GMAT Classic Mock Test - 8 - GMAT MCQ

Key Data from the Question Stem

1. a ≤ b ≤ c
2. a, b, and c are positive integers.
3. Average of the three integers = 20
4. Sum of all the three integers = 60
5. Median = b = a + 11

Check for the possible values of c

Theoretically, the least value of c is when c = b.
Therefore, a + (a + 11) + (a + 11) = 60 (b and c are equal and b, the median, is a + 11)
Or 3a = 38 or a = 12.66
So, b = c = 12.66 + 11 = 23.66

However, we know that these numbers are all integers.
Therefore, a, b, and c cannot take these values.
So, the least value for c with this constraint is NOT likely to be when c = b.

Let us increment c by 1. Let c = (b + 1)
In this scenario, a + (a + 11) + (a + 12) = 60
Or 3a = 37. The value of the numbers is not an integer in this scenario as well.

Let us increment c again by 1. i.e., c = b + 2
Now, a + (a + 11) + (a + 13) = 60
Or 3a = 36 or a = 12.
If a = 12, b = 23 and c = 25.
The least value for c that satisfies all these conditions is 25.

Choice C is the correct answer.

What will be the form of such 4-digit numbers?

The 4-digit numbers should contain the block 25.
The required 4-digit numbers will be of the form:
a. 25_ _
b. _ 25 _
c. _ _ 25

What is the test of divisibility by 75?

If a number is divisible by 75, then it will be divisible by 25 and 3.

Count the number of 4-digit numbers for three possiblities

a. Numbers of the form 25 _ _ that are divisible by 75
A number of the form 25_ _ is divisible by 25 if its rightmost 2 digits are 00, 25, 50, or 75.

Check which of these numbers are also multiples of 3
Only one number, 2550 satisfies the condition.

b. Numbers of the form _ 25 _ that are divisible by 75
A number of the form _ 25 _ is divisible by 25 if its unit digit is 0.
The 4-digit number will be of the form _ 250

What options exist for the left most digit so that the number is also divisible by 3?
The sum of the right most 3 digits of the number = 2 + 5 + 0 = 7.
If the first digit is 2 or 5 or 8, the sum of the 4 digits will be divisible by 3.
There are three 4-digit numbers that match the form _ 25 _ and are divisible by 75.

c. Numbers of the form _ _ 25 that are divisible by 75
All numbesr of the form _ _ 25 is divisible by 25.

What options exist for the first 2 digits so that the number is also divisible by 3?
We already have a 2 and 5 whose sum is 7. 7 is a multiple of 3 plus 1.
We have a (3k + 1) with us. If we add a (3m + 2), the sum will be 3(k + m) + 1 + 2 = 3(k + m) + 3, which is divisible by 3.
The least 2 digit number that is of the form (3m + 2) is 11.
For example, if 11 takes the 1st 2 places, the number is divisible by 3

11 is not the only such number.
All 2-digit numbers of the form (3m + 2) will work

How many are there? The largest 2-digit number that is of the form 3m + 2 is 98.
And all of these numbers are in arithmetic progression with a common difference of 3
So, apply the arithmetic progression formula to compute the nth term: 98 = 11 + (n - 1)3
3(n - 1) = 87
(n - 1) = 29
Or n = 30
30 such 4-digit numbers exist

Add the count of all three possiblities

1 + 3 + 30 = 34
34 such 4-digit numbers exist

Choice C is the correct answer.

Assign Variables | Frame Equations

Let ‘x’ and ‘y’ be the number of tickets sold at $11 and $7 respectively.
Then, total number of tickets sold = x + y

Collection by selling 'x' $11 tickets = 11x
Collection by selling 'y' $7 tickets = 7y
Total collection by selling (x + y) tickets = 11x + 7y

A fourth of all those who bought a ticket also spent $4 each on refreshments at the movie hall.
i.e., 1414(x + y) spent $4 on refreshments.
∴ Collections from sale of refreshments = 1414(x + y) × 4 = (x + y)

Total collection from tickets and refreshments = $124
i.e., 11x + 7y + (x + y) = 124
Or 12x + 8y = 124
Divide the equation by 4: 3x + 2y = 31

Solve 3x + 2y = 31 for x and y using these 2 facts

Fact 1: x and y are positive integers.
Fact 2: 1414 (x + y) is an integer because a fourth of total tickets sold should be an integer. i.e., (x + y) should be a multiple of 4.

