Profit, loss, discount, simple interest — basic applications: Exercises

Level 1 — Recognition

(Can you plug into the formula and name the base?)

L1·Q1 — Profit amount and profit%

A trader buys a fan for $C P = ₹750$ and sells it for $S P = ₹900$ . Find the profit and the profit%.

Recall Solution

WHAT we compare: $S P$ to $C P$ . Since $900 > 750$ , it is a gain. $Profit = S P - C P = 900 - 750 = ₹150$ WHY divide by 750: you invested the $C P$ ; profit is measured per rupee invested. $Profit % = \frac{150}{750} \times 100 = 20%$

L1·Q2 — Discount amount and discount%

A jacket is tagged $M P = ₹2000$ and sold for $S P = ₹1700$ . Find the discount and the discount%.

Recall Solution

WHAT we compare: $S P$ to $M P$ (the tag the customer saw first). $Discount = M P - S P = 2000 - 1700 = ₹300$ WHY divide by 2000: discount is a reduction from the tag, so the base is $M P$ . $Discount % = \frac{300}{2000} \times 100 = 15%$

L1·Q3 — Plain simple interest

$P = ₹6000$ is deposited at $R = 5%$ per year for $T = 4$ years. Find the simple interest $S I$ and the final amount $A$ .

Recall Solution

WHY this formula: simple interest charges the same percentage of the original $P$ each year, so we just multiply by $T$ . $S I = \frac{P \times R \times T}{100} = \frac{6000 \times 5 \times 4}{100} = ₹1200$ The /100 is there because $R$ is a percent (5 out of 100). $A = P + S I = 6000 + 1200 = ₹7200$

Level 2 — Application

(One rearrangement needed — solve for the hidden quantity.)

L2·Q1 — Find SP from profit%

$C P = ₹1600$ , desired profit $p = 35%$ . Find $S P$ .

Recall Solution

WHAT the symbols mean: here $p$ is the profit percentage ( $p = 35$ ). WHAT the formula says: you keep the whole $C P$ (that's the "1") plus $\frac{p}{100} = 0.35$ extra. $S P = C P (1 + \frac{p}{100}) = 1600 (1 + 0.35) = 1600 \times 1.35 = ₹2160$

L2·Q2 — Back out CP (the divide-not-multiply case)

A cycle is sold for $S P = ₹4200$ at a profit of $p = 20%$ . Find $C P$ .

Recall Solution

WHY divide: the $20%$ sits on the unknown $C P$ , not on $S P$ . With $p = 20$ , $S P = C P (1 + \frac{p}{100}) \Rightarrow 4200 = C P \times 1.20$ $C P = \frac{4200}{1.20} = ₹3500$ Check: $3500 \times 1.20 = 4200$ ✓.

L2·Q3 — Find the rate

$P = ₹8000$ earns $S I = ₹1440$ in $T = 3$ years. Find $R$ .

Recall Solution

Start from $S I = \frac{P R T}{100}$ and isolate $R$ : $R = \frac{100 \times S I}{P \times T} = \frac{100 \times 1440}{8000 \times 3} = \frac{144000}{24000} = 6%$ Check: $\frac{8000 \times 6 \times 3}{100} = ₹1440$ ✓.

L2·Q4 — Time given as months

$P = ₹9000$ , $R = 8%$ per year, $T = 8$ months. Find $S I$ .

Recall Solution

WHY convert: $R$ is per year, so $T$ must be in years. $T = \frac{8}{12} = \frac{2}{3} years$ $S I = \frac{9000 \times 8 \times \frac{2}{3}}{100} = \frac{9000 \times 8 \times 2}{100 \times 3} = \frac{144000}{300} = ₹480$

Level 3 — Analysis

(Multiple bases in one chain — keep them straight.)

L3·Q1 — CP → MP → discount → SP → profit

A shopkeeper buys a lamp at $C P = ₹500$ , marks it up to $M P = ₹750$ , then gives a $12%$ discount. Find the final $S P$ and the shopkeeper's profit%.

