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The Discreet Charm of the DS [#permalink]
02 Feb 2012, 03:15

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I'm posting 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. Good luck!

1. Bonnie can paint a stolen car in x hours, and Clyde can paint the same car in y hours. They start working simultaneously and independently at their respective constant rates at 9:45am. If both x and y are odd integers, is x=y? (1) x^2+y^2<12 (2) Bonnie and Clyde complete the painting of the car at 10:30am

4. How many numbers of 5 consecutive positive integers is divisible by 4? (1) The median of these numbers is odd (2) The average (arithmetic mean) of these numbers is a prime number

7. A certain fruit stand sold total of 76 oranges to 19 customers. How many of them bought only one orange? (1) None of the customers bought more than 4 oranges (2) The difference between the number of oranges bought by any two customers is even

10. The function f is defined for all positive integers a and b by the following rule: f(a,b)=(a+b)/GCF(a,b), where GCF(a,b) is the greatest common factor of a and b. If f(10,x)=11, what is the value of x? (1) x is a square of an integer (2) The sum of the distinct prime factors of x is a prime number.

Re: The Discreet Charm of the DS [#permalink]
05 Feb 2012, 06:11

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10. The function f is defined for all positive integers a and b by the following rule: f(a,b)=(a+b)/GCF(a,b), where GCF(a,b) is the greatest common factor of a and b. If f(10,x)=11, what is the value of x?

Notice that the greatest common factor of 10 and x, GCF(10,x), naturally must be a factor of 10: 1, 2, 5, and 10. Thus from f(10,x)=11 we can get four different values of x:

(1) x is a square of an integer --> x can be 1 or 100. Not sufficient.

(2) The sum of the distinct prime factors of x is a prime number ---> distinct primes of 12 are 2 and 3: 2+3=5=prime, distinct primes of 45 are 3 and 5: 3+5=8\neq{prime} and distinct primes of 100 are 2 and 5: 2+5=7=prime. x can be 12 or 100. Not sufficient.

Re: The Discreet Charm of the DS [#permalink]
05 Feb 2012, 03:31

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SOLUTIONS:

1. Bonnie can paint a stolen car in x hours, and Clyde can paint the same car in y hours. They start working simultaneously and independently at their respective constant rates at 9:45am. If both x and y are odd integers, is x=y?

Bonnie and Clyde when working together complete the painting of the car ins \frac{xy}{x+y} hours (sum of the rates equal to the combined rate or reciprocal of total time: \frac{1}{x}+\frac{1}{y}=\frac{1}{T} --> T=\frac{xy}{x+y}). Now, if x=y then the total time would be: \frac{x^2}{2x}=\frac{x}{2}, since x is odd then this time would be odd/2: 0.5 hours, 1.5 hours, 2.5 hours, ....

(1) x^2+y^2<12 --> it's possible x and y to be odd and equal to each other if x=y=1 but it's also possible that x=1 and y=3 (or vise-versa). Not sufficient.

(2) Bonnie and Clyde complete the painting of the car at 10:30am --> they complete the job in 3/4 of an hour (45 minutes), since it's not odd/2 then x and y are not equal. Sufficient.

Re: The Discreet Charm of the DS [#permalink]
05 Feb 2012, 04:43

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7. A certain fruit stand sold total of 76 oranges to 19 customers. How many of them bought only one orange?

(1) None of the customers bought more than 4 oranges --> this basically means that all customers bought exactly 4 oranges (76/19=4), because if even one customer bought less than 4, the sum will be less than 76. Hence, no one bought only one orange. Sufficient.

(2) The difference between the number of oranges bought by any two customers is even --> in order the difference between ANY number of oranges bought to be even, either all customers must have bought odd number of oranges or all customers must have bough even number of oranges. But the first case is not possible: the sum of 19 odd numbers is odd and not even like 76. Hence, again no one bought only one=odd orange. Sufficient.

Re: The Discreet Charm of the DS [#permalink]
05 Feb 2012, 03:59

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3. If a, b and c are integers, is abc an even integer?

In order the product of the integers to be even at leas on of them must be even

(1) b is halfway between a and c --> on the GMAT we often see such statement and it can ALWAYS be expressed algebraically as b=\frac{a+c}{2}. Now, does that mean that at leas on of them is be even? Not necessarily, consider a=1, b=3 and c=5. Of course it's also possible that b=even, for example if a=1 and b=7. Not sufficient.

