find the no of four digit nos whose sum of first two digits is equal to last two???
i m tryng by finding the multiples of 11...seniors help

sol: my take ->615 nos

N=abcd ; a+b =c+d

a+b+c+d =36

cases: a+b =c+d

=18; => c+d=18

=>no.of ways = 1 corresponding ways for a+b =1

total ways =1*1

=17 ; =>ways =2*2

=16; =>ways=3*3.....and so on till c+d=10 we get

1^2 +2^2 +3^2 + 4^2........9^2= 9*10*19/6=285 nos.
for c+d =a+b=9 we get,

C(9+2-1,2-1)=10cases ,since each of c,d can take any value from 0to9. but corresponding no. of ways for a+b =9 since a+b=0+9 is not acceptable.

so total ways will be 10*9;

similarly for sum=8=> 9*8;

hence total nos.= 1*2 +2*3 +.......9*10
tn=n*(n+1)=n^2+n

Sn= n*(n+1)(2n+1)/6 + n(n+1)/2.....

Sn= 285+45=330 nos.

so total number of 4 digit nos possible= 285+330 =615 nos.

mani0303 Says

It's 325 by finding out that K^2 = a where a is the first natural# in the series

please elaborate a bit

Q 1.) The number 27000001 has exactly four prime factors. Find the sum of all prime factors of the number.
Q 2.) Find all pairs of prime numbers p, q such that p^3 q^5 = (p + q)^2.

answer to second question is (7,3)

its only one set i suppose..

find the no of four digit nos whose sum of first two digits is equal to last two???
i m tryng by finding the multiples of 11...seniors help

Q 1.) The number 27000001 has exactly four prime factors. Find the sum of all prime factors of the number.


Q 2.) Find all pairs of prime numbers p, q such that p^3 q^5 = (p + q)^2.

652( i am waitin 4 easy ques. and here it comes )

Q 1.) The number 27000001 has exactly four prime factors. Find the sum of all prime factors of the number.


Q 2.) Find all pairs of prime numbers p, q such that p^3 q^5 = (p + q)^2.

in a triangle abc, d & e are points on ab and ac respectively,such that de is parallel to bc. be and cd intersect at m.if the area of triangle ade is 7/3 times area of mde,find bc/de.
pls provide a simple approach and some suggestions as i always hav problem with such questions.

Let the height of triangle ADE = x
Let the height of triangle MDE = y
Let the height of triangle BMC = z

Given x/y = 7/3
Take y = 3 units and x = 7 units

Triangle ADE ~ Triangle ABC
=> DE/BC = x/(x+y+z) = 7/(10+z) ----1

Triangle MDE ~ Triangle MBC
=> DE/BC = y/z = 3/z ----2

From 1 and 2
=> 3/z = 7/(10 +z)
=> 4z = 30
=> z = 15/2 -----3

=> BC/DE = z/y = 15/(2*3) = 5/2

anujgupta21 Says

plz sirji xplain how did u got the 2nd step.....koi formule use kiya kya aapne??:banghead:

Yaar sum of cubes of first n natural numbers = ^2 use kiya maine to :)

Sum of cubes of x to (x+7) = Sum of cubes of 1 to (x+7) - Sum of cubes of 1 to (x-1)
= (1/4)*{^2 - ^2}

anujgupta21 Says

kindly frndz..plz help me out with this problem..!!!!

QUestion 5. answer is 130 ............ i used options bro .. so get 130 and then analyze whether it is ( 9,7) or (11,3)

question 6 . Answer is 3 ... No of factors are 45 which can be written as 3*3*5=

so it can be prime 1 ^ 2 * prime 2^2* prime3^4= 3 prime number can be there

x^3 + (x+1)^3 + (x+2)^3 + .... (x+7)^3 = y^3
=> ^2 - ^2 = 4y^3

=> 4*(2x+7)*(x^2 + 7x + 28 ) = y^3

(2x+7) is odd
(x^2 + 7x + 28 ) must be even so that one extra 2 makes 4*2 = 8 (cube root of 2)

