Maharashtra Board Class 11 Maths Part 2 Chapter 9 Commercial Mathematics Miscellaneous Solutions

Question 1. A man buys a house for Rs. 10 lakh and rents it. He puts 10% of the annual rent aside for repairs, pays Rs. 1,000 as annual taxes, and realizes 8% on his investment thereafter. Find the annual rent of the house.
Answer:
Let the annual rent of the house be \( R \).
Cost of the house (Investment) = Rs. 10,00,000
Amount set aside for repairs = 10% of \( R = 0.1R \)
Annual taxes paid = Rs. 1,000
Net income realized = Annual Rent \( - \) Repairs \( - \) Taxes
Net income = \( R - 0.1R - 1,000 = 0.9R - 1,000 \)
The man realizes an 8% return on his investment.
Realized return = 8% of Rs. 10,00,000
\( \implies \) Realized return = \( \frac{8}{100} \times 10,00,000 = 80,000 \)
According to the given condition:
Net income = Realized return
\( \implies 0.9R - 1,000 = 80,000 \)
\( \implies 0.9R = 80,000 + 1,000 \)
\( \implies 0.9R = 81,000 \)
\( \implies R = \frac{81,000}{0.9} \)
\( \implies R = 90,000 \)
Therefore, the annual rent of the house is Rs. 90,000. This calculation helps property owners determine the minimum rent required to achieve their desired rate of return.
In simple words: The man wants to make an 8% profit on his Rs. 10 lakh investment, which is Rs. 80,000. After paying for repairs and taxes from the rent, the remaining amount must equal this profit. By calculating backward, we find he needs to charge Rs. 90,000 in rent each year.

🎯 Exam Tip: Clearly define the variable for annual rent at the beginning and show step-by-step algebraic simplification to secure full marks.

Question 1. Find the annual rent of the house if a man keeps 10% of the annual rent aside for repairs, pays Rs. 1000 as annual taxes, and is left with an amount which is 8% of his Rs. 10,00,000 investment for the house.
Answer:
Let Rs. 'x' be the annual rent of the house.
The man keeps 10% of the annual rent aside for repairs.
i.e., \( \frac{10}{100} \times x \) or Rs. \( \frac{x}{10} \) aside for repairs.
In addition, he pays Rs. 1000 as annual taxes.
After incurring these expenses he is left with an amount which is 8% of his investment for the house.
i.e., \( \frac{8}{100} \times 10,00,000 \)
i.e., Rs. 80,000
\( \implies x - \left(\frac{x}{10} + 1000\right) = 80,000 \)
\( \implies x - \frac{x}{10} = 81,000 \)
\( \implies \frac{10x - x}{10} = 81,000 \)
\( \implies 9x = 8,10,000 \)
\( \implies x = 90,000 \)
\( \implies \) The annual rent of the house is Rs. 90,000.
In simple words: We find the annual rent by subtracting the repair costs and taxes from the total rent, and setting it equal to the remaining savings which is 8% of the investment.

🎯 Exam Tip: Always convert percentages to fractions carefully and keep track of all expenses like taxes and repairs to avoid calculation errors.

Question 2. Rose got 30% of the maximum marks in an examination and failed by 10 marks. However, Lily who appeared for the same examination got 40% of the total marks and got 15 marks more than the passing marks. What were the passing marks in the examination?
Answer:
Let maximum marks be x
Rose scored 30% of maximum marks
i.e. Rose scored \( \frac{30}{100}x \)
Rose failed by 10 marks
\( \implies \) passing marks = \( \frac{30}{100}x + 10 \) ……(i)
Lily scored 40% of maximum marks
i.e. Lily scored \( \frac{40}{100}x \)
Lily scored 15 marks more than passing marks
\( \implies \) passing marks = \( \frac{40}{100}x - 15 \) ……(ii)
equating (i) and (ii),
\( \frac{30x}{100} + 10 = \frac{40x}{100} - 15 \)
\( \implies 10 + 15 = \frac{40x - 30x}{100} \)
\( \implies 10x = (25)(100) \)
\( \implies x = 250 \)
Substituting the value of \( x \) back into equation (i) gives the passing marks as \( \frac{30}{100}(250) + 10 = 75 + 10 = 85 \) marks.
In simple words: By setting up equations for both students' marks relative to the passing mark, we find the total marks of the exam first, and then use it to calculate the exact passing score.

