The sum of the roots of a quadratic equation ax² + bx + c is - b / a
⇒  -(2m - 1)/m = 6/(3m + 1)
⇒ 6m² - m -1 = -6m
⇒ 6m² + 5m -1 = 0
⇒ m = 1/6 or -1
Thus, the sum of the values of m: 1/6 -1 = -5/6

The number of days between 31^st
December 1899 and 31^st December 1903 = 365 x 4 = 1460
The number of days between 31^st December 1903 and 21^st April 1904 = 31 + 29 + 31+21 = 112
Total number of days = 1572

Dividing 1572 by 7 the remainder = 4
Hence, 21^st April 1904 falls on Thursday.
Hence, option 2.

Alternatively,

We know that in a non-leap year there are 52 weeks and 1 odd day.
There is no leap year between 1899 to 1903. Hence, there are 4 odd days.

Number of odd days between 31^st December 1903 and 21^st April 1904 = 3 + 1 + 3 + 0 = 7

Thus, there are total 4 odd days between the given time frame. 21^st April 1904 will be a Tuesday.
Hence, 21^st April 1904 falls on Thursday. Hence, option 2.

The maximum volume will be obtained when it is rotated about the smaller side rather than the hypotenuse.
Radius = 8 units, Height = 6 units
Thus, the volume = (1/3) x (π x 64 x 6) = 128π units.

Hence, option 4.

Let the price of one pencil be x and the price of one pen be y.

⇒ (5y - x) = (3x + 7y)/5
⇒ 25y - 5x = 3x + 7y
⇒ x = 2.25y

Thus, the price of the pencil as a percentage of the price of the pen = (2.25y /y) x 100 = 225%

For a given set containing N elements, the number of non-zero subsets = 2^N - 1
Number of non-zero subsets in S = 2⁴⁰⁰ - 1

The product of the elements of a subset will be odd when all the elements of the subset are odd.

The odd elements in S are S_odd = (1, 3, 5 , ..., 399)
Number of non-zero subsets in S_odd = 2²⁰⁰-1

Number of subsets where product of its elements is even = (2⁴⁰⁰ - 1) -(2²⁰⁰-1) = 2⁴⁰⁰ - 2²⁰⁰

Hence, option 1.

The maximum area of trapezium will be obtained when the two sides are on the opposite sides of the diameter.

The height of the trapezium = OE + OF
OE = √OC² - √EC² = √3
OF = √OB² - √FB² = 2√2

∴ the area of the trapezium = (1 / 2) x (2√2 + √3) x (6 + 4) = 5(2√2 + √3)

Hence, option 2.

The increase in price in 2004-05 session = 80 x 0.2 = Rs.16.
2A + 5B = 16 ...(i)

The increase in price in 2005-06 session = 96 x 0.2 = Rs.19.2
4A + B = 19.2 ...(ii)

Solving (i) and (ii),
A = 3.125B 
Hence, option 3

We know that,
Distance remaining = Total distance - Distance travelled
⇒ d(t) = 32 - 30t

Other than blue any other can be selected.
Hence, 2 out of 13 can be selected in ¹³C₂ ways and total ways of selecting 2 balls out of 16 is ¹⁶C₂ ways.

The required probability = ¹³C₂/¹⁶C₂ = 13/20 = 78/120

⇒ 3M + 9 + M + 5 + 140= 180
⇒ 4M = 26
⇒ M = 6.5°

In a triangle ABC, if AB, AC and the included angle A are known, Area = (1/2) x AB x AC x sin A
= (1/2) x 18 x 24 x sin 150 = 216 x 0.5 = 108 sq.m

The amount obtained in 2014 by investing 100000 in 2010 = 100000 x (1 + 0.24) = 124000

Let the rate of interest offered in 2011 be r.
∴ 100000 x (1 + 3r/100) = 124000
⇒ r = 8%

The increase percentage in the offered rate = (2/6) x 100 = 33.33 %

The rate of interest offered in 2015 = 8 x (1 + 1/3)⁴ = 25.28 %

If we draw a figure according to the information provided in the question then we can note that:
In the question it is good that each girl is facing the center of the circle.

