1 A Complete Manual To Allied Publication Sitapaila, Kathmandu Ph.No.: 01-5388827
2 A Complete Manual To The Leading Maths Class 9 ( Ashok Dangol ) Contributions: Ashim Maskey Dileep Kumar Maharjan Chandra Bahadur Dangol Hari Maharjan Special Thanks Dilkumar Maharjan Santosh Gupta Publisher: Allied Publication Sitapaila, Kathmandu Ph.No.: 01-5388827
3 Contents Unit Chapter Lesson/Practice Page I. SETS 1. Sets 1.1 Union and Intersection of Sets/Practice 1.1 1 1.2 Difference and Complement of Sets /Practice 1.2 9 1.3 Cardinality of Sets/ Practice 1.3 22 Additional Practice I 36 II. ARITHMETIC 2. Tax 2.1 Income TAX / Practice 2.1 39 2.2 Value Added TAX (VAT) / Practice 2.2 51 3. Commission and Dividend 3.1 Commission Practice 3.1 59 3.2 Dividend and Bonus / Practice 3.2 68 4. Home Arithmetic 4.1 Household Expenses for Use of Electricity / Practice 4.1 77 4.2 Household Expenses for Use of Water/ Practice 4.2 89 4.3 Household Expenses for Use of Telephone/ Practice 4.3 98 4.4 Calculation of Taxi Fare/ Practice 4.4 105 Additional Practice II 110 III. MENSURATION 5. Area 5.1 Area of Scalene Triangle / Practice 5.1 119 5.2 Surface Area of Room and Cost Estimation / Practice 5.2 129 6. Prism and Cylinder 6.1 Surface Area and Volume of Prism/Practice 6.1 138 6.2 Surface Area and Volume of Cylinder/Practice 6.2 145 7. Sphere 7.1 Surface Area and Volume of Sphere/Practice 7.1 157 Additional Practice III 169 IV. ALGEGRA 8. Sequence and Series 8.1 Introduction to Sequence and Its General Terms/Practice 8.1 174 8.2 Introduction to Series/Practice 8.2 182 8.3 General Term of Arithmetic Sequence/Practice 8.3 187 8.4 General Term of Geometric Sequence/Practice 8.4 194 9. Factorization 9.1 Factorization in the Form of (a ± b)3 and a3 ± b3 /Practice 9.1 202 9.2 Factorization in Form of (a4 + a2 b2 + b4 ) /Practice 9.2 206 10. HCF and LCM 10.1 Highest Common Factor (HCF) /Practice 10.1 210 10.2 Lowest Common Multiple (LCM) /Practice 10.2 214 11. Linear Equations 11.1 Solving Simultaneous Equations by Substitution Method/Practice 11.1 219
4 11.2 Solving Simultaneous Equations by Elimination Method/Practice11.2 231 11.3 Verbal Problems on Simultaneous Equations in Two Variables/Practice11.3 238 12. Indices 12.1 Simplification of Indices By Using Their Laws/Practice 12.1 251 Additional Practice – IV 259 V. GEOMETRY 13. Triangles 13.1 Angles of Triangle/Practice 13.1 266 13.2 Properties of Isosceles Triangle/Practice 13.2 269 13.3 Sides and Angles of a Triangle/Practice 13.3 277 13.4 Similar Triangles/Practice 13.4 281 14. Parallelogram 14.1 Properties of Parallelogram/Practice 14.1 290 14.2 Application of Theorems of Parallelogram/ Practice 14.2 301 15. Construction 15.1 Construction of Scalene Quadrilaterals/ Practice 15.1 307 15.2 Construction of Trapeziums/Practice 15.2 313 15.3 Construction of Rhombus/Practice 15.3 319 16. Circle 16.1 Circle/Practice 16.1 324 Additional Practice V 332 VI. STATISTICS AND PROBABILITY 17. Classification and Representation of Data 17.1 Collection of Data and Frequency Table/ Practice 17.1 339 17.2 Histogram/Practice 17.2 345 17.3 Frequency Polygon/Practice 17.3 348 17.4 Cumulative Frequency Curve (Ogive) / Practice 17.4 350 18. Measurement of Central Tendency 18.1 Arithmetic Mean/Practice18.1 356 18.2 Median/Practice 18.2 360 18.3 Quartiles/Practice 18.3 365 18.4 Mode/Practice 18.4 371 19. Probability 19.1 Introduction to Probability/Practice 19.1 376 19.2 Empirical and Classical Probabilities/Practice 19.2 377 Additional Practice VI 384 VII. TRIGONOMETRY 20. Trigonometry 20.1 Trigonometric Ratios/Practice 20.1 391 20.2 Trigonometric Ratios of Standard Angles/ Practice 20.2 403 Additional Practice VII 409