🎯 Exam Tip: Remember to read the final question carefully; here, finding x (maximum marks) is just the first step, and you must substitute it back to find the actual passing marks.

Question 3. Ankita’s Salary was reduced by 50%. Again the reduced salary was increased by 50%. Find loss in terms of percentage.
Answer:
Let Ankita’s initial salary be Rs. \( x \).
Her salary was reduced by 50%.

\( \implies \) Ankita’s salary after reduction = \( x \left(1 - \frac{50}{100}\right) \)
= \( x \left(1 - \frac{1}{2}\right) \)
= \( \frac{x}{2} \)
Ankita’s reduced salary was then increased by 50%.

\( \implies \) Ankita’s final salary after the increase = \( \frac{x}{2} \left(1 + \frac{50}{100}\right) \)
= \( \frac{x}{2} \left(1 + \frac{1}{2}\right) \)
= \( \frac{x}{2} \times \frac{3}{2} \)
= \( \frac{3x}{4} \)

\( \implies \) Loss in Ankita’s salary after the decrease and increase = \( x - \frac{3x}{4} = \frac{x}{4} \)
Percentage loss = \( \frac{\text{Loss in salary}}{\text{Initial salary}} \times 100 \)
= \( \frac{\frac{x}{4}}{x} \times 100 \)
= \( \frac{1}{4} \times 100 \)
= 25%
Therefore, Ankita lost 25% of her salary.
In simple words: When a value is decreased by 50% and then increased by 50%, it does not return to the original amount because the 50% increase is calculated on the new, smaller value. This results in an overall loss of 25%.

🎯 Exam Tip: Remember that percentage increases and decreases are not simply additive; always calculate them sequentially on the updated values to avoid common calculation errors.

Question 4. By selling 300 lunch boxes, a shopkeeper gains the selling price of 100 lunch boxes. Find his gain percent.
Answer:
Let Rs. \( x \) be the selling price (S.P.) of one lunch box.
Selling price of 300 lunch boxes = Rs. \( 300x \)
Gain = Selling price of 100 lunch boxes = Rs. \( 100x \)
We know that, Cost Price (C.P.) = Selling Price (S.P.) - Gain

\( \implies \) C.P. of 300 lunch boxes = Rs. \( 300x - 100x \) = Rs. \( 200x \)
Gain Percentage = \( \frac{\text{Gain}}{\text{C.P.}} \times 100 \)
= \( \frac{100x}{200x} \times 100 \)
= \( \frac{1}{2} \times 100 \)
= 50%
Therefore, his gain percent is 50%.
In simple words: The shopkeeper made a profit equal to the value of 100 boxes while selling 300 boxes, meaning the actual cost of making those boxes was equivalent to 200 boxes. Since 100 is half of 200, his profit margin is exactly 50%.

🎯 Exam Tip: Express both Cost Price and Gain in terms of the variable \( x \) to easily cancel it out when calculating the final percentage.

Question 5. A salesman sold an article at a loss of 10%. If the selling price has been increased by Rs. 80, there would have been a gain of 10%. What was the cost of the article?
Answer:
Let Rs. \( x \) be the cost price of the article.
S.P. of the article = \( x - \frac{10}{100}x = \frac{9x}{10} \) ......(i)
Given that, S.P. increased by Rs. 80 would have given 10% gain.
i.e. S.P. + 80 = \( x + \frac{10}{100}x \)
\( \implies \) \( \frac{9x}{10} + 80 = \frac{11x}{10} \) ......[From (i)]
\( \implies \) \( \frac{(11 - 9)x}{10} = 80 \)
\( \implies \) \( x = \frac{80 \times 10}{2} = 40 \times 10 = 400 \)
∴ The cost price of the article is Rs. 400. This calculation clearly shows how the price change directly relates to the percentage shift.
In simple words: If you sell something for 10% less than it cost, and then raise the price by Rs. 80 to make a 10% profit, that Rs. 80 difference represents exactly 20% of the original cost. Since 20% of the cost is Rs. 80, the total cost must be Rs. 400.

