AFCAT Numerical Ability · Profit and Loss

Profit Counted in Articles, and Rates

When a cost or a profit is measured in articles, take the price of one article as 1; the cost of x equalling the sale of y, or a gain worth n articles, becomes a simple fraction.

All Profit and Loss notes

Why this matters

The largest page of the chapter, with one AFCAT past-year question from 2019. Every row is a fraction once one article's price is set to 1, and the marks go on taking that fraction over the wrong base: the number sold instead of the cost, or the selling price instead of the cost price.

Concept 1 of 4: Cost of x articles equals the selling price of y

If 13 articles cost what 10 sell for, the seller receives the cost of 13 for every 10 he hands over. He keeps the cost of 3 on a cost of 10, a profit of

30\%

Definition

CP of $x$ = SP of $y$ with $x > y$ : profit % $= \dfrac{x - y}{y} \times 100$ .
With $y > x$ it is a loss: loss % $= \dfrac{y - x}{y} \times 100$ .
The base is $y$ , the number actually sold.
The percentage holds for any quantity, so the profit on a whole stock is that percentage of its cost.
Working back: for a profit of $p\%$ , $x : y = (100 + p) : 100$ .

Cost of x equals sale of y

\text{Profit}\% = \dfrac{x - y}{y} \times 100

Worked example

The cost price of 21 pens equals the selling price of 15 pens. Find the gain per cent.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

AFCAT · 2019 · AFCAT (I) 2019 — Numerical Ability · 02-203Easy

Example 1 · Profit and Loss · Articles and Rates

The cost price of 25 articles is equal to the selling price of 20 of them. The gain or loss per cent is given by

Divide by the number sold

The cost of 21 equals the sale of 15: the gain is 6 on 15,

40\%

. Dividing by 21 gives

28.6\%

, the wrong base.

Which side is larger decides profit or loss

If the cost of fewer articles equals the sale of more, each article sells below cost: a loss, however the question words it.

Drill 23 more on cost of x articles equals the selling price of y

Concept 2 of 4: A gain equal to the cost of n articles

If selling 40 articles earns the cost of 8, the profit is 8 parts on a cost of 40 parts: one fifth, a profit of

20\%

. The profit per article is the same fraction of one article's cost.

Definition

Selling $N$ articles gains the cost of $n$ : profit % $= \dfrac nN \times 100$ . Losing the cost of $n$ gives the same fraction as a loss.
Selling $M$ articles then gains the cost of $M \times \dfrac nN$ articles.
To gain the cost of $k$ articles, sell $k \times \dfrac Nn$ articles.
$SP : CP = (N + n) : N$ .

Gain worth the cost of n articles

\text{Profit}\% = \dfrac{n}{N} \times 100

Worked example

By selling 60 pens, a shopkeeper gains the cost price of 12 pens. Find his gain per cent, and the gain on selling 90 pens.

Practice this conceptself-check · 4 quick reps

The same idea in a practice question:

Practice questionModerate

Example 2 · Profit and Loss · Articles and Rates

By selling

75

mangoes, a fruit vendor gains the cost price of

25

mangoes. If he sells

150

mangoes, what will be his profit?

Scale the count, do not add to it

If 60 pens gain the cost of 12, then 120 pens gain the cost of 24. The gain grows with the number sold; it is not 12 more each time.

The cost of n is not the selling price of n

A gain worth the cost of n articles and one worth their selling price give different answers. Read which one the question names; the next concept handles the second.

Drill 23 more on a gain equal to the cost of n articles

Concept 3 of 4: A gain equal to the selling price of n articles

If selling 30 articles gains the selling price of 6, then those 30 cost only the selling price of 24. The profit is 6 on 24.

