David and Joseph have a total of 328 marbles. Matthew and David have 178 marbles. Joseph has 5 times as many marbles as Matthew. How many marbles does David have?
kvnmurty:
perhaps joseph as 6 times as many as mathew
Answers
Answered by
1
lets assume David as D,Joseph as J and Matthew as M
D+J=328 ..........given
M+D=178..........given
and j=6XM ...........(its maybe 6 times as many as marbles as matthew, 5 gives fractional values which is not true for counting object , we always used natural number for counting.....nobody count like 1,1.1 etc....its creates problem )
6XM+D=328
M+D=178
- - - (minus sign)
---------------------
6M-M+D-D=328-178
5M=150
M=30
J=6M=6*30
J=180
D=328-J=328-180=148
David=148 Marbles
D+J=328 ..........given
M+D=178..........given
and j=6XM ...........(its maybe 6 times as many as marbles as matthew, 5 gives fractional values which is not true for counting object , we always used natural number for counting.....nobody count like 1,1.1 etc....its creates problem )
6XM+D=328
M+D=178
- - - (minus sign)
---------------------
6M-M+D-D=328-178
5M=150
M=30
J=6M=6*30
J=180
D=328-J=328-180=148
David=148 Marbles
Answered by
1
Let david & joseph has x & y marbles , let mathew has z ,
now ,
ATQ :-
x + y = 328 ---(1)
z + x = 178 ----(2)
also , y = 5z
Putting y = 5z in eq.(1)
x + 5z = 328
x + z = 178
- - -
4z = 150
z = 150/4 ,
Now putting z in eq.2
150/4 + x = 178
x = 140.5 ~ 141 marbles
now ,
ATQ :-
x + y = 328 ---(1)
z + x = 178 ----(2)
also , y = 5z
Putting y = 5z in eq.(1)
x + 5z = 328
x + z = 178
- - -
4z = 150
z = 150/4 ,
Now putting z in eq.2
150/4 + x = 178
x = 140.5 ~ 141 marbles
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