Steel Express runs between Tatanagar and Howrah and has five stoppages in between. Find the number of different kinds of one-way second class ticket that Indian Railways will have to print to service all types of passengers who might travel by Steel Express?
(A) 49 (B) 42
(C) 21 (D) 7

Kal raat ko decide toh hua tha answer is 42..

For all those who have a difficulty in understanding why it would be 42, look at this.
One way matlab from the station of boarding to the destination station,
and it isnt written that the train cannot travel from howrah to tatanagar. It runs BETWEEN and not from or to to any place.
Hence, it would be 42.
Plus chill sir also confirmed, so kisi ko refute karna hai? chill sir ke answer ke against?

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Steel Express runs between Tatanagar and Howrah and has five stoppages in between. Find the number of different kinds of one-way second class ticket that Indian Railways will have to print to service all types of passengers who might travel by Steel Express?
(A) 49 (B) 42
(C) 21 (D) 7

opt c=21.
tot 7 stations..from tatanagar we cn purchase 6 diff tickets,similarly from sta(right after tatanagar) we cn purchase 5 diff tickets...and so on hence tot tickets=6+5+4+3+2+1=21

Given that a^5 + b^5 + c^5 = 91849, where a, b, and c are distinct digits. What is the remainder when a
six-digit number 2a5b1c is divided by 11?
a. 0 b. 6 c. 2 d. 1

My take is 0

We have unit digits of (a^5 + b^5 + c^5) and (a+b+c) as same.
=> a+b+c ends in 9
=> a+b+c = 9 or 19 .... can't be 29 or more as sum of 3 digits can be at most 27

=> a+b+c = 19 .... can't be 9 as 9^5 < 91849

=> Remainder by 11 = (a+b+c) - (2+5+1) = 19-8 = 11 => 0

Given that a^5 + b^5 + c^5 = 91849, where a, b, and c are distinct digits. What is the remainder when a
six-digit number 2a5b1c is divided by 11?
a. 0 b. 6 c. 2 d. 1

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Steel Express runs between Tatanagar and Howrah and has five stoppages in between. Find the number of different kinds of one-way second class ticket that Indian Railways will have to print to service all types of passengers who might travel by Steel Express?
(A) 49 (B) 42
(C) 21 (D) 7

From Tatanagar to Howrah (one-way) = C(7,2) = 21
From Howrah to Tatanagar (another one-way) = C(7,2) = 21

So, Total one-way tickets must be 42


Nachiket bhai :w00t:
Jain bhai bhala ho tumhara jo Nachiket bhai ke darshan karwaye tumne :clap:

am getting 3584, however 7000 is possible when
7777-7-70-700
i dont think this is correct because the number of two digit numbers = 77-7 = 70 is wrong because 77-7 is 70 in base 10. I dont know if i am wrong here. Someone please clarify this.

I think the answer must be 7*8*8*8 = 3584

OA was given to be 7000
qsn was abt base 7....
if the qsn was abt decimal numbers represented in base 7 ..I think 3854 would have been the answer......

Shawn90 Says

think bhai agar aap 0 lete to sab perfect squares to hoga 1 se 100 tak...??

Exactly..

In general, you can keep follwowing points in mind:

(1) Any odd number can be expressed as difference of two perfect sqaures.
=> 1 = 1-0 then 3 = 4-1 then 5 = 9-4 then 7 = 16-9 and so on..

(2) Any even number of from 4n can be represented as difference of two perfect sqaures
=> 4 = 4-0 then 8 = 9-1 and so on..

(3) An even number of form 4n+2 cannot expressed as difference of two perfect sqaures. 2, 6, 10, 14, 18, ... cannot be expressed as difference of two perfect sqaures

So, out of 1 to 100; a (3/4)th of numbers can be expressed as difference of two perfect sqaures.

at last i found GOD of quant = ThinkAce

Arre.... Nahi bhai.. :)
Here is photo of Chill sir.. Who has that honour.. :)

http://www.pagalguy.com/forum/pagalguy-community-events-pg-meets/67932-14th-june-happy-birthday-chillfactor-2.html#post2785370

at last i found GOD of quant = ThinkAce

4 = 2^2 - 0^2



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think bahi agar aap 0 lete hain to sab squares 0 to 100 tak to hoga...???

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Steel Express runs between Tatanagar and Howrah and has five stoppages in between. Find the number of different kinds of one-way second class ticket that Indian Railways will have to print to service all types of passengers who might travel by Steel Express?
(A) 49 (B) 42
(C) 21 (D) 7

it as c. 21 as On 1st stoppage 6 kinds can be given 5 on second and so on . On addition it comes as 21.

4 = 2^2 - 0^2



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think bhai agar aap 0 lete to sab perfect squares to hoga 1 se 100 tak...??

