[shivam07usit]

Originally Posted by lemasolai

Hi Puys,

If sumone cud elaborate on the funda of the foll. ques :

Q) A box contains 2 red balls, 8 green balls and 10 blue balls. Find the maximum number of balls to be drawn from the box to ensure that at least four balls of the same colour are obtained.

a) 7 b) 10 c) 8 d) 11 e) 9

OA : option e, but I think it shud be option c, pls comment

SINCE WE HAVE TO TAKE OUT MAXIMUM NO OF BALLS
FIRST TAKE OUT 2 REDS
THEN TAKE OUT 3 GREENS AND 3 BLUES
NOW WHEN WE TAKE OUT 1 MORE BALL WE R SURE THAT WE HAVE ATLEAST 4 BALLS OF SAME COLOUR
HENCE 2+3+3+1=9

[samkris]

Hi Puys
Can someone pls help me in solving these.

I dont have the O.A as it was given to me by my friend

1. find the remainder when 6^7 + 8^7 is divided by

i)49

a)6 b)0 c)13 d)4


ii)21

a)0 b)7 c)14 d)20


2.find the hundredth's place digit of 9^77

a)1 b)9 c)4 d)6

[uhhh...]

Originally Posted by sachinverma

A circle is drawn on a sphere whose centre is also the centre of the sphere. There are 10 circles drawn on the sphere, no three of which meet at any point. How many regions do they divide the sphere?
(a) 142 (b) 56 (c) 92 (d) 320
Source:quantexpert(dot)in

i used some form of weird mapping (not suitable for discussion).. i have no idea abt the sphere and how it will look

suppose R(n) = number of regions for n -circles

then R(n+1) - R(n) = 2n

R(10) - R(1) = 9*10.. R(1) = 2

R(10) = 92.. i am not so sure.. not that good at abstract algebra

[shivam07usit]

. find the remainder when 6^7 + 8^7 is divided by

i)49

(7-1)^7+(7+1)^7
=2(7C0 7^7 +7C2 7^5 + 7C4 7^3 +7C6 7^1)

LAST TERM IS
7C6 7^1
=7x7=49

HENCE REM IS 0


i)21

(7-1)^7+(7+1)^7
=2(7C0 7^7 +7C2 7^5 + 7C4 7^3 +7C6 7^1)
=2( 7^7 +7X3 7^5 + 7X5 7^3 + 7 7^1 )
REM IS ==2( 7^7 + 7X5 7^3 + 7 7^1 )
7^7%21=7(7^6 % 3)=7(1)=7
7X5X7^3 %21=7(5x343 %3)=7(2)=14
7 X 7^1 =49%21=7
2(7+14+7)=56%21=14
HENCE REM IS 14

[uhhh...]

Originally Posted by samkris

Hi Puys
Can someone pls help me in solving these.

I dont have the O.A as it was given to me by my friend

1. find the remainder when 6^7 + 8^7 is divided by

i)49

a)6 b)0 c)13 d)4


ii)21

a)0 b)7 c)14 d)20


2.find the hundredth's place digit of 9^77

a)1 b)9 c)4 d)6

1 i) use binomial... 8 = 7+1, 6 = 7-1
ii) number is divisible by 14, find the remainder with 3 for 8^7; 8 = 9-1

2) binomial 9 = 10-1

[bhupi3149]

Originally Posted by samkris

Hi Puys
Can someone pls help me in solving these.

I dont have the O.A as it was given to me by my friend

1. find the remainder when 6^7 + 8^7 is divided by

i)49

a)6 b)0 c)13 d)4


ii)21

a)0 b)7 c)14 d)20


2.find the hundredth's place digit of 9^77

a)1 b)9 c)4 d)6

for the first one its 0

using binomial theorom (7-1)^7 +(7+1)^ divided by 49 ..... will be 0

for second one divide 9^77/100

hundredth digit of remainder will be 1

[Ankurb]

Originally Posted by samkris

Hi Puys
Can someone pls help me in solving these.

