Ass Math

BT 07-01 Basic mathematics

Q1. U={a, b, c, d, e, f, g, h, k}, A={a, b, c, g}, B={d, e, f, g}, C={a, c, f}&D={f, h, k}. a. Compute 1) A ∪b = {a, b, c, d, e, f, g}. 2) B∪c = {a, c, d, e, f, g}. 3) A∩c = {a, c}. 4) B∩d = {f}. b. Compute 1) A-b = {a, b, c}. 2) Ā = {d, e, f, h, k } 3) A∪b∪c. = {a, b, c, d, e, f, g}. 4) A∩(b∪c). = {A, c, g} c. 1) 2) 3)

Compute (A∩b) c = {a, b, c, f}. (ac∪bc) => ac={d, e, f, h, k}, bc={a, b,c,h,k}, (ac∪bc)={a,b,c,d,e,f,h,k}. Verify that (a∪b)c=ac∩bc . A ∪b = {a, b, c, d, e, f, g}. (a∪b)c= {h.k}. ac={d, e,f, h,k}. bc={a, b,c,h,k}. (ac∩bc)={h.k}. Hence, it is verified that (a∪b)c=ac∩bc.

Q2. U={1,2,3,4,5,6,7,8,9}

A={1,2,4,6,8}, B={2,4,5,9},

C={x|x is a pointing integer and x2 =16}

D={7,8}, a. Compute 1) a∪b ={1,2,4,5,6,8,9}. a∪c, c={4,-4}. a∪c={1,2,4,-4,6,8} 2) a-b = {1,6,8}. 3) a∩c = {4}. b. Compute 1) c-d= {-4,4}. 2) (a∪b)c=> (a∪b)={1,2,4,5,6,8,9} 3) (a∩(b∪c)=b∪c={2,4,-4,5,9} (a∩(b∪c)={2,4}. 4) (a∪(b∪c)=

b∪c={{2,4,-4,5,9} (a∪(b∪c)={1,2,4,-4,5,6,8,9}.

c. Verify the theorems. 1) (a∩b)c= ac ∪ be let u={1,2,3,4,5,6,7,8,9} a={1,2,4,6,8}, b={2,4,5,9}, then a∩b={2,4}, and so (a∩b)c={1,3,5,6,7,8,9}. Ac={3,5,7,9}, Be={1,3,6,7,8}, Ac∪be={1,3,5,6,7,8,9}.

Hence it is proved that (a∩b)c=ac∪be. 2) (a∪b)c=ac∩be.

Let u={1,2,3,4,5,6,7,8,9}, A={1,2,4,6,8}, B={2,4,5,9},

Then a∪b={1,2,4,5,6,8,9} And so (a∪b)c={3,7}. Ac={3,,5,7,9} Be={1,3,6,7,8} Ac∩be={3,7}.

(a∪b)c={3,7}.

Hence it is prove that (a∪b)c=ac∩bc. Q3. A) Define negation, conjunction, disjunction of the statement with examples. Ans. Negation, conjunction, disjunction are in fact the compound proposition of connectives. Like and not or if-then, if and only if, these are connectives and there compound preposition is conjuction, disjunction, and negation.

Negation= The negation of a proposition is generally form by introducing the word “not” at proper place in the preposition or by prefixing the preposition with the phrases “it is not the case that”.

Truth table for negation P T F

-p F T

Conjunction= The conjunction of two proposition p and q is the proposition p/\q which is read as p and q.

Truth table for conjunction P

Q

P/\q

T

T

T

F

T

F

T

F

F

F

F

F

Note: the truth value of p/\q is same as truth value q/\p. Disjunction: The disjunction of two proposition p and q is the proposition p\/q which is read as p or q. the proposition p\/q has the truth value f only when p and q have the truth value f; otherwise it is true. The disjunction is defined by truth table:

Truth table for disjunction. P T F T F

Q T T F F

P\/q T T T F

Note: the truth value of p\/q is the same as that of q\/p. b) what is truth table? Write the table for the statement p\/ (q /\ r). truth table = a truth table presents compactly the truth values of one or more than one preposition for all the possible combination of the truth values of each primary proposition. P\/(q/\r) P

Q

R

(q/\r)

P\/(q/\r)

T T T T F F F F

T T F F T T F F

T F T F T F T F

T F F F T F F F

T T T T T F F f

Q4) a. 1) prove that sum of the degrees of the points of a graph is twice the number of the lines. Ans. Let a be a (p,q) graph. Every line of g is incident with two points. Hence every line gives 2 to the sum of the degrees of the points. Hence ∑deg vi = 2q Example in the given figure a be a (5,10) Graph. In which the degree of each point is 4 . there are 5 points. Hence, ∑deg vi =20 I=1

2) Cubic graph with six points.

b). define i)

Cubic graph.

