sahilmadan Says

If a 4 digit natural no is 7083 more the number formed by reversing the digits.How many such 4 digit numbers are there?

let no abcd
1000a+100b+10c+d- 1000d-100c-10b-a = 999a+90b-90c-999d = 7083
999 (a-d) +90(b-c) = 9*787
111(a-d) +10(b-c) = 787
=> only possible when 111*7 so last digit is divisible by 10
a-d =7
a=9,8,7 is possible
b-c = 1
b=9,8,7,6,5,4,3,2,1 pssible so total 27 values possible

Ye nahi ho raha mujse....koi batao plz ye...answer is 3.8

Neeraj has agreed to mow a lawn, which is a 20 m 40 m rectangle. He mows it with 1 m wide strip. If Neeraj starts at one corner and mows around the lawn toward the centre, about how many times would he go round before he has mowed half the lawn?

In a class of 12 students , 7 boys and 5 girls.The class has 4 sessions each day,one each of arithmetic ,algebra,geometry and probability.These classes are to be held one after the other in 4 distinct time slots and can be in any sequence .Further there are 2 teachers available and they can teach any topic.
In how many distinct ways can a session be scheduled for two consecutive days..??

classes can be held in 4! ways ,teachers can take classes in 2^4 ways
so total ways =(4!* 2^4)^2

In a class of 12 students , 7 boys and 5 girls.The class has 4 sessions each day,one each of arithmetic ,algebra,geometry and probability.These classes are to be held one after the other in 4 distinct time slots and can be in any sequence .Further there are 2 teachers available and they can teach any topic.
In how many distinct ways can a session be scheduled for two consecutive days..??

My take is 4! * 4! * 2^8 (Assuming that class remains same, no divsions in class)

On each day, we have 4 sessions and hence there can be 4! ways to schedule them.
On second day also, we will have 4! arrangements.

And for 8 subjects, we have 2 teachers. Each subject can be be taught by 2 teachers, so 2 choices for eac one.
So, total choices of teachers = 2*2*2.. 8 times = 2^8

So, total schedules = 4! * 4! * 2^8

A certain sum is invested in simple interest.It becomes k times itself in 16 years and 2k time itself in 40 years.In how many years it will become 4k times itself?

@Chill sir : Sorry for the typo in last one

ans 64 years hai kya?
it become k in 16 years..phir doubled in another 24 years..
so again it will double in another 24 years..so 64 years..

not sure though

A certain sum is invested in simple interest.It becomes k times itself in 16 years and 2k time itself in 40 years.In how many years it will become 4k times itself?

@Chill sir : Sorry for the typo in last one

agar koi calc mistake nahi hai toh 88 years hoga ye

In a class of 12 students , 7 boys and 5 girls.The class has 4 sessions each day,one each of arithmetic ,algebra,geometry and probability.These classes are to be held one after the other in 4 distinct time slots and can be in any sequence .Further there are 2 teachers available and they can teach any topic.
In how many distinct ways can a session be scheduled for two consecutive days..??

A certain sum is invested in simple interest.It becomes k times itself in 16 years and 2k time itself in 40 years.In how many years it will become 4k times itself?

@Chill sir : Sorry for the typo in last one

88 mein hoga ,agar galat hua toh tamatar maarna

Sanchit04421 Says

a subset of the set {1, 2, 3, , 49} is called good if the sum of its elements is greater than 612. How many good subsets are there?

My take is 48.

Sum of natural numbers till 49 = 1225
1225/2 = 612.5

So, if we make subsets of a set of first 49 natural numabers; half of them will have sum less than 612.5 and half of them will have sum more than 612.5. This is becuase, for every set A; there will be a counter set A'.
One of A and A' will have sum of elements as more than 612.5 and one will have less than 612.5.

So, we just need to find out, (maximim possible number of sets)/2

As we have 49 elements, each element has two choices: Participate in set or do not participate in set.

So, total ways = 2^49.

Sets with sum more than 612.5 = (2^49)/2 = 2^48

Let A take 1 step/sec

So A will take 10 steps in 10 sec.
B will take 90 steps in 10 sec
So speed of B is 9 steps/sec.

