Capgemini
Company
Numerical Ability
Algebra
if (p+q)=3 then what is the value of (p3 + q3 ), when it is given that p=1/q?
a. 123 b. 143 c. 111 d. 132
Read Solution (Total 9)
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- (p+q)^3 = p^3 +q^3+3pq(p+q)
p^3 +q^3 = (p+q)^3 - 3pq(p+q)
=27-3(3)
= 18 - 9 years agoHelpfull: Yes(28) No(3)
- (p+q)^3=p^3+q^3+3p^2q+3pq^2
(3)^3=p^3+q^3+3(pq)p+3(pq)q
Since p=1/q we get pq=1
27=p^3+q^3+3p+3q
27=p^3+q^3+3(p+q)
27=p^3+q^3+3(3)
27-9=p^3+q^3
18=p^3+q^3
Ans:18
But i don't know the correct answer.Admin please verify it. - 9 years agoHelpfull: Yes(9) No(2)
- p+q=3
p^3+q^3= [p+q][p sq -pq + q sq]
= 3[p sq-1 + q sq]
=3[(p+q)^2- 2pq+1]
=3[3^2 - 2 +1]
=3[9-2+1]
=24
= - 9 years agoHelpfull: Yes(3) No(3)
- p+q=3and p=1/q=>(q+1/q)=3
(q+1/q)(q+1/q)=3*3
=q^2+1/q^2+2=9
q^3+1/q^3=24
p=1/q=>p^3+1/q^3=24
- 8 years agoHelpfull: Yes(2) No(0)
- 18 is correct answer but options are wrong
- 9 years agoHelpfull: Yes(0) No(0)
- 18 is crct answer
- 9 years agoHelpfull: Yes(0) No(0)
- 18 will be the ans
- 9 years agoHelpfull: Yes(0) No(0)
- 18. Iz correct answer
- 8 years agoHelpfull: Yes(0) No(0)
- p+q=3
P=1/q
sub p=1/q
(1/q)+q=3
(1+q^2)/q=3
1+q^2=3q
q^2-3q+1=0
to find root
q=(3+or-(9-4*1*1)^(1/2))/2(1)
q=(3+-2.236)/2
q=2.3 - 8 years agoHelpfull: Yes(0) No(1)
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