1!+2!+3!+4! will end in 3, rest of the sum will have 0
so units digit of x is 3
by cyclicity units digit of 3^(4x+1) will always end in 1

Let x = 1! + 2! + 3! + 4! + ... + 100!. Unit's digit of
X^x^x^.... upto infinity is
(a) 3
(b) 1
(c) or 3 depending upon the number of times
x appears in the power.
(d) Cant be determined.
mere according option (2)
i don't have oa of this....

X^x^x^.... will be X^4k+1

1+2+6+24=3 (after this last digit will be 0 only)and 3^(4k+1) will be 3 only.

There were some people invited to a party. All but one person arrived in time and everyone shook hands with everyone else present in the party. However, John reached 30 minutes late, and so he was only able to shake hands with some of other guests. If there were exactly 99 handshakes in total, then what is the number of hands John did not shook?
(a) 6
(b) 7
(c) 8
(d) 9
(e) None of the above

suppose there are 'n' people. Initially only n-1 people arrive on time and the number of handshakes would be (n-1)C2, john shakes hands with x people. we need n-x-1
when n=15 we get 14C2 = 91 and john shakes hands with 8 people so that total number of handshakes is 99.
n=15 and x = 8
number of hands john did not shook6

Let x = 1! + 2! + 3! + 4! + ... + 100!. Unit's digit of
X^x^x^.... upto infinity is
(a) 3
(b) 1
(c) or 3 depending upon the number of times
x appears in the power.
(d) Cant be determined.
mere according option (2)
i don't have oa of this....

There were some people invited to a party. All but one person arrived in time and everyone shook hands with everyone else present in the party. However, John reached 30 minutes late, and so he was only able to shake hands with some of other guests. If there were exactly 99 handshakes in total, then what is the number of hands John did not shook?
(a) 6
(b) 7
(c) 8
(d) 9
(e) None of the above

there are two people A and B and there work efficiency is 2:5 , now if B takes 90 days less than A than how many days A takes to compelte the task

Work efficiency means amount of work done in a single day....Let A does x units of work in one day and B does y units of work in one day. It is given that

x/y = 2/5

When A does 2 units of the work , B does 5 units of work in one day. Let the total work be 10 units.
A takes 5 days......B takes 2 days to complete 10 units of work. We can see that the ratio of the number of days is just the reverse of the ratio of the efficiency.

5x - 2x = 90
3x = 90
x = 30
5x = 5*30 = 150 days.

I hope it is clear now!!

How can this be solved ?
How many zeros will be there in 26!

Calculate the number of 5 in that
26/5+26/25 = 5+1 = 6

bad bad i do not know when will i start reading correctly

there are two people A and B and there work efficiency is 2:5 , now if B takes 90 days less than A than how many days A takes to compelte the task

3x = 90
so 5X = 150

Can you please xplain it in more detailed way ???

:-(:-(:-(

See edited post please, mene kuch pada kuch aur samja tha

days will be in 5:2 ratio
so if B takes 90, A = 5*45 = 225

Can you please xplain it in more detailed way ???

:-(:-(:-(

retry Says

each of the 25 persons shake hands with 24 people. since double counting is done, the number would be 25.24/2 = 300 .. ?

Here it is number of handshake but the ways (order ) of handshake

there are two people A and B and there work efficiency is 2:5 , now if B takes 90 days less than A than how many days A takes to compelte the task

no of days taken by a:no of days taken by b=a:b=5:2

b takes x-90 and a takes x
x/(x-90)=5/2

2x=5x-450
x=150

there are two people A and B and there work efficiency is 2:5 , now if B takes 90 days less than A than how many days A takes to compelte the task

days will be in 5:2 ratio
days will be in 5:2 ratio
A-B = 90
3x=90
X=30
so A =150

there are two people A and B and there work efficiency is 2:5 , now if B takes 90 days less than A than how many days A takes to compelte the task

How can this be solved ?
How many zeros will be there in 26!

Exactly How many 5 will be there

26/5+26/25 = 5+1 = 6

How can this be solved ?
How many zeros will be there in 26!