Let us list down the different possibilities that satisfy the equation and fact 1

Of all the different positive integer values of x and y that satisfy the equation, the only combination (x, y) = (7, 5) is divisible by 4

So, the movie hall sold five $7 tickets.

Choice E is the correct answer.

Sides AB and CD are parallel to x-axis.
So, AD and BC will be parallel to y-axis.

The x-coordinates take values from 12 to 28.
We can draw lines parallel to y-axis corresponding to each of these values.
So, we will be able to draw 17 vertical lines.

The y-coordinates take values from 5 to 13.
We can draw lines parallel to x-axis corresponding to each of these values.
So, we will be able to draw 9 horizontal lines.

2 horizontal lines and two vertical lines will form a rectangle
Number of ways of selecting 2 horizontal lines from 9 horizontal lines = 9C₂
9C₂ = 9×8 / 2= 36

Number of ways of selecting 2 veritcal lines from 17 vertical lines = 17C₂
9C₂ = 17×16 / 2 = 136

Number of rectangles that can be formed

Product of the number of ways of selecting 2 horizontal lines and number of ways of selecting two vertical lines
= 36 × 136 = 4896

Choice C is the correct answer.

Key Data

Total dollar value of all gift certificates in the initial configuration : 13 hampers × 12 dollars = $156
Total dollar value of all gift certificates in the new configuration = 12 hampers × 13 dollars = $156
No change in total value of gift certificates.

Susan started with 13 hampers.
So, she has 13 of $3, 13 of $4, and 13 of $5 certificates.

What are the possible combination of certificates in the new hampers?

In the new configuration, $13 can be achieved as follows: {3, 3, 3, 4}, {4, 4, 5}, and {5, 5, 3}
The hampers that contain the configuration {3, 3, 3, 4} will not have $5 certificates.

Assign Variables and Frame Linear Equations

Let Susan make x hampers of {3, 3, 3, 4}, y hampers of {4, 4, 5}, and z hampers of {5, 5, 3}.
The answer to ‘x’ is the answer to the question.

Number of $3 certificates will be 3x + z = 13 ....(1)
Number of $4 certificates will be x + 2y = 13 ....(2)
Number of $5 certificates will be y + 2z = 13 ....(3)

Solve the 3 Equations

Multiply (3) by 2: 2y + 4z = 26
Subtract (2) from (3) × 2
2y + 4z = 26
-x - 2y = -13
------------------
4z – x = 13 ....(4)
------------------

Multiply (1) by 4: 12x + 4z = 52
Subtract (4) and (1) × 4
12x + 4z = 52
-4z + x = -13
----------------
13x = 39
-----------------
Or x = 3

'x' is the number of gift hampers without $5 certificates.
As x = 3, three gift hampers did not contain $5 gift certificates.

Choice D is the correct answer.

Step 1: What are the factors of 36?

1, 2, 3, 4, 6, 9, 12, 18, and 36.
Of these, 1, 2, 3, 4, 6, and 9 are single digit factors and can therefore, be digits of the 3 digit numbers.

Step 2: List Down Possibilities and Count

Possibility 1: Let 9 be one of the 3 digits.
The product of the remaining 2 digits will, therefore, be 4.

Possibility 2: Let 6 be one of the 3 digits.
The product of the remaining 2 digits will, therefore, be 6.

Possibility 3: 4 is one of the three digits
The product of the remaining 2 digits is 9.

Possibility 4: Let 3 be one of the three digits
The product of the remaining 2 digits is 12.

Possibility 5: Let 2 be one of the three digits
The product of the remaining 2 digits is 18.

Possibility 6: Let 1 be one of the three digits
The product of the remaining 2 digits is 36.

Number of such 3-digit positive integers is calculated by adding all the outcomes.
The outcomes from possibilities 4, 5, and 6 should not be counted because they have already been counted in the earlier possibilities.
the total Number of such 3-digit positive integers are 6 + 3 + 3 + 6 + 3 = 21 Numbers

Choice A is the correct answer.

Step 1: List Down Possibilities

The student is required to solve 6 out of 10 questions.
Questions are divided into 2 sections of 5 questions each.
Not more than 4 questions can be selected from any section.

