GRE Mock Test (New Pattern) - 2 - GRE MCQ

Let us try to understand the meaning of the sentence from the partial information.

The sentence talks about an element that doesn't justify violence itself but does something to its use. This element is the belief that death, if for a sacred cause, can be considered a proper end of life. The structure and reasoning of the sentence implies that the element is trying to justify the use of violence. In other words, justify can be a good fill-in for the sentence.

Now, let us look at the meaning of the verbs in the options.

A. incriminates – to make (someone) appear guilty of a crime or wrongdoing
B. valorizes – to give or ascribe value or validity to
C. indicts – to formally accuse of or charge with a crime
D. abrogates – to abolish by authoritative action
E. rescinds – to take away
'Valorizes,' which means to give validity to something, is the correct answer. 'Incriminates,' 'indicts,' and 'abrogates' are opposite to what the context requires. 'Rescinds' does not go with the intended meaning of the sentence.

The correct answer is B.

The first blank describes the “relationship” of patients with their doctors, and the second blank describes these patients’ “reactions” to Insurance companies. 
so, correct sentence be like: The (i) Laconic relationship between patients and their doctors is evidenced by patients’ (ii) Antagonistic reactions to the attempts by insurance companies to force patients to see new doctors in different health care networks.

The word “as” signals a continuation of ideas—the vocals were hard to hear because of the way the recordings were mastered. The answer is (A) maladroitly, which means “unskillfully” or “bunglingly.” Choices (B) copiously, (C) ingeniously, and (D) shrewdly are not in line with what you’re looking for—these words suggest more time and care than was spent. Choice (E) maliciously is too harsh. It’s doubtful the recordings were mastered with intent to harm.

The first blank is a good one to start with here. It describes something that’s happened to the guitar over its long history. Since the given example is the invention of the electric guitar, you can predict that the guitar has “grown” or “changed.” (B) evolved is a match and is correct. (A) stagnated, or “remained the same,” doesn’t fit, as an innovation is specifically mentioned in the first sentence. (C) regressed, or “moved backward,” does imply a change, but there’s no indication that the guitar has changed in a negative or regressive fashion.

The second blank explains the relationship between a modern guitar and a 15th-century instrument. “While” is a detour road sign; it indicates that there’s a contrast between this relationship and the evolution of the guitar. Furthermore, the second sentence indicates that a skilled guitarist would be able to “competently” play the vihuela. A good prediction would be that even though the guitar has gone through changes over the centuries, it is still similar to the vihuela. This idea is reinforced by the next sentence, which says that a modern guitarist could probably play the old instrument. (E) an affinity to is the correct answer; an affinity is a “similarity” or a “likeness.” (D) an incongruity with and (F) a divergence from both indicate a dissimilarity with the vihuela, which is not supported.

The third blank is directly connected to the vihuela; since all that’s known about it from this text is that it’s from the 15th century, predict that this blank means “old.” That’s (H) archaic, which means “outmoded” or “antiquated.” (G) inferior (“of lesser quality”) is wrong because there’s no support for the idea that the vihuela is worse than the guitar or any other instrument. (I) avant-garde may be tempting because it’s a term that’s o
en used in the context of music and other arts; however, it relates to “new” or “experimental” concepts, so it doesn’t fit here.

The key to this sentence is the description of the wine aficionado (or expert) as a “purist,” or someone who insists on the purity of things.
Since he rejects the first blank, you can assume it will be an antonym of pure. You can predict mixture, which fits in the context. However, he prefers the second blank, so this should be a synonym of pure and the contextual word “simple.”
For the first blank, the best match for your prediction is choice (A) amalgamation. This is an exact synonym of your prediction, mixture.
Choice (B) dissonance, which means “a disagreeable combination,” seems plausible. A purist would definitely find a mixture of two wines to be dissonant. However, we need a word that states this mixture has occurred for the sentence to make sense. Similarly, choice (C) enigma means “puzzle,” which may also be applicable to such a mixture but, again, lacks the necessary meaning of “mixture.” For the second blank, choice (E) unadulterated matches your prediction best. Choice (D) pragmatic means “practical” and does not fit as well as unadulterated in context. The wine aficionado does not prefer something for its practicality but for the taste; therefore, this choice would be misleading in the sentence. Choice (F) opaque, which means “impossible to see through,” may similarly be applicable to the wine; however, nowhere in the sentence is the color implied or relevant.

