Math, asked by Triok, 1 year ago

in an examination, rahul secured 45 %, marks which were 30 more than the minimum pass marks. in the same examination, john secured 24 %marks which were 96 less than the minimum pass marks. find the maximum marks and the minimum pass marks of the examination




please its urgent

Answers

Answered by Sanskriti141
12
ANSWER
========

let \: the \: maximum \: marks \: be \: x \\  \\percentge \: of \: marks \: obtained \: by \: rahul \:  = 45\% \\  \\ marks \: obtained \: by \: him \:  =  \:  \frac{45}{100}x \\  \\ since \: the \: marks \: obtained by \: him \: are \:30 \:  more \: than \: minimum \: marks \\  \\ minimum \: marks \:  =  \frac{45}{100} x - 30 -  -  -  -  - (1) \\  \\ percentage \: of \: marks \: obtained \: by \: john = 24\% \\  \\ marks \: obtained \: by \: him =  \frac{24}{100}x \\  \\ since \: the \: marks \: obtained by \: him \: are \:96 \:  less \: than \: the \: minimum \: marks \\  \\ minimum \: marks \:  =  \frac{24}{100} x + 96 \:  -  -  -  -  -  - (2) \\  \\ from \: equations \: (1) \: and \: (2) \: we \: get \\  \\  \  \frac{45}{100}  x - 30 =  \frac{24}{100} x + 96 \\  \\  =  >  \frac{45}{100} x -  \frac{24}{100} x = 96 + 30 \\  \\  =  >  \frac{21}{100} x = 126 \\  \\  =  > x = 126 \times  \frac{100}{21}  \\  \\  =  > x = 6 \times 100 \\  \\  =  > x = 600 \\  \\ ans) \: the \: maximum \: maks \: to \: be \: obtained \\ are \: 600 \: marks. \\  \\ minimum \: marks \:  =  \:  \frac{45}{100}  \times 600 - 30 \\  \\  =  > 270 - 30 \\  \\  =  > 240 \\  \\ ans \: ) \: the \: minimum \: marks \: to \: be \: \\  obtained \: are \: 240 \: marks.

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Answered by niharikalko09
0

Step-by-step explanation:

ANSWER

========

\begin{gathered}let \: the \: maximum \: marks \: be \: x \\ \\percentge \: of \: marks \: obtained \: by \: rahul \: = 45\% \\ \\ marks \: obtained \: by \: him \: = \: \frac{45}{100}x \\ \\ since \: the \: marks \: obtained by \: him \: are \:30 \: more \: than \: minimum \: marks \\ \\ minimum \: marks \: = \frac{45}{100} x - 30 - - - - - (1) \\ \\ percentage \: of \: marks \: obtained \: by \: john = 24\% \\ \\ marks \: obtained \: by \: him = \frac{24}{100}x \\ \\ since \: the \: marks \: obtained by \: him \: are \:96 \: less \: than \: the \: minimum \: marks \\ \\ minimum \: marks \: = \frac{24}{100} x + 96 \: - - - - - - (2) \\ \\ from \: equations \: (1) \: and \: (2) \: we \: get \\ \\ \ \frac{45}{100} x - 30 = \frac{24}{100} x + 96 \\ \\ = > \frac{45}{100} x - \frac{24}{100} x = 96 + 30 \\ \\ = > \frac{21}{100} x = 126 \\ \\ = > x = 126 \times \frac{100}{21} \\ \\ = > x = 6 \times 100 \\ \\ = > x = 600 \\ \\ ans) \: the \: maximum \: maks \: to \: be \: obtained \\ are \: 600 \: marks. \\ \\ minimum \: marks \: = \: \frac{45}{100} \times 600 - 30 \\ \\ = > 270 - 30 \\ \\ = > 240 \\ \\ ans \: ) \: the \: minimum \: marks \: to \: be \: \\ obtained \: are \: 240 \: marks.\end{gathered}

letthemaximummarksbex

percentgeofmarksobtainedbyrahul=45%

marksobtainedbyhim=

100

45

x

sincethemarksobtainedbyhimare30morethanminimummarks

minimummarks=

100

45

x−30−−−−−(1)

percentageofmarksobtainedbyjohn=24%

marksobtainedbyhim=

100

24

x

sincethemarksobtainedbyhimare96lessthantheminimummarks

minimummarks=

100

24

x+96−−−−−−(2)

fromequations(1)and(2)weget

100

45

x−30=

100

24

x+96

=>

100

45

x−

100

24

x=96+30

=>

100

21

x=126

=>x=126×

21

100

=>x=6×100

=>x=600

ans)themaximummakstobeobtained

are600marks.

minimummarks=

100

45

×600−30

=>270−30

=>240

ans)theminimummarkstobe

obtainedare240marks.

__________________________________

#BE BRAINLY

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