Step 2: List Down Possibilities

Step 3: Count Number of Outcomes for each Possibility and Add

Possibility 1: Section 1: 4 Questions | Section 2: 2 Questions
This can be done in 5C₄ × 5C₂ = 5 × 10 = 50 ways

Possibility 2: Section 2: 3 Questions | Section 2: 3 Questions
This can be done in 5C₃ × 5C₃ = 10 × 10 = 100 ways

Possibility 3: Section 1: 2 Questions | Section 2: 4 Questions
This can be done in 5C₂ × 5C₄ = 10 × 5 = 50 ways

Total number of ways = 50 + 100 + 50 = 200 ways

Choice D is the correct answer.

Key Data

1. 6-digit numbers.
2. Formed using digits {1, 2, 3,..., 9}. Note: Does not include zero.
3. Any digit that appears should appear at least twice.

Examples: Those that satisfy and those that do not

Some 6-digit numbers that satisfy the condition: 555555, 223344, 111999, etc.,
Some 6-digit numbers that do not satisfy the condition: 123456, 123444, 558812, etc.,

List Down Possibilities and Count

Possibility 1: All 6 digits are same
Example: 111111
9 such numbers possible.

Possibility 2: 4 digits show one value and 2 digits show another value. Example: 373777
Step 1: We are selecting 2 digits from 9 numbers. This can be done in 9C₂ ways.
Step 2: For example, if the digits are 3 and 7, either 3 appears 4 times and 7 appears twice or vice versa.
So, there are 2 possibilities.
Step 3: Reordering of 6 digits can be done in 6! / 4! × 2! = 15
Number of such numbers = Product of values obtained in the above 3 steps. i.e., 9C₂ × 2 × 15

= 

Possibility 3: 3 digits show one value and another 3 digits show a second value.
Example: 444777
Step 1: We are selecting 2 digits from 9 numbers. This can be done in 9C₂ 

=  ways
Step 2: Reordering of 6 digits can be done in 6!3!×3!6!3!×3! = 6×5×4×3!3!×3!6×5×4×3!3!×3! = 20 ways
Number of such numbers = 36 × 20 = 720

Possibility 4: 3 different digits, each appearing twice. Example: 234234
Step 1: We are selecting 3 digits from 9 numbers. This can be done in 9C₃ 

= = 84 ways.
Step 2: Reordering of 6 digits can be done in  = 90 ways
Number of such numbers = 84 × 90 = 7560

Total such numbers = (9 + 1080 + 720 + 7560) = 9369

Choice C is the correct answer.

Approach to find the highest power of a composite number that divides n!

Step 1: Prime factorize the divisor, ‘x’ in this case.
Step 2: Compute the highest power of each of the prime factors that divides n!
Step 3: The highest power of x that divides n! is determined by the power of that prime factor which available in the least number.

Key Inference: The value of 'y' will be the highest for such an x whose highest prime factor is the smallest.

Prime Factorize numbers given in the 5 answer options

Option A: 111 = 3 × 37. The highest prime factor of 111 is 37.
Option B: 462 = 2 × 3 × 7 × 11. The highest prime factor of 462 is 11.
Option C: 74 = 2 × 37. The highest prime factor of 74 is 37.
Option D: 33 = 3 × 11. The highest prime factor of 33 is 11.
Option E: 210 = 2 × 3 × 5 × 7. The highest prime factor of 210 is 7.

210 is the number which has the smallest value of the highest prime factor among the 5 given options.
So, the value of y will be highest for 210.

Choice E is the correct answer.

Because square of real numbers cannot be negative, sum of even powers of real numbers can be zero only if each of the terms is zero.

Therefore, (a - 1) = 0, (b - 2) = 0, ..., (j - 10) = 0
Or a = 1, b = 2, c = 3, .... j = 10

b × d × f × h × j = 2 × 4 × 6 × 8 × 10
= 2⁵ × (1 × 2 × 3 × 4 × 5)
= 32 × 120
= 3840

Choice C is the correct answer.

Constraint: All the digits of the number are even. Therefore, the digits of each of the numbers are from the set {0, 2, 4, 6, 8}

Step 1: Compute the total possible numbers

Hundreds place: 4 possible values {2, 4, 6, 8}. Hundreds place cannot be zero.
Tens place and units place: All 5 values possible.
Therefore, total possible 3-digit positive integers such that all the digits are even = 4 × 5 × 5 = 100

Step 2: Compute the value of sum of units and tens digits

Each of the 5 digits is equally likely to appear in the units place.
Therefore 100 / 5 = 20 is the number of times each digit will appear in the units place.
Therefore, value of sum of digits in units place = 20 × (0 + 2 + 4 + 6 + 8 ) = 400.

