### Discussion :: Average - General Questions (Q.No.15)

Android said: (Aug 30, 2010) | |

I didn't understand the logic. Please explain. |

Praveen said: (Nov 30, 2010) | |

Please explain. |

Arun said: (Dec 2, 2010) | |

Can you explain it briefly please ? |

Arunoid said: (Dec 2, 2010) | |

Please help the needy and me too. |

Krishna Kanth said: (Dec 13, 2010) | |

I didn't understand anything please please please some body explain. |

Rakesh said: (Jan 16, 2011) | |

If mark=63,then total mark(mark1+mark2+---+63) =(average mark*x)-(1) but in the qst given that enterd mark is 83 instead of 63,then avg is increased by 1/2(half), then total mark(mark1+mark2+----+83)=(avg mark+1/2)*x -(2) eq(2)-eq(1) 83-63=1/2*x; x=20*2=40 |

Swathy said: (Jan 21, 2011) | |

Rakesh I cant understand your logic can you explain clearly?. |

Nik said: (Feb 8, 2011) | |

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 |

Sachin Bende said: (Feb 12, 2011) | |

@ swathy rakesh's logic is absolutely correct.. explanation: =(avg mark+1/2)*x - (avg mark*x) taking "x" common we get, =x(avg mark + 1/2 - avg mark) cancelling avg mark hence,we get = x(1/2)= 1/2*x therefore, 83-63=1/2*x x=20*2=40 ok got it. |

Habib said: (Feb 24, 2011) | |

Frnz....dnt get confused boy that 1/2 factor !!!! see, (total marks)/(no. of pupils)= avg OR total marks = avg X no. of pupils Given, increase is of 1/2 , no. of pupils=x total marks increased : 83-63 = 20 Hence, for 1/2 mark avg increase, we use 20 = (1/2) * x OR x=40 |

Savinder said: (Jun 10, 2011) | |

In an exam, the average was found to be 50 marks. After detecting computer errors, the marks of 100 candidates had to be changed from 90 to 60 each and the average came down to 45. The total number of candidates who took the exam was:. Please solve. |

Nehanshu Raj Badholiya said: (Jul 15, 2011) | |

Rakesh's and habi bbhaijan both are correct so don't get confused. |

Ajay said: (Sep 3, 2011) | |

Ans of Savinder question Let total no of cand-x Dec in aver- 50-45=5 Total dec in numbers= 5x then, 5x= 100(90-60) x=600 ( Ans) |

M.V.Krishna/Palvoncha said: (Sep 9, 2011) | |

Hai nik. Your explanation is nice. |

Ravi said: (Sep 20, 2011) | |

Thanks habib. |

Priya said: (Sep 28, 2011) | |

Please trace where I went wrong. Let n be total number of students When marks was entered as 63 -----(x+63)/n=1 . . eqn1 When enterd as 83--------(x+83)/n=1.5. . . .eqn2 eqn1 --->n=x+63 eqn2---->n=(x+83)/1.5 equatin both x+63=(x+83)/1.5 1.5x+94.5=x+83 .5x=11.5 x=23 bak sub in eqn1 23+63=n n=86 Please help me. |

Yuvraj.. said: (Oct 21, 2011) | |

@priya: Increase in avg. marks of class * no. of students = increase in marks of a student---->eqn 1 Now; increase in avg marks of class = 1/2= 1.5 and increase in marks of a student = 83-63= 20 therefore from eqn 1 no. of students = increase in marks of a student/ increase in avg marks of class So, No. of students=20/1.5= 40 |

Sahithi said: (Dec 14, 2011) | |

Short cut method: Here given avg incresed 1/2 means 0.5 Taken 83 instead of 63 the differnce of avg incresed 83-63=20 Therefore how many 1/2 s in 20 ? for detailed 1 means 1/2+1/2 contain 2 halfs 20 contain 20*2=40 halfs Therefore n=40 pupils. |

