Learn Quantitative Aptitude Weight Height Marks, SBI Clerk Exam Quantitative Aptitude Weight Height Marks Exercise, Free Online Weight Height Marks Lessons

SBI Clerk Exam Quantitative Aptitude Study Material

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Weight Height Marks

In a history examination, the average for the entire class was 80% marks. If 10% of the students scored 95 marks an 20 % scored 90 marks, then what were the average marks of the remaining students of the class?

a) 65.5
b) 85
c) 75
d) 72.5

✔Answer

✔Solution

Correct Answer Is : 75

Solution Is : (C ) Let the total number of students be 100. Therefore, Required be 100. = (80100-1095-20*90)/70 = (8000-950-1800)/70 =5250/70 = 75

Neha` s weight is 140% of Tina`s weight. Mina`s weight is 90% of weight. Lina weight twice as much as Tina. If Neha`s weight is x% of Mina`s weight then x is eqal to

a) 64 2/9
b) `77 7/9
c) 90
d) 128 4/7

✔Answer

✔Solution

Correct Answer Is : `77 7/9

Solution Is :

A cricketer has a mean score of 60 runs in 10 innings. find out how many are runs to be scored in the eleventh innings to raise the mean score to 62?

a) 80
b) 81
c) 83
d) 82

✔Answer

✔Solution

Correct Answer Is : 82

Solution Is : Total runs scored in 10 innings = 60x10=600
lets runs scored in 11th innings = x
now total runs in 11 innings
600+x=11x62
600+x=682
x=682-600=82

In a history examination, the average for the entire class was 80% marks. If 10% of the students scored 95 marks an 20 % scored 90 marks, then what were the average marks of the remaining students of the class?

a) 65.5
b) 85
c) 75
d) 72.5

✔Answer

✔Solution

Correct Answer Is : 75

Solution Is : (C ) Let the total number of students be 100. Therefore, Required be 100. = (80100-1095-20*90)/70 = (8000-950-1800)/70 =5250/70 = 75

Neha` s weight is 140% of Tina`s weight. Mina`s weight is 90% of weight. Lina weight twice as much as Tina. If Neha`s weight is x% of Mina`s weight then x is eqal to

a) 64 2/9
b) `77 7/9
c) 90
d) 128 4/7

✔Answer

✔Solution

Correct Answer Is : `77 7/9

Solution Is :

A cricketer has a mean score of 60 runs in 10 innings. find out how many are runs to be scored in the eleventh innings to raise the mean score to 62?

a) 80
b) 81
c) 83
d) 82

✔Answer

✔Solution

Correct Answer Is : 82

Solution Is : Total runs scored in 10 innings = 60x10=600
lets runs scored in 11th innings = x
now total runs in 11 innings
600+x=11x62
600+x=682
x=682-600=82

In a history examination, the average for the entire class was 80% marks. If 10% of the students scored 95 marks an 20 % scored 90 marks, then what were the average marks of the remaining students of the class?

a) 65.5
b) 85
c) 75
d) 72.5

✔Answer

✔Solution

Correct Answer Is : 75

Solution Is : (C ) Let the total number of students be 100. Therefore, Required be 100. = (80100-1095-20*90)/70 = (8000-950-1800)/70 =5250/70 = 75

Neha` s weight is 140% of Tina`s weight. Mina`s weight is 90% of weight. Lina weight twice as much as Tina. If Neha`s weight is x% of Mina`s weight then x is eqal to

a) 64 2/9
b) `77 7/9
c) 90
d) 128 4/7

✔Answer

✔Solution

Correct Answer Is : `77 7/9

Solution Is :

A cricketer has a mean score of 60 runs in 10 innings. find out how many are runs to be scored in the eleventh innings to raise the mean score to 62?

a) 80
b) 81
c) 83
d) 82

✔Answer

✔Solution

Correct Answer Is : 82

Solution Is : Total runs scored in 10 innings = 60x10=600
lets runs scored in 11th innings = x
now total runs in 11 innings
600+x=11x62
600+x=682
x=682-600=82

In a history examination, the average for the entire class was 80% marks. If 10% of the students scored 95 marks an 20 % scored 90 marks, then what were the average marks of the remaining students of the class?

a) 65.5
b) 85
c) 75
d) 72.5

✔Answer

✔Solution

SBI Clerk Exam Quantitative Aptitude Study Material

Weight Height Marks

In a history examination, the average for the entire class was 80% marks. If 10% of the students scored 95 marks an 20 % scored 90 marks, then what were the average marks of the remaining students of the class?

