Some interesting word problems :)

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Jan 30th, 2002, 09:49 PM
struntz

Some interesting word problems :)

Hello everyone! :D

i can't figure these word problems!

this is not graded or anything, my pre-calc teacher handed htem out so we can pratice but he didn't give us the solutions or anything to show us how to od htem if we don't understand. SO i was wondering if any of u would like to take a "whack" at it, and show ur work so i can see how it was done it would be great. Thanks for your time. Here are some of them.....

Oh and if u care to share ur thinking techniques that would be helpful also so i can devlope better problem solving skills.

Code:

The sum of the lenghts of the three sides of a right triangle is 18 units. The sum of the squares of the lengths of the three sides is 128 units. What is the area of the triangle? answers: A. 5 B. 8 C. 9 D. 18 E. 32

Code:

If the Product 3/2 x 4/3 x 5/4 x 6/5. ....a/b = 9 what is the sum of a and b? answers: A. 11 B. 13 C. 17 D. 35 E. 37

Code:

The sum of the positive square roots of two numbers is 5. The two numbers also differ by 5. Waht is the smaller of the two numbers? answers: A. 0 B. 1 C. 4 D. 5 E. 9

Code:

The positive interger 4 can be represented as a sum of 1's or 2's in five ways. They are: 1+1+1+1, 1+1+2, 1+2+1, 2+1+1, and 2+2. In how many ways can the postive integer 5 be represnted as a sum of 1's or 2's? answers: A. 6 B. 7 C. 8 D. 9 E. 10

Code:

A cicular target with radius of 12 inches is to have a cicular bull's-eye painkted in teh center, so that the area of bull's eye is 1% of the total area of the target. What should the radius of the bull's-eye be in inches? answers: A. 0.12 B. 1.2 C. 1/sqr(pie) D. 12/sqr(145) E. 0.0

Code:

A two-inch cube (2x2x2) of silver weights 3 pounds and is worth $200. How much is a trhee-inch cube of silver worth? answers: A. $300 B. 375 C. 450 D. 560 E. 675

Code:

Given that 2/3 of the members of a committee uses 3/4 of the charis in a room, what is the least possible number of members on the committee? answers: A. 3 B. 6 C. 9 D. 12 E. 15

Code:

All the even numbers from 2 to 98 inclusive, except those ending in 0, are multiplied together. What is the rightmost digit, that is, the untis digit, of the product? answers: A. 0 B. 2 C. 4 D. 6 E. 8

Thanks for your time.

:D :D
Jan 30th, 2002, 10:34 PM
prog_tom

Quiet challenging

These questions sure are good. I may not solve it, but give me 400 hours, and I'll give a really good shot.
Jan 30th, 2002, 11:03 PM
crptcblade

Quote:

The sum of the positive square roots of two numbers is 5.
The two numbers also differ by 5.
Waht is the smaller of the two numbers?
answers:
A. 0 B. 1 C. 4 D. 5 E. 9

I say C.

sqrt(4) = 2
sqrt(9) = 3

3+2=5
9-4=5

:)
Jan 30th, 2002, 11:10 PM
prog_tom

Quote:

If the Product 3/2 x 4/3 x 5/4 x 6/5. ....a/b = 9
what is the sum of a and b?
answers:
A. 11 B. 13 C. 17 D. 35 E. 37

I don't know, but I try to use System of Linear Equations in 2 variables.

{a/b=9
{a+b=x

a=-x/10+x
b=x/10

Which is really really hard, but is solvable for me
Jan 31st, 2002, 06:36 AM
NotLKH

Re: Some interesting word problems :)

Quote:

Originally posted by struntz
Hello everyone! :D

Code:

If the Product 3/2 x 4/3 x 5/4 x 6/5. ....a/b = 9 what is the sum of a and b? answers: A. 11 B. 13 C. 17 D. 35 E. 37

There's magic going on there:
#1) b = a - 1

#2) 3/2 x 4/3 = 4/2,

3/2 * 4/3 * 5/4 = 5/2

so the pattern is a = 18, for the entire equation to = 9.

Therefore b = 17 and 18 + 17 = D
Jan 31st, 2002, 12:40 PM
Starman

whee! I got the triangle one.

call the sides a, b and c (c is the hypotenuse)

a^2 + b^2 = c^2 (right triangle)

a^2 + b^2 + c^2 = 128 (given)

a^2 + b^2 = 128 - c^2

so c^2 = 128 - c^2

128 = 2(c^2)

64 = c^2

c = 8

plug this back into the right triangle formula

a^2 + b^2 = 64

also given is a + b + c = 18

so a + b = 10

square this to get

a^2 + 2ab + b^2 = 100

subtract the plugged in bit

(a^2 + 2ab + b^2 = 100) - (a^2 + b^2 = 64)

gives 2ab = 36

formula for area of triangle is 1/2 * base * height, or 1/2(ab)

2ab = 36, so 1/2(ab) = 36/4 = 9

answer (c).
Jan 31st, 2002, 06:43 PM
Gandalf_Grey_

A two-inch cube (2x2x2) of silver weights 3 pounds and is worth $200.
How much is a trhee-inch cube of silver worth?
answers:
A. $300 B. 375 C. 450 D. 560 E. 675

the answer is E 675 because 2^3 = 8 = $200 so 3^3 = 27 = $X 27 is 3.378 times more than 8 so 3.375 * 200 (the worth of the original piece of silver) equals 675
Jan 31st, 2002, 07:15 PM
Gandalf_Grey_

A cicular target with radius of 12 inches is to have a
cicular bull's-eye painkted in teh center,
so that the area of bull's eye is 1% of the total area of the target.
What should the radius of the bull's-eye be in inches?

