The Discreet Charm of the DS

2. Is \(xy \le \frac{1}{2}\)?

(1) \(x^2+y^2=1\).

Recall that \((x-y)^2 \ge 0\) (since the square of any number is always non-negative). If we expand this, we get \(x^2-2xy+y^2 \ge 0\). Given that \(x^2+y^2=1\), we can substitute this into our equation, resulting in \(1-2xy \ge 0\). From this, we can deduce that \(xy \le \frac{1}{2}\). Sufficient.

(2) \(x^2-y^2=0\).

If we rearrange this and take the square root of both sides, we get \(|x|=|y|\). However, this does not provide us with enough information to answer whether \(xy \le \frac{1}{2}\). Not sufficient.

Answer: A.

5. What is the value of integer \(x\)?

(1) \(2x^2+9 < 9x\).

First, rearrange the inequality: \(2x^2 - 9x + 9 < 0\). Now, factor the quadratic: \((2x - 3)(x - 3) < 0\). The roots are \(\frac{3}{2}\) and 3. Since the inequality sign is "\( < \)", the solution lies between the roots: \(1.5 < x < 3\). The only integer in this range is 2, so \(x=2\). Sufficient.

(2) \(|x+10|=2x+8\).

The left-hand side (LHS) is an absolute value, which is always non-negative. Thus, the right-hand side (RHS) must also be non-negative: \(2x+8 \ge 0\), which gives us \(x \ge -4\). For this range, \(x+10\) is positive, so \(|x+10|=x+10\). Therefore, \(|x+10|=2x+8\) can be rewritten as \(x+10=2x+8\). Solving for \(x\) gives \(x=2\). Sufficient.

Answer: D.

Check this for more on solving inequalities like the one in the first statement:
https://gmatclub.com/forum/x2-4x-94661.html#p731476
https://gmatclub.com/forum/inequalities ... 91482.html
https://gmatclub.com/forum/everything-i ... me#p868863
https://gmatclub.com/forum/xy-plane-714 ... ic#p841486

Hope it helps.

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

(1) \(3x + 4 < 2y + 3\).

Re-arrange to: \(3x < 2y-1\).

Express \(2y\) as \(3y -y\) to get: \(3x < 3y - y - 1\), which can also be written as: \(3x < 3y - (y + 1)\).

Divide each term by 3: \(x < y -\frac{y+1}{3}\).

Now, \(\frac{y+1}{3}\) can be positive if \(y > -1\), and negative if \(y < -1\).

When \(\frac{y+1}{3}\) is positive, \(y -\frac{y+1}{3}\) will be less than \(y\), so we'd have: \(x < y -\frac{y+1}{3} < y\).

When \(\frac{y+1}{3}\) is negative, \(y -\frac{y+1}{3}\) will be greater than y, so we'd have: \(y < y -\frac{y+1}{3}\). In this case, \(x\) could be between \(y\) and \(y -\frac{y+1}{3}\), thus being more than \(y\): \(y < x < y -\frac{y+1}{3}\), as well as to the left of \(y\), thus being less than \(y\): \(x < y < y -\frac{y+1}{3}\).

Not sufficient.

Alternatively, after getting \(3x < 2y-1\), we could plug in numbers. If \(x\) is a very small number, for instance -100, and \(y\) is -1, then \(x < y\) and the answer is YES. However, if \(x=-2\) and \(y=-2\), then \(x=y\) and the answer is NO.

(2) \(2x - 3 < 3y - 4\).

Re-arrange to: \(2x < 3y - 1\).

Express \(3y\) as \(2y + y\) to get: \(2x < 2y + y - 1\).

Divide each term by 2: \(x < y + \frac{y-1}{2}\).

Since \(y\) is negative, \(\frac{y-1}{2}\) will also be negative, thus \(y + \frac{y-1}{2}\) will be less than \(y\). Therefore, we have \(x < y + \frac{y-1}{2} < y\).

Sufficient.


Answer: B

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 9:45 am. If both \(x\) and \(y\) are odd integers, does \(x\) equal \(y\)?

When Bonnie and Clyde work together, they complete the painting of the car in \(\frac{xy}{x+y}\) hours (the sum of the rates equals the combined rate, which is the reciprocal of the total time: \(\frac{1}{x}+\frac{1}{y}=\frac{1}{T}\). From this, we can deduce that \(T=\frac{xy}{x+y}\)). Now, if \(x=y\), then the total time would be: \(\frac{x^2}{2x}=\frac{x}{2}\). Since \(x\) is odd, this time would be \(\frac{odd}{2}\): 0.5 hours, 1.5 hours, 2.5 hours, ...

(1) \(x^2+y^2 \lt 12\).

It's possible that \(x\) and \(y\) are odd and equal to each other if \(x=y=1\), but it's also possible that \(x=1\) and \(y=3\) (or vice versa). This information is not sufficient.

(2) Bonnie and Clyde finish painting the car together at 10:30 am.

They complete the job in \(\frac{3}{4}\) of an hour (45 minutes). Since this time is not \(\frac{odd}{2}\), \(x\) and \(y\) cannot be equal. This information is sufficient.

Answer: B.