In how many ways 25 person present in a room can handshake ?

each of the 25 persons shake hands with 24 people. since double counting is done, the number would be 25.24/2 = 300 .. ?

In how many ways 25 person present in a room can handshake ?

(24+23+.........1) 24! ways
P.S.: Me confused Hu

In how many ways 25 person present in a room can handshake ?

2/1 + 3/2! + 6/3! + 11/4! + 18/5!

ka ek aur alternate kaam chalau solution ..

2 + 1.5 + 1 + 11/24 + 18/120

4.5 + 11*.04 + 18*.009

4.5 + .44 + .162 ~ 5.1

going by options .. e = 2.73 => 3e = 6+2.1+.09 = 8.2
3e -3 is the closest match ..

find the infinite sum of the foll seuence
1+(4/7)+(9/(7^2))+(16/(7^3))+.....

if options are suitable, this can also be solved by calculating first 3/4 terms

4/7 = 4*.14 = 0.56 (actual value .571 )
9/49 ~ 9/50 = 9*.02 = .18 (actual value .183 )
16/343 ~ 16/333 = 16*.003 = .048 (actual value .046 )

add to get

1+.560+.180+.048=1.788


OA shayad 49/27 hai .. 49/27 ko division se 2 decimals tak nikal lo .. comes out to ~ 1.8

can someone post the other options .. im sure dividing them will give decently spaced numbers ..

find the infinite sum of the foll seuence
1+(4/7)+(9/(7^2))+(16/(7^3))+.....

=> S = 1 + (4/7) + (9/7^2) + ...
=> S/7 = 1/7 + (4/7^2) + (9/7^3) + ...
Subtract them we get
=> 6S/7 = 1 + 3/7 + 5/49 + ...
=> 6S/49 = 1/7 + 3/49 + 5/343 + ...
Subtract them again
=> 36S/49 = 1 + 2/7 + 2/49 + ...
=> 36S/49 = 1 + (2/7) / (6/7)
=> S = 49/27

find the infinite sum of the foll seuence
1+(4/7)+(9/(7^2))+(16/(7^3))+.....

S= 1+(4/7)+(9/(7^2))+(16/(7^3))+.....

S/7 = 1/7 +(4/(7^2))+(9/(7^3))+.....
6S /7 = 1+3/7+5/7^2+.......

Now this series is in A.P. & G.P.

S = ab/1-r+dab/(1-r)^2

here a = first number in A.P.
B= first number in G.P.

1/1-1/7+2/(1-1/7)^2 = 70/18

so sm is: 7/6*70/18

P.S.: Kisi ko formula pata ho to chk kar lo, I am confused betn (1-r)^2 otr 1-r^2

find the infinite sum of the foll seuence
1+(4/7)+(9/(7^2))+(16/(7^3))+.....

TT School of Management institute involved in teaching, training and research. Currently it has 37 faculty
members. They are involved in three jobs: teaching, training and research. Each faculty member working
with TT School of Management has to be involved in at least one of the three job mentioned above:

-
A maximum number of faculty members are involved in training. Among them, a number of faculty
members are having additional involvement in the research.
-

The number of faculty members in research alone is double the number of faculty members involved
in all the three jobs.
-

17 faculty members are involved in teaching.
-

The number of faculty members involved in teaching alone is less than the number of faculty members
involved in research alone.
-

Ten faculty members involved in the teaching are also involved in at least one more job.

80. After some time, the faculty members who were involved in all the three tasks were asked to with
drawl from one task. As a result, one of the faculty members each opted out of teaching and
research, while remaining ones involved in all the three tasks opted out of training. Which one of the
following statements, then necessarily follows:
(A) The least number of faculty members is now involved in teaching.
(B) More faculty members are now associated with training as compared to research.
(C) More faculty members are now involved in teaching as compared to research.
(D) None of the above.

81. Based on the information given above, the minimum number of faculty members involved in both
training and teaching, but not in research is:
(A) 1 (B) 3 (C) 4 (D) 5


Approach plz ?