This Inference question asks you to identify statements about people's reaction to the artwork that follow from the passage. According to the critics mentioned in the passage, knowing how the art was created is potentially related to the audience’s interest in it. The reaction is mainly reliant on the concept, not the product, of the art. Thus, (B) is correct; if the audience doesn’t appreciate how the art was made, they may be less interested in it. Choice (A) doesn’t follow from the passage. Nothing suggests that the beauty or clarity of the images produced affects the success of the project. If (C) were true, it would likely increase the popularity of the performance and get people thinking about it. This has the opposite effect called for by the question stem.

Given: xy > 0.
⇒ xy is a positive quantity.
There can be two cases here.
Case 1: Both x and y are positive
Let's analyze the values of Quantity A (x²y⁴) and Quantity B (x³y⁶).
Case 1.1: Say x = y = 1
Quantity A: x²y⁴ = 12 × 14 = 1
Quantity B: x³y⁶ = 13 × 16 = 1
We see that Quantity A equals Quantity B.
Case 1.2: Say x = y = 2
Quantity A: x²y⁴ = 2² × 2⁴ = 2⁶
Quantity B: x³y⁶ = 2³ × 2⁶ = 2⁹
We see that Quantity B is greater than Quantity A. No unique answer from the two cases discussed so far, thus the correct answer is option D.
Though we concluded that the correct answer is option D, we can analyze Case 1.3 and Case 2 also for the sake of understanding.

We see that Quantity A is greater than Quantity B.
Thus, there are all the possible answers A, B and C possible when we consider that both x and y are positive.
Let's discuss the case when both x and y are negative.
Case 2: Both x and y are negative
Quantity A: x²y⁴
Since the exponents of x and y are even, the value of x² & y⁴ would be positive irrespective of whether their values are negative. So, x²y⁴ = ∣∣x²∣∣ × ∣∣y⁴∣∣ = a positive number>0.
Quantity B: x3y6
Since the exponent of x is odd, and x is negative, the value of x³ would be negative. As discussed above that since the exponents of y is even, the value y6 would be positive irrespective of whether its value is negative.
So, x³y⁶ = −∣∣x³∣∣ × ∣∣y⁶∣∣ = a negative number < 0.
Thus, quantity A is greater than B for Case 2.

Given:

• BC = 16 cm
• Rectangle FABE is a square
• The area of rectangular region FACD = 612 cm²

We have to compare the Area FABE and Area EBCD.
Say the side of the square FABE is xcm. Thus, AF = x and AC = AB + BS = x + 16
Thus, Area of FACD = x ×(x + 16) = 612 (given)
So, we have x(x + 16) = 612
⇒ x² + 16x − 612 = 0
x² + 34x − 18x − 612 = 0
x(x + 34) − 18(x + 34) = 0
(x + 34)(x − 18) = 0
⇒ x = 18 since x = −34 (not possible; side cannot be negative)
Quantity A: Area FABE = FA × AB = x × x = x² = 18² cm²
Quantity B: Area EBCD = EB × BC = x × 16 = 18 × 16cm²
It is clear that Quantity A (= 18²) is greater than Quantity B (18 × 16).

Given:

• Machine A working ALONE can complete a work in 3.5 hours
• Machine B working ALONE can complete the same work in x hours
• Machine A & B working together can complete the same work in 1.5 hours

We have to find the value of x and compare it with 3.
The part of work Machine A working ALONE completes in one hour = 1/.35 = 2/7
The part of work Machine B working ALONE completes in one hour = 1/x
Thus, the part of work Machine A & B working together completes in one hour 


x = 2.625 hours.
Thus, Quantity A (=3) is greater than Quantity B (= x = 2.625).

