Two numbers differ by 40, when each number is increased by 8, the bigger becomes thrice the lesser number. If one number is x, then find the other number.
Answers
Step-by-step explanation:
Given :-
Two numbers differ by 40, when each number is increased by 8, the bigger becomes thrice the lesser number. One number is x.
To Find:-
Find the other number ?
Solution :-
Given that
Two numbers differ by 40
One of the two numbers = X
Let the other number = 40+X
If the two numbers increased by 8 then
They will be (X+8) and (40+X+8) = 48+X
The two numbers = X+8 and X+48
The bigger number = X+48
The smaller number = X+8
Given that
Bigger number = Thrice the smaller number
=> Bigger number = 3×Smaller number
=>X+48 = 3×(X+8)
=> X+48 = (3×X)+(3×8)
=> X+48 = 3X+24
=> 3X+24 = X+48
=> 3X-X = 48-24
=> 2X = 24
=> X = 24/2
=> X = 12
One of the numbers = X = 12
Other number = X+40 = 12+40 = 52
Answer:-
The other number for the given problem is 52
Check:-
One number = 12
Other number = 52
Their difference = 52-12 = 40
If they are increased by 8 then they will be 12+8 = 20 and 52+8 = 60
The smaller number = 20
The bigger number = 60
=> 3×20
=> Thrice the smaller
Bigger number is thrice the smaller number when they are increased by 8 each.
Verified the given relations in the given problem.