Math Ncert Exemplar 2019 Solutions for Class 8 Maths Chapter 9 - Comparing Quantities

Answer:

Compound interest is the interest calculated on the principal and the interest accumulated over the previous period. Whereas, simple interest is the interest calculated only on principal. Thus, other than the first year, the interest compounded annually is always greater than that in simple interest. And, in first year the compound interest and the simple interest are equal.
i.e. if T > 1 then C > S and if T = 1 then C = S. Thus, either C > S or C = S.
Hence, the correct answer is option (b).

Answer:

As for time > 1 year compound interest is always greater than simple interest at the same rate.
Therefore for compound interest to be equal to the simple interest, rate at which compound interest is evaluated should be less than that of the rate at which simple interest is being evaluated.
Thus, R < r.
Hence, the correct answer is option (b).

Answer:

Principal (P) = ₹50000
Rate (R%) = 4%
Time (T) = 2 years 
As, A = $\mathrm{P}{\left(1+\frac{\mathrm{R}}{100}\right)}^{\mathrm{T}}$
$$\begin{gathered} \therefore \mathrm{A}=50000{\left(1+\frac{4}{100}\right)}^2 \\ =50000{\left(\frac{104}{100}\right)}^2 \\ =50000\times \frac{104}{100}\times \frac{104}{100} \\ =₹54080 \end{gathered}$$
Thus, compound interest (CI) = A – P
                                                = ₹54080 – ₹50000
                                                = 4080
Hence, the correct answer is option (b).

Answer:

Answer:

$$\begin{gathered} \frac{90}{100}\mathrm{of} x=315 \mathrm{km} \\ \Rightarrow \frac{90}{100}\times \mathrm{x}=315 \\ \Rightarrow \mathrm{x}=\frac{315\times 100}{90}=350 \mathrm{km} \end{gathered}$$
Hence, the correct answer is option (b).

Answer:

Let the marked price of the article be x.
Given that the price of the article is ₹360 and the discount given by the shopkeeper on the article is of 10%.
Thus, accordingly to the question,
The selling price of the article =$x-\left(\frac{10}{100} \mathrm{of} x\right)$
                                                 $$\begin{gathered} =x-\frac{x}{10} \\ =₹\frac{9x}{10} ...\left(1\right) \end{gathered}$$
Now Gain% = 25%
$$\begin{gathered} \therefore 25\%=\frac{\mathrm{SP}-\mathrm{CP}}{\mathrm{CP}} \\ \Rightarrow \frac{25}{100}=\frac{\mathrm{SP}-360}{360} \left(\because \mathrm{Cost} \mathrm{price}=₹360\right) \\ \Rightarrow \frac{1}{4}=\frac{{\displaystyle \frac{9x}{10}}–360}{360} \left[\mathrm{From} \left(1\right)\right] \\ \Rightarrow 90+360=\frac{9x}{10} \\ \Rightarrow \frac{4500}{9}=x \\ \Rightarrow x=₹500 \end{gathered}$$
So, the marked price is ₹500.
Hence, the correct answer is option (a).

Answer:

Marked price = x
Discount% = a%
$\begin{array}{rcl}\therefore \mathrm{Discoiunt} & =& \frac{a}{100}\times x\\ & =& x\times \frac{a}{100}\end{array}$
Hence, the correct answer is option(c).

Answer:

Given: Principal(P) = ₹100000
            Rate(R) = 12% per annum compounded half yearly
     Amount (A) = ₹112360
Let 'n' be the time period.
As, A $=\mathrm{P}{\left(1+\frac{\mathrm{R}}{100}\right)}^{2\mathrm{n}}$
$$\begin{gathered} \Rightarrow 112360=100000{\left(1+\frac{12}{200}\right)}^{2n} \\ \Rightarrow \frac{112360}{100000}={\left(\frac{212}{200}\right)}^{2n} \\ \Rightarrow {\left(\frac{53}{50}\right)}^2={\left(\frac{53}{50}\right)}^{2n} \\ \Rightarrow 2n=2 \\ \Rightarrow n=1 \end{gathered}$$

So, the time period is 1 year.
Hence, the correct answer is option (b).

Answer:

If interest is compounded half-yearly, then the rate is half the given annual rate and time is double the given numbers of years.
Hence, the correct answer is option (d).

Answer:

Since, the sales tax is charged on the sale of an item by the government and is added to the bill amount.
Therefore, amount paid by Shyama on the purchase of a scooter costing ₹36450
$$\begin{gathered} =36450+\left(\frac{9}{100} \mathrm{of} 36450\right) \\ =36450+\left(9\times 364.5\right) \\ =36450+3280.5 \\ =₹39730.5 \end{gathered}$$
Hence, the correct answer is option (b).

Answer:

Given: The marked price of an article = ₹80
            The selling price of an article = ₹76
$$\begin{gathered} \therefore \mathrm{Discount}\%=\frac{\mathrm{Marked} \mathrm{price}-\mathrm{selling} \mathrm{price}}{\mathrm{Marked} \mathrm{price}} \\ \Rightarrow \mathrm{Discount}\%=\left(\frac{80-76}{80}\right)\times 100 \\ \Rightarrow \mathrm{Discount}\%=\frac{4}{80}\times 100 \\ \Rightarrow \mathrm{Discount}\%=5\% \end{gathered}$$
Hence, the correct answer is option (a).

Answer:

Cost price of taped recorded for A = ₹8000
Selling price of tape recorded for A = $8000+\frac{20}{1000}$of 8000
                                                         $$\begin{gathered} =8000\left(1+\frac{20}{100}\right) \\ =8000\times \frac{120}{100} \\ =₹9600 \end{gathered}$$

So, cost price of tape recorded for B = ₹9600
Selling price of tape recorder for B = $9600+\frac{20}{100} \mathrm{of} 9600$
                                                         = $$\begin{gathered} =9600\left(1+\frac{20}{100}\right) \\ =9600\times \frac{120}{100} \\ =₹11520 \end{gathered}$$

As, ₹11520 > ₹9600
Therefore, selling price of tape recorder for B is greater than the selling price of tape recorded for A.
Thus, A earns less profit than B.
Hence, the correct answer is option (c).

Answer:

Cost price of teapot = ₹120
Selling price of teapot = $120+\left(\frac{5}{100} \mathrm{of} 120\right) \left(\because \mathrm{profit}\%=5\%\right)$
                                    $$\begin{gathered} =120+\left(\frac{5}{100}\times 120\right) \\ =120+6 \\ =₹126 \end{gathered}$$
Cost price of set of cups = ₹400
Selling price of set of cups = $400-\frac{5}{100}\mathrm{of} 400\left(\because \mathrm{loss}\%=5\%\right)$
                                           = 400 – 20
                                           = ₹380
Thus, the total amount received by her 
= 126 + 380
= ₹506
Hence, the correct answer is option (c).

Answer:

Let the marked price of the jacket be x.
Discount % on marked price = 20%
Selling price of jacket = ₹1120
Then, $x-\left(\frac{20}{100} \mathrm{of} x\right)=1120$
$$\begin{gathered} \Rightarrow x-\frac{20x}{100}=1120 \\ \Rightarrow \frac{80x}{100}=1120 \\ \Rightarrow x=\frac{1120\times 100}{80} \\ \Rightarrow x=1400 \end{gathered}$$

So, marked price of the jacket is ₹1400.
Hence, the correct answer is (b).

Answer:

Since, the rate of interest is calculated after every three months. So, the time period for amount in a year will be 4 times. If amount is taken for 2 years then compound interest will be charged 8 times.
Hence, the correct answer is option (a).

Answer:

The price of a washing machine = ₹13500
VAT is included in the price = 8%
∴ The original price of the washing machine  including 8% VAT $=13500-\left(13500\times \frac{8}{100}\right)$
                                                                                                        = 13500 – 1080
                                                                                                        = ₹12,420
Hence, the correct answer is option (a).