Given:

• Average speed for the first 20 miles of a 40-mile trip =50miles per hour
• Average speed for the second 20 miles of a 40-mile trip =xmiles per hour
• Average speed for the entire 40-mile trip =60miles per hour

We have to find the value of x and compare it with (x−60).
Time taken to complete first 20 miles of a 40-mile trip 

Time taken to complete second 20 miles of a 40-mile trip

Time taken to complete the entire 40-mile trip


We have, Quantity A = x − 60 = 75 − 60 = 15
Thus, Quantity A (=15) is greater than Quantity B (=10).

Given:
The roots of the equation x² − 16x − 612 = 0 are a and b.
Let's use hit and trial and split −16x into two parts such that their product equals to the product of x² and −162 i.e. −162x².
We have x² − 16x − 612 = 0
⇒ x² −34x + 18x − 612 = 0
x(x − 34) − 18(x − 34) = 0
(x − 34)(x + 18) = 0
x = 34 or −18
The two roots are 34 and −18. We can consider either of them as a and b. Thus, the sum of roots = a + b = 34 − 18 = 16.
Note that for a quadratic equation x² + Ax + B = 0, −A shows the sum of the roots and B shows the product of the roots of the equation.
Here, we had the equation x² − 16x − 612 = 0. Manipulating it to x² − Ax + B =0 , we get x² + (−16)x + (−612) = 0. Here A = −16 and B = −162.
If the quadratic equation is of the form ax² + bx + c = 0, then we must bring it in the form of x² + Ax + B = 0. Dividing the equation ax² + bx + c = 0 by a, we get . Here, sum of the roots  and the product of the roots = c/a.
In another way, a quadratic equation can be represented as:
x² − (Sum of roots)x + (Product of roots) = 0

To determine the median price, we first need to arrange the items per ascending or descending order of their price. We chose to do it in ascending order. Thus, we have the following table:

Since the total number of cell phones sold is 450, the median value

= Price of the item/cell phone

= Price of the(225.5)^th item/cell phone

= Average of the price of the 225^th item and the price of the226^th item

We see that the 225^th item lies in 225^th cumulative number of units' column in the table, thus, its price would be the price of the brand that lies in the same row of the 225^th cumulative item. We find that it is $250.

Similarly, we see that the 226^th item lies in 300^th cumulative number of units' column in the table, thus, its price would be the price of the brand that lies in the same row of the 300^th cumulative item. We find that it is $350.

Thus, the median price per cell phone = 

Note that the median price per phone (= $250) would not be the price of the middle-most brand if the prices are arranged in ascending order: 150, 200, 250, 350, and 400. This is because the logic is flawed. In this case the number of units per brand of cell phone sold is ignored.

To determine the mean price, we need to find the Total sales. Thus, we have the following table:

Mean price per item/cell phone  = Total sales / Total number of units sold = 128750 / 450

= $286.11
= ∼$286

Possible solutions of a^b = 1 are as follows:

• a = 1; for any value of b
• a = −1 and b is even
• b=  0; for any value of a other than a = 0
• From a): a = x − 2 = 1 ⇒ x = 3: one of the possible values of x
• From b): a=  x − 2 = −1 ⇒x = 1; we see that b = 2(x + 1) = 2(1 + 1) = 4 is an even integer, thus x = 1 is one of the possible values of x
• From c): b = 2(x + 1) = 0 ⇒ x = −1; at x = −1, we see that a = x − 2 = −1 − 2 = −3 ≠ 0; thus, x = −1 is one of the possible values of x

There are three possible values of x: -1, 1, and 3.

Let's cross-check these values for the sake of better understanding.

