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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.

I have seen references to use your guides in many threads, Bunuel. Now I can see why. I will be sure to use your challenge sets in the coming weeks to hopefully boost my quant score into the 48-50 range. Thank you so much for your invaluable contributions for relative GMAT newbies, like myself.

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

In respect to the question above, I assumed that any two of those 19 customer might have bought 5 & 3 oranges and hence I, marked the option insufficient. Bunnel have equated and treated the option in totally different way. I ,lack the skill to convert these sort of condition in to equation.

please can some post or point to the list of similar Word translation sentences and how to convert them in to equation. Im very new to GMAT club so please forgie me if this is the repeated posting.

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

In respect to the question above, I assumed that any two of those 19 customer might have bought 5 & 3 oranges and hence I, marked the option insufficient. Bunnel have equated and treated the option in totally different way. I ,lack the skill to convert these sort of condition in to equation.

please can some post or point to the list of similar Word translation sentences and how to convert them in to equation. Im very new to GMAT club so please forgie me if this is the repeated posting.

Thanks, Vids

I did not use any equation for this question.

Statement (2) says: 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.

Now, the sum of 19 odd integers is odd and we have that fruit stand sold total of 76, so even number of oranges, which means that the case where all customers buy odd number of oranges is not possible. And since 1 is odd then no one bought only one orange. 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?

(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 neither an integer nor integer/2 then \(x\) and \(y\) are not equal. Sufficient.

Answer: B.

please consider adding "working together" to the stmt 2- as I deciphered they both worked separately and ended at the same time so cool enough (though same answer but got carried away by the wording)

Thanks once again for nice collection!!
_________________

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?

(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 neither an integer nor integer/2 then \(x\) and \(y\) are not equal. Sufficient.

Answer: B.

please consider adding "working together" to the stmt 2- as I deciphered they both worked separately and ended at the same time so cool enough (though same answer but got carried away by the wording)

(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.

1) How did you factor 2x^2+9<9x (ie 2x^2-9x+9<0) so quickly? I always struggle with factoring polynomials in which a coefficient other than 1 is on the x^2. Did you use the quadratic formula? I am interested in knowing if there is a quicker way than the quadratic formula method.

2) Once you determined that 1.5 and 3 were the roots of the equation, how did you figure that the solution was in between 1.5 and 3 from just looking at the sign "<"?? I used the dumb method of just plugging values that lie from (-infinity , 1.5), (1.5, 3) and (3, +infinity). How did you know the sign "<" told you the solution was in in between (1.5,3)?

(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.

1) How did you factor 2x^2+9<9x (ie 2x^2-9x+9<0) so quickly? I always struggle with factoring polynomials in which a coefficient other than 1 is on the x^2. Did you use the quadratic formula? I am interested in knowing if there is a quicker way than the quadratic formula method.

2) Once you determined that 1.5 and 3 were the roots of the equation, how did you figure that the solution was in between 1.5 and 3 from just looking at the sign "<"?? I used the dumb method of just plugging values that lie from (-infinity , 1.5), (1.5, 3) and (3, +infinity). How did you know the sign "<" told you the solution was in in between (1.5,3)?

(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.

Answer: D.

Hope it helps.

Hi bunuel, Isn't |x+10|=2x+8 be written as Either x+10=2x+8 or x+10=-(2x+8) ? and then this should be solved? Please help on this one.

(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.

Answer: D.

Hope it helps.

Hi bunuel, Isn't |x+10|=2x+8 be written as Either x+10=2x+8 or x+10=-(2x+8) ? and then this should be solved? Please help on this one.

We goth that x is more than or equal to 4. Now, for this range x+10>0 so |x+10| expands only as x+10 (|x+10|=x+10).
_________________

(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) 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.

Hi bunuel, Did not got how u solved option 2.Could you please explain in detail. thanks.

(2) 2x - 3 < 3y - 4 --> \(x<1.5y-\frac{1}{2}\) --> \(x<y+(0.5y-\frac{1}{2})\). Now, since \(y\) is a negative number then \(0.5y-\frac{1}{2}=negative\) so, we have that: \(x<y+negative\). \(y+negative\) is less then \(y\) and if \(x\) is less than \(y+negative\) then it must also be less than \(y\) itself: \(x<y\).

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 also 2 and 3: \(2+3=5=prime\). \(x\) can be 12 or 100. Not sufficient.

(1)+(2) \(x\) can only be 100. Sufficient.

Answer: C.

hey..., can sm1 pls explain how primes of 100 can be 2 and 3?...(2nd last line),,.. thanx..

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 also 2 and 3: \(2+3=5=prime\). \(x\) can be 12 or 100. Not sufficient.

(1)+(2) \(x\) can only be 100. Sufficient.

Answer: C.

hey..., can sm1 pls explain how primes of 100 can be 2 and 3?...(2nd last line),,.. thanx..

It should be: "... distinct primes of 100 are 2 and 5: \(2+5=7=prime\). \(x\) can be 12 or 100".
_________________

Any numbers that I could think of really met the inequasion.

But do you have an algebric way of showing this rule?

Please read the thread. Solutions to ALL the questions are given on the previous pages.

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.

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: \(a=1\), \(b=5\) and \(c=3\), of course it's also possible that for example \(b=even\), for \(a=1\) 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.

Answer: B.

What about the case when all a,b,c are zero. In this case, abc = 0 and 0 is neither odd nor even. Hence 'E'.

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: \(a=1\), \(b=5\) and \(c=3\), of course it's also possible that for example \(b=even\), for \(a=1\) 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.

Answer: B.