1 884 UNIT I SETS CHAPTER SETS 1 1.1 Union and Intersection of Sets PRACTICE 1.1 Read / Understand / Think / Do Keeping Skill Sharp 1. (a) Define union of sets. (b) What is the intersection of two sets ? Define it. (c) Why the union of sets is different from their intersection ? Give reason. (d) Write the union of the sets M = { m, a, p} and N = {n, e, p, a, l}. (e) What is the intersection of the sets P = {2, 3, 5} and E = {2, 4}? (f) If the given sets are disjoint, what is their intersection ? Solution: (a) A set containing all elements of the given sets is called union of sets. (b) A set containing the elements of both given two sets is called intersection of two sets. (c) The union of sets is a set formed by all the elements of the given sets, but the intersection of sets is a set formed by only common elements of the given sets. (d) M N = {m, a, p} {n, e, p, a, l} = {a, p, m, n, e, l}. (e) M N = {2, 3, 5} {2, 4} = {2}. (f) If the given sets are disjoint, their intersection is { }. 2. Circle (±) the correct answer. (a) The union of two overlapping sets is a set of ............ . (i) all the elements of the given two sets. (ii) common elements of the given two sets. (iii) all the non-repeated elements of the given two sets. (iv) all the repeated elements of the given two sets. (b) The intersection of sets A and B is defined as (i) {x : x ∈ A and x ∉ B} (ii) {x : x ∈ A or x ∈ B or both} (iii) {x : x ∈ A and x ∈ B} (iv) {x : x ∈ A or x ∈ B}
( ( ( ( Soluti ( ( Chec Answ 3. S ( ( ( ( ( ( Soluti ( ( ( ( ( (c) Which is th (i) 0 (iii) {0} (d) Which is th (i) {m, a, (iii) {m, n, (e) A is the sub .......... . (i) {1, 4} (iii) {1, 5} (f) Which is tr (i) A (B (ii) A (B (iii) A (B (iv) A (B ion: (a) (iii) (d) (i) ck Your Perfo er the given que Study the types o (a) List the nam (b) Find the na (c) List the item (d) List the item (e) List the item (f) Show all th ion: (a) From the giv the name of A = {strawb B = {apples (b) The name of i.e. A B (c) The items on i.e. A B (d) The items on A – B = {str (e) The items on he intersection of he union of the se n, g, o, r, p, e, s} o} bset of B, where ue property for B C) = (A B B C) = (A B B C) = (A B B C) = (A B (b) (ii (e) (ii ormance estions for each p of fruits in the sh me of fruits of ea ame of all the fru ms only in both s ms only in the sh ms only in the sh e items of the sh ven figure alongs f fruits of each sho berries, cherries, r , oranges, grapes f all fruits in A or = {strawberrie water-melo nly in both shops = {oranges} nly in shop A but rawberries, cherri nly in shop B, bu 2 f the sets C = {c, ets M = {m, a, n, } B = {1, 3, 5, 7}. the sets A, B an B) (A C) B) (A C) B) (A C) B) (A C) ii) ii) problem. hops A and B al ach shop in set n uits in A or B or shops in set nota hop A, but not in hop B, but not in hops A and B in side, op in set notation raspberries, goos s, mangoes} r B or both shops s, cherries, raspb on, pears, apples, s in set notation, t not in B, ies, raspberries, g ut not in A, B – A , a, m, e, l} and D (ii) ϕ (iv) {c, a, m, e, , g, o} and G = { (ii) {a, g} (iv) {r, p, e, s} If A ∪ {1, 3, 5, 7 (ii) {1, 2, 3} (iv) {5, 6, 7} nd C? (c) (ii) (f) (iv) longside and ans notation. both shops. ation. n B. n A. a Venn diagram n: eberry, oranges, s, berries, gooseberr grapes, mangoes gooseberries, ban A = {apples, mang D = {d, o, g} ? , l, d, o, g} g, r, a, p, e, s} ? 7} = {1, 3, 5, 7}, swer the followin m. bananas, water-m ries, oranges, ban s} nanas, water-melo goes, grapes} then the set A = ng questions: melon, pears} nanas, on, pears} =
