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

# ALGEBRA PRACTICEï¿½ win 10 frih\$

ALGEBRA PRACTICEï¿½

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......

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
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

so only 1 solution
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..
endless
thats the problem with people who studied algebra, they do it too detailed, keep it simple
~
(maybe it's because im 14 that i think simple but )
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

so only 1 solution

OOPS!! I missed that statement!!

You are right.