A no. has exactly 32 factors of which 4 are not composite. Product of these factors is 30.How many such no.s are possible.
@Cat.Aspirant123 said:circumcentre of triangle ABC is O. /_BAC=85 & /_BCA=75 den /_OAC ? 40607090
70
@Cat.Aspirant123 said:circumcentre of triangle ABC is O. /_BAC=85 & /_BCA=75 den /_OAC ? 40607090
angle CBA = 20 => angle AOC = 40
Thus answer is [180-40]/2 = 70
@Cat.Aspirant123 said:circumcentre of triangle ABC is O. /_BAC=85 & /_BCA=75 den /_OAC ? 40607090
70?
@Cat.Aspirant123 said:circumcentre of triangle ABC is O. /_BAC=85 & /_BCA=75 den /_OAC ? 40607090
70 ??
the first term of an infinite gp series is 3 . find the sum upto infinite terms of ther series if the difference between the 4th and the 7th term is 21/64 . given that the common ration is a rational number ?
1 >
No of integral solutions of x^2 + 12 =y^4
2>
No of ordered pairs (a,b) a^2 = b^3 +1 where a and b are Integers
3>
Unit digits of 7^15^16^17
@maroof10 said:A no. has exactly 32 factors of which 4 are not composite. Product of these factors is 30.How many such no.s are possible.
20?
@Cat.Aspirant123 said:circumcentre of triangle ABC is O. /_BAC=85 & /_BCA=75 den /_OAC ? 40607090
70 ?
angle CBA = 180 - (85 + 75) = 20
angle COA = 2*20 = 40
OCA = OAC = x
OCA + OAC + COA = 180
x + x + 14 = 180
2x = 140
x = 70
@meenu05 said:the first term of an infinite gp series is 3 . find the sum upto infinite terms of ther series if the difference between the 4th and the 7th term is 21/64 . given that the common ration is a rational number ?
6?
@meenu05 said:the first term of an infinite gp series is 3 . find the sum upto infinite terms of ther series if the difference between the 4th and the 7th term is 21/64 . given that the common ration is a rational number ?
6 ?
sum = 3/1 - r
a*r^3 - a*r^6 = 21/64
r^3 - r^6 = 7/64
put r = 1/2
so sum = 3/(1 - 1/2) = 6 ?
@meenu05 said:the first term of an infinite gp series is 3 . find the sum upto infinite terms of ther series if the difference between the 4th and the 7th term is 21/64 . given that the common ration is a rational number ?
ar^3-ar^6=21/64
r^3(1-r^3)=7/64
let r^3=y
y^2-y+7/64=0
64y^2-64y+7=0
(8y-1)(8y-7)=0
r^3=1/8
r=1/2
3/1/2 = 6
If you form a subset of integers chosen from between 1 to 3000, such that no two integers add up to a multiple of 9, what can be the maximum number of elements in the subset (include both 1 and 3000).
1) 1668
2) 1332
3) 1333
4) 1336
Plz give answer alongwith explanation.
1) 1668
2) 1332
3) 1333
4) 1336
Plz give answer alongwith explanation.
new question:
@Exodia said:If you form a subset of integers chosen from between 1 to 3000, such that no two integers add up to a multiple of 9, what can be the maximum number of elements in the subset (include both 1 and 3000).1) 16682) 13323) 13334) 1336Plz give answer alongwith explanation.
9k+1 = 334
9k+2 = 334
9k+3 = 334
9k+4 = 333
9k+5 = 333
9k+6 = 333
9k+7 = 333
9k+8 = 333
9k=333
334*3+333+1
=1336
@meenu05 said:the first term of an infinite gp series is 3 . find the sum upto infinite terms of ther series if the difference between the 4th and the 7th term is 21/64 . given that the common ration is a rational number ?
6 right.??
|ar^3 - ar^6| = 21/64. Assume r^3 = k., solve quadratic eqn in k, hence find r as well.! r = 1/2
Hence a/1-r = 6
@Exodia said:If you form a subset of integers chosen from between 1 to 3000, such that no two integers add up to a multiple of 9, what can be the maximum number of elements in the subset (include both 1 and 3000).1) 16682) 13323) 13334) 1336Plz give answer alongwith explanation.
u pasted same question again.??
this was solved earlier also