Question

Rohit secured 42% of the total marks and secured 10 marks more than the passing marks. In the same exam, Mohit scored 29% of the total marks and failed by 16 marks. What was the passing percentage?

Answer 1

Answer:

37%

Step-by-step explanation:

TM = Total Marks, PM = Passing Marks

By information on Rohit,

TM*42/100 = 10+PM ...(1)

By information on Mohit

TM*29/100 = PM-16..... (2)

Subtraction of (2) from (1) gives

TM*13/100 = 26 => TM = 200

PM = 74

Passing Percentage = (PM/TM) *100 = 37%

Answer 2

Answer:

Let the Total Number be n.

◗ Rohit : 42% – 10 = Pass

◗ Mohit : 29% + 16 = Pass

• We will always Subtract Percent.
• when there is Same Terms (Pass ; Pass, Fail ; Fail) then we Subtract the Marks.
• when there is Opposite Terms (Pass ; Fail or, Fail ; Pass) then we Add the Marks.

☯ According to the Question Now :

⇴ Total Marks × Percentage = Marks

⇴ n × (42% – 29%) = (10 + 16)

⇴ n × 13% = 26

⇴ n × 13/100 = 26

⇴ n × 1/100 = 2

⇴ n = 100 × 2

⇴ n = 200 (Total Marks)

Or, By Alternative

⇴ Total × 42% – 10 = Total × 29% + 16

• As Both will be Equal to Pass

⇴ n × 42% – 10 = n × 29% + 16

⇴ 42n% – 29n% = 16 + 10

⇴ 13n% = 26

⇴ 13n/100 = 26

⇴ n/100 = 2

⇴ n = 200 ⠀(Total Marks)

_________________________

☢ Pass Percentage :

⇢ Pass = 42% – 10 Marks

• Changing Marks into Percentage

⇢ Pass = 42% – (10/200 × 100)%

⇢ Pass = 42% – (10/2)%

⇢ Pass = 42% – 5%

⇢ Pass Percentage = 37%

Or, By Alternative :

⇢ Pass = 29% + 16 Marks

• Changing Marks into Percentage

⇢ Pass = 29% + (16/200 × 100)%

⇢ Pass = 29% + (16/2)%

⇢ Pass = 29% + 8%

⇢ Pass Percentage = 37%

∴ Passing Percentage in Exam was 37%.