SOLUTION: Divided 56 into two parts such that three times the first part exceed one third of second by 48. The parts are
a. 25, 31
b.20,36
c.24, 32
Algebra.Com
Question 872738: Divided 56 into two parts such that three times the first part exceed one third of second by 48. The parts are
a. 25, 31
b.20,36
c.24, 32
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
Divided 56 into two parts such that three times the first part exceed one third of second by 48.
---------
Part one:: x
Part two:: 56-x
-----
Equation:
3(56-x)-(1/3)x = 48
--------------------------
168 - 3x - (1/3)x = 48
-10/3 x = - 120
(1/3)x = 12
x = 36
------------
56-x = 20
---------------------
Cheers,
Stan H.
-------------------
The parts are
a. 25, 31
b.20,36
c.24, 32
RELATED QUESTIONS
Divide 56 into two parts such that three times the first part exceeds one third of the... (answered by Boreal)
Divide 56 into two parts such that three times the first part exceeds one third of the... (answered by josgarithmetic)
Divide 56 into two parts such that three times the first part exceeds one third of the... (answered by math_helper)
Divide 56 Into Two Parts Such That Three Times The First Parts Exceeds One Third Of The... (answered by lynnlo)
Divide 56 into two equal parts such that 3 times the first part exceeds one third of the... (answered by ikleyn)
42 is divided into two parts such that five times the second part exceeds four time the... (answered by josgarithmetic)
divide 184 into two parts such that one third of one part may exceed one seventh of the... (answered by ankor@dixie-net.com)
divide 184 into two parts such that one-third of one part may exceed one-seventh of the... (answered by sachi)
Divide $128 into two parts such that one-third of the first part is less than
1
5
of... (answered by ewatrrr)