Speed ration A:B :: 1:9
Time ratio A:B :: 9:1
Let the speed of escalator be n steps/sec.

A has taken 50 steps to reach bottom taking 50 sec, B will take 50/9
So 50 + 50n = 90 + 50/9n
50n - 50n/9 = 40
50*(8/9)n = 40
n = 9/10 steps/sec

so visible steps of escalator : 50 + 50n : 95 steps


For the first time i am not sure in escalator question.

100 steps i guess...correct me if i am wrong 50+ 50n =90+10n so n=1 so total no of steps =100...where n is speed of escalator

ksagarwa Says

Neeraj has agreed to mow a lawn, which is a 20 m 40 m rectangle. He mows it with 1 m wide strip. If Neeraj starts at one corner and mows around the lawn toward the centre, about how many times would he go round before he has mowed half the lawn?

iska answer approx 4 times (oh yaar apna confidence level toh kam ho rha hain
)

A certain sum is invested in simple interest.It becomes k times itself in 16 years and 2k time itself in 40 years.In how many years it will become 4k times itself?

@Chill sir : Sorry for the typo in last one

sahilmadan Says

If a 4 digit natural no is 7083 times more the number formed by reversing the digits.How many such 4 digit numbers are there?

I think this times will not be there.

abcd - dcba = 7083
999(a - d) + 90(b - c) = 7083

111(a - d) + 10(b - c) = 787

=> (a - d) = 7, so 3 ways
and (b - c) = 9, so 9 ways

So, 3*9 = 27 such numbers

If a 4 digit natural no is 7083 more the number formed by reversing the digits.How many such 4 digit numbers are there?

45(base 5) = 25 (base 10)
11(base 5) = 6 (base 10)

45*11(base 5) = 25*6(base 10)

25*6 = 150 (base 10)

Now if we convert 150 from base 10 to base 5 we should get 3414.

But we are getting 1100 here,

5 does not satisfy.

25 in base 10 is equal to 100 in base 5 bhai...

My take is 103.

We need to check for power of 97 and 2; higest and smallest prime number in 100!.

Power of 97:
10000 => 103 + 1 => 104

Power of 2:
10000 => 5000 + 2500 + 1250 + 625 + 312 + 156 + 78 + 39 + 19 + 9 + 4 + 2 + 1 = 9994

Now, 100! has (50 + 25 + 12 + 6 + 3 + 1 = 97 times 2)
Now, 9994/97 = 103.xx

So, maximum power of 2^97 in 10000! is 103.

So, k can be maximum of 103.

jesus ,i making idiotic errors 10000! mein 97 ki highest power 3!!!
thanks ,ace bhai

victory2010 Says

10000! = (100!)^K P, where P and K are integers. What can be the maximum value of K?

My take is 103.

We need to check for power of 97 and 2; higest and smallest prime number in 100!.

Power of 97:
10000 => 103 + 1 => 104

Power of 2:
10000 => 5000 + 2500 + 1250 + 625 + 312 + 156 + 78 + 39 + 19 + 9 + 4 + 2 + 1 = 9994

Now, 100! has (50 + 25 + 12 + 6 + 3 + 1 = 97 times 2)
Now, 9994/97 = 103.xx

So, maximum power of 2^97 in 10000! is 103.

So, k can be maximum of 103.

FloatingBy Says

bhai yeh question sahi hain (4n+4)(n+1)=3n^3+4n^2+n+4,isse n natural number toh ni aa rha!!!

product of 44 and 11 given is 484 in decimal..is ko base 5 main convert karo toh 3414 milega..
so base 5 sahi hai..ab (3111)5 ko decimal main convert karo = 1+5+25+375=406

P.S.-koi is type ke questions ki fixed method batado yaar

k got it upar yeh galti hui hain ki humne 44 aur 11 ko bi base n mei liya ,inhe humein decimal system mein hi lena tha.aur jo result aata ,use base n mein change karna tha jisse 3414 aata.

ksagarwa Says

bhai ques sahi hai..aapne kya kiya hai batao...may be i can help

meversusme Says

Bhai question recheck karna answer to nikal hi nahi raha iska.........

