Percentage
- A candidate who scores 30 percent fails by 5 marks, while another candidate who scores 40 per cent marks gets 10 more than minimum pass marks. The minimum marks required to pass are
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Let the full marks in that examination were x.
According to the question,= 30x + 5 = 40x - 10 100 100 = 3x - 4x = 10 + 5 10 10 ⇒ x = 15 10
∴ x = 150
∴ Minimum pass marks⇒ 30 × 150 + 5 = 50 100
Aliter : Using Rule 22,
m = 30%, n = 40%,
p = 5, q = 10.
Maximum marks= 100 × (p + q) (n - m) = 100 × (5 + 10) = 150 (40 - 30)
∴ Minimum passing marks= 150 × 30 + 5 = 45 + 5 = 50 100 Correct Option: A
Let the full marks in that examination were x.
According to the question,= 30x + 5 = 40x - 10 100 100 = 3x - 4x = 10 + 5 10 10 ⇒ x = 15 10
∴ x = 150
∴ Minimum pass marks⇒ 30 × 150 + 5 = 50 100
Aliter : Using Rule 22,
m = 30%, n = 40%,
p = 5, q = 10.
Maximum marks= 100 × (p + q) (n - m) = 100 × (5 + 10) = 150 (40 - 30)
∴ Minimum passing marks= 150 × 30 + 5 = 45 + 5 = 50 100
- In an examination, 60% of the candidates passed in English and 70% of the candidates passed in Mathematics, but 20% failed in both of these subjects. If 2500 candidates passed in both the subjects, the number of candidates who appeared at the examination was
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Let the total number of candidates = x
∴ Number of candidates passed in English = 0.6x
Number of candidates passed in Maths = 0.7x
Number of candidates failed in both subjects = 0.2x
Number of candidates passed in at least one subject
= x – 0.2x = 0.8x
∴ 0.6 x + 0.7x – 2500 = 0.8 x
⇒ 1.3x – 0.8x = 2500
⇒ 0.5x = 2500⇒ x = 2500 = 5000 0.5 Correct Option: D
Let the total number of candidates = x
∴ Number of candidates passed in English = 0.6x
Number of candidates passed in Maths = 0.7x
Number of candidates failed in both subjects = 0.2x
Number of candidates passed in at least one subject
= x – 0.2x = 0.8x
∴ 0.6 x + 0.7x – 2500 = 0.8 x
⇒ 1.3x – 0.8x = 2500
⇒ 0.5x = 2500⇒ x = 2500 = 5000 0.5
- In a test a student got 30% marks and failed by 25 marks. In the same test another student got 40% marks and secured 25 marks more than the essential minimum pass marks. The maximum marks for the test were
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Let the maximum marks in the examination = x.
According to the question,= 40x - 30x = 50 100 100 = 10x = 50 100 ⇒ x = 50 × 100 = 500 10
Aliter : Using Rule 22,
m = 30%, n = 40%, p = 25 and q = 25
∴ Maximum marks= 100 × (p + q) (n - m) = 100 × (25 + 25) = 500 (40 - 30) Correct Option: C
Let the maximum marks in the examination = x.
According to the question,= 40x - 30x = 50 100 100 = 10x = 50 100 ⇒ x = 50 × 100 = 500 10
Aliter : Using Rule 22,
m = 30%, n = 40%, p = 25 and q = 25
∴ Maximum marks= 100 × (p + q) (n - m) = 100 × (25 + 25) = 500 (40 - 30)
- In an examination 80% candidates passed in English and 85% candidates passed in Mathematics. If 73% candidates passed in both these subjects, then what per cent of candidates failed in both the subjects ?
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Let total candidates be 'x'
Percentage of the candidates passing in English or Mathematics or both
= n(E) + n(M) – n (E ∩ M)
= 80 + 85 – 73 = 92
⇒ Percentage of candidates who failed in both the subjects
= 100 – 92 = 8 or 8%Correct Option: A
Let total candidates be 'x'
Percentage of the candidates passing in English or Mathematics or both
= n(E) + n(M) – n (E ∩ M)
= 80 + 85 – 73 = 92
⇒ Percentage of candidates who failed in both the subjects
= 100 – 92 = 8 or 8%
- In an examination, 35% of the candidates failed in Mathematics and 25% in English. If 10% failed in both Mathematics and English, then how much percent of candidates passed in both the subjects ?
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Percentage of students who failed in Maths or English or both
= (25 + 35 – 10)% = 50%
∴ Required percentage
= (100 – 50)% = 50%
Aliter : Using Rule 23,
a = 35%, b = 25% and c = 10%
∴ Passed candidates in both the subjects.
= 100 – (a + b – c)%
= 100 – (35 + 25 – 10)%
= 100 – 50 = 50%Correct Option: A
Percentage of students who failed in Maths or English or both
= (25 + 35 – 10)% = 50%
∴ Required percentage
= (100 – 50)% = 50%
Aliter : Using Rule 23,
a = 35%, b = 25% and c = 10%
∴ Passed candidates in both the subjects.
= 100 – (a + b – c)%
= 100 – (35 + 25 – 10)%
= 100 – 50 = 50%
