Quant by Arun Sharma

MBA CAT 2024
5046

subscribed......helps one to stay focussed

5047

At a particular time in the 21st cebtury there were seven bowlers in the indian cricket team's list of 16 players shortlisted to play the next wc . Statisticians discovered that if u looked at the no of wickets taken by any of the 7 bowlers in the current indian cricket team , the number of wickets taken by them had a strange property . The numbers were such that for any team selection of 11 players (having 1 to 7 bowlers) by using the number of wickets taken by each bowler and attaching coefficients 0f +1 , 0 , -1 to eachvalue available and adding the resultant values , any number from 1 to 1093 both included could be formed . If we denote W1,W2,W3,W4,W5,W6,W7 as the 7 values in the ascending order what would be the answer to the following questions :

47) Find the value of W1+2W2+3W3+4W4+5W5+6W6
a) 2005 b) 1995
c)1985 d)NOT

4 Find the index of the largest power of 3 contained in the product W1 W2 W3 W4 W5 W6 W7
a) 15 b) 10
c) 21 d) 6

49) If the sum of the seven coefficients is 0 , find the smallest number that can be obtained

a) -1067 b)-729
c) -1040 d) -1053

>> There was a solution posted for this question by nitrjswcoke .... but that solution is itself confusing ..... can any1 be more explanatory on how to arrive onto that GP that nitrjswcoke has explained ????
Solution by nitrjswcoke on this page >> http://www.pagalguy.com/forum/quantitative-questions-and-answers/23813-quant-by-arun-sharma-382.html#post2873459


Somebody please reply to this !!!!
5048

how you find that the coin is faced head? can you please provide the explanation.

also for the 3 digit no. question also can you provide the explanation

5049


SOL:

FOR THESE TYPE OF QUESTIONS(MY WAY)

MULTIPLY LAST TWO TERMS OF BOTH NOS.

SO 81 * 64 = YX84

SO DIGIT AT TEN'S PLACE =8
SO NONE OF THESE





AFTER SIMPLIFYING THE EQUATION;
4X- 17Y = 1
4X = 1+ 17Y

VALUES OF X AND Y SATISFYING THE EQUATION

X = COMMON DIFFERENCE=17

Y = COMMON DIFFERENCE=4

{NOTE:TO GET LAST VALUE OF X AS 999

DIVIDE 1000/17 = 58.82 (58*17=986)

BUT X IS STARTING FROM 13 (13 IS 4 LESS THAN 17)
AND SERIES OF X IS GOING IN THE SIMILAR FASHION.

SO THE 58 THE NO. =982
AND THE 59TH NO.= 999

REQUIRED ANSWER = 59

Review 4







AM GETTING 760 AS ANSWER .



DIDN'T GET THE QUESTION


THERE IS FORMULA FOR THIS :

P^N-1/N =R(1)

IS IT NONE OF THESE??????????/


Can you please provide the explanation for the answer???
5050

Originally Posted by hemant89 View Post
Block 1 Test Review

Review 3
Q3 . Find the digit at ten's place of the no. N = 7281 * 3264

  1. 0
  2. 1
  3. 6

  1. 5

  1. None of these


SOL:

FOR THESE TYPE OF QUESTIONS(MY WAY)

MULTIPLY LAST TWO TERMS OF BOTH NOS.

SO 81 * 64 = YX84

SO DIGIT AT TEN'S PLACE =8
SO NONE OF THESE


Sorry but the question is

Q3 . Find the digit at ten's place of the no. N = 7^281 * 3^264

  1. 0
  2. 1
  3. 6
  4. 5
  5. None of these
5051
@rakesh the last two digits of 2^134
(2^10)^13*16
24^13*16
24*16=84
so ans should be 84
plz correct me :)

yes it should be 84...

:)
5052
Originally Posted by hemant89 View Post
Block 1 Test Review

Review 3
Q3 .


