Chapter 9 Comparing Quantities (Class 8 - Maths NCERT Exemplar Solutions)

Given:

Marked Price (M.P.) = $\textsf{₹}$ 800

Selling Price (S.P.) = $\textsf{₹}$ 680

To Find:

Rate of discount.

Solution:

First, we calculate the amount of discount:

$\text{Discount} = \text{Marked Price} - \text{Selling Price}$

$\text{Discount} = 800 - 680 = \textsf{₹} 120$

Now, we find the rate of discount (Discount %):

$\text{Discount %} = \frac{\text{Discount}}{\text{Marked Price}} \times 100$

$\text{Discount %} = \frac{120}{800} \times 100$

$\text{Discount %} = \frac{120}{8} = 15\%$

Hence, the correct option is (b).

Given:

$\frac{7}{3}\%$ of a number = 42

Solution:

Let the required number be $x$.

According to the question:

$\frac{7}{3}\% \times x = 42$

$\frac{7}{3 \times 100} \times x = 42$

$\frac{7}{300} \times x = 42$

$x = \frac{42 \times 300}{7}$

$x = 6 \times 300$

$x = 1800$

Hence, the correct option is (c).

Given:

Cost Price (C.P.) of 10 shirts = Selling Price (S.P.) of 8 shirts.

Solution:

Let the C.P. of 1 shirt be $\textsf{₹}$ 1.

Then, C.P. of 10 shirts = $\textsf{₹}$ 10.

As per the question, S.P. of 8 shirts = C.P. of 10 shirts = $\textsf{₹}$ 10.

Now, we find the C.P. of 8 shirts = $\textsf{₹}$ 8.

Since S.P. ($10$) > C.P. ($8$), there is a profit.

$\text{Profit} = \text{S.P.} - \text{C.P.} = 10 - 8 = \textsf{₹} 2$

$\text{Profit %} = \frac{\text{Profit}}{\text{C.P.}} \times 100$

$\text{Profit %} = \frac{2}{8} \times 100 = \frac{1}{4} \times 100 = 25\%$

Hence, the correct option is (a).

Given:

Principal ($P$) = $\textsf{₹}$ 1600

Rate ($R$) = $5\%$ per annum

Time ($n$) = $1$ year

Solution:

Since the interest is compounded half-yearly:

Rate ($r$) = $\frac{5}{2} = 2.5\%$ per half-year.

Number of conversion periods ($n$) = $1 \times 2 = 2$ half-years.

$\text{Amount (A)} = P \left(1 + \frac{r}{100}\right)^n$

$A = 1600 \left(1 + \frac{2.5}{100}\right)^2$

$A = 1600 \left(1 + \frac{25}{1000}\right)^2 = 1600 \left(1 + \frac{1}{40}\right)^2$

$A = 1600 \left(\frac{41}{40}\right)^2 = 1600 \times \frac{1681}{1600}$

$A = \textsf{₹} 1681$

Hence, the correct option is (c).

Solution:

Let the Selling Price (S.P.) of 1 pen be $\textsf{₹}$ $y$.

S.P. of 50 pens = $50y$

Loss = S.P. of 10 pens = $10y$

We know that $\text{Cost Price (C.P.)} = \text{S.P.} + \text{Loss}$

C.P. of 50 pens = $50y + 10y = 60y$

$\text{Loss %} = \frac{\text{Loss}}{\text{C.P.}} \times 100$

$\text{Loss %} = \frac{10y}{60y} \times 100 = \frac{100}{6} = 16\frac{2}{3}\%$

Answer: 16$\frac{2}{3}$%

Solution:

Discount is a reduction offered on the price printed on the article or the price tagged by the seller. This price is known as the Marked Price.

Answer: Marked

Given:

Total Price (including tax) = $\textsf{₹}$ 660

Tax rate = $10\%$

Solution:

Let the Selling Price (before tax) be $\textsf{₹}$ $x$.

$\text{Tax} = 10\% \text{ of } x = 0.1x$

$\text{Total Price} = x + 0.1x = 1.1x$

$1.1x = 660$

$x = \frac{660}{1.1} = \frac{6600}{11}$

$x = 600$

Answer: $\textsf{₹}$ 600

Solution:

When interest is compounded half-yearly, it is calculated every 6 months. Since there are 12 months in a year, there are $12 / 6 = 2$ conversion periods in a year.

Answer: False (F)

Solution:

$\text{Cost of book} = \textsf{₹} 600$

$\text{Sales Tax} = 7\% \text{ of } 600 = \frac{7}{100} \times 600 = \textsf{₹} 42$

$\text{Total Amount} = \text{Cost} + \text{Tax} = 600 + 42 = \textsf{₹} 642$

The statement is correct.

Answer: True (T)

Solution:

$\text{Selling Price (S.P.)} = \text{Marked Price (M.P.)} \times (1 - \text{Discount \%})$

If M.P. = $800$ and Discount = $15\%$, then:

$\text{S.P.} = 800 \times (1 - 0.15) = 800 \times 0.85$

$\text{S.P.} = 680.00$

The statement matches the calculation.

Answer: True (T)

Solution:

In mathematical and commercial terminology used in textbooks, there is a clear distinction between the Cost Price (C.P.) and the Total Cost Price.

The Cost Price refers specifically to the price at which an article is purchased from a seller. Overhead charges, such as transportation, repairs, or labor costs, are expenses incurred after the purchase. These charges are added to the Cost Price to determine the final investment, known as the Total Cost Price.

Since overhead charges are distinct expenses that are added to the cost price rather than being "sometimes included" within the definition of the cost price itself, the statement is technically incorrect from a terminological perspective.

Answer: False (F)

To Find:

The net increase or decrease percentage after two successive changes.

Solution:

Let the original number be $100$.

First, the number is increased by $20\%$.

$\text{Increase} = 20\% \text{ of } 100 = 20$

$\text{Number after increase} = 100 + 20 = 120$

Now, this new number ($120$) is decreased by $20\%$.

$\text{Decrease} = 20\% \text{ of } 120 = \frac{20}{100} \times 120 = 24$

$\text{Final number} = 120 - 24 = 96$

Now, we find the net change from the original number:

$\text{Net change} = \text{Original number} - \text{Final number}$

$\text{Net change} = 100 - 96 = 4$

Since the final number is less than the original, there is a net decrease.

$\text{Net decrease %} = \frac{4}{100} \times 100 = 4\%$

Ans: Net decrease of 4%

Given:

Marked Price (M.P.) = $\textsf{₹}$ $280$

Discount rate = $20\%$

Profit rate = $12\%$

To Find:

Cost Price (C.P.) of the article.

Solution:

First, we find the Selling Price (S.P.) after the discount is applied to the Marked Price.

$\text{Discount} = 20\% \text{ of M.P.}$

$\text{Discount} = \frac{20}{100} \times 280 = \textsf{₹} 56$

$\text{S.P.} = \text{M.P.} - \text{Discount}$

$\text{S.P.} = 280 - 56 = \textsf{₹} 224$

Now, we know that the shopkeeper makes a profit of $12\%$ on the Cost Price.

$\text{C.P.} = \frac{\text{S.P.} \times 100}{100 + \text{Profit \%}}$

$\text{C.P.} = \frac{224 \times 100}{100 + 12}$

$\text{C.P.} = \frac{224 \times 100}{112}$

$\text{C.P.} = 2 \times 100 = \textsf{₹} 200$

Ans: $\textsf{₹}$ 200

Given:

Principal ($P$) = $\textsf{₹}$ $48,000$

Rate of interest ($R$) = $8\%$ per annum

Time ($n$) = $1$ year

Solution:

Since the interest is compounded half-yearly:

Rate ($r$) = $\frac{8}{2} = 4\%$ per half-year.

Number of conversion periods ($n$) = $1 \times 2 = 2$ half-years.

$\text{Amount (A)} = P \left(1 + \frac{r}{100}\right)^n$

$A = 48000 \left(1 + \frac{4}{100}\right)^2$

$A = 48000 \left(1 + \frac{1}{25}\right)^2$

$A = 48000 \left(\frac{26}{25}\right)^2$

$A = 48000 \times \frac{676}{625}$

$A = 76.8 \times 676 = \textsf{₹} 51,916.80$

Now, we find the Compound Interest (C.I.):

$\text{C.I.} = \text{Amount} - \text{Principal}$

$\text{C.I.} = 51916.80 - 48000$

$\text{C.I.} = \textsf{₹} 3,916.80$

Ans: $\textsf{₹}$ 3,916.80

Given:

Cost Price (C.P.) of $12$ lemons (1 dozen) = $\textsf{₹}$ $60$

Selling Price (S.P.) of $10$ lemons = $\textsf{₹}$ $40$

Solution:

To compare the prices, let's find the C.P. and S.P. for one lemon.

