Test: Problem Solving- 1 - GRE MCQ

Before figuring out how many balls you have of each integer value, consider what the question is asking: the “interquartile range” of a group of 100 integers. To find this range, split the 100 integers into two groups, a lower 50 and an upper 50. Then find the median of each of those groups. The median of the lower group is the first quartile (Q1), while the median of the upper group is the third quartile (Q3). Finally, Q3-Q1 is the interquartile range.
The median of a group of 50 integers is the average (arithmetic mean) of the 25th and the 26th integers when ordered from smallest to largest. Out of the ordered list of 100 integers from smallest to largest, then, find #25 and #26 and average them to get the first quartile. Likewise, find #75 and #76 and average them to get the third quartile. Then perform the subtraction.
Each ball has an integer value painted on the side — either 1, 2, 3, 4, 5, 6, 7, or 8. Figure out how many balls there are for each integer by applying the given formula, starting with the lowest integer in the list (1) and going up from there.
Number of balls labeled number 1 = 18 - (1 - 4)2 = 18 - (-3)2 = 18 - 9 = 9 balls. These represent balls #1 through #9.
Number of balls labeled number 2 = 18 - (2 - 4)2 = 18 - (-2)2 = 18 - 4 = 14 balls, representing balls #10 through #23.
Be careful when counting; the 14th ball is #23, not #24, because #10 is the first, #11 is the second, and so on.
Number of balls labeled number 3 = 18 - (3 - 4)2 = 18 - (-1)2 = 18 - 1 = 17 balls, representing #24 through #40.
At this point, you can tell that balls #25 and #26 both have a 3 on them. So the first quartile Q1 is the average of 3 and 3, namely 3. Now keep going!
Number of balls labeled number 4 = 18 - (4 - 4)2 = 18 - (0)2 = 18 balls, representing balls #41 through #58.
Number of balls labeled number 5 = 18 - (5 - 4)2 = 18 - (1)2 = 18 - 1 = 17 balls, representing #59 through #75.
You can stop here. Ball #75 has a 5 on it (in fact, the last 5), while ball #76 must have a 6 on it (since 6 is the next integer in the list). Thus, the third quartile Q1 is the average of 5 and 6, or 5.5. Notice that you have to count carefully — if you are off by even just one either way, you’ll get a different number for the third quartile.
Finally, Q3-Q1 = 5.5 - 3 = 2.5, the interquartile range of this list of integers

A prime number greater than 5 can have only the following four units digits: 1, 3, 7, or 9.

If the units digit of p is 1 then the units digit of 2p+1 would be 3, which is a possible units digit for a prime. For example consider p=11=prime --> 2p+1=23=prime;

If the units digit of p is 3 then the units digit of 2p+1 would be 7, which is a possible units digit for a prime. For example consider p=23=prime --> 2p+1=47=prime;

If the units digit of p is 7 then the units digit of 2p+1 would be 5, which is NOT a possible units digit for a prime;

If the units digit of p is 9 then the units digit of 2p+1 would be 9, which is a possible units digit for a prime. For example consider p=29=prime --> 2p+1=59=prime.

The product of all the possible units digits of Sophie Germain primes greater than 5 is 1*3*9=27.

5 < Purple < 11
 

88000=(8)(11)(10³)=(2³)(11)(2³)(5³)=(1)(5³)(2⁶)(11)=(1)(5³)(8²)(11)

Amount saved per unit = $x
=> Amount saved for n units = $(nx)
Extra cost per unit per day = $y
=> Extra cost for n units per day = $(ny)
Let the required number of days in storage be d
=> Extra cost for n units in d days = $(nyd)
Thus, for storage costs to remain less than or be equal to the savings on the excess units, we have:

nyd ≤ nx
=>d ≤ x/y
Thus, maximum number of days = x/y

Now the question is interesting as they have written the quotient as a decimal number. Now, lets understand that using a simple example first
20 when divided by 4 gives 5 as quotient, that's straight forward right.
20 when divided by 3 gives 6 as quotient and 2 as remainder, that's easy too
Now, 20 when divided by 3 gives 6.67 (approx.) as quotient. Now this is what they mean when they say that x/y quotient is 15.15 in option 1.
Lets go bit deeper into this, this means that 20 when divided by 3 gives 0.67 * 3 as remainder, which is nothing but 2
So, 0.67*3 will be the remainder

Now, with that context lets attempt the problem.

So, I and II are correct
So, Answer will be D

So we have a right triangle, with a hypotenuse of z, the long leg as y, and the short leg as x.

This has to satisfy 

xy = 2

Notice that if x=y, making the triangle an isosceles triangle, then 
But we know this can't be true. y is the longer leg.
This means we need to make x smaller than √2 and y bigger than √2. The only answer that gives us 

I. ac=b2
This is not necessarily true. If A & C are non zero integers and B=0 then this is not true- OUT
II. b=0
If a=b=c=1 then, the above is false.-OUT
III. ac=1
Similarly from statement 1 a & c could be any infinite value. OUT

Therefore, all the above statements are not necessarily true.

APPROACH 1: Plug in each answer choice to see which value does NOT satisfy the equation (slow but...)
For example, C) x = 3
We get: (3² + 3 - 20)² - 2(3² + 3 - 20) - 63 = 17
Evaluate: (-8)² - 2(-8) - 63 = 17
Simplify: 64 - 16 - 63 = 17
Works! So, x = 3 is a valid solution

Try D) x = 4
We get: (4² + 4 - 20)² - 2(4² + 4 - 20) - 63 = 17
Evaluate: (0)² - 2(0) - 63 = 17
Simplify: -63 = 17
Works! So, x = 4 is NOT a valid solution

GIVEN: 2^8x = 640000 
Raise both sides to the power of 

Simplify to get: 

Raise both sides to the power of 

Simplify to get: 
Divide both sides by 

Take |x+4| = y, we will get
y²−10y−24 = 0
(y-12)(y+2) = 0
y = 12 / -2
|x+4| = 12 / -2
Here, mode value can never be -ve. So |x+4| = 12
x + 4 = 12 / -12
x = 8 / -16
Sum = -8
 

225=5²∗3²  and 216=2³∗3³
Since 225 and 216 are both divisors of k, it must be that k is a multiple of the LCM of 225 and 216

=> k = Multiple of 5²∗3³∗2³
Since k=2^a×3^b×5^c, and we need the minimum values of a,b,c, we take:

P (getting a 3 = 1/6
P (NOT getting a 3= 5/6
P (getting a Head) = 1/2
P (NOT getting a Head) = 1/2
P (getting a 3 OR a Head) = P (getting a 3) + P (getting a Head) - P (both occur together) =

n=4p
Since p is a prime number greater than 2, then p is odd.
Also since p is prime the only factors of p are p and 1.

factors of 4p are: 1,2,4,p,2p,4p
Therefore
2,4,2p,4p are all even factors
The fact that p is not 2 and not 1, tells us 2p is not 4 and 4p is not 1 correspondingly. So they are all distinct.

Final Answer: 4

Same tankful resulted in car driving 462 – 336 = 126 miles more on the highway compared to the city.

It is also stated that on the highway the car traveled 6 more miles per gallon. From this we can calculate the tankful as 126 miles / (6 miles/gallons) = 21 gallons.

Number of miles in the city using tankful is 336/21 = 16

P.S.: To double check: the number of miles on highway using tankful is 462/21 = 22 which is 6 more.