FloatingBy Says

bhai yeh question sahi hain (4n+4)(n+1)=3n^3+4n^2+n+4,isse n natural number toh ni aa rha!!!

ans 406 aa raha hain..
base system is 5..no proper method..hit and trial se kiya..
iski koi fixed method ho toh batao?

bhaiyo dekho...

44*11 = 484....
now 484 = 3414 in base n
ab inke decimal rep ko equate kar do to get n as 5.....answer is not too far after that...

PS: Yaar this question is a bit ambiguous in its language...but it came in CAT as it is...

ans 406 aa raha hain..
base system is 5..no proper method..hit and trial se kiya..
iski koi fixed method ho toh batao?

45(base 5) = 25 (base 10)
11(base 5) = 6 (base 10)

45*11(base 5) = 25*6(base 10)

25*6 = 150 (base 10)

Now if we convert 150 from base 10 to base 5 we should get 3414.

But we are getting 1100 here,

5 does not satisfy.

FloatingBy Says

jara elaborate kar do bhai

product of 44 and 11 given is 484 in decimal..is ko base 5 main convert karo toh 3414 milega..
so base 5 sahi hai..ab (3111)5 ko decimal main convert karo = 1+5+25+375=406

P.S.-koi is type ke questions ki fixed method batado yaar

ashishkakkar123 Says

Find the remainder when 99999....9(54 times) is divided by 19.

rem 0...nay no repeated n-1 times vl b div by n provided nis prime...since 54 is a multiple of 18...rem is 0

ans 406 aa raha hain..
base system is 5..no proper method..hit and trial se kiya..
iski koi fixed method ho toh batao?

jara elaborate kar do bhai

Options are :-
102
103
104
105

@ pkaman :- dude plz xplain....

100! mein 97 ki power 1 aati hain aur 10000! mein 97^3 aata hain maximum
toh agar hum 100 ko 100+ power se raise karte hain toh 97 ki power b utne se raise ho jayegi ,jo 10000! mein ni hain ,isliye mujhe toh options galat lag rhe hain ,correct me if i m wrong

PS:yaar mein hi question mein kahin galti kar rha hoon ,mujhe pata ni chal rha ,ya question hi galat hain

ksagarwa Says

bhai ques sahi hai..aapne kya kiya hai batao...may be i can help

FloatingBy Says

bhai yeh question sahi hain (4n+4)(n+1)=3n^3+4n^2+n+4,isse n natural number toh ni aa rha!!!

Sagar bhai mene bhi aise hi kiya but isse to integer value nahi aa rahi base ki.....

In a number system the product of 44 and 11 is 3414. The number 3111 of this system, when converted to the decimal number system, becomes

ans 406 aa raha hain..
base system is 5..no proper method..hit and trial se kiya..
iski koi fixed method ho toh batao?

In a number system the product of 44 and 11 is 3414. The number 3111 of this system, when converted to the decimal number system, becomes

bhai yeh question sahi hain (4n+4)(n+1)=3n^3+4n^2+n+4,isse n natural number toh ni aa rha!!!

meversusme Says

Bhai question recheck karna answer to nikal hi nahi raha iska.........

bhai ques sahi hai..aapne kya kiya hai batao...may be i can help

pkaman Says

I think 104 hogaa.

Eek concept hai isme i guess:

here we have to take N = 10000
Take root of N = 100

Find the number (prime) smaller then this one 97.

Now 10000! will be divisible by 97^97.

So my Take is 97

Options are :-
102
103
104
105

@ pkaman :- dude plz xplain....

solve the inequality
24+p(t-3p-10)/p(p+6)+9 <-1 , where p=sq root q, and t=q

In a number system the product of 44 and 11 is 3414. The number 3111 of this system, when converted to the decimal number system, becomes

Bhai question recheck karna answer to nikal hi nahi raha iska.........

So number of additional lines created through intersection :

= 1/8 * [ n! / (n-4)!]

Where 'n' is the number of straight lines. In this case n=6

so 45

bhai id it some formula or what.....if so..plz tell wat is the thought behind.....