$\text{C.P. of 1 lemon} = \frac{60}{12} = \textsf{₹} 5$

$\text{S.P. of 1 lemon} = \frac{40}{10} = \textsf{₹} 4$

Since C.P. ($5$) is greater than S.P. ($4$), there is a loss.

$\text{Loss} = \text{C.P.} - \text{S.P.}$

$\text{Loss} = 5 - 4 = \textsf{₹} 1$

Now, we calculate the loss percent:

$\text{Loss %} = \frac{\text{Loss}}{\text{C.P.}} \times 100$

$\text{Loss %} = \frac{1}{5} \times 100$

$\text{Loss %} = 20\%$

Ans: Loss of 20%

Solution:

For a given principal, rate, and time:

1. If the time is exactly 1 year (and interest is compounded annually), the Simple Interest ($S$) and Compound Interest ($C$) are exactly equal. Thus, $C = S$.

2. If the time is more than 1 year, the interest starts earning interest in the compound interest method. Therefore, the compound interest will be greater than the simple interest. Thus, $C > S$.

3. Compound interest can never be less than simple interest for the same parameters. Thus, $C < S$ is impossible.

Therefore, either (i) or (ii) can be correct depending on the time period.

Hence, the correct option is (b).

Solution:

Let the principal be $P$. Since the sum doubles, the Amount ($A$) $= 2P$.

Case 1: Simple Interest (r%)

$\text{Interest} = A - P = 2P - P = P$

$P = \frac{P \times r \times 2}{100}$

$1 = \frac{r}{50} \implies r = 50\%$

Case 2: Compound Interest (R%)

$2P = P \left(1 + \frac{R}{100}\right)^2$

$2 = \left(1 + \frac{R}{100}\right)^2$

$\sqrt{2} = 1 + \frac{R}{100}$

$1.414 = 1 + \frac{R}{100}$

$\frac{R}{100} = 0.414 \implies R = 41.4\%$

Comparing the two rates: $41.4\% < 50\%$, so $R < r$.

Hence, the correct option is (b).

Given:

Principal ($P$) = $\textsf{₹}$ 50,000

Rate ($R$) = 4% p.a.

Time ($n$) = 2 years

Solution:

First, we find the Amount ($A$):

$A = P \left(1 + \frac{R}{100}\right)^n$

$A = 50000 \left(1 + \frac{4}{100}\right)^2 = 50000 \left(1 + \frac{1}{25}\right)^2$

$A = 50000 \times \frac{26}{25} \times \frac{26}{25}$

$A = 80 \times 26 \times 26$

$A = 80 \times 676 = \textsf{₹} 54,080$

Now, calculate the Compound Interest ($CI$):

$CI = A - P$

$CI = 54080 - 50000 = \textsf{₹} 4,080$

Hence, the correct option is (b).

Given:

Marked Price (M.P.) = $\textsf{₹}$ 1,200

Discount = 12%

Solution:

$\text{Discount Amount} = 12\% \text{ of } 1200$

$\text{Discount Amount} = \frac{12}{100} \times 1200 = \textsf{₹} 144$

$\text{Selling Price (S.P.)} = \text{Marked Price} - \text{Discount Amount}$

$\text{S.P.} = 1200 - 144 = \textsf{₹} 1,056$

Hence, the correct option is (a).

Given:

$90\%$ of $x = 315\text{ km}$

Solution:

$\frac{90}{100} \times x = 315$

$x = \frac{315 \times 100}{90}$

$x = \frac{315 \times 10}{9}$

$x = 35 \times 10 = 350$

Since the unit given is km, $x = 350\text{ km}$.

Hence, the correct option is (b).

Given:

Cost Price (C.P.) = $\textsf{₹}$ 360

Profit Required = 25%

Discount Offered = 10%

Solution:

First, find the required Selling Price (S.P.):

$\text{S.P.} = \text{C.P.} \times \left(1 + \frac{\text{Profit %}}{100}\right)$

$\text{S.P.} = 360 \times \frac{125}{100} = 360 \times \frac{5}{4} = 90 \times 5 = \textsf{₹} 450$

Now, this S.P. is obtained after a 10% discount on the Marked Price (M.P.).

$\text{S.P.} = \text{M.P.} \times \left(1 - \frac{\text{Discount %}}{100}\right)$

$450 = \text{M.P.} \times \frac{90}{100}$

$\text{M.P.} = \frac{450 \times 100}{90} = 5 \times 100 = \textsf{₹} 500$

Hence, the correct option is (a).

Solution:

The discount is calculated as a percentage of the marked price.

$\text{Discount} = \text{Discount Rate} \times \text{Marked Price}$

$\text{Discount} = a\% \text{ of } x$

$\text{Discount} = \frac{a}{100} \times x = x \times \frac{a}{100}$

Hence, the correct option is (c).

Given:

Principal ($P$) = $\textsf{₹}$ 1,00,000

Amount ($A$) = $\textsf{₹}$ 1,12,360

Rate ($R$) = 12% p.a. (6% per half-year)

Solution:

$A = P \left(1 + \frac{r}{100}\right)^n$

$112360 = 100000 \left(1 + \frac{6}{100}\right)^n$

$\frac{112360}{100000} = (1.06)^n$

$1.1236 = (1.06)^n$

Since it is given that $1.1236 = (1.06)^2$, we have:

$(1.06)^2 = (1.06)^n \implies n = 2$

Here, $n$ represents the number of conversion periods (half-years). Since there are 2 half-years, the time is 1 year.

Hence, the correct option is (b).

Solution:

When the interest is compounded half-yearly, the interest is calculated twice in a year (every 6 months). To adjust the standard formula:

1. The annual rate of interest ($R$) is divided by 2 because there are two half-years in a year. So, the new rate becomes $\frac{R}{2}\%$.

2. The number of years ($n$) is multiplied by 2 to find the total number of conversion periods. So, the time becomes $2n$.

Hence, the correct option is (d).

Given:

Cost of scooter = $\textsf{₹} 36,450$

Sales tax rate = $9\%$

Solution:

First, calculate the sales tax amount:

$\text{Sales Tax} = 9\% \text{ of } 36,450$

$\text{Sales Tax} = \frac{9}{100} \times 36450 = 0.09 \times 36450$

$\text{Sales Tax} = \textsf{₹} 3,280.50$

Now, calculate the total amount paid:

$\text{Total Amount} = \text{Cost Price} + \text{Sales Tax}$

$\text{Total Amount} = 36450 + 3280.50$

$\text{Total Amount} = \textsf{₹} 39,730.50$

Hence, the correct option is (b).

Given:

Marked Price (M.P.) = $\textsf{₹} 80$

Selling Price (S.P.) = $\textsf{₹} 76$

Solution:

First, find the discount amount:

$\text{Discount} = \text{M.P.} - \text{S.P.}$

$\text{Discount} = 80 - 76 = \textsf{₹} 4$

Now, find the discount rate:

$\text{Discount Rate} = \frac{\text{Discount}}{\text{M.P.}} \times 100$

$\text{Discount Rate} = \frac{4}{80} \times 100$

$\text{Discount Rate} = \frac{1}{20} \times 100 = 5\%$

Hence, the correct option is (a).

Given:

A's Cost Price (C.P.) = $\textsf{₹} 8,000$

A's profit rate = $20\%$

B's profit rate = $20\%$

Solution:

For A:

$\text{Profit for A} = 20\% \text{ of } 8000 = \frac{20}{100} \times 8000 = \textsf{₹} 1,600$

$\text{A's Selling Price (which is B's C.P.)} = 8000 + 1600 = \textsf{₹} 9,600$

For B:

$\text{Profit for B} = 20\% \text{ of } 9600 = \frac{20}{100} \times 9600 = \textsf{₹} 1,920$

Comparing the profits, $\textsf{₹} 1,600 < \textsf{₹} 1,920$. Thus, A earns less profit than B.

Hence, the correct option is (c).

Given:

C.P. of teapot = $\textsf{₹} 120$

C.P. of cups = $\textsf{₹} 400$

Solution:

$\text{S.P. of teapot (at 5 % profit)} = 120 + (5\% \text{ of } 120)$

$\text{S.P. of teapot} = 120 + 6 = \textsf{₹} 126$

$\text{S.P. of cups (at 5 % loss)} = 400 - (5\% \text{ of } 400)$

$\text{S.P. of cups} = 400 - 20 = \textsf{₹} 380$

$\text{Total amount received} = \text{Total S.P.} = 126 + 380 = \textsf{₹} 506$

Hence, the correct option is (c).

Given:

Selling Price (S.P.) = $\textsf{₹} 1,120$

Discount = $20\%$

Solution:

Let the Marked Price be $x$.

$\text{S.P.} = \text{Marked Price} - \text{Discount}$

$1120 = x - (20\% \text{ of } x)$

$1120 = x - 0.2x$

$1120 = 0.8x$

$x = \frac{1120}{0.8} = \frac{11200}{8}$

$x = \textsf{₹} 1,400$

Hence, the correct option is (b).

