Question

A candidate who gets 20% of the marks fails by 30 marks. But another candidate who gets 32%
marks get 42 marks more than are necessary for passing. Find maximum number of marks
18.
1. 400
2. 500
3. 800
4. 700
5.600​

Answer 1

Answer:- 5. 600

Step-by-step explanation:

Let total marks be = X

1st candidate = 20% of X

Pass marks = (20% of X) + 30

2nd candidate = 32% of X

Pass marks = (32% of X) – 42

ATQ:-

(20% of X) + 30 = (32% of X) – 42

20/100X + 30 = 32/100X – 42

X/5 + 30 = 8X/25 – 42

X/5 – 8X/25 = – 42 –30

–3X/25 = – 72

X = 72×25/3

X = 600

THEREFORE PASS MARKS = 600

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Answer 2

Answer:

✈Question✈

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A candidate who gets 20% of the marks fails by 30 marks. But another candidate who gets 32% marks get 42 marks more than are necessary for passing. Find maximum number of marks.

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✈Answer✈

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✍GIVEN✍

Let the Total marks be x

Minimum marks be y

• 20% of x = y - 30
• 32% of x = y + 42

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✍TO FIND✍

• Total marks=????

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✍SOLUTION✍

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According to given conditions;

32% of x-20% of x=y+42-(y-30)

=>12% of x=$\cancel{y}+42\ \cancel{-y}+30$

=>$\frac{12}{100}×x$=72

=>12x=72×100

=>x=$\frac{7200}{12}$

:. x= 600.

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So, the total marks is 600.