Problem 1 :
The sum of two numbers is 137 and their difference is 43. Find the numbers.
Solution:
Let us consider, first number = x and second number = y
x + y = 137 --- > (1)
x - y = 43 --- > (2)
Adding both the equation, we get
2x = 180
x = 90
By applying x = 90 in (1)
90 + y = 137
y = 137 - 90
y = 47
So, the numbers are 90 and 47.
Problem 2 :
The sum of thrice the first and the second is 142 and four times the first exceeds the second by 138, then the numbers.
Solution:
Let us consider, first number = x and second number = y
3x + y = 142 ---> (1)
4x - y = 138 --- > (2)
Add (1) and (2), we get
7x = 280
x = 40
By applying x = 40 in (1),
3(40) + y = 142
120 + y = 142
y = 142 - 120
y = 22
So, the numbers are 40 and 22.
Problem 3 :
Sum of two numbers is 50 and their difference is 10, then find the numbers.
Solution:
Let us consider, first number = x and second number = y
x + y = 50 --- > (1)
x - y = 10 --- > (2)
Adding both the equation, we get
2x = 60
x = 30
By applying x = 30 in (1)
30 + y = 50
y = 50 - 30
y = 20
So, the numbers are 30 and 20.
Problem 4 :
The sum of twice the first and thrice the second is 92 and four times the first exceeds seven times the second by 2, then find the numbers.
Solution:
Let us consider, first number = x and second number = y
2x + 3y = 92 --- > (1)
4x - 7y = 2 --- > (2)
(1) × 2 ==> 4x + 6y = 184
Subtract (1) from (2)
13y = 182
y = 182/13
y = 14
By applying y = 14 in (1),
2x + 3(14) = 92
2x + 42 = 92
2x = 92 - 42
2x = 50
x = 50/2
x = 25
So, the numbers are 25 and 14.
Problem 5 :
The sum of two numbers is 1000 and the difference between their squares is 256000, then find the numbers.
Solution:
Let the numbers are x and y.
x + y = 1000 --- > (1)
x^{2} - y^{2} = 256000
x^{2} - y^{2} = (x + y) (x - y)
1000 × (x - y) = 256000
x - y = 256 --- > (2)
Adding (1) and (2),
2x = 1256
x = 1256/2
x = 628
By applying x = 628 in (1),
628 + y = 1000
y = 1000 - 628
y = 372
Problem 6 :
The difference between two numbers is 14 and the difference between their squares is 448, then find the numbers.
Solution:
Let us consider, first number = x and second number = y
x - y = 14 --- > (1)
x^{2} - y^{2} = 448
(x + y) (x - y) = 448
(x + y) × 14 = 448
x + y = 448/14
x + y = 32 --- > (2)
Adding (1) and (2)
2x = 46
x = 46/2
x = 23
By applying x = 23 in (1)
23 - y = 14
-y = 14 - 23
y = 9
So, the numbers are 23 and 9.
Problem 7 :
The sum of two natural numbers is 8 and the sum of their reciprocals is 8/15. Find the numbers.
Solution:
Let the number be x and y.
x + y = 8 --- > (1)
1/x + 1/y = 8/15
(y + x) / xy = 8/15
8/xy = 8/15
xy = 15
y = 15/x
By applying y = 15/x in (1)
x + 15/x = 8
x^{2} + 15 = 8x
x^{2} - 8x + 15 = 0
x^{2} - 3x - 5x + 15 = 0
x (x - 3) - 5 (x - 3) = 0
(x - 5) (x - 3) = 0
x = 5 or x = 3
So, the two natural numbers are 5 and 3.
May 21, 24 08:51 PM
May 21, 24 08:51 AM
May 20, 24 10:45 PM