@hiteshkhurana82 said:If the largest angle in a triangle is 70o, what is least possible value of the smallest angle of the triangle?69o1o40o39o41o
40?
@hiteshkhurana82 said:The equation of four lines is given below:x + 2y - 3 = 0, 3x + 4y - 7 = 0, 2x + 3y - 4 = 0 and 4x + 5y - 6 = 0 (1) Concurrent(2) The sides of a parallelogram(3) The sides of a square(4) None of these
4?
y = -x/2 + 3/2
y = -3x/4 + 7/4
y = -2x/3 + 4/3
y = -4/5x + 6/5
Since slopes are different, we can eliminate 2 and 3.
Solving first and second equations:
3x + 6y - 9 = 0
3x + 4y - 7 = 0
y = 1, x = 1
But these values do not satisfy the third and fourth equation.
So 1 is also eliminated.
y = -x/2 + 3/2
y = -3x/4 + 7/4
y = -2x/3 + 4/3
y = -4/5x + 6/5
Since slopes are different, we can eliminate 2 and 3.
Solving first and second equations:
3x + 6y - 9 = 0
3x + 4y - 7 = 0
y = 1, x = 1
But these values do not satisfy the third and fourth equation.
So 1 is also eliminated.
hiteshkhurana: 41?
@hiteshkhurana82 said:If the largest angle in a triangle is 70o, what is least possible value of the smallest angle of the triangle?69o1o40o39o41o
40
@hiteshkhurana82
The equation of four lines is given below:
x + 2y - 3 = 0, 3x + 4y - 7 = 0, 2x + 3y - 4 = 0 and 4x + 5y - 6 = 0
x + 2y - 3 = 0, 3x + 4y - 7 = 0, 2x + 3y - 4 = 0 and 4x + 5y - 6 = 0
(1) Concurrent
(2) The sides of a parallelogram
(3) The sides of a square
(4) None of these
(2) The sides of a parallelogram
(3) The sides of a square
(4) None of these
Iska shayad None of these hai..
@hiteshkhurana82 said:If the largest angle in a triangle is 70o, what is least possible value of the smallest angle of the triangle?69o1o40o39o41o
40o?
180 - 70 = 110
Since 70 is the largest angle,
110 = 70 + 40
180 - 70 = 110
Since 70 is the largest angle,
110 = 70 + 40
@hiteshkhurana82 said:The equation of four lines is given below:x + 2y - 3 = 0, 3x + 4y - 7 = 0, 2x + 3y - 4 = 0 and 4x + 5y - 6 = 0 (1) Concurrent(2) The sides of a parallelogram(3) The sides of a square(4) None of these
none?
Question left unsolved: what is the sum of the digits of the largest 2 digit prime number which divides 200c100?
Let N be the number of ways in which 2010 can be written in the form,
2010 = 1000a3 + 100a2 + 10a1 + a0, where ai, for 0 ≤ i ≤ 3, is an integer and 0 ≤ ai ≤99.
An example of such representation is, 2010 = 1 × 10^3 + 9 × 10^2 + 11 × 10^1 + 0 × 10^0
Find N.
OPTIONS
1) 2
2) 3
3) 52
4) 202
5) None of the above
2010 = 1000a3 + 100a2 + 10a1 + a0, where ai, for 0 ≤ i ≤ 3, is an integer and 0 ≤ ai ≤99.
An example of such representation is, 2010 = 1 × 10^3 + 9 × 10^2 + 11 × 10^1 + 0 × 10^0
Find N.
OPTIONS
1) 2
2) 3
3) 52
4) 202
5) None of the above
@anantn
Should be 7, the prime number in question been 61..
Here. 200c100=200!/100!*100! = (101*102*103*......*200)/(1*2*3*....*100)
=>Thus, the prime number should be less than 100, and should have multiples of it atleast twice(2k) and thrice(3k) in the numerator..
Thus, 3k
@jain4444 said:@techsurge nice bet let sum of parent's ages = N average age of children's = x
The polynomial x^2-2x+7 divides the polynomial x^4+px^2+q. What is the value of q?
p=4,q=49
What will be the the unit's digit of (1273)^122! ?
Soln (given in the Arun Sharma: 122! is a number of the form 4n. Hence the answer should be 1.
I don't undersand this..... Plz expalin.
@hiteshkhurana82 said:If the largest angle in a triangle is 70o, what is least possible value of the smallest angle of the triangle?69o1o40o39o41o
40?
@tmohan02 said:What will be the the unit's digit of (1273)^122! ?Soln (given in the Arun Sharma: 122! is a number of the form 4n. Hence the answer should be 1.I don't undersand this..... Plzexpalin.
122! = 1*2*3*4*5*...*122 = 4n
Now the any number with the units digit as 3 follows the pattern
N = ...3
N^2 = ...9
N^3 = ...7
N^4 = ...1
N^5 = ...3
N^6 = ...9
.
and so on..
As you can see, the units digit has a pattern (3,9,7,1,3,9,7,1,...) which has a cyclicity of 4.
So the units digit of 1273^221! will be 1.
Now the any number with the units digit as 3 follows the pattern
N = ...3
N^2 = ...9
N^3 = ...7
N^4 = ...1
N^5 = ...3
N^6 = ...9
.
and so on..
As you can see, the units digit has a pattern (3,9,7,1,3,9,7,1,...) which has a cyclicity of 4.
So the units digit of 1273^221! will be 1.
@sujamait said:Let N be the number of ways in which 2010 can be written in the form,2010 = 1000a3 + 100a2 + 10a1 + a0, where ai, for 0 ≤ i ≤ 3, is an integer and 0 ≤ ai ≤99.An example of such representation is, 2010 = 1 × 10^3 + 9 × 10^2 + 11 × 10^1 + 0 × 10^0Find N.OPTIONS1) 2 2) 3 3) 52 4) 202 5) None of the above
None of the above hai kya?
@Phoenix_Sohom said:I am posting abt aCoaching Center in Kolkata...PIONEER CAREER ....They have opened their center in Kolkata in collaboration with Career Forum, Pune.Its located just opposite to NAVINA cinema Hall.2nd Floor,16/1, Prince Anwar Shah Road, Kol-33Can anybody give me a feedback abt this place. I have visited the place and the guy told me that they only allow 15guys in a batch and will be taught by an author who was with CL previously. The Course Fee is also affordable..PUYS -- I am not so versed with these Coaching Centers, so IF any one of the experienced PUYS can go and have a look.
Sorry, this is not the right thread for this discussion.