(2) a = b - c --> a+c=b. Since it's not possible that the sum of two odd integers to be odd then the case of 3 odd numbers is ruled out, hence at least on of them must be even. Sufficient.

Re: The Discreet Charm of the DS [#permalink]
05 Feb 2012, 04:32

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5. What is the value of integer x?

(1) 2x^2+9<9x --> factor qudratics: (x-\frac{3}{2})(x-3)<0 --> roots are \frac{3}{2} and 3 --> "<" sign indicates that the solution lies between the roots: 1.5<x<3 --> since there only integer in this range is 2 then x=2. Sufficient.

(2) |x+10|=2x+8 --> LHS is an absolute value, which is always non negative, hence RHS must also be non-negative: 2x+8\geq{0} --> x\geq{-4}, for this range x+10 is positive hence |x+10|=x+10 --> x+10=2x+8 --> x=2. Sufficient.

Re: The Discreet Charm of the DS [#permalink]
05 Feb 2012, 05:35

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9. If x and y are negative numbers, is x<y?

(1) 3x + 4 < 2y + 3 --> 3x<2y-1. x can be some very small number for instance -100 and y some large enough number for instance -3 and the answer would be YES, x<y BUT if x=-2 and y=-2.1 then the answer would be NO, x>y. Not sufficient.

Re: The Discreet Charm of the DS [#permalink]
03 Feb 2012, 13:56

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5. What is the value of integer x? (1) 2x^2+9<9x (2) |x+10|=2x+8

Stmt 1 - 2x^2+9<9x Put the values -5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5 So the only value that satifies the equation is 2. Sufficient

Stmt 2 |x+10|=2x+8 In number line take one number X = - 10. Condition one X > - 10 which will result in positive equation. x+10=2x+8 x = 2 x = 2 greater than -10 so satifies the equation.

Condition two X < - 10 which will result in negative equation. x+10 = - 2x - 8 x = 6 which is is not less than -10 which cannot the solution of inequality.

Re: The Discreet Charm of the DS [#permalink]
05 Feb 2012, 03:37

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2. Is xy<=1/2?

(1) x^2+y^2=1. Recall that (x-y)^2\geq{0} (square of any number is more than or equal to zero) --> x^2-2xy+y^2\geq{0} --> since x^2+y^2=1 then: 1-2xy\geq{0} --> xy\leq{\frac{1}{2}}. Sufficient.

Re: The Discreet Charm of the DS [#permalink]
05 Feb 2012, 06:38

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11. If x and y are integers, is x a positive integer?

(1) x*|y| is a prime number --> since only positive numbers can be primes, then: x*|y|=positive --> x=positive. Sufficient

(2) x*|y| is non-negative integer. Notice that we are told that x*|y| is non-negative, not that it's positive, so x can be positive as well as zero. Not sufficient.

Re: The Discreet Charm of the DS [#permalink]
03 Feb 2012, 07:01

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3. If a, b and c are integers, is abc an even integer? (1) b is halfway between a and c (2) a = b - c

Funda for product abc to be even, if any one of them even then product will be even.

Stmt 1 - says b = (a+c)/2 means a+c is some even number. E + E also results in even O + O also results in Even and b can be anything even or odd so not sufficient.

Stmt 2 - says a = b - c say worst condition b an c are odd . will results in a even. or lets says any one among b or c is even then a off but since one number is even the product will be even so sufficient.

Re: The Discreet Charm of the DS [#permalink]
03 Feb 2012, 13:56

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4. How many numbers of 5 consecutive positive integers is divisible by 4? (1) The median of these numbers is odd (2) The average (arithmetic mean) of these numbers is a prime number

out of 5 consecutive intergers there are only 2 option either there will 1 or 2 numbers which will divide by 4. There will 2 numbers if the first number fo sequence will be divible by 4. So we have to find out if first number is divisible by 4.

Stmt 1 - Median is odd means first number is odd ( odd , even , odd, even, odd) So sufficuent to tell that there will only 1 number which be divible by 4. Stmt 2 - In case of consecutive numbers median is always equal to average. Numbers are positive so first numbers cannot be 0 to make 2 as median/average. Which means average/mean is odd, Again going by logic in stmt 1 first number should be odd which is also sufficient

Re: The Discreet Charm of the DS [#permalink]
03 Feb 2012, 13:56

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7. A certain fruit stand sold total of 76 oranges to 19 customers. How many of them bought only one orange? (1) None of the customers bought more than 4 oranges (2) The difference between the number of oranges bought by any two customers is even

76 oranges / 19 customers.

