1. If CAT 2005 contains 150 questions and each question contains 4 options find the total number of ways in which a student can answer atleast 2 out of 150 questions.

2.  highest power of 3 in 1! x 2! x 3! x 4! x .......... x 98! x 99! x 100!

3. Find the remainder when 104^303 is divided by 101.

4. i have infinite no of 1 Re, 5 Re, 10 Re and 25 Re notes...in how many ways can i give the change for Rs 100???(pg 3)a+5b+10c+25d=100i'll xplaine it for one value of d=0 and same process can be used for other values of d....d=0 or a+5b+10c = 100 or a+5b = 100-10cfor solving such equations we can first find a solution to a+5b=1 which in this case can be found very easily because the coefficients i.e. 1 and 5 are easier to deal with .....for difficult looking coeffs. we can use the simple continued fractions....for this case a=-4 and b=1 satisfies the equation a+5b=1.....therefore a solution to the equation a+5b=100-10c will be a=-4(100-10c) and b=1(100-10c) which can be understood with a little reasoning..if a solution to equation type ax+by=c is known (lets say x=A,y=B satisfies it) then whole series of solutions is given by (A+bt) and (B-at) for any positive or negative integral value of t.....therefore in this case the general solution is given by a=-400+40c+5t and b=100-10c-t.....since a>=0 and b>=0,putting the values of a and b we get 80-8c<=t<=100-10c for all integral t's satisfying this conditionwe will get different solutions for the basic equation....also 0<=c<=10 that makes the series 21+19+17.......i hope this helps to find a definitive solution...

5. Find the total number of ways in which two sqares (1x1) can be selected from a chess board such that they will have a side in common.(pg 3)(  For a m * n n & required of size p=1 -> [m-(p-1)] * [n-(p-1)] + [n -(p-1]*[m-(p-1)].)I think it will not work out when I am selecting 2 squares of side 2 x 2 from a chess board.

If you had followed the generalisation which I made for the chess boardthinking in the same way the generalisation would be:

First the no. lines on a m x n dimensional board would be m+1 vertical lines and n+1 horizontal lines.

Now the no. vertical lines that can be the common side is given by (m+1-2s) where s is the side of the squares that we are selecting.Now from each line I can select two squares that share this line in(m-s+1) ways.Similary no. of horizontal lines that can be the common side is given by (n+1-2s).Now from each line I can select two squares that share this line in(n-s+1) ways.

So the total no. ways is given by 

(m+1-2s)(m-s+1) + (n+1-2s)(n-s+1)

1. Find the last two digits in the expansion of 2^ 999(pg 6)

2. Find the remainder when 2 ^ 1990 is divided by 1990What Sagar said is absolutely right. I have something different from that. Just have a look at that. But befor going into this I think all of u know the following basic funda.

Let R be the remainder when N is divided by D1.Now the remainder when N is divided by D2, where D2 is a factor of D1 is either1. R ( if R< D2) or

2. R1 where R1 is the ramainder obtained when R is divided by D2 ( if R>D2)Remember the converse is also true.

Let's look at an example here. ( let me give the converse example)

Let 23 be the remainder obtained when a number is N divded by 30.Now the remainder obtained when the number is divided 90 could be either23 or 30x1+23 or 30x2+23 i.e 23 or 53 or 83. ( I think this is clear to everyone)

Applying this rules to the above 2 queries

1. last two digits of N = 2^999.All of us know that the remainder obtained when a number isa) divided by 10 the remainder is the unit digitb) divided by 100 the remainder is the last two digits of the numberc) divided by 1000 the remainder is the last four digits of the number and so on.

So let's divide 2^ 999 with 100 and find the ramainder.First find the remainder obtained when it is divided by 25 ( since 25 is factor of 100).