3 A B U 2 7 8 10 3 5 9 1 4 6 A B U 2 7 8 10 3 5 9 1 4 6 A B U 3 7 2 4 8 12 5 11 6 10 14 (f) Representing all the items of the shop A and B in a Venn diagram as below, 4. Study the given Venn diagram and answer the following questions: (a) Write the elements of the sets A and B. (b) Find the set A B and shade it. (c) Find the set A B and shade it in the next diagram. (d) Write the relation between A B and A B. Solution: (a) From the given Venn diagram alongside, A = {2, 7, 8, 10}, B = {3, 5, 8, 9, 10} (b) A B = {2, 3, 5, 7, 8, 9, 10} Shading A B in the Venn diagram. (c) A B = {8, 10} Shading A B in the Venn diagram. (d) The set A B is a subset of the set A B. 5. It is given that the sets A = {Prime numbers less than 12} and B = {Multiples of 2 less than 16}. Answer the following questions: (a) List the above information in the set notation. (b) Show the above sets in a Venn diagram. (c) What is the union of A and B ? Find it. (d) Find B A and shade it in the same Venn diagram. (e) Why are A B and B A equal? Give suitable reason. Solution: (a) Given, A = {Prime numbers less than 12}, B = {Multiples of 2 less than 16} Writing the above information in set notation A = {2, 3, 5, 7, 11}, B = {2, 4, 6, 8, 10, 12, 14} (b) Representing the sets A and B in a Venn diagram. (c) A B = {2, 3, 5, 7, 11} {2, 4, 6, 8, 10, 12, 14} = {2, 3, 4, 5, 6, 7, 8, 10, 11, 12, 14} strawberries cherries raspberries goose-berries bananas water-melon pears apples mangoes grapes U A B U 2 7 8 10 3 5 9 1 4 6
4 (d) B A = {2, 4, 6, 8, 10, 12, 14} {2, 3, 5, 7, 11} = {2, 3, 4, 5, 6, 7, 8, 10, 11, 12, 14} A B or B A (e) A B and B A are equal because they are commutative property. 6. The sets E and F are defined as E is the set of the letters in the word 'education' and F is the letters in the word 'equation'}. (a) List the given facts in the set notation. (b) Show the above sets in a Venn diagram. (c) Find the intersection of E and F ? Shade it in the same Venn diagram. (d) In which condition does E F equal to E ? Solution: (a) Listing the above given facts in the set notation: E = {e, d, u, c, a, t, i, o, n}, F = {e, q, u, a, t, i, o, n} (b) Showing the above sets in a Venn diagram alongside: (c) E F = {e, d, u, c, a, t, i, o, n} {e, q, u, a, t, i, o, n} = {e, u, a, t, i, o, n} E F is shown by shading in the above Venn diagram. (d) E F equals to E when E is subset of F. 7. Observe the given Venn diagrams. (a) Write the name of the shaded portions from the given Venn diagrams. (b) What are the differences between the shaded portions in both Venn diagrams ? Write. Solution: (a) In figure 7. (a) the shaded portion is P Q while in figure 7. (b) it is A B. (b) In the Venn diagram 7. (a), all portions of P and Q are shaded while in the Venn diagram 7. (b), common parts of A and B are only shaded. A B U 3 7 2 4 8 12 5 11 6 10 14 u t o e a i n d c q E F U EF P Q U 7. (a) A B U 7. (b)
5 A B U 1 4 9 6 10 14 8 12 15 2 3 5 7 11 13 A B U 1 4 9 6 10 14 8 12 15 2 3 5 7 11 13 8. Study the given Venn diagrams. (a) Write the name of the shaded portions in the given Venn diagrams. (b) Write the relation between the shaded sets in the Venn diagrams. Solution: (a) In the Venn diagram 8. (a), the shaded portion is (A B) (B C) (C A) while in the Venn diagram 8. (b), the shaded portion is C (A B). (b) The shaded set of the Venn diagram 8. (a) is a proper subset of the set shown by the shaded portion in the Venn diagram 8. (b). 