Mathematics

Given:

• Principal in fixed deposit, (P = \text{Rs. } 1,50,000)
• Annual interest rate, (R = 10%)
• Time, (T = 2) years
• Compounded semi-annually, so number of periods (n = 2 \times 2 = 4)
• Semi-annual interest rate (= \frac{10%}{2} = 5%)
• Depreciation rate of tractor (= 5%) per year
• Depreciation period (= 2) years

Step 1: Amount received from fixed deposit

For compound interest,

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

Here, semi-annual rate (=5%), so

[ A = 1,50,000\left(1+\frac{5}{100}\right)^4 ]

[ A = 1,50,000(1.05)^4 ]

[ A = 1,50,000 \times 1.21550625 ]

[ A = \text{Rs. } 1,82,325.94 \text{ approximately} ]

So the farmer buys the tractor for about Rs. 1,82,325.94.

Step 2: Value of tractor after 2 years at 5% depreciation

Depreciated value,

[ V = A\left(1-\frac{5}{100}\right)^2 ]

[ V = 1,82,325.94 \times (0.95)^2 ]

[ V = 1,82,325.94 \times 0.9025 ]

[ V = \text{Rs. } 1,64,549.16 \text{ approximately} ]

Answer:

The value of the tractor after the next 2 years will be approximately:

[ \boxed{\text{Rs. } 1,64,549.16} ]

Let the diameter be $d$ and the height be $h$. Take $\pi = \frac{22}{7}$.

Given:

$h + d = 250 \qquad\text{and}\qquad \pi d h = 704$

Step 1: Find $dh$.

$\frac{22}{7}\, dh = 704 \;\Rightarrow\; dh = 704\times\frac{7}{22} = 224$

Step 2: Find $d$ and $h$.

Their sum is 250 and product is 224, so they are roots of

$t^2 - 250t + 224 = 0 \;\Rightarrow\; t = \frac{250 \pm \sqrt{61604}}{2} \approx 249.10 \text{ or } 0.90$

The data are therefore unusual (one dimension is almost a point compared with the other); the numbers in the question appear to be mis-set, but the question can be solved as given.

Step 3: Total surface area.

$\text{TSA} = \text{CSA} + 2\pi r^2 = 704 + \frac{\pi d^2}{2}$

• If $d \approx 0.90$ cm and $h \approx 249.10$ cm (a long thin rod):

$\text{TSA} \approx 704 + \frac{1}{2}\times\frac{22}{7}\times(0.8992)^2 \approx 704 + 1.27 = \mathbf{705.27\ cm^2}$

• If instead $d \approx 249.10$ cm and $h \approx 0.90$ cm (a thin disc), $\text{TSA} \approx 98{,}213\ \text{cm}^2$.

The first case, a tall thin cylinder, is the realistic reading, so the total surface area is about 705.27 cm².

Given:

• Marked price (MP) = Rs 50,000
• Discount = 10%
• VAT = 13%

Solution:

Price after discount: [ = 50{,}000 - 10% \text{ of } 50{,}000 ] [ = 50{,}000 - \frac{10}{100}\times 50{,}000 ] [ = 50{,}000 - 5{,}000 = Rs\ 45{,}000 ]

VAT amount: [ = 13% \text{ of } 45{,}000 ] [ = \frac{13}{100}\times 45{,}000 = Rs\ 5{,}850 ]

Net selling price: [ = 45{,}000 + 5{,}850 = Rs\ 50{,}850 ]

Therefore, the net selling price of the mobile phone is Rs 50,850.

Let the assumed mean (A=25) and class width (h=10).

1. Mean

[ \bar{x}=A+\frac{\sum fd}{N}\times h ]

[ \bar{x}=25+\frac{-4}{74}\times 10 ]

[ \bar{x}=25-0.5405=24.4595 ]

[ \boxed{\bar{x}\approx 24.46\text{ kg}} ]

2. Standard Deviation

[ \sigma=h\sqrt{\frac{\sum fd^2}{N}-\left(\frac{\sum fd}{N}\right)^2} ]

[ \sigma=10\sqrt{\frac{120}{74}-\left(\frac{-4}{74}\right)^2} ]

[ \sigma=10\sqrt{1.6216-0.0029} ]

[ \sigma=10\sqrt{1.6187} ]

[ \sigma=10\times 1.2723=12.723 ]

[ \boxed{\sigma\approx 12.72\text{ kg}} ]

3. Coefficient of Variation

[ CV=\frac{\sigma}{\bar{x}}\times 100% ]

[ CV=\frac{12.723}{24.4595}\times 100% ]

[ CV=52.02% ]

[ \boxed{CV\approx 52.02%} ]

Final Answer:

[ \boxed{\text{Mean}=24.46\text{ kg},\quad \text{S.D.}=12.72\text{ kg},\quad \text{C.V.}=52.02%} ]

Let tap A be turned off after (x) hours.
Tap B runs for a total of (12) hours.

Rates:

• A: (\frac{1}{30}) tank/hour
• B: (\frac{1}{15}) tank/hour

Filled by both for (x) hours + filled by B alone for ((12-x)) hours = 1 tank:

[ x\left(\frac{1}{30}+\frac{1}{15}\right)+(12-x)\frac{1}{15}=1 ]

[ \frac{x}{10}+\frac{12-x}{15}=1 ]

Multiply by 30:

[ 3x+2(12-x)=30 ]

[ 3x+24-2x=30 ]

[ x=6 ]

So, tap A was turned off after 6 hours.

Given:

• Marked price (MP) = Rs. 75,000
• Discount = 15%
• VAT = 15%

Selling price after discount: [ \text{Discounted price} = 75{,}000 \times \left(1-\frac{15}{100}\right) ] [ = 75{,}000 \times 0.85 = Rs.\ 63{,}750 ]

Amount of VAT: [ \text{VAT} = 63{,}750 \times \frac{15}{100} = Rs.\ 9{,}562.50 ]

Total amount paid by customer: [ = 63{,}750 + 9{,}562.50 = Rs.\ 73{,}312.50 ]

Answer: The customer has to pay Rs. 73,312.50.

Let the borrowed (and invested) sum be Rs $P$.

Simple interest for 2 years at 5%:

$SI = \frac{P \times 5 \times 2}{100} = \frac{P}{10}$

Compound interest for 2 years at 5%:

$CI = P\left[\left(1+\frac{5}{100}\right)^2 - 1\right] = P\left[(1.05)^2 - 1\right] = 0.1025P$

Given gain:

$CI - SI = 20$

$0.1025P - 0.1P = 20$

$0.0025P = 20$

$P = \frac{20}{0.0025} = 8000$

Hence, the amount borrowed was Rs 8,000.

Let the assumed mean be (A=10).

[ \bar x=A+\frac{\sum fd}{N}=10+\frac{30}{300}=10.1 ]

[ \sigma=\sqrt{\frac{\sum fd^2}{N}-\left(\frac{\sum fd}{N}\right)^2} ]

[ \sigma=\sqrt{\frac{1780}{300}-\left(\frac{30}{300}\right)^2} ]

[ \sigma=\sqrt{5.9333-0.01}=\sqrt{5.9233}=2.4338 ]

[ \boxed{\sigma \approx 2.43} ]

So, the standard deviation is approximately 2.43.