2^999 % 25= (1024^99) x 2^9 % 25Remainder obtained when 1024 is divided by 25 is 24 but we can take it as -1 also( always take the smallest remainder irrespective of sign)So finally the problem becomes -1 x 2^9 %25= -512 %25= -12 =13 ( finally give the positive remainder)So when our number is divided by 25 the remainder is 13So when it is divided by 100 the remainders could be13 or 38 or 63 or 88.It is evident that the given number is exactly divisible 4.I am taking 4 here because the other factor of 100 is 4.So from the above three remainders there is only one remainder which is exactly divisible by 4 i.e 88.So the remainder obtained when it is divided by 100 is 88 and thus the last two digits are 88.Remember here once I decide the remainders using 25 in the exam I can make use of the multiple options and then proceed.So folks remember when the divisor is big make use of its factors and options given in the exam

Now let,s come to our next problem.Remainder obtained when 2^1990 is divided by 1990.Again the factor here are 10 and 199We are lucky here since 199 is prime and let's use our FERMAT'S rule.

So 2^198 when divided by 199 the remainder is 1( Fermat's rule).So 2^199 will leave a remainder 2So 2^ 1990 = (2^199)^10 % 199= (2^10) % 199 = 29.Now when 2^1990 is divided by 10 the remainder is nothing but the unit's digit and we know we can calculat unit's digit by using the cycylicity of 2( i.e after every 2^4 the unit's digit will repeat.) So obviously the unit's digit is nothing but 2^2 =4.

Now since we got the remainder as 29 when divided by 199the remainders possible when divided by 1990 are29 or 199x1+29 or 199x2+29.........But since the unit's digit is 4 the only possible remainder is 199x5+ 29 =1024

That's it folks. These two examples tell u how powerfull our rule is. I think from now however big the divisor is I think we should be able to give the remainders becauseIf the divisor is big and prime please phone a friend (i.e FERMAT)if it's not prime don't call me please make use of the factors of the divisors and ur options in the exam.

Remainder obtained when 777777777....7777 (101 digits) is divided by 440.1. 337 2. 277 

3. 377 4. 227

Folks, I think most of u are not concentrating on the options given to you.

Now N = 777777.........77777(101) digits is the given number.It is clear that the above no. leaves a remainder 1 when divided by 4 (simply divide 77 by 4 and take the remainder (that's the divisibility rule of 4))

It is also clear that the above no. leaves a remainder 7 when divided by 11.

Since 11 and 4 are factors of 440 the remainder obtained when N is divided by 440 should be such that if it is further divided by 11 and 4 it should leave remainders 7 and 1 respectively.In the given options here I think there's only one option i.e 337 which satisfy the above condition.So the answer has to be 337.

NOW LET'S TRY WITHOUT USING THE OPTIONS.So as I told u earlier we need one more basic funda here.Let me explain that first.All of us know that if a number N is exactly divisible by D1 and D2 then the number N will be exactly divisible by L.C.M (D1,D2).Now did u ever thought of the case where If N is not exactly divisible by D1 and D2 i.eLet R1 be the remainder when N is divided by D1and R2 be the remainder when N is divided by D2.In this case the remainder obtained when N is divided by L.C.M (D1,D2)is the obtained by dividing the least possible value of N by L.C.M(D1, D2)

Let me make it more clear by using an example.Let N be a number which leaves remainder 4 when divided by 10 and remainder 1 when divided by 9.If I want the remainder when N is divided by 90, first calculat the least value of N (i.e a number leaving remainder 4 when divided by 10 and 1 when divided bt 9)In this case it is 64 ( I explained how to calculate min val of N in our problem below).So the remainder obtained when N (whatever be the value of N)is divided by 90 is nothing but 64.

Now let's apply this to our problemN = 77777777.....7777(101digits)Our divisor is 440 whose factors are 40 and 11.Clearly when N is divided by 11 the remainder is 7.Let's calculate the remainder obtained when N is divided by40.We can do this fast using our previous funda again.It is clear that the remainder obtained when N is divided by 1000 is 777.Since 40 is a factor of 1000 the remainder obtained when divided by 40 is obtained by dividing 777 by 40 which is 17.So finally when N is divided by(1) 11 the remainder obtained is 7 and(2) 40 the remainder obtained is 17Let's find least value of N now.From (2) N= 40k + 17...................(a)From (1) since N leaves a remainder 7 when divided by 11N-7 must be exaclty divisible by 11i.e 40k+17-7 must be exactly divisible by 11i.e 40k+10 must be exactly divisible by 11i.e 7k+10 must be exactly divisible by 11Verifying the value for K the first value is 8 (7 x 8 + 10=66)So the least value of N is N= 40k+17= 320+17=337Now when N is divided by L.C.M(40,11) i.e 440 the remainder has to be 337

