ATAR Notes: Forum
VCE Stuff => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Mathematics => Topic started by: chuckjefster90 on April 18, 2010, 05:37:59 pm
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Hey Guys, I'm unable to solve this problems. Would you be able to help me solve them? The answers are at the bottom, I just need to know how to get them. Thanks Heaps.
1. After spending 5/8 of his salary, Tom has $315 left. What is his salary?
2. 11,12,13 are consexutive integers. Find another 3 consecutive integers which add to 114.
3. A square and an equilateral triangle have the same perimeter. The sides of the triangle are 4cm longer than the sides of the square.
How long is one side of the square?
4. If Mrs Nguyen is 38 years old and her daughter is 8 years old, in how many years will Mrs Nguyen be 3 times as old as her daughter?
5. Tim runs between 2 places, A and B at 10km/hr and cucles back again at 20 km/hr. The total time taken to run and cucle is 3 hours. If the disance between A and B is x kim, form an equation in x and solve it to find the distance between A and B.
6. THe sum of the interior angles of a polygon with 'n' sides is (2n-4) right angles. If each interior angle is a regular polygon is 135 degrees, how many sides has the polygon?
7. The hypotenuse of a right angles triangle is (x + 5) cm in length and the other 2 sides are 5 cm and (x + 4) cm. Use the Theorem of Pythagoras to write an equation in x and solve it to find the lengths of the unknown sides of the triangle.
Answers:
1) $840 2) 37,38,39 3) 12 cm 4) 7 Years 5) 20 km 6) 8 (an octagon) 7) 12 and 13 cm
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1)

2) Let x be the middle number
 + x + (x+1) = 114)


So the numbers are 37, 38 and 39
3) )

4) )


I'm going out now, someone wanna take over? :P
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q2) n+n+1+n+2=114 3n+3 =114 n=37 so 37,38,39
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5.





6.  \times 90^\circ)
Note that the number of sides is the number of angles.
)


7. By Pythagoras for a right angled triangle 
So, ^2=(x+5)^2)



So lengths are 5cm, 12cm and 13cm.
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7. The hypotenuse of a right angles triangle is (x + 5) cm in length and the other 2 sides are 5 cm and (x + 4) cm. Use the Theorem of Pythagoras to write an equation in x and solve it to find the lengths of the unknown sides of the triangle.
(x+5)2 = 52 + (x+4)2
x2 + 10x + 25 = 25 + x2 + 8x +16
2x = 16
x= 8
Sub x into the sides,
8+5 = 13 and 8+4 = 12
:)
Try to remember that 3, 4, 5 and 5, 12, 13 form right angle triangles.
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3) )

Hey watchman, is that some kind of formula? :D
I go about doing it this way:
let side length of square be x
4x = 3(x+4)
And then solve for x.
And oops, sorry, just found out that matty did qn 7 lol
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Nope, it's exactly the same as what you did, except I couldn't be bothered defining a variable :P
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you guys are legends!
Thanks heaps!!! :)
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yeahh further haha
forgot all of it haha
hey can you like explain q3?
im still a bit confused on that
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wait no worries i got it now :)
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hey watchman, sorry third comment but for question four what is 'n' suppose to stand for?
thanksss
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It stands for the no. of years
... after which the conditions set out by the question are met :P
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thankss this all helped me so mcuh :)
ill be asking you guys defs if i need help!
thanks again :D
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thankss this all helped me so mcuh :)
ill be asking you guys defs if i need help!
thanks again :D
No probs, happy to help :)
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What further module is this?
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I wish I knew VN last year, you are one lucky person OP.
(think you're in year 11?)
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I wish I knew VN last year
Lol, I'm glad I found this place for yr 11 :)