Given:

Time = 2 years

Compounding period = every 3 months

Solution:

Compounding every 3 months means the interest is charged quarterly.

Number of quarters in 1 year = $\frac{12}{3} = 4$

Number of quarters in 2 years = $2 \times 4 = 8$

Hence, the correct option is (a).

Given:

Price including VAT = $\textsf{₹} 13,500$

VAT rate = $8\%$

Solution:

Let the original price be $x$.

$\text{Price with VAT} = x + (8\% \text{ of } x)$

$13500 = x + 0.08x$

$13500 = 1.08x$

$x = \frac{13500}{1.08} = \frac{1350000}{108}$

$x = \textsf{₹} 12,500$

Hence, the correct option is (c).

Given:

For the first iron: Cost Price (C.P.1) = $\textsf{₹} 900$, Gain% = $10\%$

For the second iron: Cost Price (C.P.2) = $\textsf{₹} 1200$, Loss% = $5\%$

To Find:

Overall profit or loss in the transaction.

Solution:

First, we calculate the gain or loss on each individual item.

For the first iron:

$\text{Gain} = 10\% \text{ of } \textsf{₹} 900 = \frac{10}{100} \times 900 = \textsf{₹} 90$

For the second iron:

$\text{Loss} = 5\% \text{ of } \textsf{₹} 1200 = \frac{5}{100} \times 1200 = \textsf{₹} 60$

Now, we find the net result of the transaction:

$\text{Net Profit/Loss} = \text{Total Gain} - \text{Total Loss}$

$\text{Net Profit} = 90 - 60 = \textsf{₹} 30$

Since the gain is more than the loss, Avinash has a profit of $\textsf{₹} 30$.

Hence, the correct option is (c).

Given:

Price of TV (including VAT) = $\textsf{₹} 26,250$

VAT rate = $5\%$

To Find:

Original price (price before VAT) of the TV set.

Solution:

Let the original price of the TV set be $\textsf{₹} x$.

According to the question, the price including VAT is equal to the original price plus 5% of the original price.

$x + 5\% \text{ of } x = 26250$

$x + 0.05x = 26250$

$1.05x = 26250$

$x = \frac{26250}{1.05}$

$x = \frac{2625000}{105}$

$x = 25000$

Therefore, the original price of the TV set is $\textsf{₹} 25,000$.

Hence, the correct option is (b).

To Find:

The value of the expression $40\% \text{ of } [100 - 20\% \text{ of } 300]$.

Solution:

First, we solve the inner part of the bracket using the BODMAS rule:

Step 1: Calculate $20\% \text{ of } 300$

$20\% \text{ of } 300 = \frac{20}{100} \times 300 = 60$

Step 2: Calculate the value inside the square bracket

$[100 - 60] = 40$

Step 3: Calculate $40\% \text{ of } 40$

$40\% \text{ of } 40 = \frac{40}{100} \times 40 = 16$

Hence, the correct option is (b).

Given:

Original price of the car = $\textsf{₹} 2,50,000$

Decrease in first year = $10\%$

Decrease in second year = $12\%$

To Find:

Overall decrease percent over two years.

Solution:

Let the initial price be $100$ units.

Price after 1st year:

$\text{Decrease} = 10\% \text{ of } 100 = 10$

$\text{Price} = 100 - 10 = 90$ units

Price after 2nd year:

$\text{Decrease} = 12\% \text{ of } 90 = \frac{12}{100} \times 90 = 10.8$ units

$\text{Final Price} = 90 - 10.8 = 79.2$ units

Net Decrease Calculation:

$\text{Total Decrease} = \text{Initial Price} - \text{Final Price}$

$\text{Total Decrease} = 100 - 79.2 = 20.8$ units

$\text{Overall Decrease %} = \frac{20.8}{100} \times 100 = 20.8\%$

Hence, the correct option is (c).

Discount is a reduction on the marked price of the article.

Explanation:

In commercial arithmetic, the price at which an article is listed or printed is called the Marked Price (M.P.). To increase sales or clear old stock, shopkeepers often offer a rebate on this price. This reduction is known as the Discount.

Given:

Original Number = $150$

New Number = $162$

To Find:

Percentage increase.

Solution:

First, we find the absolute increase in the number:

Increase = New Number $-$ Original Number

Increase = $162 - 150 = 12$

Now, we calculate the percentage increase using the formula:

$\text{Percentage Increase} = \left( \frac{\text{Increase}}{\text{Original Number}} \times 100 \right) \%$

$\text{Percentage Increase} = \left( \frac{12}{150} \times 100 \right) \%$

$\text{Percentage Increase} = \frac{12 \times \cancel{100}^2}{\cancel{150}_3} \%$

$\text{Percentage Increase} = \frac{\cancel{12}^4 \times 2}{\cancel{3}_1} \%$

$\text{Percentage Increase} = 4 \times 2 = 8\%$

Hence, the increase is $8$ per cent.

Given:

Original Price of the article = $\textsf{₹} 1,620$

Increase Percentage = $15\%$

To Find:

The absolute increase in price (in $\textsf{₹}$).

Solution:

The increase in price is calculated as $15\%$ of the original price:

$\text{Increase} = 15\% \text{ of } \textsf{₹} 1,620$

$\text{Increase} = \frac{15}{100} \times 1620$

$\text{Increase} = \frac{\cancel{15}^3}{\cancel{100}_{20}} \times 1620$

$\text{Increase} = \frac{3 \times \cancel{1620}^{81}}{\cancel{20}_1}$

$\text{Increase} = 3 \times 81 = 243$

Hence, the increase is $\textsf{₹} 243$.

Discount = Marked Price $-$ Sale Price.

Explanation:

The Marked Price (M.P.) is the price printed on the label of an article. The Sale Price (S.P.) is the actual price at which the item is sold after reducing the discount. Therefore, the difference between the M.P. and S.P. is the discount offered.

Discount = Discount % of Marked Price.

Explanation:

In financial transactions, discount is always calculated on the Marked Price (M.P.) or List Price of an article, unless stated otherwise. This is similar to how profit or loss percentage is usually calculated on the Cost Price.

Sales Tax (or GST) is charged on the sale of an item by the government and is added to the bill amount.

In India, previously Sales Tax and Value Added Tax (VAT) were charged. Currently, the Goods and Services Tax (GST) is the indirect tax levied on the supply of goods and services. It is collected by the shopkeeper from the customer and paid to the government.

Amount when interest is compounded annually is given by the formula $A = P \left( 1 + \frac{R}{100} \right)^n$.

Where:

$A$ = Total Amount after $n$ years

$P$ = Principal amount (Initial Investment)

$R$ = Rate of interest per annum

$n$ = Number of years (Time period)

Sales tax = tax % of Selling Price.

Explanation:

Sales tax (or GST) is always calculated on the net Selling Price (S.P.) of an item after any discounts have been applied. The final bill amount is the sum of the Selling Price and the calculated Sales Tax.

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

Explanation:

In compound interest, interest is not paid out but is added back to the principal at fixed intervals. These intervals (like annually, half-yearly, or quarterly) are known as conversion periods. At the end of each conversion period, the interest earned is added to the previous principal to form the new principal for the next period.

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

Explanation:

When an item is purchased, sometimes extra money is spent on its transportation, repairs, labor charges, or insurance. These are called Overhead Expenses and are always added to the purchase price to calculate the total Effective Cost Price.

$\text{Total Cost Price} = \text{Purchase Price} + \text{Overhead Expenses}$

The discount on an item for sale is calculated on the Marked Price.

Explanation:

The Marked Price (M.P.) is the price printed on an article. Since a discount is a reduction offered to the customer on this printed price, the calculation is always based on the Marked Price unless otherwise specified.

Amount = $P \left( 1 + \frac{r}{200} \right)^{2t}$.

Explanation:

When interest is compounded semi-annually (half-yearly):

1. The annual rate of interest ($r$) is halved, becoming $\frac{r}{2}\%$.

2. The time period ($t$) is doubled, as there are two conversion periods in one year, becoming $2t$.

Substituting these into the standard compound interest formula $A = P \left( 1 + \frac{R}{100} \right)^n$, we get:

$A = P \left( 1 + \frac{r/2}{100} \right)^{2t} = P \left( 1 + \frac{r}{200} \right)^{2t}$

Percentages are numerators to fractions with denominator equal to 100.

Explanation:

The word "percent" means "per hundred". Mathematically, $x\%$ represents the fraction $\frac{x}{100}$. Here, $x$ is the numerator and $100$ is the fixed denominator.

Given:

Selling Price ($S.P.$) = $\textsf{₹} 880$

Discount Percentage ($D\%$) = $12\%$

To Find:

Marked Price ($M.P.$).