At what rate were the oranges sold per dozen if by bargaining to get one orange per rupee more, I saved Rs 16 per dozen?
1. Rs.8 2. Rs.4 3. Rs.5 4. Rs.12

My take is 8

Say we purchased 12 oranges at Rs. 'x'.

=> We got 'x' orages as bonus
=> We paid 'x' for (12-x) orages

=> Rs. x for (12-x) oranges and Rs. (16+x) for 12 orages
=> x/(12-x) = (16+x)/12
=> x = 8

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Steel Express runs between Tatanagar and Howrah and has five stoppages in between. Find the number of different kinds of one-way second class ticket that Indian Railways will have to print to service all types of passengers who might travel by Steel Express?
(A) 49 (B) 42
(C) 21 (D) 7

(Selection of two different stations among 7)*2 = 7C2*2=>42

65*29*37*63*71*87*62 last two digit plz tell me a short cut to solve it????

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Steel Express runs between Tatanagar and Howrah and has five stoppages in between. Find the number of different kinds of one-way second class ticket that Indian Railways will have to print to service all types of passengers who might travel by Steel Express?
(A) 49 (B) 42
(C) 21 (D) 7

7c2=21
bhaiya one way hai to 21*2=42 nahi hogaaaaaaa kal cat prep main bada bawala hua tha is par

techsavvy123 Says

Starting my preps for the 2012 and here is my subscription. Hope, my office doesn't ban PG and i remain active here throughout!

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Steel Express runs between Tatanagar and Howrah and has five stoppages in between. Find the number of different kinds of one-way second class ticket that Indian Railways will have to print to service all types of passengers who might travel by Steel Express?
(A) 49 (B) 42
(C) 21 (D) 7

At what rate were the oranges sold per dozen if by bargaining to get one orange per rupee more, I saved Rs 16 per dozen?
1. Rs.8 2. Rs.4 3. Rs.5 4. Rs.12

Shawn90 Says

jain bhai how can 4 be expressed as the difference o f two squares....??

4 = 2^2 - 0^2

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(a-b)(a+b) = k


when both are odd , consider all prime number below 100 except 2

so , 24 such numbers.

when both are even,

possibilities = 4*(prime number)

9 cases : 8, 12, 20, 28, 44, 52, 68, 76, 92

Apart from above we can also count 1 and 4 , so


total = 24+9+2 = 35

jain bhai how can 4 be expressed as the difference o f two squares....??

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allan89 Says

sorry jain bhai, par how are u gettng 4 as a difference??

because we need even number on both side

for example :- 4*13 = 26*2

if you take 8 then it can be further classified in to more even cases

for example :- 8*13 = 26*4 = 52*2

What will be the last digit of the expression??
11^1 + 12^2 + 13^3 + 14^4 + 15^5 + 16^6

my take is 9...is that the oa?

yeah i think we won't consider 1 here

so , -1 (1) and +1(16)

answer remain 35

sorry jain bhai, par how are u gettng 4 as a difference??

1001 = 7*11*13

because of square we need at least 2 power of 7,11,13

so , in 26! we have 2-2 power of all prime factors

mayank malik Says

Can you please elaborate this

1*2*3......6*7*8......10*11*12*13...

now see in 13! we have 1-1 power of all three but in 1001^2 we need at least 2 power of all three

1*2*3......6*7*8......10*11*12*13*14(7*2)....21*22(11*2)*23....25*26(13*2)

now in 26! we have 2-2 power of all three

but jain bhai wat about the no 16??

16=16*1

16=8*2;

16=4*4;

out of these there is only 1 way. 8*2, i.e 25-9=16

yeah i think we won't consider 1 here

so , -1 (1) and +1(16)

answer remain 35

1001 = 7*11*13

because of square we need at least 2 power of 7,11,13

so , in 26! we have 2-2 power of all prime factors

Can you please elaborate this

(a-b)(a+b) = k


when both are odd , consider all prime number below 100 except 2

so , 24 such numbers.

when both are even,

possibilities = 4*(prime number)

9 cases : 8, 12, 20, 28, 44, 52, 68, 76, 92

Apart from above we can also count 1 and 4 , so


total = 24+9+2 = 35

but jain bhai wat about the no 16??

16=16*1

16=8*2;

16=4*4;

out of these there is only 1 way. 8*2, i.e 25-9=16

What will be the last digit of the expression??
11^1 + 12^2 + 13^3 + 14^4 + 15^5 + 16^6

My take - 9

What will be the last digit of the expression??

11^1 + 12^2 + 13^3 + 14^4 + 15^5 + 16^6

we just need to find remainder by 10

1+4+7+6+5+6 mod 10 = 9 = unit digit

What will be the last digit of the expression??
11^1 + 12^2 + 13^3 + 14^4 + 15^5 + 16^6

1+4+7+6+5+6=299

What will be the last digit of the expression??

