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In an examination, six subjects - A, B, C, D, E, and F have equal maximum marks. The number of marks scored by one particular candidate in subject A is 20% less than his marks in subject F. The ratio of marks scored by the same candidate in subject B to that in C is 4 : 5 and that in D to E is 3 : 4. The number of marks scored by this candidate in subject E is 25% more than that of F. He scored 65 marks in the subject C. He scored 436/9 % in the examination and the average of marks scored by him in all the subjects is 218/3.

Important for :

1

D

From the following common explanation, we getThe total marks = 900

Hence, option D is correct.

Let the marks scored by the candidate in the subject F = 10x then

The marks scored by the candidate in the subject A = 80% of 10x = 8x

The ratio of marks scored in B to that in C is 4 : 5 and that D to E is 3 : 4. The number of marks scored by the candidate in E is 25% more than that of F

In E, the marks obtained = 125% of 10x = | 25x | = 4a |

2 |

Then, the marks obtained in D = | 3 × 25x | = | 75x |

2 × 4 | 8 |

The marks obtained in C = 5y = 65

The marks obtained in B = 4y = | 65 × 4 | = 52 |

5 |

According to the question,

A + B + C + D + E + F = | 218 × 6 |

3 |

8x + 52 + 65 + | 75x | + | 25x | + 10x = 436 |

8 | 2 |

8x + | 75x | + | 25x | + 10x= 436 – 52 – 65 = 319 |

8 | 2 |

(64x + 75x + 100x + 80x) = 319 × 8

x = 8

The total marks = z then 48 | 4 | % of z = 436 |

9 |

By solving, z = | 436 × 900 | = 900 |

436 |

2

D

From the following common explanation, we getThe marks of each subject = | 900 | = 150 |

6 |

The marks obtained by the candidate in the subject E

= | 25x | = 100 |

2 |

The reqd. % = | 100 × 100 | = 66.66% |

150 |

Hence, option D is correct.

Let the marks scored by the candidate in the subject F = 10x then

The marks scored by the candidate in the subject A = 80% of 10x = 8x

The ratio of marks scored in B to that in C is 4 : 5 and that D to E is 3 : 4. The number of marks scored by the candidate in E is 25% more than that of F

In E, the marks obtained = 125% of 10x = | 25x | = 4a |

2 |

Then, the marks obtained in D = | 3 × 25 | = | 75x |

2 × 4 | 8 |

The marks obtained in C = 5y = 65

The marks obtained in B = 4y = | 65 × 4 | = 52 |

5 |

According to the question,

A + B + C + D + E + F = | 218 × 6 |

3 |

8x + 52 + 65 + | 75x | + | 25x | + 10x = 436 |

8 | 2 |

8x + | 75x | + | 25x | + 10x= 436 – 52 – 65 = 319 |

8 | 2 |

(64x + 75x + 100x + 80x) = 319 × 8

x = 8

The total marks = z then 48 | 4 | % of z = 436 |

9 |

By solving, z = | 436 × 900 | = 900 |

436 |

3

B

From the following common explanation, we getThe number of marks obtained by the candidate in the subject B = 52

The number of marks obtained by the candidate in the subject D

= | 75x | = 75 |

8 |

The required difference = 75 – 52 = 23

Hence, option B is correct.

Let the marks scored by the candidate in the subject F = 10x then

The marks scored by the candidate in the subject A = 80% of 10x = 8x

The ratio of marks scored in B to that in C is 4 : 5 and that D to E is 3 : 4. The number of marks scored by the candidate in E is 25% more than that of F

In E, the marks obtained = 125% of 10x = | 25x | = 4a |

2 |

Then, the marks obtained in D = | 3 × 25 | = | 75x |

2 × 4 | 8 |

The marks obtained in C = 5y = 65

The marks obtained in B = 4y = | 65 × 4 | = 52 |

5 |

According to the question,

A + B + C + D + E + F = | 218 × 6 |

3 |

8x + 52 + 65 + | 75x | + | 25x | + 10x = 436 |

8 | 2 |

8x + | 75x | + | 25x | + 10x= 436 – 52 – 65 = 319 |

8 | 2 |

(64x + 75x + 100x + 80x) = 319 × 8

x = 8

The total marks = z then 48 | 4 | % of z = 436 |

9 |

By solving, z = | 436 × 900 | = 900 |

436 |

4

A

From the following common explanation, we getThe sum of the marks obtained by the candidate in the subject E and F together = 100 + 80

The reqd. average = | 180 | = 90 |

2 |

Hence, option A is correct.

Let the marks scored by the candidate in the subject F = 10x then

The marks scored by the candidate in the subject A = 80% of 10x = 8x

The ratio of marks scored in B to that in C is 4 : 5 and that D to E is 3 : 4. The number of marks scored by the candidate in E is 25% more than that of F

In E, the marks obtained = 125% of 10x = | 25x | = 4a |

2 |

Then, the marks obtained in D = | 3 × 25 | = | 75x |

2 × 4 | 8 |

The marks obtained in C = 5y = 65

The marks obtained in B = 4y = | 65 × 4 | = 52 |

5 |

According to the question,

A + B + C + D + E + F = | 218 × 6 |

3 |

8x + 52 + 65 + | 75x | + | 25x | + 10x = 436 |

8 | 2 |

8x + | 75x | + | 25x | + 10x= 436 – 52 – 65 = 319 |

8 | 2 |

(64x + 75x + 100x + 80x) = 319 × 8

x = 8

The total marks = z then 48 | 4 | % of z = 436 |

9 |

By solving, z = | 436 × 900 | = 900 |

436 |

5

D

From the following common explanation, we get
The number of marks obtained by the candidate in the subject C = 65

The number of marks obtained by the candidate in the subject E = 100

The reqd. % = | (100 – 65) × 100 | = 35% |

100 |

Hence, option D is correct.

Let the marks scored by the candidate in the subject F = 10x then

The marks scored by the candidate in the subject A = 80% of 10x = 8x

The ratio of marks scored in B to that in C is 4 : 5 and that D to E is 3 : 4. The number of marks scored by the candidate in E is 25% more than that of F

In E, the marks obtained = 125% of 10x = | 25x | = 4a |

2 |

Then, the marks obtained in D = | 3 × 25 | = | 75x |

2 × 4 | 8 |

The marks obtained in C = 5y = 65

The marks obtained in B = 4y = | 65 × 4 | = 52 |

5 |

According to the question,

A + B + C + D + E + F = | 218 × 6 |

3 |

8x + 52 + 65 + | 75x | + | 25x | + 10x = 436 |

8 | 2 |

8x + | 75x | + | 25x | + 10x= 436 – 52 – 65 = 319 |

8 | 2 |

(64x + 75x + 100x + 80x) = 319 × 8

x = 8

The total marks = z then 48 | 4 | % of z = 436 |

9 |

By solving, z = | 436 × 900 | = 900 |

436 |

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