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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.

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.

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.

(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.

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.

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.

(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.

Given: \(6a=3b=7c\) --> least common multiple of 6, 3, and 7 is 42 hence we ca write: \(6a=3b=7c=42x\), for some number \(x\) --> \(a=7x\), \(b=14x\) and \(c=6x\).

(1) ac=6b --> \(7x*6x=6*14x\) --> \(x^2=2x\) --> \(x=0\) or \(x=2\). Not sufficient.

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.

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.

4. How many numbers of 5 consecutive positive integers is divisible by 4?

(1) The median of these numbers is odd --> the median of the set with odd number of terms is just a middle term, thus our set of 5 consecutive numbers is: {Odd, Even, Odd, Even, Odd}. Out of 2 consecutive even integers only one is a multiple of 4. Sufficient.

(2) The average (arithmetic mean) of these numbers is a prime number --> in any evenly spaced set the arithmetic mean (average) is equal to the median --> mean=median=prime. Since it's not possible that median=2=even, (in this case not all 5 numbers will be positive), then median=odd prime, and we have the same case as above. Sufficient.

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.

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

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.

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

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