For the same reason, sum of tens digits = 400
Hence, value of the sum of digits in tens place = 400 × 10 = 4000

Step 3: Compute the value of sum of hundreds digits

Each of the 4 digits is equally likely to appear in the hundreds place.
Therefore 100 / 4 = 25 numbers will begin with each of {2, 4, 6, and 8}
Therefore sum of digits in hundreds place = 25(2 + 4 + 6 + 8) = 500
Value of the sum of digits in 100s place = 500 × 100 = 50,000

Step 4: Compute the sum of such 3-digit numbers

Required Sum = Sum of units place + Sum of tens place + Sum of hundreds place
= 400 + 4000 + 50000 = 54,400

Choice E is the correct answer.

Let N be the given number.
N leaves a remainder 31 when divided by 44, i.e., For N / 44, remainder is 31
So, N + 13 will be divisible by 44.
N leaves a remainder 43 when divided by 56, i.e., For N / 56, remainder is 43
So, N + 13 will be divisible by 56.
N leaves a remainder 19 when divided by 32, i.e., For N / 32, remainder is 19
So, N + 13 will be divisible by 32.
Hence, N + 13 will be divisible by 44, 56, 32. i.e., N + 13 is a multiple of 44, 56, 32.

The least value of (N + 13) is the LCM of 44, 56, and 32.
Find the LCM of 44, 56, 32

Step 1 of Computing LCM: Prime Factorize 44, 56, and 32
44 = 2² × 11
56 = 2³ × 7
32 = 2⁵

Step 2 of Computing LCM: LCM is the product of highest power of all primes.
LCM (44, 56, 32) = 2⁵ × 7 × 11 = 2464
Therefore, N + 13 = 2464
Hence, N = 2464 – 13 = 2451

Choice C is the correct answer.

Key Data
'n' is a positive integer.
Product of the factors of n² is n³.

If the product of the factors of n² = n³, the only factors of n² are 1, n, and n².
So, we can infer that n does not have any factor other than 1 and itself.
Therefore, n is a prime number.

Factors of n³ if n is a prime number are 1, n, n² and n³.
So, the product of the factors of n³ = 1 × n × n² × n³ = n⁶

Choice B is the correct answer.

What are the Constraints?
1. The digits of the 3-digit number should be distinct.
2. The numbers are even numbers.

Possibility 1: 3-digit numbers an with odd number in hundreds place

Numbers such as 104, 308, 510, 724, and 942 meet this criterion.
The hundreds place can be any of the 5 values viz., 1, 3, 5, 7, or 9.
Because the numbers are even, the unit digit has to be even.
The units place can be any of the 5 values viz., 0, 2, 4, 6, or 8
The tens place should be different from the values that appear in the hundreds and units place.
So, it has 10 - 2 = 8 possibilities.
Number of such 3-digit numbers = 5 × 8 × 5 = 200

Possibility 2: 3-digit numbers with even number in hundreds place

Numbers such as 246, 482, 674, and 892 meet this criterion.
The hundreds place can be one of the 4 even numbers other than 0 viz., 2, 4, 6, or 8.
The units place has to be even and should be different from the digit in the hundreds place.
So, 4 possibilities out of the 5 even values are possible.
The tens place should be different from the values that appear in the hundreds and units place.
So, it has 10 - 2 = 8 possibilities.
Number of such 3-digit numbers = 4 × 8 × 4 = 128

Total Number of such 3-digit positive integers = count of step 1 + count of step 2 = 200 + 128 = 328

Choice C is the correct answer.

Given data:
x and y are non-negative integers.
So, both x and y can take 0 or positive integer values.
4x + 7y = 68
⇒ 4x = 68 - 7y
Because x is an integer, 4x is divisible by 4. So, (68 - 7y) is divisible by 4.
68 is divisible by 4. So, 7y should also be divisible by 4.

x and y are non-negative integers.
So, the least possible value for y is 0.
So, 7y = 0. Note: 0 is divisible by 4.
Subsequently, let us plug in multiples of 4 for y till such time x remains non-negative.
When y = 0, x = 17
When y = 4, x = 10
When y = 8, x = 3
When y = 12, x = -4 (x is negative)
For values of y that are multiples of 4 and are greater than 8, x will be negative.

