@bs0409
n=1 and p =5.
@Logrhythm said:n^41^4 = 12^4 = 63^4 = 14^4 = 65^5 = 56^4 = 67^4 = 18^4 = 69^4 = 10^4 = 0so only n = 1 is the solution.......rest all wld either be divisible by 2 or 5..
Wht abt 5^5. 5+4=9 so it can be possible as well. prime number might end with 9.....:p
@bs0409 said:Wht abt 5^5. 5+4=9 so it can be possible as well. prime number might end with 9.....
han but 5^4 + 4 is not a solution...so (5k)^4 + 4 wld also not be na....thoda assume karna padta hai.. :p
@iLoveTorres said:ans n=1 p=5?
@Logrhythm said:han but 5^4 + 4 is not a solution...so (5k)^4 + 4 wld also not be na....thoda assume karna padta hai..
OA-1
N=1,p=5
@techgeek2050 said:the work done by a man ,woman and a child are in the ratio 3:2:1. If daily wages for 20 men,30 women and 36 children amt to rs 78,what will b the wages of 15 men,21 women and 30 children for 18 weeks?
a^4 + b^4 + c^2 >= Kabc for all positive values of a,b,c . then the greatest value of K is ???
a) 2(sq root2)
b) 4(sq root2)
c) 2
d) 4
a^4 + b^4 + c^2 >= Kabc for all positive values of a,b,c . then the greatest value of K is ???
a) 2(sq root2)
b) 4(sq root2)
c) 2
d) 4
@swapnil4ever2u said:a^4 + b^4 + c^2 >= Kabc for all positive values of a,b,c . then the greatest value of K is ???a) 2(sq root2)b) 4(sq root2)c) 2d) 4
4root 2 hoga kya?
A sweet shop sells laddus boxes of different sizes. The laddus are priced at Rs10 per laddu upto 400 laddus. For every 10 additional laddus, the price of entire lot goes down by 10 paise per laddu. Find the size of the box that would have the maximum cost.
600
500
700
800
600
500
700
800
explain plz
there are three equal containers that are completely filled with different water-alcohol mixtures with water and alcohol in the ratio 2:3,3:4,4:5 respectively,they are emptied into a bigger container .what fraction of the mixture in the bigger container should be replaced by water so that the resulting mixture has equal quantities of water and alcohol ??
@dream_IIMA said:A sweet shop sells laddus boxes of different sizes. The laddus are priced at Rs10 per laddu upto 400 laddus. For every 10 additional laddus, the price of entire lot goes down by 10 paise per laddu. Find the size of the box that would have the maximum cost. 600500700800explain plz
Price of the entire box = (400 + 10n)(10-0.1n)
going by options.. n can be 20,10,30,40
this is max for 30.
hence .. 700?
going by options.. n can be 20,10,30,40
this is max for 30.
hence .. 700?
143/945
@dream_IIMA said:A sweet shop sells laddus boxes of different sizes. The laddus are priced at Rs10 per laddu upto 400 laddus. For every 10 additional laddus, the price of entire lot goes down by 10 paise per laddu. Find the size of the box that would have the maximum cost. 600500700800explain plz
box--->400 laddus
Rs 10 per laddu
for every additional 10 laddus the price will decrease by 10paise
let 10x be the additional laddus placed in the box
size=400+10x
price per laddu = 5.00-0.10x
Selling price= (400+10x) (5-0.10x)
=2000-40x-x^2+50x
=2000+10x-x^2
to maximise the revenue
differentiate
dSP/dx=10-2x=0
x=5
therefore maximum laddus =400+50
450 
can go by options too
@dream_IIMA said:A sweet shop sells laddus boxes of different sizes. The laddus are priced at Rs10 per laddu upto 400 laddus. For every 10 additional laddus, the price of entire lot goes down by 10 paise per laddu. Find the size of the box that would have the maximum cost. 600500700800explain plz
i dint get the ans@dream_IIMA said:A sweet shop sells laddus boxes of different sizes. The laddus are priced at Rs10 per laddu upto 400 laddus. For every 10 additional laddus, the price of entire lot goes down by 10 paise per laddu. Find the size of the box that would have the maximum cost. 600500700800explain plz
cos of entire box is given by = (400+10n)(10-0.1n) = 400+60n-n^2
to get the maximum we need to differentiate
so -2n+60 = 0
n = 30
so it should be 400+30*10 = 700(edited)
@chandrakant.k said:cos of entire box is given by = (400+10n)(10-0.1n) = 400+60n-n^2to get the maximum we need to differentiate so -2n+60 = 0n = 30so it should be 400+30 = 430please check the options @pavimai why did you take 5???
yes 
got it..but 430 not in options
got it..but 430 not in options
box--->400 laddus
Rs 10 per laddu
for every additional 10 laddus the price will decrease by 10paise
let 10x be the additional laddus placed in the box
size=400+10x
price per laddu = 10.00-0.10x
Selling price= (400+10x) (10-0.10x)
=4000-40x-x^2+100x
=4000+60x-x^2
to maximise the revenue
differentiatedSP/dx=60-2x=0
x=30
therefore maximum laddus =400+30
430
Rs 10 per laddu
for every additional 10 laddus the price will decrease by 10paise
let 10x be the additional laddus placed in the box
size=400+10x
price per laddu = 10.00-0.10x
Selling price= (400+10x) (10-0.10x)
=4000-40x-x^2+100x
=4000+60x-x^2
to maximise the revenue
differentiatedSP/dx=60-2x=0
x=30
therefore maximum laddus =400+30
430