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Lady bird [3.3K]
3 years ago
11

A farmer buys a tractor for rs.70000 and after using it for some time, he sells it at a loss of 5% to another farmer. He also se

lls it after an year to the third farmer at a loss of 10%. Find the price paid by the third farmer. ​
Mathematics
2 answers:
Vedmedyk [2.9K]3 years ago
7 0
  • Total loss %=10+5=15%

  • CP=70000
  • loss%=15%

Now

Loss:-

\\ \sf\longmapsto 15\%\:of\:70000

\\ \sf\longmapsto \dfrac{15}{100}(70000)

\\ \sf\longmapsto 10500

Again

\\ \sf\longmapsto SP=CP-Loss

\\ \sf\longmapsto SP=70000-10500

\\ \sf\longmapsto SP=59500

Third farmer will pay 59500

Veronika [31]3 years ago
3 0

Answer:

<h2>The 3rd farmer will pay ₹ 59,500.</h2>

Step-by-step explanation:

C.P of the tractor = ₹70,000

Now,

Percentage of Loss suffered by the farmer after selling it to another farmer = 5%

Percentage of Loss suffered by the 2nd farmer while selling it to the 3rd farmer = 10%

Therefore,

total \: loss \: suffered \:  by \: them = 10\% + 5\%

= 15\%

Therefore,

Amount of loss suffered by both the farmers

= 15\% \: of \: 70000

=   \frac{15}{100}  \times 70000

= 15 \times 700

= 10500

Now,

Therefore,

Price paid by the 3rd farmer to buy the tractor = C.P - loss

= 70000 - 10500

= 59500

So the 3rd farmer will pay ₹59500 to buy the tractor.

Hope it helps you!!

#IndianMurgaツ

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Tree casts a shadow 30 feet long. A MHS student standing near the tree casts a shadow 9 feet long. The  student is 6 feet tall. What is the height of the tree? Show all work

<em><u>Answer:</u></em>

Option D

The height of tree is 20 feet tall

<em><u>Solution:</u></em>

From given question,

Shadow of tree = 30 feet

Height of tree = ?

Height of student = 6 feet

Shadow of student = 9 feet

We have to find the height of tree

We can solve the sum by proportion

\frac{\text{height of tree}}{\text{shadow of tree}} = \frac{\text{height of student}}{\text{shadow of tree}}

This forms a proportion and we can solve the sum by cross multiplying

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Answer:

a)  \displaystyle \frac{dy}{dx} \bigg| \limits_{x = 0} = -1

b)  \displaystyle \frac{dy}{dx} \bigg| \limits_{x = \frac{\pi}{2}} = -1

General Formulas and Concepts:

<u>Pre-Calculus</u>

  • Unit Circle

<u>Calculus</u>

Differentiation

  • Derivatives
  • Derivative Notation

Derivative Property [Multiplied Constant]:                                                           \displaystyle \frac{d}{dx} [cf(x)] = c \cdot f'(x)

Derivative Property [Addition/Subtraction]:                                                         \displaystyle \frac{d}{dx}[f(x) + g(x)] = \frac{d}{dx}[f(x)] + \frac{d}{dx}[g(x)]  

Basic Power Rule:

  1. f(x) = cxⁿ
  2. f’(x) = c·nxⁿ⁻¹  

Derivative Rule [Product Rule]:                                                                             \displaystyle \frac{d}{dx} [f(x)g(x)]=f'(x)g(x) + g'(x)f(x)

Derivative Rule [Quotient Rule]:                                                                           \displaystyle \frac{d}{dx} [\frac{f(x)}{g(x)} ]=\frac{g(x)f'(x)-g'(x)f(x)}{g^2(x)}

Derivative Rule [Chain Rule]:                                                                                 \displaystyle \frac{d}{dx}[f(g(x))] =f'(g(x)) \cdot g'(x)

Trigonometric Differentiation

Logarithmic Differentiation

Step-by-step explanation:

a)