Possible values for x + y:
17 + 0 = 17
10 + 4 = 14
8 + 3 = 11
3 values are possible

Choice B is the correct answer.

Evaluate Statement 1 ALONE

Statement 1: x(1 – x²) < 0
i.e., x - x³ < 0 or x < x³

For what values of x will x < x³ ?
Interval 1: (1 < x < ∞)
x < x³. Here, x is greater than zero. Answer to the question - YES
Interval 2: (-1 < x < 0)
x < x³. Here, x is lesser than zero. Answer to the question - NO

We are not able to get a conclusive answer using Statement 1.
Hence, statement 1 is not sufficient.
Eliminate answer option A and D.

Evaluate Statement 2 ALONE

Statement 2: x(1 - x) < 0
i.e., x – x² < 0 or x < x²

For what values of x will x < x² ?
Interval 1: (1 < x < ∞)
x < x². Here, x is greater than zero. Answer to the question - YES
Interval 2: (-1 < x < 0)
x < x². Here, x is lesser than zero. Answer to the question - NO

We are not able to get a conclusive answer using Statement 2.
Hence, statement 2 is not sufficient.
Eliminate answer option B.

Evaluate Statements TOGETHER

Statements: From Statement 1: x < x³
From Statement 2: x < x²

Both conditions hold good in the following intervals,
Interval 1: (1 < x < ∞)
x < x³ and x < x². Here, x is greater than zero. Answer to the question - YES
Interval 2: (-∞ < x < 0)
x < x³ and x < x². Here, x is lesser than zero. Answer to the question - NO

Despite combining the statements, we are not able to get a conclusive answer.
Eliminate answer option C.

Choice E is the correct answer.

Evaluate Statement 1 ALONE

Statement 1: Their median is 6.
So, a, b, 6, d, e are the 5 numbers. Therefore, a + b + d + e = 19

Because d and e are greater than 6, the following possibilities exist : (d, e) could be (7, 8), (7, 9), and (8, 9)

Possibility 1: If (d, e) = (7, 8): a + b + 7 + 8 = 19 or a + b = 4
The only value that (a, b) can take is (1, 3)
Range of the 5 numbers is 8 - 1 = 7

Possibility 2: If (d, e) = (7,9): a + b + 7 + 9 = 19 or a + b = 3
The only value that (a, b) can take is (1, 2).
Range of the 5 numbers is 9 - 1 = 8

Possibility 3: If (d, e) = (8, 9): a + b + 8 + 9 = 19 or a + b = 2
No values of (a, b) that are distinct positive integers will satisfy this case.
So, possibility 3 is infeasible.

We are not able to find a UNIQUE value for the range using Statement 1.
Hence, statement 1 is not sufficient.
Eliminate answer options A and D.

Evaluate Statement 2 ALONE

Statement 2: The average of the 3 largest among the 5 numbers is 7.
c, d, and e are the 3 largest of the 5 numbers.
Therefore, c + d + e = 21 and a + b = 4

Only possible value for (a, b) = (1, 3). So, a = 1.
However, c, d, and e can take different values. Let us list down possibilities.
Possibility 1: c = 5, d = 7, e = 9. Range is 9 - 1 = 8
Possibility 2: c = 6, d = 7, e = 8. Range is 8 - 1 = 7
Possibility 3: c = 4, d = 8, e = 9. Range is 8 - 1 = 7

We are not able to find a UNIQUE value for the range using Statement 2.
Hence, statement 2 is not sufficient.
Eliminate answer option B.

Evaluate Statements TOGETHER

Statements: From Statement 1: Their median is 6.
From Statement 2: The average of the 3 largest among the 5 numbers is 7.

Key inferences: From statement 1: 'c' has to be 6.
From statement 2: (a, b) has to be (1, 3)
So, 1 + 3 + 6 + d + e = 25 or d + e = 15
d > 6 and e > d. The only possible values are d = 7 and e = 8.
Hence, the range is 8 - 1 = 7.

We are able to find a UNIQUE value for the range using the statements together.
Statements together are sufficient.
Eliminate answer option E.

Choice C is the correct answer.

Evaluate Statement 2 ALONE

Statement 2: a + 8, b, and c, in that order are in AP.

