FRIHOSTFORUMSSEARCHFAQTOSBLOGSCOMPETITIONS
You are invited to Log in or Register a free Frihost Account!


ALGEBRA PRACTICE� win 10 frih$





airh3ad
ALGEBRA PRACTICE�

Only skilled people can open this file.......once you succeed to open this
file, you will find names of the people who have managed to open this......
Now it is your turn!

A man wanted to get into his work building, but he had forgotten his code.
However, he did remember five clues. These are what those clues were:

The fifth number plus the third number equals fourteen.

The fourth number is one more than the second number.

The first number is twice the second number less than one.

The second number plus the third number equals ten.

The sum of all five numbers is 30.

What were the five numbers and in what order?
endless
is it

7 - 4 - 6 - 5 - 8
airh3ad
You got it endless hehehe good job man..congrats! A7 - B4 -C 6 - D5 - E8 but i need also the solotion additional 5frih$ points.
endless
The fifth number plus the third number equals fourteen.

The fourth number is one more than the second number.

The first number is twice the second number less than one.

The second number plus the third number equals ten.

The sum of all five numbers is 30.

What were the five numbers and in what order?

how i found it:

  • in the first hint it says that it no 3 and no 5 equals 14
  • the last hint says that the sum of all number is 30
  • so i know that no 1 2 and 4 are 16 together
  • with the 2nd and 3th hint i just experimented with some digitys
  • with 4 you get:
    no 1 = 7
    no 2 = 4
    no 4 = 5
    all together: 16
  • the 4th hint says 2nd and 3th equal 10 so no 3 is 6
  • 3 and 5 equal 16 so i know 5 is 8

is that what you meant with solution?
hunnyhiteshseth
Well I do not think thats the correct answer. What he gave was one of the solution but there exist more than one solution to this problem. Let me elaborate:


Let the 5 numbers be:-
a,b,c,d,e in order.

So, according to given statements:-

c+e=14----------------------------------(i)
d=b+1-----------------------------------(ii)
a=2b-1----------------------------------(iii)
a+b+c+d+e=30------------------------(iv)

Putting value of (i),(ii) & (iii) in (iv)

We get 4b=16
=> b=4

=> d=5

=> a=7

Now, there can be multiple pairs of c and e

Since c + e = 14 and c and e are single digit numbers. So, there can be following pairs possible:-


(9,5), (5,9)
(8,6), (6,8 )
(7,7)
endless
hunnyhiteshseth wrote:
Well I do not think thats the correct answer. What he gave was one of the solution but there exist more than one solution to this problem. Let me elaborate:


Let the 5 numbers be:-
a,b,c,d,e in order.

So, according to given statements:-

c+e=14----------------------------------(i)
d=b+1-----------------------------------(ii)
a=2b-1----------------------------------(iii)
a+b+c+d+e=30------------------------(iv)

Putting value of (i),(ii) & (iii) in (iv)

We get 4b=16
=> b=4

=> d=5

=> a=7

Now, there can be multiple pairs of c and e

Since c + e = 14 and c and e are single digit numbers. So, there can be following pairs possible:-


(9,5), (5,9)
(8,6), (6,8 )
(7,7)


didn't you read the other hitn?
The second number plus the third number equals ten.

so b + c must equal 10
Rolling Eyes

so only 1 solution Very Happy
airh3ad
Here is my own solution

Let a = 1st number
b = 2nd number
c = 3rd number
d = 4th number
e = 5th number


from ist clue :

e + c + 14 eq. 1

from 2nd clue

d = 1 + b eq. 2

from 3rd clue

a + 2b - 1 eq.3

from 4rt clue

b + c = 10 eq. 4

from 5th clue

a+b+c+d+e = 30 eq.5

from eq. 1

e+c = 14

e=14-c eq.6

from eq.4

b+c = 10

c=10-b eq.7

e = 14 -c

e=14 (10-b)

e= 4 + b eq.8

By Subt.

from eq 5

a+b+c+d+e = 30

(2b-1)+b+(10-b)+(1+b)+(4+b)=30

2b-1+b+10-b+1+b+4+b=30

2b+b-b+b+b=30-4-1-10+1

4b= 16
b=4

from eq. 3 b=4
a=2b-1
a=2(4)-1
a=7

from eq.7 b=4
c=10-b
c=10-4
c=6

from eq.2 ; b=4

d= 1 +b
d=1+4
d=5

from eq.8: b=4
e=4+b
e=4+4
e=8

Combination code

A B C D E
1 4 6 5 8

thats great and detailed solotion of mine..
Razz
endless
thats the problem with people who studied algebra, they do it too detailed, keep it simple Razz
~
(maybe it's because im 14 that i think simple but Razz)
hunnyhiteshseth
endless wrote:
hunnyhiteshseth wrote:
Well I do not think thats the correct answer. What he gave was one of the solution but there exist more than one solution to this problem. Let me elaborate:


Let the 5 numbers be:-
a,b,c,d,e in order.

So, according to given statements:-

c+e=14----------------------------------(i)
d=b+1-----------------------------------(ii)
a=2b-1----------------------------------(iii)
a+b+c+d+e=30------------------------(iv)

Putting value of (i),(ii) & (iii) in (iv)

We get 4b=16
=> b=4

=> d=5

=> a=7

Now, there can be multiple pairs of c and e

Since c + e = 14 and c and e are single digit numbers. So, there can be following pairs possible:-


(9,5), (5,9)
(8,6), (6,8 )
(7,7)


didn't you read the other hitn?
The second number plus the third number equals ten.

so b + c must equal 10
Rolling Eyes

so only 1 solution Very Happy



OOPS!! I missed that statement!! Embarassed

You are right.
airh3ad
thats part of curriculum i think algebra is the subject in schools.
Related topics
Next 11 members who joins comicstrips.be get 10 frih$
5frih$ For first 10 members! Chance to win 10 more! [new!]
10 frih$ for membership
Problem: last letter of my username got cut off up-10 frih$
Sigs for sale 10 FRIH$ each!!
Your video converted into Flash 8 for 10 FRIH$
Need brushes [offer 10 FRIH$]
10 Frih$ to the first person to explain this riddle
selling sigs @ 5 FRIH$ each
Layouts!~ 10 Frih and up
What Font for 10 FRIH$ and another 5 FRIH$
Get 2 FRIH$ just for Trying to win 10 Frih$ (whats to lose?)
earn 10 frih just for trying to win a .info
The Presidential Sweepstakes: WIN 500 Frih$ !
Reply to topic    Frihost Forum Index -> Miscellaneous -> Contests

FRIHOST HOME | FAQ | TOS | ABOUT US | CONTACT US | SITE MAP
© 2005-2011 Frihost, forums powered by phpBB.