@audiq7 said:how many 8 digit no.s are there for which the sum of digits is 4?
10C7
a+b+c+d+e+f+g+h=4
a'+b+c+d+e+f+g+h=3
a+b+c+d+e+f+g+h=4
a'+b+c+d+e+f+g+h=3
@jain4444 my approach
=Let the actual price be 100 Rs. if he cheats while buying,
then the faulty cost price becomes 110 Rs.
& now he keeps a profit of 10% to this.
so he is selling at 121 Rs. so its a profit of 21% :)
=Let the actual price be 100 Rs. if he cheats while buying,
then the faulty cost price becomes 110 Rs.
& now he keeps a profit of 10% to this.
so he is selling at 121 Rs. so its a profit of 21% :)
@jain4444 said:A grain dealer cheats to the extent of 10% while buying as well as selling by using false weights.His total profit percentage is:1.20%2.21%3.23%4.25%
21%
Let P(x) be a polynomial with integer coefficients such that P(17) = 10 and P(24) = 17.
If P(n) = n + 3 has two distinct integer solutions n1 and n2, then find the sum n1 + n2.
1) 41
2) 27
3) 30
4) 34
5) 53
Itna Sannata kyun hai bhai
If P(n) = n + 3 has two distinct integer solutions n1 and n2, then find the sum n1 + n2.
1) 41
2) 27
3) 30
4) 34
5) 53
Itna Sannata kyun hai bhai
@ScareCrow28 said:Let P(x) be a polynomial with integer coefficients such that P(17) = 10 and P(24) = 17.If P(n) = n + 3 has two distinct integer solutions n1 and n2, then find the sum n1 + n2. 1) 41 2) 27 3) 30 4) 34 5) 53Itna Sannata kyun hai bhai
41
In a college of 525 studs, each stud takes at least 2 items from among idly, dosa,puri and chapati. If 375 studs take idly, 275 take dosa, 375 take puri and 375 chapati, then the no of studs who take all the items is at most?
275
300
225
325
275
300
225
325
@Brooklyn said:kaise kiya??
@ScareCrow28 said:Dare to Share (the approach)
Let the polynomial be an^2+bn+c (we can safely assume quadratic polynomial coz 2 roots r given)
Now
P(n) = n + 3
=>an^2+n(b-1)+c-3=0
Sum of the roots=-(b-1)/a
P(17) = 10=>289+17b+c=10
P(24) = 17=>576a+24b+c=17
Subtract both
(b-1)/a=41
@sonamaries7 said:In a college of 525 studs, each stud takes at least 2 items from among idly, dosa,puri and chapati. If 375 studs take idly, 275 take dosa, 375 take puri and 375 chapati, then the no of studs who take all the items is at most?275300225325
275??? maximum can be those who take dosa take all items...
it is actually a 4 variable venn diagram problem which i wont do 😛 😛
@sonamaries7 said:In a college of 525 studs, each stud takes at least 2 items from among idly, dosa,puri and chapati. If 375 studs take idly, 275 take dosa, 375 take puri and 375 chapati, then the no of studs who take all the items is at most?275300225325
275?
@sonamaries7 said:In a college of 525 studs, each stud takes at least 2 items from among idly, dosa,puri and chapati. If 375 studs take idly, 275 take dosa, 375 take puri and 375 chapati, then the no of studs who take all the items is at most?275300225325
275
@chandrakant.k said:275??? maximum can be those who take dosa take all items...it is actually a 4 variable venn diagram problem which i wont do
@ScareCrow28 said:275?
@deedeedudu said:275
@sonamaries7 said:In a college of 525 studs, each stud takes at least 2 items from among idly, dosa,puri and chapati. If 375 studs take idly, 275 take dosa, 375 take puri and 375 chapati, then the no of studs who take all the items is at most?275300225325
225?
Has to be 225!
Has to be 225!
@chandrakant.k said:den solve by venn diagram you will get 225 i dont dare to do
No need for venn diagram..(If options are given) :P
@ScareCrow28 said:No need for venn diagram..(If options are given)
thoda prakash dalo... options tho diya hua hai na!!! by writing equations??
@sonamaries7 said:In a college of 525 studs, each stud takes at least 2 items from among idly, dosa,puri and chapati. If 375 studs take idly, 275 take dosa, 375 take puri and 375 chapati, then the no of studs who take all the items is at most?275300225325
mera ans opt me hi nai hai 
2*II + 3*III+4*IV = 1400....(1)
II+III+IV = 525...(2)
(1) - 2*(2)
III + 2*IV = 350
so max value of IV will be 175
@sonamaries7 said:ryt hai..kaise?
Let N2, N3, N4 be exactly 2, exactly 3 and exactly 4
Now, N2+N3+N4= 525
Also, 2(N2) + 3(N3) + 4(N4) = 375+275+375+375 = 1400
Put N4 = 375..It won't satisfy..
Put N4= 275
=> 2(N2)+3(N3)=400 and N2+N3= 250 (But this is also not possible) Since 2(N2)+2(N3) =500
So N4=225
@chandrakant.k
Now, N2+N3+N4= 525
Also, 2(N2) + 3(N3) + 4(N4) = 375+275+375+375 = 1400
Put N4 = 375..It won't satisfy..
Put N4= 275
=> 2(N2)+3(N3)=400 and N2+N3= 250 (But this is also not possible) Since 2(N2)+2(N3) =500
So N4=225
@chandrakant.k