Definition

Selling $N$ gains the SP of $n$ : CP of $N$ = SP of $(N - n)$ , and profit % $= \dfrac{n}{N - n} \times 100$ .
Losing the SP of $n$ : CP of $N$ = SP of $(N + n)$ , and loss % $= \dfrac{n}{N + n} \times 100$ .
Profit : SP $= n : N$ , and CP : SP $= (N - n) : N$ .
Given the price of one article, find the total sale first and subtract the gain to get the total cost.

Gain worth the sale of n articles

\text{Profit}\% = \dfrac{n}{N - n} \times 100

Worked example

By selling 30 articles, a trader gains the selling price of 6. Find his profit per cent.

Practice this conceptself-check · 4 quick reps

The same idea in a practice question:

Practice questionModerate

Example 3 · Profit and Loss · Articles and Rates

A trader sells

117

items and gains the selling price of

13

items. What is his profit percentage?

Subtract before dividing

A gain of the selling price of 6 on 30 articles is 6 on a cost of 24, that is

25\%

. Dividing 6 by 30 gives

20\%

, which is the profit as a share of the selling price.

A loss adds to the count

A loss of the selling price of

n

means the cost is the selling price of

N + n

articles, not

N - n

Drill 16 more on a gain equal to the selling price of n articles

Concept 4 of 4: Bought and sold at so many for a price

'8 for ₹10' is a price of ₹1.25 an article. Turn both rates into a price per article, or buy and sell a number of articles that both counts divide, and compare.

Definition

Price per article = rupees ÷ count.
LCM method: take the LCM of the two counts as the number bought and sold; both totals come out whole.
Bought $a$ for ₹ $b$ and sold $b$ for ₹ $a$ (counts and prices swapped): $\dfrac{SP}{CP} = \left(\dfrac ab\right)^2$ .
Per dozen and per piece: put both in the same unit first (a dozen is 12).
'How many for ₹ $x$ to gain $y\%$ ?': find the new selling price per article, then divide $x$ by it.

Profit from two rates

\text{Profit}\% = \left(\dfrac{\text{SP per article}}{\text{CP per article}} - 1\right) \times 100

Worked example

Oranges are bought at 8 for ₹10 and sold at 4 for ₹6. Find the gain per cent.

Practice this conceptself-check · 4 quick reps

The same idea in a practice question:

Practice questionModerate

Example 4 · Profit and Loss · Articles and Rates

Bananas are bought at the rate of 6 for ₹5 and sold at the rate of 5 for ₹6. What is the profit percent?

Compare prices per article, not counts

'8 for ₹10' against '4 for ₹6' compares nothing until each is a price for one article (or for the same number of articles).

Keep the fractions

A cost of

\tfrac 23

rupee rounded to 0.67 can turn an exact

20\%

into

19.4\%

. Work in fractions, or with an LCM number of articles.

Drill 11 more on bought and sold at so many for a price

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (4)

Cost of x articles equals the selling price of y
Cost of x equals sale of y
$\text{Profit}\% = \dfrac{x - y}{y} \times 100$
A gain equal to the cost of n articles
Gain worth the cost of n articles
$\text{Profit}\% = \dfrac{n}{N} \times 100$
A gain equal to the selling price of n articles
Gain worth the sale of n articles
$\text{Profit}\% = \dfrac{n}{N - n} \times 100$
Bought and sold at so many for a price
Profit from two rates
$\text{Profit}\% = \left(\dfrac{\text{SP per article}}{\text{CP per article}} - 1\right) \times 100$

Watch out for (8)

Divide by the number sold
→ Cost of x articles equals the selling price of y
Which side is larger decides profit or loss
→ Cost of x articles equals the selling price of y
Scale the count, do not add to it
→ A gain equal to the cost of n articles
The cost of n is not the selling price of n
→ A gain equal to the cost of n articles
Subtract before dividing
→ A gain equal to the selling price of n articles
A loss adds to the count
→ A gain equal to the selling price of n articles
Compare prices per article, not counts
→ Bought and sold at so many for a price
Keep the fractions
→ Bought and sold at so many for a price

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