle, the hypotenuse of the triangle must be diameter of circle and the median extending to the hypotenuse of the triangle must be radius of the circle -->Median is equal to half of the hypotenuse --> AC = 24. Sufficient.

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Re: New DS set!!!  [#permalink]

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11 Apr 2013, 21:48
2
1
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.

Re: New DS set!!!   [#permalink] 11 Apr 2013, 21:48

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