A regular graph of degrees 3 is called a cubic graph.

2) complete graph-> a graph in which any two distinct points are adjacent is called complete graph.

3) Regular graph. For any graph g we define б(g)=min{deg v/v(g)Єv(g)} and (g)=max{deg v/v∈v(g)} If all the points of g have the same degree r then δ(a)=(g)=r and in this case g is called a regular graph of degree r. V1 v6 V3 V2

v5 v4

05. a. prove that, 1) in any graph “g” the number of points of odd degree is even. Ie=∑deg vi =2q ProofLet g be a (p,q) graph. Every line of g is incident with two points. hence line gives 2 to the sum of degrees g the points. Hence ∑deg vi =2q In the fig a be a (5,10)graph. In which the degree of each point is 4. there are 5 points hence 5. ∑ deg vi =20 I=1.

V1 v2

V5

v3

v4

2) Every cubic graph has an even no of points.

Proof# Let g be a cubic graph with p points then, ∑deg.v=3(no of points in g)=3p We know that ∑deg v is always even by theorem, hence 3p is even Then p must be even. Hence the number of points in a cubic graph is even. Q5) b. let (A*) be a group. Show that (A*) is an abelian group. If and only if (a*b)2 = a2+b2 for any a,b in A. Ans. Taking binary operation (a+b)2 = a2 +b2 of adds. a2 + b2 +2ab=a2 + b2 2ab=0 ab=0 (b*a)2 = b2*a2 ba=0 therefore, ab=ba. Hence, it is an abilian group. Proved.

Q6. How many different members committees can be formed each containing 4 female members from available set of 20 females and 5 male from available set of 30 males? Ans. Here order does not matter in the individual choices. Females can be performed in 20 c4 = !20/!4!(20-4)=!20/!4!16 20*19*18*17/4*3*2*1=4845 male can be performed in: 30 c5=!30/!5!(30-5)=142506 by the fundamental theorem there are (4845)(142506)= 690441570 way a members committee can be formed. 50

c9 = !5/!9!(50-9)=50*49*48*47*46*44*43*42/9*8*7*6*5*4*3*2*1=2505433700.ans

from20c4*30c5/50c5. b. How many ‘words’ can be formed from the alphabet {a, b, c, d, f). Ans.

The number is 5 p4=! 5/! (5-4)=! 5/1=120ans.

BT O7-02 Basic mathematics

Q. 1) a. A survey of 500 hundred T. V. watchers produced the following information 285 watch the football games 115 watch basketball games 45 watch basketball and football games 70 watch football and hockey games 50 watch hockey and basketball games 50 do not watch any of three games

i)

How many people in this survey watch all the games?

ii)

How many people watch exactly one of the three games? b. A = { 1, 2, 3, 4 },

B = { 5, 6, 7 },

C = { 3, 4 }

Compute i) A X B, ii) B X A, iii) C X A, iv) B X C c. A = { a, b, c, d, e },

B = { 1, 2, 3 },

C = { 4, 5, 3 }

Verify that i) A X B ≠ B X A, ii) BC ≠ C – B , iii) A ∩ B = B ∩ A

Ans. Let assume that a=for football. B=hockey. C=basket. Therefore, p(a)=285. B=195 C=115. a∩c=45. a∩b=70. b∩c=50. a∪b∪c=500-50. P(a∪b)=p(a)+p(b)-p(a∩b) (a∪b∪c)=p(a)+p(b)+p(c)-p(a∩c)+p(a∩b∩c) 450=285+195+115-45-70-45+(a∩b∩c) 15=a∩b∩c 450-[45+70+45+15] 450-[175]=275 ans.

