Question

Malcolm and Peter each have some $50, $10 and $2-notes. The total number of notes they have is the same. The number of $2-notes Peter has is 1/2 of the number of $2-notes Malcolm has. The number of $10-notes Peter has is 4/5 of the number of $10-notes Malcolm has. Peter has 8 more $50-notes than Malcolm. The total number of $2 and $10-notes they have is 42.
(a) Who has more money?
(b) How many $2-notes do they have altogether?

Answer 1

It consider a 3 variable equation as follows.

ax1 + bx2 + cx3 = d — (1)
ex1 + fx2 + gx3 = h — (2)
ix1 + jx2 + kx3 = l — (3)

if common solution to equation (1), (2) and (3) are x1 = x2
x2 = B & x3 = 7. Then these values of x1, x2 and x3
will satisfy the equation (1), (2) & (3)

Let The total Number of notes Malcolm and Peter have are (a + b + c) and (x + y + z) respectively.
a and x are $50 Notes.
b and y are $10 Notes.
c and z are 52 Notes
As per the problem total Number of notes for both are
Same.
a + b + c = x + y + z — (1)
z = c/z — (2)
y = 4/5b — (3)
x = a + 8 — (4)
From equation (1), (2), (3) and (4) we can write:
a + b + c = a + 8 + 4/5b + c/2 => b(1 - 4/5) + c(1 - 1/2) = 8

b/5 + c/2 = 8 — (5)
Since b and c are integers
Now total Number $2 and $10 Notes they have
z + c + y + b = 42 — (6)
From (2) & (3)
c + c/2 + 4/5 + b = 42 => 3/2c + 9/5b = 42 — (7)
From equation (5) we can write 3c/2 = 24 - 3b/5
=> From (5) & (7)
24 - 3b/5 = 42 - 9/5b => 6/5b = 18
b = 15, y = 4/5b = 12
Since c/2 = 8 - b/5 = 8 - 15/5
c/2 = 5 or c = 10, z = c/2 = 5

a) Now total money for Malcolm and Peter are CM and CP respectively.
CM = 50a + 10b + 2c = 50a + 10 * 15 + 10 * 2
CM = 50a + 170
Similarly
CP = 50x + 10y + 2z = 50(a + 8) + 12 * 10 + 5 * 2
CP = 50a + 400 + 120 + 10 = 50a + 530
Clearly (CP > CM) Peter have more money than Malcolm.

b) The total Number of $2 Notes are
N = Z + C = 10 + 5
N = 15