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What is the result of evaluating the following two expressions using three-digit floating point arithmetic with rounding?

(113. + -111.) + 7.51

113. + (-111. + 7.51)

1. 9.51 and 10.0 respectively

2. 10.0 and 9.51 respectively

3. 9.51 and 9.51 respectively

4. 10.0 and 10.0 respectively

Rewrite the smaller number such that its exponent matches with the exponent of the larger number.

$(113. + -111.) = 1.13 * 10^2 + -1.11 * 10^2 = 0.02 * 10^2 = 2.0 * 10^0$

$2.0 * 10^0 + 7.51 * 10^0 = 9.51 * 10^0$

$(-111. + 7.51) = -1.11 * 10^2 + 7.51 * 10^0 = -1.1 * 10^2 + 0.08 * 10^2 = -1.03 * 10^2$

$113. + -1.03 * 10^2 = 1.13 * 10^2 + -1.03 * 10^2 = 0.1 * 10^2 = 10.0$

selected
sir Is it not same thing as i had posted ?
yes, I just gave an intermediate step :)
@Mithilesh I had told you not to edit others answer. Everyone won't be thinking like you.
there is difference in the answers rt?

+7.51∗10^0

why 7.51 is not rounded of to 8.00 in first part addition?
Why it should be rounded? Rounding happens only when it is really required (as we loose precision) or when explicitly told. Also, this question is regarding fixed point arithmetic which is in GATE syllabus but not well covered in most books.

@Arjun Sir,

Here in the following step:
(−111.+7.51) = −1.11∗102+7.51∗100 = −1.1∗102+0.08∗102
The rule is to convert the exponent of smaller number as same as of the larger. But, here lager number is 7.51, whose exponent is zero. But, here you have converted the exponent of larger number to that of smaller's (i.e., exponent of 7.51 (larger) to that of -1.11 (smaller))?

So floating point additions are not associative right?
Thanks for the link @Arjun sir thanks a lot ...really it was life saving ...
3 digit floating point arithmetic is used..
(113.+-111.)+7.51 = 2.00 + 7.51 = 9.51
113.+(-111.+7.51) = 113. + (-111. + 8.00) //rounding off to make compatible 7.51 and 111. with respect  3 digit floating point arithmetic
113. - 103. = 10.0
First one is 9.51
Second evaluates to 10.49(let 10.4 as per 3digit )
As per option ..select A coz 10.4 can be rounded to 10 not to 9.51

10.51 ???
is it three-digit floating point arithmetic ???

As it is  three-digit floating point arithmetic with rounding

so Rounding off should be done on (111.+7.51) ie, 103.49 to 103.

Though ans will remain same..But approach should be this,in my view.103.103.49 is rounded to 103.0

to apply operation ur operand should be THREE FLOATING POINT ARITHMATIC ..

Please explain the meaning "three-digit floating point arithmetic with rounding"

2nd part:  113. + (-111. + 7.51)

(-111. + 7.51) = - .111 x 10+ .751 x 10

= 10( - .111 x 102 + .751)

= 10(-11.1 + .751)

= 10(- 10.349)

= - 103.49

now, 113. + (- 103.49) = .113 x 103 - .10349 x 103

= 103(.113 - .103)  // ignor 49 bcz  three-digit floating point arithmetic with rounding

= 103 x .010

= 10