# Quant

rename
andsy's
version from
2018-05-09 20:18

## Section 1

Question | Answer |
---|---|

If A working Alone takes a days more than A & B, & B working Alone takes b days more than A & B. Then , Number of days, taken by A & B working together to finish a job is = | √ab |

If A & B working together can finish a work in x days & B is k times efficient than A, then the time taken by, A working Alone will take ⇒ | (k + 1) x |

If A & B working together can finish a work in x days & B is k times efficient than A, then the time taken by B working Alone will take ⇒ | ((k+1)/k)x |

If A can finish a work in x days and B is k times efficient than A, then the time taken by both A and B, working together to complete the work is | x/(1+k) |

If A & B working together, can finish a piece of work is x days, B & C in 4 days, C & A in z days. Then, A + B + C working together will finish the job is ⇒ | 2xyz/(xy+yz+zx) |

Two persons A & B, working together, can complete a piece of work in x days. If A, working alone, can complete the work in y days, then B, working alone, will complete the work in ⇒ | xy/(y-x) |

If A, B & C will working alone, can complete a work in x, y and z days, respectively, then they will together complete the work in | xyz/(xy+yz+zx) |

If A can do a piece of work is x days and B can do a piece of work in 4 days, then both of them working together will do the same work in | xy/(x+y) days |

If working efficiency of A & B is → x:y. Then, the time taken by A & B to finish the work is in the ratio → | y:x |

## Section 2

cuboid

Question | Answer |
---|---|

Lateral surface Area(पार्श्व पृष्ठ का क्षेत्रफल ) = | Perimeter of Base × Height Base = 2(l + b) × h |

Total surface area (कुल पृष्ठ का क्षेत्रफल )= | Lateral surface Area + 2 × Area of base = 2 (lh + bh + lb) |

Diagonal (विकर्ण)= | √(l²+b²+h² ) |

V = | √( LB×BH×HL) OR LBH |

Total surface area(advance)= | (sum of all three side)² – (Diagonal)² |

Volume of the cube (घन का आयतन)= | (√(total surface area)/6)³ |

In Radius of cube(घन की अन्तः त्रिज्या)= | a/2 |

Circumradius of cube(घन की परित्रिज्या) = | √3/2 a |

## Section 3

Right circular coneQuestion | Answer |
---|---|

Volume (वक्र पृष्ठ का क्षेत्रफल)= | 1/3×area of base×height = 1/3 πr² h |

Curved surface area(कुल पृष्ठ का क्षेत्रफल)= | 1/2 (Perimeter of base) × slant height = 1/2 × 2πr×l=πrl=πr√(r²+h² ) |

Total surface area(कुल पृष्ठ का क्षेत्रफल) = | C.S.A + Area of base = πrl+πr²=πr(l+r) |

Radius of maximum size sphere in a cone(एक शंकु के भीतर अधिकतम आकार के गोले की त्रिज्या) | (h × r)/(l + r) |

If cone is cut parallel to its base and ratio of heights, radius or slant height of both parts is given as →x∶y Then Ratio of their volume | x³ ∶y³ |

Volume of pyramid | 1/3 (Area of base) × height |

Lateral Surface area | 1/2 (Perimeter of base) × Slant height. |

Average of n natural no's= | (n+1)/2 |

Average of even No'= | (n+1) |

Average of odd No'= | n |

Question | Answer |
---|---|

Suppose A invests Rs. x for p months and B invests Rs. y for q months then,(A's share of profit):(B's share of profit)= | xp:yq |

Let a container contains x units of liquid from which y units are taken out and replaced by water.The quantity of pure liquid after 'n' operations is equal to | [ x ( 1 - y/x}^n] |

Gain% | {(Gain*100)/C.P.} |

S.P. = | {(100+Gain%) / 100} * C.P. |

C.P. = | {(100 / (100+Gain%)} * S.P. |

When a person sells two similar items, one at a gain of say x%, and the other at a loss of x%, then in this transaction the seller always incurs a loss given | {x^2/100}% |

A single discount equivalent to discount series of x% and y% given by the seller is equal to | {x +y - xy/100}% |

If a trader professes to sell his goods at cost price, but uses false weights, then Gain% | {Error/(True value - Error) x 100}% |

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