1. If x = (0.12345678910111213............) and y = 1/x find x + y 

Please give answer in the form of p/q

2. A pious rich man goes to a temple everyday and distributes certain amount of money to the poor people. However he follows the following two rules

Rule1: He always takes money in the denominations of 1 Rupee only. Rule 2: He distributes the money equally among the poor people irrespective of the no. of people he distributes. 

On a particular day if he has taken with him Rs 8640/ find the probability of money being distributed among 27 people.

Please give answer in the form of p/qSol: First let's calculate the total no.of ways in which he can divide the money among different no. of people such that each receives an integral amount.

If P persons are given Y Rs each then P x Y =8640.i.e the total no.of ways in which the money can be divided is nothing but the total no. of ways in which 8640 can be written as product of two numbers.

We know that the total no. of ways in which a no. can be written as product of two numbers is 1/2(total factors).

N= 8640 = 2^6 x 3^3 x 5.Total factors = (6+1)(3+1)(1+1) = 56.So total no. ways of writing N as product of two factors is 28.

But we should take it as 56 and the simple reason is this.

Now 8640 = 864 x 10as well as 8640 = 10 x 864.

Since in the normal case we do not take both of them we get 28 ways.But here we need to take both of them because8640 = 864 x 10 means 10 Rs each to 864 people whereas8640 = 10 x 864 means 864 Rs each to 10 people which are two different ways in our problem.

So the total no. of ways in which 8640 Rs can be distributed is 56 out of which there will be only one way of distributing the money among 27 people( i.e 27 people getting 320 Rs each)

So the probability is 1/56-----------------Find the remainder when 1212121212..........1212 (300 digits) is divided by 999I have a different approach here.But before that let's look at a small problem here.

If I want the remainder when 5 x 36 / 34

Normally we can calculate the rem of 36 when divided by 34 and we get it as 2So the final remainder is 5x2 = 10

Now 5x36/ 34Since 2 is common in 36 and 34 let me cacel that 2 in 36 and 34

Now my problem becomes 5x 18/17I think there will be no change in the quotient but my remainder will be altered.

Now the remainer of 5x18/17 is ofcourse 5x1 =5Since I cancelled 2 earlier in both 36 and 34 multiply this remainder with 2 i.ethe final remainder is going to be 5x2=10

Another example:

Rem of (23 x 75)/ 45 It is reduced to (23 x 5) /3 ( cancelling 15 in both 75 and 45)So the remainder becomes 1.But since we cancelled 15 earlier the final remainder will 1x15=15

Now let's look at our problem 

1212121212121212.................1212121212( 300 digits) /999

all of us know that a number of the form ABCD can be written as A x 10^3 + B x 10^2 + C x 10^1 + D x 10^0 ( here our base is 10)Similarly if I have no. of the form ABABABAB instead writting it in the base 10 let's write it in thebase 100 since two digits are same i.e

ABABABAB = AB X 100^3 + AB X 100^2 + AB X 100^1 + AB X 100^0

S our no. 12121212.......121212 (300 digits) can be written as 

12 x 100^149 + 12 x 100^148 + .................+ 12 x 100^1 + 12 x 100^0.[simply treat it as 150 digited no. where each digit is 12 now i.e 12,12,12,12,12,.....12 (150 digits)]

= 12 x (100^149 + 100^148 +.............+ 100^1 + 100^0)= 12 x ( 10^298 + 10^ 296 +..............+ 10^2 + 10^0= 12 x (10^150 -1)/(10^2-1) ( above sum is in GP where the first term is 10^0 i.e 1 and the common ration is 10^2 and no. of terms is 150)

= 12 x ( 10^150-1)/99= 12 x [9999..........9999 ( 150 digits)]/99= 12 x [10101010...........1010101 (149 digits)] (please remember this generalisation,don't deduce iteverytime)i.e ABABABAB.....................ABAB ( 2n digited number)is equal to AB x [ 101010...........101(2n-1 digits)]

Now remainder of 12 x [101010.....10101(149 digits)]/999Again express 101010-------10101 and proceeding further 

we get remainder as 666

This may seem little bit hectic but i want all of u to get the concepts.

