# Something based on Perentage. [closed]

A candidate who gets 20% marks in an examination fails by 30 marks but another candidate who gets 32% gets 42 marks more than the passing marks. Then the percentage of pass marks is:

1. 52%
2. 50%
3. 33%
4. 25%
• Note that you don't even have to do any calculations here; only one choice could possibly be right. – pjs36 Dec 28 '16 at 17:17
• You should explain what you don't get and explain what you have tried. This is insulting. – The Count Dec 28 '16 at 17:19
• Haha, nice observation @pjs36. Anyways Shubham, if you wanna solve it nonetheless, it is a simple case of a system of two linear equations with two variables, passing marks and total marks. – Shraddheya Shendre Dec 28 '16 at 17:21
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• Sorry people. I think I have ended up at the wrong place. I was simply looking for a site where any question can be discussed without considering the asker's ability and capability. This place is good, but, not for what I am looking for. Anyways, I got an explanation and an insight view into my question. And, I was not copy pasting my homework or something. I was simply looking for someone, may be at the wrong place, who could have simply explained this to me without any extra comments. Wrong Place Wrong Time – Shubham Raj Dec 28 '16 at 19:09

Passing marks = 20% of x + 30 ....(1)

As he needs 30 more marks wo added.

Passing marks = 32% of x - 42 .....(2)

As he has 42 more marks then passing so subtracted.

From (1) and (2)

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

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

12% of x = 72

x = $\frac{72 * 100}{12}$

x = 600

So total marks 600.

From equation (1)

Passing marks = 20% of x + 30

= $\frac{20}{100}$ * 600 + 30 = 120 + 30 = 150

Passing percentage = $\frac{150}{600} * 100$ = 25%

• good explanation. Thanx – Shubham Raj Dec 29 '16 at 12:27

If you fail at 20% and pass at 32% the percentage at which you pass is between these. Only one option satisfies this.

Otherwise solve the system:

$$0.2x = y - 30 \\ 0.32x = y+42$$