Answer / nash

1434




STARTING ADDRESS IS 1258.
EACH ELEMENT IS OF 4 BYTES.
SO (5,8)=44 I.E. POSITION OF ELEMENT
44*4=176
1258+176=1434
DO REPLY.

Answer / jyothi

STARTING ADDRESS IS 1258.
FIRST ELEMENT ADDRESS FROM 1258 TO 1261.
LIKE THAT 5,8 MATRIX ADDRESS FROM 1430 TO 1433
SO LAST BYTE ADDRESS OF 5,8 IS 1433

Answer / snehal

(1,1)=1258
after tat for 1 it contain 8
for 2=9
for 3=9
for 4=9
for 5=8
total=43
43*4=172
n 1258+172=1430

Answer / rohit

to find starting byte for matrix a(5,1):
1258+(4*(9*4))-(1)=1401

then to find last byte for matrix a(5,8):

1401+(8*4)=1433

Answer / vijay

plz post clr question..?

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

ans is 1433
In column address increments by (8x4+4)36=clmn-gap
In row address increments by 4=row-gap
So
add of (5,8) element (4 clmn-gaps & 7 row-gaps )
=1258+4x36+4x7=1430
add of last byte=[1430,1431,1432,1433]
ans 1433

Answer / shobha

1258+5*8*4bytes=1418

Answer / sona

Question is confusing we are unable to decide wheather (x,y)
is a matrix or a position in the matrix

Answer / manish panwar

1258-1261 1262-1265 1266-1269 1270-1273 1274-1277 1278-1281 1282-1285 1286-1289

1290-1293 1294-1297 1298-1301 1302-1305 1306-1309 1310-1313 1314-1317 1318-1321

1322-1325 1326-1329 1330-1333 1334-1337 1338-1341 1342-1345 1346-1349 1350-1353

1354-1357 1358-1361 1362-1365 1366-1369 1370-1373 1374-1377 1378-1381 1382-1385

1386-1389 1390-1393 1394-1397 1398-1401 1402-1405 1406-1409 1410-1413 1414-1417



(5,8) matrix

4 4 4 4 4 4 4 4 = (4*7)+3=31 1258+31 = 1289
4 4 4 4 4 4 4 4 = 4*8=32 1289+32 = 1321
4 4 4 4 4 4 4 4 = 4*8=32 1321+32 = 1353
4 4 4 4 4 4 4 4 = 4*8=32 1353+32 = 1385
4 4 4 4 4 4 4 4 = 4*8=32 1385+32 = 1417

(32*5)-1 = (160 -1)= 159
1258+159 = 1417


So answer 1410

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