Rav said: (Jan 7, 2012) | |

In the question above is "A pupil's" mean "A student" ? |

Karthy said: (Jan 25, 2012) | |

The average increase for one person is 0.5 Total increase is 20 Therefore Total no of pupil =20/0.5 = 40. |

Adnan said: (Apr 9, 2012) | |

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 |

Sowjanya said: (May 15, 2012) | |

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. |

Ash said: (Aug 17, 2012) | |

Karthy is right.... |

Vishal said: (Aug 20, 2012) | |

@ sowjanya... 7x/4y +2x/9y=1/1 we get 2:3 |

Tukaram D. Yedale said: (Oct 11, 2012) | |

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 ? |

Mohammed Aijaz said: (Jan 31, 2013) | |

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. |

Megha said: (Jul 4, 2013) | |

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. |

Ravi said: (Jul 6, 2013) | |

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. |

Anonymous said: (Jul 16, 2013) | |

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. |

Surbhi Thakur said: (Oct 1, 2013) | |

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. |

Phani said: (Oct 26, 2013) | |

Hi. I have come up with one more simple solution: 1st case: Lets say avg is X when marks are 63. So X = 63/n. 2nd case: When marks were noted as 83, the avg is increased by 0.5 so, X+0.5 = 83/n. Now if you subtract 2nd case - 1st case then (83-63) /n = X+0.5-X. So, 20/n = 0.5 and n = 20/0.5 so n = 40. Have a great day dudes. ! |

Adi said: (Jan 31, 2014) | |

Let total number be x. Then it is given that average for 63 is A. i.e, sum/x = A. When it is wrongly entered as 83 then. (sum-63+83) /x = A + 1/2. Then (sum- 20) /x =A + 1/2. (sum/x) - (20/x) =A+1/2. Now pick from above that sum/x=A. Now A- (20/x) =A + 1/2. Thus, it brings x=40. |

Midhun J said: (Feb 20, 2014) | |

It is actually easy if you get the proper steps, actually I kind of struggled a bit for myself to understand with the limited steps given in answer but I kind of figured out the steps. Note : I am just an aspiring student and so I don't know if my explanation is correct but its just method I found quite easy to understand Do it using following steps: Let the total marks without 1 pupils mark be = y. Let the total no. of pupil be = x. It is given that 1 pupil marks entered wrong as 83 and thus increases avg. marks by 1/2. Therefore, (y+83)/x = avg.marks + 1/2 -->(i). But, The pupil's actual mark is 63. (y+63)/x = avg.marks --> (ii). Sub (ii) for avg.marks in (i), then, (y+83)/x = (y+63)/x +1/2. => (y+83)/x - (y+63)/x =1/2. => (y + 83 - y - 63)/x = 1/2 [ i have put in brackets to show whole divided by 'x' ]. => (83- 63)/x =1/2. => 20/x = 1/2. => x = 40. Therefore, there are 40 pupils in class. |

Aayush said: (Jun 14, 2014) | |

I hope this helps you all. It's quite easy if u just go step by step according to the question. Let no.of pupils be x. Let total marks be y. Earlier y/x = Avg...eq(1). Later Marks increased by 20 (as 83-63=20) and the average increases by 1/2 Hence, (y+20)/x = (1/2+Avg)....eq(2). Now clearly we see the relation in the above 2 equations so hence we try to divide them and let's see what do we get, Dividing eq(1)/eq(2). (y/x)*(x/y+20) = (Avg)/(1/2+Avg). On solving. We get, y=40*Avg.....eq(3). No substitute this value of 'y' eq(3) in eq(1). i.e., y/x=Avg....eq(1). So after putting value of y it becomes, (40*Avg)/x=Avg. Hence, x=40. This is the easiest method I have found so far... |

Saji said: (Jul 5, 2014) | |

let x be the no.of students. When it is 83, the avg is increased by 1/2. So we get, x(avg+1/2) = 83 --(1). Also given that 63 is the correct mark so we get an equ like, x(avg) = 63 --(2). Now (1)-(2)we get, x(avg)+x(1/2)-x(avg) = 83-63. Therefore, x/2 = 20. x = 40. |

