@chandrakant.k yar mere posted Q bhi solve kar do.
@pyashraj said:Shld be Option(4). S>1Here, AM > HM=> (100+101+....+1000)/991 > 991/[(1/100) + (1/101) +.....+ (1/1000)]=>[1001*(1000/2) - 99*(100/2)]/991 > 991/S=>(9911*50)/991 > 991/S=>S > (991*991)/(9911*50)=> S > 1 [Aprroximately]
why 991??? 
number of terms = 901 right??
number of terms = 901 right??@Unternehmer said:Guys...please help me with these doubts.......
1)A regular polygon has each interior angle equal to twice its exterior angle. Find the number of diagonals in it.
a.5
b.8
c.6
d.9
2)A circular track is 400 m long. A, B and C run in the same direction along the track. They take 50 seconds to meet each other for the first time. If the ratio of their speeds is 4: 2 :1 find the time taken by C to complete a round.
a.100 seconds
b.75 seconds
c.50 seconds
d.60 seconds
3)In a machine a series of words is fed in which then gives an output after following a certain number of steps.
Input: Sale Fake Hence Gate His
Step 1: Sale His Fake Hence Gate
Step 2: Sale His Hence Fake Gate
Step 3: Sale His Hence Gate Fake
Step 3 is the output.
What will be the second step if the input to the same machine is
Input: Play Hope Greece Soul Watch
a.Soul Watch Play Hope Greece
b.Watch Soul Play Hope Greece
c.Watch Play Hope Greece Soul
d.Watch Soul Play Greece Hope
5)How many such pair of letters are there in the word PAMPHLET which have as many letters between them in the word as in the English alphabet?
a.0
b.1
c.2
d.3
Q1)
Sum of Internal angles in regular polygon = (n-2)*180.
And exterior angle = 360/n
So, (n-2)180/n = 360/n. Solve for n. n = 6 ( 6 Sides).
Diagonals = nC2 - n = 9..
@vbhvgupta said:@chandrakant.k yar mere posted Q bhi solve kar do.
konsa?? i guess i i have missed it can you paste the link??
can you paste the link??ab khane ka time. so will solve once i am back 😁 :D
@chandrakant.k said:why 991??? number of terms = 901 right??
Thnxs for pointing out on that Bro..Yeah woh 901 hi hoga..Typo tha.. :)
@Exodia said:Fourteen fruits and twenty two flowers are to be distributed among 10 people in such a way thateach person gets something. Anyone who gets more than two flowers cannot get more than onefruit and anyone who gets more than one fruit cannot get more than three flowers. What is themaximum number of flowers that one can get?(a) 3 (b) 5 (c) 19 (d) 22Plz give answer alongwith explanation
option d
if any1 is gettin 22 flowers he will get 1 fruit
now remainig 13 fruits among 9 people
simplest way distribute 3 friuts each to 2 people nd then 1 each to remainig 7
if any1 is gettin 22 flowers he will get 1 fruit
now remainig 13 fruits among 9 people
simplest way distribute 3 friuts each to 2 people nd then 1 each to remainig 7
If abc=177/3 and a,b,c are all real and positive.What is the least possible value of the expression (a+b+c)(ab+bc+ca)?
a)177
b)354/3
c)354
d)531
Please post the approach along with solution.
The number 123456789012...890 is four thousand digit long. first u removed all the digit in odd numbered place starting at left most place. next u removed the digits in odd numbered place in the remaining 2000 digits. you perform same operation untill no digit remain. what digit is to be last removed? expln plzďťż
@Exodia said:Fourteen fruits and twenty two flowers are to be distributed among 10 people in such a way thateach person gets something. Anyone who gets more than two flowers cannot get more than onefruit and anyone who gets more than one fruit cannot get more than three flowers. What is themaximum number of flowers that one can get?(a) 3 (b) 5 (c) 19 (d) 22Plz give answer alongwith explanation
OA : D) 22
Explanation: All the flowers can be given to one person and then the fruits distributed among all others in such a way that all of them get at least one fruit.
Explanation: All the flowers can be given to one person and then the fruits distributed among all others in such a way that all of them get at least one fruit.
@saurabhlumarrai said:If abc=177/3 and a,b,c are all real and positive.What is the least possible value of the expression (a+b+c)(ab+bc+ca)?a)177b)354/3c)354d)531Please post the approach along with solution.
apply AM>=GM
a+b+c>=3(a+b+c)^1/3
ab+bc+ac>=3(ab*bc*ca)^1/3
multiply both
(a+b+c)(ab+bc+ca)>=9(abc)
9*177/3==== 531
x2 €“ 3y2 = 1376
How many integer solutions exist for the given equation?
