Latest TCS Aptitude Question SOLUTION: The matrix of a(7,9) was given.the address of the first byte of a(1,1)=1258.it takes 4 bytes to store the nuymber.then calculate the address of the last byte of a(5,8)

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(#M40021733) TCS Logical Reasoning question Keep an EYE

The matrix of a(7,9) was given.the address of the first byte of a(1,1)=1258.it takes 4 bytes to store the nuymber.then calculate the address of the last byte of a(5,8)

Asked In TCS Papri Bhowmick (11 years ago)

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aman kesharwani  (11 years ago)

1430 i.e. 1258+(3*36+2*32)=1430

Is this solution Helpfull? Yes (11) | No (1)  

Avinash Aryan  (11 years ago)

1430.
a=1258
d=4
n=44
so nth term=1258+(44-1)*4=1430.

Is this solution Helpfull? Yes (11) | No (1)  

reddy siri  (11 years ago)

1434 is the answer..we have to calaculate the address of a(5,8)
i.e from a(1,1)+a(2,2)+------a(1,9) i.e 9 terms
a(2,1)+-------------a(2,9) i.e 9 terms
a(3,1)+-------------a(3,9) 9 terms
a(4,1)+-------------a(4,9) 9 terms
a(5,1)-------------upto a(5,8) 8 terms
so total 44 terms..so address calaculation is given by
starting address+(no. of terms)(no. of bytes)=1258+44(4)=1434

Is this solution Helpfull? Yes (11) | No (4)  

@rthy  (11 years ago)

1434... (4*36)+(4*8) =176..add it with 1258+176=1434.

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smith kumar saha  (11 years ago)

1430 is correct

Is this solution Helpfull? Yes (2) | No   

shahbaz ahmad  (11 years ago)

number will be start storing from 1430 but question has asked abt the last byt of storage, so after storing (5.8),last byt will be 1434.
address calculation formula= base address+ no.of col * (no. of rows-1)*size of storing 1 number + (no. of col - 1)* size of storing 1 no.
1258+ 9*(5-1)*4+(8-1)*4=1430,
but last byt will be at 1434.

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Amit Gupta  (11 years ago)

1434 is correct answer.

Is this solution Helpfull? Yes (1) | No (3)  

Rudra  (11 years ago)

Plzzzz anybody explain in details...........

Is this solution Helpfull? Yes (1) | No   

Rudra  (11 years ago)

@aman plzzz explain in detail

Is this solution Helpfull? Yes | No (3)  

gitanjali chandel  (11 years ago)

i didnt get it..plz explain it

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KUMAR VIVEK  (11 years ago)

a(1,1)=1258 ; (given)
we have to calculate the address of the last byte of a(5,8) ,i.e
44th term in the given matrices have to find
and if 1 term takes 4 bytes to store the number
then 44th term will be 1258 + (44*4) = 1258 + 176 = 1434

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ashish chikate  (7 years ago)

1433....last byte

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