• At x = −1, we have (x − 2)^{2(x + 1)} = (−1 − 2)^{2(−1 + 1)} = −30 = 1 = RHS
• At x = 1, we have (x − 2)^{2(x + 1)} = (1 − 2)^{2(1 + 1)} = −14 = 1 = RHS
• At x = 3, we have (x − 2)^{2(x + 1)} = (3 − 2)^{2(3 + 1)} = 18 = 1 = RHS

Given:

• ABC is a straight line
• Angles x & y are integer multiples of 20
  

We have to find out the value of x.
We have:
2x + 3y = 180
Let x = 20p and y = 20q, where p and q are positive integers
⇒ 40p + 60q = 180
⇒ 2p + 3q = 9
The only possible ordered pair of positive integers for p & q are {3,1}.
Thus, x = 20p = 20 × 3 = 60°.

Let the number of pens and notebooks purchased be x and y, respectively.

⇒ 13x + 19y = 58
The number of pens and notebooks purchased must be positive integers.

Let us try with a few positive integer values of x and calculate the qualified values of y:

x = 1: 13 + 19y = 58 ⇒ y = 45/19 = not an integer - Not a valid solution

x = 2: 26 + 19y = 58 ⇒ y = 32/19 = not an integer - Not a valid solution

x = 3: 39 + 19y = 58 ⇒ y = 1 - A valid solution

Thus, for all other higher values of x, the value of y would be either less than 1 or negative and hence can be ignored.

Total number of items purchased by Suzy  = x + y = 3 + 1 = 4

Since the number leaves a remainder 2 when divided by 15 and by 9, the number would leave the same remainder '2' when divided by the Least common multiple (LCM) of 15 and 9, i.e. 45.

We need to find the numbers between 180 and 300, inclusive, that leave the remainder 2 that are in the format (45n + 2); where n is a positive integer.

We know that 180 ≤ 45n + 2 ≤ 300
180 − 2 ≤ 45n ≤ 300 − 2
= 178 ≤ 45n ≤ 298
= 17845 ≤ n ≤ 29845
= 3.96 ≤ n ≤ 6.62
= n = 4,5,or 6
Thus, the required numbers are:

45n + 2 = 45 × 4 + 2 = 182
45n + 2 = 45 × 5 + 2 = 227
45n + 2 = 45 × 6 + 2 = 272

Given:
a₁ = −2, a₂ = −5, a₃ = 4, a₄ = 3,and a_n = a_n−4 forn > 4
We have to compare the sum of the first 64 terms and the first 98 terms of the sequence.
Since the question involves the sum of the terms up to 98 terms, which is a big number, you must anticipate that there is some kind of pattern in the sequence.
We have the values for the first 4 terms: a₁ = −2, a₂ = −5, a₃ = 4,and a₄ = 3.
It is given that for n > 4, an = a_n−4
Thus,

• a₅ = a_{5 − 4} = a₁ = −2
• a₆ = a_{6 − 4} = a₂ = −5
• a₇ = a_{7 − 4} = a₃ = 4
• a₈ = a_{8 − 4} = a₄ = 3

You would observe that there is a pattern of 4 terms in the sequence. 5th term onwards of the sequence repeats.
Thus, Quantity A: the sum of the first 64 terms of the sequence
= (a₁ + a₂ + a₃ + a₄)+(a₅ + a₆ + a₇ + a₈)+...+(a₆₁ + a₆₂ + a₆₃ + a₆₄)
= (−2−5 + 4 + 3) + (−2−5 + 4 + 3)+...+(−2−5 + 4 + 3)
 there would be 64/4 = 16 groups with each having their value equal to 0.

Similarly, Quantity B: the sum of the first 98 terms of the sequence
= (a₁ + a₂ + a₃ + a₄) + (a₅ + a₆ + a₇ + a₈)+...+(a₉₃ + a₉₄ + a₉₅ + a₉₆)+(a₉₇ + a₉₈) 
=(−2−5 + 4 + 3)+(−2−5 + 4 + 3)+...+(−2−5 + 4 + 3)+(−2−5)

We see that Quantity A (= 0) is greater than Quantity B (= −7).