Stmt 1 - Let's say everyone got maximum 4 oranges to 76 distributed. And if someone got less than 4 and someone would have more than 4 which is not possible. So sufficient to tell that noone got 1 orange.

Stmt 2 - Difference will be even if, even - even OR odd - odd.

Say suppose they got odd, the nall of them have to recieve odd number else the difference will not even. Which is not possible as there 76 oranges which wont divide oddly. But if its even if someone gets 2 then someone can get 6 which will make the difference even. All evens will evenly divide 79 between 19 of them.

Re: The Discreet Charm of the DS [#permalink]
03 Feb 2012, 13:56

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8. If x=0.abcd, where a, b, c and d are digits from 0 to 9, inclusive, is x>7/9? (1) a+b>14 (2) a-c>6

7/9 = .7777777 so question is asking if a > 7, if a = 7 then is b > 7 , if b = 7 then is c > 7 and so forth.

stmt 1 - a+b > 14 so possible values of (a,b) = (8,7) , (7,8), (9,6), (6,9) So a can be = 7 or even > 7 so insufficient.

stmt 2 - a - c > 6 possible values of a,c = (7,0), (8,1), (9,2) so a can take multiple values so insufficient.

Combining stmt 1 and stmt 2. a can taken values as 7,8,9 you will see that if a = 7 then b is 8 which is greater than 7/9 All other values will result yes X < 7/9 which is sufficient

Re: The Discreet Charm of the DS [#permalink]
03 Feb 2012, 14:00

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sourabhsoni wrote:

1. Bonnie can paint a stolen car in x hours, and Clyde can paint the same car in y hours. They start working simultaneously and independently at their respective constant rates at 9:45am. If both x and y are odd integers, is x=y? (1) x^2+y^2<12 (2) Bonnie and Clyde complete the painting of the car at 10:30am

The catch is that they are working independently.

Answer B

Just to clarify more, working independently does not mean they are painting different cars. They are still painting the same car. Only that, the events are mutually exclusive, as you say in probabilistic terms, so that rates are not affected when they work simultaneously.

Another way : 1) more than one solution possible. 2) let's say 1)x=y=1. Work will be completed in 1/2 hour i.e. 10.15 am. 2)x=y=3. Work will be completed in 3/2 hour i.e. 11.15 am. Not possible. B. _________________

Re: The Discreet Charm of the DS [#permalink]
03 Feb 2012, 16:37

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Bunuel wrote:

sourabhsoni wrote:

2. Is xy<=1/2? (1) x^2+y^2=1 (2) x^2-y^2=0

My funda - Area of square is largest among all the quadilateral with same perimeter. Stmt 1 - Only possible values of x and y are 1/Sqrt(2). So sufficient as xy = 1/2 Stmt 2 - Only says x and y are equal. Not sufficient Answer A

You are close to correct reasoning for (1), though from it you can not say that xy=1/2 and the only possible value for x and y are 1/Sqrt(2). Consider the following example: 0^1+1^2=1.

As for (2): x^2-y^2=0 doesn't mean that x=y.

1) x2 + y2 = 1 is a circle with radius 1, and origin as center. We are only concerned with 1st and 3rd quad (think why? ), Draw a line y=x which intersects circle at x=y=1/sqrt(2). We can observe, as x becomes greater than 1/sqrt2, y gets lesser than 1/sqrt2 moving on the circle (below y=x line). Hence xy<= 1/2. Similarly, as y becomes greater than 1/sqrt2, x gets lesser than 1/sqrt2 moving on the circle (above y=x line). Hence xy<= 1/2. Hence sufficient.

For more mathematically inclined, draw graph of xy=1/2. It has only one point of contact with the circle, tangential at x=y=1/sqrt(2).

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Re: The Discreet Charm of the DS [#permalink]
03 Feb 2012, 17:08

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vailad wrote:

IMO, if you can imagine drawing the circle and y=x line, it takes less than 30 sec. to figure it out. It only takes so much longer to explain in text. And ofcourse, no need to draw the xy graph. Will wait though for the even faster method, if any.

If you answered this question in less than 30 sec using this approach then all I can say is great job!

As for the easier/faster solution, little hint: it involves simplest algebraic manipulation. _________________

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