The other faster way for this is as we have discussed earlier split 999 as 111 and 9 Clearly the given no. is exactly divisible by 9 as well as 111.Now use your options given in the exam.---------------------FIND THE SINGLE DIGIT SUM OF THE NUMBER N = 4444^3333Here comes our next basic funda.

All of us know that the remainder obtained when a number is divided by10 is nothing but the unit's digit of that no.

Had u ever thought of the remainder that is obtained when divided by 9. It

is nothing but the single digit sum of that number.

For example the single digit sum of 4658 = 4 + 6 + 5 + 8 = 23 = 2 + 3 =5.Now when 4658 is divided by 9 the remainder is nothing but the remainder obtained when sum of the digits of 4658 i.e 23 is divided by 9 which in turn is nothing but the remainder obtained when 2+3 = 5 is divided by 9.

So that's the funda here folks.

Finally the remainder obtained when 4444^3333 is divided by 9 is nothing but the single digit sum. 4444^3333 % 9= 7^ 3333 % 9= 343^ 111 % 9= 1.Since the remainder obtained when N is divided by 9 is 1 the single digit sum is 1.

Remember if the remainder obtained is 0 then the single digit sum is not 0 but ofcourse it is 9Is the five digited number ABCDE divisible by 13?

I. The number DE is divisible by 13.

II. 10A + B + 4C + 3D - E is divisible by 13.

a) Statement I alone is sufficient.

b) Statement II alone is sufficient.

c) Both the statements together are sufficient.

d) Both the statements together are not sufficient.

MOST OF THE PEOPLE WHO ANSWERED THIS USED THE DIVISIBILITY RULE OF 13 OR SOME OTHER ASSUMPTIONS TAKING EXAMPLES.

WHAT ABOUT IF THE DIVISOR IS 23 / 29 / 31/.......................So let's get into the discussion now.

All of u know that ABCDE = A x 10^4 + B x 10^3 + C x 10^2 + D X 10^1 + E= 10000A + 1000B + 100C + 10D +E

So N = 10000A + 1000B + 100C + 10D +ENow the remainder obtained when N is divided by 13 is nothing but the remainder obtained when 10000A + 1000B + 100C + 10D + E is divided by 13i.e 3A + 12B + 9C + 10D + E 

NOW IF I CONSIDER 3A, THIS IS REMAINDER OBTAINED WHEN IT IS DIVIDED BY 13I CAN TAKE IT AS -10A INSTEAD OF 3ASIMILARLY 12B CAN BE TAKEN AS -BAND 9C CAN BE TAKEN AS -4CAND 10D CAN BE TAKEN AS -3C

SO FINALLY 10A + 12B + 9C + 10D + E BECOMES(-10A) + ( -B) + (-4C) + (-3D) + E

WHICH IS NOTHING BUT-[ 10A+ B + 4C + 3D - E]SO WHEN ABCDE IS DIVIDED BY 13 THE REMAINDER IS -[ 10A+ B + 4C + 3D - E]SINCE IN THE SECOND STATEMENT IT IS GIVEN THAT [ 10A+ B + 4C + 3D - E] IS DIVISIBLE BY 13 OUR NUMBER IS ALSO DIVISIBLE BY 13.

SO STATEMENT II ALONE IS SUFFICIENT

THINKING IN THIS WAY WE CAN ANSWER WHATEVER THE DIVISOR IS 23,29,31......

I THINK THIS IS LUCID TO EVERYONE.

GUYS IF THIS IS NOT CLEAR PLEASE LET ME KNOW BECAUSE U NEED GET INTO THIS BECAUSE THIS IS AGAIN A GENERALISATION--------------prove that N= 244 ^ 1500 - 1 is divisible by 1001.