I dont have the O.A as it was given to me by my friend

1. find the remainder when 6^7 + 8^7 is divided by

i)49

a)6 b)0 c)13 d)4


ii)21

a)0 b)7 c)14 d)20


2.find the hundredth's place digit of 9^77

a)1 b)9 c)4 d)6

1.
i)option b)0
(-1+7)^7 + (1+7)^7 = -1 + 1 =0
remainder = 0
ii)option c)14
6^7 + 8^7 mod 3 = -1
6^7 + 8^7 mod 7 = 0
remainder = 14

2.
option a)1
9^77 mod 8 = 1
(-1+10)^77 mod 125 =
-1 + 770 - 77*38*100 mod 125 =
-1 + 20 - 100 mod 125 = -81

125p - 81 = 8q +1
q = (125p - 82)/8,==>p=2
remainder = 250 - 81 = 169
100th digit = 1

[naga25french]

Originally Posted by raj8626

@jaidev, madnikhil and apocalyptic.. 16 is the right answer

A tricky question.. Euler's number is not always the only number which gives a remainder 1. Sometimes its factors also give a remainder 1.. You need to check for those cases.. We hav to remember this

Originally Posted by apocalyptic

The divisiblity rule of 13 states devide the number into parts by starting from the last 3 digits then add/subtract alternately.

The first number of type 10^x that leaves a remainder 1 is therefore 10^6
The second is 10^12.

Total 16

Originally Posted by reshma15

yes mate i saw the sol bt still i don get t part 678*400/37
can you pls post ur approach?
thnx

u will realise this funda come handy for such prob..

http://www.pagalguy.com/forum/quanti...ml#post1719020 (Official Quant thread for CAT 09-III (September onwards))

[sachinverma]

Hi PUYS,
HAS ANYONE SOLVED THE QUANTEXPERT NOVEMBER CONTEST BOOKLET.
THE 5 QUESTIONS ARE AS FOLLOWS:
QUESTION 1:
ABC is a ∆ right-angled at A with length of inradius 1cm. Given AB=4cm, find the area of ∆ABC?

QUESTION 2:
The sum of the 1st 10 terms of an AP with common difference (d) is 155 and the sum of 1st 2 terms of a GP with common ratio (r) is 9. If r = 1st term of the AP and d = 1st term of the GP, then find the 1st term of the AP?

QUESTION 3:
A circle is drawn on a sphere whose centre is also the centre of the sphere. There are 10 circles drawn on the sphere, no three of which meet at any point. How many regions do they divide the sphere?

QUESTION 4:
Calculate the sum
[√1] +[√2] +[√3] +[√4] +…+[√99] where the symbol [x] means the greatest integer less than or equal to x?


QUESTION 5:
For how many values of ‘x’ is the expression
[(p^(x+2)+q^(x+2)) / (p^(x+1)+q^(x+1))], the harmonic mean of p and q?


MY ANSWERS:
I AM GETTING THE ANSWERS AS FOLLOWS:
Q1:6
Q2:2,25/2
Q3:COULD NOT SOLVE
Q4:615
Q5:-2
=============
PLZ HELP IN SOLVING THE 3rd QUESTION.
THE BOOKLET CAN BE DOWNLOADED FROM THE SITE OF QUANTEXPERT.
JUST SEARCH QUANTEXPERT IN GOOGLE.
THANX
==============
quantexpert(dot)in

[Vineetverma]

Originally Posted by sachinverma

Hi PUYS,
QUESTION 2:
The sum of the 1st 10 terms of an AP with common difference (d) is 155 and the sum of 1st 2 terms of a GP with common ratio (r) is 9. If r = 1st term of the AP and d = 1st term of the GP, then find the 1st term of the AP?


==============
quantexpert(dot)in

2) 1st term of ap = 2 with D = 3

EDIT:
4)
for 1-3 it's 1
2-8 = 2
9-15 = 3
16-24 = 4
25-35 = 5
36-48 = 6
49 - 63 = 7
64-80 = 8
81 - 99 = 9
hence 3 + 14 + 21 + 36 + 55 + 78 + 105 + 136 + 171