Comparing Quantities — NCERT Solutions, AI Tutor & Practice

Number of girls $= 24$, so number of boys $= 40 - 24 = 16$.

Ratio of girls to boys: $24 : 16 = \frac{24}{16} = \frac{3}{2} = 3 : 2$

Girls as a percentage of the class: $\frac{24}{40} \times 100 = 60\%$

So the ratio is $3:2$ and girls form $60\%$ of the class.

Marked Price (MP) $= ₹1500$, Discount $= 20\%$.

Discount amount: $= 20\% \text{ of } 1500 = \frac{20}{100} \times 1500 = ₹300$

Selling Price (SP): $= \text{MP} - \text{Discount} = 1500 - 300 = ₹1200$

The selling price is ₹1200.

Original price $= ₹800$, New price $= ₹920$.

Increase in price: $= 920 - 800 = ₹120$

Percentage increase: $= \frac{\text{Increase}}{\text{Original price}} \times 100 = \frac{120}{800} \times 100 = 15\%$

The price increased by 15%.

Cost Price (CP) $= ₹2400$, Selling Price (SP) $= ₹2700$.

Profit: $= \text{SP} - \text{CP} = 2700 - 2400 = ₹300$

Profit per cent: $= \frac{\text{Profit}}{\text{CP}} \times 100 = \frac{300}{2400} \times 100 = 12.5\%$

Profit is ₹300 and profit per cent is 12.5%.

Marked Price (MP) $= ₹700$, Selling Price (SP) $= ₹560$.

Discount: $= \text{MP} - \text{SP} = 700 - 560 = ₹140$

Discount per cent: $= \frac{\text{Discount}}{\text{MP}} \times 100 = \frac{140}{700} \times 100 = 20\%$

The discount is 20%.

Price before tax $= ₹1200$, GST $= 5\%$.

GST amount: $= 5\% \text{ of } 1200 = \frac{5}{100} \times 1200 = ₹60$

Total amount paid: $= 1200 + 60 = ₹1260$

The customer pays ₹1260.

Principal $P = ₹5000$, Rate $R = 8\%$ per annum, Time $T = 2$ years.

Simple Interest formula: $\text{SI} = \frac{P \times R \times T}{100}$

Substituting: $= \frac{5000 \times 8 \times 2}{100} = \frac{80000}{100} = ₹800$

The simple interest is ₹800.

Principal $P = ₹8000$, Rate $R = 10\%$, Time $n = 2$ years.

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

Substituting: $A = 8000\left(1 + \frac{10}{100}\right)^2 = 8000 \times \left(\frac{11}{10}\right)^2$ $= 8000 \times \frac{121}{100} = ₹9680$

Compound Interest: $= A - P = 9680 - 8000 = ₹1680$

The compound interest is ₹1680.

Present population $P = 50000$, Rate $R = 4\%$, Time $n = 2$ years.

Formula (like compound growth): $P_n = P\left(1 + \frac{R}{100}\right)^n$

Substituting: $= 50000\left(1 + \frac{4}{100}\right)^2 = 50000 \times \left(\frac{26}{25}\right)^2$ $= 50000 \times \frac{676}{625} = 54080$

The population after 2 years will be 54,080.

Principal $P = ₹6250$, Amount $A = ₹7290$, Time $n = 2$ years.

Using the formula: $A = P\left(1 + \frac{R}{100}\right)^2$ $7290 = 6250\left(1 + \frac{R}{100}\right)^2$

Divide both sides by 6250: $\left(1 + \frac{R}{100}\right)^2 = \frac{7290}{6250} = \frac{729}{625}$

Take the square root: $1 + \frac{R}{100} = \frac{27}{25}$ $\frac{R}{100} = \frac{27}{25} - 1 = \frac{2}{25}$ $R = \frac{2}{25} \times 100 = 8\%$

The rate of interest is 8% per annum.

Present value $P = ₹40000$, Rate of depreciation $R = 10\%$, Time $n = 2$ years.

For depreciation, use a minus sign: $\text{Value} = P\left(1 - \frac{R}{100}\right)^n$

Substituting: $= 40000\left(1 - \frac{10}{100}\right)^2 = 40000 \times \left(\frac{9}{10}\right)^2$ $= 40000 \times \frac{81}{100} = ₹32400$

The value of the machine after 2 years is ₹32,400.

Principal $P = ₹10000$, Annual rate $= 8\%$, Time $= 1$ year.

Since interest is compounded half-yearly:

• Rate per half-year $= \frac{8}{2} = 4\%$
• Number of periods $n = 1 \times 2 = 2$

Amount: $A = P\left(1 + \frac{4}{100}\right)^2 = 10000 \times \left(\frac{26}{25}\right)^2$ $= 10000 \times \frac{676}{625} = ₹10816$

Compound Interest: $= A - P = 10816 - 10000 = ₹816$

The compound interest is ₹816.