Rohan is asked to figure out the marks scored by Sunil in three different subjects with the help of
certain clues. He is told that the product of the marks obtained by Sunil is 72 and the sum of the
marks obtained by Sunil is equal to the Rohan's current age (in completed years). Rohan could not
answer the question with this information. When he was also told that Sunil got the highest marks
in Physics among the three subjects, he immediately answered the question correctly. What is the
sum of the marks scored by Sunil in the two subjects other than Physics?
(a) 6 (b) 8 (c) 10 (d) Cannot be determined

Okay, the solution goes like this
possible combinations 3,3,8 and 2,6,6 both add up to 14
But only 3,3,8 is possible as 8 is the highest in Physics so answer is 6

BUT, I say, the combination can be 1,8,9 and 4,6,3 why aren't these considered? Shouldn't the answer be CBD?

Shabnam is considering three alternatives to invest her surplus cash for a week. She wishes to guarantee maximum returns on her investment. She has three options, each of which can be utilized fully or partially in conjunction with others.

Option A: Invest in a public sector bank. It promises a return of +0.10%
Option B: Invest in mutual funds of ABC Ltd. A rise in the stock market will result in a return of +5%, while a fall will entail a return of - 3%.
Option C: Invest in mutual funds of CBA Ltd. A rise in the stock market will result in a return of - 2.5%, while a fall will entail a return of + 2%.

13. The maximum guaranteed return to Shabnam is

(1) 0.25% (2) 0.10% (3) 0.20% (4) 0.15% (5) 0.30%


14. What strategy will maximize the guaranteed return to Shabnam?

(1) 100% in option A
(2) 36% in option B and 64% in option C
(3) 64% in option B and 36% in option C
(4) 1/3 in each of the three options
(5) 30% in option A, 32% in option B and 38% in option C


Please post your answers along with the APPROACH..

Neeraj has agreed to mow a lawn, which is a 20 m 40 m rectangle. He mows it with 1 m wide strip. If Neeraj starts at one corner and mows around the lawn toward the centre, about how many times would he go round before he has mowed half the lawn?

ricky9 Says

bhai plz thoda explain kardo...

intersection points :6c2=15, lines that can be drawn joining these 15 points=15C2
but 5 points of intersection of a line(with other lines) lie on that same straight line only ,so we have to remove those case ,i.e 5C2 ,as there are 6 lines so we have to subtract these points for all of them ,so we get 6* 5C2
SO TOTAL NUMBER OF WAYS=15C2-6*5C2=45

There are 6 straight lines in a plane, no 2 of which are parallel and no 3 of which pass through the same point. If their points of intersection are joined, then the number of additional lines thus introduced is
(a) 45 (b) 78 (c) 105 (d) none of the above

So number of additional lines created through intersection :

= 1/8 * [ n! / (n-4)!]

Where 'n' is the number of straight lines. In this case n=6

so 45

FloatingBy Says

my take :45 additional lines need to be drawn

bhai plz thoda explain kardo...

FloatingBy Says

my take :45 additional lines need to be drawn

bhai plz thoda explain kardo...

Eek concept hai isme i guess:

here we have to take N = 10000
Take root of N = 100

Find the number (prime) smaller then this one 97.

Now 10000! will be divisible by 97^97.

So my Take is 97

sir 10000! mein 97^ 97 factor ni hota (or i m missing any point here)

If x, y and z are real numbers such that x + y + z = 5 and xy + yz + zx = 3, what is the largest value that x can have?

my ans is 3..
x+y+z=5 ..squaring both sides..
x^2+y^2+z^2=19..
x^2=19 not possible as x+y+z=5..
put x = 3, y = 3, z=-1..it will satisfy all conditions..

There are 6 straight lines in a plane, no 2 of which are parallel and no 3 of which pass through the same point. If their points of intersection are joined, then the number of additional lines thus introduced is
(a) 45 (b) 78 (c) 105 (d) none of the above

my take :45 additional lines need to be drawn

Let S be the set of integers x such that
I. 100<=x<=200
II. x is odd and
III. x is divisible by 3 but not by 7.
How many elements does S contain?