Solution:

We use the formula relating Marked Price, Selling Price, and Discount Percentage:

$M.P. = \frac{S.P. \times 100}{100 - D\%}$

$M.P. = \textsf{₹} \frac{880 \times 100}{100 - 12}$

$M.P. = \textsf{₹} \frac{\cancel{880}^{10} \times 100}{\cancel{88}_{1}}$

$M.P. = 10 \times 100 = 1000$

The marked price of the article is $\textsf{₹} 1,000$.

Given:

Principal ($P$) = $\textsf{₹} 8,000$

Rate ($R$) = $16\%$ p.a.

Time ($t$) = $1$ year

Interest is compounded half-yearly.

To Find:

Compound Interest ($C.I.$).

Solution:

Since the interest is compounded half-yearly:

New Rate ($r$) = $\frac{16}{2} = 8\%$ per half-year

Number of periods ($n$) = $1 \times 2 = 2$ half-years

Amount ($A$) = $P \left( 1 + \frac{r}{100} \right)^n$

$A = 8000 \left( 1 + \frac{8}{100} \right)^2$

$A = 8000 \times (1.08)^2$

$A = 9331.20$

Now, Compound Interest ($C.I.$) = $A - P$

$C.I. = 9331.20 - 8000 = 1331.20$

The compound interest is $\textsf{₹} 1,331.20$.

Given:

Initial Investment = $\textsf{₹} 6,00,000$

Loss in 1st year = $5\%$

Gain in 2nd year = $10\%$

To Find:

The net result.

Solution:

Value after 1st year (Loss of 5%):

Value = $6,00,000 - (5\% \text{ of } 6,00,000)$

Value = $6,00,000 - 30,000 = \textsf{₹} 5,70,000$

Value after 2nd year (Gain of 10% on $\textsf{₹} 5,70,000$):

Value = $5,70,000 + (10\% \text{ of } 5,70,000)$

Value = $5,70,000 + 57,000 = \textsf{₹} 6,27,000$

Net Result = Final Value $-$ Initial Investment

Net Result = $6,27,000 - 6,00,000 = 27,000$

The net result is a gain of $\textsf{₹} 27,000$.

Given:

Amount expression: $A = 6000 \left( 1 + \frac{5}{100} \right)^3$

Compounding type: Half-yearly

To Find:

Annual Rate of interest ($R$) and Time in years ($t$).

Solution:

The standard formula for half-yearly compounding is:

$A = P \left( 1 + \frac{R/2}{100} \right)^{2t}$

By comparing the given expression with the standard formula:

1. The term $\frac{R}{2} = 5$

$R = 5 \times 2 = 10\%$ p.a.

2. The power $2t = 3$

$t = \frac{3}{2} = 1.5$ years.

The rate of interest is $10\%$ p.a. and time is $1.5$ years.

Given:

Selling Price ($S.P.$) = $\textsf{₹} 1,12,000$

Gain Percentage = $40\%$

To Find:

Cost Price ($C.P.$).

Solution:

We use the formula:

$C.P. = \frac{S.P. \times 100}{100 + Gain\%}$

$C.P. = \frac{112000 \times 100}{100 + 40}$

$C.P. = \frac{112000 \times 100}{140}$

$C.P. = \frac{\cancel{112000}^{800} \times 100}{\cancel{140}_{1}}$

$C.P. = 800 \times 100 = 80,000$

The cost price of the article was $\textsf{₹} 80,000$.

To Find:

Loss percentage.

Solution:

Let the Selling Price ($S.P.$) of $1$ geometry box be $x$.

Total $S.P.$ of $140$ boxes = $140x$

Loss = $S.P.$ of $10$ boxes = $10x$

We know that: $C.P. = S.P. + Loss$

$C.P. = 140x + 10x = 150x$

$\text{Loss Percentage} = \left( \frac{\text{Loss}}{C.P.} \times 100 \right) \%$

$\text{Loss Percentage} = \left( \frac{10x}{150x} \times 100 \right) \%$

$\text{Loss Percentage} = \frac{\cancel{10}}{\cancel{150}_{15}} \times 100 \%$

$\text{Loss Percentage} = \frac{\cancel{100}^{20}}{\cancel{15}_{3}} \% = \frac{20}{3} \% = 6\frac{2}{3} \%$

The loss per cent is $6\frac{2}{3}\%$ (or approximately $6.67\%$).

Given:

$10 \times C.P. = 5 \times S.P.$

To Find:

Profit percentage.

Solution:

From the given condition:

$\frac{S.P.}{C.P.} = \frac{10}{5} = \frac{2}{1}$

This means if Cost Price ($C.P.$) is $1$ unit, then Selling Price ($S.P.$) is $2$ units.

Profit = $S.P. - C.P. = 2 - 1 = 1$ unit.

$\text{Profit Percentage} = \left( \frac{\text{Profit}}{C.P.} \times 100 \right) \%$

$\text{Profit Percentage} = \left( \frac{1}{1} \times 100 \right) \% = 100 \%$

The profit per cent in this transaction is $100\%$.

Given:

Number of pens = $100$

Rate per pen = $\textsf{₹} 3.50$

Sales Tax = $4\%$

To Find:

Total amount paid by Abida.

Solution:

First, calculate the total Cost Price ($C.P.$) of the pens before tax:

$\text{Total C.P.} = 100 \times 3.50 = \textsf{₹} 350$

Now, calculate the Sales Tax amount:

$\text{Sales Tax} = 4\% \text{ of } \textsf{₹} 350$

$\text{Sales Tax} = \frac{4}{100} \times 350 = \textsf{₹} 14$

Total Amount Paid = $\text{Total C.P.} + \text{Sales Tax}$

$\text{Total Amount} = 350 + 14 = 364$

The total amount paid by Abida is $\textsf{₹} 364$.

Given:

Total cost (inclusive of tax) = $\textsf{₹} 10,800$

Sales Tax rate = $8\%$

To Find:

The price before sales tax was charged.

Solution:

Let the price before sales tax be $\textsf{₹} x$.

According to the question, the total cost is the price plus $8\%$ tax on that price.

$x + (8\% \text{ of } x) = 10,800$

$x + \frac{8x}{100} = 10,800$

$\frac{108x}{100} = 10,800$

$x = \frac{10800 \times 100}{108}$

$x = \frac{\cancel{10800}^{100} \times 100}{\cancel{108}_{1}}$

$x = 10,000$

The price of the tape-recorder before sales tax was charged is $\textsf{₹} 10,000$.

Given:

Initial value = $500$

Final value = $2500$

To Find:

Percentage increase.

Solution:

Difference (increase) = $2500 - 500 = 2000$

Now, calculate the percentage by which it is greater than the base value ($500$):

$\text{Percentage Increase} = \left( \frac{\text{Difference}}{\text{Base Value}} \times 100 \right) \%$

$\text{Percentage Increase} = \left( \frac{2000}{500} \times 100 \right) \%$

$\text{Percentage Increase} = \frac{\cancel{2000}^{4}}{\cancel{500}_{1}} \times 100\%$

$\text{Percentage Increase} = 400\%$

$2500$ is greater than $500$ by $400\%$.

Solution:

Let the original number be $x$.

Four times the number = $4x$.

Increase in the number = $4x - x = 3x$.

Percentage Calculation:

$\text{Percentage Increase} = \left( \frac{\text{Increase}}{\text{Original Number}} \times 100 \right) \%$

$\text{Percentage Increase} = \left( \frac{3x}{x} \times 100 \right) \%$

$\text{Percentage Increase} = 3 \times 100 = 300\%$

Four times a number is a $300\%$ increase in the number.

Given:

Marked Price ($M.P.$) = $\textsf{₹} 200$

Discount = $5\%$

Sales Tax = $5\%$

Solution:

First, find the Selling Price after the discount is applied to the Marked Price:

$\text{Discount} = 5\% \text{ of } 200 = \textsf{₹} 10$

$\text{Selling Price (S.P.)} = 200 - 10 = \textsf{₹} 190$

Now, calculate the Sales Tax on the Selling Price:

$\text{Sales Tax} = 5\% \text{ of } 190 = \frac{5}{100} \times 190 = \textsf{₹} 9.50$

$\text{Total Amount Payable} = S.P. + \text{Sales Tax}$

$\text{Amount Payable} = 190 + 9.50 = 199.50$

The amount payable is $\textsf{₹} 199.50$.

True (T)

Explanation:

The compound interest formula $A = P(1 + \frac{r}{100})^n$ is used for any quantity that increases or decreases at a constant percentage rate over time, which accurately describes population growth or bacterial growth.

False (F)

Explanation:

Additional expenses like repairs, transportation, and labor are known as Overhead Expenses. Value Added Tax (VAT) is a government tax levied on the sale of goods.

False (F)

Explanation:

Discount is a reduction offered on the Marked Price (the printed price) of an article to attract customers.

True (T)

Explanation:

In compound interest, the interest for the next period is calculated on the principal plus any accumulated interest from previous periods (which is the previous year's amount).