Approach : Counter example
Example : a = 1, b = 10, and c = 11. a + 8 = 9
So, 9, 10, 11 are in AP. Range = 11 - 1 = 10
Counter Example : a = 1, b = 11, c = 13. a + 8 = 9
So, 9, 11, 13 are in AP. Range = 13 - 1 = 12
Counter example exists.

We are not able to find a UNIQUE value for the range using Statement 2.
Hence, statement 2 is not sufficient.
Eliminate answer option B.

Evaluate Statements TOGETHER

Statements: Statement 1: 21a = 9b = 7c
Statement 2: a + 8, b, and c in that order are in AP.

Key inferences:
From statement 1: we know a = 3k, b = 7k and c = 9k
From statement 2: we know c - b = b - (a + 8) (because a + 8, b, and c are in AP) ... (1)
Substiute c = 9k, b = 7k and a = 3k in equation (1)
So, 9k - 7k = 7k - (3k + 8)
2k = 4k - 8
or k = 4
Therefore, the range (9k - 3k) = 6k = 6(4) = 24

We are able to find a UNIQUE value for range using the statements together.
Statements together are sufficient.
Eliminate answer option E.

Choice C is the correct answer.

Evaluate Statement 1 ALONE

Statement 1: The number comprises only the digits 0, 1, and 2, each written four times.
Therefore, sum of the digits of the number is 4(0) + 4(1) + 4(2) = 12

Sum of digits is divisible by 3. So, the number is divisible by 3.
If the number is a perfect square, if it is divisible by 3 it will also be divisible by 3² = 9.
However, the sum of the digits of the number is 12, which is not divisible by 9.
So, the number is divisible by a prime but not its square.
So, the given number is not a perfect square.

We are able to answer the question with a DEFNITE NO using Statement 1.
Hence, statement 1 alone is sufficient.
Eliminate answer option B, C, and E.

Evaluate Statement 2 ALONE

Statement 2: The sum of the digits of the twelve digit number is 12.
The analysis is the same as that for statement 1.

The sum of the digits is given as 12. So, the number is divisible by 3.
If the number is a perfect square, if it is divisible by 3 it will also be divisible by 3² = 9.
However, the sum of the digits of the number is 12, which is not divisible by 9.
So, the number is divisible by a prime but not its square.
So, the given number is not a perfect square.

We are able to answer the question with a DEFNITE NO using Statement 2.
Hence, statement 2 alone is sufficient.
Eliminate answer option A.

Statements are independently sufficient.

Choice D is the correct answer.

Evaluate Statement 1 ALONE

Statement 1: 13a = 43b
a : b :: 43 : 13
So, a = 43x and b = 13x
a + b = 43x + 13x = 56x
56 is not prime. Therefore, 56x cannot be prime.

We are able to answer the question with a DEFNITE NO.
Hence, statement 1 alone is sufficient.
Eliminate answer option B, C, and E.

Evaluate Statement 2 ALONE

Statement 2: 8a = 15b
a : b :: 15 : 8
So, a = 15x and b = 8x
a + b = 15x + 8x = 23x
23 is prime.
If x is 1, a + b will be prime. For other values of x, a + b will not be prime.

We are not able to answer the question with a DEFNITE Yes or No.
Hence, statement 2 alone is not sufficient.
Eliminate answer option D.

Statement 1 alone is sufficient. Statement 2 is NOT sufficient.

Choice A is the correct answer.

Evaluate Statement 1 ALONE

Statement 1: The curve cuts the Y-axis at -20.
(0, -20) is a point on the curve. The point will satisfy y = (x - p)(x - q)
-20 = (0 - p)(0 - q)
pq = -20

Approach: Counter example
Example: p = 10, q = -2; pq = -20
p + q = 10 + (-2) = 8
Answer to the question is YES.

Counter Example: p = -10, q = 2; pq = -20
p + q = -10 + 2 = -8
Answer to the question is NO.
Counter Example exists.

We are not able to answer the question with a DEFNITE Yes or No.
Hence, statement 1 alone is not sufficient.
Eliminate answer option A and D.

 Evaluate Statement 2 ALONE

Statement 2: Minimum value of y is -36.
We know that y = (x - p) (x - q), where p and q are the roots of the equation.

Approach: Counter example
Example: Both p and q are positive for y_min = -36

Now, the sum of p and q is positive.
Answer to the question is YES.

