Math, asked by Anonymous, 1 year ago

 \large \textbf {A \: student \: rank \: 19th\: in\:the\:class\:of\:50\\what \: is \: his\: rank\:from\:the\:last?}

Answers

Answered by Anonymous
200

◀ HEY THERE!◀




◀  Question: ◀



→  A student is ranked 19th in a class of 50.what Is his rank from the last?



Method of Solution:



→ Given:



⇒  Total Student in the Class = 50



⇒  Student rank = 19th



→  To Find:



⇒  Rank From the last = ?



There are two method to solve this type of Questions.



→  Method l : (Solving by Alternative Method)




→  Let the rank from last or end be x ,



Rank From Start + Rank From Last = Total Number of Students + 1




  ⇒ 19 + x = 50+1



⇒  19+x = 51



⇒ x =51-19



⇒  x = 32



→  Therefore, 32th rank from from the last.



◀  Method ll: ◀



Given:



⇒  Total Student in the Class = 50



⇒  Student rank = 19th



Lets find How many students behind him/she,



Total Student- Student Rank



⇒  = 50- 19



⇒  31



→ Concept: It's means 31 Student behind that Student.



Student from end + 1

⇒  31 + 1

→   ◀ Thus ,it meant 32 Student from the last.◀



sanjana854: hey aman
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Anonymous: Great answer!
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Answered by Anonymous
69
As 19 + x( rest students) = 50

x= 31

So rank of 19 th student from the last is 31 + 1 = 32 th rank

OR

A student is at 19 th position

and total 50 students

As we know if x1,x2,.........xn students are there , obviously there are n students OR

so to find no of students it's ( xn - x1 +1)

So to find position of 19 th student from last

simply calculate no of students from 19 th position

19, 20,.........50

So it's (50-19) + 1 = 31+ 1 = 32



⚡⚡⚡⚡Dhruv⚡⚡⚡⚡⚡⚡⚡

sprao534: i have one doubt. please don't think otherwise. why his rank changes. his rank will be given as per decreaseing order. from the ending we have to take in increasing order
sprao534: please understand my question. it is not the question of even or odd
sprao534: sorry, it is question of rank, not the question of good or bad. some times first rank holder may be bad.
sprao534: ok, leave it
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