Example 1 :
The sum of a two digit number and the number formed by interchanging the digits is 110. If 10 is subtracted from the first number, the new number is 4 more than 5 times the sums of the digits of the first number. Find the first number
Solution :
Let "xy" be the two digit number.
xy + yx = 110
10x + y + 10y + 1x = 110
11x + 11y = 110
Divide it by 11,
x + y = 10 --------(1)
xy - 10 = 5(x + y) + 4
10x + y - 10 = 5x + 5y + 4
10x - 5x + y - 5y = 4 + 10
5x - 4y = 14 --------(2)
Let us use elimination method to solve these equations.
(1) x 4 + 2 ==> 4x + 4y = 40
5x - 4y = 14
----------------
9x = 54
x = 54/9 = 6
By applying the value of x in (1), we get
6 + y = 10
y = 10 - 6 = 4
Hence the required two digit number is 64.
Example 2 :
The sum of the numerator and denominator of a fraction is 12. If the denominator is increased by 3, the fraction becomes 1/2. Find the fraction.
Solution :
Let "x/y" be the fraction
x + y = 12 ------(1)
x/(y + 3) = 1/2
2x = y + 3
2x - y = 3 ------(2)
(1) + (2)
x + y = 12
2x - y = 3
---------------
3x = 15
x = 5
By applying the value of x in (1), we get
5 + y = 12
y = 12 - 5
y = 7
Hence the required fraction is 5/7.
Example 3 :
ABCD is a cyclic quadrilateral such that ∠A = (4y + 20)° , ∠B = (3y –5)°, ∠C =(4x) ° and ∠D = (7x + 5)° . Find the four angles.
Solution :
The sum of interior angles of a cyclic quadrilateral = 360
4y + 20 + 3y - 5 + 4x + 7x + 5 = 360
11x + 10y = 360 - 20
11x + 10y = 340 ----(1)
<A + <C = 180
4y + 20 + 4x = 180
4x + 4y = 160
x + y = 40 -------(1)
∠B = (3y –5)° ∠D = (7x + 5)°
<B + <D = 180
3y - 5 + 7x + 5 = 180
7x + 3y = 180 ----------(2)
(1) x 3 - (2)
3x + 3y = 120
7x + 3y = 180
(-) (-) (-)
-----------------
-4x = -60
x = 15
By applying the value of x in (1), we get
15 + y = 40
y = 40 - 15
y = 25
∠A = (4y + 20)° = 120
∠B = (3y –5)° = 70
∠C =(4x)° = 60
∠D = (7x + 5)° = 110
Hence the required angles are 120, 70, 60 and 110.
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