Highest possible marks : 60

Lowest possible marks : -20

So a total off 80 x 3+1=241 possible scores but scores such as 59+2/3, 59 + 1/3 and 58 + 1/3 are not possible to get Hence , 241 - 3 = 238 possibilities.

**Q-20**

solution please.

what is the difference between 1st and 4th option?

solve Q20.

Find the last three digits of 7^{9999}

The quicker way to calculate last three digits of a number is to calculate the remainder by 1000.

1000 = 125 x 8

7^{9999}/8 = (-1)^{9999} = -1 or 7.

7^{9999}/125

Phi(125) = 100

7^{100k} gives a remainder of 1 when divided by 125 .

so 7^{10000} gives the remainder 1.

Lets say 7^{9999} gives remainder R when divided by 125.

So , 7^{9999} x 7 /125 => 1

7R/125 => 1

R = 18.

Using Chinese Remainder Theorem:

125a + 18 = 8b + 7 gives 143.

To win the game A has to leave exactly one coin on the table so that B has to pick it up and would lose the game .

So A make his strategy to remove 9 coins from the table

So , 9 - 4 - 4 i.e 1 coin.

if u leave pick matchstick there is chance that other player will get chance to leave one coin for u. .. yes i don't know the ans but 1 can't be the ans this is certainly true

the maximum possible value of y = min(1/2 - 3x^2/4 , 5x^2/4) for the range 0<x<1 is ? (cat 1993)

a. 1/3

b. 1/2

c. 5/27

d. 5/ 16

Hello Tina,

Please find the solution,

r + (1/r) = 2007 + (1/2007) and

s + (1/s) = 2006 + (1/2006)

=> **r = 2007 or (1/2007) and**

s = 2006 or (1/2006)

We can verify these results by solving the two quadratic equations also

2007r^{2} - (2007^{2} + 1)r + 2007 = 0 and

2006s^{2} - (2006^{2} + 1)s + 2006 = 0

Now (r - s) is maximum when r is maximum and s is smallest.

So **(r - s)**** _{maximum}** = r

_{max}- s

_{min}= 2007 - (1/2006) =

**(4026041/2006) = 2006.9999**

Hello, sir please solve this questions.

If x + 1 = x^{2} and x > 0, then 2x^{4} is:

Hi Tina,

x + 1 = x^{2}

x^{2} – x – 1 =0

x=(1+sqrt(1+4))/2

x = (1+sqrt5)/2

x^{2} =(6 + 2sqrt5)/4

x^{2} =(3+sqrt5)/2

x^{4} =(14+6sqrt5)/4

x^{4} =(7+3sqrt5)/2

2x^{4} =7+ 3sqrt5

This question is my doubt from copy cat 03-

A box contains 300 matches , Rahul and Sachin take turns removing no more than half the matches in the box. The player who cannot move looses. What should be Rahul's first move to ensure his win if he is starting the game?

If f(x) = (x-1)/(x-2), what is f invesrse (4) ?

f(x) = ( x - 1)/( x - 2)

y =( x - 1 )/(x -2)

yx - 2y = x - 1

yx - x = 2y - 1

x ( y-1) = 2y -1

x = (2y -1)/( y-1)

f^{-1}(x) = (2x -1 )/( x -1)

f^{-1}(4) = (8 -1)/( 4-1)= 7/3

Hello sir , this is a doubt from copycat 05 , please explain the solution of this problem -

Find the sum of the given infinite series : 3 + 5/(1+2^{2} ) + 7/(1+2^{2}+3^{2}) +..............

In miscellaneous type of series find nth term and than try to solve ..

Series can be written as

(1-2/3) + (2/3-2/5) + (2/5-2/7) + (2/7-2/9) + (2/9-2/11)

= 1-2/11

= 9/11 option (4)

a , ar , ar^{2} , ar^{3} , ar^{4 }, …..

a + ar = 15

a ( 1 +r) = 15

a = 15/ ( 1+ r) …. (1)

Now ,

3(ar + ar^{2} + ar^{3} + ar^{4} +….) = a

3 [ar / (1-r)] = a

1 - r = 3r

r = ¼

from (1)

a = 12

Sum of the series : a , ar , ar^{2} , ar^{3} , ar^{4 }, …..

a / ( 1 –r) = 12 ( 1-1/3) = 16