11^1 + 12^2 + 13^3 + 14^4 + 15^5 + 16^6

How many different four digit numbers are there in the octal(Base 8 ) system,expressed in that system?
3584
2058
6000
7000

OA:7000

am getting 3584, however 7000 is possible when
7777-7-70-700
i dont think this is correct because the number of two digit numbers = 77-7 = 70 is wrong because 77-7 is 70 in base 10. I dont know if i am wrong here. Someone please clarify this.

I think the answer must be 7*8*8*8 = 3584

how man numbers below 100 can be expressed as a difference of 2 perfect squares in only 1 way?
15
25
35
45

(a-b)(a+b) = k


when both are odd , consider all prime number below 100 except 2

so , 24 such numbers.

when both are even,

possibilities = 4*(prime number)

9 cases : 8, 12, 20, 28, 44, 52, 68, 76, 92

Apart from above we can also count 1 and 4 , so


total = 24+9+2 = 35

we need to look form backwards

40,39,38,37.......14,13

so , 40-13+1 numbers = 28

den OA must be wrong

28 it is den

snipped........

we need to look form backwards

40,39,38,37.......14,13

so , 40-13+1 numbers = 28

srry puys i had the wrong OA....

answer must be 28 then

how man numbers below 100 can be expressed as a difference of 2 perfect squares in only 1 way?
15
25
35
45

SOL: P^2-Q^2= N

=>(P+Q)(P-Q)=N

=>P+Q AND P-Q Both must either be odd or both even. also since they can be expressed in only 1 way so numbers satisfying would be:

ALL Primes between 1 to 100 except 2. =24

All nos of the form 4k where k is a prime number and 1<4k<100=>k=2,3,5,7,11,13,17,19,23 =9 numbers

All nos of the form 8k where k is a prime number and 1<8k<100

but only k=2 only satisfy.

total 34numbers i'm GETTING. kahan miss ho raha hain

Let S={1,2,3,....N} be a set of N natural numbers.Let T be a subset of S such that sum of any 3 elements of T is greater than or equal to N.
Find the maximum number of elements in an subset T for N=40
26
27
28
None of these

OA:None of these

we need to look form backwards

40,39,38,37.......14,13

so , 40-13+1 numbers = 28

On similar lines::

if (1001)^2 exactly divides N! .then what is the lowest value N can assume??

1001 = 7*11*13

because of square we need at least 2 power of 7,11,13

so , in 26! we have 2-2 power of all prime factors

m getting 28
T=(13,14,15...........,40)
TOTAL 28 terms are there

I am also getting the same........OA is none of these

how man numbers below 100 can be expressed as a difference of 2 perfect squares in only 1 way?
15
25
35
45

On similar lines::

if (1001)^2 exactly divides N! .then what is the lowest value N can assume??

the lowest value N = 26

Let S={1,2,3,....N} be a set of N natural numbers.Let T be a subset of S such that sum of any 3 elements of T is greater than or equal to N.
Find the maximum number of elements in an subset T for N=40
26
27
28
None of these

OA:None of these

m getting 28
T=(13,14,15...........,40)
TOTAL 28 terms are there

frnds help me on following quest.

1 Find the maximum value of n such that 157! is perfectly divisible by 18^n ?

On similar lines::

if (1001)^2 exactly divides N! .then what is the lowest value N can assume??

How many different four digit numbers are there in the octal(Base 8 ) system,expressed in that system?
3584
2058
6000
7000

OA:7000

Last four digit number is 7777

so,= 7777-7(total single digit #'s) -70(total two digit #'s) - 770(total three digit #'s)

=>7000

Let S={1,2,3,....N} be a set of N natural numbers.Let T be a subset of S such that sum of any 3 elements of T is greater than or equal to N.
Find the maximum number of elements in an subset T for N=40
26
27
28
None of these

OA:None of these

How many different four digit numbers are there in the octal(Base 8 ) system,expressed in that system?
3584
2058
6000
7000

OA:7000

frnds help me on following quest.

1 Find the maximum value of n such that 157! is perfectly divisible by 18^n ?

2. What will be the remainder when 51^203 is divided by 7 ?

using euler ..51/7 =2
so rem is 2^203/7
e(7)=6
2^6k+5/7
2^5/7

32/7
ans =4:)

1)37(not sure )

sol 1:-18=3^2 * 2;

highest power of 3 in 157= /2=/2=37

highest power of 2 in 157= 78 + 39 + 19 + 9 + 4 + 2 + 1=142

HIGHEST Power of 18 will be decided by whichever is lowest;

ANS=37

2: REM= 51^203mod

=>2^203mod

=>*2^2mod

=> *4mod

=>*4 mod

=> 1*4 mod

=>4 ....good morning puys....sabhi ko mera ______/\_______

frnds help me on following quest.