Radius of circle =10m, Diameter of circle =20m

We need to find the shortest distance between any two girls faces each other.

We can see from the image that oppositely standing girls will only face each other so the shortest distance between any two girls facing each other will be equal to the diameter of the circle ie 20 m.

Since the sum of two middle term s is asked, the series has an even number of terms.
Now for even number of terms we know that, Sum of the middle two term of an A.P. will be equal to the sum of the first and last terms of the A.P.
i.e. - 17 + 19 = 2

Alternatively,

First term = a = -17
Commond difference = d = -13 - (-17) = 4
Last term = 19 = -17 + (n - 1)4
⇒ n = 10

Middle term will be 5th and 6th terms.
Their sum will be = a + 4d + a + 5d = 2a + 9d
i.e. 2 (-17) + 9 (4) = 2

3 girls can be arranged in 3! Ways.

Now there are 4 gaps remaining where 4 boys can be arranged in 4! ways.

Total number of ways = 3! x 4! = 6 x 24 = 144

Let the numbers be x, y and z.
y = 2x and y = 3z
As the average = 77
⇒ x + y + z = 231
⇒ y/2 + y + y/3 = 231
⇒ y = 126

Hence, option 2.

Solving the quadratic equations we get,
x = -8 or -5 and y = -5 or -3.
⇒ x ≤ y

Hence, option 4.

Case 1:
Let the marked price be m.
The selling price = 0.9m
⇒ 0.9m = 540 
⇒ m = 600
Cost price = 540 -90 = Rs. 450

Case 2:
Discount = Rs. 60, Marked price = Rs. 650
Selling price = 650 - 60 = Rs. 590
Profit = 590 - 450 = Rs. 140
Profit percentage = (140/450) x 100 = 31.11%

When the policeman arrives, the thief has traveled 6 meters.
So, the time taken by the policeman to catch the thief along the circular track = 6/1.5 = 4 seconds.

The total distance covered by the policeman = 12 meters.
Thus, the minimum distance is when the policeman takes the route via the center i.e. he travels a length of 2 x 5 = 10 meters.

The minimum time taken = 10/3 seconds.

The minimum distance covered by the thief = 1.5 x (10/3) = 5 meters

Hence, option 3.

Let the cost price of 1 kg of pure sugar be Rs. x.
The cost price of 1 kg sugar-like substance = Rs. 2x / 3

The selling price of one kg of mixture = Rs. x.

Let the amount of pure sugar in 1 kg of mixture be y.
The amount of sugar like substance = (1 - y)

The cost price of mixture = xy + (2x / 3) x (1 - y) = (x / 3) x (2 + y)
⇒ x / [(x / 3) x (2 + y)] = 1.3
⇒ y = 4/13

Thus, the ratio of sugar-like substance to pure sugar = (1-y):y = 9:4 

Hence, option 3.

 x = 4⁴ x 5⁶ x 6⁵ x 7¹⁰ x 8⁸
⇒ x = 2³⁷ x 3⁵ x 5⁶ x 7¹⁰ = (2 x 3) (2³⁶ x 3⁴ x 5⁶ x 7¹⁰)

Power of 2 can take values of 0, 2 , 4 , . . . , 36 = 19
Power of 3 can take values of 0, 2,4 = 3
Power of 5 can take values of 0, 2,4, 6 = 4
Power of 7 can take values of 0, 2,4, 6, 8, 10 = 6

Total number of factors = 19 x 3 x 4 x 6 = 1368 

Let the percentage total number of candidates who passed in both the subjects be x.

Let the sets M and H represent students who passed in Mathematics and History respectively.

65 - x + x + 55 - x = 100 - 22
⇒ x = 42%

Now, 42% i.e. 48909 students passed in both the subjects.
∴ Total number of students = 48909 x (100/42) = 116450

If x is a prime number greater than 1, then x is not a factor of (x - 1)!