Mohit, here's the answer

1001 =7 x 11 x 13 

now 244 ^ 1500 is divided by 7 the remainder is 

[(244^7)^ 214 x 244^ 2]/7= [1 x 244^2]/7 (FERMAT'S RULE)= 1 X (-1)^2 =1 SO WHEN 244^1500 IS DIVIDED BY 7 THE REMAINDER IS 1HENCE N IS DIVISIBLE BY 7SIMILARY SINCE 11 AND 13 ARE ALSO PRIME WE CAN USE FERMAT'S RULE IN THESE CASES AND WE WILL GET THAT N IS DIVISIBLE BY 11 AND 13

SINCE N IS DIVSIBLE BY 7,11 AND 13 N IS DIVISIBLE BY L.C.M (7,11,13) = 1001

Find the last four digits in the binary expansion of the number 

7654823986734573975876926.Let me make it more clear.

Let me take a number 476463If i need the last digit of this number in binary expansion we divide it by 2 and the remainder 1 will be the last digit.

If i need the last two digits of this number in binary expansion divide it by 2^2= 4 and the remainder here is 3. Now express this 3 in the binary form which is 11. So the last two digits will be 11. 

If i need the last three digits of this number in binary expansion divide it by 2^3 = 8 and remainder here is 7. Now express this 7 in the binary form which is 111. So the last two digits will be 111. 

So if i need the last n digits of any number in binary expansison divide it by 2^n and express the remainder obtained in the binar form which will be last n digits of that number in the binary form.

Now coming to our number the last 4 digits of this number in the binary expansion is nothing but the binary form of the remainder obtained when the no. is divided by 2^4=16

7654823986734573975876926

The remaider obtained when the above no. is divided by 16 is nothing but the remainder obtianed when the last four digits i.e 6926 is divided by 16 which is 14. Now the binary form of 14 is 1110.

Hence the last four digits are 1110.------------------------

Three teachers respectively distributed 370, 430 and 580 chocolates equally among their three classes. The number of students in all the three classes was the same. That evening three students A, B, C one from each of the three classes respectively got together and shared their chocolates equally after pooling them first. The teachers also pooled their remaining chocolates, which happened to be the same number for all of them and distributed these total remaining chocolates among students of the first class on the next day. The teachers also found that the strength of the class is maximum satisfying all the above conditions.

1. Find the total no. chocolates received by student A of the first class.

2. Find the total no. chocolates distributed among the students of first class.The strength of the class should be such that when it divides 370 430 and 580 the remainder has to be same and it has to be maximum.

So THE STRENGTH OF THE CLASS IS A MAXIMUM NUMBER WHICH LEAVES SAME REMAINDER WHEN IT DIVIDES 370 430 AND 580.

SINCE IT LEAVES SAME REMAINDER WITH EACH NUMBER IT MUST DIVIDE THE DIFFERENCE OF THESE NUMBERS EXACTLY.

i.e GCD(430-370, 580-430) = 30.

SO THE STRENGTH OF EACH CLASS IS 30.

Now 370 chocolates are divided among students of I class So each student gets 12 chocolates and the teacher will be left with 10.

Similarly students of second class gets 14 chocolates each and the teacher will be left with 10 and students of third class gets 19 chocolates each and the teacher will be left with 10 .

So A gets 12 B gets 14 and C gets 19.Now (12+ 14+ 19)/3= 15So finally A got 15 chocolates.

Now coming to the second question 

Initially 360 chocolates are distributed among the students of first class and then the remaining chocolates with all the three teachers i.e 30 are also distributed among the students of first class. Totally 390 and in addition to this A recieved 3 extra chocolates earlier by pooling his chocolates with B and C .

So totally 393 chocolates were received by the students of first class.------------------------A rectangle is such that it can be perfectly cut into smaller squares of a maximum possible side of length 12 units. It is also known that the perimeter of such a rectangle is 384 units.

1.The maximum possible number of squares of side 5 units that can be cut

out of any such rectangle is

2. How many distinct rectangles are there which satisfy the criteria given

3. Suresh takes all such rectangles possible and cuts each of them completely and perfectly into small squares. What is the minimum number of total squares that he can end up with?