False (F)

Explanation:

The correct formula is S.P. = M.P. $-$ Discount. The Cost Price ($C.P.$) is the price paid by the seller to acquire the item, not necessarily the result of subtracting discount from Marked Price.

False (F)

Given:

Cost Price ($C.P.$) of the bicycle = $\textsf{₹} 1,040$

Selling Price ($S.P.$) of the bicycle = $\textsf{₹} 800$

To Find:

Whether there is a gain or loss, and its percentage.

Solution:

Since the Selling Price ($\textsf{₹} 800$) is less than the Cost Price ($\textsf{₹} 1,040$), the man incurred a loss, not a gain.

$\text{Loss} = 1040 - 800 = \textsf{₹} 240$

$\text{Loss Percentage} = \left( \frac{\text{Loss}}{C.P.} \times 100 \right) \%$

$\text{Loss Percentage} = \left( \frac{240}{1040} \times 100 \right) \% \approx 23.08\%$

Since the transaction resulted in a loss and the percentage is not 30%, the statement is false.

False (F)

Solution:

Let the original number be $x$.

Case 1: Three times the number

New number = $3x$

Increase = $3x - x = 2x$

Percentage Increase = $\left( \frac{2x}{x} \times 100 \right) \% = 200\%$

This part of the statement is true.

Case 2: One-third of the number

New number = $\frac{1}{3}x$

Decrease = $x - \frac{1}{3}x = \frac{2}{3}x$

Percentage Decrease = $\left( \frac{2x/3}{x} \times 100 \right) \% = \frac{200}{3} \% \approx 66.67\%$

Since a decrease of $200\%$ is mathematically impossible (it would result in a negative number), and the actual decrease is $66.67\%$, the statement is false.

True (T)

Reasoning:

For the first conversion period (usually the first year), Simple Interest ($S.I.$) and Compound Interest ($C.I.$) are equal. From the second period onwards, $C.I.$ is calculated on the principal plus the previous interest, making it greater than $S.I.$, which is always calculated on the original principal only.

True (T)

Given:

Principal ($P$) = $\textsf{₹} 7,000$

Rate of depreciation ($R$) = $8\%$ p.a.

Time ($n$) = $2$ years

To Find:

The value after 2 years.

Solution:

Value after $n$ years = $P \left( 1 - \frac{R}{100} \right)^n$

Value = $7000 \left( 1 - \frac{8}{100} \right)^2$

Value = $7000 \times \left( \frac{92}{100} \right)^2$

Value = $7000 \times 0.92 \times 0.92$

Value = $7000 \times 0.8464 = 5924.80$

The calculated value matches the statement, so it is true.

False (F)

Reasoning:

The formula for discount percentage is:

$\text{Discount Percentage} = \left( \frac{\text{Discount}}{\text{Marked Price}} \times 100 \right) \%$

Given Discount = $y$ and Marked Price = $x$, the correct formula should be:

$\text{Discount Percentage} = \left( \frac{y}{x} \times 100 \right) \%$

Since the expression in the statement is inverted ($\frac{x}{y}$), it is false.

True (T)

Given:

Original Number = $91,422$

New Number = $1,16,054$

Solution:

Increase = $1,16,054 - 91,422 = 24,632$

$\text{Percentage Increase} = \left( \frac{24632}{91422} \times 100 \right) \%$

$\text{Percentage Increase} \approx 0.2694 \times 100 \% \approx 26.94\%$

Since $26.94\%$ is approximately $27\%$, the statement is true.

True (T)

Given:

$9 \times S.P. = 15 \times C.P.$

Solution:

$\frac{S.P.}{C.P.} = \frac{15}{9} = \frac{5}{3}$

Let $C.P. = 3$ and $S.P. = 5$

Profit = $S.P. - C.P. = 5 - 3 = 2$

$\text{Profit Percentage} = \left( \frac{2}{3} \times 100 \right) \% = \frac{200}{3} \% = 66 \frac{2}{3} \%$

The statement is true.

False (F)

Reasoning:

The formula for Amount ($A$) after $T$ years is $P \left( 1 + \frac{R}{100} \right)^T$.

The formula for Compound Interest ($C.I.$) is $A - P$, which is:

$C.I. = P \left[ \left( 1 + \frac{R}{100} \right)^T - 1 \right]$

The expression provided in the statement is neither the correct formula for Amount (missing power $T$) nor for Compound Interest.

True (T)

Reasoning:

When there is a gain, the Selling Price is higher than the Cost Price. The gain is calculated as a percentage of the Cost Price. The standard formula for $S.P.$ in terms of $C.P.$ and Gain% is indeed:

$S.P. = C.P. + \left( \frac{Gain\%}{100} \times C.P. \right) = C.P. \left( 1 + \frac{Gain\%}{100} \right) = \frac{(100 + Gain\%) \times C.P.}{100}$

False (F)

Reasoning:

In case of loss, the correct formula for Cost Price in terms of Selling Price is:

$C.P. = \frac{100 \times S.P.}{100 - Loss\%}$

The statement uses a plus sign ($+$) in the denominator, which would apply to gain, not loss. For loss, the percentage must be subtracted from 100.

False (F)

Given:

Original Value ($P$) = $\textsf{₹} 4,40,000$

Rate of depreciation ($R$) = $10\%$ p.a.

Time ($n$) = $3$ years

Solution:

The value after depreciation is calculated using the formula:

$\text{Value after } n \text{ years} = P \left( 1 - \frac{R}{100} \right)^n$

Value after 3 years = $4,40,000 \left( 1 - \frac{10}{100} \right)^3$

Value after 3 years = $4,40,000 \times \left( \frac{9}{10} \right)^3$

Value after 3 years = $4,40,000 \times \frac{729}{1000}$

Value after 3 years = $440 \times 729 = 3,20,760$

Since the calculated value is $\textsf{₹} 3,20,760$ and not $\textsf{₹} 3,08,000$, the statement is false.

False (F)

Given:

Marked Price ($M.P.$) = $\textsf{₹} 190$

Sales Tax = $2\%$

Solution:

$\text{Sales Tax Amount} = 2\% \text{ of } 190$

$\text{Sales Tax Amount} = \frac{2}{100} \times 190 = \textsf{₹} 3.80$

$\text{Total Cost} = M.P. + \text{Sales Tax}$

$\text{Total Cost} = 190 + 3.80 = 193.80$

The actual cost is $\textsf{₹} 193.80$, not $\textsf{₹} 192$, so the statement is false.

True (T)

Given:

Quantity of flour = $5\text{ kg}$

Rate of flour = $\textsf{₹} 20\text{ per kg}$

Sales Tax (ST) = $5\%$

Solution:

To determine if the statement is true, we calculate the buying price per unit (per kg) inclusive of sales tax.

Sales Tax on $1\text{ kg}$ of flour = $5\% \text{ of } \textsf{₹} 20$

$\text{Sales Tax} = \frac{5}{100} \times 20 = \textsf{₹} 1$

Now, we find the buying price per kg:

$\text{Buying Price per kg} = \text{Rate} + \text{Sales Tax}$

$\text{Buying Price per kg} = 20 + 1 = \textsf{₹} 21$

Since the resulting buying price per kg matches the value mentioned in the statement ($\textsf{₹} 21$), the statement is considered True.

Note:

In this context, the "buying price" refers to the effective rate at which the item is purchased after adding the tax component to the original rate per kg.

True (T)

Given:

Price inclusive of VAT = $\textsf{₹} 324$

VAT rate = $8\%$

Solution:

Let the original price be $\textsf{₹} x$.

$x + (8\% \text{ of } x) = 324$

$1.08x = 324$

$x = \frac{324}{1.08} = 300$

The original price is indeed $\textsf{₹} 300$, so the statement is true.

False (F)

Reasoning:

Sales tax is calculated on the Selling Price ($S.P.$) of the item (the price at which the shopkeeper sells it to the customer), not on the Cost Price ($C.P.$) (the price at which the shopkeeper bought it).

Given:

Percentage of women workers = $35\%$

Difference between number of men and women = $252$

To Find:

Total number of workers.

Solution:

Let the total number of workers in the factory be $x$.

Number of women workers = $35\% \text{ of } x = 0.35x$

Percentage of men workers = $(100 - 35)\% = 65\%$

Number of men workers = $65\% \text{ of } x = 0.65x$

According to the question, the number of men exceeds women by $252$:

$\text{Number of Men} - \text{Number of Women} = 252$

$0.65x - 0.35x = 252$

$0.30x = 252$

$x = \frac{252}{0.30}$

$x = \frac{25200}{30} = 840$

The total number of workers in the factory is $840$.

Given:

Total weight of sugar in 3 bags = $64.2\text{ kg}$

Bag 2 = $\frac{4}{5} \times \text{Bag 1}$

Bag 3 = $45\frac{1}{2}\% \times \text{Bag 2}$

To Find:

The weight of sugar in each bag.

