Posts and Comments

@vivek417
169

Posted 12 Oct '08

> prateek7563 Says
>
> A bag contains a total of 105 coins of rs1,50p,25p. find total no. of 1rs coins if there are rs50.5 in bag and it is known that number of 25p coins is 133.33% more than the number of 1rs coins.
U have...the following eqns
x+y+z = 105
x+0.5y+0.25z = 50.5
& z...

prateek7563 SaysA bag contains a total of 105 coins of rs1,50p,25p. find total no. of 1rs coins if there are rs50.5 in bag and it is known that number of 25p coins is 133.33% more than the number of 1rs coins.

U have...the following eqns

x+y+z = 105

x+0.5y+0.25z = 50.5

& z = (233.33/100) x

Solve these in terms of x & y..to get the values..thats all

Thanks

Vivek

@vivek417
169

Posted 03 Oct '08

> 1\. Seventeen Century englandwas much like the Third World War today, in that a government office was a recognized route to wealth.
2\. This parchment shows that the great fortune of that time derived more from what we would now call corruption than from commerce.
3\. By the nineteenth ccen...

1. Seventeen Century englandwas much like the Third World War today, in that a government office was a recognized route to wealth.

2. This parchment shows that the great fortune of that time derived more from what we would now call corruption than from commerce.

3. By the nineteenth ccentury,that had changed:technology had made it possible to create wealth faster than one could steal it.

4.The prototypical rich man of nineteenth century was not a courtier but an industrialist.

My take:-

1) J

2) I

3) J

4) J

Thanks

Vivek

@vivek417
169

Posted 01 Oct '08

> > *Q2) Find the complete range of values of x for which (x^2-7x+13)^x < 1 *
x<3 ,x< 4
*=> ( - infinity , 4 ) ??*
>
>
Your answer is wrong....check for x = 1
The answer should be 3<4
(x^2-7x+13)^x < 1
=> (x^2-7x+13)^x < 1 ^x
So x^2-7x+13 < 1
x^2-7x+12 <0
...

Q2) Find the complete range of values of x for which (x^2-7x+13)^x < 1

x<3 ,x< 4

=> ( - infinity , 4 ) ??

Your answer is wrong....check for x = 1

The answer should be 3<4

(x^2-7x+13)^x < 1

=> (x^2-7x+13)^x < 1 ^x

So x^2-7x+13 < 1

x^2-7x+12 <0

(x-3)(x-4) <0

Hence 3<4

Thanks

Vivek

- 1 Like

@vivek417
169

Posted 01 Oct '08

> hiiii plzzzzzz solve these problems
For how many integral values of M is M2( it is M square) +100M+1000 a perfect square ?
ans; 8
m^2 + 100m + 1000 = k^2
(m+50)^2 = k^2 +1500
(m+50-k)(m+50+k) = 1500
Now 1500 = 2^2*3^1*5^3...
So we have 3*2*4 = 24 factors
Now m+5...

hiiii plzzzzzz solve these problems

For how many integral values of M is M2( it is M square) +100M+1000 a perfect square ?

ans; 8

m^2 + 100m + 1000 = k^2

(m+50)^2 = k^2 +1500

(m+50-k)(m+50+k) = 1500

Now 1500 = 2^2*3^1*5^3...

So we have 3*2*4 = 24 factors

Now m+50+k should always be greater than 50

& m+50-k should always be less than 50

Use these criteria to filter out the values from 24 factors.....we will get the answer

Thanks

Vivek

@vivek417
169

Posted 01 Oct '08

>
*Q) If sin q and cos q are the roots of the equation ax^2+bx+c=0,then a,b,c are related as:*
a) a^2+b^2+2ac=0
b) a^2-b^2+2ac=0
c) b^2+c^2-2ac=0
d) a^2-c^2+2ab=0
Here we have sin q + cos q = -b/a
& sin q * cos q = c/a
Now for the options....divide throughout by a^2......

Q) If sin q and cos q are the roots of the equation ax^2+bx+c=0,then a,b,c are related as:

a) a^2+b^2+2ac=0

b) a^2-b^2+2ac=0

c) b^2+c^2-2ac=0

d) a^2-c^2+2ab=0

Here we have sin q + cos q = -b/a

& sin q * cos q = c/a

Now for the options....divide throughout by a^2....

like a^2-b^2+2ac=0

divide by a^2 then we have 1 - (b/a)^2 + 2*(c/a) = 0....now put the values & check.....thats it

Thus answer is b.