Solve and find the pair of non negative integers (x, y) such that
(x+ y - 5)^2 = 9xy.

rohitpal Says

hi plz solve the attached problems.

Doing Second: log(4)a-b = log(4) (_/ a-_/b)^2+log(4)4
log(base 4)a-b = log(base 4) 4*(_/ a-_/b)^2
a-b = 4*(_/ a-_/b)^2
taking sqaure root
_/ (a-b) = 2_/a-2_/b
iI think it should be 1 only keeping option in mind

ind the unit digit : 12^55/3^11 + 8^48/16^18

is ans 0?

(12^5/3) ^11 % 3

obviously this will be 0 .

8^48/16^18 = 2^144/ 2^72 : remainder will again 0 so my take 0

Shyam, Gopal and Madhur are three partners in a business. Their capitals are respectively
Rs 400, Rs 8000 and Rs 6000. Shyam gets 20% of total profit for managing the business.
The remaining profit is divided among the three in the ratio of their capitals. At the end of the
year, the profit of Shyam is Rs 2200 less than the sum of the profit of Gopal and Madhur.
How much profit, Madhur will get?

could anyone expalin me that?

remaining whatever it is divided into

S:G:M = 4: 80: 60 = 1:20:15

Now let profit y then 80% of the same will be distributed and S will get y/36*80/100+20/100Y

G = 20/36*80/100y

M=15/36*80/100y

so the final ratio will be 800:1600:1200

8:16:12 = 2:4:3

2x 4x 3x

5x = 2200

x= 440

so 3x = 440 =1320 my take

Q. A card is drawn at random from a well shuffled pack of 52 cards.

X: The card drawn is black or a king.
Y: The card drawn is a club or a heart or a jack.
Z: The card drawn is an ace or a diamond or a queen.
Then which of the following is correct?
A. P(X) > P(Y) > P(Z)
B. P(X) >= P(Y) = P(Z)
C. P(X) = P(Y) > P(Z)
D. P(X) = P(Y) = P(Z)

IIFT - 2009

p(x) = 26/52+6/52 =32/52
p(y) = 13/52 +13/52+2/52 =28/52
p(z) = 13/52+3/52+3/52 = 19/52

so p(x)> p(y) > p(z)

option A

P.S.: Pata nahi me tash k patto ko thik se samaj payi hu ya nai, I always got confused betn "Ace" and "club" or "jack"

thanks dude..i tried the same eqn to solve,but i couldnt solve it..can u pls show the solving procedure..
x^2 + y^2 = 2.5(x+y) --(i)
=> x^2 + y^2 - (3xy) = 1 -- (ii)

x^2 + y^2 = 2.5(x+y) --(i)
=> x^2 + y^2 - (3xy) = 1 -- (ii)

2.5x+2.5y = 1+3xy

let x=0 y =1/2.5
y=0 X= 1/2.5

I think soln can be more than 1

If we add the square of the digit in the tens place of the positive two digit number to the product of the digits of that number we get 52,and if we add the square of the digit in the unit's place to the same product of the digits we get 117.Find the two digit number??
i got till this step
x^2+xy=52
x(x+y)=52--(1)
y^2+xy=117
y(y+x)=117--(2)
after this hw to proceed the given answer is 30..pls explain in details with steps.
thanks in advance

10x+y is the number

x*y+x^2 =52
x*y+y^2 =117

y^2-x^2 = 65

y+x = 13
y-x= 5

aading 2y = 18 y= 9
x= 4

so value is 49

Ani1308 Says

hi..a little help in remainder part...find the rem of (3^5^7^9)/41

5^7^9 = 5k

so it is 4k+1

3*3*3*3*3 /41 remainder 38

jain4444 Says

The sum of digits of a natural number (10^n 1) is 4707, where n is a natural number. The value of n is ?