Solution:

Let the weight of sugar in the first bag be $x\text{ kg}$.

Weight of sugar in the second bag = $\frac{4}{5}x = 0.8x\text{ kg}$

Weight of sugar in the third bag = $45.5\% \text{ of } 0.8x$

Weight of sugar in the third bag = $\frac{45.5}{100} \times 0.8x $$ = 0.455 \times 0.8x $$ = 0.364x\text{ kg}$

Total weight = Sum of weights in all three bags:

$x + 0.8x + 0.364x = 64.2$

$2.164x = 64.2$

$x = \frac{64.2}{2.164} \approx 29.67$

Weight in each bag:

Bag 1 = $29.67\text{ kg}$

Bag 2 = $0.8 \times 29.67 = \mathbf{23.74\text{ kg}}$

Bag 3 = $0.364 \times 29.67 = \mathbf{10.79\text{ kg}}$

(Values are rounded to two decimal places).

(a) Given:

Marked Price ($M.P.$) = $\textsf{₹} 5450$

Discount = $5\%$

Solution:

Discount Amount = $5\% \text{ of } 5450 = \frac{5}{100} \times 5450 = \textsf{₹} 272.50$

Selling Price ($S.P.$) = $M.P. - \text{Discount}$

$S.P. = 5450 - 272.50 = 5177.50$

The Selling Price is $\textsf{₹} 5,177.50$.

(b) Given:

Marked Price ($M.P.$) = $\textsf{₹} 1300$

Discount = $1.5\%$

Solution:

Discount Amount = $1.5\% \text{ of } 1300 = \frac{1.5}{100} \times 1300 = \textsf{₹} 19.50$

Selling Price ($S.P.$) = $M.P. - \text{Discount}$

$S.P. = 1300 - 19.50 = 1280.50$

The Selling Price is $\textsf{₹} 1,280.50$.

(a) Given:

Selling Price ($S.P.$) = $\textsf{₹} 495$

Discount = $1\%$

Solution:

We use the formula: $M.P. = \frac{S.P. \times 100}{100 - \text{Discount}\%}$

$M.P. = \frac{495 \times 100}{100 - 1} = \frac{495 \times 100}{99}$

$M.P. = 5 \times 100 = 500$

The Marked Price is $\textsf{₹} 500$.

(b) Given:

Selling Price ($S.P.$) = $\textsf{₹} 9,250$

Discount = $7.5\%$

Solution:

$M.P. = \frac{9250 \times 100}{100 - 7.5} = \frac{9250 \times 100}{92.5}$

$M.P. = \frac{9250 \times 1000}{925}$

$M.P. = 10 \times 1000 = 10000$

The Marked Price is $\textsf{₹} 10,000$.

(a) Given:

Marked Price ($M.P.$) = $\textsf{₹} 625$

Selling Price ($S.P.$) = $\textsf{₹} 562.50$

Solution:

Discount = $M.P. - S.P. = 625 - 562.50 = \textsf{₹} 62.50$

$\text{Discount}\% = \left( \frac{\text{Discount}}{M.P.} \times 100 \right) \%$

$\text{Discount}\% = \left( \frac{62.50}{625} \times 100 \right) \% = 10\%$

The discount is $10\%$.

(b) Given:

Marked Price ($M.P.$) = $\textsf{₹} 900$

Selling Price ($S.P.$) = $\textsf{₹} 873$

Solution:

Discount = $M.P. - S.P. = 900 - 873 = \textsf{₹} 27$

$\text{Discount}\% = \left( \frac{27}{900} \times 100 \right) \% = 3\%$

The discount is $3\%$.

Given:

Marked Price ($M.P.$) = $\textsf{₹} 500$

Discount = $5\%$

Profit = $25\%$

To Find:

Cost Price ($C.P.$) of the article.

Solution:

First, calculate the Selling Price ($S.P.$):

$\text{Discount Amount} = 5\% \text{ of } 500 = \textsf{₹} 25$

$S.P. = M.P. - \text{Discount} = 500 - 25 = \textsf{₹} 475$

Now, we have $S.P. = \textsf{₹} 475$ and $\text{Profit} = 25\%$. We find $C.P.$:

$C.P. = \frac{S.P. \times 100}{100 + \text{Profit}\%}$

$C.P. = \frac{475 \times 100}{100 + 25} = \frac{475 \times 100}{125}$

$C.P. = \frac{\cancel{475}^{95} \times \cancel{100}^{4}}{\cancel{125}_{\cancel{5}_1}}$ results in:

$C.P. = 95 \times 4 = 380$

The cost price of the article is $\textsf{₹} 380$.

Given:

For 2007-08: Appeared = $1,05,332$; Passed = $88,151$

For 2008-09: Appeared = $1,16,054$; Passed = $1,03,804$

Solution:

Pass % in 2007-08:

$\text{Pass}\% = \left( \frac{88151}{105332} \times 100 \right) \% \approx 83.69\%$

Pass % in 2008-09:

$\text{Pass}\% = \left( \frac{103804}{116054} \times 100 \right) \% \approx 89.44\%$

Comparison:

The pass percentage has increased.

$\text{Increase in Pass}\% = 89.44\% - 83.69\% = 5.75\%$

There is an increase of approximately $5.75\%$ in the pass result.

Given:

Worth of watch ($M.P.$) = $\textsf{₹} 5400$

Sale Price ($S.P.$) = $\textsf{₹} 4500$

To Find:

Discount percentage.

Solution:

Discount Amount = $M.P. - S.P. = 5400 - 4500 = \textsf{₹} 900$

$\text{Discount}\% = \left( \frac{\text{Discount}}{M.P.} \times 100 \right) \%$

$\text{Discount}\% = \left( \frac{900}{5400} \times 100 \right) \%$

$\text{Discount}\% = \frac{100}{6}\% = 16.67\%$

The discount offered is $16.67\%$ (or $16\frac{2}{3}\%$).

Given:

Initial number of patients in 2001 ($P$) = $4,375$

Rate of decrease ($R$) = $8\%$ p.a.

Time period ($n$) = $2003 - 2001 = 2$ years

To Find:

Number of patients in 2003.

Solution:

This is a case of depreciation (decrease). The formula is:

$\text{Final Count} = P \left( 1 - \frac{R}{100} \right)^n$

$\text{Patients in 2003} = 4375 \left( 1 - \frac{8}{100} \right)^2$

$\text{Patients in 2003} = 4375 \times \left( \frac{92}{100} \right)^2$

$\text{Patients in 2003} = 4375 \times 0.92 \times 0.92$

$\text{Patients in 2003} = 4375 \times 0.8464 = 3703$

The number of malaria patients in 2003 was $3,703$.

Given:

Total amount paid (inclusive of tax) = $\textsf{₹} 3,155$

Sales Tax Rate = $4.5\%$

To Find:

Price of the product before tax.

Solution:

Let the price before tax be $\textsf{₹} x$.

The total price is calculated as the original price plus the sales tax on that price.

$\text{Price} + \text{Sales Tax} = \text{Total Amount}$

$x + (4.5\% \text{ of } x) = 3,155$

$x + \frac{4.5x}{100} = 3,155$

$x + 0.045x = 3,155$

$1.045x = 3,155$

$x = \frac{3155}{1.045}$

$x \approx 3,019.14$

The price before tax was added is $\textsf{₹} 3,019.14$.

Given Data:

(a) Bathing and Sanitation Together:

Water for Bathing = $20\text{ litres}$

Water for Sanitation = $40\text{ litres}$

Total for both = $20 + 40 = 60\text{ litres}$

$\text{Percentage} = \left( \frac{60}{150} \times 100 \right) \% = 40\%$

The percentage of water used for bathing and sanitation together is $40\%$.

(b) Comparison: Cooking and Bathing:

Water for Bathing = $20\text{ litres}$

Water for Cooking = $4\text{ litres}$

Difference in usage = $20 - 4 = 16\text{ litres}$

$\text{Less percentage} = \left( \frac{\text{Difference}}{\text{Bathing usage}} \times 100 \right) \%$

$\text{Percentage} = \left( \frac{16}{20} \times 100 \right) \% = 80\%$

Water used for cooking is $80\%$ less than that used for bathing.

(c) Drinking, Cooking and Gardening Together:

Water for Drinking = $3\text{ litres}$

Water for Cooking = $4\text{ litres}$

Water for Gardening = $23\text{ litres}$

Total for these three = $3 + 4 + 23 = 30\text{ litres}$

$\text{Percentage} = \left( \frac{30}{150} \times 100 \right) \% = 20\%$

The percentage of water used for drinking, cooking and gardening together is $20\%$.

