Math, asked by anirudhmehra1225, 6 days ago

Q11 Exactly 1/5 of the clients who entered Fl's shop yesterday were
women. If exactly one third of the women were above 5'6" tall
, what is
the minimum possible number of clients that entered the shop
yesterday?​

Answers

Answered by amitnrw
0

Given :  Exactly 1/5 of the clients who entered Fl's shop yesterday were

women.

Exactly one third of the women were above 5'6" tall

To Find : what is the minimum possible number of clients that entered the shop  

Solution:

Let say number of clients that entered the shop   = N

N must be  natural number

Exactly 1/5 of the clients who entered Fl's shop yesterday were

women.

=> Women = N/5

Exactly one third of the women were above 5'6" tall

=> Above 5'6" tall = (1/3) (N/5)

= N/15

N/15  must be natural number

least  natural number   = 1    

N/15 = 1

=> N = 15

Hence  minimum possible number of clients that entered the shop = 15

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Answered by brainlyvirat187006
2

Answer:

Given :  Exactly 1/5 of the clients who entered Fl's shop yesterday were

women.

Exactly one third of the women were above 5'6" tall

To Find : what is the minimum possible number of clients that entered the shop  

Solution:

Let say number of clients that entered the shop   = N

N must be  natural number

Exactly 1/5 of the clients who entered Fl's shop yesterday were

women.

=> Women = N/5

Exactly one third of the women were above 5'6" tall

=> Above 5'6" tall = (1/3) (N/5)

= N/15

N/15  must be natural number

least  natural number   = 1    

N/15 = 1

=> N = 15

Hence  minimum possible number of clients that entered the shop = 15

Learn More:

If the numerator of a fraction is decreased by 40% and the ...

brainly.in/question/16964835

If the numerator and denominator of a proper fraction are increased ...

brainly.in/question/11141630

꧁༒BRAINLYVIRAT187006༒꧂

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