The sum of digits of a natural number (10^n 1) is 4707, where n is a natural number. The value of n is ?
10^n-1 is obviously multiple of 9 and 11
so 4707/9 =523 (sum will be 9*n where n is no. by which 9 is repeated)
so value is 9999999999999999............523 times

P.S.:- jain sahab itna easy kaise de sakte hai

hi sry man...i dont have the answers bt i have one more question on the seme pattern...if u liketo practice dn..
5^25^125^3125 /11..

thanks all...

The remainder is 1.
E(11) = 10
25^anything has unit digit 5
5^(10K+5) % 11 = 5^5 % 11 = 25*25*5 = 45 = 1

hi sry man...i dont have the answers bt i have one more question on the seme pattern...if u liketo practice dn..
5^25^125^3125 /11..

thanks all...

E(41) = 40

We need to find 5^7^9 % 40

E(40) = 16

7^9%16 = 7

5^(16K+7) % 40 = 5 * 5^(16K+6)%8

Now, E(8 ) = 4

Therefore => (5^16K + 6) % 8 = 1
=> 5^(16K+7) % 40 = 5
3^(40K + 5) % 41 = (3*81) % 41 = -3 = 38

The remainder is 38

Ani1308 Says

but dude u cant apply euler in bold part...5 and 40 are nt co prime

I have not applied Euler in that part...I just calculated the remainder of 7^9 with 16....Then I represented 7^9 as (16K + 7), then I segregated common multiple from numerator and denominator. After that I applied Euler of 8....
What is the O.A.??

Ani1308 Says

hi..a little help in remainder part...find the rem of (3^5^7^9)/41

(3^5^7^9)/41
3 and 41 are coprimes.
=> E(41) = 40
5^7^9/40 => 40 = 5*8

=> 5^7^9/8 => E(8 ) = 4
=> 7^9/4 = 3^9/4 => -1 or 3 remainder

=> 5^3/8 = 125/8 => 5 remainder

5^7^9 => 5x = 8y+5 => 5 remainder

3^5/41 = 3*81/41 = 120/41 = 38 remainder

Sorry for x blunder.:-(

No problem.

E(41) = 40

We need to find 5^7^9 % 40

E(40) = 16

7^9%16 = 7

5^(16K+7) % 40 = 5 * 5^(16K+6)%8

Now, E(8 ) = 4

Therefore => (5^16K + 6) % 8 = 1
=> 5^(16K+7) % 40 = 5
3^(40K + 5) % 41 = (3*81) % 41 = -3 = 38

The remainder is 38

but dude u cant apply euler in bold part...5 and 40 are nt co prime

Ani1308 Says

hi..a little help in remainder part...find the rem of (3^5^7^9)/41

E(41) = 40

We need to find 5^7^9 % 40

E(40) = 16

7^9%16 = 7

5^(16K+7) % 40 = 5 * 5^(16K+6)%8

Now, E(8 ) = 4

Therefore => (5^16K + 6) % 8 = 1
=> 5^(16K+7) % 40 = 5
3^(40K + 5) % 41 = (3*81) % 41 = -3 = 38

The remainder is 38

Ani1308 Says

hi..a little help in remainder part...find the rem of (3^5^7^9)/41

i am getting 38 ... oa?


C(41)=40

checking for 5^7^9 mod 40

C(40) = 4

7^9 mod 4= 3

so... 5^3 mod 40=5

and 3^5 mod 41 = 38

4x+7y = 3
When x = 6; y=-3 .. x=13, y=-7
=> x increases with a common difference of 7 and y decreases with a common difference of 4. As x is positive, we'll consider only x now.
=> 6,13,20 ..., Last term (Tn) = a+(n-1)d = 1000 => 6+7n-7 = 1000
=> 7n = 1001 => n=143 terms
=> last term = 6+142*7 = 6+994 = 1000. But we want x less than 1000. Hence only 142 terms will be considered.