Counter Example: Both p and q are negative for y_min = -36

Now, the sum of p and q is negative.
Answer to the question is NO.
Counter Example exists.

We are not able to answer the question with a DEFNITE Yes or No.
Hence, statement 2 alone is not sufficient.
Eliminate answer option B.

Evaluate Statements TOGETHER

Statements: From Statement 1: The curve cuts the Y-axis at -20.
From Statement 2: Minimum value of y is -36.

From the statements: pq = -20 and y_min = -36
In a parabola, the minimum value of y happens when x = (r₁+r₂)/2, where r₁ and r₂ are the roots of the equation.

Example: p = -10, q = 2
y = (x + 10)(x – 2) ....(1)
y will be minimum when x = (p + q)2(p + q)2 = (−10 + 2)2(−10 + 2)2 = -4
Substitute x = -4 in equation (1)
y = (-4 + 10)(-4 - 2) = 6 × -6 = -36
Now, p + q = -10 + 2 = -8
Answer to the question is NO.

Counter Example:p = 10, q = -2
y = (x - 10) (x + 2) ....(2)
y will be minimum when x = (p + q)2(p + q)2 = (10 - 2)2(10 - 2)2 = 4
Substitute x = 4 in equation (2) y = (4 - 10)(4 + 2) = -6 × 6 = -36
Now, p + q = 10 – 2 = 8
Answer to the question is YES.

Despite combining both the statements, we are not able to answer the question with a definite Yes or No.
Statement TOGETHER are NOT sufficient.
Eliminate answer option C.

Choice E is the correct answer.

Evaluate Statement 1 ALONE

Statement 1: BC = 2BD

ABC is a right triangle. So are BDC and BDA.
In right triangle BDC, BC is hypotenuse.
If BC = 

= √3 BD
So, the sides are in the ratio 1 : √3 : 2

In ΔBDC, angles opposite BD, CD, and BC will be 30°, 60° and 90°
So, we can deduce that ∠BAD = 60° and ∠DBA = 30°

ΔBDA is also a 30 – 60 – 90 triangle.
If BD = a, CD = √3 a and AD = a√3
CD : AD = √3 a : a√3
CD : AD = 3 : 1

We are able to find a UNIQUE answer to the question.
Hence, statement 1 alone is sufficient.
Eliminate answer option B, C, and E.

Evaluate Statement 2 ALONE

Statement 2: AC = 2AB

If AC = 2 AB, sides of ΔABC will be in the ratio, AB : BC : AC = 1 : √3 : 2

So angles opposite AB, BC, and AC will be 30°, 60° and 90°
So, we can deduce that angle DBC = 60° and angle DBA = 30°

Because ΔBDA and ΔCDB are 30 – 60 – 90 triangles.
If BD = a, CD = √3 a and AD = a√3
CD : AD = √3 a : a√3
CD : AD = 3 : 1

We are able to find a UNIQUE answer to the question.
Hence, statement 2 alone is also sufficient.
Eliminate answer option A.

Statements are independently sufficient.

Choice D is the correct answer.

Evaluate Statement 1 ALONE

Statement 1: m² – n² = 105
m and n are positive integers.
m² – n² can be written as (m – n)(m + n). Therefore, (m – n)(m + n) = 105
Both (m – n) and (m + n) have to be positive integers. (Why? Because (m + n) is positive and the product of (m + n) and (m - n) is positive.)
The ways of expressing 105 as a product of 2 positive integers is listed in the table given below.

(m - n)(m + n)1105335521715

So, (m + n) could be 105, 35, 21, or 15.
We are not able to answer the question with a UNIQUE value.
Hence, statement 1 alone is not sufficient.
Eliminate answer options A and D.

Evaluate Statement 2 ALONE

Statement 2: Neither m nor n is divisible by 8.
Infinite possibilities exist for this condition.

We are not able to answer the question with a UNIQUE value.
Hence, statement 2 alone is not sufficient.
Eliminate answer option B.

Evaluate Statements TOGETHER

Statement 1: m² – n² = 105
Statement 2: Neither m nor n is divisible by 8.

From statement 1: (m – n) (m + n) = 105 and both (m – n) and (m + n) are positive integers.

(m - n)(m + n)mn110553523351916521138715114

From Statement 2: We can eliminate (19, 16) and (13, 8).
We still have two values for (m, n): (53, 52) and (11, 4)
Therefore, (m + n) could be 105 or 15.