Sorry but the question is

Q3 . Find the digit at ten's place of the no. N = 7^281 * 3^264
  1. 0
  2. 1
  3. 6
  4. 5
  5. None of these

7^281.. the last two digits will be 07 cyclicity of 7 i s 4

and for 3^264 = it will be 3^4... cyclicity of 3 i s 20
.
07*81=67 = answer is 6
5053
hemant89 Says
Can you please provide the explanation for the answer???


i will provide the detailed solution,if the answers given by me is correct.
5054

1) 3(x-3)+4(y+4)=6
3x-9+4y+16=6
3x+4y+7-6=0
3x+4y+1=0
2) 2(x-3)+5(y-3)=4
2x-6+5y-15=4
2x+5y-25=0

5055
Originally Posted by hemant89 View Post
Block 1 Test Review
Review 3
Q3 . Find the digit at ten's place of the no. N = 7281 * 3264

  1. 0
  2. 1
  3. 6

  1. 5

  1. None of these

SOL:

FOR THESE TYPE OF QUESTIONS(MY WAY)

MULTIPLY LAST TWO TERMS OF BOTH NOS.

SO 81 * 64 = YX84

SO DIGIT AT TEN'S PLACE =8
SO NONE OF THESE


Sorry but the question is

Q3 . Find the digit at ten's place of the no. N = 7^281 * 3^264
  1. 0
  2. 1
  3. 6
  4. 5
  5. None of these

Is the ans 6

Sol:

We can use remainder theorem

for 10th place divide the no by 100

7^281/3^264/100
=49*(7^3)^93 * 3^4(3^5)^52/100
=49*43^93*81*43^52/100
=49*81*43^145/100
=81*43*49^73/100
=81*43*49/100
=18667/100

So 10th digit will be 6.:cheers:
5056
hello!
i needed help with a pattern in number system.
the question types are:
1)find the maximum value of n such that 50! is perfectly divisible by 2520^n?
here the largest prime factor(i.e 7) was taken as restricting factor..
2)) find the maximum value of n such that 50! is perfectly divisible by 12600^n?
here instead of the biggest prime no.(i.e 7), 3 was taken as restricting prime factor.
so,
can somebody plz help me with this concept? i know that the longer way would be to calculate it for all the prime factors and see which one will be the smallest...but thus can be really lengthy (like in case of 504^n if we take 2 as the prime factor).. so can sumbody plz tell dat how is the limiting factor considered?
thanks..


I think for question 2.
The limiting factor is 7.
the power of 7 in 50! is 8
if we have 12600^8 then we need 16 5s in 50! but there are 12 5s hence n=6
5057

yes both the answers are correct, can you please explain the process



sol:
At first round -69 matched to be played (with 138 palyers)


at second round -35 matches to be played(with 69 players from 1st round and 1 player who got bye at 1st round)


at third round - 17 matches to be played and one will get bye (with 35 players)


at 4th round - 9 matches to b played


at 5 th round - 4 matches to be played and one will get bye


at 6 th round - 2 matches to be played and one will get bye


at 7th round - 1 match to played and one will get bye


at 8th round - 1 matched to be played with 2 bowlers


now adding all matches = 69 + 35 + 17 + 9 + 4 + 2 + 1 + 1 = 138 matches






is it head??????////




is it 4 ??//////////

5058

Hi All,

Here are a few questions from time,speed and distance...

1. A wall clock gains 2 mins in 12 hrs,while a table clock loses 2 mins in 36 hrs.both are set right at noon on tuesday.The correct time when they both show the same time next would be...
a.12.30 night
b.12 noon
c.1.30 night
d.12 night

2.A naughty bird is sitting on top of a car.It sees another car approaching it at a distance of 12 km.THe speed of the two cars is 60 kmph each.The bird starts flying from the first car and moves towards the second car,reaches the second car and come back to the first car and so on.If the speed at which the bird flies is 120kmph,then answer the foll questions :(assume the two cars have a crash)
a)the total distance traveled by the bird before the crash is :
1) 6km 2)12km 3)18km 4)none

b)the total distance travelled by the bird before it reaches the second car for the second time is :
1)10.55km 2)11.55km 3)12.33km 4)none

c)the total number of times the bird reaches the second bonnet is (theoretically) :
1)12 times 2)18 times 3)Infinite times 4)cannot be determined


Please give ur asnswers and explanations as well.

