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 5 of 13.
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.
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.
Durga said:
7 years ago
Avg = (sum of observations)/(no. of observations).
Given,
Avg = (x+63)/n ------> (1)
Avg + 1/2 = (x+83)/n ------> (2) (Avg increased by half).
Substituting eqn 1 in eqn 2,
(x+63)/n + 1/2 = (x+83)/n.
(x+83)/n - (x+63)/n = 1/2.
n = 40.
Given,
Avg = (x+63)/n ------> (1)
Avg + 1/2 = (x+83)/n ------> (2) (Avg increased by half).
Substituting eqn 1 in eqn 2,
(x+63)/n + 1/2 = (x+83)/n.
(x+83)/n - (x+63)/n = 1/2.
n = 40.
Ram said:
5 years ago
Average = total marks/number of pupils.
Average=x+63/number of pupils -------------> (1)
1.5 average=x+83/number of pupils ---------> (2)
eqn(2)-eqn(1).
0.5 = 20/number of pupils.
Number of pupils =20/0.5,
The number of pupils = 40.
Average=x+63/number of pupils -------------> (1)
1.5 average=x+83/number of pupils ---------> (2)
eqn(2)-eqn(1).
0.5 = 20/number of pupils.
Number of pupils =20/0.5,
The number of pupils = 40.
Selva ganesh said:
4 years ago
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.
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.
(1)
Selva ganesh said:
4 years ago
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.
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.
(1)
MAHESH BHOI said:
9 years ago
LET, X=TOTAL MARKS AND N=NUMBER OF STUDENT.
AS GIVEN INCREASE AVG+1/2 WHEN WRONGLY ADDED.
AVG + 1/2 = X/N.........1
LET CORRECT IT,
AVG = X - 63 + 83/N........2
FROM 1 AND 2.
X/N-1/2 = X/N-20/N,
1/2 = 20/N,
N = 40.
AS GIVEN INCREASE AVG+1/2 WHEN WRONGLY ADDED.
AVG + 1/2 = X/N.........1
LET CORRECT IT,
AVG = X - 63 + 83/N........2
FROM 1 AND 2.
X/N-1/2 = X/N-20/N,
1/2 = 20/N,
N = 40.
Nik said:
1 decade ago
Avg = sum of marks of each studnt/total no of studnts
= s/n
according to the given condition
(s+20)/n = (s/n)+0.5 ....(here 1/2 is considered as 0.5)
cross multiplying we get,
s+20 = s + 0.5n
so, n = 40
= s/n
according to the given condition
(s+20)/n = (s/n)+0.5 ....(here 1/2 is considered as 0.5)
cross multiplying we get,
s+20 = s + 0.5n
so, n = 40
Parth said:
6 years ago
s= other student.
n = number of student.
avg (when its 83) = avg(when its 63)+1/2.
(s+83)/n = (s+63)/n + 1/2,
(s+83)/n = [(s+63) + 0.5n]/n,
s+83 = s+63+0.5n,
83 = 63+0.5n,
0.5n = 83-63,
n = 20/0.5.
n=20.
n = number of student.
avg (when its 83) = avg(when its 63)+1/2.
(s+83)/n = (s+63)/n + 1/2,
(s+83)/n = [(s+63) + 0.5n]/n,
s+83 = s+63+0.5n,
83 = 63+0.5n,
0.5n = 83-63,
n = 20/0.5.
n=20.
Shubham Singh said:
8 years ago
The simplest logic is;
If by increasing 20 marks(83-63) in the class the average increases by it's half = x+x/2.
Average of total marks = x
Average of total marks + 20 = x + x/2
Thus 20 = x/2;
=> x = 40.
If by increasing 20 marks(83-63) in the class the average increases by it's half = x+x/2.
Average of total marks = x
Average of total marks + 20 = x + x/2
Thus 20 = x/2;
=> x = 40.
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