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Eight candidate take a competition exam two times. The increase in average point of all the candidate from first exam to second exam is 35%. The difference of the total point to first seven candidate in two exam is 20% of their total point in second exam. If in the first exam the point of the 8th candidate were mistakenly written 1/9 times more than the original. What is the actual percentage increase in his points ? 
1. 127.77%
2. 152%
3. 27.77%
4. 97.6%

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Correct Answer - Option 1 : 127.77%

Given:

The increase in average point of all the candidate from first exam to second exam is 35%.

The difference of the total point to first seven candidate in two exam is 20% of their total point in second exam. 

Formula used:

Average = sum of observation/Number of observation

Calculation:

The increase in average point of all the candidate from first exam to second exam is 35%.

The difference of the total point to first seven candidate in two exam = 20% of their total point in second exam

For first seven candidate

Total point in second exam – Total point in first exam = 20% of total point in second exam 

Or, Total points in the first exam = 0.8 × total point in second exam

Increase the total points of 7 candidate is 25%

Let the increase in the point of 8th candidate from first exam to second exam = Z%

⇒ 35% = (7 × 25% + 1 × Z%)/8

⇒ 280 – 175 = Z%

⇒ z = 105%

Let point of 8th candidate in first exam = 100p

Then, point in second exam = 100p + 105P

⇒ 205p

Point in first test miscalculated as 1/9 more, which means

⇒ (1 + 1/9)

⇒ 10/9 (Actual point)

⇒ 100p = 10/9(Actual point)

Actual point = 90p

Final point = 205p

Actual increase in point = (205 – 90)

⇒ 115p

Increase in percentage 

⇒ (115p/90p)× 100

⇒ 127.77%

∴ The actual percentage increase in his points is 127.77%.

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