Answer:

Cost price of the electric icon = ₹900
Gain% = 10%
So, selling price of the electric iron
$$\begin{gathered} =900+\left(\frac{10}{100} \mathrm{of} 900\right) \\ =900+90 \\ =₹990 \end{gathered}$$

Loss% on another electric iron = 5%
Cost price of the another electric iron = ₹1200
So, the selling price of the another electric iron
$$\begin{gathered} =1200-\left(\frac{5}{100} \mathrm{of} 1200\right) \\ =1200-60 \\ =₹1140 \end{gathered}$$

Total amount paid by Avinash for purchasing electric irons = 900 + 1200
                                                                                               = ₹2100

Total amount received by Avinash for irons = 990 + 1140
                                                                      = ₹2130
 Thus, the profit = 2130 – 2100
                           = ₹30

Hence, the correct answer is option (c).

Answer:

Cost price of TV set = ₹26250
VAT of 5% is included in the cost price of the TV set.
∴ Original price $=26250-\left(\frac{5}{100} \mathrm{of} 26250\right)$
                           $$\begin{gathered} =26250-\left(\frac{5}{100}\times 26250\right) \\ =26250-1312.5 \\ =₹24937.5 \end{gathered}$$

Hence, the correct answer is option (c).

Answer:

40% of (100 – 20% of 300)
$$\begin{gathered} =\frac{40}{100}\times \left(100-\frac{20}{100}\times 300\right) \\ =\frac{40}{100}\times \left(100-60\right) \\ =\frac{40}{100}\times 40 \\ =16 \end{gathered}$$

Hence, the correct answer is option (c).

Answer:

Cost price of car bought by Radhika = ₹250000
Its price decreased by 10% next year so now the price becomes
$$\begin{gathered} =250000-\left(\frac{10}{100}\mathrm{of} 250000\right) \\ =250000-\left(\frac{10}{100}\times 250000\right) \\ =250000-25000 \\ =₹225000 \end{gathered}$$

Further next year, its price decreased by 12%, then the price will became.
$$\begin{gathered} =225000=\frac{12}{100}\mathrm{of} 225000 \\ =225000-\left(\frac{12}{100}\times 225000\right) \\ =225000-27000 \\ =₹198000 \end{gathered}$$

In two years, overall decrease percentage
$$\begin{gathered} =\frac{250000-198000}{250000} \\ =\frac{52000}{250000}\times 100 \\ =\frac{520}{25} \\ =20.8\% \end{gathered}$$

Hence, the correct answer is option (c).

Answer:

Discount is a reduction on the marked price.

Answer:

Initial number = 150
final number = 162
Increased in number = 162 – 150 = 12
Percentage increase in number
$$\begin{gathered} =\frac{12}{150}\times 100 \\ =8\% \end{gathered}$$
Hence, the increase of a number from 150 to 162 is equal to increase of 8 percent.

Answer:

Let the price of an article be x.
Then, after the increase of 15% in the price, the price became ₹1620
So, 1620 = $x+\left(x\times \frac{15}{100}\right)$
$$\begin{gathered} \Rightarrow 1620=\frac{115x}{100} \\ \Rightarrow x=\frac{1620\times 100}{115} \\ \Rightarrow x=₹1408 \end{gathered}$$
Hence, the increase in the price
= ₹1620 – ₹1408
= ₹212

Answer:

Discount = Marked price – Selling price

Answer:

As, discount is a reduction given on marked price.
∴ Discount = Discount% of Marked price

Answer:

Sales tax is charged on the sale o fan item by the government and is added to the bill amount.

Answer:

Amount when interest is compounded annually, is given by the formula
$\mathrm{A}=\mathrm{P}\left(1+\frac{\mathrm{R}}{100}\right)\mathrm{T}$
Where A is the amount, P is the principal, R is the rate per annum and T is the time period.

Answer:

Sales tax is charged on the sale of on item by the government and is added to the bill amount.
∴ Sale tax = Tax % of bill amount.

Answer:

The time period after which the interest is added each time to form a new principal, is called the conversion period.

Answer:

Overhead expenses are the additional expenses inurned by a buyer for an item over and above it cost of purchase. 

Answer:

The discount on a item for sale is calculated on the market price. 

Answer:

When principal P is compounded semi-annually at r % per annum for t year.
i.e. Rate = $\frac{r}{2}$ and time = 2t
Then, amount = $\mathrm{P}{\left(1+\frac{r}{200}\right)}^{2t}$.
 

Answer:

Percentages are equal to fractions with denominator equal to 100.
For example : 8% = $\frac{8}{100}$

Answer:

Selling price of an article = â‚¹880
Discount % = 12%
Let the marked price of the article be $x$.
So, $x-\left(12\% \mathrm{of} x\right)=880$
$$\begin{gathered} \Rightarrow x-\left(\frac{12}{100} \mathrm{of} x\right) \\ \Rightarrow x-\frac{12x}{100}=880 \\ \Rightarrow \frac{88x}{100}=880 \\ \Rightarrow x=\frac{880\times 100}{88} \\ \Rightarrow x=1000 \end{gathered}$$
So, the marked price of the article is â‚¹1000.

Answer:

Given: Principle (P) = â‚¹8000
Rate per annum (R%) = 16%
Time period (t) = 1 year
Now, as the compound interest is applied half yearly.
$$\begin{gathered} \therefore \mathrm{A}=\mathrm{P}{\left(1+\frac{\mathrm{R}\%}{2}\right)}^{2\mathrm{t}} \\ \Rightarrow \mathrm{A}=8000 {\left(1+\frac{16\%}{2}\right)}^{2\times 1} \\ \Rightarrow \mathrm{A}=8000 {\left(1+\frac{16}{200}\right)}^2 \\ \Rightarrow \mathrm{A}=8000 {\left(1.08\right)}^2 \\ \Rightarrow \mathrm{A}=8000\times 1.1664 \\ \Rightarrow \mathrm{A}=₹9331.2 \end{gathered}$$

Answer:

Given :  The investment amount = 6,00,000
In first year, the loss % is 5%
$\therefore$ Resultant amount is first year = 6,00,000 $-\left(\frac{5}{100} \mathrm{of} 6,00,000\right)$
                                                  = 6,00,000 $-$ 30,000
                                                  = 5,70,000
In 2^nd year, the gain is 10%
So, net result 
$$\begin{gathered} =570000+\left(10\% \mathrm{of} 570000\right) \\ =570000+\left(\frac{10}{100} \mathrm{of} 570000\right) \\ =570000+57000 \\ =₹627000 \end{gathered}$$
 

Answer:

Amount = 6000${\left(1+\frac{5}{100}\right)}^3$
where â‚¹6000 is the principle and compound interest payable half yearly, then rate of interest per half year is 5% and compound interest is charged 3 times after every half year.
As, rate of interest per annum is double the rate of interest per half year.
$\therefore$ rate of interest per annum = $2\times 5\%$
                                             = 10%
And, time in year = $\frac{3}{2}=1\frac{1}{2}$ years.
Hence, if amount on the principle of â‚¹6000 ${\left(1+\frac{5}{100}\right)}^3$ and compound interest payable half - yearly, then rate interest per annum is 10% and $1\frac{1}{2}$ is the time is years.

Answer:

Selling price of an article = â‚¹112000
Gain % = 40%
Let $x$ be the cost price of the article.
Therefore, $x+\left(40\% \mathrm{of} x\right)=112000$
$$\begin{gathered} \Rightarrow x+\left(\frac{40}{100} \mathrm{of} x\right)=112000 \\ \Rightarrow x+\frac{40x}{100}=112000 \\ \Rightarrow \frac{140x}{100}=112000 \\ \Rightarrow x=\frac{112000\times 100}{140} \\ \Rightarrow x=₹80,000 \end{gathered}$$
Hence, the cost price of the article was â‚¹80,000.
 

