Let the ten's digit be x. Then, number = 10x + 3 and sum of digits = ( x + 3 )
So ,(x+3) = 1/7(10x + 3)
? 7x + 21 = 10x + 3
? 3x = 18
? x=6
Hence, the number = 10x + 3 = (10 x 6) + 3 = 63.

Let the number be N.
According to question,
A number is doubled and 9 is added then 2N + 9
Again According to question,
If the resultant is tripled, it becomes 75.
3 ( 2N + 9 ) = 75
? 2N + 9 = 25
?2N =16
? N = 8

Let us assume the original fraction be x/y.
According to question,
200x/100 / 200y/100 = 14/5
2x/2y = 14/5
?x/y = 14/5

Let the number be 10x + y where x > y
According to the question.
x + y = 15 ............(1)
and x - y = 3 .........(2)
Add the Equation (1) and (2) , We will get
x + y + x - y = 15 + 3
? 2x = 18
? x = 9
Put the value of x in equation (1). We will get
9 + y = 15
? y = 15 - 9
? y = 6
So we have digits of number x = 9, y = 6
? x * y = 9 * 6 = 54

According to question,
40 notebooks at the rate of ? 18 per notebook and 55 pencils at the rate of ? 8 per pencil.
Amount paid = ? (40 x 18 + 55 x 8)
Amount paid = ? (720 + 440)
Amount paid = ? 1160

Let us assume the first part is A then second part will be 50 - A.
According to question,
Sum of their reciprocals is 1/2.
1/A + 1/(50 - A) = 1/12
? (50 - A + A)/(50 - A)A = 1/12
? (50)/(50 - A)A = 1/12
? (50) x 12 = 1 x (50 - A)A
? 600 = (50 - A)A
? 600 = 50A - A²
? A² - 50A + 600 = 0
? A² - 30A - 20A + 600 = 0
? A(A - 30) - 20(A - 30) = 0
? (A - 30)(A - 20) = 0
? (A - 30) = 0 or (A - 20) = 0
? A = 20 , 30

Let us assume the abscissa coordinate be x and Ordinate be y.
According to question,
ordinate four times its abscissa
y = 4x
? y - 4x = 0

Let cost of one saree and shirt be x and y, respectively.
2x + 4y = 16000 ....(i)
x + 6y = 16000 ....(ii)

On multiplying Eq. (ii) by 2 and subtracting from Eq. (i). we get
2x + 4y = 16000
2x + 12y = 32000
-------------------------
-8y = -16000
? y = 2000

Putting the value of y in Eq. (ii), we get
x +6 x 2000 = 16000
? x = 4000
? Cost of 12 shirts = 12y
= 12 x 2000 = ? 24000

Given equations of system
3x + y = 4 ...(i)
x + 2y = 8 ...(ii)
Here, a₁ = 3 , b₂ = 2 and c₂ = B
? a₁/a₂ = b₁/b₂ = c₁/c₂ = 1/2
So, the system of equations has infinite solutions, because it represents a parallel line.

Given, 6x - 10y = 10 ..........(i)
and x/(x + y) = 5/7
? 7x = 5x + 5y
? 2x - 5y = 0 ...(ii)

On multiplying Eq. (ii) by 2 and subtracting from Ed.(i), we get
6x - 10y = 10
4x - 10y = 0
---------------------
2x = 10
? x = 5
Putting the value of x in Eq. (i), we get
30 - 10y = 10
? 10y = 20
? y = 2
? (x - y) = 5 - 2 = 3