🎯 Exam Tip: Clearly define the variable \( x \) for the cost price and show both the loss and gain equations step-by-step to avoid calculation errors.

Question 6. Find the single discount equivalent to a series discount of 10%, 20%, and 15%.
Answer:
Let the marked price be Rs. 100.
After 1st discount the price = \( 100 \left(1 - \frac{10}{100}\right) = 90 \)
After 2nd discount the price = \( 90 \left(1 - \frac{20}{100}\right) = 72 \)
After 3rd discount the price = \( 72 \left(1 - \frac{15}{100}\right) = 61.2 \)
∴ The selling price after 3 discounts is Rs. 61.2.
∴ Single equivalent discount = marked price - selling price
= 100 - 61.2
= Rs. 38.8. This means a single discount of 38.8% is identical to the three successive discounts combined.
In simple words: Instead of taking 10%, 20%, and then 15% off one after another, giving a single discount of 38.8% on the original price will result in the exact same final price.

🎯 Exam Tip: Always assume the initial marked price is Rs. 100 for percentage discount problems, as it makes successive calculations much simpler and less prone to errors.

Question 7. Reshma put an amount at simple interest at a certain rate for 3 years. Had it been put at 2% higher rate, she would have received Rs. 360 more. Find the sum.
Answer: Let P and R represent the principal amount and rate of interest p.a. respectively.
Given duration (T) = 3 years
Simple Interest = \( \frac{\text{PRT}}{100} = \frac{3\text{PR}}{100} \)
Given that, had the amount been kept at 2% higher rate, then the gain would have been Rs. 360 more.
i.e. \( \frac{\text{P}(\text{R} + 2)(3)}{100} = \frac{3\text{PR}}{100} + 360 \)
\( \implies \frac{3\text{PR} + 6\text{P}}{100} = \frac{3\text{PR}}{100} + 360 \)
\( \implies \frac{3\text{PR}}{100} + \frac{6\text{P}}{100} = \frac{3\text{PR}}{100} + 360 \)
\( \implies \frac{6\text{P}}{100} = 360 \)
\( \implies \text{P} = \frac{360 \times 100}{6} \)
\( \implies \text{P} = 6000 \)
Therefore, the sum of money is Rs. 6,000. This shows how a small change in the interest rate can significantly impact the total interest earned over time.
In simple words: If the interest rate goes up by 2% each year for 3 years, the total interest increases by 6% of the principal. Since this 6% equals Rs. 360, the starting principal amount must be Rs. 6,000.

🎯 Exam Tip: Always write down the formula for simple interest clearly before substituting the values to avoid calculation errors.

Question 8. The compound interest on Rs. 30000 at 7% p.a. is Rs. 4347. What is the period in years?
Answer: Given that,
Principal (P) = Rs. 30,000
Rate of interest (R) = 7% p.a.
Compound interest (CI) = Rs. 4,347
Amount (A) = Principal (P) + Compound Interest (CI)
\( \implies \text{A} = \text{Rs. } 30,000 + \text{Rs. } 4,347 = \text{Rs. } 34,347 \)
We know that,
\( \text{A} = \text{P}\left(1 + \frac{\text{R}}{100}\right)^\text{n} \)
\( \implies 34,347 = 30,000\left(1 + \frac{7}{100}\right)^\text{n} \)
\( \implies \frac{34,347}{30,000} = \left(\frac{107}{100}\right)^\text{n} \)
\( \implies \frac{11,449}{10,000} = (1.07)^\text{n} \)
\( \implies (1.07)^2 = (1.07)^\text{n} \)
\( \implies \text{n} = 2 \)
Therefore, the period is 2 years. Calculating the time period helps us understand how long an investment takes to grow to a specific target amount.
In simple words: We find the total amount by adding the principal and the interest, then use the compound interest formula to see how many years it takes for Rs. 30,000 to grow to Rs. 34,347 at 7% interest.

