The numerator of a fraction is 6x + 1 and the denominator is 7 - 4x, x can have any value between -2 and 2, both included. The values of x for which the numerator is greater

Let the number be x and y.
Then, according to the question,
x + y = 15 ...(1)
x - y = 3 ...(ii)
on adding Eqs. (i) and (ii), we get
2x = 18
? x = 9
On putting the value of in Eq. (i), we get
y = 6
? Product = xy = 54

Let hens = H, goats = G
According to the question,
H + G = 90 ...(i)
2H + 4G = 248 ...(ii)
On multiplying Eq. (i) by 2 and subtracting from Eq. (ii) , we get
2H + 2G = 180
2H + 4G = 248
-------------------
-2G = -68
? G = 34

Given equation are 6x + 3y = 7xy
? (6/y) + (3/x) = 7 .....(i)

And 3x + 9y = 11xy
? (3/y) + (9/x) = 11 .....(ii)

On multiplying Eq. (ii) by 2 and subtracting from Eq. (i), we get
(6/y) + (3/x) = 7
(6/y) + (18/x) = 22
-------------------------
- 15/x = - 15
? x = 1

On putting this value in Eq. (i), we get
(6/y) + 3 = 7
? 6/y = 4
? y = 6/4 = 3/2

Given, 3^{x + y} = 81
? 3^{x + y} = 3⁴
? x + y = 4 ...(i)
and 81^{x - y} = 3 or (3⁴)^{x - y} = 3
? x - y = 1/4
On solving the Eqs. (i) and (ii), we get
2x = 17/4 ? x = 17/8

we have, x + y - 7 = 0
? x + y = 7 ...(i)
and 3x + y - 13 = 0
? 3x + y = 13 ..(ii)
By subtracting Eq. (i) from Eq. (ii), we get
3x + y = 13
x + y = 7
----------------------
2x = 6 ...(ii)
? x = 3

On putting the value of x in Eq. (i), we get
3 + y = 7
? y = 4

Now, 4x² + y² + 4xy
= 4 x (3)² + (4)² + 4 x 3 x 4
= 4 x 9 + 16 + 48
= 36 + 16 + 48 = 100

Given, (x/4) + (y/3) = 10/24
? (3x + 4y)/12 = 10/24
? 3x + 4y = 5 ...(i)
and (x/2) + y = 1
x + 2y = 2 ...(ii)
On multiplying Eq. (ii) by 2 and subtracting from Eq. (i).
3x + 4y = 5
2x + 4y = 4
- - -
--------------------------
x = 1
On putting the value of x in Eq. (ii), we get
x + 2y = 2
? 1 + 2y = 2
? y = 1/2

? x + y = 1 + 1/2 = 3/2

Let cost of 1 pencil and 1 clipper be p and c, respectively.
Now, according to the question,
21p + 9c = ? 819
? 3(7p + 3c) = ? 819
? 7p + 3c = ? 273
Cost of 7 pencil and 3 clippers = ? 273

For infinite solution
a₁/a₂ = b₁/₂ = c₁/c₂
? K/12 = 3/K = (-K + 3)/ -K
? K/12 = 3/K
? K² = 36
? K = ?36 = 6

Let the number of correct answer be x and number of wrong answer be y.
Then, 4x - y = 200 ..(i)
and x + y = 200 ..(ii)

On adding Eqs. (i) and (ii), we get
4x - y = 200
x + y = 200
----------------
5x = 400
? x = 80

Let us assume N be the given number.
According to the question,
Sum of third , fourth and fifth part of a number exceeds half of the number by 34.
N/3 + N/4 + N/5 - 34 = N/2
N/3 + N/4 + N/5 - N/2 = 34
(20N + 15N + 12N - 30N)/60 = 34
17N/60 = 34
N = 34 x 60/17
N = 2 x 60
N = 120