Find minimum value of a + b + c if a, b, c are distinct positive integers such that 1/a + 1/b + 1/c = 1.
6
7
9
9 ?
Solution please
By A.M >= H.M
we have three nos. i.e a,b,c
(a+b+c)/3 >= 3/{(1/a)+(1/b)+(1/c)}
Since (1/a)+(1/b)+(1/c) =1
a+b+c>=9
So the minimum Value is 9
In a regular polygon of 2012 sides, distance from centre of polygon to a vertex is 5cm. Find the area of the polygon approximately.
44
31
78
If we find out interior angle it will be
(n-2)180/n where n is the no. Of sides i.e 2012
2010*180/2012 approx value will be 180 so it will be almost a circle
Area = π×5*5 = 22*25/7
= 78.5 hence option third
Three men make the following statements regarding a murder that they are suspected of. Two of the men are lying, and one of them is telling the truth. Only one of the men committed the crime.
A says: I didn’t do it.
B says: C did it.
C says: A did it.
Who is certainly lying?
A
B
C
CBD
Which of the three is most likely to be innocent?
A
B
C
CBD
We have to take three cases
Case 1 if a says truth
Since a is saying true so he didn't do it, it would be b or c
B is lying so c didnt do it
And c is lying so a didnt do it which was already told by a so b is the criminal
Case 2 if b is saying true
Since a is saying false means he did it
B said c did it which is true so that means c did it since question says only one has done the crime so this case isnt true
3rd case if c is saying true
A is lying so he did it
B is lying so c didnt do it and C is saying true which we got earlier by A
So in all three cases b is certainly lying and c is most likely to be innocent
Keshav and Krishna start a game with one pile of 40 coins. They take turns. On each turn, a player must take 1, 3, or 4 coins from the pile. The player, who takes the last coin from the pile of 40 coins, wins the game.
If Keshav plays first, what number of coins must he take from the pile on his first turn in order to guarantee that he can win the game?
1
3
4
Keshav can never win
If Krishna starts the game by taking 2 coins from pile, then who has the winning strategy?
Krishna
Keshav
No one has winning strategy
data insufficient
?
hi sir, please help
Puttning x=0 and y=402 we have
f(402)= f(f(402))
f-1(f(402))=f(402)
since f-1(f(x)) =x { property}
402=f(402)
option 3rd
Thankyou Aniket 🙂
Please tell the approach
Ans- option (A)
Hi Richa PFA the solution
Let roots be p,q,r
p+q+r= a.........(1)
pq+qr+pr=b
pqr=a …..........(2)
If there are three no. p,q,r
By A.M>= G.M
(p+q+r)/3>=(pqr)^(1/3)
Putting value from 1 and 2
a^3/27>=a
a^2 >=27
Hence a can't be equal to 0 is the answer
x³ - ax² + bx -a = 0
x² ( x - a) + b ( x-a) = 0
if , b = 1 , then the equation reduces to
x²( x - a) + (x-a) = 0
(x-a)(x² + 1) = 0
it has one real root a and two imaginary roots . hence b ≠ 1 .
Option D.
Que.10
In annual dealer meet of Lam Organics, all 101 dealers are gathered and they shook hands with each other. It was known later that each dealer shook hands with at least 68 other dealers. Find the minimum number of dealers who shook hands with same number of other dealers.
0
2
3
4
solution please
Que.12
Diagonals of a quadrilateral intersect in 1 point, and that of a pentagon intersect in at most 5 distinct points. What is the maximum number of points of intersections of diagonals in a hexagon?
6
9
10
15
Total points 15
is there any general formula?
Select any four points in 6C4 = 15 ways .
Find the number of integral triplets (a, b, c) such that 1 ≤ a ≤ b ≤ c ≤ 10.
120
180
220
286
All distinct a < b < c .
10C3 => 120
two same 10C1 x 9C1 => 90 cases
a=b = c
10 cases
Total : 120 + 90 + 10 = 220.
What are the last two digits of 9119 – 1991?
62
38
82
18
last two digits of 91^19 => 11
last two digits of 19^91 => 19
last two digits of 91^19 - 19^91 => 08
options are wrong.
Find the sum of squares of roots of the equation (x2 - 1)2 = 4x + 1.
0
-2
-4
4
Two real roots of a quadratic polynomial (x + a)(x + b) + 4 are p and q. What are the roots of the quadratic polynomial (x - p)(x - q) – 4?
a, b
–a, -b
a-2, b-2
–a-2, -b-2
Find x such that |x + a| - |x - b| > 0 where a < b.
x > b
x < (b - a)/2
x > (b - a)/2
None of these
Que.26
Find the values of k for which x2 – 2kx + 2k – 1 is always positive for any real number x.
k > 0
k > 1
all real numbers
Ans is all real nos,?
k>1
x2 -2kx + 2k -1 > 0.
Coefficient of x2 is positive.
Discriminant should be less than 0.
(-2k)2 – 4 . 1 . ( 2k -1) < 0
4k2 -8k + 4 < 0
( k-1)2 < 0.
which will never be negative Hence, no solution .
From a deck of 52 cards, a card is picked and its value is recorded (say 10 of spades) and it is placed back in the deck. This process is repeated 100 times. Let there are x different possible outcomes, then find the remainder of x with 5.
1
2
3
?
