Latest Infosys Aptitude Question SOLUTION: ab+a+b=250 bc+b+c=300 ca+c+a=216 find a+b+c

• Placement Papers
• /
• Infosys
• /
• Numerical Ability
• /
• Algebra

Infosys Question Papers: Infosys Interview Experience(10) INFOSYS Company Profile

(#M40024490) INFOSYS Numerical Ability question Keep an EYE

ab+a+b=250 bc+b+c=300 ca+c+a=216
find a+b+c

Asked In Infosys asxa asxaxax (10 years ago)

Unsolved Read Solution (4)

Is this Puzzle helpful?   (9)   (5) Submit Your Solution Algebra

Read Solution (4) :

hitesh kapila  (10 years ago)

ab+a+b=250
10a+b+a+b=250
11a+2b=250----(1)
like wise
11b+2c=300----(2)
11c+2a=216----(3)
adding 1,2,3
we get 13(a+b+c)=766
a+b+c=766/13=58.92----ans

Is this solution Helpfull? Yes (10) | No (4)  

Puja Pujari  (10 years ago)

ab+ a+ b = 250---1
bc+b+c = 300---2
ca+ c+ a= 216----3
Adding 1 2 3
2(a+b+c) + (ab+bc+ca) = 766
2(a+b+C) = 766 - ab - bc- ca
2(a+ b+ c) = 766 -(a + b - 250) - (b+c -300) -(c+ a -216)
2(a+b+c) = 766 + 250+300+216 -2(a +b+c)
4(a+b+c) = 766+766
a+b+c = 383

Is this solution Helpfull? Yes (9) | No (5)  

varaprasad  (10 years ago)

given,
ab + a + b = 250
a(1+b) + b = 250

a = (250−b)/(1+b)--------(1)

ac + a + c = 300
c(1+a) + a = 300
c = (300−a)/(1+a)
substitute eqn(1) in d above eqn, we get

c=(50+301b)/251 -------(2)

given bc+b+c=216
substitute eqn(2) in d above eqn, we get
43b^2 +86b -7738=0
and we get
b=12.45 or -14.45 ( we have [-b +or- sqrt(b^2 - 4ac)]/2a )
substitue b=12.45 in eqn(1), we get
a=17.66
substitue b=12.45 in eqn(2), we get
c=15.13
so,
a=17.66
b=12.45
c=15.13

so
a+b+c= 17.66+12.45+15.13 = 45.24

Is this solution Helpfull? Yes (5) | No   

neelimamalineni  (10 years ago)

ab + a + b = 250
a(1+b) + b = 250

a = (250−b)(1+b)

ac + a + c = 300
c(1+a) + a = 300

c = (300−a)(1+a)

=300−(250−b)(1+b)1+(250−b)(1+b)=300(1+b)−(250−b)1+b+250−b

=300(1+b)−(250−b)251=(50+301b)251=(50+301b)251

bc + b + c = 216

(50+301b)b251+b+(50+301b)251=216

(50 + 301b)b + 251b + 50 + 301b = 54216

301b2 +602b−54216 =0

b=12.46

a =(250−b)(1+b)=(250−12.46)(1+12.46) = 17.65
c = (300−a)(1+a)=(300−17.65)(1+17.65) = 15.14

These values of a,b and c statisfies the given conditions,as given below
ab + a + b = 17.65 *12.46 + 17.65 + 12.46 ≈ 250
ac + a + c = 17.65 *15.14 + 17.65 + 15.14 ≈ 300.21
bc + b + c = 12.46 *15.14 + 12.46 + 15.14 ≈ 216

Hence,

Is this solution Helpfull? Yes | No (8)  

Please Login to Read Solution.