=> 7 = 1+6= 2+5 = 3+4

Case 1 : 7x = x + 6x
=> x^2 + 7x + 6 => No integer values of x


Case 2: 7x = 5x + 2x
=> x^2 + 7x + 10 = (x+2)*(x+5) => x = -2 and -5
=> x^2 + 7x + 28 = 18

so y^3 = 4*3*18 = 6^3
Also, y^3 = 4*(-3)*18 = (-6)^3


Case 3: 7x = 4x + 3x
=> x^2 + 7x + 12 = (x+3)*(x+4) => x = -3, -4

=> y^3 = 4*1*(16) = 4^3
Also, y^3 = 4*(-1)*(16) = (-4)^3

So 4 values of x will satisfy it

plz sirji xplain how did u got the 2nd step.....koi formule use kiya kya aapne??:banghead:

anujgupta21 Says

kindly frndz..plz help me out with this problem..!!!!

1.) a+b = k ----1
a^2 - b^2 = 2k -----2
ab+1 = 4k -----3

From 1 and 2
=> a-b = 2

Therefore a^2 + b^2 + 2ab = k^2 and a^2 + b^2 - 2ab = 4
=> 4ab = k^2 - 4
=> 4*(4k - 1) = k^2 - 4

=> k = 16

=> a = 9 and b = 7
=> a^2 + b^2 = 9^2 + 7^2 = 81 + 49 = 130



2.) 45 = 3*3*5
=> Number can be of the form = a^2 * b^2 * c^4
=> It is divisible by maximum three prime numbers

anujgupta21 Says

kindly frndz..plz help me out with this problem..!!!!

No.5

Let the numbers be a,b
a+b:a^2-b^2:ab+1=1:2:4

a+b/a^2-b^2=1/2

gives a-b=2 a=b+2

Also, a+b/(1+ab)=1/4

putting a=b+2 here we get b=7 a=9

so sum of squares=49+81=130

in a triangle abc, d & e are points on ab and ac respectively,such that de is parallel to bc. be and cd intersect at m.if the area of triangle ade is 7/3 times area of mde,find bc/de.
pls provide a simple approach and some suggestions as i always hav problem with such questions.

right...........

Let a, b, c be non-zero real numbers such that a^2 + b^2 + c^2 = 1 and
a(1/b + 1/c) + b(1/c + 1/a) + c(1/a + 1/b) = -3, then a+b+c equals

(a) -1 (b) 0 (c) 1 (d) all of the foregoing (e) at least 2 of the foregoing

a(b+c)/bc + b(a+c)/ac + c*(a+b)/bc = -3

=> a^2*(b+c) + b^2*(a+c) + c^2*(a+b) = -3abc ----1

If a+b+c = 0 then equation 1 becomes

=> a^3 + b^3 + c^3 = 3abc => which is true when a+b+c = 0

so a+b+c = 0

P.S. It can be wrong, meanwhile koi aur approach sochta hoon

kindly frndz..plz help me out with this problem..!!!!

Find no. of solutions in integers of the equation-
x^3 + (x + 1)^3 + (x + 2)^3 +.................. + (x + 7)^3 = y^3

1. 1
2. 2
3. 3
4. 4


_/\_ Varun bhai,Sood bhai,chillbhai and rest Champs also......................

x^3 + (x+1)^3 + (x+2)^3 + .... (x+7)^3 = y^3
=> ^2 - ^2 = 4y^3

=> 4*(2x+7)*(x^2 + 7x + 28 ) = y^3

Now, (2x+7) is odd
=> (x^2 + 7x + 28 ) must be even so that one extra 2 makes 4*2 = 8 (cube root of 2)

=> 7 = 1+6= 2+5 = 3+4

Case 1 : 7x = x + 6x
=> x^2 + 7x + 6 = 0 => x = 6 and 1

=> y^3 = 4*19*22 => Not a cube
Also, y^3 = 4*9*22 => Not a cube


Case 2: 7x = 5x + 2x
=> x^2 + 7x + 10 = (x+2)*(x+5) => x = -2 and -5
=> x^2 + 7x + 28 = 18

so y^3 = 4*3*18 = 6^3
Also, y^3 = 4*(-3)*18 = (-6)^3


Case 3: 7x = 4x + 3x
=> x^2 + 7x + 12 = (x+3)*(x+4) => x = -3, -4

=> y^3 = 4*1*(16) = 4^3
Also, y^3 = 4*(-1)*(16) = (-4)^3

So 4 values of x will satisfy it

mani0303 Says

It's 325 by finding out that K^2 = a where a is the first natural# in the series

right...........