(a) One (b) Two (c) Four (d) Zero
Plz give answer alongwith explanation.
How many integer solutions exist for the given equation?
(a) One (b) Two (c) Four (d) Zero
Plz give answer alongwith explanation.
Function f(x) is a continuous function defined for all real values of x, such that f(x) = 0 only for two distinct real values of x. It is also known that
f(6) + f(8) = 0
f(7).f(9) > 0
f(6).f(10) f(0) > 0 and f(1)
How many of the following statements must be true?
I. f(1).f(2).f(3) II. f(3).f(5).f(7).f(9) > 0
III. f(7).f(8) IV. f(0) + f(1) + f(9) + f(10) > 0
(a) 1 (b) 2 (c) 3 (d) 4
Plz give answer alongwith explanation.
f(6) + f(8) = 0
f(7).f(9) > 0
f(6).f(10) f(0) > 0 and f(1)
How many of the following statements must be true?
I. f(1).f(2).f(3) II. f(3).f(5).f(7).f(9) > 0
III. f(7).f(8) IV. f(0) + f(1) + f(9) + f(10) > 0
(a) 1 (b) 2 (c) 3 (d) 4
Plz give answer alongwith explanation.
ABCD is a rectangle with BC = a units and DC = 3 a units. The perpendicular dropped from point A meets BD at point F. The diagonals AC and BD intersect at point G. What is the area (in squareunits) of ΔAFG?
(a) [root(3) x a2]/12
(b) [root(3) x a2]/6
(c) [root(3) x a2]/10
(d) [root(3) x a2]/8
Plz give answer along with explanation.
(a) [root(3) x a2]/12
(b) [root(3) x a2]/6
(c) [root(3) x a2]/10
(d) [root(3) x a2]/8
Plz give answer along with explanation.
@Exodia said:x2 €“ 3y2 = 1376How many integer solutions exist for the given equation?(a) One (b) Two (c) Four (d) Zero
Plz give answer alongwith explanation.
A square is always of form 3*n or 3*n+1
3*y^2 = 3*n type, 1376 = 3*n+2 type
x^2 = 1376 + 3*y^2
=> x^2 = 3*n + 3*n+2= 3*n+2 type = cannot be a square
=> zero solutions
Toss a fair coin 43 times. What is the number of cases where number
of 'Head'> number of 'Tail'?
(A) 2^43
(B) (2^43)-43
(C) 2^42
(D) None of the above.
of 'Head'> number of 'Tail'?
(A) 2^43
(B) (2^43)-43
(C) 2^42
(D) None of the above.
Toss a fair coin 43 times. What is the number of cases where number
of 'Head'> number of 'Tail'?
(A) 2^43
(B) (2^43)-43
(C) 2^42
(D) None of the above.
of 'Head'> number of 'Tail'?
(A) 2^43
(B) (2^43)-43
(C) 2^42
(D) None of the above.
@ishu1991 said:Toss a fair coin 43 times. What is the number of cases where numberof 'Head'> number of 'Tail'?(A) 2^43(B) (2^43)-43(C) 2^42(D) None of the above.
Ans: (C) 2^42
=43C22 + 43C23 + .......................43C43
= 2^43/2
=2^42
=43C22 + 43C23 + .......................43C43
= 2^43/2
=2^42
@ishu1991 said:Toss a fair coin 43 times. What is the number of cases where numberof 'Head'> number of 'Tail'?(A) 2^43(B) (2^43)-43(C) 2^42(D) None of the above.
No. of heads > No. of tails
=> No. of heads can be 22, 23, 24... 43
=> 43C22 + 43C23 + 43C24 .. 43C43
=> We know that 43C0 + 43C1 + 43C2 ... 43C43 = 2^43
=> We also know know that our answer has 22 terms which are 43C22 + 43C23 + 43C24 .. 43C43
=> These 22 terms are equal to = 43C0 + 43C1 + 43C2 .. 43C21
=> Both of them are equal and add up to 2^43
=> So each one of them is 1/2 of 2^43 = 2^42
=> 43C22 + 43C23 + 43C24 .. 43C43 = 2^42. Option C
@ishu1991 said:Toss a fair coin 43 times. What is the number of cases where number
of 'Head'> number of 'Tail'?
(A) 2^43
(B) (2^43)-43
(C) 2^42
(D) None of the above.
1/2*2^43 = 2^42