🎯 Exam Tip: When solving for the exponent 'n', simplify the fraction on the left side to see if it can be expressed as a power of the base on the right side.

Question 9. The value of the machine depreciates at the rate of 15% p.a. It was purchased 2 years ago. Its present value is Rs. 7,225. What was the purchase price of the machine?
Answer:
Given,
Rate of depreciation \( (r) = 15\% \) p.a.
Number of years \( (n) = 2 \) years
Present value of machine \( (\text{P.V.}) = \text{Rs. } 7,225 \)
Let the purchase price of the machine be \( V \).
The purchase price of the machine can be found using the depreciation formula:
\( \text{P.V.} = V \left(1 - \frac{r}{100}\right)^n \)
Substituting the given values into the formula:
\( 7225 = V \left(1 - \frac{15}{100}\right)^2 \)
\( \implies 7225 = V \left(\frac{85}{100}\right)^2 \)
\( \implies 7225 = V (0.85)^2 \)
\( \implies 7225 = V \times 0.7225 \)
\( \implies V = \frac{7225}{0.7225} \)
\( \implies V = 10,000 \)
Thus, the original purchase price of the machine was Rs. 10,000.
In simple words: Since the machine loses 15% of its value every year, we work backward from its current value of Rs. 7,225 over 2 years to find that it originally cost Rs. 10,000.

🎯 Exam Tip: Always use the subtraction sign in the formula for depreciation problems, as the value of the asset decreases over time.

Question 10. A tree increases annually by \( \frac{1}{8} \) of its height. By how much will it increase after \( 2\frac{1}{2} \) years if its length today is 8 m?
Answer:
Given:
Height of the tree today \( = 8\text{ m} \)
Annual growth rate \( = \frac{1}{8} \) of its height.

At the end of the 1st year, the height of the tree will be:
\( = 8 + \left(\frac{1}{8} \times 8\right) = 8 + 1 = 9\text{ m} \)

At the end of the 2nd year, the height of the tree will be:
\( = 9 + \left(\frac{1}{8} \times 9\right) \)
\( = 9\left(1 + \frac{1}{8}\right) \)
\( = 9 \times \frac{9}{8} \)
\( = \frac{81}{8}\text{ m} \)

For the remaining half year (6 months), the growth rate will be half of the annual rate, which is \( \frac{1}{2} \times \frac{1}{8} = \frac{1}{16} \).
After six more months, the height of the tree will be:
\( = \frac{81}{8} + \frac{1}{16}\left(\frac{81}{8}\right) \)
\( = \frac{81}{8}\left(1 + \frac{1}{16}\right) \)
\( = \frac{81}{8} \times \frac{17}{16} \)
\( = \frac{1377}{128} \approx 10.75\text{ m} \)

Therefore, the increase in the height of the tree after \( 2\frac{1}{2} \) years is:
\( = 10.75 - 8 = 2.75\text{ m} \). This steady growth demonstrates how compound increases accumulate significantly over time.
In simple words: Every year, the tree grows by one-eighth of its current height. After two and a half years of compounding this growth, the tree's height reaches 10.75 meters, meaning it has grown by 2.75 meters.

🎯 Exam Tip: For fractional years like \( 2\frac{1}{2} \) years, calculate the compound growth for the whole years first, and then apply half the growth rate for the remaining 6 months.

Question 11. A building worth Rs. 1,21,000 is constructed on land worth Rs. 81,000. After how many years will the value of both be the same if land appreciates at 10% p.a. and buildings depreciate at 10% p.a.?
Answer:
Let the value of both land and building become equal after \( n \) years.

