Maths questions... can you solve these?

Question

its your wish to try.....

1) A Total of 50,000$ is to be distributed among 200 persons as prize. A prize is either 500$ or 100$. Find the number of each type of prize.

2) Diagonals of a Quadrilateral are of lenght 10cm and 24cm. If the diagonals bisect each other at right angles, Find the lenght of each side of the quadrilateral.

Answer 1

Hi,

x + y = 200

500x + 100y = 50000

Multiply the first equation by -100 and add the equations together to solve.

-100(x + y = 200)

500x + 100y = 50000

-100x - 100y = -20000

500x + 100y = 50000

---------------------------------

400x = 30,000

x = 75

y = 125

There were 75 "500$" prizes and 125 "100$" prizes. <==ANSWER

If diagonals of a quadrilateral bisect each other forming right angles, then the quadrilateral is a rhombus. Find the length of each side of the rhombus by using half the length of each diagonal as the legs of a right triangle so you can solve for the hypotenuse using the Pythagorean Theorem. this length is also the length of a side.

With diagonals of 10 cm and 24 cm, half of those lengths are 5 cm and 12 cm as the legs of the right triangle.

5² + 12² = c²

25 + 144 + c²

169 = c²

13 = c

The length of each side of the quadrilateral is 13 cm. <==ANSWER

I hope that helps!! :-)

Answer 2

why did u ask these??? i am curious...

1.   let the no of 500 $ prize= X and the no of 100$ prize=Y

now,

     500X + 100Y= 50000

          X + Y= 200

so, equating both , X = 75 and Y= 125... bingo!!!

2.  each side of the length will be= 13 cm..

    after bisecting the diagonals constitute 4 equal right angle triangles

with the lengths 5 and 12 cms. by pythogorus theorem u can easily calculate the other one as 13 cm which is our desired length of qadrilateral...voila!!!!

glad i could help...keep writing

Answer 3

1)

a=number of 500 $ prize

b=number of 100 $ prize

a+b=200

500a+100b=50,000

a=200-b

500(200-b)+100b=50,000

100,000-500b+100b=50,000

50,000=400b

b=50,000/400=125 b=125 and a=200-125=75  a=75

2)

you have 4 right triangles which the sides of normal angle are 10/2=5 and 24/2=12

Side lenght is (5^2+12^2)sqrt=169sqrt=13

Answer 4

1)If x persons get $500 and y get $100 then x+y = 200 and 500x + 100y = 50000. Solve this to get 75 get $500 and 125 get $100.

2) Info is insufficient. The diagonals intersect in what ratio must be given. If a,bc,d are sides and parts of diagonals are x and 10-x for one and y and 24 -y for second, then we have 6 unknowns and not 6 eqns to find them

Answer 5

1)

Let x= number of $500 prizes

Let y= number of $100 prizes

x+y=200...................Eq.1

500x+100y=50000.....Eq.2

From Eq. 1, y=200-x

Substitute this value for y in Eq. 2, to get

500x+100(200-x)=50000

Solve

500x+20000-100x=50000

400x=50000-20000

400x=30000

4x=300

x=75

From x+y=200

75+y=200

y=200-75

y=125

2)

The length of each side is 13cm.

Sketch 2 intersecting lines that are bisected at 90 degrees. One

line is 10cm, the other 24cm.

Because of the bisection, each half of the lines are 5cm and 12cm

respectively.

Now join the 4 corners.

Because of the contained 90 degrees in each of the 4 triangles that

result from joining the 4 corners, you have 4 congruent triangles of

sides 5 and 12 with the contained angle being 90 degrees.

Find one side and you've found all 4 sides.

Using the Pythagorean Formula,

H^2=5^2+12^2

H^2=25+144

H^2=169

H=13

All 4 sides 13cm