When x = -1, y=1; x=-8, y=5 Here y is increasing with a common difference of 4 and is positive, So, we'll consider only y here.
=> 1,5,9,13,.. (1+4n-4 = 1000 => 4n=1003 => n = 250.xx or 250 terms)

Had to google .. the actual question is x < 1000 and |y < 1000
So, Again x will reach 1000 first .. 1,8,14, .. (1+7n-7 = 1000 => 7n = 1006 => n=143) Hence 143 terms => 143rd term = 1+142*7 = 1+994 = 995

Hence, total terms = 142+143 = 285 terms.

Thanks a lot.:)
Sorry for |x blunder.:-(

hi..a little help in remainder part...find the rem of (3^5^7^9)/41

Kindly help me solve this.
How many integer values of x and y satisfy the expression 4x+7y=3 where x<1000 & y<1000?
a).284
b).285
c).286

4x+7y = 3
When x = 6; y=-3 .. x=13, y=-7
=> x increases with a common difference of 7 and y decreases with a common difference of 4. As x is positive, we'll consider only x now.
=> 6,13,20 ..., Last term (Tn) = a+(n-1)d = 1000 => 6+7n-7 = 1000
=> 7n = 1001 => n=143 terms
=> last term = 6+142*7 = 6+994 = 1000. But we want x less than 1000. Hence only 142 terms will be considered.

When x = -1, y=1; x=-8, y=5 Here y is increasing with a common difference of 4 and is positive, So, we'll consider only y here.
=> 1,5,9,13,.. (1+4n-4 = 1000 => 4n=1003 => n = 250.xx or 250 terms)

Had to google .. the actual question is x < 1000 and y < 1000
So, Again x will reach 1000 first .. 1,8,14, .. (1+7n-7 = 1000 => 7n = 1006 => n=143) Hence 143 terms => 143rd term = 1+142*7 = 1+994 = 995

Hence, total terms = 142+143 = 285 terms.

Kindly help me solve this.
How many integer values of x and y satisfy the expression 4x+7y=3 where x<1000 & y<1000?
a).284
b).285
c).286

The equation is x = (3-7y)/4
y = 1,5,9,....,569
Total 143 values

Also y = -3,-7,.....,-567
Total values = 142

My take is 143+142 = 285

Kindly help me solve this.
How many integer values of x and y satisfy the expression 4x+7y=3 where x<1000 & y<1000?
a).284
b).285
c).286

jain4444 Says

The sum of digits of a natural number (10^n 1) is 4707, where n is a natural number. The value of n is ?

kya jain bhai surra ball dal rahe ho....
523

jain4444 Says

The sum of digits of a natural number (10^n 1) is 4707, where n is a natural number. The value of n is ?

4707/9 = 523

hence n = 523

jain4444 Says

the sum of digits of a natural number (10^n 1) is 4707, where n is a natural number. The value of n is ?

4707/9=523

jain4444 Says

the sum of digits of a natural number (10^n - 1) is 4707, where n is a natural number. The value of n is ?

523???????

The sum of digits of a natural number (10^n - 1) is 4707, where n is a natural number. The value of n is ?

thanks dude..i tried the same eqn to solve,but i couldnt solve it..can u pls show the solving procedure..
x^2 + y^2 = 2.5(x+y) --(i)
=> x^2 + y^2 - (3xy) = 1 -- (ii)

Sure..
See
we can write it as 2.5(x+y) = 1 + 3xy
=> 5x + 5y = 2 + 6xy
Hit and trial..see LHS will either end with 5 or 0 and RHS needs to be same too..so for 0 we can try RHS = 20..
=> x=1,y=3 or x=3,y=1

thanks dude..i tried the same eqn to solve,but i couldnt solve it..can u pls show the solving procedure..
x^2 + y^2 = 2.5(x+y) --(i)
=> x^2 + y^2 - (3xy) = 1 -- (ii)

1)The sum of the squares of the digits constituting a two- digit positive number is 2.5 times as large as the sum of its digits and is larger by unity than the trebled product..