Let a, b, c be non-zero real numbers such that a^2 + b^2 + c^2 = 1 and
a(1/b + 1/c) + b(1/c + 1/a) + c(1/a + 1/b) = -3, then a+b+c equals

(a) -1 (b) 0 (c) 1 (d) all of the foregoing (e) at least 2 of the foregoing

The sum of 2k+1 consecutive natural numbers is 2n such that the sum of first k+1 natural numbers among these (2k+1) numbers is same as next k numbers. Which of the following cannot be the 7th least natural number among the 2k+1 natural numbers, if k>=3
1) 231 2) 175 3) 535 4) 325 5) none of these

It's 325 by finding out that K^2 = a where a is the first natural# in the series

arumugadas Says

a train and car start at same time ...car at rear end reaches front end and come back to rear end . in the mean time train travels 1 km and speed of train is 1 km /hr ...wat is distance covered by car:banghead:

The distance covered by the car = 2*(length of the train) + Distance traveled by the train
=> Distance = 2*(length of the train) + 1

arumugadas Says

length is 1 km but ans is 1+root 2 as per book:banghead:

Oh yes it should be 1 + _/2

Let the speed of the car be x
=> We have 1/(c - 1) + 1/(c+1) = 1
=> c^2 - 2c - 1 = 0

=> c = (1 + _/2) km/hr

So in one hour car would travel = (1 + _/2) k.m.

The sum of 2k+1 consecutive natural numbers is 2n such that the sum of first k+1 natural numbers among these (2k+1) numbers is same as next k numbers. Which of the following cannot be the 7th least natural number among the 2k+1 natural numbers, if k>=3
1) 231 2) 175 3) 535 4) 325 5) none of these

my take 4

x=-5 y=-6
x=-4 y=-4
x=-3 y=4
x=-2 y=6

Thats stupendous bro.......................keep it up

Find no. of solutions in integers of the equation-
x^3 + (x + 1)^3 + (x + 2)^3 +.................. + (x + 7)^3 = y^3

1. 1
2. 2
3. 3
4. 4


_/\_ Varun bhai,Sood bhai,chillbhai and rest Champs also......................

my take 4

x=-5 y=-6
x=-4 y=-4
x=-3 y=4
x=-2 y=6

The distance covered by the car = 2*(length of the train) + Distance traveled by the train
=> Distance = 2*(length of the train) + 1

length is 1 km but ans is 1+root 2 as per book:banghead:

sumit99 Says

Yaa the ans. is not 2 Try again bro.......

shit man ... Ho hi nahi raha... i m Not Trying ....


P.S : I Got 98 In Quant ... and This iS my Level shame on me...:(

If last digit of (N^2) is twice of its second last digit and 1<= N <= 200.

Then find the total number of elements in set 'N

Taking non zero last two digits of N^2 into consideration,here are the possibilities for N:

N= 32,132,82,182,44,144,94,194,06,106,56,156,18,118,68,168

I am getting 2 .... i dont think that it is 2

-4,-4
-3,4

plzz check...

Yaa the ans. is not 2 Try again bro.......

Find no. of solutions in integers of the equation-
x^3 + (x + 1)^3 + (x + 2)^3 +.................. + (x + 7)^3 = y^3

1. 1
2. 2
3. 3
4. 4


_/\_ Varun bhai,Sood bhai,chillbhai and rest Champs also......................

I am getting 2 .... i dont think that it is 2

-4,0
-3,4

plzz check...