Dinesh said: (Jul 15, 2014) | |

Its very simple given difference of avg is half hence, (83/x)-(63/x) = 0.5. Hence by solve x = 40. |

Danny Blac said: (Oct 31, 2014) | |

@Midhun J. You are right. I can so much relate with your answer. |

Tasleem said: (Dec 13, 2014) | |

Average of the total as it is increased by 20 can be written as: = s+20/n. = s+20/n = s/n*1/2 (as the original avg is increased by 1/2). = 2(s+20) = s. Hence s = 40. |

Pritisha said: (Feb 9, 2015) | |

Let the total marks be x, Let the total no of students be x, and let the average marks of all the students be x. Increase in the total marks = 83-63 = 20. So, total marks/total students = avg marks. => x+40/x = x+1/2. x gets cancelled on both sides of the equation. Then the remaining terms are, => 20/x = 1/2. => x = 40. |

Yamuna said: (Feb 13, 2015) | |

Let x be the no. of pupil, y is the avg. 63/x = y, 83/x = y+1/2, By dividing these 2 equations you will get y value. Substitute y in 1st equation, you get x as 40. Simple. |

Rituparna Sengupta said: (Apr 24, 2015) | |

Let, average marks of class = x & also let, total no. of pupil in the class = A. When the mark is 63. Then, the total marks of all pupil = Ax. When, 83 is entered instead of 63, Then avg marks of the class = A(x+1/2). Then, A(x+1/2)-Ax = 83-63. => A/2 = 20. => A = 40. Option(C). |

Vignesh Bs said: (Apr 30, 2015) | |

Correct mark = 63. Wrongly entered = 83. The average marks for the class got increased = 1/2 => 0.5. A = For single pupil extra mark = 1*1/2 = 0.5. B = Extra added mark = 83 - 63 => 20. The number of pupils in the class "X"= B/A. Therefore x = 20/0.5; 0.5 =1/2. x = 20/(1/2). x = 20* 2. x = 40. The number of pupils in the class is 40. |

Shahin said: (Jun 24, 2015) | |

Very simple. Let no. of. pupil be x. New mark (wrongly entered) = 83. Replaced mark (correct mark) = 63. Increased avg = 0.5(1/2). Formula: New mark=Replaced mark+(Number of pupil*inc avg). We can apply this formula for this type of question. Now, 83 = 63+(x*0.5). = 83-63 = x*0.5. = 20 = x*0.5. = 20/(1/2) = x. = 20*2 = x. = 40 = x. Number of pupils in the class is 40. Hope it clear your doubts. |

San_B said: (Jul 20, 2015) | |

Let N be the total number of students. Total mark increase => 83-63 = 20. (because Old Sum-New Sum of marks will be affected only due to change of 83 to 63) __ (1). Now, average marks are increased by 1/2. It means each student's mark is increased by 1/2. (Average is equally applicable to all the students). Hence, total mark increased => 1/2*N __(2). From (1) and (2) , 20 = (1/2)*N. N = 40. |

Nitish Guleria said: (Aug 28, 2015) | |

First we have average formula = Sum of all the marks/No of students. Then for this question let be assume average as = A. And no of students = X. Now A.T.Q = A = 63 +mark1+. /X. equation no. 1 means average formula. No second he say that when marks increase by 83 then again A = 83+ mark1+. /x. But he said that average increased by 0.5. So A+0.5 = 83 + marks1+. /x. equation no 2. Now equate these equations you will find exact answer. |

