Arranging the equation, we know that there are no solutions when the lines are parallel:

for that the condition had to $\frac{p}{3}=-\frac{4}{k}$ ≠ $\frac{2}{a}$

Checking through options: 

Option A: using the first and last terms in our relation, we see that ap must not be equal to 6 for the lines to be parallel. This option puts no conditions on that and thus is not relevant. 

Option C: This question is the opposite of what we want; if this is true, the lines can never be parallel. 

Option D: Using the first and second terms of the relation, we see that we want kp = -12, or kp-12 = 0. Hence, this statement is not what we want. 

Option B: Using the second and third terms, we see that we do not want k equals to -2a, or we do not want k+2a to be equal to 0

Therefore, B is a condition that is necessary for the lines to be parallel and have no solution. 

Therefore, Option B is the correct answer. 

It is given that for some real numbers a and b, the system of equations x+y=4 and (a + 5)x + ($b^2$ − 15)y= 8b has infinitely many solutions for x and y.

Hence, we can say that 

=>  $\frac{a + 5}{1}$ = $\frac{b^2 - 15}{1}$ = $\frac{8b}{4}$​

This equation can be used to find the value of a, and b.

Firstly, we will determine the value of b. 

=>   $\frac{b^2 - 15}{1}$ = $\frac{8b}{4}$​ => $b^2 - 2b - 15$ = 0

=> (b - 5)(b + 3) = 0

Hence, the values of b are 5, and -3, respectively. 

The value of a can be expressed in terms of b, which is a + 5 = $b^2 − 15$ => a = $b^2 − 20$

When b=5, a = $5^2$ − 20 = 5

When b=−3, a = $3^2$ − 20 = −11

The maximum value of ab = (−3)⋅(−11) = 33

Let us assume the initial stock of all the fruits is S.

Let us take we have 'b' and 'a' mangoes initially.

Stock of Mangoes = 40% of S = 2S/5

The total number of fruits sold are Mangoes Sold + Apples Sold + Bananas Sold

= 2S/10 + 96 + 4a/10 = S/2 (Given)

=> S/5 + 96 + 2a/5 = S/2

=> S = $\frac{(4a+960)}{3}$

=> $\frac{4a}{3} + 320$

'a' has to be a multiple of 3 for the above term to be an integer.

But 'a' has to be a multiple of 5 for 4a/10 to be an integer.

=> The smallest value of 'a' satisfying both conditions is 15.

=> $\frac{4a}{3}$ + 320 = $\frac{4(15)}{3}$ + 320 = 340

xy + yz = 19

y(x + z) = 19

Observe that 19 is prime and can’t be factorized further. Since x, y and z are all natural

numbers, the value of (x + z) can’t be 1. (It should at least be 2).

Therefore, y = 1 and ( x + z) = 19

yz + xz = 51

z(y + x) = 51

z(x + 1) = 51

51 = 17 * 3

Case 1: z = 17, x + 1 = 3

x = 2, y = 1, z = 17

xyz = 2 * 1 * 17 = 34

Case 2: z = 3, x + 1 = 17

x = 16, y = 1, z = 3

xyz = 16 * 1 * 3 = 48

Hence the minimum value of x * y * z = 34

Hence, the answer is '34'

In the question, it is given that the equation ∣x+a∣+∣x−1∣=2

∣x+a∣+∣x−1∣=2 has an infinite number of solutions for any value of x. This is possible when x in |x+a| and x in |x-1| cancels out.

Case 1:

x + a < 0, x - 1 ≥ 0

- a - x + x - 1 = 2

a = -3

Case 2:

x + a ≥ 0 and x - 1 < 0

x + a - x + 1 = 2

a = 1

Largest value of a is 1.

The answer is option C.

a(a + b + 1) = 14 …… (1)

b(a + b + 1) = 28 …… (2)

$\frac{a}{b} = \frac{1}{2}$

b = 2a

Substituting in (1), we get

a(3a + 1) = 14

$3a^2 + a - 14 = 0$

$3a^2 - 6a + 7a - 14 = 0$

$3a(a - 2) + 7(a - 2) = 0$

Given, a and b are natural numbers.

Therefore, a = 2 and b = 4

2a + b = $2(2) + 4 = 8$

Let the number of questions attempted be x+y out of which x are correct and y are incorrect and the number of questions unattempted be z.

