Question

A candidate who gets 36% marks in an examination fails by 24 marks but another candidate, who gets 43% marks , get 18 more marks than the minimum pass marks . Find the maximum marks and percentage of pass marks
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Answer 1

Step-by-step explanation:

Given:-

• A candidate gets 36% marks in an examination fails by 24 marks.

• Another candidate gets 43% marks, get 18 more marks than the minimum pass marks.

To Find:-

• The maximum marks

• The pass marks.

Solution:-

$\sf Let \: the \: maximum \: marks \: be x$

$\sf And \: pass \: marks \: be y$

$\sf \underline{For \: first \: candidate:-}$

$\sf He\: scores \: 36\% \: marks \: and \: fails \: by\: 24 marks$

$\sf \implies 36\% \: of \: total \: marks = Pass \: marks - 24$

$\sf \implies 36\% \: of \: x = y - 24$

$\sf \implies \dfrac{36}{100} times x + 24 = y$

$\sf \implies \dfrac{36x}{100} + 24 = y......(i)$

$\sf \underline{For \: second \: candidate:-}$

$\sf He\: scores \: 43\% \: marks \: and \: gets\: 24\: marks\: more \: than \: pass \: marks$

$\sf \implies 43\% \: of \: total \: marks = Pass \: marks + 18$

$\sf \implies 43\% \: of \: x = y +18$

$\sf \implies \dfrac{43}{100} times x - 18 = y$

$\sf \implies \dfrac{43x}{100} -18= y......(ii)$

$\sf \underline{From\: equations \: (i) \: and \: (ii) :-}$

$\sf\dfrac{36x}{100} + 24 \: \: = \: \: \dfrac{43x}{100} - 18 \: \: = \: \: y$

$\sf \implies24 + 18 = \dfrac{43x}{100} - \dfrac{36x}{100}$

$\sf \implies42 = \dfrac{43x - 36x}{100}$

$\sf \implies \dfrac{7x}{100} = 42$

$\sf \implies x = 42 \times \dfrac{100}{7}$

$\sf \implies x = 6\times 100$

$\sf \implies x = 600$

$\sf Substituting \: x = 600\: in \: equation\: (i)$

$\sf \implies \dfrac{36x}{100} + 24 = y$

$\sf \implies \dfrac{36\times600}{100} + 24 = y$

$\sf \implies \dfrac{36x}{100} + 24 = y$

$\sf \implies 216 + 24 = y$

$\sf \implies y= 240$

We have:-

• x = 600

• y = 240

$\underline{\boxed{\pink{\rm \therefore Maximum \: marks = 600}}}$

$\underline{\boxed{\purple{\rm \therefore Pass \: marks = 240}}}$

Answer 2

$\large { \underline{ \sf { \red{ Given:- }}}}$

• A candidate who gets 36% marks in an examination fails by 24 marks but another candidate, who gets 43% marks , get 18 more marks than the minimum pass marks .

$\large { \underline{ \sf { \green{ To \: Find:- }}}}$

• Find the maximum marks and percentage of pass marks.

$\large { \underline{ \sf { \orange{ Directions:- }}}}$

$\sf \purple{Let:- }$

• Let the passing marks be 'x'.
• Let the Total marks be 'y'.

$\sf \purple{According \: to \: Question:- }$

• First candidate:- 36% of y + 24 = x
• Second candidate:- 43% of y - 18 = x

• Where, First Candidate = Second Candidate

$\sf \purple{So \: now:- }$

$\implies \mathtt{ \frac{36y}{100} + 24 = \frac{46y}{100} - 18}$

$\implies \mathtt{24 + 18 = \frac{46y}{100} - \frac{36y}{100} }$

$\implies \mathtt{42 = \frac{7y}{100} }$

$\implies \mathtt{42 \times 100 = 7y} \implies \mathtt{\frac{4200}{7} = y}$

$\implies \pink{ \mathtt{y = 600}}{ \bigstar}$

$\sf \purple{So \: further :- }$

$\implies \mathtt{ \frac{36y}{100} + 24 = x}$

$\implies \mathtt{ \frac{36}{100} \times 600+ 24 = x}$

$\implies \mathtt{ 36 \times 6+ 24 =x}$

$\implies \mathtt{ 216+ 24 =x}$

$\implies \mathtt \pink{ 240=x}{ \bigstar}$

$\large \pink { \boxed{ \sf{Hence, \: The \: total \: marks \: are \: 600. }}}$

$\large \purple { \boxed{ \sf{And \: the \: passing \: marks \: are \: 240. }}}$