For Land (Appreciation):
Present Value (\( P \)) \( = \text{Rs. } 81,000 \)
Rate of appreciation (\( r \)) \( = 10\% \text{ p.a.} \)
Value of land after \( n \) years (\( A_1 \)) \( = P\left(1 + \frac{r}{100}\right)^n \)
\( = 81000\left(1 + \frac{10}{100}\right)^n \)
\( = 81000\left(\frac{11}{10}\right)^n \)

For Building (Depreciation):
Present Value (\( P \)) \( = \text{Rs. } 1,21,000 \)
Rate of depreciation (\( r \)) \( = 10\% \text{ p.a.} \)
Value of building after \( n \) years (\( A_2 \)) \( = P\left(1 - \frac{r}{100}\right)^n \)
\( = 121000\left(1 - \frac{10}{100}\right)^n \)
\( = 121000\left(\frac{9}{10}\right)^n \)

According to the given condition, the values will be equal after \( n \) years:
\( A_1 = A_2 \)

\( \implies 81000\left(\frac{11}{10}\right)^n = 121000\left(\frac{9}{10}\right)^n \)

\( \implies \frac{\left(\frac{11}{10}\right)^n}{\left(\frac{9}{10}\right)^n} = \frac{121000}{81000} \)

\( \implies \left(\frac{11}{9}\right)^n = \frac{121}{81} \)

\( \implies \left(\frac{11}{9}\right)^n = \left(\frac{11}{9}\right)^2 \)

Comparing the powers on both sides, we get:
\( n = 2 \) years.
Thus, the value of both the land and the building will be the same after 2 years. This shows how appreciation and depreciation balance each other over time.
In simple words: Since the land value is increasing by 10% each year and the building value is decreasing by 10% each year, their values will meet and become exactly equal after 2 years.

🎯 Exam Tip: Remember to use the plus sign \( (+) \) for appreciation and the minus sign \( (-) \) for depreciation in the compound interest formula to avoid calculation errors.

Question 11. The present value of a building is Rs. 1,21,000 and that of land is Rs. 81,000. If the value of the land appreciates at 10% per annum and that of the building depreciates at 10% per annum, after how many years will both have the same value?
Answer: Given,
Value of the building = V.B. = Rs. 1,21,000
Value of land = V.L. = Rs. 81,000
Rate of appreciation of land = rate of depreciation of building = \( r = 10\% \).
For the value of building and land to be the same:
\( \text{V.L.} \left(1 + \frac{r}{100}\right)^n = \text{V.B.} \left(1 - \frac{r}{100}\right)^n \)
\( \implies 81,000 \left(1 + \frac{10}{100}\right)^n = 1,21,000 \left(1 - \frac{10}{100}\right)^n \)
\( \implies 81,000 \left(\frac{11}{10}\right)^n = 1,21,000 \left(\frac{9}{10}\right)^n \)
\( \implies \frac{11^n}{10^n} \times \frac{10^n}{9^n} = \frac{1,21,000}{81,000} \)
\( \implies \frac{11^n}{9^n} = \frac{121}{81} \)
\( \implies \left(\frac{11}{9}\right)^n = \left(\frac{11}{9}\right)^2 \)
\( \implies n = 2 \) years.
Therefore, after two years, the value of the building and land will be the same. This calculation assumes that the rates of appreciation and depreciation remain constant throughout the period.
In simple words: The value of the land is going up by 10% each year, while the value of the building is going down by 10% each year. By using the compound interest formula for growth and decay, we find that their values will become exactly equal in 2 years.

🎯 Exam Tip: When solving appreciation and depreciation problems together, set up the equations on both sides and simplify the powers of 10 first to make calculations much faster.