Vivek Kumar said: (Sep 5, 2015) | |

Its simple, Let us assume that the total number of pupils in class be x+1, Now Let the sum of the marks of x students (whose marks have been entered correctly be S) , and let original average be A, then (S+63)/(x+1) = A; Equation 1. And (S+83)/(x+1) = (A+1/2); Equation 2. Divide Equation 1 by Equation 2; (S+63)/(S+83) = (2*A)/(2*A+1) ; Solving it, we get: S+63=40*A; Equation 3. Now, put value of Equation 3 in Equation 1, we get: (40*A)/(x+1) = A; Solve it, we get: x+1=40. And we have assumed the total students to be x+1. Hence total students = x+1 = 40. |

Er Dharmendra Kumar said: (Mar 19, 2016) | |

83-63/(1/2) = 2*20 = 40. |

Mohit Tomar said: (Apr 10, 2016) | |

Let as take, m is actual total marks of students and p is number of the pupil. Difference between average = 1/2. m/p - (m+20)/p = 1/2. p = 40. |

Maria Ninan said: (Apr 13, 2016) | |

But the question says increased by half, so shouldn't it mean the incorrect average was increased by half of the correct average? Thanks in advance |

Kanmani said: (Jun 23, 2016) | |

I too have that same doubt @Maria Ninan. Can anyone clarify it? |

Div said: (Jul 13, 2016) | |

Once read the question, it's stated that. Instead of 83, it's wrongly entered as 63. Now. As we know sum/num=avg ------->1 (when mark entered is correct). 83 - 63 = 20. This is the new difference when the mark is entered wrongly. So, sum+20/num=avg+avg (1/2) ------->2 (when mark entered is the wrong Avg is increased by 1/2). Sum + 20/num = avg (1 + 1/2). Sum + 20/num = avg (3/2). Substituting 1 in 2; Sum + 20/num = sum/num (3/2). Sum + 20 = 3/2sum. Sum = 40. |

Jayshree said: (Jul 18, 2016) | |

Let suppose average marks A. Number of students be X. If the marks entered correctly i.e. 63 then. A = (Total_1) / X -----------> (1). Total_1= A * X -------------> (2). If the marks entered is wrong, ie 83 then, A+ (1/2) = (Total_2) /X -----------> (3). Total_2 = [A + (1/2)] * X-------------> (4). Total_2 = A * X + X/2 -----------> (5). And the difference between Total_1 and total_2 is (83 - 63 = 20). Therefore Eq (5) - (2). A * X + X/2 - A * X = 20. X/2 = 20. X = 40. |

Jayshree said: (Jul 18, 2016) | |

The Correct average increased by half. |

Thiru said: (Aug 14, 2016) | |

(83 - 63) = 20, X * 20 = 1/2, = 40. |

Akash Soni said: (Oct 19, 2016) | |

Avg = sum * total number. 1 if all are correct then the eqn is, avg = sum/number. sum = avg * number----> 1st eqn. 2 if wrong 83 instead of 60 then eqn is, sum + (83 - 63) = (avg + 0.5) * number. sum = (avg + 0.5) * number - 20----> 2nd eqn. let divide 1st eqn by 2nd eqn. sum/sum = 1. Avg * number = (avg + 0.5)number - 20. In solving this avg canceled by avg, So, avg - avg = 0. 0.5 - (20/number) = 0 0.5 * number = 20 Number = 20/0.5 Number =40 |

Sumaira said: (Nov 17, 2016) | |

Thanks for your explanation @Nik. |

Bitu said: (Dec 21, 2016) | |

Let the number of pupils = x. Avg of mistaken marks = 83/x, Avg of correct marks = 63/x, By question; 83/x = 63/x + 1/2, 83/x - 63/x = 1/2, 20/x = 1/2, x = 40. |

Mahesh Bhoi said: (Jan 5, 2017) | |

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. |

Siddhesh said: (Mar 31, 2017) | |

Let x is average marks when total marks are 63. Let n=total no.of pupils then, 63/n =x --<1> 83/n = x + 0.5. From 1, 83/n=63/n+0.5, 20/n = 0.5. n = 40. |

Deepak Kumar said: (May 14, 2017) | |

Let x is average marks when total marks are 63. Replacement, 83 - 63 = 20. Then, x + 20 = x\2. X = 40. |

