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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 = (233.33/100) x

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

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

Q2) Find the complete range of values of x for which (x^2-7x+13)^x
=> ( - infinity , 4 ) ??

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

The answer should be 3
(x^2-7x+13)^x => (x^2-7x+13)^x
So x^2-7x+13 x^2-7x+12
Hence 3

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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


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) = put the values & check.....thats it

Thus answer is b.

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.


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' total sum 79992/2 = 39996


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-2
& y= x+5

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

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.

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

(2x/3^0.5) + x = 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