Answer:

Let the selling price of 1 geometry box be â‚¹1
So, the selling price of 140 geometry boxes 
$$\begin{gathered} =₹1\times 140 \\ =₹140 \end{gathered}$$
Similarly, selling price of 10 geometry boxes 
$$\begin{gathered} =₹1\times 10 \\ =₹10 \end{gathered}$$
Loss = selling price of 10 geometry boxes 
$=₹10$
Total cost price of the geometry box = 140 + 10 = â‚¹150
$$\begin{gathered} \therefore \mathrm{Loss} \mathrm{percentage} =\frac{\mathrm{Loss}}{\mathrm{CP}}\times 100 \\ =\frac{10}{150}\times 100 \\ =6\frac{2}{3}\% \end{gathered}$$

Answer:

Let the cost price of 1 table be â‚¹1.
The cost price of 10 tables = Sales price of 5 tables
$\Rightarrow ₹10$ = sales price of 5 tables 
Also, the cost price of 5 tables is â‚¹5.
$\begin{array}{rcl}\therefore \mathrm{Profit} \%& =& \frac{\mathrm{Profit}}{\mathrm{CP}}\times 100\\ & =& \frac{\left(10-5\right)}{5}\times 100\\ & =& \frac{5}{5}\times 100\\ & =& 100\%\end{array}$
Hence, the profit percent in this transaction is 100%.

Answer:

Number of pens bought by Abida = 100
Rate of per pen = â‚¹3.50
So, cost of 100 pens = 100 $\times$ 3.50
                                = â‚¹350
Now, sales tax on â‚¹350 is 4%.
Therefore, the total amount paid by Abida
$$\begin{gathered} =350+\left(4\% \mathrm{of} 350\right) \\ =350+\left(\frac{4}{100}\times 350\right) \\ =350+14 \\ =₹364 \end{gathered}$$

Answer:

The cost of a tape recorder is â‚¹10800 inclusive of sales tax charged at 8%.
Let the price of the tape recorder before sales tax was charged be $x$.
So, $x=\left(8\% \mathrm{of} x\right)=1080$
$$\begin{gathered} \Rightarrow x+\left(\frac{8}{100}\times x\right)=10800 \\ \Rightarrow \frac{108x}{100}=10800 \\ \Rightarrow x=\frac{10800\times 100}{108} \\ \Rightarrow ₹10,000 \end{gathered}$$
Hence, the price of the tape recorder before sales tax charged is â‚¹10,000.

Answer:

Difference between 2500 and 500
= 2500 $-$ 500
= 2000
$\Rightarrow \frac{2000}{500}\times 100=400\%$
Hence, 2500 is greater than 500 by 400%.

Answer:

Let $x$ be the number.
Then, four times '$x$' is 4$x$.
Thus, four times a number is greater than $x$ by $4x-x$ i.e. 3$x$.
$\therefore$ The percentage increase in the number.
$=\frac{3x}{x}\times 100=300\%$

Answer:

Given: The marked price of the article discount % = 5%
So, selling price of the article 
$$\begin{gathered} =200-\left(\frac{5}{100} \mathrm{of} 200\right) \\ =200-10 \\ =₹190 \end{gathered}$$
Now, given that sales tax is charged on the article at 5% rate.
$\therefore$ Selling price including 5% sales tax
$$\begin{gathered} =190+\left(\frac{5}{100} \mathrm{of} 190\right) \\ =190+9.5 \\ =₹199.5 \end{gathered}$$
Hence, the payable amount is â‚¹199.50.

Answer:

True, As, the bacteria grows exponentially. 
Therefore, for calculating the growth of a bacteria if the rate of growth is known, the formula for calculation of amount in compound interest can be used, where A = growth after $n$ year, P = Initial number of bacteria and R is the rate of growth.

Answer:

False, 
Additional expenses made after buying an article that are included in the cost price are known as overhead expenses and not VAT.

Answer:

False,
Discount is a reduction given on marked price and not on cost price.

Answer:

True,
Compound interest is the interest calculated on the previous year’s amount.

Answer:

False,
When discount is subtracted from the marked price then we get the selling price i.e. SP = MP $-$ Discount.

Answer:

False,
Cost price of the bicycle = â‚¹1040
Selling price of the bicycle = â‚¹800

$\begin{array}{rcl}\mathrm{Loss} \mathrm{percent} & =& \frac{\mathrm{CP}-\mathrm{SP}\times 100}{\mathrm{CP}}\\ & =& \frac{1040-800}{1040}\times 100\\ & =& \frac{240}{1040}\times 100\\ & =& 23.07\%\end{array}$

              
 

Answer:

False,
Let the number be $x$.
So, three times of $x$ is 3$x$
Difference between 3$x$ and $x$ is $2x$
$\begin{array}{rcl}\therefore \mathrm{Percentage} \mathrm{increase} & =& \frac{{\displaystyle \frac{2\mathrm{x}}{3}}}{\mathrm{x}}\times 100\\ & =& \frac{2}{3}\times 100\\ & =& 66.66\%\end{array}$
Thus, three times a number is 200% increase in the number but one-third of the same number is 66.66% decrease in the nymber.

Answer:

False,
For 1 year, the simple interest and compound interest for same amount on same rate of interest are equal.
But for 2 years, the simple interest is less than the compound interest for same amount on same rate of interest.

Answer:

True,
Principle (P) = â‚¹7000
Rate of  depreciation (R) = 8% per annum
Time period (n) = 2 years
$$\begin{gathered} \mathrm{So}, \mathrm{Amount} =\mathrm{P}{\left(1-\frac{\mathrm{R}}{100}\right)}^{\mathrm{n}} \\ =7000{\left(1-\frac{8}{100}\right)}^2 \\ =7000\times {\left(\frac{92}{100}\right)}^2 \\ =7000\times \frac{23}{250}\times \frac{23}{25} \\ =11.2\times 23\times 23 \\ =₹5924.8 \end{gathered}$$

Answer:

False,
Discount percentage = $\frac{\mathrm{Discount}}{\mathrm{Marked} \mathrm{price}}\times 100\%$
As, marked price is â‚¹$x$ and discount amount is â‚¹$y$.
$\therefore$ Discount percentage = $\frac{y}{x}\times 100\%$

Answer:

True,
Number of students increased from 91422 in 1999 - 2000to 116054 in 2008 - 2009.
$\therefore$ Increase in number of students
$$\begin{gathered} =116054-91422 \\ =24632 \end{gathered}$$
Percentage of increase in number of students
$$\begin{gathered} =\frac{\mathrm{Number} \mathrm{of} \mathrm{students} \mathrm{increase}}{\mathrm{Number} \mathrm{of} \mathrm{students} \mathrm{in} 1999-2000}\times 100 \\ =\frac{24632}{91422}\times 100 \\ =0.2694\times 100 \\ =27\% \end{gathered}$$

Answer:

True,
Let the cost price of 1 article be â‚¹1
So, the cost price of 15 articles will be â‚¹15 and the cost price of a articles will be â‚¹9
Given that the selling price of 9 articles is equal to the cost price of 15 articles.
$\therefore$ The selling price of a articles is â‚¹15
Thus, selling price of a articles is â‚¹15 and the cost price of a articles is â‚¹9
So, profit in selling a articles = â‚¹15 $-$ â‚¹9
                                              = â‚¹6
$\begin{array}{rcl}\mathrm{And}, \mathrm{the} \mathrm{profit} \mathrm{percentage}& =& \frac{\mathrm{profit}}{\mathrm{Cost} \mathrm{price}}\times 100\\ & =& \frac{6}{9}\times 100\\ & =& \frac{2}{3}\times 100\\ & =& 66\frac{2}{3}\%\end{array}$

