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Numerical Ability
Age Problem
If f(1)=4 and f(x+y)=f(x)+f(y)+7xy+4, then f(2)+f(5)=?
Read Solution (Total 12)
-
- f(1)=4
f(1+1)=19
f(1+2)=48
f(2+2)=70
f(2+3)=106
then f(2)+f(5)=125
- 11 years agoHelpfull: Yes(23) No(12)
- 125 ans
f(2)=f(1)+f(1)+7*1*1+4=4+4+7+4=19
f(3)=f(1)+f(2)+7*1*2+4=4+19+18=41
f(5)=f(2)+f(3)+7*2*3+4=19+41+46=106
106+19=125 - 11 years agoHelpfull: Yes(12) No(0)
- f(2)=f(1)+f(1)+7*1*1+4=19
f(5)=f(2)+f(3)+7*2*3+4
=19+[f(2)+f(1)+7*2*1+4]+7*2*3+4
=19+19+4+14+4+42+4
=106 - 11 years agoHelpfull: Yes(8) No(12)
- f(2+5)=f(2)+f(5)+7(2)(5)+4
=>f(2)+f(5)=f(7)-70-4=>f(7)-74
=>f(1)=4
f(1+1)=19
f(2+1)=41
f(2+2)=70
so f(7)=f(3+4)=199
f(2)+f(5)=199-74
i.e., 125 - 11 years agoHelpfull: Yes(8) No(0)
- f(1)=4 is given,
f(x+y)=f(x)+f(y)+7xy+4
then take to the x=1 and y=1
f(2)=f(1)+f(1)+7(1)(1)+4
=4+4+7+4
f(2)=19.
after find f(3)=f(1)+f(2)+7(1)(2)+4
=4+19+14+4
=41.
NEXT f(5), take to the x value = 3 and y value = 2
f(5)=f(3)+f(2)+7(3)(2)+4
=41+19+42+4
=106.
add f(2)+f(5)=19+106
ans is =125. - 11 years agoHelpfull: Yes(3) No(0)
- f(1)=4
f(2)=>f(1+1)=f(1)+f(1)+7(1)(1)+4=>4+4+7+4
=>19
f(3)=>f(1+2)=f(1)+f(2)+7(1)(2)+4=>4+19+14+4
=>41
f(5)=>f(2+3)=f(2)+f(3)+7(2)(3)+4=>19+41+42+4
=>106
f(2)+f(5)=19+106
=125 answer - 11 years agoHelpfull: Yes(2) No(0)
- f(2)=f(1)+f(1)+7*1*1+4
=4+4+7+4=19
f(2)=19
f(2+1)=f(2)+f(1)+7*2*1+4
=19+4+14+4
f(3)=41
f(3+2)=f(3)+f(2)+7*3*2+4
=41+19+42+4
=106
f(2)+f(5)=19+106=125
- 11 years agoHelpfull: Yes(1) No(0)
- 125
since f(2)=f(1)+f(1)+7*1*1+4=19
f(3)=f(2)+f(1)+7*2*1+4=41
f(5)=f(3)+f(2)+7*3*2+4=106
so,f(5)+f(2)=106+19=125 - 11 years agoHelpfull: Yes(1) No(0)
- f(2)=f(1+1)=f(1)+f(1)+7*1*1+4=4+4+7+4=19
f(3)=f(1)+f(2)+7*1*2+4=4+19+18=41
f(5)=f(2)+f(3)+7*2*3+4=19+41+70+4=106
so f(2)+f(5)=106+19=125 - 11 years agoHelpfull: Yes(1) No(0)
- f(2)=f(1)+f(1)+7*1*1+4
f(2)=19
f(3)=f(2)+f(1)+7*2*1+4
f(3)=19+4+14+4
f(3)=31
f(2+3)=f(5)=f(2)+f(3)+7*2*3+4
19+31+42+4
f(5)=96
f(2)+f(5)= 19+96
- 11 years agoHelpfull: Yes(0) No(0)
- f(1)=4(given)
f(2)=f(1+1)=f(1)+f(1)+7*1*1+4=19
f(3)=f(2+1)=41
similarly,f(5)=106
so,f(2)+f(5)=125 - 11 years agoHelpfull: Yes(0) No(1)
- If f(1)=4 and f(x+y)=f(x)+f(y)+7xy+4
f(1+1)=f(1)+f(1)+7(1)(1)+4
f(2)=4+4+7+4=19
f(3)=19+4+14+4=51
f(5)=f(3)+f(2)+7(3)(2)+4=51+19+42+4=116
f(2)+f(5)=19+116=135 - 11 years agoHelpfull: Yes(0) No(1)
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