Anriudh Swaminathan said: (Jun 5, 2017) | |

P1+P2+......+ let it be = x. Let Y be the number of students. Let Z be the correct avg marks. (i) due to the mistake, the equation is:- (x+83)/y =z + 0.5. (ii) now without mistake:- (x+63)/y = z. Solve equation (i) and (ii). 20/y = 0.5. Therefore Y = 40. |

Prashant Ekal said: (Jun 7, 2017) | |

Let n = no of pupils. X = avg marks. Now, For 63 marks avg is, 63/n = X ----> (1). For 83 marks avg is 83/n = X+0.5 ----> (2) Solve 1 and 2. N = 40. |

Chinmay said: (Jul 29, 2017) | |

Total marks=x, Number of pupils=y, Average=x/y. But, (x+y)/20=x/y+1/2, (x+20)/y=(2x+y)/2y, Therefore y=40. |

Mgm said: (Aug 25, 2017) | |

How about this? (83- 63)/ ? = 1/2 that is 20/ (1/2) = ? Ans 40. |

Sharma said: (Aug 29, 2017) | |

@Vishal explain it please. 7x/4y +2x/9y = 1/1. we get 2:3. |

Utsav Sinha said: (Aug 31, 2017) | |

Simple logic. Let no.of candidate = x. Their average marks =y. Generate an equation, xy+83-63=x(y+1/2). Solving we get y=40. |

Shubham Singh said: (Oct 5, 2017) | |

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. |

Deekshith.D said: (Oct 18, 2017) | |

1st condition : 83/n =av, Here av is average. 2nd one is: 63/n = 1/2 +av, So by solving these 2 eq. We have, 1/2 = 83/n - 63/n. Solving this we will get n=2*20=40. |

Abhishek Ojha said: (Jan 6, 2018) | |

@Deekshit. It's INSUFFICIENT DATA. 63/n = 1/2 +av is incorrect . the correct one should be; 63/n = 1/2av + av : because av (average) is increased by half which means avg was changed to av + av/2 . And Most of the folks have done thia mistake only. Please ignore this question, there is Insufficient data to solve this. |

Abhishek Ojha said: (Jan 6, 2018) | |

@Shubham. In your case, You calculated Average Marks Not num of the pupil. |

Abhishek said: (Jan 6, 2018) | |

@Utsav Sinha. RHS should be; X (y + 1/2y). Because average is increasing by half that means avg + 1/2 of avg. |

Nelson said: (Apr 12, 2018) | |

Let we take n be the number of students, x be the average of the students, so nx be the sum of observations, (nx+20)/n=x+0.5, nx+20=nx+0.5n, 0.5n=20, n=40. |

Yash said: (Apr 18, 2018) | |

@Nik. Thanks for your explanation. |

Zubayer said: (May 24, 2018) | |

Let, pupil=x and real avrge=63. xy+20=(y+1/2)x. |

Krishu said: (Jun 10, 2018) | |

Take average except for the person who's marks wrongly entered = x. LET, total no of students = N. AND, initial average =A. So, CORRECT AVERAGE should be ----> (x+63)/N = A { EQ. 1} and, WRONG AVERAGE calculated is----> (x+83)/N = A+(1/2) {EQ. 2} --->(i.e increase of the average marks by 1/2). NOW, put the value of A from EQ.1 in EQ.2 and solve for N. i.e, (x+83)/N=(x+63)/N + 1/2, (x+83-x-63)/N=1/2, 20/N=1/2, N = 40. |

Durga said: (Jul 6, 2018) | |

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. |

Yash said: (Jul 15, 2018) | |

Thank you @Nik. |

Ashok said: (Jul 19, 2018) | |

Please explain me, I am not getting this. |

Kush said: (Jul 22, 2018) | |

I am not getting it. Please someone explain to me briefly. |

Bharat Gupta said: (Jul 31, 2018) | |

When 63 is entered the let avg is x. x = (sum of the marks of rest of students+63)/n. Let the sum of the marks of rest of the students=y. Therefore, x=(y+63)/n. When 83 is entered avg is increased by 1/2 or 0.5. So, x+0.5= (y+83)/n. Equating both the terms we get; 0.5=20/n. Therefore, n=40. |