Answer:

False,
The compound interest on a sum of â‚¹P for T year at R% per annum compounded annually is given by the formula, compound interest = A $-$ P, where A = $\mathrm{P}{\left(1+\frac{\mathrm{R}}{100}\right)}^{\mathrm{T}}$

Answer:

True,
Gain = Selling price $-$ Cost price.................(1)
$\therefore$ Gain % = $\frac{\mathrm{Gain}}{\mathrm{Cost} \mathrm{price}}\times 100$ .................(2)
$\Rightarrow \frac{\mathrm{Gain}\%}{100}\times \mathrm{Cost} \mathrm{price}=\mathrm{Gain}$...............(3)
From (1), selling price = cost price + gain ...............(4)
From (3) and (4), we get,
$$\begin{gathered} \mathrm{Selling} \mathrm{price} = \mathrm{cost} \mathrm{price} + \frac{\mathrm{Gain}\%}{100}\times \mathrm{Cost} \mathrm{price} \\ =\left(1+\frac{\mathrm{Gain}\%}{100}\right)\times \mathrm{cost} \mathrm{price} \\ =\frac{\left(100+\mathrm{Gain}\%\right)\times \mathrm{cost} \mathrm{price}}{100} \end{gathered}$$
Hence, $\mathrm{SP}=\frac{\left(100+\mathrm{Gain}\%\right)\times \mathrm{CP}}{100}$

Answer:

False,
We know that, Loss = CP $-$ SP
As, Loss% = $\frac{\mathrm{Loss}}{\mathrm{CP}}\times 100$
$\Rightarrow$Loss% = $\frac{\mathrm{CP}-\mathrm{SP}}{\mathrm{CP}}\times 100$
$\therefore \mathrm{Cp}=\frac{100}{100-\mathrm{Loss}\%}\times \mathrm{SP}$

Answer:

False,
Principal (P) =₹440000
Rate of depreciation (R%) = 10% per annum
Time period (T) = 3 years
As, $\mathrm{A}=\mathrm{P}{\left(1-\frac{\mathrm{R}}{100}\right)}^{\mathrm{T}}$
$$\begin{gathered} \Rightarrow A=440000{\left(1-\frac{10}{100}\right)}^3 \\ =440000\times \frac{9}{10}\times \frac{9}{10}\times \frac{9}{10} \\ =440\times 729 \\ =₹320760 \end{gathered}$$
 

Answer:

False,
Marked price of a book = â‚¹190
Sakes tax = 2%
The cost price of the book after 2% sales tax
$$\begin{gathered} =190+\left(\frac{2}{100} \mathrm{of} 190\right) \\ =190+\left(\frac{2}{100}\times 190\right) \\ =190+3.8 \\ =193.8 \end{gathered}$$

Answer:

True,
Total flour bought = 5 kg
Rate of one kg flour = â‚¹20
Cost of 5 kg flour with 5% sales tax
$$\begin{gathered} =5\times 50+\frac{5}{100}\times \left(5\times 20\right) \\ =100+\frac{5}{100}\times 100 \\ =100+5 \\ =₹105 \end{gathered}$$
So, per kg flour rate after 5% sales tax
$$\begin{gathered} =\frac{105}{5} \\ =₹21 \end{gathered}$$

Answer:

False

Given that, the original price of a shampoo bottle is ₹300.

Now,
$\begin{array}{rcl}\mathrm{Cost} \mathrm{price} \mathrm{of} \mathrm{shampoo} \mathrm{bottle} \mathrm{after} 8\% \mathrm{VAT}& =& 300+\frac{8}{100}\times 300\\ & =& 300+24\\ & =& ₹324\end{array}$

Hence, the statement is false.

Answer:

False

The sales tax of an item is calculated on its selling price and is added to the value of the bill.

Hence, the statement is false.

Answer:

Given that,
Percentage of women in factory = 35%
∴ Percentage of men in factory = 100 – 35 = 65%

Let the total number of people working in the factory be x.

So,
$$\begin{gathered} \mathrm{Number} \mathrm{of} \mathrm{men}-\mathrm{Number} \mathrm{of} \mathrm{women}=252 \\ \Rightarrow x\times \frac{65}{100}-x\times \frac{35}{100}=252 \\ \Rightarrow \frac{30}{100}x=252 \\ \Rightarrow x=\frac{252\times 100}{30}=840 \end{gathered}$$

Hence, the total number of people working in the factory is 840.

Answer:

​Given that, the total weight of sugar in three bags is 64.2 kg.

Let the first bag contains x kg sugar.
So,
$\begin{array}{rcl}\mathrm{Sugar} \mathrm{in} \mathrm{the} \mathrm{second} \mathrm{bad}& =& x\times \frac{4}{5}\\ & =& \frac{4x}{5}\end{array}$

$\begin{array}{rcl}\mathrm{Sugar} \mathrm{in} \mathrm{the} \mathrm{third} \mathrm{bag}& =& 45\frac{1}{2}\%\times \frac{4x}{5}\\ & =& \frac{91}{2}\%\times \frac{4x}{5}\\ & =& \frac{4x}{5}\times \frac{{\displaystyle \frac{91}{2}}}{100}\\ & =& \frac{91x}{250}\end{array}$

Now,
$$\begin{gathered} \mathrm{Sum} \mathrm{of} \mathrm{the} \mathrm{sugar} \mathrm{in} \mathrm{three} \mathrm{bags}=x+\frac{4x}{5}+\frac{91x}{250} \\ \Rightarrow 64.2=x+\frac{4x}{5}+\frac{91x}{250} \\ \Rightarrow 64.2=\frac{250x+200x+91x}{250} \\ \Rightarrow 541x=64.2\times 250=16050 \\ \Rightarrow x=\frac{16050}{541}=29.67 \mathrm{kg} \end{gathered}$$

Therefore,
Sugar in the first bag = 29.67 kg
Sugar in the second bag = $29.67\times \frac{4}{5}=23.73 \mathrm{kg}$
Sugar in the third bag = $29.67\times \frac{91}{250}=10.8 \mathrm{kg}$

Answer:

(i) Given that, M.P. = ₹5450 and discount = 5%.
Now,
$\mathrm{SP}=\mathrm{MP}-\frac{\mathrm{Discount} \%}{100}\times \mathrm{MP}$
Therefore,
$\begin{array}{rcl}\mathrm{SP}& =& 5450-\frac{5}{100}\times 5450\\ & =& 5450-272.5\\ & =& ₹5177.5\end{array}$

Hence, the SP is â‚¹5177.5.

(ii) Given that, M.P. = ₹1300 and discount = 1.5%.
Now,
$\mathrm{SP}=\mathrm{MP}-\frac{\mathrm{Discount} \%}{100}\times \mathrm{MP}$
Therefore,
$\begin{array}{rcl}\mathrm{SP}& =& 1300-\frac{1.5}{100}\times 1300\\ & =& 1300-19.5\\ & =& ₹1280.5\end{array}$

Hence, the SP is â‚¹1280.5.

Answer:

(i) Given that, S.P. = ₹495 and discount = 1%.
Now,
$\mathrm{SP}=\mathrm{MP}-\frac{\mathrm{Discount} \%}{100}\times \mathrm{MP}$
$$\begin{gathered} \Rightarrow 495=\mathrm{MP}-\frac{1}{100}\times \mathrm{MP}=\frac{99}{100}\mathrm{MP} \\ \Rightarrow \mathrm{MP}=\frac{495\times 100}{99}=₹500 \end{gathered}$$
Hence, MP = â‚¹500.