What about the case when all a,b,c are zero. In this case, abc = 0 and 0 is neither odd nor even. Hence 'E'.

Welcome to GMAT Club. Below is an answer to your question.

Notice that zero is an even integer. An even number is an integer that is "evenly divisible" by 2, i.e., divisible by 2 without a remainder. Since 0/2=0=integer then zero is even.

Here is my approach : 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

Ans: x & y are odd integer. statement1: x=1,y=1 or x=3,y=1 not sufficient statement2: time of completion is 10.30am-9.45am =45min =3/4Hr; i.e rate and time consumed by both is same. hence, statement B is sufficient.

i didnt find any number which doesnot comply to statement 1. so, sufficient. statement2: x^2-y^2=0 ==> mod(x) = mod(y) ==> x=y & x=-y not sufficient

Ans: A

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

Ans: a,b,c are integers,not in sequence. statement 1: b is half way between a & c. a=2,b=4,c=6 abc=48 even a=2,b=3,c=6 abc=36 even a=3,b=5,c=7 abc=105 odd

statement 1 Not sufficient Statement 2: a=b-c ==> b=a+c ; we cant say that abc will be even or odd because we dont know whether a,b,c is odd or even. Not sufficient on combining both statement also, we cant say anything about abc.

Ans: E.

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

Ans: E ( No explanation)

5. What is the value of integer x? (1) 2x^2+9<9x (2) |x+10|=2x+8 Ans: statement 1: 2x^2+9<9x ==>2x^2-9x+9<0 ==>(2x-1)(x-3)<0 so, 1/2<x<3 or x>3&x<1/2 Not sufficient

Statement 2: |x+10| = 2x+8 if x>10; x+10=2x+8 ==>x=2 but (x>10) if x<10; -x-10=2x+8 ==>x= -6 and (x<10) so, x=-6 Sufficient

Ans B

6. If a and b are integers and ab=2, is a=2? (1) b+3 is not a prime number (2) a>b Ans: ab=2 ==> a=2/b

statement 1: b+3 is not a prime number i.e b+3=1,4,6,8 so, b could be = -2,1,3,5 Not sufficient Statement 2: a>b and ab=2 and a&b are integers..only possible value is a=2 & b=1

Sufficient Ans B

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 Ans: total oranges =76 No of customer =19 how many bought only 1 oranges? statement1: if none bought more than 4, then,max no of oranges bought is 19x4 =76 oranges. in short, each customer has bought 4 oranges. sufficient statement 2: customer can buy any no of oranges totaling 76. 4-4=0 even, 5-3=2 even,and many more. not sufficient

Ans : A

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 Ans: x=0.abcd 7/9=0.777777

statement 1: a+b>14 (a,b) 7,8) ,(7,9),(8,9),(8,8),(9,9) x=0.abcd ; replacing the value of a&b x=0.78cd x=0.79cd x=0.89cd x=0.88cd x=0.99cd all are greater then 0.77777 hence Sufficient

statment 2: a-c>6 (a,c): (9,2) (7,0) and many more x=0.92cd is >0.7777 ok x=0.70cd is <0.7777 not ok

Not sufficient

Ans A

9. If x and y are negative numbers, is x<y? (1) 3x + 4 < 2y + 3 (2) 2x - 3 < 3y - 4 Ans: x,y <0 statement 1: 3x+4<2y+3 ==>3x-2y+1<0 not sufficient

statement 2: 2x-3<3y-4 ==> 2x-3y +1<0 not sufficient

on combining both statement and solving for x& y x< -1/5 & y< 1/5 so, y>x for interval (-1/5 to 1/5) since both are -ve so interval should be (-1/5 to 0) and y=x for (-infinity to -1/5) Not sufficient

Ans E

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. Ans: f(10,x)=11 ==> (10+x)/GCF(10,x) =11 ==>x = GCF(10,x)-10

Statement 1: x could be =1,4,9,16,25.. GCF of (10,1) , (10,4),(10,9) will be different. Not sufficient

Statement 2: x= 2 , no of factor 2 (1&2) ok x= 4 , no of factor 3 (1,2,4) ok x= 10 , no of factor 4 (1,2,5,10) not ok not sufficient

on combining I & II we can get value like 1,4,25 which satisfy both the statement but no unique value of x can be found.

Ans E

11. If x and y are integers, is x a positive integer? (1) x*|y| is a prime number. (2) x*|y| is non-negative integer. Ans: statement 1: x*|y| is prime no no information about +ve or -ve no. Not sufficient

Satement 2: for x*|y| has to non-ve integer both x& y has to -ve or +ve simultaneously any value inside mode is always positive. mode(y) = positive to make x*|y| +ve, X has to be positive.

hence sufficient. Ans B

12. If 6a=3b=7c, what is the value of a+b+c? (1) ac=6b (2) 5b=8a+4c Ans: 6a=3b=7c= k a=k/6 b=k/3 c=k/7 a+b+c = (k/6)+(k/3)+k/7) if we can find the value of K, we wil have our answer.

(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.

(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.

\((x-\frac{3}{2})\) < 0 , (x-3)>0 => x < \((\frac{3}{2})\) , x > 3

and in that case x can have infinite values.

and if that is the case Stmt-1 alone would not be sufficient.

The second case of \((x-\frac{3}{2})<0\) and \((x-3)>0\) is not possible. This condition leads to \(x<\frac{3}{2}\) and \(x>3\), but \(x\) can not be simultaneously less than \(\frac{3}{2}\) (to make \(x-\frac{3}{2}\) negative) and more than 3 (to make \(x-3\) positive).