<u>Step 1: Define</u>

<em>Identify</em>

\displaystyle y = ln \bigg( \frac{1 - x}{\sqrt{1 + x^2}} \bigg)

<u>Step 2: Differentiate</u>

  1. Logarithmic Differentiation [Chain Rule]:                                                     \displaystyle \frac{dy}{dx} = \frac{1}{\frac{1 - x}{\sqrt{1 + x^2}}} \cdot \frac{d}{dx}[\frac{1 - x}{\sqrt{1 + x^2}}]
  2. Simplify:                                                                                                         \displaystyle \frac{dy}{dx} = \frac{-\sqrt{x^2 + 1}}{x - 1} \cdot \frac{d}{dx}[\frac{1 - x}{\sqrt{1 + x^2}}]
  3. Quotient Rule:                                                                                               \displaystyle \frac{dy}{dx} = \frac{-\sqrt{x^2 + 1}}{x - 1} \cdot \frac{(1 - x)'\sqrt{1 + x^2} - (1 - x)(\sqrt{1 + x^2})'}{(\sqrt{1 + x^2})^2}
  4. Basic Power Rule [Chain Rule]:                                                                     \displaystyle \frac{dy}{dx} = \frac{-\sqrt{x^2 + 1}}{x - 1} \cdot \frac{-\sqrt{1 + x^2} - (1 - x)(\frac{x}{\sqrt{x^2 + 1}})}{(\sqrt{1 + x^2})^2}
  5. Simplify:                                                                                                         \displaystyle \frac{dy}{dx} = \frac{-\sqrt{x^2 + 1}}{x - 1} \cdot \bigg( \frac{x(x - 1)}{(x^2 + 1)^\bigg{\frac{3}{2}}} - \frac{1}{\sqrt{x^2 + 1}} \bigg)
  6. Simplify:                                                                                                         \displaystyle \frac{dy}{dx} = \frac{x + 1}{(x - 1)(x^2 + 1)}

<u>Step 3: Find</u>

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  2. Evaluate:                                                                                                         \displaystyle \frac{dy}{dx} \bigg| \limits_{x = 0} = -1

b)

<u>Step 1: Define</u>

<em>Identify</em>

\displaystyle y = ln \bigg( \frac{1 + sinx}{1 - cosx} \bigg)

<u>Step 2: Differentiate</u>

  1. Logarithmic Differentiation [Chain Rule]:                                                     \displaystyle \frac{dy}{dx} = \frac{1}{\frac{1 + sinx}{1 - cosx}} \cdot \frac{d}{dx}[\frac{1 + sinx}{1 - cosx}]
  2. Simplify:                                                                                                         \displaystyle \frac{dy}{dx} = \frac{-[cos(x) - 1]}{sin(x) + 1} \cdot \frac{d}{dx}[\frac{1 + sinx}{1 - cosx}]
  3. Quotient Rule:                                                                                               \displaystyle \frac{dy}{dx} = \frac{-[cos(x) - 1]}{sin(x) + 1} \cdot \frac{(1 + sinx)'(1 - cosx) - (1 + sinx)(1 - cosx)'}{(1 - cosx)^2}
  4. Trigonometric Differentiation:                                                                       \displaystyle \frac{dy}{dx} = \frac{-[cos(x) - 1]}{sin(x) + 1} \cdot \frac{cos(x)(1 - cosx) - sin(x)(1 + sinx)}{(1 - cosx)^2}
  5. Simplify:                                                                                                         \displaystyle \frac{dy}{dx} = \frac{-[cos(x) - sin(x) - 1]}{[sin(x) + 1][cos(x) - 1]}

<u>Step 3: Find</u>

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  2. Evaluate [Unit Circle]:                                                                                   \displaystyle \frac{dy}{dx} \bigg| \limit_{x = \frac{\pi}{2}} = -1

Topic: AP Calculus AB/BC (Calculus I/I + II)

Unit: Differentiation

Book: College Calculus 10e

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