# Ihih

**Topics:**Numerical digit, Decimal, Positional notation

**Pages:**10 (1923 words)

**Published:**April 1, 2013

(a). Perform the following:

(i) 10000110002 to Hexadecimal

(ii) ABCDE16 to Decimal

(iii) A1F2.F16 to Octal

________

(i) 19 08 07 06 05 14 13 02 0100 2 ( base 10 ( base 16

= (29) + (24) + (23)

= 536

16|536=21816

33 8

2 1

(ii) A B C D E16 ( base 10

A4 B3 C2 D1 E 0

= (10*164) + (11*163) + (12*162) + (13*161) + (14*160)

= 70371010

(iii)A1F2.F16 (base 2 ( base 8

A3 12 F1 20 . F-1

=1010 0001 1111 0010.1111 base 2

=1 010 000 111 110 010.111 1 base 8

= 120762.748

(b) A computer stores a number of 16 bits word using floating-point arrangement. Given that the first bit is reserved for the sign and followed by 6 bits for the exponent using biased form. The remaining bits are used for the mantissa with a hidden bit.

(i) Show how the computer stores -37.87510

(ii) What is the decimal value for 0100 1111 0111 01012?

(i)-37.875

2|37100101.111

18. 1

9. 0

4. 1

2. 0

1. 0

0.875

2 X

1.750

1 -1

0.750

2 X

1.500

1 - 1

0.500

2 X

1.0001

- =100101.1111X 20 |- TE = 5| Mant = 1.001011110

= 1.00101111 X 25 stored = true expo + bios = =001011110 = 5 + 31 = 36

- convert expo to binary |- Mant = 1.001011110

2| 36= 1001002 = 001011110 ( 1.100100 001011110

18 0 sign = -

9 0 = 1

4 1

2 0

1 0

(ii) 0 100111 1011101012

Expo mant

-Expo 1001112 = 15 04 03 12 11 10 |- stored expo = true expo + bios

= (25)+(22)+(21)+(20)39= TE + 31

= 32 + 4 + 2 +1 TE= 8

= 39

-101110101+1|-18 17 06 15 14 13 02 11 00 . 1-1

=0.101110101+1 = (28)+(27)+(25)+(24)+(23)+(21)+(2-1)

=1.101110101 x 28 = 442 + (2-1)

=110111010.1 = 442.510

(c) Using BCD 8-4-2-1 representation, calculate 6789 + 7156 – 365 ______

6789 = 0110 0111 1000 1001

7156 = 0111 0001 0101 0110 +

1101 1000 1101 1111

0110 0110 0110 +

0001 0011 1001 0100 0101

365 = 0000 0000 0011 0110 0101 –

0001 0011 0101 1110 0000

0110 -

0001 0011 0101 1000 0000

1 3 5 8 0

(d) X = 4.521 is corrected to 3D, Y = 2.7 is corrected to 2SF and Z = 1.64 is corrected to 2D. Find the range of values for the true value of (X - Y)(Y +2Z) to 6SF. __________

(X-Y)(Y+2Z) = (4.521 – 2.7)(2.7+2(1.64))

= (1.821)(5.98)

= 10.88959 Ω 10.8896 (6SF)

(e) Round off 20685 to 3SF and round down 0.002786 to 5D.

_____

-20685 ( 3SF round off

=20700

-0.002786 ( 5D rounded down

=0.00278

2

(a) Suppose that 40 of students in a class read three types of magazines and at least one of them read one magazine. Given that the number of students who read educational magazines only is the same as the number of students who read health magazines only. Also,

10 students read sports magazines only.

11 students read sports and health magazines.

6 students read educational and health magazines.

7 students read educational and sports magazines.

4 students read all three magazines.

(i) Represent the above information in a Venn diagram.

(ii) Find the number of students who read two magazines only. ____

(i)

[pic]

-n(S) = 10, n(H) = n(E) = x , n(S∩H) =11, n(E∩S) =7, n(E∩H) = 6, n(S∩H∩E) = 4 N(total)= n(S) + n(H) + n(E) – n(S∩H) – n(E∩S) – n(E∩H) + n(S∩E∩H) 40. = 10 + x + x -11 – 7 – 6 – 4

40. = 2x – 10

X = 25

(ii)S ∪ H ∪ E= 7 + 2 +3

= 12

(b) Let P = {2,5}, Q ={1,2,3,4,5} and R ={2,3,5,7,11,12}. Draw and shaded a Venn diagram for (P∩Q∩R).

____

[pic]

(c) The Venn diagrams shows set L, set M and set N with the condition that U = (L∪ M ∪ N). On the diagram shade

[pic]

(i) the set M’

(ii) the set L∩(M ∪ N) .

____

(i)

[pic]

(ii)

[pic]

(d) Let A = {0, 2, 4, 6, 8, 10}, B = {0, 1, 2, 3, 4, 5, 6}...

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