Recall Solution

Step 1 — discount lives on $M P$ (here $d = 12$ ): $S P = M P (1 - \frac{d}{100}) = 750 \times 0.88 = ₹660$ Step 2 — profit lives on $C P$ (switch the base!): $Profit = S P - C P = 660 - 500 = ₹160$ $Profit % = \frac{160}{500} \times 100 = 32%$ WHY two bases: the $12%$ describes the customer's saving (base $M P$ ); the $32%$ describes how well the shopkeeper's money worked (base $C P$ ). See Marked Price and Successive Discounts.

L3·Q2 — Find the required markup

A dealer's $C P = ₹400$ . He wants a $20%$ profit after giving a $20%$ discount. What $M P$ should he mark?

Recall Solution

Step 1 — target $S P$ from profit (base $C P$ ): $S P = 400 \times 1.20 = ₹480$ Step 2 — recover $M P$ from discount (base $M P$ , so divide): $S P = M P \times 0.80 \Rightarrow M P = \frac{480}{0.80} = ₹600$ Sanity: mark ₹600, take $20%$ off → ₹480, subtract $C P$ ₹400 → ₹80 profit $= 20%$ of 400 ✓.

L3·Q3 — Find the principal from interest

A sum grows to $A = ₹6200$ in $T = 4$ years at $R = 6%$ simple interest. Find the principal $P$ .

Recall Solution

WHY not just use $S I$ ? We are given the amount $A$ , which already includes $P$ . Use: $A = P (1 + \frac{R T}{100}) = P (1 + \frac{6 \times 4}{100}) = P \times 1.24$ WHY divide: the factor $1.24$ multiplies the unknown $P$ , so to undo it we divide (never subtract a percent of $A$ — that would use the wrong base). $P = \frac{6200}{1.24} = ₹5000$ A note on messy numbers: not every $A$ gives a whole principal. If instead $A = ₹7728$ , then $P = \frac{7728}{1.24} = ₹6232.258 \dots$ In real currency you would round to the nearest paisa, $P \approx ₹6232.26$ ; exam problems are almost always rigged (like this ₹6200 case) to land on a clean rupee value.

Level 4 — Synthesis

(Combine two whole topics, or set up an equation.)

L4·Q1 — Interest lets you afford the item

Rohan borrows $P = ₹10000$ at $R = 9%$ simple interest for $T = 2$ years. With that money he buys goods and sells them at a $30%$ profit. After repaying the loan (principal + interest), how much profit is left?

Recall Solution

Step 1 — cost of borrowing (base $P$ ): $S I = \frac{10000 \times 9 \times 2}{100} = ₹1800, A = 10000 + 1800 = ₹11800$ Step 2 — sale proceeds (base = $C P = ₹10000$ , the goods bought): $S P = 10000 \times 1.30 = ₹13000$ Step 3 — net profit = money in hand − money owed: $13000 - 11800 = ₹1200$ WHY subtract $A$ not $P$ : he must return the interest too, so the true outflow is $A$ .

L4·Q2 — Two successive discounts as one

A phone is tagged $M P = ₹20000$ with successive discounts of $10%$ then $5%$ . Find the single equivalent discount% and the $S P$ .

Recall Solution

WHY multiply, not add: the second $5%$ is taken on the already-reduced price, not on $M P$ . $S P = M P \times (1 - 0.10) \times (1 - 0.05) = 20000 \times 0.90 \times 0.95$ $= 20000 \times 0.855 = ₹17100$ Single equivalent discount: $Discount % = \frac{20000 - 17100}{20000} \times 100 = \frac{2900}{20000} \times 100 = 14.5%$ Note $14.5% \neq = 15%$ : adding $10 + 5$ over-counts. This is the Marked Price and Successive Discounts idea. The figure below draws the price as a bar that gets shorter twice, so you can see that the second cut eats a smaller bar and therefore removes fewer rupees than a naive $5%$ of the full tag would.

Figure — Profit, loss, discount, simple interest — basic applications

Figure: the ₹20000 tag (white) shrinks first by $10%$ (blue bar, −₹2000), then by $5%$ of the blue bar, not the white one (yellow bar, −₹900). The two pink chunks removed total ₹2900 = $14.5%$ , not the ₹3000 a flat $15%$ would suggest.