There are 29 primes up to 110. 4 is also one such number.

Hence, option 4.

Let the initial speed of the person be x and the time taken to go from P to Q be t.
The distance from Q to R = 4xt / 3
The time taken to go from R to Q = (4xt/3) / (4x/9) = 3t

The time taken to go from Q to P = (xt) / (8x/27) = (27/8)t

Hence, the total time taken = t + 2t + 3t + (27 / 8)t = (75 / 8)t = 9.375t
∵ t = 15 minutes = 900 seconds
The total time taken = 8437.5 seconds.

Hence, option 4.

x² - 3x + 2 > 0 is true when x < 1, x > 2 ... (i)
y² - 4y + 3 > 0 is true wheny < 1, y > 3 ... (ii)

∵ xy = 16
The possible values of (x, y) are (1, 16), (2, 8), (4, 4), (8, 2) and (16, 1).

From (i) and (ii), we can see that only (4,4) satisifies both the conditions.
⇒ |x - y | = |4 - 4 | = 0

Hence, option 2.

The minimum value of a quadratic expression ax² + bx + c (when a > 0) occurs at x = -b / 2a and its minimum value is -(b² - 4ac) / 4a

The minimum value of the equation is = [4(3)(-23) - (11)²]/12 = -397 / 12

Hence, option 4.

Area of each slab = 72 / 50 m² = 1.44 m²
Length of each slab =√1.44 =1.2m =120cm

Let the number of days required by B alone to complete the project be x
The number of days required by A alone to complete it = 2x

1/x + l/2x = 1/30
⇒ x = 45

The part of work done in 10 days =1/3
Work done by A and C in 1 day = (1/90) + (1/20) = (11/180)
∴ The number of days taken by A and C = (2/3) / (11/180) = 120/11 = (10*12) / 11 days.

Hence, option 2.

Let the total population of the town be x.
The number of people who read XYZ = 0.37x - 8000
0.37x+0.37x - 8000 + 0.27x = x
⇒ x = 8,00,000

Hence, option 1.

x² + x - 2 < 0
⇒ (x - 1) (x + 2) < 0
⇒ (x - 1) < 0 and (x + 2) > 0 or (x - 1) > 0 and (x + 2) < 0
⇒ x < 1, x > -2 or x > 1, x < -2 which is not possible.

∴ -2 < x < 1
⇒ (x - 1) < 0 and (x + 2) > 0 

Hence, option 3.

When the minute hand travels 60 minutes, the hour hand travels 5 minutes.

The minute hand travels 55 minutes more than the hour hand in an hour.
In 54 minutes, the Minute hand travels = (54 x 55/ 60) = 49.5

Hence, option 3.

Let the number of hours worked by B be x.
The hours worked by A = (12 - x)
The manhours of work finished by A = (12 - x - 2) x 0.6 + 2

∵ (12 - x - 2) x 0.6 + 2 = x
⇒ x = 5

The total manhours of work done by them in a day = 2 * x = 2 * 5 = 10

Thus, the number of days required to finish the work = 120/10 =12 days

Hence, option 2.

Let the number of apples in the carton with the least number of apples be x and the second least number of apples be y.
 ∵ x +y + x + 100 = 270
⇒ 2x +y = 170...(i)

Also, the total amount spent on the two cartons with x and y apples = 4500 - 1900 = 2600
⇒ 20x + 25y = 2600
⇒ 4x + 5y = 520.. .(ii)

From (i) and (ii)
x = 55,y = 60

The lowest number of apples in a cartoon = 55.

Hence, option 3.

Let the length of the saree be ‘x’ cm.

x - (1/4) x - (2/5) x = 3850
⇒ (7/20) x = 3850
⇒ x = 11000 cm = 110 m

The total weight of (36 + 44) Students of A and B = (36×40) + (44×35) = 2980 kg

⇒ Average weight of the whole class = 2980/80 = 37.25kg