# Question

Source: Woodgrove Primary

This old question resurfaced to the top for some weird reason.

Anyway, here is a non-algebraic solution using block diagram. (Underlying principle is algebra but I think maybe P5s understand block diagrams better than equations with 2 unknowns.)

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Here’s how I do it by equations:

Let B be the cost of a blouse.
Let S be the cost of a skirt.

5 B + 4 S = 215 … (1)

4 B + 5 S = 215 + 20 = 235
=> 5 B + (5/4)5 S = (5/4)235
=> 5 B + (25/4) S = 293.75 … (2)

(2) – (1) :
(21/4) S = 293.75 – 215 = 78.75  (* Mistake here… see below *)
S = 78.75 (4/21) = \$15 #

Chief, you did your mental sums too quickly….I make such mistakes too sometimes.

4S = (16/4)S  not (4/4) S

So (25/4) S – (16/4) S = (9/4) S  not (21/4)S

Oops… so I did… * embarrassed *

Thanks for finding the mistake!

Let’s do it again:

Let B be the cost of a blouse.
Let S be the cost of a skirt.

5 B + 4 S = 215 … (1)

4 B + 5 S = 215 + 20 = 235
=> 5 B + (5/4)5 S = (5/4)235
=> 5 B + (25/4) S = 293.75 … (2)

(2) – (1) :
(9/4) S = 293.75 – 215 = 78.75
S = 78.75 (4/9) = \$35 #

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[——5B———][——-4S———]<—20—>

[——4B—–][—————5S—————–]

So,  4B+5S-5B-4S = 20  => S-B = 20 => B = S-20

5B + 4S = 215

5(S-20) + 4S = 215

5S +4S = 215+100=315

9S = 315

S = \$35

Therefore, cost of a skirt is \$35

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