Thanks

Vivek

@vivek417
169

Posted 01 Oct '08

> *Q) Which pair of point slie on either side of the line 3x+2y-6=0?*
a) (0,-7),(-7,-8]
b) (0,0),(2,3)
c) (1,1),(-1,-4)
d) (-1,1),(-2,-2)
Clearly only b satisfies the criteria. (0,0) gives negative result while (2,3) gives positive.
Hence answer is b.
Thanks
Vivek

Q) Which pair of point slie on either side of the line 3x+2y-6=0?

a) (0,-7),(-7,-8]

b) (0,0),(2,3)

c) (1,1),(-1,-4)

d) (-1,1),(-2,-2)

Clearly only b satisfies the criteria. (0,0) gives negative result while (2,3) gives positive.

Hence answer is b.

Thanks

Vivek

@vivek417
169

Posted 30 Sep '08

>
2) Find the sum of all the four didgits number that can be formed with the didgits 3,2,3,4?
Here is my solution....
Sum of all numbers formed by n digits (a,b,c,d...n) = (n-1)! * (Sum of all digits) * 11...n times
So here we have 3! *(2+3+3+4) * 1111 = 79992
But sinc...

2) Find the sum of all the four didgits number that can be formed with the didgits 3,2,3,4?

Here is my solution....

Sum of all numbers formed by n digits (a,b,c,d...n) = (n-1)! * (Sum of all digits) * 11...n times

So here we have 3! *(2+3+3+4) * 1111 = 79992

But since we have 2 3's...so total sum 79992/2 = 39996

Thanks

Vivek

@vivek417
169

Posted 26 Sep '08

> @maskedmenace
2\. If the X-Y plane, what is the area of the region bounded between the two curves
f(x)=max(x-1,1-x)-1 and g(x)=min(x+2,-2-x)+3?
1\. 2 sq units 2. 2.5sq units 3. 3 sq units 4.3.5 sq units
ans is 3.5
I am not getting the region....
I am getting answer as 3....

@maskedmenace

2. If the X-Y plane, what is the area of the region bounded between the two curves

f(x)=max(x-1,1-x)-1 and g(x)=min(x+2,-2-x)+3?

1. 2 sq units 2. 2.5sq units 3. 3 sq units 4.3.5 sq units

ans is 3.5

I am not getting the region....

I am getting answer as 3.5.

Lines are as follows:-

y=-x

y=-x+1

y= x-2

& y= x+5

So area = (7/2)*(2)^0.5 * (1/2)*(2)^0.5 = 7/2 =3.5

Thanks

Vivek

@vivek417
169

Posted 25 Sep '08

> hi bhai,
i am unable to visualise the figure...
can you please elaborate your method...
Let the triangle be ABC & square be MNOP...
Here M lies on AB while P lies on BC.
Let side of square = x
So Sin 60 = x/BM
BM = 2x/root(3)
AMN is equilateral triangle...clearl...

hi bhai,

i am unable to visualise the figure...

can you please elaborate your method...

Let the triangle be ABC & square be MNOP...

Here M lies on AB while P lies on BC.

Let side of square = x

So Sin 60 = x/BM

BM = 2x/root(3)

AMN is equilateral triangle...clearly visible

So AM = x

But BM + AM = 1

So 2x/root(3) + x = 1

Solve this to get the answer.

Thanks

Vivek

- 1 Like

@vivek417
169

Posted 25 Sep '08

> MaskedMenace Says
>
> What is the area of square whose vertices lie on sides of equilateral triangle of side 1cm ?
(2x/3^0.5) + x = 1...here x is side of square
Solve it ...we get x = 2(3)^0.5 - 3
So area of square = x^2 = 21-12(3)^0.5
Thanks
Vivek

MaskedMenace SaysWhat is the area of square whose vertices lie on sides of equilateral triangle of side 1cm ?

(2x/3^0.5) + x = 1...here x is side of square

Solve it ...we get x = 2(3)^0.5 - 3

So area of square = x^2 = 21-12(3)^0.5

Thanks

Vivek

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