Given:

• The arithmetic mean of the list {3, a, 1, 9, b, 3} = 4
• a and b are integers

We have to compare the Median and the Mean (= 4) of the list.
Let's first find out the relationship between a and b by using the information that the arithmetic mean of the list {3, a, 1, 9, b,3 } = 4

⇒ a + b = 24 − 16 ⇒ a + b = 8
We already know Quantity B (= Mean = 4); we need to calculate the value of Median.
So, we have a+b=8. Let's discus two extreme cases.
Case 1: Say a = −2. Thus, b = 8 − (−2) = 10
So, the list arranged in an ascending order is {−2,1,3,3,9,10}.

We see that Quantity A (= 3) is less than Quantity B (= 4).
Case 2: Say a = b = 4.
So, the list arranged in an ascending order is {1,3,3,4,4,9}.

We see that Quantity A (= 3.5) is less than Quantity B (= 4).
There is no need to consider the case which has b=−2 & a=10 since that would be the same as Case 1.
We cannot consider a case which has a=3.5 because that would be an invalid case since a & b both are integers.
Thus, median of the list would be either 3 or 3.5, and in either case, Quantity A (= 3 or 3.5) is less than Quantity B (= 4).

Given:

We have to compare the maximum value of −(5−x) and the maximum value of 2x.
Let's first find out the range of x.
We have 

⇒ 12 < 7 − x ; multiplying both the sides by +3; we can multiply/divide the inequality by a positive number without changing the sign of the inequality and can do so by a negative number but with reversing the sign of the inequality
= 5 < −x; subtracting both the side by 7
= − 5 > x; reversing the sign of inequality
We see that x is a negative number and can have any large negative value.
Quantity A: Maximum value of −(5 − x)
⇒ Maximum value of (x − 5)
Since x is negative, and we need to get the maximum value of (x−5), the least possible value of x in the magnitude would be eligible
Say x = −5.001
Maximum value of (x − 5) = −5.001 − 5 = −10.001
Quantity B: Maximum value of 2x at x = −5.001
Maximum value of 2x = 2 × −5.001 = −10.002
Thus, Quantity A (=-10.001) is greater than Quantity B (=-10.002).
No matter how close to -5, you choose the value of x, you would find that Quantity A is greater than Quantity B. Let's take another example.
Say x=−5.00001
Maximum value of (x − 5) = −5.00001 − 5 = −10.00001
Quantity B: Maximum value of 2x at x = −5.00001
Maximum value of 2x = 2 × −5.00001 = −10.00002
Thus, Quantity A (=-10.00001) is greater than Quantity B (=-10.00002).

Prime numbers less than 11 are 2, 3, 5, and 7. Note that 1 is not a prime number.

Possible areas of a rectangle = Length × Breadth
⇒ 2 × 2 = 4; 2 × 3 = 6; 2 × 5=10; 2 × 7 = 14; 3 × 3 = 9; 3 × 5 =  15; 3 × 7 = 21; 5 × 5 = 25; 5 × 7 = 35; note that a square is also a rectangle.

We are given that the maximum possible perimeter = 24 unit.

We should find out the maximum possible perimeter considering the maximum possible lengths and breadths given above (2, 3, 5, and 7). This will help us eliminate the unqualified values quickly.

Length and breadth: 5 and 7: Perimeter = 2(5 + 7) = 24. Since the perimeter 24, the possible value of area = 5 × 7 = 35 is not possible.
Length and breadth: 5 and 5: Perimeter = 2(5 + 5) = 20. Since the perimeter <24, the possible value of area = 5 × 5 = 25 is possible.
We need not check further as the smaller values for the lengths and the breadths would certainly qualify.

Thus, the possible values of areas = 2 × 2 = 4;  2 × 3 = 6; 2 × 5 = 10; 2 × 7 = 14; 3 × 3 = 9; 3 × 5 = 15; 3 × 7 = 21; 5 × 5 = 25

Let the number of paperback and hardcover books bought be x and y, respectively.

We know that:

x > 10, thus x ≥ 11, and y > 8, thus y ≥ 9
We have to calculate the possible values of total number of books, ie, (x + y)
Since the total cost of all the books is between $240 and $300, exclusive, we have:

⇒ 240<  8x + 20y < 300
⇒ 60 < 2x + 5y < 75
If we take x = 11 (minimum value), we will get the possible values of y or the possible number of hardcover books; similarly, if we take y = 9 (minimum value), we will get the possible values of x or the possible number of paperback books. This way, we can get the minimum and the maximum possible values of (x+y) or the minimum and the maximum possible values of total number of books.