Find no. of solutions in integers of the equation-
x^3 + (x + 1)^3 + (x + 2)^3 +.................. + (x + 7)^3 = y^3

1. 1
2. 2
3. 3
4. 4


_/\_ Varun bhai,Sood bhai,chillbhai and rest Champs also......................

arumugadas Says

a train and car start at same time ...car at rear end reaches front end and come back to rear end . in the mean time train travels 1 km and speed of train is 1 km /hr ...wat is distance covered by car:banghead:

The distance covered by the car = 2*(length of the train) + Distance traveled by the train
=> Distance = 2*(length of the train) + 1

EDIT: It should be (1 + _/2). See my post on the next page

a train and car start at same time ...car at rear end reaches front end and come back to rear end . in the mean time train travels 1 km and speed of train is 1 km /hr ...wat is distance covered by car:banghead:

Only possibilities for last two digits of N^2 are 24 and 00

So, 10, 20, 30, ..., 200 are such N
Also, N can be of form 50k 18

So, 18, 32, 68, 82, 118, 132, 168 and 182

So, 28 such numbers

why cant it be 36 for last two digit:shock:

A tractor towing 120 trolleys overtakes Sharan who is going in the same direction in 36 seconds. It travels for half an hour from the time it starts overtaking Sharan (driving the Sulsar) to before it finishes crossing Deepak (driving the Nebula) coming from the opposite direction in 24 seconds. In how much time (in minutes) after the tractor has crossed Deepak, does Sharan meet Deepak? (the lengths of Sulsar and Nebula are negligible)
(1) 59 minutes 18 seconds
(2) 60 minutes
(3) 59 minutes
(4) 59 minutes 36 seconds

I am assuming the speeds of Deepak and Sharan to be equal.
Let the speed of the tractor be u
Let the speed of Deepak = Sharan = v
Let the length of the train be 120

We have: u - v = 120*60/36 = 200
Also, u + v = 120*60/24 = 300

=> u = 250 and v = 50

Tractor traveled after overtaking Sharan for = 30 - 3/5 + 2/5 = 149/5 minutes

Distance traveled by tractor = (149/5)* 250 = 7450
Distance traveled by Sharan = 1490

=> Distance between Sharan and Deepak = 7450 - 1490 = 5960

Deepak and Sharan will meet after = Distance between them / 2v

=> Time = 5960/(2*50) = 59.6 minutes = 59 minutes and 36 seconds

Only possibilities for last two digits of N^2 are 24 and 00

So, 10, 20, 30, ..., 200 are such N
Also, N can be of form 50k 18

So, 18, 32, 68, 82, 118, 132, 168 and 182

So, 28 such numbers

sir 50k+-6 chhopdh diye aapne..

so 6,44,56,94,106,144,156,194=8 ye bi honge

so 36 such numbers honge..

and agar 00 wala case rehne de to 16..

Bhai 18^2 = 324 hota hai.
So N has 2 elements.

Also 32^2 = 1024. :p

Only possibilities for last two digits of N^2 are 24 and 00

So, 10, 20, 30, ..., 200 are such N
Also, N can be of form 50k 18

So, 18, 32, 68, 82, 118, 132, 168 and 182

So, 28 such numbers

:banghead: I thought total number of elements in N.
Perfect apprch by chill sr.
Sir but ye 50k+/-18 is mere observation?? ya fir koi reason hai?

If last digit of (N^2) is twice of its second last digit and 1<= N <= 200.

Then find the total number of elements in set 'N

My take is 36

Last digit of a square can be 0,1,4,5,6,9.

However, second last digit is half the last digit. Hence, last digit must be even.

Possible units digit of N are 0 or 4 or 6.

Last two digits can be 00 or 36 or 24.

For 24 - 18,32,68,82,118,132,168,182
For 36 - 6,56,106,156
For 00 - 10,20,30 .. 200

So, 36 values.

If last digit of (N^2) is twice of its second last digit and 1<= N <= 200.

