CBSE Class 12-commerce Answered

Table below shows the symbols considered for Male and female students of each course

Above Table shows 18 unknown variables . Hence we need 18 equations to be formed

from 18 given information. Then solve these 18 equations to get all the unknowns .

⇒  Information #1 :- total number of students (male+female ) of all courses = 2000 .

⇒ Information #2 :- Total Female students greater than total male students by 60. 

From above informations, we get the following two equations.

b + h + n + d + j + p + f + l + r = 1030 .........................(1)

a + g + m + c + i + o + e + k + q = 970 ........................(2)

⇒ information #3  :- BA students of all year = 36 % =  (36/100) × 2000 = 720

a+g+m+b+h+n = 720 ................................(3)

⇒ information #4  :- B.Com students of all year = 30.5 % =  (30.5/100) × 2000 = 610

e + k + q + f + l + r = 610  ................................(4)

⇒ information #5  :-  From information # 4 and #5 , we get B.Sc students of all year = (100-36-30.5) %  = 33.5 %

B.Sc students of all year =  (33.5/100) × 2000 = 670

c + i + o + d + j + p = 670 ................................(5)

⇒ information #6  :- 32% of students are in first year

Number of students in first year = ( 32/100) × 2000 = 640

a + b + c + d + e + f = 640 ..........................(6)

⇒ information #7  :- Number of students in second year is 100 more than those of second year .

From known value of total students , i.e. 2000 and from information #6 and #7 , we get

Number of students in second year = 730

Number of students in third year = 630

Hence we get

g + h + i + j + k + l = 730 ...............................(7)

m + n + o + p + q + r = 630 ........................(8)

⇒ information #8  :- 30 more females in BA than number of females in B.Com

b + h + n - f - l - r = 30 ..........................(9)

⇒ information #9  :- 20 less females in B.Sc than number of females in B.Com

f + l + r - d - j - p = 20  ........................(10)

by adding eqn.(9) and (10) , we get

b + h + n - d - j - p = 50 .........................(11)

By adding (1) and (9) , we get

2( b + h + n ) + d + j + p  = 1060 .........................(12)

By solving eqn.(11) and (12) , we get

(b+h+n) = 370 ...........................(13)

(d+j+p) = 320 ............................(14)

By substituting d+j+p) in (11) , we get

(f+l+r) = 340 ..........................(15)

 Using information about female students from eqn.(13), (14) and (15) , we get the following information

of male students from eqn.(3) , (4) and (5)

a + g+ m = 350 ................................(16)

e + k + q  = 270  ................................(17)

c + i + o  = 350 ................................(18)

⇒ information #10  :- Ratio of BA male students in first year to those in third year = 8 : 7

a : m = 8 : 7   or   m = (7/8) a ..............(19)

⇒ information #11  :- second year BA male students are greater by 5 tan  those in first year

g = a + 5 ..........................(20)

Using eqn.(19) and (20) , we rewrite eqn.(16) as

a + a+5 + (7/8)a = 350

From above expression, we get a = 120 .

By substituting a in eqn. (19) and (20) , we get m = 105  and g = 125 .

⇒ information #12  :- second year BA male students and second year BSC male students are equal

Hence we get , i = g = 125

⇒ information #13  :-  First year B.SC male students are greater than those of third year by 5

Hence we get ,  c = o + 5

Hence we rewrite eqn.(18) as

o+ 5 + 125 + o  = 350

From above expression we get o = 110

hence , c = 115

⇒ information #14  :-  second year femal BA  students= second year male B.Com students

Hence, we get , h = k ................... (21)

⇒ information #15  :-  second year femal BSC  students= second year male B.Sc students

Hence , we get j = i = 125 .................(22)

⇒ information #16  :-  Female students of second year B.com and second year B.Sc are equal

Hence, we get , j = l ........................(23)

From eq.(22) and eqn.(23) , we get   l = 125  ...................(24)

⇒ information #17  :-   In B.Com Each number of first year male students and female students is less than respective third year students by 5

From above information, we get

q - e = 5 ....................................(25)

r - f = 5 ....................................(26)

⇒ information #18  :- Female students in B.Sc Third year is 100

p = 100 ...................(27)

By substituting p from eqn.(26) and substituting  j from eqn.(22) , we get d from eqn.(14) as

(d+125+100) = 320

d = 95 ...................(28)

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By substituting l from eqn.(24) and f from eqn. (26) , we rewrite eqn.(15) as

(f+l+r) =  ( f+125+f+5 ) = 340

From above expression, we get  f = 105

Hence r = 110 .....................(29)

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Let us consider eqn.(7)

g + h + i + j + k + l = 730 ...............................(7)

in above expression , except h and k all other values are known .

Hence from above expression, by substituting relevant values ,

we get equation for h and k .

125 + h + 125 + 125 +k + 125 = 730

h + k = 230

From eqn.(21) h = k

Hence h = k = 115 .........................(30)

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Let us consider eqn.(17)

e + k + q  = 270  ................................(17)

if k = 115 ( eqn.30) , then q + e = 155

From eqn.(25) ,  q - e = 5

By solving above equations in e and q , we get

q = 80 ..................................(31)

e = 75  ............................(32)

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Let us consider eqn. (6)

a + b + c + d + e + f = 640 ..........................(6)

In above expression , except b all other values are known .

Hence from above expression, by substituting relevant values ,we get b

120 + b + 115 + 95 + 75 + 105 = 640

b = 130  .............................(33)

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Let us consider eqn. 913)

(b+h+n) = 370 ...........................(13)

bu substituting values of b and h , we get n

130 + 115 + n = 370

n  = 125

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numbers of students in all the courses are known and the values are tabulated below