(ii) Given that, S.P. = ₹9,250 and discount $=7\frac{1}{2}\%$.
Now,
$\mathrm{SP}=\mathrm{MP}-\frac{\mathrm{Discount} \%}{100}\times \mathrm{MP}$
$$\begin{gathered} \Rightarrow 9250=\mathrm{MP}-\frac{{\displaystyle \frac{15}{2}}}{100}\times \mathrm{MP}=\frac{185}{200}\mathrm{MP} \\ \Rightarrow \mathrm{MP}=\frac{9250\times 200}{185}=₹10000 \end{gathered}$$
Hence, MP = â‚¹10000.
 

Answer:

(i) Given that, M.P. = ₹625 and S.P. = ₹562.50.
Now,
$\mathrm{SP}=\mathrm{MP}-\frac{\mathrm{Discount} \%}{100}\times \mathrm{MP}$
Therefore,
$$\begin{gathered} 562.5=625-\frac{\mathrm{discount} \%}{100}\times 625 \\ \Rightarrow \frac{\mathrm{discount} \%}{100}\times 625=62.5 \\ \Rightarrow \mathrm{discount} \%=\frac{62.5\times 100}{625}=10\% \end{gathered}$$

Hence, the discount percentage is 10%.

(ii) Given that, M.P. = ₹900 and S.P. = ₹873.
Now,
$\mathrm{SP}=\mathrm{MP}-\frac{\mathrm{Discount} \%}{100}\times \mathrm{MP}$
Therefore,
$$\begin{gathered} 873=900-\frac{\mathrm{discount} \%}{100}\times 900 \\ \Rightarrow \frac{\mathrm{discount} \%}{100}\times 900=27 \\ \Rightarrow \mathrm{discount} \%=\frac{27\times 100}{900}=3\% \end{gathered}$$

Hence, the discount percentage is 3%.
 

Answer:

Given that, the marked price of the article is ₹500 and the shopkeeper gives a discount of 5%.
Now,
$\mathrm{SP}=\mathrm{MP}-\frac{\mathrm{Discount} \%}{100}\times \mathrm{MP}$
Therefore,
$\begin{array}{rcl}\mathrm{SP}& =& 500-\frac{5}{100}\times 500\\ & =& 500-25\\ & =& ₹475\end{array}$

Thus, the SP is â‚¹475.

Also, the shopkeeper makes a profit of 25%.
Now,
$\mathrm{SP}=\mathrm{CP}+\frac{\mathrm{Profit} \%}{100}\times \mathrm{CP}$
$\begin{array}{rcl}& \Rightarrow & 475=\mathrm{CP}+\frac{25}{100}\times \mathrm{CP}\\ & =& \mathrm{CP}+\frac{1}{4}\times \mathrm{CP}\\ & =& \frac{5}{4}\mathrm{CP}\\ & \Rightarrow & \mathrm{CP}=\frac{475\times 4}{5}\\ & =& ₹380\end{array}$

Hence, the CP is â‚¹380.
 

Answer:

Given:
Number of students appeared in 2007-08 = 105332
Number of students passed in 2007-08 = 88151
Number of students appeared in 2008-09 = 116054
Number of students passed in 2008-09 = 103804

Now,
$\begin{array}{rcl}\mathrm{Percentage} \mathrm{of} \mathrm{students} \mathrm{who} \mathrm{passed} \mathrm{in} 2007-08& =& \frac{\mathrm{Number} \mathrm{of} \mathrm{students} \mathrm{who} \mathrm{passed}}{\mathrm{Number} \mathrm{of} \mathrm{students} \mathrm{who} \mathrm{appeared}}\times 100\\ & =& \frac{88151}{105332}\times 100\\ & =& 83.68\%\end{array}$
$\begin{array}{rcl}\mathrm{Percentage} \mathrm{of} \mathrm{students} \mathrm{who} \mathrm{passed} \mathrm{in} 2008-09& =& \frac{\mathrm{Number} \mathrm{of} \mathrm{students} \mathrm{who} \mathrm{passed}}{\mathrm{Number} \mathrm{of} \mathrm{students} \mathrm{who} \mathrm{appeared}}\times 100\\ & =& \frac{103804}{116054}\times 100\\ & =& 89.44\%\end{array}$

Therefore,
Increase in percentage = 89.44 − 83.68 = 35.76%.

Answer:

Given that, a watch worth ₹5400 is offered for sale at ₹4,500.

Now,
$\mathrm{SP}=\mathrm{MP}-\frac{\mathrm{Discount} \%}{100}\times \mathrm{MP}$

Here,
SP = â‚¹4500 and MP = â‚¹5400.

$$\begin{gathered} \Rightarrow 4500=5400-\frac{\mathrm{Discount} \%}{100}\times 5400 \\ \Rightarrow \frac{\mathrm{Discount} \%}{100}\times 5400=900 \\ \Rightarrow \mathrm{Discount} \%=\frac{900}{54}=\frac{50}{3} \end{gathered}$$

Hence, the discount percentage is $\frac{50}{3}$%.

Answer:

Given that, the number of malaria patients admitted in a hospital in 2001 is 4375. Also, the rate of decrement of malaria patients is 8%.

Now,
Time period from 2001 to 2003, 2001 not included = 2 yrs

Let the number of patients in the year 2003 be A.

Since, the patients decrease at the rate of 8% every year.
Therefore,
$\begin{array}{rcl}A& =& P{\left(1-\frac{r}{100}\right)}^t\\ & =& 4375{\left(1-\frac{8}{100}\right)}^2\\ & =& 4375\times \frac{23}{25}\times \frac{23}{25}\\ & =& 7\times 23\times 23\\ & =& 3703\end{array}$

Hence, the number of patients in 2003 is 3703.

Answer:

Given that, a product bought by Jyotsana for ₹3155 includes 4.5% sales tax.

Let the price of the product before sales tax be ₹x.
Therefore,
$$\begin{gathered} 3155=x+x\times \frac{4.5}{100} \\ \Rightarrow 3155=\frac{1045x}{1000} \\ \Rightarrow x=\frac{3155\times 1000}{1045}=₹3019.138 \\ \Rightarrow x\approx ₹3019.14 \end{gathered}$$

Hence, the price of the product before sales tax is â‚¹3019.14.

Answer:

(a) Water used for bathing per day = 20 L
Water used for sanitation = 40 L
Total water used per day = 150 L
$\begin{array}{rcl}\mathrm{Percentage} \mathrm{of} \mathrm{water} \mathrm{used} \mathrm{for} \mathrm{them}& =& \frac{\mathrm{Water} \mathrm{used} \mathrm{for} \mathrm{bathing} \mathrm{and} \mathrm{sanitation}}{\mathrm{Total} \mathrm{water} \mathrm{used}}\times 100\\ & =& \frac{20+40}{150}\times 100\\ & =& \frac{60}{150}\times 100\\ & =& 40\%\end{array}$

(b) Water used for cooking per day = 4 L
Water used for bathing per day = 20 L
Therefore,
$\begin{array}{rcl}\mathrm{Difference} \mathrm{of} \mathrm{percentage} \mathrm{of} \mathrm{water} \mathrm{used} \mathrm{for} \mathrm{them}& =& \frac{\mathrm{Difference} \mathrm{between} \mathrm{the} \mathrm{water} \mathrm{used} \mathrm{for} \mathrm{them}}{\mathrm{Total} \mathrm{water} \mathrm{used}}\times 100\\ & =& \frac{20-4}{150}\times 100\\ & =& \frac{16}{150}\times 100\\ & =& \frac{32}{3}\%\end{array}$
(c) Water used for drinking per day = 3 L
Water used for cooking per day = 4 L
Water used for gardening per day = 23 L
$\begin{array}{rcl}\mathrm{Percentage} \mathrm{of} \mathrm{water} \mathrm{used} \mathrm{for} \mathrm{them}& =& \frac{\mathrm{Water} \mathrm{used} \mathrm{for} \mathrm{them}}{\mathrm{Total} \mathrm{water} \mathrm{used}}\times 100\\ & =& \frac{3+4+23}{150}\times 100\\ & =& \frac{30}{150}\times 100\\ & =& 20\%\end{array}$

