Free Practice Questions for Profit-and-loss in Maths

Let $x$ be the original price paid by $P$
Selling price for $\mathrm{P}=\mathrm{x}\times \frac{100+20}{100}=1.2\mathrm{x}$
Selling price of $Q=1.2x\times \frac{100+30}{100}=1.56x$
Given, $1.56\mathrm{x}-1.2\mathrm{x}=108$
$\Rightarrow 0.36\mathrm{x}=108$
$\Rightarrow \mathrm{x}=300$

According to question
$\begin{array}{rl}& 450\times \frac{100-10}{100}\times \frac{100-x}{100}\times \frac{100-20}{100}=267.30\\ & 450\times \frac{90}{100}\times \frac{100-x}{100}\times \frac{80}{100}=267.30\\ & 100-\mathrm{x}=\frac{267.30\times 100\times 100\times 100}{450\times 90\times 80}\\ & 100-\mathrm{x}=82.5\\ & \mathrm{x}=17.5\end{array}$

$12.5\mathrm{\%}=\frac{1}{8},15\mathrm{\%}=\frac{3}{20}$
MP of item $=805\times \frac{8}{7}=$₹ 920
When he had not offered the discount, he would have earned a profit of $15\mathrm{\%}$.

So, cost price of the item $=920\times \frac{20}{23}=$₹ 800

$15\mathrm{\%}=\frac{3}{20}$
Cost price of the article $==170\times \frac{20}{17}=$₹ 200
Required profit $\mathrm{\%}=\frac{206.40-200}{200}\times 100$
$=\frac{6.40}{200}\times 100$
= 3.2 %

MP $\times \frac{90}{100}\times \frac{95}{100}=3420$
MP $=\frac{3420\times 100\times 100}{90\times 95}$
MP $=$₹ 4000

Selling price when first discount is given $=7,500-7,500\times \frac{10}{100}$
$=7,500-750=$₹ 6750
But the article was sold for ₹ 5,805
so, second discount percent $=\frac{6750-5805}{6750}\times 100$
$=\frac{945}{6750}\times 100=14\mathrm{\%}$

Cost price of 360 packets $=360\times 15=5400$
Number of $60\mathrm{\%}$ packets $=360\times \frac{60}{100}=216$
Selling price of 216 packets $=216\times 18=3888$
Number of $40\mathrm{\%}$ packets $=360-216=144$
Selling price of 144 packets $=144\times 20=2880$
Total SP $=3888+2880=6768$
Required gain $\mathrm{\%}=\frac{6768-5400}{5400}\times 100=25.33\mathrm{\%}$

$40\mathrm{\%}=\frac{2}{5}$
Required Marked price $=900\times \frac{5}{3}=1,500$

Alternative method

$20\mathrm{\%}=\frac{1}{5},30\mathrm{\%}=\frac{3}{10}$
Cost price of 40 dozen $=1440\times \frac{5}{6}=$₹ 1200
Cost price of 1 dozen $=\frac{1200}{40}=$₹ 30
Cost price of 20 dozen $=30\times 20=$₹ 600
Required selling price of 20 dozen bananas $=600\times \frac{13}{10}=$₹ 780

There is a gain of $10.6\mathrm{\%}$ on the the article.
then, $110.6\mathrm{\%}\Rightarrow$₹ 254.38
$1\mathrm{\%}\Rightarrow 2.3$
$100\mathrm{\%}\Rightarrow 230$
So, Cost Price of article is ₹ 230.
Required loss when it is sold at ₹ 224.25 $=\frac{230-224.25}{230}\times 100$
$=\frac{5.75}{230}\times 100=2.5\mathrm{\%}$

Selling Price $=$ Cost price $+$ profit
Selling price $=100\mathrm{\%}+20\mathrm{\%}=120\mathrm{\%}$
According to question
$20\mathrm{\%}\to 30.80$
$10\mathrm{\%}\to 15.40$
Hence, $120\mathrm{\%}\to 12\times 15.4=$₹ 184.80

Total cost price $=22000+3000=25000$
Selling price $=25000\times \frac{112}{100}=$₹ 28000
New selling price $=28000-500=$₹ 27500
Profit $\mathrm{\%}=\frac{(27500-25000)}{25000}\times 100=\frac{2500}{250}=10\mathrm{\%}$

According to question
$36\mathrm{SP}=36\mathrm{CP}+12\mathrm{SP}$
$\begin{array}{rl}& 24\mathrm{SP}=36\mathrm{CP}\\ & \frac{SP}{CP}=\frac{36}{24}\\ & \frac{SP}{CP}=\frac{3}{2}\end{array}$
Profit % $=\frac{3-2}{2}\times 100$
$\begin{array}{rl}& =\frac{1}{2}\times 100\\ & =50\mathrm{\%}\end{array}$

Given, Selling price $=$₹ 720
Cost price $=$₹ 640
Market price $=720\times \frac{100}{90}=$₹ 800
Required profit $\mathrm{\%}=\frac{800-640}{640}\times 100$
$=\frac{160}{640}\times 100$
$=25\mathrm{\%}$

Successive Discount formula $=A+B-\frac{A\times B}{100}$
For $12\mathrm{\%}$ and $5\mathrm{\%}$
$12+5-\frac{12\times 5}{100}=17-0.60=16.4\mathrm{\%}$
Now for $16.4\mathrm{\%}$ and $7\mathrm{\%}$
$16.4+7-\frac{16.4\times 7}{100}=23.4-1.148=22.252\mathrm{\%}$

Selling Price $=$ Cost Price $\times \frac{(100-\text{ Loss })}{100}$
$18400=$ Cost Price $\times \frac{(100-8)}{100}$
Cost Price $=\frac{18400\times 100}{92}=$₹ 20000
New selling price $=20000\times \frac{(100+5)}{100}=200\times 105=$₹ 21000

Let total goods be 12 , then
$\frac{1}{3}\times 12=4$ goods at $15\mathrm{\%}$ profit
$\frac{25}{100}\times 12=3$ goods at $20\mathrm{\%}$ profit
$12-(4+3)=5$ goods at $10\mathrm{\%}$ loss
Overall profit/loss $\mathrm{\%}=\frac{4\times 15\mathrm{\%}+3\times 20\mathrm{\%}+5\times (-10\mathrm{\%})}{12}=\frac{60\mathrm{\%}+60\mathrm{\%}-50\mathrm{\%}}{12}=\frac{70\mathrm{\%}}{12}$
According to question
$\frac{70\mathrm{\%}}{12}=$₹ 350
$1\mathrm{\%}=$₹ 60
$100\mathrm{\%}=$₹ 6000

Selling Price $=$ Cost Price $\times \frac{(100-\text{ Loss })}{100}$
$846=$ Cost Price $\times \frac{(100-6)}{100}$
Cost Price $=\frac{846\times 100}{94}=$₹ 900
New selling price $=900\times \frac{(100+8)}{100}=9\times 108=$₹ 972

Let the original selling price be $x$.
According to question
$x\times \frac{(100-25)}{100}=$₹ 6873
$x=\frac{6873\times 4}{3}=$₹ 9164
Hence, Original selling price of a perfume is ₹ 9164 .

As per the question,
Cost price of 18 oranges $=\frac{151.2}{80}\times 100=189\mathrm{rs}$
So, Cost price of 1 orange $=\frac{189}{18}=$₹ 10.5
For $30\mathrm{\%}$ profit, selling price of 1 orange $=\frac{10.5}{100}\times 130=$₹ 13.65
No of oranges $=\frac{819}{13.65}=60$ oranges