Then find the total number of elements in set 'N

Only possibilities for last two digits of N^2 are 24 and 00

So, 10, 20, 30, ..., 200 are such N
Also, N can be of form 50k 18

So, 18, 32, 68, 82, 118, 132, 168 and 182

So, 28 such numbers

If last digit of (N^2) is twice of its second last digit and 1<= N <= 200.

Then find the total number of elements in set 'N

Bhai 18^2 = 324 hota hai.
So N has 2 elements.

Also 32^2 = 1024.

If last digit of (N^2) is twice of its second last digit and 1<= N <= 200.

Then find the total number of elements in set 'N

Anjali nd Sachin decided to meet at Barista sometime between 5pm to 6 pm.they decided that they would wait max of 10 min their and then leave.What is the probability that they met?

Flanker bhai's question ^^^

My take is 11/36.
Explanation in the figure.

repeat possible

1*1*3*3*1*3=27 hona chaiye..

12_ _ _ _

now 3rd place can be fill in 7,8,9
4th place 3,6,9
5th place is 3 times sixth so 6th place can be fiiled in 1,2,3 ways becoz after 3 i get 2 digit number

total ways =3*3*3=9

Messy_19 Says

1*1*3*3*1*3=27? is it?

Udit bhai calc mistake. 3*3*3 = 27. :D

All correct!!

Jain oye, 20 page bache the abi uss thread ke. :p

=========================

Q) A credit card number has 6 digits (between 1 to 9). The first two digits are 12 in that order, the third digit is bigger than 6, the forth is divisible by 3 and the fifth digit is 3 times the sixth. How many different credit card numbers exist?

sol:
12_ _ _ _

3rd place can take 7,8,9
4th place can take 3,6,9
5th place is 3 trimes 6th place so sixth place can take 1,2,3 after that 2 digit banege
so total values=3*3*3=27

Jain oye, 20 page bache the abi uss thread ke. :p

=========================

Q) A credit card number has 6 digits (between 1 to 9). The first two digits are 12 in that order, the third digit is bigger than 6, the forth is divisible by 3 and the fifth digit is 3 times the sixth. How many different credit card numbers exist?

1*1*3*3*1*3=27? is it?

Jain oye, 20 page bache the abi uss thread ke. :p

=========================

Q) A credit card number has 6 digits (between 1 to 9). The first two digits are 12 in that order, the third digit is bigger than 6, the forth is divisible by 3 and the fifth digit is 3 times the sixth. How many different credit card numbers exist?

12_ _ _ _

now 3rd place can be fill in 7,8,9
4th place 3,6,9
5th place is 3 times sixth so 6th place can be fiiled in 1,2,3 ways becoz after 3 i get 2 digit number

total ways =3*3*3=27

Jain oye, 20 page bache the abi uss thread ke. :p

=========================

Q) A credit card number has 6 digits (between 1 to 9). The first two digits are 12 in that order, the third digit is bigger than 6, the forth is divisible by 3 and the fifth digit is 3 times the sixth. How many different credit card numbers exist?

repeat possible

1*1*3*3*1*3=27 hona chaiye..

A tractor towing 120 trolleys overtakes Sharan who is going in the same direction in 36 seconds. It travels for half an hour from the time it starts overtaking Sharan (driving the Sulsar) to before it finishes crossing Deepak (driving the Nebula) coming from the opposite direction in 24 seconds. In how much time (in minutes) after the tractor has crossed Deepak, does Sharan meet Deepak? (the lengths of Sulsar and Nebula are negligible)
(1) 59 minutes 18 seconds
(2) 60 minutes
(3) 59 minutes
(4) 59 minutes 36 seconds

All the quant enthusiasts

please continue here.

here is the link to the previous thread:
http://www.pagalguy.com/forum/quantitative-questions-and-answers/74662-official-quant-thread-cat-2012-a.html

Jain oye, 20 page bache the abi uss thread ke. :p

=========================

Q) A credit card number has 6 digits (between 1 to 9). The first two digits are 12 in that order, the third digit is bigger than 6, the forth is divisible by 3 and the fifth digit is 3 times the sixth. How many different credit card numbers exist?