Free Practice Questions for Profit-and-loss in Maths

As per the question,
Selling price of car $=$₹ 66000
So cost price of the car $=\frac{66000}{110}\times 100=$₹ 60000
If he sold 600 more than previous selling price,
New selling price $=66000+600=$₹ 66600
Now, Profit $\mathrm{\%}=\frac{66600-60000}{60000}\times 100$
$=\frac{6600}{600}=11\mathrm{\%}$

Let the cost price of radio is $100x$ and the cost price of TV is $100\mathrm{y}$
As per the question,
$120x+115y=20850$.............(i)
$115x+120y=21450$............ (ii)
Subtracting equation(i) from equation(ii)
$5y-5x=600$
$y-x=120........(iii)$
By adding equation (ii) and (i)
$\begin{array}{rl}& 235x+235y=42300\\ & x+y=180\dots \dots \dots \dots .\dots \text{ (iv) }\end{array}$
Solving eq(iii) and eq(iv)
$y=150\text{ and }x=30$
So Cost price of TV set $=100\times 150=$₹ 15000

As per the question,
Cost price of cooler $=$₹ 5400
If he gets $25\mathrm{\%}$ profit then,
Selling price of cooler $=\frac{5400}{100}\times 125$ = ₹ 6750
If he gives $25\mathrm{\%}$ discount
Then, Marked price of cooler $=\frac{6750}{75}\times 100=$₹ 9000

Given,
Marked price of the article $=$₹ $480$
Selling price $=\frac{480}{100}\times 82=$₹ $393.6$
So, Cost price of the article $=$ Selling price $-$ Profit
$=393.6-63.6$
$=$₹ $330$
Profit $\mathrm{\%}=\frac{63.6}{330}\times 100=19\frac{3}{11}\mathrm{\%}$

Let $x$ be the required pencils.
$\begin{array}{rl}& \frac{54x}{9}-\frac{35x}{10}=350\\ & \frac{540x-315x}{90}=350\\ & 225x=350\times 90\\ & x=\frac{350\times 90}{225}=140\end{array}$

We know CP is $100\mathrm{\%}$.
So, $85\mathrm{\%}\Rightarrow 5,270$
$100\mathrm{\%}\Rightarrow 6,200$
$\mathrm{CP}=6200$
Required profit $=\frac{6,324-6200}{6200}\times 100=2\mathrm{\%}$

Three successive discounts of 10%, 15% and 20% is given.

Now, dealer marked the price of an item 50 % above the cost price

(500-459)unit = 410

41 unit = 410

1 unit   = 10

Required SP = 459 unit = 459×10 = 4590

$12.5\mathrm{\%}=\frac{1}{8}$
Market price $=1,225\times \frac{8}{7}=$₹ $1,400$
Selling price $=1,400\times \frac{100-9.5}{100}$
$=14\times 90.5=$₹ $1,267$

CP for $8$ pieces of TV sets $=24,000\times 8=$₹ $1,92,000$
Given, SP for $8$ pieces of TV set $=$₹ $2,26,560$
Profit $=2,26,560-1,92,000=$₹ $34560$
Profit $\mathrm{\%}=\frac{34560}{192000}\times 100=18\mathrm{\%}$

Let $x$ be the cost price of item,
According to question,
$\begin{array}{rl}& x\times \frac{112}{100}\times \frac{110}{100}\times \frac{115}{100}=56672\\ & x\times \frac{28}{25}\times \frac{11}{10}\times \frac{23}{20}=56672\\ & x=\frac{56672\times 25\times 10\times 20}{28\times 11\times 23}\\ & x=40000\end{array}$

Using formula 

Badri should sell scooter for $=17,000\times \frac{100}{85}\times \frac{118}{100}$
$\begin{array}{rl}& =17,000\times \frac{20}{17}\times \frac{59}{50}\\ & =Rs.23,600\end{array}$

CP of 10 chairs $=450\times 10=4500$
SP of 10 chairs $=3825$
Required discount $=\frac{4500-3825}{4500}\times 100$
$\begin{array}{rl}& =\frac{675}{4500}\times 100\\ & =15\mathrm{\%}\end{array}$
$$

$\mathrm{CP}$ is $25\mathrm{\%}$ less than MP
So, $75\mathrm{\%}=1545$
$100\mathrm{\%}=2060$
$\mathrm{MP}=2060$
When Article is sold at $18\mathrm{\%}$ discount
$\mathrm{SP}=2060\times \frac{82}{100}$
$=$ Rs. $1689.2$
Required profit $\mathrm{\%}=\frac{1689.2-1545}{1545}\times 100$

$=9.33\mathrm{\%}$

He sold the item at $25\mathrm{\%}$ loss.
$\begin{array}{rl}\text{So,}75\mathrm{\%}& =870\\ 25\mathrm{\%}& =290\\ 100\mathrm{\%}& =1160\end{array}$
CP of article is Rs. 1160.
To earn the profit of $25\mathrm{\%}$
$\mathrm{SP}=1160\times \frac{125}{100}=\text{ Rs. }1450$

Required cost of production of the table $=1100\times \frac{100}{105}\times \frac{100}{110}\times \frac{100}{112}$
$=850.34\sim 850$

$25\mathrm{\%}=\frac{1}{4}\frac{\leftarrow \text{ discount }}{\leftarrow \text{ market price }}$
We know, $\mathrm{SP}=3$ unit
3 unit $=614.25$
1 unit $=204.75$
4 unit $=819$
Required cost price $=819\times \frac{100}{117}=700$

$\begin{array}{rl}\frac{CP}{MP}=& \frac{100-D\mathrm{\%}}{100+P\mathrm{\%}}\\ =& \frac{100-20}{100+4}\\ =& \frac{80}{104}\end{array}$
Required $\mathrm{\%}=\frac{104-80}{80}\times 100$
$=\frac{24}{80}\times 100$
$=30\mathrm{\%}$

Single equivalent discount $=(a+b-\frac{ab}{100})\mathrm{\%}$
Single equivalent discount $5\mathrm{\%}$ and $10\mathrm{\%}$
$\begin{array}{rl}& \Rightarrow (5+10-\frac{5\times 10}{100})\mathrm{\%}\\ & \Rightarrow (15-0.5)\mathrm{\%}=14.5\mathrm{\%}\end{array}$
Single equivalent discount $14.5\mathrm{\%}$ and $16\mathrm{\%}$
$\begin{array}{rl}& \Rightarrow (14.5+16-\frac{14.5\times 16}{100})\mathrm{\%}\\ & \Rightarrow (30.5-2.32)\mathrm{\%}=28.18\mathrm{\%}\end{array}$

Let cost price of goods be Rs. 100
Marked price of goods $=100\times \frac{132}{100}=$ Rs. 132
Selling price when discount is $20\mathrm{\%}=132\times \frac{80}{100}=$ Rs. $105.6$
Selling price when discount is $40\mathrm{\%}=132\times \frac{60}{100}=$ Rs. $79.2$
Total selling price of goods $=3\times 132+1\times 105.6+1\times 79.2=$ Rs. $116.16$

Profit % = $\frac{116.16-100}{100}\times 100$ = 16.16 %

Cost price of the book for retailer $=312\times \frac{100}{120}=$ Rs. 260
Cost price of the book for the wholesale dealer $=260\times \frac{100}{125}=$ Rs. 208