 

Answer:

Given:
The consumption of water in 1975 = 3850 cu km/yr
The consumption of water in 2000 = 6000 cu km/yr
∴ Increase in consumption of water from 1975 to 2000 = 6000 − 3850
                                                                                        = 2150 cu km/yr

Now,
$\begin{array}{rcl}\mathrm{Percentage} \mathrm{increase} \mathrm{in} \mathrm{water} \mathrm{consumption}\left(1975-2000\right)& =& \frac{\mathrm{Increase} \mathrm{in} \mathrm{water} \mathrm{consumption}}{\mathrm{Initial} \mathrm{water} \mathrm{consumption}}\times 100\\ & =& \frac{2150}{3850}\times 100\\ & =& 55.84\%\end{array}$

Also, in a total of 25 years (= 2000 − 1975), the increase in consumption of water is 2150 cu km/yr.
Therefore,
$\mathrm{Annual} \mathrm{increase} \mathrm{in} \mathrm{water} \mathrm{consumption}=\frac{2150}{25}=86 \mathrm{cu} \mathrm{km}/\mathrm{yr}$

Now,
$\begin{array}{rcl}\mathrm{Annual} \mathrm{percentage} \mathrm{increase} \mathrm{in} \mathrm{water} \mathrm{consumption}& =& \frac{\mathrm{Increase} \mathrm{in} \mathrm{water} \mathrm{consumption} \mathrm{in} \mathrm{a} \mathrm{year}}{\mathrm{Initial} \mathrm{water} \mathrm{consumption}}\times 100\\ & =& \frac{86}{3850}\times 100\\ & =& 2.23\%\end{array}$

Answer:

(a) The total litres of engine oil used = 3.10 L
Rate of engine oil per litres = ₹178.75
∴ cost of engine oil = 3.10 × 178.75
                                = ₹554.125

Now,
$\begin{array}{rcl}\mathrm{Cost} \mathrm{of} \mathrm{engine} \mathrm{oil} \mathrm{with} 20\% \mathrm{VAT}& =& 554.125+554.125\times \frac{20}{100}\\ & =& 554.125+110.825\\ & =& ₹664.95\end{array}$

(b) Amount to be paid for all the services = â‚¹1105.12
Also, VAT at 12.5% is to be paid.
Therefore,
$\begin{array}{rcl}\mathrm{Total} \mathrm{amount} \mathrm{to} \mathrm{be} \mathrm{paid}& =& 1105.12+\frac{12.5}{100}\times 1105.12\\ & =& 1105.12+138.14\\ & =& ₹1243.26\end{array}$

(c) Labour charges = â‚¹2095.8
Also, the service tax is 10%.
Therefore,
$\begin{array}{rcl}\mathrm{Total} \mathrm{amount} \mathrm{to} \mathrm{be} \mathrm{paid}& =& 2095.80+\frac{10}{100}\times 2095.80\\ & =& 2095.80+209.580\\ & =& ₹2305.38\end{array}$

(d) Given that, there is 3% cess on the service tax. Now, the service tax is â‚¹209.58.
Therefore,
$\begin{array}{rcl}\mathrm{Cess}& =& \frac{3}{100}\times 209.58\\ & =& ₹6.285\\ & \approx & ₹6.29\end{array}$

Hence, the total bill amount = â‚¹664.95 + â‚¹1243.26 + â‚¹2305.38 + â‚¹6.29 = â‚¹4219.88.

Answer:

Given that, the principal = ₹40,000, rate of interest = 8% p.a. compounded annually.

(a) Interest if the period is 1 year
Now, $A=P{\left(1+\frac{r}{100}\right)}^t$.
$\begin{array}{rcl}& \Rightarrow & A=40000{\left(1+\frac{8}{100}\right)}^1\\ & =& 40000\times \frac{108}{100}\\ & =& ₹43200\end{array}$

Therefore,
CI for the first year = 43200 − 40000 = ₹3200.

(b) Principal for the second year will be the amount of the first year.
Hence, the principal for the second year will be â‚¹43200.

(c) Interest for the second year
For the second year,
$\begin{array}{rcl}& \Rightarrow & A=43200{\left(1+\frac{8}{100}\right)}^1\\ & =& 43200\times \frac{108}{100}\\ & =& ₹46656\end{array}$

So,
$CI=A-P=₹46656-₹43200=₹3456$

(d) Amount if the period is 2 years is â‚¹46656.

Answer:

Given:
The total number of seats available for admission in 2009-10 = 49000
Seats reserved for General Category students = 28200
Seats reserved for SC Category students = 7400
Seats reserved for ST Category students = 3700

(i) 
$\begin{array}{rcl}\mathrm{Percentage} \mathrm{of} \mathrm{general} \mathrm{students}& =& \frac{\mathrm{Number} \mathrm{of} \mathrm{general} \mathrm{students}}{\mathrm{Total} \mathrm{number} \mathrm{of} \mathrm{students}}\times 100\\ & =& \frac{28200}{49000}\times 100\\ & =& 57.55\%\end{array}$

(ii)
$\begin{array}{rcl}\mathrm{Percentage} \mathrm{of} \mathrm{SC} \mathrm{and} \mathrm{ST} \mathrm{students}& =& \frac{\mathrm{Number} \mathrm{of} \mathrm{SC} \mathrm{and} \mathrm{ST} \mathrm{students}}{\mathrm{Total} \mathrm{number} \mathrm{of} \mathrm{students}}\times 100\\ & =& \frac{7400+3700}{49000}\times 100\\ & =& \frac{11100}{49000}\times 100\\ & =& 22.65\%\end{array}$

Answer:

Given that, the price of the medicine, i.e., ₹36.40 included 4% VAT.

Now,
SP = CP + VAT

So,
$\begin{array}{rcl}36.40& =& \mathrm{CP}+\frac{4}{100}\times \mathrm{CP}\\ & =& \frac{104}{100}\mathrm{CP}\end{array}$
$\Rightarrow \mathrm{CP}=\frac{36.40\times 100}{104}=₹35$

Hence, the price before VAT was added was â‚¹35.

Answer:

Given that, she paid â‚¹387 inclusive of taxes of  â‚¹43.

So,
$\begin{array}{rcl}\mathrm{Cost} \mathrm{without} \mathrm{tax}& =& ₹387-₹43\\ & =& ₹344\end{array}$

Therefore,
$\begin{array}{rcl}\mathrm{Tax} \mathrm{Percentage}& =& \frac{43}{344}\times 100\\ & =& 12.5\%\end{array}$

Answer:

According to the given table,

$\begin{array}{rcl}\mathrm{Toal} \mathrm{amount} \mathrm{of} \mathrm{the} \mathrm{bill}& =& 200-16\% \mathrm{of} 200+371-22.10\% \mathrm{of} 371+153-18.30\% \mathrm{of} 153\\ & =& 200-\frac{16}{100}\times 200+371-\frac{22.10}{100}\times 371+153-\frac{18.30}{100}\times 153\\ & =& 200-32+371-81.991+153-27.999\\ & =& ₹582.01\end{array}$

Hence, she has to pay â‚¹582.01.

Answer:

$\begin{array}{rcl}\mathrm{Total} \mathrm{amount}& =& 199.75+12\% \mathrm{of} 199.75+599.25+10\% \mathrm{of} 599.25\\ & =& 199.75+\frac{12}{100}\times 199.75+599.25+\frac{10}{100}\times 599.25\\ & =& 799+23.97+59.925\\ & =& ₹822.895\end{array}$

Now, 3% education cess was also charged on service tax.
Therefore,
$\begin{array}{rcl}\mathrm{Final} \mathrm{bill} \mathrm{amount}& =& 822.895+\frac{3}{100}\times 822.895\\ & =& 822.895+\frac{3}{100}\times 822.895\\ & =& 822.895+26.48\\ & =& ₹909.39\end{array}$

Hence, the final bill amount is â‚¹909.39.