Given:

Consumption in $1975$ = $3850\text{ cu.km/year}$

Consumption in $2000$ = $6000\text{ cu.km/year}$

Total time period = $2000 - 1975 = 25\text{ years}$

To Find:

(i) Total percentage increase

(ii) Annual percentage increase

Solution:

First, find the total increase in consumption:

$\text{Total Increase} = 6000 - 3850 = 2150\text{ cu.km/year}$

(i) Total Percentage Increase:

$\text{Total % Increase} = \left( \frac{\text{Increase}}{\text{Initial Consumption}} \times 100 \right) \%$

$\text{Total % Increase} = \left( \frac{2150}{3850} \times 100 \right) \% = \left( \frac{21500}{385} \right) \% \approx 55.84\%$

(ii) Annual Percentage Increase:

Since consumption increases uniformly over 25 years:

$\text{Annual % Increase} = \frac{\text{Total Percentage Increase}}{\text{Number of Years}}$

$\text{Annual % Increase} = \frac{55.84\%}{25} \approx 2.23\%$

The total increase is $55.84\%$ and the annual increase is $2.23\%$.

Step-by-Step Calculation:

(a) Engine Oil:

Cost of oil = $3.10 \times 178.75 = \textsf{₹} 554.125$

VAT ($20\%$) = $20\% \text{ of } 554.125 = \textsf{₹} 110.825$

Total for Oil = $554.125 + 110.825 = \textsf{₹} 664.95$

(b) Other Services:

Cost = $\textsf{₹} 1,105.12$

VAT ($12.5\%$) = $12.5\% \text{ of } 1105.12 = \textsf{₹} 138.14$

Total for other services = $1105.12 + 138.14 = \textsf{₹} 1,243.26$

(c) Labour Charges and Service Tax:

Labour Cost = $\textsf{₹} 2,095.80$

Service Tax ($10\%$) = $10\% \text{ of } 2095.80 = \textsf{₹} 209.58$

Total for labour cost and service tax = $2,095.80 + 209.58 = \textsf{₹} 2305.38$

(d) Cess on Service Tax:

Cess = $3\% \text{ of } \text{Service Tax}$

Cess = $3\% \text{ of } 209.58 = \textsf{₹} 6.2874 \approx \textsf{₹} 6.29$

Total Bill Amount:

Total = $664.95 + 1243.26 + 2305.38 + 6.29$

$\text{Total} = \textsf{₹} 4,219.88$

The total bill amount is $\textsf{₹} 4,219.88$.

Given:

Initial Principal ($P_1$) = $\textsf{₹} 40,000$

Rate ($R$) = $8\%$ p.a.

(a) Interest for the first year:

$I_1 = \frac{P_1 \times R \times T}{100}$

$I_1 = \frac{40000 \times 8 \times 1}{100} = \textsf{₹} 3,200$

(b) Principal for the 2nd year:

In compound interest, the principal for the next year is the amount at the end of the previous year.

$P_2 = P_1 + I_1$

$P_2 = 40000 + 3200 = \textsf{₹} 43,200$

(c) Interest for the 2nd year:

$I_2 = \frac{P_2 \times R \times 1}{100}$

$I_2 = \frac{43200 \times 8 \times 1}{100} = \textsf{₹} 3,456$

(d) Amount after 2 years:

$A = P_2 + I_2$

$A = 43200 + 3456 = \textsf{₹} 46,656$

The interest for the 1st year is $\textsf{₹} 3,200$, principal for 2nd year is $\textsf{₹} 43,200$, interest for 2nd year is $\textsf{₹} 3,456$, and total amount after 2 years is $\textsf{₹} 46,656$.

Given:

Total Seats = $49,000$

General Category Seats = $28,200$

SC Category Seats = $7,400$

ST Category Seats = $3,700$

(i) Percentage of General Category Seats:

$\text{Percentage} = \left( \frac{\text{General Category Seats}}{\text{Total Seats}} \times 100 \right) \%$

$\text{Percentage} = \left( \frac{28200}{49000} \times 100 \right) \%$

$\text{Percentage} = \frac{2820}{49} \% \approx 57.55\%$

(ii) Percentage of SC and ST Category Together:

Total SC + ST Seats = $7400 + 3700 = 11,100$

$\text{Percentage} = \left( \frac{11100}{49000} \times 100 \right) \%$

$\text{Percentage} = \frac{1110}{49} \% \approx 22.65\%$

The percentage for General Category is $57.55\%$ and for SC/ST combined is $22.65\%$.

Given:

To Find:

Price before VAT was added.

Solution:

Let the price before VAT was added be $x$.

The price including VAT is calculated as:

Now, we solve for $x$:

Final Answer: The price of the medicines before VAT was added was $\textsf{₹} 35$.

Given:

To Find:

The tax percentage (Tax%).

Solution:

First, we need to find the price before taxes (Net Price):

Now, we calculate the tax percentage based on the price before tax:

Simplifying the fraction:

Final Answer: The tax percentage is $12.5\%$.

Solution:

We calculate the discount and the net amount for each item separately.

(a) Atta:

(b) Detergent:

(c) Namkeen:

Bill Summary:

Calculating the total bill amount:

Final Answer: The total amount of the bill Arunima has to pay is $\textsf{₹} 582.01$.

Solution:

First, we calculate the service tax for both periods.

Period 1 (17-02-09 to 23-02-09):

Period 2 (24-02-09 to 16-03-09):

Total Service Tax:

Education Cess:

Education cess is charged at $3\%$ on the service tax amount.

Final Bill Amount:

Final Answer: Rounding to two decimal places, the final bill amount is $\textsf{₹} 885.41$.

Given:

Since the interest is compounded half-yearly, the rate for 6 months (half year) is:

(i) Interest for 6 months:

(ii) Amount after 6 months:

(iii) Interest for next 6 months:

For compounding, the amount after the first 6 months becomes the principal for the next 6 months.

(iv) Amount after one year:

Final Answers:

(i) Interest for 6 months = $\textsf{₹} 5,000$

(ii) Amount after 6 months = $\textsf{₹} 1,05,000$

(iii) Interest for next 6 months = $\textsf{₹} 5,250$

(iv) Amount after one year = $\textsf{₹} 1,10,250$

Given:

To Find:

Gain or Loss percentage on the whole dealing.

Solution:

First, we calculate the Total Cost Price (CP):

Now, we find the quantity sold at different rates:

Now, we calculate the Total Selling Price (SP):

Since $\text{Total SP} > \text{Total CP}$, there is a Gain:

Now, we calculate the Gain Percentage:

Final Answer: Babita's gain percentage on the whole dealing is $27.08\%$.

Given:

To Find:

The final amount paid by Pragya.

Solution:

This is a case of successive discounts. First, we apply the 25% discount on the marked price:

Now, the 30% off-season discount is applied to the reduced price:

Final Answer: Pragya paid $\textsf{₹} 630$ for the skirt.

Given:

To Find:

Marked Price (MP) of the mobile phone.

Solution:

Let the Marked Price of the mobile phone be $x$.

The discount amount is calculated as a percentage of the marked price:

Solving for $x$:

Final Answer: The marked price of the mobile phone was $\textsf{₹} 7,840$.

Given:

Solution:

1. Calculation of Simple Interest (SI):

2. Calculation of Compound Interest (CI):

3. Difference between CI and SI:

Final Answer: The difference between CI and SI is $\textsf{₹} 7,305.30$.

Given:

To Find:

The cost of the computer after two years.

Solution:

Depreciation follows the compound interest formula but with a negative rate of interest:

Final Answer: The cost of the computer after two years will be $\textsf{₹} 25,000$.

Given:

To Find:

Population two years ago ($P_0$).

Solution:

The population decreases successively. The formula for the population after two years of decrease is:

Substituting the given values:

Final Answer: The population of the town two years ago was $7,00,000$.

Given:

To Find:

Gain or Loss percentage.

Solution:

First, we find the Cost Price and Selling Price of one lemon:

Since the Cost Price is equal to the Selling Price ($CP = SP$):

Final Answer: There is no gain and no loss (0%) in the whole transaction.

Solution:

The formula to calculate the Percentage Decrease is:

1. For Petrol:

2. For Diesel:

3. For LPG:

Completed Table:

Solution:

We calculate the percentage change for each party using the formula:

Party A:

Party B:

Party C:

Party D:

Final Results:

Party A: $29.61\%$ decrease

Party B: $18.97\%$ increase

Party C: $500\%$ increase

Party D: $9.09\%$ increase

Given:

To Find:

Percentage of more seats won by X as compared to Y.

Solution:

First, we find the difference in seats between X and Y:

Now, we calculate this difference as a percentage of Party Y's seats:

Final Answer: Party X won $50.48\%$ more seats as compared to Party Y.

Given:

To Find:

Overall gain or loss %.

Solution:

For the first cooler (Gain):

For the second cooler (Loss):

Total Transaction:

Final Answer: Ashima suffered an overall loss of $0.25\%$.

Alternate Solution:

When two items are sold at the same price, one at a gain of $x\%$ and another at a loss of $x\%$, there is always a loss given by:

Given:

To Find:

Gain or Loss percentage.