Let me explain it .

First let me calculate how many such different rectangles exist.

It is given that the rectangle can be completely cut into squares of max side 12 which means there must be no wastage of rectangle when I cut it into squares of side 12.

This is possible only if the length of the rectangle as well as width of the rectangle is exactly divisible by 12. Since it is given that the max side is 12 so the the greatest common divisor of length and width is 12.

Let W= 12xa and L=12xb (Remember a and b are coprimes here since the GCD is 12)

It is given perimeter is 394i.e 2(12xa + 12xb) = 394i.e a + b = 16

so the combinations are a = 1, b= 15a = 3, b= 13a = 5, b= 11a = 7, b= 9 (Only these are the combinations because a and b are coprimes)

So the rectangles possible are12 xa , 12 x b1. 12, 1802. 36, 1563. 60, 1324. 84, 108

So 4 distinct rectangles are possible.

Coming to our questions now 

The maximum possible number of squares of side 5 units that can be cut out of any such rectangle is 

Since the perimeter of the above 4 rectangles is same I can get max no. squares of side 5 units from a rectangle where L and W are close i.e rectangle 4.i.e rectangle with dimensions 84 x 108

The no. of squares of side 5 units that i can get from this rectangle is 16 x 21 = 336 . [( 84 x 10 / (5 x 5)]

Coming to the next question

Suresh takes all such rectangles possible and cuts each of them completely and perfectly into small squares. What is the minimum number

of total squares that he can end up with?

Since I need the minimum no. squares the side of the squares has to be maximum and the max side possible here is 12.

So if I take the first rectangle which is 12 x 180 (i.e12 x 1, 12 x 15)the no. squares of side that can be obtained from this rectangle is nothing but 1 x 15 =15 which is nothing but a x b

Similarly from second rectangle no. squares that can be obtained is 3 x 13 =39Similarly from third rectangle no. squares that can be obtained is 5 x 11 =55and from fourth rectangle no. squares that can be obtained will be 7 x 9 = 63

So total no. of min squares that can be obtainedis 15 + 39 + 55 + 63 = 172-------------------------------

In the above figure ABC is a rightangled triangle and B= 90 o. A circle is drawn taking AB as the diameter such that it cuts AC at P. Now a tangent is drawn to the circle at the point P. If the tangent meets BC at Q find BQ given that BC = 20cms angle B = angle Q = 90 o.angle A = angle P = x o.

So by AA property they are similarSo great friends different ways and all of them are right.Ofcourse we can use ALTERNATE SEGMENT THEOREM also.

A SQUARE IS ROTATED 45 o ABOUT ITS VERTEX . FIND THE AREA COMMON TO THE TWO POSITIONS.

Let ABCD be a square of side ‘a’. Now APQR is square that is obtained by rotating vertex A of ABCD through an angle 45. Clearly the common region is nothing but the area of the quadrilateral APTD.

Area of APTD = area of ACD – area of TPC

Now it is clear that TCP = 45 (because diagonal bisect DCP in a square)

Also TPC = 90. (Because RQP = 90 in a square)

Obviously CTP = 45 . (Sum of the angles in a triangle is 180)

So clearly TP = PC

So TPC is a right-angled isosceles triangle.

Hence area of TPC = ½ x PC x PT

We know that diagonal AC = a 2

And AP = a (side of the square)

So PC = a2 – a = a (2-1) = TP

Hence area of TPC = ½ x a (2-1) 2 Clearly area of ACD= a2 /2.So the area of APTD = a2 /2- ½ x a (2-1) 2 = a2 (2-1)

I hope everything is clear.

Regards,Suresh

Let { R1,R2,..................., Rn} - where each Ri ( for i= 1-n) represents a rectangle- denote the set of all possible distinct rectangles with integer sides and having an area of 864 sq units each. These rectangles are placed one beside the other on the positive X axis starting from the origin in the following manner.

Each rectangle Ri is placed twice on the X axis - once with one of it's longer sides lying on the X axis and a second time with one of it's shorter sides lying on the X axis. In this manner all the rectangles are placed such that the left edge of every rectangle touches the right edge of the previous rectangle.