Answer:

Given that, the principal is ₹1,00,000 and the rate of interest is 10% compounded half yearly.

(i) For compound interest,
$\begin{array}{rcl}A& =& P{\left(1+\frac{{\displaystyle \frac{r}{2}}}{100}\right)}^t\\ & =& 100000{\left(1+\frac{10}{200}\right)}^1 \left(\because t=1 \mathrm{for} 6 \mathrm{months}\right)\\ & =& 100000\left(\frac{210}{200}\right)\\ & =& ₹105000\end{array}$

Therefore,
$\begin{array}{rcl}CI& =& A-P\\ & =& 105000-100000\\ & =& ₹5000\end{array}$

(b) Amount for 6 months = â‚¹105000

(c) For interest of next 6 months,
$\begin{array}{rcl}A& =& P{\left(1+\frac{{\displaystyle \frac{r}{2}}}{100}\right)}^t\\ & =& 105000{\left(1+\frac{10}{200}\right)}^1 \left(\because t=1 \mathrm{for} 6 \mathrm{months}\right)\\ & =& 105000\left(\frac{210}{200}\right)\\ & =& ₹110250\end{array}$

Therefore,
$\begin{array}{rcl}CI& =& A-P\\ & =& 110250-105000\\ & =& ₹5250\end{array}$

(d) Amount after one year is â‚¹110250.

Answer:

Given that, her cost price is â‚¹48 per kg for 160 kg of mangoes.

Therefore,
Total CP = 48 × 160
               = ₹7680

Now, she sold 70% of the mangoes at ₹70 per kg.
70% of 160 mangoes = $\frac{70}{100}\times 160=112 \mathrm{kg}$
So,
SP of 112 kg mangoes = 70 × 112
                                     = ₹7840

Also, she sold the remaining mangoes at ₹40 per kg.
Remaining mangoes = 160 − 112 = 48 kg
So,
SP of 48 kg mangoes = 40 × 48
                                   = ₹1920

Therefore,
Total SP = 7840 + 1920 = ₹9760

Since SP > CP.
Therefore, Babita gained money out of the deal.

So,
$\begin{array}{rcl}\mathrm{Profit} \mathrm{Percentage}& =& \frac{\mathrm{Profit}}{\mathrm{CP}}\times 100\\ & =& \frac{9760-7680}{7680}\times 100\\ & =& \frac{2080}{7680}\times 100\\ & =& 27.08\%\end{array}$

Answer:

Given that, the shopkeeper gave 25% discount initially and then offered 30% discount over and above the existing discount.

Also, Pragya bought a skirt which was marked for Rs 1,200.

After 25% discount,
$\begin{array}{rcl}\mathrm{Cost} \mathrm{of} \mathrm{the} \mathrm{skirt}& =& 1200-\frac{25}{100}\times 1200\\ & =& 1200-300\\ & =& ₹900\end{array}$

Again, after 30% discount,
$\begin{array}{rcl}\mathrm{Cost} \mathrm{of} \mathrm{the} \mathrm{skirt}& =& 900-\frac{30}{100}\times 900\\ & =& 900-270\\ & =& ₹630\end{array}$

Hence, Pragya paid â‚¹630 for the dress.
 

Answer:

Let the marked price of the phone be ₹x.
Now, 25% discount was given on it and Ramandeep got it for â‚¹1960.

So,
$$\begin{gathered} 1960=x\times \frac{25}{100} \\ \Rightarrow 1960=\frac{x}{4} \\ \Rightarrow x=1960\times 4=₹7840 \end{gathered}$$

Hence, the marked price of the phone was â‚¹7840.

Answer:

Given:
P = â‚¹45000
r = 12% p.a.
t = 5 years

Now,
$\begin{array}{rcl}SI& =& \frac{P\times r\times t}{100}\\ & =& \frac{45000\times 12\times 5}{100}\\ & =& ₹27000\end{array}$

For Compound Interest,
$\begin{array}{rcl}A& =& P{\left(1+\frac{r}{100}\right)}^t\\ & =& 45000{\left(1+\frac{12}{100}\right)}^5\\ & =& 45000{\left(\frac{112}{100}\right)}^5\\ & =& 45000{\left(\frac{112}{100}\right)}^5\\ & =& ₹79305.37\end{array}$
$\begin{array}{rcl}& \Rightarrow & CI=A-P\\ & =& 79305.37-45000\\ & =& 34305.37\end{array}$

Therefore,
$\begin{array}{rcl}& \Rightarrow & CI-SI=34305.37-27000\\ & =& ₹7305.37\end{array}$
 

Answer:

Given:
Cost of the computer = â‚¹10000
Depreciation rate =  50% p.a.
Time period = 2 years

Let the cost of the computer after 2 years be â‚¹A.
Now,
$\begin{array}{rcl}A& =& P{\left(1-\frac{r}{100}\right)}^t\\ & =& 100000{\left(1-\frac{50}{100}\right)}^2\\ & =& 100000\times \frac{1}{2}\times \frac{1}{2}\\ & =& ₹25000\end{array}$

Hence, the cost of the computer after two years is â‚¹25000.

Answer:

Given that, the present population of the town is 6,31,680.

Last year the migration was 4%. Let the population of the last year be x.
Therefore,
$\begin{array}{rcl}631680& =& x-\frac{4}{100}x\\ & =& \frac{96}{100}x\end{array}$
$\Rightarrow x=\frac{631680\times 100}{96}=658000$

Also, the migration was 6% the year before it. So, let the population two years back be y.
$\begin{array}{rcl}658000& =& y-\frac{6}{100}y\\ & =& \frac{94}{100}y\end{array}$
$\Rightarrow y=\frac{658000\times 100}{94}=700000$

Hence, the population of the town two years back was 700000.
 

 

Answer:

Given that, lemons were bought at ₹48 per dozen.
Thus,
$\begin{array}{rcl}\mathrm{CP} \mathrm{of} \mathrm{a} \mathrm{lemon}& =& \frac{48}{12}\\ & =& ₹4\end{array}$

Also, lemons were sold at the rate of ₹40 per 10.
Thus,
$\begin{array}{rcl}\mathrm{SP} \mathrm{of} \mathrm{a} \mathrm{lemon}& =& \frac{40}{10}\\ & =& ₹4\end{array}$

Since, the SP = CP.
Therefore, no profit or no loss.

Answer:

(a) The price of petrol decreases from â‚¹45.62 to â‚¹40.62.
Therefore,
$\begin{array}{rcl}\mathrm{Percentage} \mathrm{decrease} \mathrm{in} \mathrm{the} \mathrm{price}& =& \frac{\mathrm{Decrease} \mathrm{in} \mathrm{the} \mathrm{price}}{\mathrm{Initial} \mathrm{price}}\times 100\\ & =& \frac{45.62-40.62}{45.62}\times 100\\ & =& \frac{5}{45.62}\times 100\\ & =& 10.96\%\end{array}$

(b) The price of diesel decreases from â‚¹32.86 to â‚¹30.86.
Therefore,
$\begin{array}{rcl}\mathrm{Percentage} \mathrm{decrease} \mathrm{in} \mathrm{the} \mathrm{price}& =& \frac{\mathrm{Decrease} \mathrm{in} \mathrm{the} \mathrm{price}}{\mathrm{Initial} \mathrm{price}}\times 100\\ & =& \frac{32.86-30.86}{32.86}\times 100\\ & =& \frac{2}{32.86}\times 100\\ & =& 6.08\%\end{array}$