12. If \(6a=3b=7c\), what is the value of \(a+b+c\)?

Given: \(6a = 3b = 7c\). The least common multiple of 6, 3, and 7 is 42, so we can write: \(6a = 3b = 7c = 42x\), for some number \(x\). Thus, \(a = 7x\), \(b = 14x\), and \(c = 6x\).

(1) \(ac=6b\)

Not sufficient.

(2) \(5b=8a+4c\)

As \(x = 0\), we find that \(a = b = c = 0\) and \(a + b + c = 0\).

Sufficient.

Answer: B.

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

In order for the product of the integers to be even, at least one of them must be even.

(1) \(b\) is halfway between \(a\) and \(c\).

On the GMAT, we often see such statements and they can ALWAYS be expressed algebraically as \(b=\frac{a+c}{2}\). But, does this mean that at least one of them is even? Not necessarily: for example, if \(a=1\), \(b=3\) and \(c=5\), they are all odd. It's also possible that for instance \(b= 4 = even\) when \(a=1\) and \(c=7\). Not sufficient.

(2) \(a = b - c\).

Re-arrange: \(a+c=b\). Since the sum of two odd integers cannot be odd, the case of three odd numbers is ruled out. Hence, at least one of them has to be even. Sufficient.


Answer: B

6. If \(a\) and \(b\) are integers and \(ab=2\), is \(a=2\)?

Note that we are not given that \(a\) and \(b\) are positive integers.

The possible integer pairs for \((a, b)\) that satisfy \(ab=2\) are: (1, 2), (-1, -2), (2, 1), and (-2, -1). Essentially, we are examining whether we have the third case, where \(a=2\) and \(b=1\).

(1) \(b+3\) is not a prime number.

This statement eliminates the 1st and 4th cases. However, the 2nd and 3rd cases still remain, so it is not sufficient to determine whether \(a=2\).

(2) \(a > b\).

This statement also eliminates the 1st and 4th options. Not sufficient.

(1)+(2) When combining both statements, there are still two possibilities: (-1, -2) and (2, 1). Not sufficient.

Answer: E.

4. How many integers in a set of 5 consecutive positive integers are divisible by 4?

(1) The median of the set is odd.

When a set has an odd number of terms, the median is simply the middle term. In this case, our set of 5 consecutive numbers follows the pattern: {Odd, Even, Odd, Even, Odd}. Among two consecutive even integers, only one will be a multiple of 4. This information is sufficient to determine that there is exactly one integer in the set of 5 consecutive positive integers that is divisible by 4.

(2) The average (arithmetic mean) of the set is a prime number.

In any evenly spaced set, the arithmetic mean (average) is equal to the median. Thus, we have mean = median = prime. Since it's not possible for the median to be equal to the even prime number 2 (in this case, not all 5 numbers will be positive), the median must be an odd prime. This results in the same pattern as the first statement: {Odd, Even, Odd, Even, Odd}. This statement is also sufficient.

Answer: D.

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, \(\frac{7}{9}\) is a recurring decimal and can be expressed as \(0.\overline{7}\) (a bar over a sequence of digits in a decimal indicates that the sequence repeats indefinitely, so \(0.\overline{7}=0.777...\)).

(1) \(a+b \gt 14\).

The least value of \(a\) that satisfies this condition is 6 (since \(6+9=15 \gt 14\)), so in this case, \(x=0.69d \lt 0.\overline{7}\). However, \(a=7\) and \(b=9\) is also possible, and in that case \(x=0.79d \gt 0.\overline{7}\). Not sufficient.

(2) \(a-c \gt 6\).

The least value of \(a\) that satisfies this condition is 7 (since \(7-0=7 \gt 6\)), but we don't know the value of \(b\). Not sufficient.

(1)+(2) The least value of \(a\) that satisfies both conditions is 7, and in this case, from (1), the least value of \(b\) is 8 (since \(7+8=15 \gt 14\)). Hence, the least value of \(x=0.78d \gt 0.\overline{7}\). Sufficient.


Answer: C

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 prime, it follows that \(x * |y|\) is positive. Thus, \(x\) must be positive. Sufficient.

(2) \(x*|y|\) is a non-negative integer.

Observe that we are told \(x * |y|\) is non-negative, not necessarily positive. Hence, \(x\) could be positive or zero. Not sufficient.

Answer: A.

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

Answer D.

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.

So we get only one value X = 2
Sufficient

Answer D

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.

Sufficient

Answer D

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

Answer C

Hey Bunuel can I ask a question for 12?

We know 6a=3b

And for statement one:

ac =6b. Can't 6b =12a

Then it becomes ac=12a ==> c=12. I know it's wrong since if a is 0 then they will be equal regardless, but can you explain why what I did was wrong?

Also,

6a = 3b = 7c

Can we say a/b= 1/2, b/c = 7/3, and a/c = 7/6
a) ac = 6b, therefore c = 6b/a
substituting this in b/c => b / (6b/a) = 7/3 => a =14, b=28, c = 12

Isnt A also sufficient? Am I ignoring something?

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.



Your doubt is partially addressed above, though there is another thing: from 6a = 3b you can not write a/b=1/2 because b can be zero and we can not divide by zero. The same for other ratios you wrote.

Hope it's clear.