Arushi Mathur said: (Aug 26, 2018) | |

Let x = total no of students, y=Avg of students when 63 is taken. Sum of students when 63 is taken=xy. So, (xy-63+83) /x=y+0.5. (xy-20) =xy-0.5x. x = 40. |

Sheik said: (Aug 29, 2018) | |

Thanks @Ravi. |

Ashwini said: (Sep 7, 2018) | |

See the difference 83 and 63, which means extra 20 marks they have given. It increased by 1/2 means we shoud divide 20 by 1/2 which is nothing but 40. |

Royal said: (Sep 13, 2018) | |

Nice explanation @Nik. |

Mudit said: (Sep 17, 2018) | |

I didn't understand why we are multiplying x with 1/2? Please tell me. |

Deepak said: (Oct 11, 2018) | |

Let avg.+1/2 = y+83/x- ------> eq1. Y=remaining student's marks. X =total no of students. Now avg.= y+63/x-----> eq2 Eq2-eq1. 83-63/x=1/2. X=40. |

Pawan Yadav said: (Oct 31, 2018) | |

Let x = total number of student. Increase 1/2 means x * 1/2 = x/2, Means x/2 = 83-63 x = 40. |

Arun said: (Nov 14, 2018) | |

83-63/.5 = 40. |

Natasha said: (Nov 16, 2018) | |

Let 'x' be the avg marks and 'y' be the no of pupils due to entering 83 instead of 63 the avg mark 'x' became 'x + 0.5'. Please note that the avg was increased by 0.5 = 1/2 and not by x/2. Since the difference between 83 answer 63 yields 20 which is divided among y pupils. y/2 =20 gives y=40. |

Vinodha said: (Dec 14, 2018) | |

I didn't understand this please explain clearly. |

Sanchit said: (Feb 11, 2019) | |

Take sum as S and number if people as N. Avg + (1/2) = (s + 83)/n----> (1) Now, Avg = (s+63)/n ----> (2) Just substitute in place of avg [(s + 63)/n] + 1/2 = [(s + 83)/n]. 1/2 = [(s + 83)/n] - [(s + 63)/n]. 1/2 = (83-63)/n, n. = 2 x 20. n = 40. |

Prashanth.A said: (Feb 12, 2019) | |

Let a Total number of students = x. Form an equation of averages from given statement, (Total marks/x)+((83-63)/x)=(total marks/x)+ (1/2). Solve for x. i.e x = 40. |

Karan S.Bisht said: (May 26, 2019) | |

Let suppose x is the total students. Average increases due to error is = x.5. Total error is 83-63 which is 20, x+20=x.5, 0.5x=20, X=40 answer. |

Rudrendra said: (Jun 14, 2019) | |

@Nik. Well explained, Thanks. |

Kabir said: (Jun 14, 2019) | |

Increased by half is old + 1/2 not (1.5 * old). |

Parth said: (Jun 24, 2019) | |

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. |

Saravanan said: (Jul 23, 2019) | |

Let x be the cumulative marks excepting the wrongly entered mark. Let y be the no of pupils, (x+83)/y = ((x+63)/y) + 0.5. x+83 = x+63+0.5y, 20 = 0.5y, y = 40. |

Sneha Agarwal said: (Jul 28, 2019) | |

Simply: x = num of pupils. y = total average. Therefore, Total marks of class=xy Total marks of class due to wrong entry = xy-63 + 83. New average of class due to wrong entry = (xy-63 + 83)/x Also, (according to question) new average = y+0.5 (or y+1/2) Thus, we get: (xy-63+83)/x = y + 0.5, x = 40. |