9. Given U = {x : x ≤ 15, x ∈ N}, P = {x : x is a square number, x ∈ U} and Q = {x : x is a composite number, x ∈ U}. (a) Write the above information in the set notation. (b) Show the above sets in a Venn diagram. (c) Find P Q and shade in the same Venn diagram. (d) Find P Q and shade in the next Venn diagram. (e) What kinds of the sets are P and Q when P Q and P Q are equal? Give suitable reason. Solution: (a) Given, U = {x : x 15, x N}, P = {x : x is a square number, x U} and Q = {x : x is a composite number, x U} Writing above information in the set notation, U = {1, 2, 3, 4, ..........., 13, 14, 15} P = {1, 4, 9} and Q = {4, 6, 8, 9, 10, 12, 14, 15} (b) The above sets are shown in a Venn diagram alongside. (c) P U Q = {1, 4, 9} {4, 6, 8, 9, 10, 12, 14, 15} = {1, 4, 6, 8, 9, 10, 12, 14, 15} Shading P Q in the Venn diagram alongside, (d) P Q = {1, 4, 9} { 4, 6, 8, 9, 10, 12, 14, 15} = {4, 9} Shading P Q in the Venn diagram alongside, (e) The sets P and Q are equal when the sets P Q and P Q are equal. A B C U 8. (a) A B C U 8. (b) A B U 1 4 9 6 10 14 8 12 15 2 3 5 7 11 13
6 10. If U = {x : 5 ≤ x < 18, x ∈ N}, A = {x : x is the number divisible by 3}, B = {x : x is the number divisible by 4} and C = {x : x is the number divisible by 5}, find the elements of the following sets and show them in the Venn diagrams separately. (a) Write the above information in the listing method. (b) Find A (B C) and shade in the same Venn diagram. (c) Find (A B) C and shade in the same Venn diagram. (d) Which part is common in the sets A (B C) and (A B) C ? Solution: (a) Given, U = {x : 5 x < 18, x N}, A = {x : x is the number divisible by 3}, B = {x : x is the number divisible by 4} and C = {x : x is the number divisible by 5}. Writing above information in the listing method, U = {5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17} A = {6, 9, 12, 15}, B = {8, 12, 16} and C = {5, 10, 15} (b) Here, B C = {8, 12, 16} {5, 10, 15} = {5, 8, 10, 12, 15, 16} A (B C) = {6, 9, 12, 15} {5, 8, 10, 12, 15, 16} = {12, 15} Shading A (B C) in the Venn diagram alongside, (c) Here, A B = {6, 9, 12, 15} {8, 12, 16} = {6, 8, 9, 12, 15, 16} (A B) C = {6, 8, 9, 12, 15, 16} {5, 10, 15} = {15} Shading (A B) C in the Venn diagram alongside: (d) Here, A C is common in the sets A (B C) and (A B) C. 11. Let A = {a, s, h, o, k}, M = {m, e, d, n, a, t, h}, R = {r, u, p, a} are the given sets. (a) Verify the following relations: (i) A M = M A (ii) R A = A R (b) Why are they equal ? Give reasons. Solution: Given, A = {a, s, h, o, k}, M = {m, e, d, n, a, t, h} and R = {r, u, p, a} (i) (a) LHS = A M = {a, s, h, o, k} {m, e, d, n, a, t, h} = {a, s, h, o, k, m, e, d, n, t} RHS = M A = {m, e, d, n, a, t, h} {a, s, h, o, k} = {m, e, d, n, a, t, h, s, o, k} Hence A M = M A, proved. U 8 16 12 15 6 9 5 10 BAC A (B C) 17 14 13 11 7 U 8 16 12 15 6 9 5 10 BAC (A B) C 17 14 13 11 7
7 (b) LHS = R A = {r, u, p, a} {a, s, h, o, k} = { } RHS = A R = {a, s, h, o, k} = { } Hence, R A = A R, proved. (ii) They are equal because according to the laws of set, two sets are commutative in both cases either their intersection or their union. 12. Let U = { a, b, c, ..., k}, A = {a, b, d, f}, B = {c, e, g. i} and C = {o, a, e, h, i} are the given sets. (a) Verify the following relations: (i) A (B C) = (A B) C) (ii) A (B C) = (A B) C) (b) Show the above relations in the Venn diagrams. (c) Why are they equal ? Give reasons. Solution: Given, U = {a, b, c, d, ......, k}, A = {a, b, d, f}, B = {c, e, g, i} and C = {a, e, h, i} (a) (i) B C = {c, e, g, i} {a, e, h, i} = {a, c, e, g, h, i} LHS = A (B C) = {a, b, d, f} {a, c, e, g, h, i} = {a, b, c, d, e, f, g, h, i} Now, A B = {a, b, d, f} {c, e, g, i} = {a, b, c, d, e, f, g, i} RHS = (A B) C = {a, b, c, d, e, f, g, i} {a, e, h, i} = {a, b, c, d, e, f, g, h, i} Since LHS = RHS A (B C) = (A B) C, proved. (ii) A C = {c, e, g, i} {a, e, h, i} = {e, i} LHS = A (B C) = {a, b, d, f} {e, i} = { } Also, A B = {a, b, d, f} {c, e, g, i} = { } RHS = (A B) C = { } {a, e, h, i} = { } Since, LHS = RHS A (B C) = (A B) C, proved. (b) Showing the above relations in the