B. i) A X B = {(1,5), (1,6), (1,7), (2,5), (2,6), (2,7), (3,5), (3,6), (3,7), (4,5), (4,6), (4,7) }

ii)

B X A = {(5,1), (5,2), (5,3), (5,4), (6,1), (6,2), (6,3), (6,4),

(7,1), (7,2), (7,3), (7,4)} iii) C X A ={(3,1), (3,2), (3,3), (3,4), (4,1), (4,2), (4,3),(4,3), (4,4)} iv) B X C = {(5,3), (5,4), (6,3), (6,4), (7,3), (7,4)} c. i) verify A X B ≠ B X A L.H.S. A X B = { (1,5), (1,6), (1,7), (2,5), (2,6), (2,7), (3,5), (3,6), (3,7), (4,5), (4,6), (4,7) } R.H.S. B X A ={ (5,1), (5,2), (5,3), (5,4), (6,1), (6,2), (6,3), (6,4), (7,1), (7,2), (7,3), (7,4) } Hence we verify that the given statement A X B ≠ B X A is true. ii) BC ≠ C – B {1,2,3,4,5} ≠ {4,5} iii) A∩ B=B∩A L.H.S. A ∩ B = {Φ} R.H.S. B ∩ B = {Φ} Hence A ∩ B = B ∩ A

Q. 2 a. Define a tautology. Prove that a statement (p ٨ q) → (p ٧ q) is also a tautology. b. Write the negation of statement i) p ٨ (q → r) ,

ii) [ ( p ٨ q ) ٨ ~ ( q ٧ r )]

Ans. a. TATULOGY:

A compound proposition which is always truth values of its components is called a tautology. We have to prove that the statement given ( p ٨ q ) → (p ٧ q ) is a tautology.

P T T F F

q T F T F

(p ٨ q) T F F F

(p ٧ q) T T T F

Hence the result of the given statement is all true that shows a Tautology.

(p ٨ q)→ (p ٧ q) T T T T

b. i)

the negation of this statement p ٨ (q → r) is given below:

p T T T T F F F F

q T T F F T T F F

r T F T F T F T F

(q → r) T F T T T F T T

P ٨ (q → r) T F T T F F F F

~[p ٨ (q →r)] F T F F T T T T

ii) the negation of the following statement [ ( p ٨ q) ٧ ~(p ٧ r) ] is given below

P

Q

R

p٨r

q٧r

~(q ٧ r)

(p ٨q) ٧ ~ (q ٧r)

T T T T F F F F

T T F F T T F F

T F T F T F T F

T T F F F F F F

T T T F T T T F

F F F T F F F T

T T F T F F F T

~[(p ٨q)٧~(q ٧r) ] F F T F T T T F

Q3. a. There are two restaurants next to each other. One has a sign that says “good food is not cheap” and the other has a sign that says, “cheap food is not good”. Are the signs saying the same thing? b. Construct a switching circuit for p ٨ (q ٧ p) and write the flow table. If possible write an equivalent simple circuit?

Ans. a. Let g: food is good c: food is cheap. The sign of first restaurant can be written as g→~c, and the sign of second be written as c→~g.

g T

c T

~g F

~c F

g→~c F

c→~g F

T F F

F T F

F T T

T F T

T T T

T T T

The above truth table shows that the two signs say the same thing.

b. we have to construct a switching circuit for p ٨ (q ٧p). the truth table for the statement p ٨ (q ٧ p) is given below

p

Q

p

q٨p

1 1 0 0

1 0 1 0

0 0 1 1

1 0 1 1

p ٨(q ٧p ) 1 0 0 0

q p

Q. 4) A. Show that following are binary operation by giving examples i) ii) iii) iv)

A={set of integers} a x b = a + b for all a, b, Є A. A={set of integers} a x b = max {a.b} for all a, b, Є A A={set of integers} a x b = a + b + 5 for all a, b , Є A A={set of integers} a x b = a x b for all a, b, Є A

b. Show that following are semi group by giving examples i) A = {set of all integers} (A, + ) is a semi group where + is an ordinary addition ii) A= {set of + integers} (A, . ) is a semi group where . is ordinary multiplication iii)_A = {set of all integers}

Δ = {a, b} where for all a, b, Є A (A, Δ) is a semi group c. A = {set of integers with zero} and + is a ordinary addition then show (A, + ) is a monoid.

Ans a. i) Consider two elements i.e.3, 2 , that belongs to a (3,2 є a). Then 3 x 2 = 3 + 2 = 5. We can observe that 5 is also є a. Hence ‘x’ is a binary operation on a. 2) Consider two elements 1,2 that belongs to a (1,2 є a). then 1 x 2 = max.{1,2} = 2 (max of two taken)

2є a hence ‘x’ is a binary operation on a. 3) Consider two elements 2, 4 that belongs to a (2, 4 є a). Now 2 x 4 = (2 + 4) + 5 = 11 11 є a Hence ‘x’ is a binary operation on a. 4) Consider two elements 1, 2 that belongs to a (1,2 є a). now 1 x 2 = 1x2 =2 2є a hence ‘x’ is a binary operation on a.

