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there are a total of 30 seats such that we have to select 6 blocks of 4 seats.

That means total of 30-{6*4} = 6 seats will be left b/w these blocks.

Let these left over seats be :

a,b,c,d,e,f now they will be arranged b/w 6 blocks like.

x....B1....y.....B2....z.....B3......p.....B4......q.......B5......r......B6.....s

now these places x,y,z,p,q,r,s can take 6 seats {a,b,c,d,e,f} all of same type.

So x+y+z+p+q+r+s = 6

whole number solutions of this equation : C(12,6)

My Take C(12,6)

Hi but as per the last eqns the number of soln of eqn

x+y+z+p+q+r+s = 6 in case of non zero soln will be

n-1^(C)r-1 is'nt it please explain i m still not clear abt it ??

Hi puys solve this im really confused abt this questn plz post detailed soln

There is row of 30 seats & there are 6 groups of 4 persons each. In how many ways can 6 blocks with 4 consecutive seats in each block, be selected to seat all the six groups ??

There is row of 30 seats & there are 6 groups of 4 persons each. In how many ways can 6 blocks with 4 consecutive seats in each block, be selected to seat all the six groups ??

n.vishal87 SaysI think if we divide the expression by 9, we get 4 as the remainder and not 6. Plz check and let me know if I am wrong

Dear its simple we have 2^2(49*50*99)/6 that will be equal to

2*49*50*33 which when divided by 9 will give 6 as remndr fr sure may be jumped some steps thts why u are getng 4 recheck it i hope this is now clear

For the frst one answr is 6 as given below:-

1>it can be written as 2^2(1^2+11^2+111^2.....& so on till 1111.....49times)

Now a number is divisible by 9 if the sum of its digit is divisible by 9 so we have

2^2(1^2+2^2+3^2.......49^2)

which will be 2^2(49*50*99/6-----(A)

using formula for (1^2+2^2...n^2)==n*(n+1)(2n+1)/6

Now if we divide the same (A)by 9 we wll get 6 as remndr hope its clr now

2>For second one it is 160

Plz help me out with these problems on Number system

1.Find the remainder of 2^2+22^2+222^2+2222^2+......(222.....49 twos)^2 is divided by 9

2.n=202*20002*200000002*20000....2(15 zeroes)*20000...2(31 zeroes).find the sum digits in this multiplication

a.112

b.160

c.144

d.cant determine

plz provide ur answers with explanation.

For the frst one answr is 6 as given below:-

1>it can be written as 2^2(1^2+11^2+111^3.....& so on till 1111.....49times)

Now a number is divisible by 9 if the sum of its digit is divisible by 9 so we have

2^2(1^2+2^2+3^2.......49^2)

which will be 2^2(49*50*9/6-----(A)

using formula for (1^2+2^2...n^2)==n*(n+1)(2n+1)/6

Now if we divide the same (A)by 9 we wll get 6 as remndr hope its clr now

- 2 Likes

two mocks in one day and both i end up in the nineties

QADI 17c 11w 40

VALR 21c 5w 58

OA 38c 16w 98

tough quant section decent DI .. lengthy LR and straight forward VA

hey ravi thts ausum score man btw hw do u manage to attempt so many questions in a paper of this level pl thrw some tips dear i badly need it as i cud do a max 8 in QA+DI section btw thnkx in advance

rkpushkar Saysanswer is 10:46 and

ya the answr is indeed 10:46 but can u expln ur apprch in detail btw thnx in advance ur di looks vry strng

rkpushkar Saysanswer is 10:46 and

ya the answr is indeed 10:46 but can u expln ur apprch in detail btw thnx in advance u di looks vry strng

hey anybdy knws hw to solve this plz explain ?? looking frwd to yr replyz

plz solve this along as given in attachment with explanations thnx in advance

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