Venn diagrams: (c) They are equal because of the associative property of sets for the union and intersection of the sets. 13. A farmer cultivates tomato, cabbage, radish, carrot, brinjal, lady's finger and peas in the farm P; tomato, cabbage, radish, garlic, brinjal, onion, turnip and potato in the farm Q and cauliflower, choko, cabbage, radish, cucumber, brinjal, pumpkin, turnip and capsicum in the farm R. List all the vegetables in each farm by supposing the bigging small letter of each vegetable. Such as c1, c2, ... (a) List the name of vegetables of each farm in the set notation. (b) Prove the following relations: (i) P (Q R) = (P Q) (P R) (ii) P (Q R) = (P Q) (P R) (c) Why are they equal ? Give reasons by using Venn diagram. U b d f a e i h c g ACB Fig. (a) j k 7 A(BC) or (AB)C U b d f a e i h c g ACB Fig. (b) j k 7 A(BC) or (AB)C
8 Solution: Let, c1 = cabbage, c2 = carrot, c3 = cauliflower, c4 = choko, c5 = cucumber, c6 = capsicum p1 = peas, p2 = potato, p3 = pumpkin, t1 = tomato and t2 = turnip Let the remaining vegetables be represented by the beginning small letter of their names, (i) Now, writing the vegetables in set notation: P = {t1, c1, r, c2, b, , p1}, Q = {t1, c1, r, g, b, o, t2, p2} and R = {c3, c4, c1, r, c5, b, p3, t2, c6} (ii) (a) To prove P (Q R) = (P Q) (P R) Q R = {t1, c1, r, g, b, o, t2, p2} {c3, c4, c1, r, c5, b, p3, t2, c6} = {t1, c1, r, g, b, o, t2, p2, c3, c4, c5, p3, c6} LHS = P (Q R) = {t1, c1, r, c2, b, , p1} {t1, c1, r, g, b, o, t2, p2, c3, c4, c5, p3, c6} = {t1, c1, r, b} P Q = {t1, c1, r, c2, b, , p1} {t1, c1, r, g, b, o, t2, p2} = {t1, c1, r, b} P R = {t1, c1, r, c2, b, , p} {c3, c4, c1, r, c5, b, p3, t2, c6} = {c1, r, b} RHS = (P Q) (P R) = {t1, c1, r, b} {c1, r, b} = {t1, c1, r, b} P (Q R) = (P Q) (P R) (b) To prove P (Q R) = (P Q) (P R) Here, Q R = {t1, c1, r, g, b, o, t2, p2} {c3, c4, c1, r, c5, b, p3, t2, c6} = {c1, r, b, t2} LHS = P (Q R) = {t1, c1, r, c2, b, , p1} {c1, r, b, t2} = {t1, c1, r, c2, b, , p1, t2} Now, P Q = {t1, c1, r, c2, b, , p1} {t1, c1, r, g, b, o, t2, p2} = {t1, c1, r, c2, b, , p1, g, o, t2, p2} P R = {t1, c1, r, c2, b, , p1} {c3, c4, c1, r, c5, b, p3, t2, c6} = {t1, c1, r, c2, b, , p1, c3, c4, c5, p3, t2, c6} RHS = (P Q) (P R) = {t1, c1, r, c2, b, , p1, g, o, t2, p2} {t1, c1, r, c2, b, , p1, c3, c4, c5, p3, t2, c6} = {t1, c1, r, c2, b, , p1, t2} P (Q R) = (P Q) (P R), proved. (iii) They are equal because of distributive properties of intersection over union and union over intersection of sets. P Q R U t1 c2 p1 c1 r b t2 12 12 p2 c3 c4 c5 p3 c6 P (Q R) P Q R U t1 c2 p1 c1 r b t2 12 12 p2 c3 c4 c5 p3 c6 P (Q R)
9 1.2 Difference and Complement of Sets PRACTICE 1.2 Read / Understand / Think / Do Keeping Skill Sharp 1. (a) Define difference of two sets. (b) What is the complement of a set ? Define it. (c) Write one difference between difference of two sets and complement of a set. (d) Write the set C – O if the sets C = {c, o, v, i, d} and O = {o, m, n, i, v, r, u, s}. (e) What is the complement of the set P = {2, 4, 6, 8, 10}, which is the subset of a universal set U = {1, 2, 3, .........., 10}? Solution: (a) The difference of two sets A and B is defined as a set of all the elements that are in the set A but not in the set B. The set notation of the difference between the two sets A and B is A − B. (b) The complement of a set A is defined as a set that contains the elements present in the universal set but not in the given set A. (c) The difference of two sets is a set containing in only one set but not in another set while the complement of a set is a set of the elements in the universal set but not in the given set. (d) C – O = {c, o, v, i, d} – {o, m, n, i, v, r, u, s} = {c, d}. (e) P' = U – P = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} – {2, 4, 6, 8, 10} = {1, 3, 5, 7, 9}. 