Regards,
Krish

5059

thank u brothr. got it

5060

hello brothr bt cud u xplen 3rd solution a little further. hw did u tek the Nos. 10 - 19, 20,20.....
thanks in advance

5061
I think for question 2.
The limiting factor is 7.
the power of 7 in 50! is 8
if we have 12600^8 then we need 16 5s in 50! but there are 12 5s hence n=6[/QUOTE

wat is d ans is dt 8.
5062

1. A wall clock gains 2 mins in 12 hrs,while a table clock loses 2 mins in 36 hrs.both are set right at noon on tuesday.The correct time when they both show the same time next would be...
a.12.30 night
b.12 noon
c.1.30 night
d.12 night

Guess but i doubt :)

one is losing and other is gaining ... so there ratio's is 1:3 for 36 hours

as both are moving towards each other in a way to meet ...so total speed will be 1+3=4

hence in 36 hour they will cover 8 min gap...

we have to find for 60*24 hour

60*24/8=180 hours hence it will meet at 180/24=7 complete +12 hours from noon which is 12 night



2.A naughty bird is sitting on top of a car.It sees another car approaching it at a distance of 12 km.THe speed of the two cars is 60 kmph each.The bird starts flying from the first car and moves towards the second car,reaches the second car and come back to the first car and so on.If the speed at which the bird flies is 120kmph,then answer the foll questions :(assume the two cars have a crash)


a)the total distance traveled by the bird before the crash is :
1) 6km 2)12km 3)18km 4)none


ohk here is the simplest way to solve it

total time the bird travel = total time for car to crash

total time to crash = total distance /total speed

12/60+60 here 60+60 becoz both are moving towards each other

therefore total time =12/120

hence distance covered by bird would be time * speed = 12/120*120=12 km



b)the total distance travelled by the bird before it reaches the second car for the second time is :
1)10.55km 2)11.55km 3)12.33km 4)none

how i will do is

first touch to the car second distance covered will be

12 ( total distance between intially ) / 120+60 ( total speed becoz of different direction * 120 speed of bird

12*120/180=8

now at the same time car 1 will cover 4 km becoz it is travelling with 60km/hr

so now distance between bird and car1 when it touch car 2 is 4 km

so distance travel will be again = 4*120/180= 8/3

and distance between car 1 and car 2 when bird reach car 1 is 4/3

so distance covered by bird finally reaching car2 second time = 4/3*2/3=8/9

total distance travel = 8+ 8/3 + 8/9 = 11.555



c)the total number of times the bird reaches the second bonnet is (theoretically) :
1)12 times 2)18 times 3)Infinite times 4)cannot be determined

infinite no becoz the bird and car speed goes on decreasing

forexample the bird distance coverred will be

8+8/3+8/9+8/27+8/81 ..... hence infinite times

5063
I think for question 2.
The limiting factor is 7.
the power of 7 in 50! is 8
if we have 12600^8 then we need 16 5s in 50! but there are 12 5s hence n=6[/QUOTE

wat is d ans is dt 8.

Highest powers
2-47
3-31
5-12
7-8

12600=7^1*2^3*3^2*5^2
So answer should be 6 as we select the least one and deciding factor must be 5 in my opinion.
Please correct me if I am wrong.
Thanks
:)
5064
Hi Guys,

Could you please help me in these questions and let me know the procedure to solve them

Topic:Numbers

Question 1:

Find the sum and number of factors Divisible by 245

Question 2:

Find the Consecutive zeros at the end of following number
1!*2!*3!*...........*50!
5065


Question 1:

Find the sum and number of factors Divisible by 245

Question 2:

Find the Consecutive zeros at the end of following number
1!*2!*3!*...........*50!

buddy
in the 1st question you just need to find out number of factors/divisors and sum of them...formulae for both is given in the book

2.)in his question you need find no of zeros....or number of 5(power of 5) present in this whole
i hope this will help