Aptitude - Profit and Loss - Discussion

Discussion Forum : Profit and Loss - General Questions (Q.No. 2)
2.
The cost price of 20 articles is the same as the selling price of x articles. If the profit is 25%, then the value of x is:
15
16
18
25
Answer: Option
Explanation:

Let C.P. of each article be Re. 1 C.P. of x articles = Rs. x.

S.P. of x articles = Rs. 20.

Profit = Rs. (20 - x).

20 - x x 100 = 25
x

2000 - 100x = 25x

125x = 2000

x = 16.

Video Explanation: https://youtu.be/sidqvjUoZxY

Discussion:
353 comments Page 3 of 36.

Vignesh.R said:   1 decade ago
Chetan! your explanation is cute and understandable.

Jacky said:   1 decade ago
Sanjeet kumar gives the nice explanation.

which is easly understandable.
thanks frnd.
if 25% profit given u can take the example of c.p is 4 and s.p is 5.
20*4=x*5
x=16

Sumit yadav said:   1 decade ago
@chetan

Its Nice. You realy did well, you make this problem very simply.

Vishal said:   1 decade ago
priya...
in general form.
let the cost of 20 article = y rs
so cost of 1 article= y/20rs
and cost of x article = xy/20
so the profit= price-sailing price/sailing rate
25=(y-x*y/20)/(x*y/20)
by solve this u got 16 answer

Yuri said:   1 decade ago
How we assume cp for 1 article is 1?

If we assume else then?

Habib said:   1 decade ago
Its very very simple....

Given, CP(20 Articles)=SP(X articles)...............A

Assume, CP of 1 article is Re.1,
CP of X artilcles is rs.X..............B
CP(20 Articles)=rs.20..................C
So A becomes,
rs.20 = SP(X articles).................D

Now, Profit = SP(X articles) - CP(X articles)
= 20 - x (from B&D)

Hence find profit %, (20-x)/x * 100

Haritha said:   1 decade ago
c.p of 20 articles= s.p of x articles .........1
s.p=((100+profit%)/100)*c.p
hence from the above values we get
s.p=(5/4)c.p
substitute in 1 we get
X=16

Prathyusha said:   1 decade ago
Sundar and sanjeet explanation was nice!.

Good work guys!.

Thanku.

Kumar said:   1 decade ago
How it was come by using 2000..i cant understand plz explain detaily

Pankaj said:   1 decade ago
How to solve ( (20/x-1) x100) /25?


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