2. Circle (±) the correct answer. (a) The complement of a set is a set of ............ . (i) elements contained in the given set. (ii) elements contained in the given set but not in the universal set. (iii) elements not contained in the first set but in the second set. (iv) elements not contained in the given set, but in the universal set. (b) The complement of the set M is defined as .......... . (i) {x : x ∈ M and x ∉ M} (ii) {x : x ∉ M and x ∈ U} (iii) {x : x ∈ M and x ∉ U} (iv) {x : x ∈ M and x ∈ N} (c) Which is the difference of the sets P = {2, 3, 5} from N = {3, 4, 5, 6}? (i) {2, 3, 4, 5, 6} (ii) {3, 5} (iii) {2} (iv) {4, 6} (d) Which is the complement of the sets M = {m, a, n, g, o} in the universal set U = {p, o, m, e, g, r, a, n, t} ? (i) {p, o, m, e, g, r, a, n, t} (ii) {p, e, r, t} (iii) {m, a, n, g, o} (iv) {r, p, e} (e) Which is true for the sets A and B? (i) A B = A B (ii) A B = A B (iii) A B = A B (iv) A B = A B Solution: (a) (iv) (b) (ii) (c) (iv) (d) (ii) (e) (i)
Chec Answ 3. O u ( ( ( ( ( Soluti ( ( ( ( ( 4. O ( ( ck Your Perfo er the given que Observe the ute utensils of Maya (a) List the ele the set nota (b) Which ute Maya that o (c) Which ute Maya that o (d) Illustrate th (e) Identify the ion: (a) M = {cooke Y = {cooker (b) Maya has bu i.e. M – Y (c) Plate, glass a i.e. Y – M (d) Illustrating t (e) The resultin i.e. M – Y Observe the give (a) Write the n set builder (b) Write one s M bucket pan spoon coo kn mu ju M – P 4. ( ormance estions for each p nsils in the adjo a by M and Yang ements of M an ation. ensils are mor of Yangri ? Find ensils are mor of Yangri ? Find he above sets in t e resulting sets in er, bucket, pan, kn r, mug, plate, kni ucket, pan, spoon = {cooker, buc = {pan, bucket and bowl are mo = {cooker, mu = {plate, glass the above sets in ng sets in (b) and Y – M because en Venn diagram name of the shad also. similar characte Y U oker nife ug ug plate bowl glass – Y Q U a) 10 problem. oining kitchen r gri by Y. Then a nd Y in re with d it. re with d it. the Venn diagra n (b) and (c) equ nife, mug, jug, sp ife, bowl, glass, ju n more than Yang cket, pan, knife, m t, spoon} re with Yangri th ug, plate, knife, bo , bowl} the Venn diagram (c) are not equal. e the difference o ms. ded portions in t r on both Venn rooms of Maya answer the quest ams. ual or not? Why poon} ug} gri mug, jug, spoon} hat of Maya. owl, glass, jug} – ms. . of two sets is not the given Venn d diagrams. M Y bucket pan spoon cooker knife mug jug Y – M M 4. (b) and Yangri. Su tions asked below y ? – {cooker, mug, bowl, glass, j – {cooker, bucket mug, jug, spoo commutative pro diagrams and w Y U plate bowl glass U uppose the set of w: , plate, knife, ug} t, pan, knife, on} operty. write them in the f
11 Solution: (a) The name of the shaded portions in the Venn diagrams of figure 4. (a) and figure 4. (b) are P – Q and M respectively. P – Q = {x : x P and x Q} and M = U – M = {x : x U and x M} (b) One similar character on both Venn diagram is both are difference of one set from another. 