Say David bought x = 11 (minimum value) paperback books, thus,

from 60 < 2x + 5y < 75
⇒ 60 < 2 × 11 + 5y < 75 ⇒ 60 − 22 < 5y < 75 − 22 ⇒ 38 < 5y < 53
⇒ 7.60 < y < 10.60 ⇒ y = 8, 9 or 10. The value y = 8 is not eligible as we know that y ≥ 9. Thus, y = 9 or 10.

Total number of books bought = x + y = 11 + (9 or 10) = 20 or 21
Say David bought y=9 (minimum value) hardcover books, thus,

from 60 < 2x + 5y < 75
⇒ 60<  2x + 5 × 9 < 75 ⇒ 60 − 45 < 2x < 75 − 45 ⇒ 15 < 2x < 30
⇒ 7.5 < x < 15 ⇒ x = 8, 9, 10, 11, 12, 13, or 14; The values for x = 8, 9 and 10 are not eligible as we know that x ≥ 11. Thus, x = 11,12,13, or 14
Total number of books bought = x + y = (11,12,13, or 14) + 9 = 20, 21, 22 or 23.

The possible values of (x + y) = 20, 21, 22 and 23

Let the number of candidates in the year 2014 be 100.

Thus, the number of candidates in the year 2017 = 100 + (100 × 40%) = 140
Number of candidates for the GMAT course in 2014 = 35% of 100 = 35; read data from the chart

Number of candidates for the GMAT course in 2017 = 45% of 140 = 63; read data from the chart

Thus, percent increase = 
This increase happened over a period of 3 years: 2014-2015, 2015-2016 and 2016-2017.

Thus, the simple annual percent increase = 80 / 3 = 26.67% ⇒ 27% as whole number rounded

The answer is not 80% since the question asked to calculate the simple annual percent increase for three years.

We know that the total number of candidates in 2014 was 500.

Since every year for the next two years, i.e., for 2014-2015 and 2015-2016, the percent increase in the number of candidates is 20%, the number of candidates in 2015 = 120% of 500 = 600.

Similarly, number of candidates in 2016 = 120% of 600 = 720.

Thus, the number of candidates for the LSAT course in 2016 = 15% of 720 = 108; from the 2016 chart, we know that there are 15% candidates for the LSAT course.

Alternatively, you can calculate it as following:

C₂₀₁₆ = C₂₀₁₄ × (1+r%)ⁿ; say C₂₀₁₆ = number of candidates in 2016, C₂₀₁₄ = number of candidates in 2014, the rate of increase = r% and the number of years = n
C₂₀₁₆ = 500 × (1 + 20%)²; given that C₂₀₁₄ = 500, r% = 20%, and there are 2 years from 2014 to 2016, thus n = 2
C₂₀₁₆ = 500 × 1.22 = 500 × 1.44 = 720
Thus, the number of candidates for the LSAT course in 2016 = 15% of 720 = 108.

Given:

• The box-and-whisker plot above shows weights for 60 kids
• 23 kilograms represents the 90th percentile value

We have to find out how many kids weight between 18 kilograms and 23 kilograms, inclusive?
A box-and-whisker plot is a convenient means of graphically representing data using quartiles, as shown below:

Thus, we have, from the given plot:

• Q₁(25th percentile value) = 11
• Q₂ = Median (50th percentile value) = 12
• Q₃(75th percentile value) = 18

We know that 23 kilograms represents the 90th percentile value on the plot above.
Thus, weights between 18 and 23 is represented by values between 75th percentile and 90th percentile, i.e., 90 - 75 = 15% of the total number of kids
= 15% of 60
= 9 kids

We have 50 guests and each has either a $40 ticket or a $60 ticket with an average ticket-receipt of more than $50. We need to include few new guests with $40 tickets so that the average ticket-receipt of all the guests becomes less than $50.

