SME
Grade 8 Math · Problem type
How many of each
Use this for “8 tickets cost $41. Adult tickets are $7 and child tickets $4. How many of each?”
Table
How many of the first kind (x)
Total cost (T)
($)
This is an example. Type your own values to start a new problem.
x
+
y
=
n
p
x
+
q
y
=
T
x
How many of the first kind
worked out (not typed)
y
How many of the second kind
worked out (not typed)
Formulas
x
=
(
T
−
q
×
n
)
÷
(
p
−
q
)
(how
many
of
the
first
kind
=
(total
cost
−
price
of
the
second
kind
×
items
in
all)
÷
(price
of
the
first
kind
−
price
of
the
second
kind))
3 = (41 − 4 × 8) ÷ (7 − 4)
y
=
n
−
x
(how
many
of
the
second
kind
=
items
in
all
−
how
many
of
the
first
kind)
5 = 8 − 3
Assumptions
•
Two equations: the count adds to n, and the cost adds to T.
•
If all were the cheaper kind, the cost would be q × n. Each of the dearer kind adds p − q.
•
A count must come out whole; if not, no mix of the two makes that total.
Step-by-step
WE KNOW
n = 8 (items in all), T = $41 (total cost), p = $7 (price of the first kind), q = $4 (price of the second kind)
FIND
how many of the first kind (x), how many of the second kind (y)
Step 1 · Find how many of the first kind (x)
Use x = (T − q × n) ÷ (p − q) (how many of the first kind = (total cost − price of the second kind × items in all) ÷ (price of the first kind − price of the second kind))
If
all
were
the
second
kind,
the
cost
is
q
×
n
.
The
extra,
shared
at
p
−
q
each,
is
the
first
kind.
x
=
41 − 4 × 8
7 − 4
x
=
41 − 32
3
x
=
9
3
x
=
3
Step 2 · Find how many of the second kind (y)
Use y = n − x (how many of the second kind = items in all − how many of the first kind)
The rest are the second kind.
y
=
8
−
3
y
=
5
Check
Put the numbers back in. Both sides should match.
3 = (41 − 4 × 8) ÷ (7 − 4) ✓
5 = 8 − 3 ✓
Topic check
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