L4·Q3 — Solve for CP with a two-condition sentence

If an article is sold for $₹960$ there is a $20%$ profit. At what price must it be sold to get a $35%$ profit?

Recall Solution

Step 1 — find the shared base $C P$ (divide): $960 = C P \times 1.20 \Rightarrow C P = \frac{960}{1.20} = ₹800$ WHY divide here: the $1.20$ is a factor multiplying the unknown $C P$ ; to peel it off and leave $C P$ alone you must do the inverse operation, division — subtracting a percent of ₹960 would wrongly use ₹960 (an $S P$ ) as the base instead of $C P$ . Step 2 — new $S P$ at $35%$ (base still $C P$ ): $S P = 800 \times 1.35 = ₹1080$ WHY find $C P$ first: both percentages hang off the same $C P$ ; once you know it, either target is one multiplication away. This is a two-step linear equation.

Level 5 — Mastery

(Reverse engineering + a subtle trap; think before writing.)

L5·Q1 — Loss turns to profit: find CP

By selling an item at $₹720$ a shopkeeper loses $10%$ . To gain $20%$ instead, at what price must he sell it?

Recall Solution

Step 1 — loss means base $C P$ is bigger; divide by $(1 - loss)$ : $720 = C P (1 - \frac{10}{100}) = C P \times 0.90 \Rightarrow C P = \frac{720}{0.90} = ₹800$ WHY $0.90$ : at a $10%$ loss he keeps only $90%$ of $C P$ . Step 2 — new $S P$ for $20%$ gain (same $C P$ ): $S P = 800 \times 1.20 = ₹960$

L5·Q2 — Marked-up + discount that still loses money

A shopkeeper marks goods $50%$ above $C P$ , then during a sale gives a $40%$ discount. Does he profit or lose, and by what percent? Take $C P = ₹100$ (a convenient base — see Percentages).

Recall Solution

WHY set $C P = 100$ : percentages are ratios, so the answer is independent of the actual price — pick 100 to make arithmetic trivial. Step 1 — mark up (base $C P$ ): $M P = 100 \times 1.50 = ₹150$ . Step 2 — discount (base $M P$ ): $S P = 150 \times (1 - 0.40) = 150 \times 0.60 = ₹90$ . Step 3 — compare $S P$ to $C P$ : $90 < 100$ , so it is a loss. $Loss % = \frac{100 - 90}{100} \times 100 = 10%$ Insight: a big markup can still lose money after a big discount, because the discount's base ( $M P = 150$ ) is larger than the markup's base ( $C P = 100$ ).

L5·Q3 — Equal SI, find the split

$₹12000$ is split into two parts. The first is lent at $R_{1} = 10%$ and the second at $R_{2} = 8%$ , both for $2$ years (simple interest). The two interests are equal. Find each part.

Recall Solution

Set up (base = each part; see Linear Equations): let the first part be $x$ , so the second is $12000 - x$ . Time and the /100 are common, so equal interest means: $x \times 10 \times 2 = (12000 - x) \times 8 \times 2$ $20 x = 16 (12000 - x)$ $20 x = 192000 - 16 x \Rightarrow 36 x = 192000 \Rightarrow x = ₹5333.33 \dots$ Hmm — not whole. Keep it exact as a ratio: the parts are in the ratio $R_{2} : R_{1} = 8 : 10 = 4 : 5$ (higher rate needs less money for the same interest). $First part = \frac{4}{9} \times 12000 = ₹5333. \overline{3}, Second = \frac{5}{9} \times 12000 = ₹6666. \overline{6}$ Check the ratio logic: equal interest ⇒ $x R_{1} = (12000 - x) R_{2}$ ⇒ $x : (12000 - x) = R_{2} : R_{1}$ .

Connections

Percentages — every solution is a percentage of a chosen base; the " $C P = 100$ " trick (L5·Q2) comes from here.
Ratio and Proportion — the inverse split in L5·Q3 is a ratio.
Compound Interest — try re-doing the SI problems where interest stacks; answers differ.
Marked Price and Successive Discounts — L4·Q2 lives here.
Linear Equations — L2·Q2, L4·Q3, L5·Q3 are one-variable equations in disguise.