1 Find the maximum value of n such that 157! is perfectly divisible by 18^n ?

18 = 2*3^2

max power of 3 in 157

157/3 = 52
52/3 = 17
17/3 = 5
5/3 = 1

total = 75

but here we have 2 powers of 3 in 18
75/2 = 37.5 = 37

max power of 18 in 157! is 37

2. What will be the remainder when 51^203 is divided by 7 ?

E(7) = 6

203 mod 6 = 5
51 mod 7 = 2

2^5 mod 7 = 4
Pagalguy song :- http://www.pagalguy.com/2009/09/listen-to-the-pagalguy-song/

in how many ways can a commitee of 4 women and 5 men be chosen from 9 women and 7 men ,if mr a refuses to serve on the commitee if ms b is a member

A)1608 b)1860 c)1806 4)1680

five balls of different colours are to be placed in three boxes of different sizes .each box can hold all five balls .the number of ways in which we can place the balls in the boxes so that no box remains empty
1)30 2)150 3)600 4)900

plz give the approach also

1)case 1---- ms b is there

8c3*6c5

case2 ---- ms b is nt there
8c4*7c5

total 1806


2) total case- when all the ball have been packed in two parcels+ when all the balls haave been packed in single parcel

3^5-3c1*2^5+3c2*1^5 = 150

in how many ways can a commitee of 4 women and 5 men be chosen from 9 women and 7 men ,if mr a refuses to serve on the commitee if ms b is a member

A)1608 b)1860 c)1806 4)1680

when ms 'b' is member :-

8C3*6C5 = 336

when mr 'a' is member :-

8C4*6C4 = 1050

when both are not a member :-

8C4*6C5 = 420

total = 1806

five balls of different colours are to be placed in three boxes of different sizes .each box can hold all five balls .the number of ways in which we can place the balls in the boxes so that no box remains empty
1)30 2)150 3)600 4)900

plz give the approach also

1 2 2 = 5C1*4C2*2C2*3!/2! = 90
1 1 3 = 5C1*4C1*3C3*3!/2! = 60
total = 90+60 = 150

30... my approach is.... fixing one color there totally 6 combinations by which the balls can be placed.. so for 5 balls 30 ways...


five balls of different colours are to be placed in three boxes of different sizes .each box can hold all five balls .the number of ways in which we can place the balls in the boxes so that no box remains empty
1)30 2)150 3)600 4)900

plz give the approach also

in how many ways can a commitee of 4 women and 5 men be chosen from 9 women and 7 men ,if mr a refuses to serve on the commitee if ms b is a member

A)1608 b)1860 c)1806 4)1680

five balls of different colours are to be placed in three boxes of different sizes .each box can hold all five balls .the number of ways in which we can place the balls in the boxes so that no box remains empty
1)30 2)150 3)600 4)900

plz give the approach also

frnds help me on following quest.

1 Find the maximum value of n such that 157! is perfectly divisible by 18^n ?

2. What will be the remainder when 51^203 is divided by 7 ?

using euler ..51/7 =2
so rem is 2^203/7
e(7)=6
2^6k+5/7
2^5/7

32/7
ans =4:)

1)37(not sure )

frnds help me on following quest.

1 Find the maximum value of n such that 157! is perfectly divisible by 18^n ?

2. What will be the remainder when 51^203 is divided by 7 ?

if A=10!+12!+....+100!, then what is the highest power of 2 in A?


7
8
9
10

sol:

A= 10!

=>A=10!

=>A=10!

=>2 will only divide 10!

=>highest power= highest power of 2 in 10!

=5+2+1=8 ... goodnight puys..kal subha milte hain

Q.) Set P = {1,5,6,25,26,30,...} denoted all possible numbers that are the sum of one or more distinct powers of 5 (in any term one power of 5 can be added only once). Find the 65th term of this set ?

O.A. 5^6 + 1

Is it 5^6+1

Corresponding to 5^k, we will add to it,5^0,5^1,...,5^k-1 and 5^0+5^1,5^1+5^2..etc.
so, any power of 5 less than 5^k can be added to it or not added to it.
so corresponding to 5^k, we will have 2^k-1+1=2^k terms

Now, sum over k=1 to 5 ,2^k=64-1=63
so 64th term will be 5^6
and 65th term will be 5^6+5^0

Q.) Set P = {1,5,6,25,26,30,...} denoted all possible numbers that are the sum of one or more distinct powers of 5 (in any term one power of 5 can be added only once). Find the 65th term of this set ?

O.A. 5^6 + 1

sumit99 Says

plz elaborate ur approach to find the general term....

I found it out by mere inspection. There is no general method......