Question 12. Varun invested 25%, 30%, and 20% of his savings in buying shares of three different companies, ‘A’, ‘B’, and ‘C’ which declared dividends, 10%, 12%, and 15% respectively. If his total income on account of dividends is Rs. 6,370/-, find the amount he invested in buying shares of company ‘B’.
Answer: Let \( T \) be Varun’s total savings.
Investment of Varun in:
Company A = \( 25\% \text{ of } T = \frac{25}{100} \times T = \frac{T}{4} \)
Company B = \( 30\% \text{ of } T = \frac{30}{100} \times T = \frac{3T}{10} \)
Company C = \( 20\% \text{ of } T = \frac{20}{100} \times T = \frac{T}{5} \)
Company A, B and C declared dividends 10%, 12% and 15% respectively.
Dividend from company A = \( 10\% \text{ of } \frac{T}{4} = \frac{10}{100} \times \frac{T}{4} = \frac{T}{40} \)
Dividend from company B = \( 12\% \text{ of } \frac{3T}{10} = \frac{12}{100} \times \frac{3T}{10} = \frac{36T}{1000} \)
Dividend from company C = \( 15\% \text{ of } \frac{T}{5} = \frac{15}{100} \times \frac{T}{5} = \frac{3T}{100} \)
Since his total income on account of dividends is Rs. 6,370, we can add these individual dividends together.
\( \implies \frac{T}{40} + \frac{36T}{1000} + \frac{3T}{100} = 6370 \)
\( \implies \frac{25T + 36T + 30T}{1000} = 6370 \)
\( \implies \frac{91T}{1000} = 6370 \)
\( \implies 91T = 6370 \times 1000 \)
\( \implies T = \frac{6370000}{91} \)
\( \implies T = 70,000 \)
So, Varun's total savings is Rs. 70,000.
Amount invested in buying shares of company B:
\( \implies \text{Investment in B} = \frac{3T}{10} = \frac{3 \times 70,000}{10} = 21,000 \)
Therefore, the amount invested in buying shares of company B is Rs. 21,000.
In simple words: We represent Varun's total savings as a variable. By calculating the dividend earned from each company as a percentage of his investment, we set up an equation equal to his total dividend income to find his total savings, and then find his specific investment in company B.

🎯 Exam Tip: Always define a single variable for total savings first, express each investment and dividend in terms of that variable, and double-check your fraction additions using a common denominator.

Question 13. Find the annual dividend received from Rs. 25,000, 8% stock at Rs. 108.
Answer: Given:
Amount invested = Rs. 25,000
Dividend rate = 8%
Market Value (M.V.) of share = Rs. 108
Assuming Face Value (F.V.) of each share = Rs. 100

Annual income (dividend) per share = \( \frac{\text{Dividend Rate}}{100} \times \text{F.V.} \)
\( = \frac{8}{100} \times 100 \)
\( = \text{Rs. } 8 \)

Number of shares purchased = \( \frac{\text{Amount Invested}}{\text{M.V.}} \)
\( = \frac{25000}{108} \)

Total annual dividend received = \( \text{Number of shares} \times \text{Annual income per share} \)
\( = \frac{25000}{108} \times 8 \)
\( = \frac{200000}{108} \)
\( = \text{Rs. } 1851.85 \)

Additionally, the rate of return on this investment can be calculated to show the efficiency of the yield.
Rate of return = \( \frac{\text{Annual income per share}}{\text{Market value}} \times 100 \)

\( \implies \text{Rate of return} = \frac{8}{108} \times 100 \)

\( \implies \text{Rate of return} = 7.4\% \)

Therefore, the annual dividend received is Rs. 1851.85.
In simple words: To find the total dividend, we first find how many shares were bought by dividing the total money by the market price of one share. Then, we multiply this number of shares by the dividend earned on a single share.

🎯 Exam Tip: Always state your assumption of the Face Value (usually Rs. 100) clearly before calculating the dividend per share to avoid losing marks.