Despite combining both the statements, we are not able to find a UNIQUE value for (m + n).
Statements TOGETHER are NOT sufficient.
Eliminate answer option C.

Choice E is the correct answer.

Evaluate Statement 1 ALONE

Statement 1: Both sets are symmetric about their respective means.

Approach: Counter example
Example: Set A = {1, 2, 3, 4, 5}; Range = 5 – 1 = 4
Set B = {6, 7, 8, 9, 10}; Range = 10 – 6 = 4
Answer to the question is YES.
Counter Example: Set A = {1, 2, 3, 4, 5}; Range = 5 – 1 = 4
Set B = {10, 20, 30, 40, 50}; Range = 50 – 10 = 40
Answer to the question is NO.
Counter Example exists.

We are not able to find a CONCLUSIVE answer to the question.
Hence, statement 1 alone is not sufficient.
Eliminate answer options A and D.

Evaluate Statement 2 ALONE

Statement 2: The median of both the sets is 50.

Approach: Counter example
Example: Set A = {47, 48, 50, 52, 53}; Range = 53 – 47 = 6
Set B = {47, 49, 50, 51, 53}; Range = 53 – 47 = 6
Answer to the question is YES.
Counter Example: Set A = {50, 50, 50}; Range = 50 – 50 = 0
Set B = {40, 50, 60}; Range = 60 – 40 = 20
Answer to the question is NO.
Counter Example exists.

We are not able to find a UNEQUIVOCAL answer to the question.
Hence, statement 2 alone is not sufficient.
Eliminate answer option B.

Evaluate Statements TOGETHER

Statements: From Statement 1: Both sets are symmetric about their respective means.
From Statement 2: The median of both the sets is 50.

Example: Set A = {48, 50, 50, 50, 52}; Range = 4
Set B = {48, 49, 50, 51, 52}; Range = 4 Answer to the question is Yes.
Counter Example:Set A = {47, 50, 53}; Range = 6
Set B = {40, 50, 60}; Range = 10
Answer to the question is NO.
Counter Example exists.

Despite combining both the statements, we are not able to find a DEFINITE answer to the question.
Statement TOGETHER are NOT sufficient.
Eliminate answer option C.

Choice E is the correct answer.

BOTH statements (1) and (2) TOGETHER are sufficient to answer the question asked

BOTH statements (1) and (2) TOGETHER are sufficient to answer the question asked

The correct answer is: (A)

(1) That (–1)^(y+2) =−1, dictates that y + 2 must be odd and therefore y is odd since only an odd exponent can produce a value < 0. Furthermore, if an odd number is divided by 2, the remainder must be 1; SUFFICIENT.

(2) That y is prime does not definitively identify y as definitively even or odd, so the remainder could be either 0 or 1; NOT SUFFICIENT.

The correct answer is A; statement 1 alone is sufficient.

The correct answer is: (A).

(1) If Chiku made $620 last week, it is possible to solve for her old pay rate, then her new pay rate, and then the number of hours she’d need to work to guarantee she makes more this week than she did last week; SUFFICIENT.

(2) That her raise was greater than that of any of her coworkers introduces a new unknown variable; NOT sufficient.

The correct answer is A; statement 1 alone is sufficient.

The correct answer is: (D).

(1) Let h be the number of hamsters at the store and r be the number of rabbits to derive the equation 40 = 2.5h from the statement and given conditions, so h = 16; SUFFICIENT.

(2) Using the same variables derive the equation 56 = 2.5h + h, so that h = 16; SUFFICIENT.

The correct answer is D; each statement alone is sufficient.

The correct answer is: C)

(1) From the function, f(t) = pe^kt, when t = 2,f(t) = 275. The equation becomes pe^2k = 275. This is one equation in two unknowns, hence, we cannot solve it. More information is required; NOT SUFFICIENT.

(2) When k = 0.16, we have f(t) = pe^0.16t. In this case, p is unknown, hence more information is required; NOT SUFFICIENT.

Considering the two cases, we have pe^2k = 275 and f(t) = pe^0.16t which reduced to pe^2(0.16) = 275. We solve the equation

We now solve the equation
When t = 8, we have

f(8) = 11.21e^0.16(8) = 11.21e^1.28 = 40.32

The correct answer is C; BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.