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ISRO-2013-1

• edited by
11,447 views
12 12 votes

Let $A(1:8, -5:5, -10:5)$ be a three dimensional array. How many elements are there in the array $A$?

  1. $1200$
  2. $1408$
  3. $33$
  4. $1050$

4 Answers

Best answer
32 32 votes
I am just attempting it not sure

Here i think they have specified the size

So 1:8 mean 8 elements ( both are inclusive  )

(-5 : 5 ) mean 11 elements

(-10 : 5 ) mean 16 elements

So no of elements will be ( just like Multidimensional array will be ) 8*11*16=1408

so option b
• selected by
5 5 votes
ul = upper limit

ll=lower limit

 

(ul3-ll3+1) x  (ul2-ll2+1) x(ul1-ll1+1) =(8-1+1) x( 5- (-5)+1)  x   (5 +10+1) =1408
2 2 votes
Length of First dimension =8-1+1=8

Length of Second dimensions =5-(-5)+1=11

Length of Third dimensions =5-(-10)+1=16

Number of elements= 8×11×16= 1408
0 0 votes
8 2d arrays
one array has 11 rows and 16 columns
so
one 2d array has 16*11 elements
so all are
8*16*11=1408
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