It is given,

x + y + z = 75 ...... (1)

3x - y + z = 97 ...... (2)

(2)-(1) -> x - y = 11

(1)+(2) -> 2x + z = 86

z > x + y

z > 75 - z

z > 37.5

Minimum possible value of z is 38

2x + 38 = 86

2x = 48

x = 24

The maximum number of correct answers is 24.

Let $\dfrac{x}{y}$ be $t$

Therefore, $c = 16t + \dfrac{49}{t}$

Applying AM ≥ GM

$\frac{16t + \frac{49}{t}}{2} \ge (16t \times \frac{49}{t})^{\frac12}$

​

$16t + \dfrac{49}{t} \ge 56$

When $t$ is positive then $c$ is greater than equal to $56$.

When $t$ is negative then $c$ is less than equal to $-56$.

Therefore $c \in (-\infty, -56] \cup [56, \infty]$

Let the number of 100 cheques, 250 cheques and 500 cheques be x, y and z respectively.

We need to find the maximum value of z.x + y + z = 100 ...... (1)

100x + 250y + 500z = 15250

2x + 5y + 10z = 305 ...... (2)

2x + 2y + 2z = 200 ....... (1)

(2) - (1), we get

3y + 8z = 105

At z = 12, x = 3

Therefore, maximum value z can take is 12.

Let the cost of an apple, an orange and a mango be a, o, and m respectively.

Then it is given that:

2a+4o+6m = a+4o+8m

or a = 2m.

Also, a+4o+8m = 8o + 7m

10m-7m = 4o

3m = 4o.

We can now express the cost of a basket in terms of mangoes only:

2a+4o+6m = 4m+3m+6m = 13m.

We need to check for all regions:

x >= 0, y >= 0

x >= 0, y < 0

x < 0, y >= 0

x < 0, y < 0

However, once we find out the answer for any one of the regions, we do not need to calculate for other regions since the options suggest that there will be a single answer.

Let us start with x >= 0, y >= 0,

3x + 3y = 7

2x + 3y = 1

Hence, x = 6 and y = -11/3

Since y > = 0, this is not satisfying the set of rules.

Next, let us test x >= 0, y < 0,

3x - y = 7

2x + 3y = 1

Hence, y = -1

x = 2.

This satisfies both the conditions. Hence, this is the correct point.

WE need the value of x + 2y

x + 2y = 2 + 2(-1) = 2 - 2 = 0.

Let the number of large shirts be l and the number of small shirts be s.

Let the price of a small shirt be x and that of a large shirt be x + 50.

Now, s + l = 64

l (x+50) = 5000

sx = 1800

Adding them, we get,

lx + sx + 50l = 6800

64x +50l = 6800

Substituting l = (6800 - 64x) / 50, in the original equation, we get

$\frac{(6800−64x)}{50}$(x+50)=5000

(6800 - 64x)(x + 50) = 250000

$6800x+340000−64x^2−3200x=250000$

$64x^2−3600x−90000=0$

Solving, we get, x= $\frac{225±375}8 = \frac{600}{8} or −\frac{150}8$

SO, x = 75

x + 50 = 125

Answer = 75 + 125 = 200.

Alternate approach: By options.

Hint: Each option gives the sum of the costs of one large and one small shirt. We know that large = small + 50

Hence, small + small + 50 = option.

SMALL = (Option - 50)/2

LARGE = Small + 50 = (Option + 50)/2

Option A and Option D gives us decimal values for SMALL and LARGE, hence we will consider them later. 

Lets start with Option B.

Large = 150 + 50 / 2 = 100

Small = 150 - 50 / 2 = 50

Now, total shirts = 5000/100 + 1800/50 = 50 + 36 = 86 (X - This is wrong)

Option C - 

Large = 200 + 50 / 2 = 125

Small = 200 - 50 / 2 = 75

Total shirts = 5000/125 + 1800/75 = 40 + 24 = 64 ( This is the right answer)

Let tom's age = x

=> Dick=3x

=> harry = 6x

Given, 

3x+1 = (x+3x+6x)/3 

=> x= 3

Hence, Harry's age = 18 years

$\text{Two linear equations } ax + by = c \text{ and } dx + ey = f \text{ have a unique solution if } \frac{a}{d}$ ≠ $\frac{b}{e}$

$\therefore \frac{k}{4}$ ≠ $\frac{1}{k} \Rightarrow k^{2}$ ≠ $4$

$\Rightarrow k$ ≠ $|2|$