2)the unit digit of the two digit number is greater than its tens digit by 2,and the product of that number by the sum of its digit is 144.find the number

pls explain the steps
thanks in advance

1) Let the 2 digit be xy
=> x^2 + y^2 = 2.5(x+y) --(i)
=> x^2 + y^2 - (3xy) = 1 -- (ii)
We can solve both the eq to get the values

2) 2 digit no is xy
=> y - x = 2 -- (i)
=> xy * (x+y) = 144 -- (ii)
No is 24

1)The sum of the squares of the digits constituting a two- digit positive number is 2.5 times as large as the sum of its digits and is larger by unity than the trebled product..

2)the unit digit of the two digit number is greater than its tens digit by 2,and the product of that number by the sum of its digit is 144.find the number

pls explain the steps
thanks in advance

find the max and min values of

f(x) = 3sinx + 4cosx


can sum1 pls explain me the sol for the same... am weak at maxima and minima...

If f(x) = aSinx + bCosx, then

f(x) = {(a + b)}{aSinx/(a + b) + bCosx/(a + b)} = (a + b) Sin(x + y)
where y = Sin inverse (b/(a + b))

So, range for f(x) is

If we add the square of the digit in the tens place of the positive two digit number to the product of the digits of that number we get 52,and if we add the square of the digit in the unit's place to the same product of the digits we get 117.Find the two digit number??
i got till this step
x^2+xy=52
x(x+y)=52--(1)
y^2+xy=117
y(y+x)=117--(2)
after this hw to proceed the given answer is 30..pls explain in details with steps.
thanks in advance

are you sure the answer is 30???
cos i proceeded this way...
x^2 + xy = 52 ............(1)
y^2 + xy = 117 ..........(2)

sub 1 and 2 gives u
y^2 - x^2 = 65
(y + x)(y-x) = 65...
further proceeding...
y+x = 13 and y-x = 5.. leaves us the option as y=9 and x=4...
so the number is 49.. it cannot be 30 dude.. 30 doesnt satisfy whatever is given the question.... answer is 49....

correct me if am wrong....

find the max and min values of

f(x) = 3sinx + 4cosx


can sum1 pls explain me the sol for the same... am weak at maxima and minima...

Anne is 26 years old and Barbara is older than Anne. After 'n' years, where n is a natural number, Anne's age and Barbara's age will both be two digit numbers and have the property that Anne's age is obtained by interchanging the digits of Barbara's age. Let d be Barbara's present age. How many ordered pairs of positive integers (d,n) are possible?

could anyone explain it clearly?

Suppose age of Anne after n years = 10a + b
=> Age of Barbara will be = 10b + a

=> b > a

So, we can choose (a, b) in C(9, 2) = 36 ways

But ab > 26
So, 12, 13, ..., 19, 23, 24, 25, 26 are not possible

So, 36 - 12 = 24 ordered pairs!!

If we add the square of the digit in the tens place of the positive two digit number to the product of the digits of that number we get 52,and if we add the square of the digit in the unit's place to the same product of the digits we get 117.Find the two digit number??
i got till this step
x^2+xy=52
x(x+y)=52--(1)
y^2+xy=117
y(y+x)=117--(2)
after this hw to proceed the given answer is 30..pls explain in details with steps.
thanks in advance

Hi Neha,

I wud just tel u d method without telling you the actual steps, try solving it on ur own

Assume profit to be 100 ( comfort sake ). Then profit ratio wud be: 2% for shyam, 55 % for Gopal and 43 % for madhur. Now shyam has to get 20 % of total profit and the remainig profit ( 80%) wud be divided in the ratio : 2:55:43 percentages.

Now shyam= (Gopal + Madhur) minus 2200.

Put it in equations and voila !! u get d answer
Note: remember to divide only the 80% of the total profit in the ratio of their capitals bcos shyam has to get 20 % of total profit :)

Shyam, Gopal and Madhur are three partners in a business. Their capitals are respectively
Rs 400, Rs 8000 and Rs 6000. Shyam gets 20% of total profit for managing the business.
The remaining profit is divided among the three in the ratio of their capitals. At the end of the
year, the profit of Shyam is Rs 2200 less than the sum of the profit of Gopal and Madhur.
How much profit, Madhur will get?

could anyone expalin me that?