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Important Formulas

Vedic/Quicker Math

Alligation or Mixture

Area

Average

Banker's Discount

Boats and Streams

Calendar

Chain Rule

Clock

Compound Interest

Decimal Fraction

Logarithm

Numbers

Odd Man Out and Series

Partnership

Percentage

Permutation & Combination

Pipes and Cistern

Probability

Problems on Ages

Problems on H.C.F & L.C.M

Problems on Numbers

Problems on Trains

Profit & Loss

Races and Games

Ratio and Proportion

Simple Interest

Simplification

Square Root and Cube Root

Stocks and Shares

Surds and Indices

Time and Distance

Time and Work

True Discount

Volume and Surface Area

Practice Questions

:- You can add tough questions into
your favourite list for second revision. Note |

Q1. | The angle of elevation of a ladder leaning against a wall is 60° and the foot of the ladder is 4.6 m away from the wall. The length of the ladder is : |

A. | 2.3 m [Wrong Answer] |

B. | 4.6 m [Wrong Answer] |

C. | 7.8 m [Wrong Answer] |

D. | 9.2 m [Correct Answer] |

View Answer
Explanation:-
Answer : DDiscuss it below : Let AB be the wall and BC be the ladder. Then, ^{AC}/_{BC}^{1}/_{2}Click here for more comments |

Q2. | Two ships are sailing in the sea on the two sides of a lighthouse. The angles of elevation of the top of the lighthouse as observed from the two ships are 30° and 45° respectively. If the lighthouse is 100 m high, the distance between the two ships is : |

A. | 173 m [Wrong Answer] |

B. | 200 m [Wrong Answer] |

C. | 273 m [Correct Answer] |

D. | 300 m [Wrong Answer] |

View Answer
Explanation:-
Answer : CDiscuss it below : Let AB be the lighthouse and C and D be the positions of the ships. Then, AB = 100 m, ^{AB}/_{AC}^{1}/_{√3}AC = AB X √3 = 100√3 m. ^{AB}/_{AD}CD = (AC + AD) = (100√3 + 100) m = 100 (√3 +1) m = (100 X 2.73) m = 273 m. Click here for more comments |

Q3. | If the height of a pole is ^{2}√3 metres and the length of its shadow is 2 metres,
find the angle of elevation of the sun. |

A. | 50° [Wrong Answer] |

B. | 60° [Correct Answer] |

C. | 70° [Wrong Answer] |

D. | 80° [Wrong Answer] |

View Answer
Explanation:-
Answer : BDiscuss it below : Let AB be the pole and AC be its shadow. Let angle of elevation, ^{AB}/_{AC}^{2√3}/_{2}Click here for more comments |

Q4. | From a point P on a level ground, the angle of elevation of the top of a tower is 30°. If the tower is 100 m high, the distance of point P from the foot of the tower is : |

A. | 149 m [Wrong Answer] |

B. | 156 m [Wrong Answer] |

C. | 173 m [Correct Answer] |

D. | 200 m [Wrong Answer] |

View Answer
Explanation:-
Answer : CDiscuss it below : Let AB be the tower. Then, ^{AB}/_{AP}^{1}/_{√3}^{100}√3 m.
= (100 X 1.73) m = 173 m.Click here for more comments |

Q5. | The angle of elevation of the sun, when the length of the shadow of a tree is √3 times the height of the tree, is : |

A. | 30° [Correct Answer] |

B. | 45° [Wrong Answer] |

C. | 60° [Wrong Answer] |

D. | 90° [Wrong Answer] |

View Answer
Explanation:-
Answer : ADiscuss it below : Let AB be the tree and AC be its shadow. Let Then, ^{AC}/_{AB}Cot θ = √3 θ = 30° Click here for more comments |

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