Prithika said: (Sep 24, 2019) | |

I am not understanding this, please explain it. |

Chetan Shahi said: (Oct 3, 2019) | |

1/2 means 0.5. Because of difference of 20 this avg 0.5 is increased. So distributing 0.5 equally to get total 20. We divide 20/0.5 will give no.of student. i.e. 20/0.5 = 40 student. |

Himani said: (Aug 24, 2020) | |

Avg = total marks/No. Of pupil. 1/2 = 20/x, X = 40. |

Ram said: (Oct 23, 2020) | |

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. |

K Chandan Achary said: (Nov 8, 2020) | |

Let x be the sum of marks of all the students except the incorrect mark. and y be the no of students. Now, Avg mark of all the students with correct mark =(x+63)/y. Avg mark of all the students with incorrect mark =(x+83)/y/ Acc to the ques; (x+83)/y= (x+63)/y+1/2, (x+83)/y= (2x+126+y)/2y. (taking LCM of denominators) 2(x+83)= 2x +126+y, 2x+166 = 2x+126+y, y= 40. |

Tomoriba Shira said: (Dec 9, 2020) | |

According to the property of Average, if each quantity is increased by certain valu "X" then the new average is increase by X. Now in the question, it is given that average has been increased by 1/2 due to the marks entered wrongly. Therefore we can say that the marks of each n every student should increase by 1/2 as per the property. Let x be no of students. Therefore (x *1/2) = x/2. Now, increase value = 83 - 63 = 20. Therefore x/2 = 20. x = 40. |

S.Rohini said: (Feb 24, 2021) | |

@Rakesh. Thank you. |

Akash Ghadage said: (Jul 19, 2021) | |

Increase/Decrease In(Average) =Increase/Decrease In(sum of entity)//Increase/Decrease In(number of entity). Therefore => Increase/Decrease In(number of entity ) =Increase/Decrease In(sum of entity) //Increase/Decrease In(Average). In this Example => Decrease In(sum of entity) =83-63 =20, Increase In(Average) =1/2 and, therefore; Number of pupil's =20/1/2 ==> 20*2 = 40. |

Selva Ganesh said: (Sep 13, 2021) | |

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. |

Selva Ganesh said: (Sep 13, 2021) | |

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. |

Jithendra said: (Sep 23, 2021) | |

Here the simple solution; A pupil's marks -> one student marks entered wrong. Here 's1' means student mark. Let (S1+S2+...63) = S total marks, 63 is the one student marks. So we don't know about remain students marks.thats why I take it as S1,S2. Like this; Marks entered wrong here that's 83. (S1+S2+....83) = (S1+S2+....63) + 20 we can write like this isn't it. (S1+S2+....83) = S +20 we can also write this. Avg = total marks/total students. Avg = S/X -------> (1) when correct marks(63). Avg + 1/2 = S+20/X ----> (2) when 83 marks. Substitute (1) in (2). S/X + 1/2 = (S+20)/X. ( 2(S) +X ) / 2X = ( S+20 )/X Now cross multiple. X ( 2(S) + X ) = 2X ( S + 20). Now cancel one 'x' from both sides. 2(S) + X = 2 ( S + 20 ) => 2S + X = 2S + 40. Now, 2S will cancel. Then, X = 40. Hope it helps. |

Priyanka said: (Oct 2, 2021) | |

Let's take the actual avg marks = X and the number of pupil as Y. Given, 63 is the actual mark obtained by Y pupil. Which gives us, X = 63/Y (This should have been the actual avg) So, XY = 63 ---------> (1). But instead, due to the error, it was replaced to 83 which increased the actual avg by half So, the new eqn for the error should be, X + 1/2 = 83/ Y (Since the number of pupil remains the same) Solve: (2X+1)/2 = 83/Y. 2X+1 = 166/Y. 2XY+Y = 166 -------> (2) (1) in (2). 2(63)+Y = 166. Y = 40 (Number of pupil). |

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