MATHS M.C.Q. [1 Marks Each]

Let $C.P$ of one article $= 1\ Rs.$
Then $C.P$ of $40$ article $= 40\ Rs.$
According to the question
$S.P$ of $50$ article $= C.P$ of $40$ article $= 40\ Rs.$
We assume that $C.P$ of one article $= 1\ Rs.$
then $C.P$ of $50$ article $= 50\ Rs.$
Loss $= 50 - 40 = 10\ Rs.$
Loss $\%=\frac{10}{50}\times100=20\%$

$40\%$ of students out of $40$ students will be
$=\big(\frac{40}{100}\big)\times40$
$=\frac{2}{5}\times40$
$= 16$

$Rs. 1 = 100$ paise
$50$ paise $: 100$ paise
$50 : 100$
$1 : 2$

The given ratio $= 4 : 7$
It can be expressed in percentage as:
$\frac{4}{7}\times100\%=\frac{400}{7}$
$=57\frac{1}{7}\%$

$1km = 1000m$
$10m : 1000m$ or $10 : 1000$ or $1 : 100$

Discount on $Rs. 100 = 25$
$\therefore$ Sale price $= 100 - 25 = Rs. 75$

 Given, marked price of an article $= Rs. 1200$
Discount $\% = 12\%$ Discount = Discount $\%$ on marked price $= A × 1200 = 12 × 12 = Rs. 144$
Selling price $=$ Marked price $–$ Discount Selling price
$= Rs. 11200 – Rs. 144 = Rs. 1056$
Hence, option $(a)$ is correct.

Let the total number of examinees be $100$
Percentage of examinees passed $= 65\%$
Percentage of examinees faield $= (100 - 65)\% = 35\%$
$\therefore\frac{35}{100}\times\text{x}=420$
$\Rightarrow35\text{x}=42000$
$\Rightarrow\text{x}=1200$

Number of science student
$=340\times\frac{55}{100}=187$
$\therefore$ Number of students of Commerce
$= 340 - 187 = 153$

 The marked price of an article $= Rs. 80$ Sold price of the article $= Rs. 76$ We know that,
Selling price = Marked price $–$ Discount price
Discount $= Rs. 80 − Rs. 76 = Rs. 4$
Discount$\% = \frac{4}{80}\times100=\frac{40}{80}=5\%$
Hence, option $(a)$ is correct.

$ SP = Rs. 100$
Gain $= Rs. 20$
$CP = Rs. (100 - 20)$
$= Rs. 80$
$\text{Gain percentsge}=\Big(\frac{\text{gain}}{\text{CP}}\times100\Big)\%$
$=\Big(\frac{20}{80}\times100\Big)\%$
$=25\%$

 $\frac{2000\times4\times\text{R}}{100}=400$
$\Rightarrow\text{R}=\frac{400\times100}{2000\times4}=5$

 $50\%$ of students out of $20$ students $= 50\%$ of $20$
$=\Big(\frac{50}{100}\Big)\times20$
$=\frac{1}{2}\times20$
$=10$

 Present population $= 4000$
Rate of growth $= 5\%$
To find the population of the town after $2$ years, we have:
$\text{A}=\text{P}\times\Big(1+\frac{\text{R}}{100}\Big)^\text{n}$
$=4000\times\Big(1+\frac{5}{100}\Big)^2$
$=4000\times\Big(\frac{5}{100}\Big)^2$
$=4000\times\Big(\frac{21}{20}\Big)\times\Big(\frac{21}{20}\Big)$
$=(10\times21\times21)$
$=4410$

 $4 : 25=\frac{4}{25}\times100\%=16\%$

$CP. = Rs. 2,160$ and Rate of $VAT = 8\%$
Let $C.P$ without $VAT$ is $Rs. 100$ then price including $VAT = 100 + 8 = Rs. 108$
$\therefore$ When price including $VAT$ is $Rs. 108,$ then original price $=\frac{100}{108}$
$\therefore$ When price including $VAT$ is $Rs. 2,160$
The original price $=\frac{100}{108}\times2160 =\text{Rs. }2,000$

Let the $CP$ be $Rs. x$
Then, we have:
$625 - x = x - 435$
$⇒ x + x = 625 + 435$
$⇒ 2x = 1060$
$\therefore x = Rs. 530$

Let the $CP$ be $Rs .x.$
Then, we have:
$(105\%$ of $x) - (95\%$ of $x) = 5$
$\Rightarrow\Big(\frac{105}{100}\times\text{x}\Big)-\Big(\frac{95}{100}\times\text{x}\Big)=5$
$\Rightarrow\Big(\frac{105\times\text{x}}{100}\Big)-\Big(\frac{95\text{x}}{100}\Big)=5$
$\Rightarrow\Big(\frac{105\text{x}}{100}-\frac{95\text{x}}{100}\Big)=5$
$\Rightarrow\frac{10\text{x}}{100}=5$
$\Rightarrow\frac{\text{x}}{100}=5$
$\Rightarrow\text{x}=50$
$\therefore\text{CP}=\text{Rs. }50$

 Present population $(P) = 24000$
Rate of increase $(R) = 5\%$ p.a
Period $(n) = 2$ years
$\therefore$ Population after $2$ years $=\text{P}\Big(1+\frac{\text{R}}{100}\Big)^{\text{n}}$
$=\text{Rs. }24000\Big(1+\frac{5}{100}\Big)^2$
$=\text{Rs. }24000\times\frac{21}{20}\times\frac{21}{20}$
$=26460$

 Since, sales tax is charged on the sale of an item by the government and is added to the bill amount.
Shyama purchase a scooter of costing $= Rs. 36450$
Sales tax paid $= 9\%$
So, total amount paid by her $= Rs. 36450$ of $9\% + Rs. 36450$
$=\frac{9}{100}\times36450+36450$
$= 9 × 364.5 + 36450 = 3280.5 + 36450 = Rs. 39730.5$
Hence, option $(b)$ is correct.

 $4:25=\frac{4}{25}\times100\%=16\%$