Here options would have helped but anyways:

All multiples of 3 in 100-200 {that are odd}
105,111,......195 : 16 numbers.

Now multiple of 7 and 3 that are odd : 105,147,189.

So required answer : 16-3 : 13

In a number system the product of 44 and 11 is 3414. The number 3111 of this system, when converted to the decimal number system, becomes

If x, y and z are real numbers such that x + y + z = 5 and xy + yz + zx = 3, what is the largest value that x can have?

victory2010 Says

10000! = (100!)^K P, where P and K are integers. What can be the maximum value of K?

Eek concept hai isme i guess:

here we have to take N = 10000
Take root of N = 100

Find the number (prime) smaller then this one 97.

Now 10000! will be divisible by 97^97.

So my Take is 97

There are 6 straight lines in a plane, no 2 of which are parallel and no 3 of which pass through the same point. If their points of intersection are joined, then the number of additional lines thus introduced is
(a) 45 (b) 78 (c) 105 (d) none of the above

Yaar Harsh bhai, ismein aapne 2^4 ke liye good sets kaise determine kiye ?
I mean what sum did you consider and why ?

Did you take the sum as 7 ? i.e., 15/2 ?

{1,2,3,4...49} Sum = 1225

we are asked to find 1225/2 = 662

similarly , {1,2,3,4,5} = 15/2 = 7

so any set which has sum greater then 7 will be good sets .

{1,2,3,4,5}

Not good sets :

{},{1},{2},{3},{4},{5},{1,2},{1,3},{1,4},{1,5},{2,3},{2,4},{2,5},{3,4},{1,2,3},{1,2,4}---- Total 16 --2^4

Total sets of {1,2,3,4,5} ---2^5

So, good sets 2^5-2^4 = 2^4

Hence for {1,2,3,4...49}, it should be 2^48

Let S be the set of integers x such that
I. 100<=x<=200
II. x is odd and
III. x is divisible by 3 but not by 7.
How many elements does S contain?

iska answer 13 aa rha hain

victory2010 Says

10000! = (100!)^K P, where P and K are integers. What can be the maximum value of K?

i think the answer is 3

Could you please help me with a proper solution for this problem?

An escalator is descending at constant speed. A walks down and takes 50 steps to reach the bottom. B runs down and takes 90 steps in the same time as A takes 10 steps. How many steps are visible when the escalator is not operating?

Let A take 1 step/sec

So A will take 10 steps in 10 sec.
B will take 90 steps in 10 sec
So speed of B is 9 steps/sec.

Speed ration A:B :: 1:9
Time ratio A:B :: 9:1
Let the speed of escalator be n steps/sec.

A has taken 50 steps to reach bottom taking 50 sec, B will take 50/9
So 50 + 50n = 90 + 50/9n
50n - 50n/9 = 40
50*(8/9)n = 40
n = 9/10 steps/sec

so visible steps of escalator : 50 + 50n : 95 steps


For the first time i am not sure in escalator question.

Let S be the set of integers x such that
I. 100<=x<=200
II. x is odd and
III. x is divisible by 3 but not by 7.
How many elements does S contain?

odd nos-50
divisible by 3-16
both 3 and 7-5 of which 3 are odd.
13 hoga just a silly mistake!!

1. In an examination the maximum mark for each of the three paper is 50 each.Maximum mark for fourth paper is 100. Find the no. of ways in which a candidate can score 60% mark in the aggregate.
2. A student has 3 books (B1,B2,B3) and his book-rack has 3 shelves.He has to arrange the books in the shelves such that he puts 2 books together in one shelf and the remaining 1 book in another shelf.

A2:-FIRST select 2 books can be selected in 3c2 ways.. for these 2 books 1 shelf can be selected in 3c1 ways..and remaining book can be placed in any of the remaining two shelves in 2c1 ways...so ans= 3c2 * 3c1 * 2c1=18

Neetujain1987 Says

IS k =0?????