Question 14. A, B, and C enter into a partnership. A invests 3 times as much as B invests and B invests two-thirds of what ‘C’ invests. At the end of the year, the profit earned is Rs. 8,800. What is the share of ‘B’?
Answer:
Let ‘a’, ‘b’ and ‘c’ be the amounts invested by A, B and C respectively.
Given that, A invests 3 times as much as B and B invests two-thirds of what ‘C’ invests.
\( \therefore a = 3b \) and \( b = \frac{2}{3}c \)
\( \therefore \frac{a}{b} = \frac{3}{1} \) and \( \frac{b}{c} = \frac{2}{3} \)
or \( \frac{a}{b} = \frac{6}{2} \) and \( \frac{b}{c} = \frac{2}{3} \)
\( \therefore a : b = 6 : 2 \) and \( b : c = 2 : 3 \)
\( \dots a : b : c = 6 : 2 : 3 \)
This ratio helps us divide the total profit proportionally among the partners.
Given that profit earned = Rs. 8,800
\( \therefore \) Share of ‘B’ in profit = \( \frac{2}{11} \times 8800 = \text{Rs. } 1600 \)
\( \dots \) B's share in profit is Rs. 1,600.
In simple words: To find B's share, we first find the ratio of investments of A, B, and C, which is 6:2:3. Then, we divide the total profit of Rs. 8,800 according to B's share of 2 parts out of 11 total parts.

🎯 Exam Tip: Always convert individual ratios into a combined ratio (like a:b:c) by making the common term (b) equal before dividing the profit.

Question 15. The ratio of investment of two partners Santa and Banta is 11 : 12 and the ratio of their profits is 2 : 3. If Santa invested the money for 8 months, then for how much time did Banta his money?
Answer:
Let ‘x’ be the time in months for which Banta invested his money.
Santa and Banta invested their money in the ratio 11 : 12.
Santa invested his money for 8 months and the ratio of their profits is 2 : 3.
This calculation shows how investment duration directly affects the distribution of profits.
\( \therefore 11 \times 8 : 12 \times x = 2 : 3 \)
\( \therefore \frac{88}{12x} = \frac{2}{3} \)
\( \implies 88 \times 3 = 2 \times 12x \)
\( \implies 264 = 24x \)
\( \implies x = \frac{264}{24} \)
\( \implies x = 11 \)
\( \therefore \) Banta invested his money for 11 months.
In simple words: Profit is shared based on both the amount invested and the time it was invested for. By setting up a ratio of their total investment-months, we can easily solve for Banta's investment time, which is 11 months.

🎯 Exam Tip: Remember that Profit Ratio = (Investment of A × Time of A) : (Investment of B × Time of B). Use this formula directly to solve for any missing variable.

Question 16. Akash, Sameer, and Sid took a house on rent for one year for Rs. 16,236. They stayed together for 4 months and then Sid left the house. After 5 more months, Sameer also left the house. How much rent should each pay?
Answer:
Let 'R' be the rent per month to be paid to the landlord.
Given that, Sid left the house after 4 months.
During these 4 months, all three stayed together, so they shared the rent equally.

\( \implies \) Rent paid by Sid = \( \frac{R}{3} \times 4 = \frac{4R}{3} \)
Sameer left the house after another 5 months (totaling 9 months of stay). For the first 4 months, he shared with three people, and for the next 5 months, he shared with only Akash.

\( \implies \) Rent paid by Sameer = \( \frac{R}{2} \times 5 + \frac{R}{3} \times 4 \)

\( \implies R\left(\frac{5}{2} + \frac{4}{3}\right) \)

\( \implies \frac{23R}{6} \)
Akash stayed in the house for the entire year (12 months). He shared with three people for 4 months, with one person for 5 months, and stayed alone for the remaining 3 months.

\( \implies \) Rent paid by Akash = \( 3R + \frac{R}{2} \times 5 + \frac{R}{3} \times 4 \)

\( \implies R\left(3 + \frac{5}{2} + \frac{4}{3}\right) \)

\( \implies \frac{41R}{6} \)
The rent paid by the three of them over the period of one year must be in proportion to their usage:
\( \frac{41R}{6} : \frac{23R}{6} : \frac{4R}{3} \)
Multiplying throughout by \( \frac{6}{R} \), we get the ratio:
\( 41 : 23 : 8 \)
Let \( x \) be the constant of proportionality.
Rent to be paid by Akash = Rs. \( 41x \)
Rent to be paid by Sameer = Rs. \( 23x \)
Rent to be paid by Sid = Rs. \( 8x \)
The total rent for the house was Rs. 16,236.

