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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.

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.

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.

In the 1st condition if we take ac = 6b and hence multiply the the given expression 6a=3b=7c by 2 we get 12a=6b=14c.

Now substituting 12a=ac=14c

I get 12a=ac --> c = 12 and similarly a = 14 and hence b= 28.

Can you please point where am I going wrong?

Regards

ac=12a (here you cannot reduce by a and write c=12 as you exclude possibility of a=0) --> a(c-12)=0 --> either a=0 OR c=12. So, we get either a=b=c=0 or a=14, b=28 and c=12.

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.

In the 1st condition if we take ac = 6b and hence multiply the the given expression 6a=3b=7c by 2 we get 12a=6b=14c.

Now substituting 12a=ac=14c

I get 12a=ac --> c = 12 and similarly a = 14 and hence b= 28.

Can you please point where am I going wrong?

Regards

ac=12a (here you can not reduce by a and write c=12 as you exclude possibility of a=0) --> a(c-12)=0 --> either a=0 OR c=12. So, we get either a=b=c=0 or a=14, b=28 and c=12.

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.

Answer: B.

Hi Bunuel is there any reason to discard zero from statement #2, i mean if you consider that a could be zero b=c and then axbxc would be zero giving an additional result. It never says that these are positive integres. Thanks in advance i could be wrong it wouldn't be the first time

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.

Answer: B.

Hi Bunuel is there any reason to discard zero from statement #2, i mean if you consider that a could be zero b=c and then axbxc would be zero giving an additional result. It never says that these are positive integres. Thanks in advance i could be wrong it wouldn't be the first time

abc could be zero, but zero is even, so the answer is still YES.
_________________

(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.

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.

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.

8. If x=0.abcd, where a, b, c and d are digits from 0 to 9, inclusive, is x>7/9?

First of all 7/9 is a recurring decimal =0.77(7). For more on converting Converting Decimals to Fractions see: math-number-theory-88376.html

(1) a+b>14 --> the least value of a is 6 (6+9=15>14), so in this case x=0.69d<0.77(7) but a=7 and b=9 is also possible, and in this case x=0.79d>0.77(7). Not sufficient.

(2) a-c>6 --> the least value of a is 7 (7-0=7>6), but we don't know the value of b. Not sufficient.

(1)+(2) The least value of a is 7 and in this case from (1) least value of b is 8 (7+8=15>14), hence the least value of x=0.78d>0.77(7). Sufficient.

Answer: C.

Why have you taken value on rhs till second place of decimal. Thereby concentrating only on a and b? Please explain

(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.

How did you decide which approach to use in A, one could have solved like the way you did in B 3X < 2y - 1 or x < .67y - 0.33 , so x < y + negative , so x is more negative, so x < Y.

8. If x=0.abcd, where a, b, c and d are digits from 0 to 9, inclusive, is x>7/9?

First of all 7/9 is a recurring decimal =0.77(7). For more on converting Converting Decimals to Fractions see: math-number-theory-88376.html

(1) a+b>14 --> the least value of a is 6 (6+9=15>14), so in this case x=0.69d<0.77(7) but a=7 and b=9 is also possible, and in this case x=0.79d>0.77(7). Not sufficient.

(2) a-c>6 --> the least value of a is 7 (7-0=7>6), but we don't know the value of b. Not sufficient.

(1)+(2) The least value of a is 7 and in this case from (1) least value of b is 8 (7+8=15>14), hence the least value of x=0.78d>0.77(7). Sufficient.

Answer: C.

Why have you taken value on rhs till second place of decimal. Thereby concentrating only on a and b? Please explain

Please elaborate your question. Where did I do that?
_________________

(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.

How did you decide which approach to use in A, one could have solved like the way you did in B 3X < 2y - 1 or x < .67y - 0.33 , so x < y + negative , so x is more negative, so x < Y.

Is this is a 700 question ?

The approach used for the second statement does not work for the first:

(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.

I don't understand something. Sometimes, abs of "x" will be checked to equal -x if x is negative, or x if x is positive. So why isn't there a check for x-10 and -(x-10)?

(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.

I don't understand something. Sometimes, abs of "x" will be checked to equal -x if x is negative, or x if x is positive. So why isn't there a check for x-10 and -(x-10)?

There can be more than one way to solve a question.
_________________