I think 104 hogaa.

hi

ques-1)a+b+c=6, a^2+b^2+c^2=26 , a^3+b^3+c^3=90 , a^4+b^4+c^4=?????

ans 338
(give shortest approach....)

ques-2) a bag contains 6 balls of one or more colors. a ball is picked and is found to be red.what is the probability that the bag initially had exactly 6 red balls?

ans- BAAD mai


ques-3)find the number of 5 letter words that can be formed using letters a ,e,b,c,d ,where ae can occur any number of times and the other letter can occur at most once.

ans- pta nai.

""

fastest way is hit nd trial method ...v c 4,3 -1 satisfy da eqn...so 338

-----snipped------

victory2010 Says

10000! = (100!)^K P, where P and K are integers. What can be the maximum value of K?

IS k =0?????

Should be 2^48

Taken a small set {1,2,3,4,5} and then generalized
Getting 2^4 good sets:

5 elements ----2^4 good sets
so, we having here 49 elements , so, it should be 2^48

Yaar Harsh bhai, ismein aapne 2^4 ke liye good sets kaise determine kiye ?
I mean what sum did you consider and why ?

Did you take the sum as 7 ? i.e., 15/2 ?

Its an LR set
@Nachieket bhai , Gaurav bhai..or any puys!! plzz help
Post approach too..!!


Consider A tournament having 4 teams A,B,C,D...The Table given below Shows the proceedings of the tournament. The Table shows NRR of the team after that match....

Team_____First Match_____Second Match____Third Match
A__________+0.8____________A2_____________+0.6____
B__________B1______________-0.4___________+0.066__
C__________+1______________+0.25__________C3_____
D__________D1______________-0.5____________D3____


A team Plays its second match, once every team has played its first match..Similarly, it plays 3rd match only after every team has played its 2nd match.
If A team was allout within the stipulated 50 overs, it was deemed to have batted 50 overs.
All matches ended in a victory..except 1..which was a tie...In these victories, all the teams that batted first were the winners in all the matches except one which happened to be the last match played by C.

Q1. Which Teams were Involved In a tie ?
1. A and D
2. A and B
3. B and D
4. CBD

Q2. What is Value of A2 ?
1. +1.3
2. +1.05
3. +0.65
4. +0.525

Q3. By how many runs A won the match against C
1. 50
2. 90
3. 45
4. 25

answers are
1) B and D
2) 0.65
3) 25

my approach -

first of all
net run rate = (total runs scored/total balls faced) - ( total runs conceded/total overs bowled)

given - If a team was all out within the stipulated 50 overs, it was deemed to have batted 50 overs..
thus total overs faced = total overs bowled

if a team plays 3 matches, then it's final NRR would be average of NRR of all 3 matches...
thus column showing net run rate in the 2nd round is average of the NRRs of 1st two matches..same for 3rd column..

when two teams play a match, net run rate of one team is equal to that of other team but sign would be opposite i.e. winning team will have +ve NRR while losing team will have -ve NRR (same value)

in case of a tie, NRR of both teams would be 0.00

now NRR of C is given for 1st 2 matches..
NRR after 1st and 2nd match are 1 and 0.25
thus NRR in 2nd match = 1+ x/2 = 0.25
x = -0.5..NRR of C -2nd match is -0.5..thus match played by C in 2nd round is not a tie.

now for B and D - NRR after 1st match is any of -0.8 and -1.0 .
so no tie in 1st round.

for a tie to take place in 2nd round - NRR after 2 matches would be half of the NRR after 1st round..as (x+0)/2 = x/2..

from data we can see that for B and D, NRR after 2 matches is -.4 and -.5 which is half of their NRR after 1st match..

So tie took place between B and D

2) as a tie took place between B and D, A played against C in that round..
as we found that NRR of C in 2nd match is -0.5..
So NRR of A is 0.5
thus A2 = (0.5 + 0. )/2 = 0.65

3) NRR between A and C in 2nd match is 0.5..
for eg. if RR of A is 6..then RR of C is 5.5..
then net difference in runs = 300 - 275 = 25 runs
same would be for other RRs..

P.S. - yeh maine LR-DI ki thread main solve kiya tha bahut pehle..abhi copy paste kar raha hun

sanchit04421 Says

a subset of the set {1, 2, 3, , 49} is called good if the sum of its elements is greater than 612. How many good subsets are there?

********** too low a value***********