b) i) A= {set of integers}(a,+) is a semi group where + is a ordinary addition Rule 1: closed property: for any a, b Є A; then a * b Є A; Consider two elements 1, 2 Now 1 * 2 = 1 + 2 =3 3 Є A , Hence it holds the closed property. Rule 2: associative property: For a, b, c, є A; Then (a * b) * c = a* (b * c) Consider three elements 1, 2, 3 Now (1 * 2) * 3 = 1* ( 2 * 3) (3) * 3 = 1 * (5) 6 = 6 hence it holds the associative property. ii) A = {set of integer }(a,.)is a semi group where . is a ordinary multiplication. rule1: closed property: for any a, b Є A; Then (a* b) Є A; Consider two elements 3, 5 Now 3. 5 = 15 15 Є A, hence it holds the closed property. Rule2: associative property: For any a, b, c, Є A; Then a* (b * c ) = ( a * b ) * c Consider three elements 1, 2, 3 Є A 1 * (2 * 3 ) = ( 1 * 2 ) * 3 1 * (2. 3) = (1.2) * 3 1 * (6) = (2) * 3 1.6= 2.3

6

=

6 hence it holds the associative property.

c) A ={ set of integers with zero} and + is an ordinary addition then show (a,+)is monid. rule 1 : closed property: for any a, b Є A then a * b Є A ; e.g. consider two elements 1, 2 Є A now 1 * 2 = 1 + 2 Є A; =3 3 Є A; Hence it holds the closed property. Rule2: associative property: For a, b, c Є A then ( a * b ) * c = a * ( b * c) e.g. consider three elements 1, 2, 3 Є A now ( 1 * 2) * 3 = 1 * (2 * 3 ); (1 + 2 ) * 3 = 1 * ( 2 + 3); 3 * 3 = 1 * 5; 3 + 3 = 1 + 5; 6 = 6; hence it holds the associative property. Rule 3: identity property; For any a Є A ; a + 0 = 0 + a; there fore 0 is an identity element in A. hence (A, + ) is a MONOD.

Q. 5) a. i) find the mean of the following Marks 0 – 10 No. of students 5

10 – 20 8

20 – 30 8

30 – 40 4

ii) calculate the mean, median, mode of the following data: Weight (in gm.) 0–10 10–20 20-30 30-40 40-50 50-60 No. of articles Ans.

a. Marks 0-10 10-20

14

17

22

i) Midpoint 5 15

no. student fi xi 5 25 8 120

26

23

18

20-30 30-40

25 35

mean =

8 200 4 140 ∑fi = 25 ∑fi xi = 485 = 485 / 25 = 19.4

∑ fi xi

ii) Weight no. of articles cumulative frequency mid pt. ∑fixi 0-1 14 14 5 10-20 17 31 15 20-30 22 53 25 30-40 26 79 35 40-50 23 102 45 50-60 18 120 55 N=120

3810

MEAN: ∑ fixi ∑ fi = 3810 / 120 = 31.75 MEDIAN: Here N = 120 Therefore n / 2 = 120 / 2 = 60 Hence the median class will be 30 – 40 Here L = 30 h = 10 C = 53 f = 26 Therefore median = L + ( h / f ) ( ( N / 2 ) – C ) = 30 + (10 / 26 ) (60 – 53 ) = 32.6922 here maximum frequency will be 26 and the model class will be 30 – 40 L = 30 fo= 26 h = 10 f-1 = 22 f1 = 23

MODE:

h ( fo – f –1 ) Mode =

L

+ 2 fo - f–1 - f1

10 (26 – 22) =

30 +

70 255 550 910 1035 990

2. 26 – 22 – 23 = 33.07

Q. 6) a. An urn contains 10 black and 10 white balls. Find the probability of drawing two balls of the same color? b. A problem in statistics is given to three students A, B and C whose chances of solving it are ½ , ¾ and ¼ respectively. What is the probability that the problem will be solved? Ans. a. There are 10 black balls and 10 white balls Total balls will be 20 Out of 20 , we draw 2 balls Sample space = 20 C 2 If the balls are black, therefore the events will be E1 = 10 C 2 If the balls are white, therefore the events will be E2 = 10 C 2 10 Total Probability will be C2 + 10 C2 20

10 . 9. 8! / 8 !

=

C2

× 2

20. 19. 18 ! / 18! = = 9 / 19

10. 9. 2 20. 19

b. chances of solving a problem by A = ½ chances of solving a problem by B = ¾ chances of solving a problem by C = ¼ Therefore the probability that the problem will be solved =½+¾+¼ =3/2 =1.5 Ans.