Latest TCS Aptitude Question SOLUTION: a,b,c are positive numbers such that a+b+ab= 8, b+c+bc=15 and c+a+ca = 35 What is the value of a+b+c+abc?

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(#M40165549) TCS Numerical Ability question Numbers Keep an EYE

a,b,c are positive numbers such that a+b+ab= 8, b+c+bc=15 and c+a+ca = 35
What is the value of a+b+c+abc?

Asked In TCS SAITEJ (5 years ago)

Unsolved Read Solution (3)

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SAITEJ  (5 years ago)

add 1 on both sides to all the equations and take common elements
now the equations become
(a+1)(b+1)=9
(b+1)(c+1)=16
(c+1)(a+1)=36
solve the equations for a , b, c
so we get
a=3.5 b=1 and c=7
so the value of a+b+c+abc= 36

Is this solution Helpfull? Yes (19) | No (9)  

sathiya devi  (5 years ago)

GIVEN:-


a + b + ab = 8


b + a(1 + b) = 8


a = {8 - b}/{1 + b}}-------( 1 )


b + c + bc = 15


b + c(1 + b) = 15

c = {15 - b}/{1 + b}} ------( 2 )


a + c + ac = 35

{8 - b}/{1 + b} + {15 - b}/{1 + b} +{8 - b}/{1 + b}*{15 - b}/{1 + b} = 35


From----------( 1 ) &-----------( 2 )


{8 - b + 15 - b}/{1 + b} + {120 - 8b - 15b + b^2}/{(1 + b)^2} = 35


{(23 - 2b)(1 + b) + (120 - 23b + b^2) = 35(1 + b)^2}


{23 + 23b - 2b - 2b^2 + 120 - 23b + b^2 = 35 + 35b^2 + 70b}


{143 - 35 - 2b - b^2 = 35b^2 + 70b}


{108 - 72b - 36b^2 = 0}


{b^2 + 2b - 3 = 0}


{b^2 + 3b - b - 3 = 0}


{b(b + 3) - 1(b + 3)}


{(b + 3)(b - 1)}


so, b = 1 , b = -3[neglect]


because a , b , c ,are positive,


now,


from----------( 1 )


a = (8 - b)(1 + b)


a = (8 - 1)(1 + 1)


a = 7/2


from-----------( 2 )


c = (15 - b)(1 + b)


c = (15 - 1)(1 + 1)


c = 14/2


c = 7


thus, the value of "a + b + c + abc" is


7/2 + 1 + 7 + (7 * 7/2 * 1)


{(7 + 2 + 14 )/2 + 49/2}


(23/2 + 49/2)


(23 + 49)/2


72/2


36


HENCE, the value of (a + b + c + abc) = 36

Is this solution Helpfull? Yes (6) | No (4)  

Saritha  (5 years ago)

a-b=25.9
a=6b
Sub value of a in 1st eqn
b=5.18
a=30.08
a+b=38.26

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