Nigel's Final Exam Tests

$\frac{m+98}{n}=90$

$\frac{m}{n-1}=88$

$n=5$

let the number of subjects excluding English be x

total marks excluding english = 88x

total marks including english = 90(x+1)

thus, 90(x+1) - 88x=98

that gives us x=4

so, total number of subjects= x+1= 5

m--total no. of mark

n-- no. of subject

m/n =90

m - 90n = 0 ---(1)

m-98/n-1 = 88

m -88n =10 ---- (2)

m - 90n = 0-------(1)

m - 88n = 10-------(2)

By solving (1) and (2) we get

    - 2n     =  -10

       n     =   5

let x be no.of subjects taken without English

let y be the total score of subjects excluding english .

(98+y )/(x+1)=90................... (1)

y/x = 88...........................(2)

y=88x ..... (3)

substitute (3) in (1)

98+88x = 90x+90

x=4

n= x+1 = 5

Hello,as for this problem,

let a = total marks, b = number of subjects,

as for the average mean (without english) is 88,

a / b = 88

a = 88b(1st),

as for the average mean(with english) is 90,

a + 98 / b + 1 = 90

a + 98 = 90 (b + 1)

a = 90b + 90 -98

a = 90b - 8(2nd),

by substituting (1st) into (2nd),

88b = 90b -8

-2b = -8

b =4,

therefore total marks is a = 88b =88(4) =352 and the total subjects taken = b + 1 = 4 + 1 =5

thanks....

Consider the number of subjects as 'n', marks not including english as 'k', k/(n-1)=88,(k+98)/n=90. so the value of n is 5, said SHERLOCK HOLMES.