5. Study the given Venn diagrams. (a) Write the name of the shaded portions in the given Venn diagrams. (b) Write the relation between the shaded sets in both Venn diagrams. Solution: (a) The shaded portion in figure 5. (a) is (B C) – A while in figure 5. (b) is A – (B C). (b) The relation of both sets are: (B C) – A = (B – A) (C – A) and A – (B C) = (A – B) (A – C) The relation between them {(B C) – A} {A – (B C)} = 6. Suppose, U = {English months of a year}, J = {Months beginning with 'J'}, S = {6th to 8th months}, W = {1st three months}. (a) Write the given information in the set notation. (b) List the elements of each of the following sets. (i) J – W (ii) J – S (iii) W Δ S (iv) J W (v) J S (vi) J S (vii) J – (S W) (viii) J S W – W (ix) (J – S) (J – W) (c) Show each of the above sets in the Venn diagrams. (d) Which sets are equal in (b) ? Solution: Given, U = {English months of a year}, J = {Months beginning with J}, S = {6th to 8th months}, W = {1st three months} (a) The above information can be written in the set notation as, U = {January, February, March, April, May, June, July, August, September, October, November, December} J = {January, June, July} S = {June, July, August} W = {January, February, March} A B C U 5. (a) A B C U 5. (b)
12 (b) (i) J – W = {January, June, July} – {January, February, March} = {June, July} (ii) J – S = {January, June, July} – {June, July, August} = {January} (iii) W – S = {January, February, March} – {June, July, August} = {January, February, March} (iv) J W = U – (J W) Now, J W = {January, June, July} {January, February, March} = {January, February, March, June, July} J W = U – (J W) = {January, ..... , November, December} – {January, February, March, June, July} = {April, May, August, September, October, November, December} (v) J S = U – (J S) Now, J S = {January, June, July} {June, July, August} = {June, July} J S = U – (J S) = {January, ..... , November, December} – {June, July} = {January, February, March, April, May, August, September, October, November, December} (vi) J S = (U – J) (U – S) Now, U – J = {January, ..... , November, December} – {January, June, July} = {February, March, April, May, August, September, October, November, December} U – S = {January, ..... , November, December} – {June, July, August} = {January, February, March, April, May, September, October, November, December} J S = (U – J) (U – S) = {February, March, April, May, August, September, October, November, December} {January, February, March, April, May, September, October, November, December} = {January, February, March, April, May, August, September, October, November, December} (vii) S W = {Aug, June, July} {Jan, Feb, March} = {Jan, Feb, March, Aug, June, July} J – (S W) = {Jan, June, July} – {Jan, Feb, March, Aug, June, July} = { } (viii) J S W – W = {U – (J S W)} – W Now, J S W = {Jan, June, July} {June, July, Aug} {Jan, Feb, Mar} = J S W = U – (J S W) = {Jan, ...., Nov, Dec} – { } = {Jan, ...., Nov, Dec} J S W – W = {Jan, Feb, ........., Nov, Dec} – {Jan, Feb, March} = {April, May, June, July, Aug, Sep, Oct, Nov, Dec} (ix) J – S = {Jan, June, July} – {June, July, Aug} = {Jan} J – W = {Jan, June, July} – {Jan, Feb, March} = {June, July} (J – S) (J – W) = {Jan} {June, July} = { } (c) Representing the above sets in the Venn diagrams,
13 7. Suppose, P = {The letters in the word of a lung disease caused by inhalation of very fine silicate or quartz dust 'pneumonoultramicroscopicsilicovolcanoconiosis'} and its subsets C = {The letters in the word of a heart artery disease 'coronary'}, H = {The letters in the word of a lever disease 'hemochromatosis'} and D = {The letters in the word of a brain disease 'Dementias'}. (a) Write the given information in the set notation. (b) List the members of each of the following sets. (i) C (H D) (ii) C (H D) (iii) (C D) (H D) (iv) C H D (v) C – (H D) (vi) (C – H) (C – D) (c) Show each of the above sets in the Venn diagrams. (d) Which sets are equal in (b) ? Why ? Give reasons. June July Jan Feb March Aug Oct Nov Dec Apr May Sep SJW U (i) J – W June July Jan Feb March Aug Oct Nov Dec Apr May Sep SJW U (ii) J – S June July Jan Feb March Aug Oct Nov Dec Apr May Sep SJW U (iii) W S June July Jan Feb March Aug Oct Nov Dec Apr May Sep SJW U (iv) J W June July Jan Feb March Aug Oct Nov Dec Apr May Sep SJW U (v) J S (vi) J S June July Jan Feb March Aug Oct Nov Dec Apr May Sep SJW U June July Jan Feb March Aug Oct Nov Dec Apr May Sep SJW U (vii) J – (S W) June July Jan Feb March Aug Oct Nov Dec Apr May Sep SJW U (viii) J S W – W June July Jan Feb March Aug Oct Nov Dec Apr May Sep SJW U (ix) (J – S) (J – W)
