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# Tutor profile: Rishi Tarkesh G.

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Rishi Tarkesh G.
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## Questions

### Subject:Trigonometry

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Question:

1. Two ships are sailing in the sea on the two sides of a lighthouse. The angle of elevation of the top of the lighthouse is observed from the ships are 30º and 45º respectively. If the lighthouse is 100 m high, the distance between the two ships is: A. 300 m B. 173 m C. 273 m D. 200 m

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Rishi Tarkesh G.

Assume a triangle with ACD, Let BD be the lighthouse and A and C be the positions of the ships. Then, BD = 100 m, angle BAD = 30° , angle BCD = 45° tan 30° = BD/BA ⇒ 1/√3 = 100/BA ⇒ BA = 100√3 tan 45° = BD/BC ⇒ 1 = 100/BC ⇒ BC = 100 Distance between the two ships = AC = BA + BC = 100√3 + 100 = 100(√3+1) = 100√3+100=100(√3+1) = 100(1.73+1) = 100 × 2.73 = 273 m Answer is Option C

### Subject:GMAT

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Question:

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Rishi Tarkesh G.

Statement 1 alone is not sufficient because by knowing the product a.b we are not sure about the values of a or b There are multiple values of a and b which fits in here a=1, b=8 a=2, b=4 which is why we can not arrive at a unique solution for our problem Now look at Statement 2 alone, a/b=3 With this we can derive the value of b/a which is a/3 Now that we have unique values for a/b and b/a we can solve the problem using statement 2 alone So the answer for this question is B

### Subject:Algebra

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Question:

Evaluate: |5 - 7(3 - 9)| - |4 - 19|

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Rishi Tarkesh G.

To solve these type of questions, one should keep in mind BODMAS rule Solving math with in brackets first |5-7(-6)|-|-15| Multiplication of 7 & -6 will be next step |5-(-42)|-|-15| = |5+42|-|-15| = |47|-|-15| Solving modulus will modify this equation as = 47-15 = 32 Solution is 32

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