\( \implies 41x + 23x + 8x = 16236 \)

\( \implies 72x = 16236 \)

\( \implies x = 225.5 \)
This proportional division ensures that each person pays exactly for the duration and sharing conditions of their stay.

\( \implies \) Akash should pay \( 41 \times 225.5 = \text{Rs. } 9245.5 \)

\( \implies \) Sameer should pay \( 23 \times 225.5 = \text{Rs. } 5186.5 \)

\( \implies \) Sid should pay \( 8 \times 225.5 = \text{Rs. } 1804 \)
In simple words: Since the three friends lived in the house for different lengths of time, the total rent is divided based on how long each person stayed and how many people shared the house during those months. Sid pays the least because he left first, while Akash pays the most because he stayed the whole year.

🎯 Exam Tip: When solving partnership or shared expense problems, always calculate the individual share ratio based on the time period each person utilized the resource to avoid calculation errors.

Question 17. Ashwin Auto Automobiles sold 10 motorcycles. Total sales amount was Rs. 6,80,000. 18% GST is applicable. Calculate how much CGST and SGST the firm has to pay.
Answer: Given, total sales amount for Ashwin Automobiles was Rs. 6,80,000. 18% GST is applicable.
\( \therefore \) GST payable = 18% of 6,80,000
\( = \frac{18}{100} \times 6,80,000 \)
\( = \text{Rs. } 1,22,400 \)
Now, CGST = SGST = 9%
\( = \frac{\text{GST payable}}{2} \)
\( = \frac{1,22,400}{2} \)
\( = \text{Rs. } 61,200 \)
\( \therefore \) CGST = SGST = Rs. 61,200. This calculation ensures that the tax burden is shared equally between the central and state governments.
In simple words: The total tax (GST) is 18% of the sales. Since CGST and SGST are always equal halves of the total GST, we just divide the total tax of Rs. 1,22,400 by 2 to get Rs. 61,200 each.

🎯 Exam Tip: Remember that CGST and SGST are always equal and each represents exactly half of the total GST rate. Clearly showing this division step helps secure full marks.

Question 18. ‘Sweet 16’ A ready made garments shop for Women’s garments, purchased stock for Rs. 4,00,000 and sold that stock for Rs. 5,50,000 (12% GST is applicable) Find,
(i) Input Tax Credit
(ii) CGST and SGST paid by the firm.
Answer: Given that, stock purchased by ‘Sweet 16’ was worth Rs. 4,00,000. GST applicable is 12%.
(i) Input tax = 12% of 4,00,000
\( = \frac{12}{100} \times 4,00,000 \)
\( = \text{Rs. } 48,000 \)
\( \therefore \) Input Tax Credit (ITC) = Rs. 48,000. This mechanism prevents double taxation on the same goods as they pass through the supply chain.
(ii) The garment stock was sold for Rs. 5,50,000.
Output tax = 12% of 5,50,000
\( = \frac{12}{100} \times 5,50,000 \)
\( = \text{Rs. } 66,000 \)
\( \dots \) GST payable = Output tax \( - \) ITC
\( = 66,000 - 48,000 \)
\( = \text{Rs. } 18,000 \)
\( \therefore \) CGST = SGST = \( \frac{\text{GST payable}}{2} = \frac{18,000}{2} = \text{Rs. } 9,000 \)
In simple words: Input Tax Credit (ITC) is the tax paid on purchases, which is Rs. 48,000. The shopkeeper collects Rs. 66,000 tax on sales (Output Tax). The actual GST paid to the government is the difference between these two, which is Rs. 18,000, split equally into Rs. 9,000 each for CGST and SGST.

🎯 Exam Tip: Always clearly define Input Tax Credit (ITC) as the tax paid on purchases. When calculating the final GST payable, remember the formula: GST Payable = Output Tax - ITC.