Correct Answer Is : 75

Solution Is : (C ) Let the total number of students be 100. Therefore, Required be 100. = (80*100-10*95-20*90)/70 = (8000-950-1800)/70 =5250/70 = 75

Neha` s weight is 140% of Tina`s weight. Mina`s weight is 90% of weight. Lina weight twice as much as Tina. If Neha`s weight is x% of Mina`s weight then x is eqal to

Correct Answer Is : `77 7/9

Solution Is :

A cricketer has a mean score of 60 runs in 10 innings. find out how many are runs to be scored in the eleventh innings to raise the mean score to 62?

Correct Answer Is : 82

Solution Is : Total runs scored in 10 innings = 60x10=600 lets runs scored in 11th innings = x now total runs in 11 innings 600+x=11x62 600+x=682 x=682-600=82

In a history examination, the average for the entire class was 80% marks. If 10% of the students scored 95 marks an 20 % scored 90 marks, then what were the average marks of the remaining students of the class?

Correct Answer Is : 75

Solution Is : (C ) Let the total number of students be 100. Therefore, Required be 100. = (80*100-10*95-20*90)/70 = (8000-950-1800)/70 =5250/70 = 75

Neha` s weight is 140% of Tina`s weight. Mina`s weight is 90% of weight. Lina weight twice as much as Tina. If Neha`s weight is x% of Mina`s weight then x is eqal to

Correct Answer Is : `77 7/9

Solution Is :

A cricketer has a mean score of 60 runs in 10 innings. find out how many are runs to be scored in the eleventh innings to raise the mean score to 62?

Correct Answer Is : 82

Solution Is : Total runs scored in 10 innings = 60x10=600 lets runs scored in 11th innings = x now total runs in 11 innings 600+x=11x62 600+x=682 x=682-600=82

In a history examination, the average for the entire class was 80% marks. If 10% of the students scored 95 marks an 20 % scored 90 marks, then what were the average marks of the remaining students of the class?

Correct Answer Is : 75

Solution Is : (C ) Let the total number of students be 100. Therefore, Required be 100. = (80*100-10*95-20*90)/70 = (8000-950-1800)/70 =5250/70 = 75

Neha` s weight is 140% of Tina`s weight. Mina`s weight is 90% of weight. Lina weight twice as much as Tina. If Neha`s weight is x% of Mina`s weight then x is eqal to

Correct Answer Is : `77 7/9

Solution Is :

A cricketer has a mean score of 60 runs in 10 innings. find out how many are runs to be scored in the eleventh innings to raise the mean score to 62?

Correct Answer Is : 82

Solution Is : Total runs scored in 10 innings = 60x10=600 lets runs scored in 11th innings = x now total runs in 11 innings 600+x=11x62 600+x=682 x=682-600=82

In a history examination, the average for the entire class was 80% marks. If 10% of the students scored 95 marks an 20 % scored 90 marks, then what were the average marks of the remaining students of the class?

Correct Answer Is : 75

Solution Is : (C ) Let the total number of students be 100. Therefore, Required be 100. = (80*100-10*95-20*90)/70 = (8000-950-1800)/70 =5250/70 = 75

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Solution Is : (C ) Let the total number of students be 100. Therefore, Required be 100. = (80100-1095-20*90)/70 = (8000-950-1800)/70 =5250/70 = 75

Solution Is : Total runs scored in 10 innings = 60x10=600
lets runs scored in 11th innings = x
now total runs in 11 innings
600+x=11x62
600+x=682
x=682-600=82

Solution Is : (C ) Let the total number of students be 100. Therefore, Required be 100. = (80100-1095-20*90)/70 = (8000-950-1800)/70 =5250/70 = 75

Solution Is : Total runs scored in 10 innings = 60x10=600
lets runs scored in 11th innings = x
now total runs in 11 innings
600+x=11x62
600+x=682
x=682-600=82

Solution Is : (C ) Let the total number of students be 100. Therefore, Required be 100. = (80100-1095-20*90)/70 = (8000-950-1800)/70 =5250/70 = 75

Solution Is : Total runs scored in 10 innings = 60x10=600
lets runs scored in 11th innings = x
now total runs in 11 innings
600+x=11x62
600+x=682
x=682-600=82

Solution Is : (C ) Let the total number of students be 100. Therefore, Required be 100. = (80100-1095-20*90)/70 = (8000-950-1800)/70 =5250/70 = 75