1. Find the coordinates of the top right corner of the right most rectangle if it is known that the last rectangle has the least possible height.

2. If all rectangles placed on the X axis have to traced out using a pencil, and one pencil can last only for a length of 100 units, find the minimum no. of pencils required.( common boundaries between any two rectangles will be traced twice- once with each rectangle)

3. Find the area of all rectangles put together.

The answers are 1. (2520,1)2. 1013. 20,736

1. A SQUARE IS ROTATED ABOUT AT IT'S CENTRE THROUGH AN ANGLE OF 45 degrees. FIND THE AREA COMMON TO THE TWO POSITIONS OF THE SQUARE.

2. ABCD IS A PARALLELOGRAM. THE ANGULAR BISECTOR OF ANGLE ABC INTERSECTS AD AT P. IF DP =5 cms, CP = 6cms AND BP = 6 cms FIND AB? 

2. ---------------------------A paper ABCD is in the form of a rectangle of dimensions 18 x 12 cms. The vertex A is placed on the vertex C and is pressed. Find the length of the crease(i.e the line fomed in the middle of the paper) formed.1. TWO CIRCLES OF RADII 2 CMS ARE PLACED ON A PLANE SUCH THAT THE DISTANCE BETWEEN THEIR CENTRES IS 2 CMS. FIND THE AREA COMMON TO THE TWO CIRCLES.

2. IN A TRIANGLE ABC THE PERPENDICULAR DROPPED FROM ITS INCENTRE MEETS AB, BC AND AC AT D, E AND F RESPECTIVELY. FIND THE LENGTHS OF AD, BE AND CF RESPECTIVELY if AB=10 cms BC=12 cms AND AC= 16 cms(pg 23)

1.ABC is right angled triangle with right angled vertex B. AB=6 cms BC=8 cms AC=10 cms. A perpendicular dropped from B meets AC at D. Now a circle is drawn by taking BD as raidus and B as centre. This circle cuts AB at P and BC at Q. Find AP : CQ

2. Mr. X has a triangle whose area is 150 sqcms. The length of the greatest side is 50cms and length of the other side is 10 cms. Find the length of the third side. (pg 23)

Q1.

Given that AB= 6cms BC= 8 cms and AC = 10

By taking BC as the base and AB as the height we get area of the ABCSimilarly if we take AC as the base BD will be the height since BD is AC

So AB x BC = AC x BD

Hence we get BD = 4.8 cms = BQ = BP ( radius of the circle)So AP = AB- BP = 6 - 4.8 = 1.2 & QC = BC- BQ = 8 - 4.8 = 3.2

Finally AP: QC = 3: 8

Q2.

Given greatest side is 50 and other side is 10 cms. Let BC = 50 cms and AC = 10 cms.Since area is given as 150 we have ½ x BC x AD = 150

So ½ x 50 x AD = 150. Hence AD = 6Now ADC is a right-angled triangle in which AC =10 and AD= 6. So DC = 8 cmsHence BD = 42 cms and AD = 6 cmsSince ABD is right angled AB = [(42) 2+(6) 2] = 1800 = 302 cms

Guys I think everything is clear now.

\-------------A square ABCD of side ‘ a ’ units is drawn. With the vertex A, B, C and D as centers, circles of radius ‘ a ’ units are drawn. Find the area that is common to all four circles and the square.

1. For n N, a n = (1000 n)/ n! is greatest when n =?

2. A = 50 99 B= 99! . Which is greater A or B?

3. Find the least value of a x b x c if a, b, c R+ and ab + bc + ac = 122. For n N, a n = (1000 n)/ n! is greatest when n =?

2. A = 50 99 B= 99! . Which is greater A or B?

3. Find the least value of a x b x c if a, b, c R+ and ab + bc + ac = 12

Solutions

1. a n / a n+1 = n+1 / 1000

Clearly a) a n / a n+1 < 1 for 1 n < 999b) a n / a n+1 = 1 for n = 999c) a n / a n+1 > 1 for n > 999

Hence a 1 < a 2 < a 3……………..< a 999 = a 1000 > a 1001 >…………….

