The value of k for which kx + 3y - k + 3 = 0 and 12x + ky = k, have infinite solutions, is

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

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

6x + 1 > 7 - 4x
? x > 3/5
? 3/5 < x ? 2

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

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

Let us assume the number of Hens are H and Number of Goats are G.
According to question,
The total number of animal heads are 81. Since one animal has one head.
Number of head of animals = Number of total animals
H + G = 81..........................(1)
The total number of animal legs are 234. Since hen's have 2 legs and goat's have 4 legs.
2H + 4G = 234...................(2)
Multiply 2 with equation (1) and subtract from equation (2), we will get.
? 2H + 4G - 2 x (H + G ) = 234 - 2 x 81
? 2H + 4G - 2H - 2G = 234 - 162
? 2G = 72
? G = 36
Put the value of G in equation (1), we will get.
H + 36 = 81
H = 81 - 36
H = 45
Krishna have number of goats = G = 36

Let us assume the present age of father = F year and Son's present age = S years
According to question, 5 years ago,
Father's age = F - 5 and Son's age = S - 5.
According to the question,
The age of the father 5 years ago was 5 times the age of his son.
F - 5 = 5(S - 5)
F - 5 = 5S - 25.....................(1)
At present the father's age is 3 times that of his son.
F = 3S.................................(2)
Put the value of F from equation (2) in equation (1), we will get
? 3S - 5 = 5S - 25
? 25 - 5 = 5S - 3S
? 20 = 2S
? 10 = S
? S = 10.
Put the value of S in Equation (2). we will get,
F = 3S = 3 x 10 = 30
So the present Age of Father = 30.

Let us assume the digits of the original number are unit's digit a and ten's digit b.
The Original Number will be 10a + b.
After interchanging the digits the new number will be 10b + a.
According to question,
The number obtained by interchanging the two digits of a two-digit number is lesser than the original number by 54.
New Number = Original Number - 54
10b + a = 10a + b - 54
? 10b + a - 10a - b = -54
? 9b - 9a = -54
? a - b = 6....................................(1)
Again according to question,
Sum of the digits of original number = 10
a + b = 10..................................................(2)
Add the equation (1) and (2), we will get
a - b + a + b = 10 + 6
2a = 16
a = 8
Put the value of a in Equation (2) , we will get
8 + b = 10
b = 10 - 8
b = 2
Put the value of a and b for original number, we will get
10a + b = 10 x 8 + 2 = 80 + 2 = 82