Aptitude - Average - Discussion
Discussion Forum : Average - General Questions (Q.No. 15)
15.
A pupil's marks were wrongly entered as 83 instead of 63. Due to that the average marks for the class got increased by half (1/2). The number of pupils in the class is:
Answer: Option
Explanation:
Let there be x pupils in the class.
| Total increase in marks = |  |
x x | 1 |  |
= | x |
| 2 | 2 |
 |
x | = (83 - 63)  |
x | = 20 x= 40. |
| 2 | 2 |
Discussion:
123 comments Page 3 of 13.
Adnan said:
1 decade ago
Let avg=y, no. of pupils=x, sum of marks=s
Case 1) y= s/x -------------(i)
Case 2) y + 1/2= (s+20)/x ----------(ii)
Solving eq. i & ii gives,x=40
Case 1) y= s/x -------------(i)
Case 2) y + 1/2= (s+20)/x ----------(ii)
Solving eq. i & ii gives,x=40
Sowjanya said:
1 decade ago
If solution A which has milk to water in the ratio 7:4 is mixed with solution B which has the ratio of water to milk as 9:2 such that the ratio becomes 1:1, then in what ratio were they mixed?
A) 2:3 B) 7:3 C) 1:1 D) 3:4.
Please solve this.
A) 2:3 B) 7:3 C) 1:1 D) 3:4.
Please solve this.
Ash said:
1 decade ago
Karthy is right....
Vishal said:
1 decade ago
@ sowjanya...
7x/4y +2x/9y=1/1
we get 2:3
7x/4y +2x/9y=1/1
we get 2:3
Tukaram D. Yedale said:
1 decade ago
Hi guys !
Let no.of people = x (83/x) - (63/x) = 1/2 after solving this we get x= 40 it is very easy method to understand is't it ?
Let no.of people = x (83/x) - (63/x) = 1/2 after solving this we get x= 40 it is very easy method to understand is't it ?
Mohammed Aijaz said:
1 decade ago
Hi friends.
Let us consider p be the no. of pupil, x be the average.
Case1: original marks 63/p=x ----- average of case 1.
Case2: new marks 83/p=x+1/2 ----average of case 2.
2-1 => (83/p-63/p)=x+1/2-x
=> 20/p=1/2
=> p=40.
Let us consider p be the no. of pupil, x be the average.
Case1: original marks 63/p=x ----- average of case 1.
Case2: new marks 83/p=x+1/2 ----average of case 2.
2-1 => (83/p-63/p)=x+1/2-x
=> 20/p=1/2
=> p=40.
Megha said:
1 decade ago
Hai friends,
Let average marks = x.
Total number of students = n, then
total mark = x*n.
According to the question the difference in the marks is = 83-63 = 20.
x*n/n=x.......(Can you remember that)
Here the difference in the marks is 20 and the average is increased by half so that x+1/2.
So, ((x*n)+20)/n = x+1/2.
Cross multiply we get,
2xn+40 = 2xn+n.
From that we get total number of students,
n=40.
Let average marks = x.
Total number of students = n, then
total mark = x*n.
According to the question the difference in the marks is = 83-63 = 20.
x*n/n=x.......(Can you remember that)
Here the difference in the marks is 20 and the average is increased by half so that x+1/2.
So, ((x*n)+20)/n = x+1/2.
Cross multiply we get,
2xn+40 = 2xn+n.
From that we get total number of students,
n=40.
Ravi said:
1 decade ago
Let total of other students be x and number of students be n,
(x+83)/n - (x+63)/n = 1/2.
(x+83-x-63)/n = 1/2.
20/n = 1/2.
Therefore n = 40.
(x+83)/n - (x+63)/n = 1/2.
(x+83-x-63)/n = 1/2.
20/n = 1/2.
Therefore n = 40.
Anonymous said:
1 decade ago
Before..
Av = Sum/num ------(1).
After..
Av+1/2 = (Sum+20)/num -------(2).
Putting value of (1) in (2).
Sum/num + 1/2 = Sum/num + 20/num.
num = 40.
Av = Sum/num ------(1).
After..
Av+1/2 = (Sum+20)/num -------(2).
Putting value of (1) in (2).
Sum/num + 1/2 = Sum/num + 20/num.
num = 40.
Surbhi thakur said:
1 decade ago
Let the total avg is x and it's given that avg increase by 0.5 then,
x: (3x/2)-83+63
-(x/2): -20
So, x:40 Answer, in question 63 marks are correct so we use +sign and 83 are incorrect so we use -ve sign.
x: (3x/2)-83+63
-(x/2): -20
So, x:40 Answer, in question 63 marks are correct so we use +sign and 83 are incorrect so we use -ve sign.
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