An algebra problem by Ahmed Abdelbasit

Algebra Level 2

you have (100$) and wanna buy 100 pen .. but you found out that there are three types of pens .. and you wanna buy 100 pen containing all three types and also spend the all 100 dollars ..

the price of the $1^{st}$ type is (5.00) $

the price of the $2^{nd}$ type is (1.00) $

the price of the $3^{rd}$ type is (0.05) $

if we consider that :

the number of the $1^{st}$ type is A

the number of the $2^{nd}$ type is B

the number of the $3^{rd}$ type is C

so that A+B+C = 100

what is the value of $\frac{C+10}{A-10B}$

The answer is 10.

3 solutions

Ajit Panada
Jun 12, 2014

A+B+C=100 and 5A+B+(1/20)C=100 so,....99A+19B=1900 BUT 19*99=1881 ....THATS WHY WHEN A=19 => B=1 THIS COSTS 20(19+1) PENS FOR 96$(65$+1$) AND WE NEED 4 DOLLARS......so,...C=80 COSTS 4$ FINALLY SUBSTITIUE A=19,B=1,C=80 TO GET REQUIRED ANSWER OF 10 :)

Oh .. that is an excellent analogy for thinking .. Congratulation on your notes ... :)

Ahmed Abdelbasit - 6 years, 12 months ago

Unstable Chickoy
Jun 13, 2014

A diophantine equation

$A + B + C = 100$

$5A + B + .05C = 100$

$A = 19 , B = 1 , C = 80$

$\frac{80 + 10}{19 - 10(1)} = \boxed{10}$

that is all right

Ahmed Abdelbasit - 6 years, 9 months ago

Ahmed Abdelbasit
Jun 4, 2014

the two main operating equations are :..... A+B+C=100 ..... & ..... 5A+B+0.05C=100 ......... if you wanna spend the whole of 100 dollars , the price of pens of third type must be integer ... and that can be acheived by five possibilities : ( 20 or 40 or 60 or 80 ) pens .. so , in the two operating equations you have the possibilities of C .. then, substituting for C by each possibility and solving the equations together gives only one solution .. that is: ....... A=19 ..... B=1 ....... C=80 ...... so, the required value is : $\frac{C+10}{A-10B}$ = $\frac{80+10}{19-10*1}$ = $\frac{90}{9}$ = 10 ......

An algebra problem by Ahmed Abdelbasit

The answer is 10.

3 solutions

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