We see that if the number of guests with $40 tickets and the number of guests with $50 tickets initially were equal in number, i.e. 50/2 = 25 each, the average ticket-receipt of all the guests would be

= 

= $50  (i.e. the simple average of $40 and $60)

Since it is given that the average ticket-receipt is more than $50, we can conclude that the number of guests with $60 tickets is greater than the number of guests with $40 tickets.

Thus, currently, there are 26 or greater guests with $60 tickets and 24 or fewer guests with $40 tickets.

Since we finally need to reduce the average ticket-receipt of all the guests to less than $50, we must have a greater number of guests with $40 ticket than the number of guests with $60 ticket.

This means that the guests with $40 tickets must be greater than 26 or at least 27. Presently, there are 24 guests with $40 tickets; thus, 3 or more new guests with $40 tickets must be included to reduce the average ticket-receipt of all the guests to less than $50.

Thus, we can see that if including either 1 or 2 number of guests with $40 tickets would definitely not be sufficient.

We know that two of the interior angles are 90° each.

Since the sum of the angles of a quadrilateral is 360°, the sum of the other two angles

=360° − (90° + 90°) = 180°

We know that:

∠ABC = 2 × ∠BCD

Let us consider all possible cases for the relation ∠ABC = 2 × ∠BCD

• Say ∠BCD = 90° ⇒ ∠ABC=2 × ∠BCD = 180°

  Since one more angle is 90°, the measure of the fourth angle

  = 360° − (90° + 90° + 180°) = 0° - This is not possible since this will not render a quadrilateral.

• Say ∠ABC = 90° ⇒ ∠BCD = 90°/2 = 45°

  Since one more angle is 90°, the measure of the fourth angle

  =360°− (90° + 90° + 45°) = 135° 

  Thus, the largest interior angle among 90°, 90°, 45°, and 135° is 135° -Option D is correct

• Neither ∠ABC∠ABC nor ∠BCD∠BCD is equal to 90°

  Thus, we have:

  ∠ABC + ∠BCD = 360° − (90° + 90°) = 180°

  Since one angle between ∠ABC and ∠BCD is twice the other, we have:

  Smaller angle = (1 /1 + 2) × 180° = 60°

  Larger angle = 180° − 60°=120°

  Thus, the largest interior angle among 90°, 90°, 60°, and 120° is 120° - Option C is correct

Let the number of students in the class before the group of students joined = x
Let the number of students who joined the class = y
It is given that the initial average age of the class = 18years
The average age of the group of students who joined the class = 16years
The average age of all the students after the group of students joined  = 17years
Thus, we have:

⇒ 18x + 16y = 17x + 17y

⇒ y / x = 1

Volume of the original mixture = 120ml
Volume of Chemical X present = 40% of 120ml = 48ml
Let x ml of the mixture be removed.

Percent of Chemical X present in x ml is also 40%
Thus, the volume of Chemical X removed = 40% of x ml = 2 / 5x ml
Thus, volume of Chemical X remaining =

Total volume of the mixture left = (120 − x)ml
Volume of water added = x ml
Thus, total volume of the final mixture

= (120 − x) + x = 120ml
Final volume of Chemical X = 10% of 120 = 12ml
Thus, from (i), we have:

48 − 2 / 5x = 12
⇒ x = 90ml

Given:
x and y are non-negative integers such that 2x + 3y = 8 and z = x² + y².
We have to find out the maximum value of z.
We have 2x + 3y = 8

Since x and yx are positive integers, (4 − x) must be a multiple of 3. Only two eligible values of x are possible, and they are 1 and 4.
The set of values of x & y are {1,2} and {4,0}.

• Value of z = x² + y² at {1, 2}: z = x² + y² = 1² + 2² = 1 + 4 = 5
• Value of z = x² + y² at {4, 0}: z = x² + y² = 4² + 0² = 16 + 0 = 16

The maximum of the two values 5 and 16 is 16 and that is the correct answer.