model 7 marks scored in examinations Section-Wise Topic Notes With Detailed Explanation And Example Questions

Answer: (a)

Let the number of students in the class be 100.

Number of students in Biology = 72 and number of students in Maths = 44.

Number of students opting for both subjects = 72 + 44 – 100 = 16

Since, When 16 students opt for both subjects, total number of students = 100

When 40 students opt for both subjects,

total number of students = $100/16 × 40$ = 250

Answer: (b)

46% of 500 = ${500 × 46}/100 = 230$

32% of 300 = ${300 × 32}/100 = 96$

Required marks = 230 – 96 = 134

Let x% of 200 = 134

${200 × x}/100 = 134$

2x = 134 ⇒ x = $134/2$ = 67%

Answer: (c)

Difference of percentages of maximum marks obtained by two candidates = 32% – 20% = 12%

Difference of scores between two candidates = 30 +42 = 72

12% of maximum marks = 72

Maximum marks = ${72 × 100}/12$ = 600

Pass marks = 20% of 600 + 30

= 120 + 30 = 150

Required percentage = $150/600 × 100 = 25%$

n = 32%, m=20%, p=30, q = 42.

Full Marks = $100/{n – m} × (p + q)$

= $100/{(32 - 20)} × (30 + 42)$

= $100/32 × 72$ = 600

Pass marks =20% of 600 + 30

= 120 + 30 = 150

Required percentage = $150/600 × 100$ = 25%

Answer: (a)

Let the maximum marks be x.

According to the question,

40% of x = 90 + 10

x = ${100 × 100}/40 = 250$

Using Rule 24,

a = 40%, b = 90, c = 10

Maximum marks = ${(b + c)}/a × 100$

= ${(90 + 10)}/40 × 100 = 250$

Answer: (b)

Let maximum marks be x, then,

${36 × x}/100 = 113 + 85 = 198$

$x = {198 × 100}/36 = 550$

Using Rule 24,

a = 36%, b = 113, c = 85

Maximum marks = ${(b + c)}/a × 100$

= ${(113 + 85)}/36 × 100$

= $198/36 × 100$ = 550

Answer: (a)

Let the total marks be x.

According to the question,

25% of x + 40 = 33% of x

(33 – 25)% of x = 40

8% of x = 40

x = ${40 × 100}/8$ = 500