A. 0.12 B. 1.2 C. 1/sqr(pie) D. 12/sqr(145) E. 0.0

the bulls eye should be 1.2 ".

A = pi(r)^2
A = pi(12)^2
A = 452.3 the area of the target

452.3 * 0.01 = 4.524 the area of the bulls eye

A = pi(r)^2
4.524 = pi(r)^2

divide this by pi

4.524 / pi = r^2

then square root it

sqrt(4.524 / pi) = r

so r finally works out to 1.2
Feb 1st, 2002, 09:47 AM
thinktank2

Here is my way of doing it.

Starting From question 2

If the Product 3/2 x 4/3 x 5/4 x 6/5. ....a/b = 9
what is the sum of a and b?

Note the pattern, Numerator of a term and the denominator of the next term are equal, so on multiplication they will cancel out. So the eq. reduces to

a/2 = 9 or a = 18 , b should be the previous term's numerator which is 17.
______________________________________________________________________________________________

The sum of the positive square roots of two numbers is 5.
The two numbers also differ by 5.
Waht is the smaller of the two numbers?
A. 0 B. 1 C. 4 D. 5 E. 9

Sqrt(x) + Sqrt(y) = 5
x - y = 5

We know, (x - y) = [Sqrt(x) + Sqrt(y)] * [Sqrt(x) - Sqrt(y)]

=> 5 = 5 * [Sqrt(x) - Sqrt(y)]

=> Sqrt(x) - Sqrt(y) = 1 or Sqrt(x) = Sqrt(y) + 1

=> The roots of the two numbers y,x are successive integers

so they are 9 and 4, 4 being the smallest
______________________________________________________________________________________________

The positive interger 4 can be represented as a sum of 1's or 2's in five ways.
They are: 1+1+1+1, 1+1+2, 1+2+1, 2+1+1, and 2+2.
In how many ways can the postive integer 5 be represnted as a sum of 1's or 2's?

There are,

1 combination of 1+1+1+1+1
4 combinations of 1+1+1+2
3 combinations of 1+2+2

so there are 1+4+3 = 8 ways totally
______________________________________________________________________________________________

A cicular target with radius of 12 inches is to have a
cicular bull's-eye painkted in teh center,
so that the area of bull's eye is 1% of the total area of the target.
What should the radius of the bull's-eye be in inches?

if R be the radius of Circular Target, r be the radius of bull's eye

Then (Pi*r^2)/(Pi*R^2) = 0.01

r^2 = 0.01 * R^2 = 0.01 * 144 = 1.44

or r = 1.2
______________________________________________________________________________________________

A two-inch cube (2x2x2) of silver weights 3 pounds and is worth $200.
How much is a trhee-inch cube of silver worth?

2^3 = 8 Cubic Inches Cost $200 , That is $25 per cubic Inch
Therfore 3^3 = 27 Cubic Inches cost $25 * 27 = $675
______________________________________________________________________________________________

Given that 2/3 of the members of a committee uses
3/4 of the chairs in a room,
what is the least possible number of members on the committee?

Let the number of members in the committee be = m
the number of chairs be = n

The number of members sitting is equal to the number of chairs used.
then (2/3) * m = (3/4) * n

or m = (9/8) * n

But both m and n are Integers, so for m to be an Integer n must be a multiple
of 8, the least value of m is 9 , when n=8

so, the least possible number of members in the committee is 9
______________________________________________________________________________________________

All the even numbers from 2 to 98 inclusive,
except those ending in 0, are multiplied together.
What is the rightmost digit, that is, the untis digit, of the product?

The product is something like

2 * 4 * 6 * 8 * 12 * 14 * 16 * 18 * 22 ........... * 98

we have a pattern here repeating

(x + 2)(x + 4)(x + 6)(x + 8)

Where x takes the values 0,10,20,30....90
That is there are 10 values of x

(x + 2)(x + 4)(x + 6)(x + 8) = x^4 + 20 x^3 + 140 x^2 + 400 x + 384

now powers of 0,10,20,30...90 are all going to end in zeros
and multiples of these also end in zeros.

So, the last digits are determined by the term 384

so the last terms of the product 2*4*6*8*...98
is determined by the last digits of 384^10
(since the pattern repeats 10 times)

384^10 = (380+4)^10 = ((380+4)^2)^5 = (380^2 + 2*380*4 + 16)^5

again here 380^2 ends in zero, 2*380*4 ends in zero
so the last digit is determined by 16^5

6 is such a number that when you multiply it by 6, the last digit will be six.

so the last digit of 16^5 {16*16*16*16*16}

is definitely 6

So the last digit of the product

2 * 4 * 6 * 8 * 12 * 14 * 16 * 18 * 22 ........... * 98

is 6