Author Topic: Fun questions :) (Read 114326 times) Tweet Share

kamil9876 · « **Reply #105 on:** April 28, 2009, 09:20:38 pm »

lol yes. at an exponential rate like adrian please

TrueTears · « **Reply #106 on:** April 29, 2009, 06:58:52 pm »

Quote from: kamil9876 on April 28, 2009, 09:20:38 pm

lol yes. at an exponential rate like adrian please

lol that would flood the world

19, 24, 27, 29, 30, 32, 33, 42

Over9000 · « **Reply #107 on:** April 29, 2009, 08:44:35 pm »

33. When a school marching band lines up in 4 columns, the last row is not complete. When it lines up in 3 columns, the last row is not complete, but there are 3 more rows. When it lines up in 2 columns, the last row is not complete and there are 5 more rows than there were with 3 columns. The number of students in the band is between?
A. 10-19 B. 20-29 C. 30-39 D. 40-49 E. 50-59

Let amount of people in band be n
Let amount of row's be x
Let the amount of spots remaining in first situation be y, where y is a positive integer
And let the amount of spots remaining in second situaution be z, where z is a positive integer

$4x - y = n$ , where $y ={1,2,3}$ is first situation (4 rows times x columns - amount of unfilled spots in last row and y can be 1 2 or 3, because there are 4 spots in each row, and one must be filled, so 1 2 or 3 can be empty)
$3 (x+3) - z = n$ , where $z ={1,2}$ is second situation (z can be 1 or 2 for much the same reasoning as above)
$2 (x+8) - 1 = n$ is third situation (we know there can only be one unfilled spot because there are only 2 spots per row, and one must be taken)

$3 (x+3) - z = 2 (x+8) - 1$
$3x + 9 - z = 2x + 15$
$x - 6 - z = 0$
$x - z = 6$
let $z = 1$
$x = 7$
let $z = 2$
$x = 8$

when $x = 7$
sub $x = 7$ into $2 (x+8) - 1 = n$
$n = 29$

when $x = 8$
sub $x = 8$ into $2 (x+8) - 1 = n$
$n = 31$

however when we sub $x = 7$ into $4x - y = n$
we get $28 - y = 29$
which makes $y = -1$ , however y is a positive integer
so x cannot equal 7

when x = 8
we get 32 - y = 31
which makes y = 1, this is entirely possible
Therefore n = 31, answer is C

EDIT: I SUCK AT LATEX

TrueTears · « **Reply #108 on:** April 29, 2009, 08:51:37 pm »

correct

TrueTears · « **Reply #109 on:** April 30, 2009, 04:42:54 pm »

A TOTAL OF 50 SOLVED/UNSOLVED QUESTIONS NOW TO ENTERTAIN YOURSELF GUYS!

19, 24, 27, 29, 30, 32, 33

NEW QUESTIONS FROM 42 - 50

Mao · « **Reply #110 on:** April 30, 2009, 05:00:08 pm »

19:

Probability of both being australian: $\frac{n}{N} \cdot \frac{n-1}{N-1}$

Probability of at least one being australian: $^2C_1 \frac{n}{N}\cdot \frac{N-n}{N-1} + \frac{n(n-1)}{N(N-1)} = \frac{2nN-n^2 - n}{N(N-1)} = \frac{n(2N - n - 1)}{N(N-1)}$

Probability of both given at least one: $\frac{\mbox{Both}}{\mbox{At least one}} = \frac{n-1}{2N - n - 1}$

i think...?

Mao · « **Reply #111 on:** April 30, 2009, 05:05:31 pm »

24, when two circles 'just touch', the distance between the centers is the sum of radii (because of the common tangent at the point of contact being perpendicular to both radii).

That is, the fifth sphere's center is equidistant to the existing four spheres, forming a regular tetrahedron.

A diagonal is $\sqrt{20^2 + 20^2} = 20\sqrt{2}$ , hence, using pythagoras

$h = \sqrt{20^2 - (10\sqrt{2})^2} = 10\sqrt{2}$

TrueTears · « **Reply #112 on:** April 30, 2009, 05:13:49 pm »

Good work Mao both correct

19, 24, 27, 29, 30, 32, 33

Mao · « **Reply #113 on:** April 30, 2009, 05:44:54 pm »

27

let $PR = \vec{a},\; PQ = \vec{b}$

$\implies QR = \vec{a} - \vec{b}$

$\implies PS = 2\vec{b}$

$\implies PU = \frac{3}{5}\vec{a}$

$\implies US = UP + PS = -\frac{3}{5} \vec{a} + 2\vec{b}$

$QT = k\cdot QR = QP + PU + UT = QP + PU + \lambda US$

$\implies k(\vec{a} -\vec{b}) = -\vec{b} + \frac{3}{5}\vec{a} + \lambda \left( -\frac{3}{5}\vec{a} + 2\vec{b}\right)$

$\implies k\vec{a} - k\vec{b} = \frac{3}{5} (1-\lambda) \vec{a} - (1-2\lambda)\vec{b}$

$\implies 1-\frac{5}{3}k = \lambda = \frac{1}{2}(1-k)$

$\implies k = \frac{3}{7}$

$\implies QT = \frac{3}{7}QR$

$\implies \frac{QT}{QR} = \frac{3}{7}$

TrueTears · « **Reply #114 on:** April 30, 2009, 05:47:16 pm »

nice! correct 8-)

TrueTears · « **Reply #115 on:** April 30, 2009, 05:49:26 pm »

27. Using Menelaus' Theorem

$\frac{PS}{QS} \times \frac{QT}{RT} \times \frac{RU}{PU} = 1$

$2 \times \frac{QT}{RT} \times \frac{2}{3} = 1$

$\frac{QT}{RT} = \frac{3}{4}$

$\therefore \frac{QT}{RQ} = \frac{3}{7}$

Mao · « **Reply #116 on:** April 30, 2009, 05:55:10 pm »

32

'combined' New age - old age = 909

$909 = (x+9)^2 - x^2 \implies x=46$

$46^2 = 2116$

sum of original age is therefore 37

TrueTears · « **Reply #117 on:** April 30, 2009, 05:58:06 pm »

correct

yay all questions before 42 is answered!!!

new questions now, have funs ~!

kamil9876 · « **Reply #118 on:** April 30, 2009, 06:29:27 pm »

44.) let the square have side a. because the two remaining triangles are similair(since its a right angle triangle since the converse of pythagoras theorem is true), their rise over run must be the same.

hence:

$\frac{3-a}{a}=\frac{a}{4-a}$
$a^2=12-7a+a^2$
$0=12-7a$
$a=\frac{12}{7}$

hence area is $\frac{144}{49}$

TrueTears · « **Reply #119 on:** April 30, 2009, 06:31:47 pm »

Check working again kamil

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