Thus the greatest value of a n = a 999 or a 1000

Hence n =999 or 1000

2. A = 50 99 B= 99!

50 can be written as the A.M of the first 99 natural numbers

Hence A=50 99 = [(1+2+3+…………….+97+98+99)/99] 99

Similarly 99! Can be written as (G.M of first 99 natural numbers) 99

Hence B= 99! = [(1 x 2 x 3 x …………x 97 x 98 x 99) 1/99] 99

Since A.M > G.M

A > B

2. Least value of a x b x c if a, b, c R+ and ab + bc + ac = 12

Let me tell you a small funda here.

If a 1+ a 2+ a 3+ a 4+…………+ a n= k (constant)

Then we can give the max value of a 1x a 2x a 3x a 4x ……………x a n

It will be max when a 1= a 2= a 3=……………. =a n

Since ab + bc + ac =12So the max value of ab x bc x ac occurs when ab = bc = ac i.e. a = b = cSo now given expression becomes 3a 2 = 12 i.e. a= 2 (aR+)Hence the max value of a x b x c = a 3 = 8

Now the least value of a x b x c cannot be given exactly but it has to be greater 0 (since a, b, c R+)

So the interval in which a x b x c will lie is (0, 8]

Guys I think everything is clear

Regards,Suresh

1.Find the max value of the x3y4 subject to the condition 2x + 3y = 7 & x 0, y 0

 2. Find the min value of 3x + 4y given that x2y3=6 and x 0,y 0 3. A = Sin2 + Cos4 then for real values of , A ?

Pg 27

1. A pious rich man distributes certain amount of money on every Sunday to poor people. He always carries money in the denominations of 1 Rs and he also distributes equal amount of money to each and every person. On a particular Sunday if he takes 18,720 Rs with him find the probability that the money is divided among 24 people pg 40

Guys, the answer is 72. I am sorry for my previous post saying that there's a small mistake for the people who gave the answer 72, because I took the total money as 20,720 Rs.

Now here comes the explanationTotal money that is distributed is 18,720 Rs.Now we have to calculate the total facators of this number for the following reason.

Now 18,720 =24 * 780

i.e we can interpret this two ways1. 24 people each receiving 780 Rs2. 780 people each receiving 24 Rs.

So total number of chances to distribute the money is nothing but it's total factors i.e 72 in this case out of which there is only one favourable chance

Hence the probability of distributing 18,720 Rs among 24 people is 1/72

1. A cow was standing on a bridge, 5feet away from the middle of the bridge. suddenly a lightning express with 90 miles/hr was coming towards the bridge from nearest end of the cow.seeeing this the cow ran towards the express and managed to escape when the train is one feet away from the bridge. if it would have ran to opposite direction(ie away from train) it would have been hit by the train one ft away from the end of the bridge. Calculate the length of bridge.

2. There is only a single track between two stations P and Q. Exactly one hour after train A leaves P for Q , tain B arrives at P which started at Q. On a partcular day tain A started late by 20 and inorder to manage the schedule it has doubled its speed. Train B also changed its speed and finally reached P on time. If their initial speed ratios are 4:1 respectively find the ratio of their modified speeds.

3. Mr. A starts at a point P at 3 a.m and moves towards Q. Mr.B starts at point Q at 8a.m and moves towards P. They met at 12 noon. After meeting if they took equal amout of time to reach their respective destinations, at what time did they reach their destinations.

4. Certain men can do a work in certain number of days. After they worked for 12 days 1/4 of them have left. To complete the remaining work the remaining have taken as many days as the initial number of would have taken to complete the entire work. In how many days is the work completed.

Remainder when (56)^493 is divided by 126. pg 51

1.A set of six different positive integers a<b<c<d<e<f has a range that is equal to its mean. What is the smallest possible value of a? 

(A) 1 (B) 2 (C) 3 (D) 4

2.If the sum of four consecutive positive integers a three digit multiple of 50, the mean of the these integers must be one of k possible values, where k= 

(A) 7 (B) 8 (C) 9 (D) 10