Solution:

Since $\text{CP} > \text{SP}$, there is a loss:

Calculating the loss percentage:

Final Answer: The lady suffered a loss of $22.22\%$.

Given:

To Find:

New Selling Price ($\text{SP}_2$).

Solution:

First, we find the Cost Price (CP) of the chair:

Now, we calculate the Selling Price to achieve an $8\%$ gain:

Final Answer: The shopkeeper should sell the chair for $\textsf{₹} 864$ to gain 8%.

Given:

To Find:

Total amount paid by the customer.

Solution:

First, we find the Selling Price (SP) before tax:

Now, we calculate the sales tax on this selling price:

Total amount paid by the customer:

Final Answer: The customer pays $\textsf{₹} 3,561.60$ in all.

Given:

To Find:

Profit percentage.

Solution:

First, calculate the Total Cost Price (CP):

Now, calculate the profit:

Calculating the profit percentage:

Final Answer: Achal's profit percentage is $30\%$.

Given:

To Find:

The Selling Price (SP) of the air-conditioner.

Solution:

First, we calculate the Total Cost Price (CP) by adding the overhead expenses:

Now, we calculate the Selling Price to achieve a gain of $15\%$:

Final Answer: The lady should sell the air-conditioner for $\textsf{₹} 18,400$.

Given:

To Find:

Marked Price (MP) of the article.

Solution:

First, we find the Selling Price (SP) that the shopkeeper needs to achieve the desired gain:

The Selling Price is also the amount after applying the discount to the Marked Price:

Solving for MP:

Final Answer: The shopkeeper should mark the article at $\textsf{₹} 800$.

Given:

To Find:

1. Cost of one coat to the customer.

2. Total profit earned by Brinda.

Solution:

I. Calculation for one coat:

II. Calculation of Total Profit:

Final Answer: The cost of one coat to the customer is $\textsf{₹} 1,653.60$ and the total profit earned by Brinda is $\textsf{₹} 1,620$.

Given:

Solution:

Since the interest is compounded half-yearly:

We use the compound interest formula for Amount ($A$):

Calculating Interest Paid:

Final Answer: Rahim will pay $\textsf{₹} 10,75,840$ in total, and the interest paid is $\textsf{₹} 51,840$.

Given:

To Find:

The total cost of all items including sales tax.

Solution:

First, calculate the total price of all items before tax:

Now, calculate the sales tax on the total price:

Finally, calculate the total cost for Shikha:

Final Answer: The items will cost Shikha $\textsf{₹} 5,243$.

(a) Solution:

The total amount of fat in Cream of Tomato soup can be measured more accurately. This is because the label provides a more precise measurement for its components, such as Saturated Fat being $1.5$g (expressed as a decimal), whereas the Sweet Corn label uses only whole numbers ($0$g), indicating a lower level of precision in reporting.

(b) Given:

To Find:

The recommended daily value of sodium in milligrams.

Solution:

According to the provided data, the sodium content in one serving is $690$ mg. To find the total recommended daily value (100%):

Final Answer: The recommended daily value of sodium is $690$ mg.

(c) Solution:

Given:

Calculation:

Final Answer: The increase per cent of sugar is $120\%$.

(d) Solution:

Given:

Calculation:

Dividing both numerator and denominator by their common factor $3$:

Final Answer: The ratio of calories from fat is $3 : 7$.

Given:

Method 1 (Sonia's Method):

Find $25\%$ of the original price and subtract it from the original price.

Method 2 (Rahul's Method):

If the discount is $25\%$, the customer pays $100\% - 25\% = 75\%$ of the original price.

Final Answer: The Selling Price (S.P.) of the music CD is $\textsf{₹} 90$.

Given:

To Find:

Which store offers the game at a lower price?

Solution:

Store A:

Offers a discount of $25\%$. We will use Rahul's method to find the Sale Price:

Store B:

Comparison:

Final Answer: The game is less expensive at Store A.

Given:

(a) Solution:

The sale price is calculated as $66\%$ of the original price:

(b) Solution:

The savings is the difference between the original price and the sale price:

Final Answer: (a) The sale price is $\textsf{₹} 59.40$. (b) The amount saved is $\textsf{₹} 30.60$.

Given:

(a) Solution:

No, both methods will not give the same result. Method 2 will be more beneficial for Sheela because the $25\%$ discount is applied to the larger original amount, leading to a higher total discount before the voucher is deducted.

(b) Calculation:

Method 1: Subtract $\textsf{₹} 50$ first, then apply $25\%$ discount.

Method 2: Apply $25\%$ discount first, then subtract $\textsf{₹} 50$.

(c) Solution:

Most stores use Method 1. By subtracting the voucher amount first, the store reduces the base price on which the percentage discount is calculated, meaning the store gives away less money in the form of a percentage discount.

Final Answer: Sheela pays $\textsf{₹} 202.50$ by Method 1 and $\textsf{₹} 190$ by Method 2. Method 2 is better for her.

Solution:

Neelgiri Apartments:

Savana Apartments:

Comparison:

Final Answer: Neelgiri Apartments will be cheaper by $\textsf{₹} 900$.

Explanation:

Let the original amount be $x$.

1. After a $20\%$ increase:

2. After a $20\%$ decrease on the new amount:

The decrease is now calculated on $1.20x$, not the original $x$.

Since $0.96x < x$, the final amount is less than the original amount. This happens because the $20\%$ decrease is applied to a larger base ($1.20x$) than the $20\%$ increase was applied to ($x$), resulting in a larger absolute subtraction than the initial addition.

1. Solution:

If the sunscreen allows $\frac{1}{15}$ of rays, it "aborts" (blocks) the rest:

2. Solution:

(a) If it allows $25\%$, it blocks $100\% - 25\% = 75\%$.

(b) SPF is the reciprocal of the fraction of rays allowed.

3. Solution:

For SPF 30, the fraction of rays allowed is $\frac{1}{30}$.

Since $96.67\%$ is approximately $97\%$, the claim is true.

Given:

To Find:

Selling Price (SP) of the property.

Solution:

Let the Selling Price of the property be $x$.

Solving for $x$:

Final Answer: The agent sells the property for $\textsf{₹} 12,50,000$.

Given:

To Find:

The reduced price and the original price of tea.

Solution:

First, we calculate the money saved due to the $15\%$ reduction in price:

With this saved amount of $\textsf{₹} 6.75$, Tonu is able to buy $2$ kg more tea at the reduced price. Therefore:

Now, let the original price per kg be $x$. Since the reduced price is $85\%$ of the original price ($100\% - 15\%$):

Final Answer: The reduced price of tea is $\textsf{₹} 3.375$ per kg and the original price was approximately $\textsf{₹} 3.97$ per kg.

Solution:

To find the Final %, we use the formula: $\text{Final % } = \left( \frac{\text{Total Marks Obtained}}{\text{Maximum Total Marks}} \right) \times 100$.

Observation: Vidit performed best in English Language (90.4%) and Physics (90%). He needs improvement in English Literature (81.6%) and Chemistry (82%).

Given:

To Find:

Success rate in percentage ($\text{Success %}$).

Solution:

Final Answer: Sita's success rate is approximately $91.43\%$.

Solution:

To compare their progress, we express both as percentages.

1. Neha:

2. Minakshi:

Comparison:

Comparing the remaining work:

Final Answer: Minakshi must finish a greater per cent of her homework.

Given:

To Find:

Percentage of species found in rain forests.

Solution:

Final Answer: $36\%$ of the world’s identified plant species are found in rain forests.

Given:

To Find:

Percentage of the floor covered by the carpet.

Solution:

First, we calculate the total area of the floor:

Now, we calculate the percentage of the floor covered:

Final Answer: The carpet covers $44.44\%$ of the floor.

Given:

To Find:

Weight of water in Jyoti's body.

Solution:

Final Answer: The weight of water in Jyoti's body is $37.52$ kg.

Given:

To Find:

The amount of pure gold in grams.

Solution:

Final Answer: There are $4.4308$ grams of pure gold in the ring.

Solution:

The student's error lies in the incorrect setup of the proportion. To find a percentage of a number, the standard proportion is:

In the problem "Find $5\%$ of $32$":

- The Percent is $5$.

- The Whole is $32$.

- The Part ($n$) is what we need to find.

Therefore, the correct proportion should have been:

What the student did wrong: The student placed $32$ as the "Whole" in the denominator of the percent side and $100$ in the denominator of the part side. By using $\frac{n}{100} = \frac{5}{32}$, the student is actually calculating what value $n$ is out of $100$ if $5$ is a part of $32$.

Solution:

To find the sales tax rate, we first find the tax amount and then calculate its percentage relative to the original cost.

For Brand X:

For Brand Y:

Comparison:

Since $7.14\% > 4.84\%$, Brand X has a greater sales tax rate.

Final Answer: Brand X has a greater sales tax rate. The rates are $7.14\%$ for Brand X and $4.84\%$ for Brand Y.