(c) The price of LPG decreases from â‚¹304.70 to â‚¹279.70.
Therefore,
$\begin{array}{rcl}\mathrm{Percentage} \mathrm{decrease} \mathrm{in} \mathrm{the} \mathrm{price}& =& \frac{\mathrm{Decrease} \mathrm{in} \mathrm{the} \mathrm{price}}{\mathrm{Initial} \mathrm{price}}\times 100\\ & =& \frac{304.70-279.70}{304.70}\times 100\\ & =& \frac{25}{304.70}\times 100\\ & =& 8.20\%\end{array}$

 

Answer:

The number of seats won by A in 2004 is 206 and it decreased to 145 in 2009.
Therefore,
$\begin{array}{rcl}\mathrm{Percentage} \mathrm{decrease} \mathrm{in} \mathrm{seats} \mathrm{won} \mathrm{by} \mathrm{A}& =& \frac{\mathrm{Decrease} \mathrm{in} \mathrm{the} \mathrm{number} \mathrm{of} \mathrm{seats}}{\mathrm{Initial} \mathrm{number} \mathrm{of} \mathrm{seats} \mathrm{won}}\times 100\\ & =& \frac{206-145}{206}\times 100\\ & =& \frac{61}{206}\times 100\\ & =& 29.61\%\end{array}$

The number of seats won by B in 2004 is 116 and it increased to 138 in 2009.
Therefore,
$\begin{array}{rcl}\mathrm{Percentage} \mathrm{increase} \mathrm{in} \mathrm{seats} \mathrm{won} \mathrm{by} \mathrm{B}& =& \frac{\mathrm{Increase} \mathrm{in} \mathrm{the} \mathrm{number} \mathrm{of} \mathrm{seats}}{\mathrm{Initial} \mathrm{number} \mathrm{of} \mathrm{seats} \mathrm{won}}\times 100\\ & =& \frac{138-116}{116}\times 100\\ & =& \frac{22}{116}\times 100\\ & =& 18.96\%\end{array}$

The number of seats won by C in 2004 is 4 and it increased to 24 in 2009.
Therefore,
$\begin{array}{rcl}\mathrm{Percentage} \mathrm{increase} \mathrm{in} \mathrm{seats} \mathrm{won} \mathrm{by} \mathrm{C}& =& \frac{\mathrm{Increase} \mathrm{in} \mathrm{the} \mathrm{number} \mathrm{of} \mathrm{seats}}{\mathrm{Initial} \mathrm{number} \mathrm{of} \mathrm{seats} \mathrm{won}}\times 100\\ & =& \frac{24-4}{4}\times 100\\ & =& \frac{20}{4}\times 100\\ & =& 500\%\end{array}$

The number of seats won by D in 2004 is 11 and it increased to 12 in 2009.
Therefore,
$\begin{array}{rcl}\mathrm{Percentage} \mathrm{increase} \mathrm{in} \mathrm{seats} \mathrm{won} \mathrm{by} \mathrm{D}& =& \frac{\mathrm{Increase} \mathrm{in} \mathrm{the} \mathrm{number} \mathrm{of} \mathrm{seats}}{\mathrm{Initial} \mathrm{number} \mathrm{of} \mathrm{seats} \mathrm{won}}\times 100\\ & =& \frac{12-11}{11}\times 100\\ & =& \frac{1}{11}\times 100\\ & =& 9.09\%\end{array}$

Disclaimer: Answers given in the NCERT Exemplar are incorrect. The correct answers are as per the calculations given above.

Answer:

Given that, the number of seats wone by X in the election was 158 and that won by Y was 105.
$\begin{array}{rcl}\mathrm{Percentage} \mathrm{of} \mathrm{seats} \mathrm{won} \mathrm{by} \mathrm{X}& =& \frac{\mathrm{Number} \mathrm{of} \mathrm{seats} \mathrm{won} \mathrm{by} \mathrm{X}}{\mathrm{Total} \mathrm{number} \mathrm{of} \mathrm{seats}}\times 100\\ & =& \frac{158}{294}\times 100\\ & =& 53.74\%\end{array}$

And
$\begin{array}{rcl}\mathrm{Percentage} \mathrm{of} \mathrm{seats} \mathrm{won} \mathrm{by} \mathrm{Y}& =& \frac{\mathrm{Number} \mathrm{of} \mathrm{seats} \mathrm{won} \mathrm{by} \mathrm{Y}}{\mathrm{Total} \mathrm{number} \mathrm{of} \mathrm{seats}}\times 100\\ & =& \frac{105}{294}\times 100\\ & =& 35.71\%\end{array}$

Therefore,
$\begin{array}{rcl}\mathrm{Difference} \mathrm{in} \mathrm{percentage} \mathrm{of} \mathrm{seats}& =& 53.74-35.71\\ & =& 18.03\%\end{array}$

Answer:

Given that, Ashima sold two coolers for Rs 3,990 each and gained 5% on one and suffered a loss of 5% on the other.

Now, for the first scooter, let x be the CP.
$$\begin{gathered} 3990=x+x\times \frac{5}{100} \\ \Rightarrow 3990=\frac{105x}{100} \\ \Rightarrow x=\frac{3990\times 100}{105}=₹3800 \end{gathered}$$

Similarly, for the second scooter, let y be the CP.
$$\begin{gathered} 3990=y-y\times \frac{5}{100} \\ \Rightarrow 3990=\frac{95y}{100} \\ \Rightarrow y=\frac{3990\times 100}{95}=₹4200 \end{gathered}$$

So, the total CP of the scooters = â‚¹3800 + â‚¹4200 = â‚¹8000.
And she sold the scooters for â‚¹7980. (= 3990 + 3990).
Thus, SP < CP. So, she suffered a loss in the transaction.

Now,
$\begin{array}{rcl}\mathrm{Loss} \mathrm{Percentage}& =& \frac{\mathrm{Loss}}{\mathrm{CP}}\times 100\\ & =& \frac{8000-7980}{8000}\times 100\\ & =& \frac{20}{8000}\times 100\\ & =& 0.25\%\end{array}$
Hence, she suffered a loss of 0.25%.

Answer:

Consider that she buys x number of pencils.

Since, she buys x number of pencils for â‚¹3. Also, she buys x number of pencils for â‚¹6.

Now, SP of 2x pencils is â‚¹7 and their CP is ₹9.
$\Rightarrow \mathrm{SP}<\mathrm{CP}$
Thus, there is a loss.

Now,
$\begin{array}{rcl}\mathrm{Loss} \mathrm{percentage}& =& \frac{\mathrm{Loss}}{\mathrm{CP}}\times 100\\ & =& \frac{9-7}{9}\times 100\\ & =& \frac{2}{9}\times 100\\ & =& 22.22\%\end{array}$

Answer:

Given that, on selling a chair for Rs 736, a shopkeeper suffers a loss of 8%.

Therefore,
$$\begin{gathered} 8=\frac{\mathrm{CP}-736}{\mathrm{CP}}\times 100 \left(\because \mathrm{Loss} \mathrm{Percentage}=\frac{\mathrm{CP}-\mathrm{SP}}{\mathrm{CP}}\times 100\right) \\ \Rightarrow \frac{8}{100}=1-\frac{736}{\mathrm{CP}} \\ \Rightarrow \frac{736}{\mathrm{CP}}=\frac{92}{100} \\ \Rightarrow \mathrm{CP}=\frac{736\times 100}{92}=₹800 \end{gathered}$$

Now, he wants to gain 8% by selling it.
Therefore,
$\begin{array}{rcl}\mathrm{SP}& =& \mathrm{CP}+\frac{\mathrm{Profit}}{100}\times CP\\ & =& 800+\frac{8}{100}\times 800\\ & =& 800+64\\ & =& ₹864\end{array}$

Hence, he should sell it at â‚¹864 to gain 8%.