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If10^x is divided by 13 and remainder is 1....we know a^(p-1)/p remainder is 1 if a and p are relatively prime --- this is a property, i dont know the name...therefore for x= 12, the equality would hold, similarly for 10^24..it would hold again as (10^12)^2 == (mod1)^2....therefore net anser would be 100/12 == 8 values

Some correction

also check for 10^6/13 which would make all multiples of 6 as the answer, hence net answer would be 8*2 =16

i got the answer as 16.

yeah its 12

mn bhai mei apne kamjoor student ka hath beech mein nahi chodta :cheerio::cheerio::cheerio::cheerio:

i also got it as 12..

1 - If the roots of the equation (a^2 + b^2)x^2 - 2(ac+bd)x + (c^2 + d^2) = 0 are equal then which of the following is true?

1 - ab = cd

2 - ad = bc

3 - ad = bc^1/2( square root of bc)

4 - ab = cd^1/2( square root of cd)

5 - abc = d^1/2( square root of d)

2 - Find the value of the expression

(sqrt x +( sqrt x+ ( sqrt x... +)))

3 - Find the maximum value of the function 1/ ( x^2 - 3x + 2)

a - 11/4

b - 1/4

c - 0

d - -4

e - none of these

( guys plz tell me the approach to sovling such sums - max and min value types)

4 - Find the minimu m v alue of the function

f(x) = log (x^2 - 2x + 5 ) to the base 2 .

a - -4

b - 2

c - 4

d - -2

e - negative infinity

these are my answers.....

)ad=bc

2)2sqrt(x)

3)-4

4)-4

these are my answers.....:splat::splat::splat:

1)ad=bc

2)2sqrt(x)

3)-4

4)-4

answers in bold??

@ninad bhai kuch missing lagta hai??

i have a method.....

consider the speed of A be 3x and that of B be x.

time at which they meet for the first time will be = distance/relative speed.

relative speed = 3x+x(opposite direction in circular tracks)=4x

distance =2pie r=2*22/7*7=44km

so they will meet after =44/4x=11/s minutes

distance travelled by A will be 3x*11/x=33km and by b will be 11 km at M1

so their next meeting will be after 22km from m0 and 11km from M1

1)11 km ---ans

2) i din get the question

3) each time they meet in a differnce of 11km so their fourth meeting will be after 44km from starting point...M4

4)firstly A travels with a speed of 3x....ie 3x*11/s=33km....then it will travel again with a speed of x km/hr= x*11/x=11Km.....again it will travel with 3x speed so 33 km

total=33+33+11=77.

i have a method.....

consider the speed of A be 3x and that of B be x.

time at which they meet for the first time will be = distance/relative speed.

relative speed = 3x+x(opposite direction in circular tracks)=4x

distance =2pie r=2*22/7*7=44km

so they will meet after =44/4x=11/s minutes

distance travelled by A will be 3x*11/x=33km and by b will be 11 km at M1

so their next meeting will be after 22km from m0 and 11km from M1

1)11 km ---ans

2) i din get the question

3) each time they meet in a differnce of 11km so their fourth meeting will be after 44km from starting point...M4

4)firstly A travels with a speed of 3x....ie 3x*11/s=33km....then it will travel again with a speed of x km/hr= x*11/x=11Km.....again it will travel with 3x speed so 33 km

total=33+33+11=77.

consider the speed of A be 3x and that of B be x.

time at which they meet for the first time will be = distance/relative speed.

relative speed = 3x+x(opposite direction in circular tracks)=4x

distance =2pie r=2*22/7*7=44km

so they will meet after =44/4x=11/s minutes

distance travelled by A will be 3x*11/x=33km and by b will be 11 km at M1

so their next meeting will be after 22km from m0 and 11km from M1

1)11 km ---ans

2) i din get the question

3) each time they meet in a differnce of 11km so their fourth meeting will be after 44km from starting point...M4

4)firstly A travels with a speed of 3x....ie 3x*11/s=33km....then it will travel again with a speed of x km/hr= x*11/x=11Km.....again it will travel with 3x speed so 33 km

total=33+33+11=77.

One-fourth portion of each of the two solid spheres having radii 4 units and 6 units are joined together such that the centres of the two spheres coincide . What is the total surface area (in sq. units) of the solid thus formed?

88 pie , 83 pie, 75 pie, 73 pie

Is it 83????

1:2 i also followed the same approach of shivam sundaram....

i din get the mail

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