14 Solution: (a) Here, P = {p, n, e, u, m, o, l, t, r, a, i, c, s, v} C = {c, o, r, n, a, y}, H = {h, e, m, o, c, r, a, t, s, i}, D = {d, e, m, n, t, i, a, s} (b) (i) C (H D) = {c, o, r, n, a, y} ({h, e, m, o, c, r, a, t, s, i} {d, e, m, n, t, i, a, s}) = {c, o, r, n, a, y} {h, e, m, o, c, r, a, t, i, s, d, n} = {c, o, r, n, a} (ii) C (H D) = {c, o, r, n, a, y} ({h, e, m, o, c, r, a, t, s, i} {d, e, m, n, t, i, a, s}) = {c, o, r, n, a, y} {e, m, a, t, i, s} = {c, o, r, n, a, y, e, m, t, i, s} (iii) (C D) (H D) = ({c, o, r, n, a, y} {d, e, m, n, t, i, a, s}) ({h, e, m, o, c, r, a, t, s, i} {d, e, m, n, t, i, a, s}) = {n, a} {e, m, a, t, s, i} = {n, a, e, m, t, s, i} (iv) C H D = P – (C H D) = {p, n, e, u, m, o, l, t, r, a, i, c, s, v} – ({c, o, r, n, a, y} {h, e, m, o, c, r, a, t, s, i} {d, e, m, n, t, i, a, s}) = {p, n, e, u, m, o, l, t, r, a, i, c, s, v} – {c, o, r, n, a, y, h, e, m, t, s, i, d} = {p, u, l, v} (v) C – (H D) = {c, o, r, n, a, y} – ({h, e, m, o, c, r, a, t, s, i} {d, e, m, n, t, i, a, s}) = {c, o, r, n, a, y} – {e, m, n, a, t, s, i} = {c, o, r, y} (vi) (C – H) (C – D) = ({c, o, r, n, a, y} – {h, e, m, o, c, r, a, t, i, s}) ({c, o, r, n, a, y} – {d, e, m, n, t, i, a, s}) = {n, y} {c, o, r, y} = {n, y, c, o, r} (c) Representing the above information in the Venn diagrams, (d) All are unequal. C H D U y c h a n i d p o r s t m e v l u (i) C (H D) C H D U y c h a n i d p o r s t m e v l u (ii) C (H D) C H D U y c h a n i d p o r s t m e v l u (iii) (C D) (H D) C H D U y c h a n i d p o r s t m e v l u (iv) C H D C H D U y c h a n i d p o r s t m e v l u (v) C – (H D) C H D U y c h a n i d p o r s t m e v l u (vi) (C – H) (H – D)
15 8. Suppose, U = {x: x is a whole number up to 15}, M = {x: x is a multiple of 3 from 1 to 15}, O = {x: x is an odd number from 1 to 15} and F = {x: x is a factor of 12}. (a) Write the given information in set notation. (b) List the elements containing each of the following. (i) at least one set. (ii) at most two sets. (iii) neither all sets. (iv) M, but not in O F. (v) M F, but not in M F. (vi) not in F, but not in M O. (c) Shade each of the above sets in the Venn diagrams. Solution: (a) Given, U = {x : x is a whole number up to 15}, M = {x: x is a multiple of 3 from 1 to 15}, O = {x: x is an odd number from 1 to 15} and F = {x: x is a factor of 12}. Writing the above information in set notation. U = {0, 1, 2, 3, 4, 5, 6, ..., 13, 14, 15}, M = {3, 6, 9, 12, 15} O = {1, 3, 5, 7, 9, 11, 13, 15} & F = {1, 2, 3, 4, 6, 12} (b) (i) A set of elements containing at least one set. i.e. M O F = {3, 6, 9, 12, 15} {1, 3, 5, 7, 9, 11, 13, 15} {1, 2, 3, 4, 6, 12} = {1, 2, 3, 4, 5, 6, 7, 9, 11, 12, 13, 15} (ii) A set of elements containing at most two set, ( M O F ) = ? M O F = {3} M O F = U – (M O F) = {1, 2, 4, 5, ......, 15} – { 3} = {1, 2, 4, 5, ..., 15} (iii) A set of elements containing neither all sets, i.e. M O F = U – (M O F) = {1, 2, ..., 14, 15} – {1, 2, 3, 4, 5, 6, 7, 9, 11, 12, 13, 15} = {8, 10, 14} (iv) A set of elements containing M, but not in O F. i.e. M – (O F) = {3, 6, 9, 12, 15} – [{1, 3, 5, 7, 9, 11, 13, 15} {1, 2, 3, 4, 6, 12}] = {3, 6, 9, 12, 15} – {1, 3} = {6, 9, 12, 15} (v) A set of elements containing M F, but not in M F i.e. (M F) – (M F) = [{3, 6, 9, 12, 15} {1, 2, 3, 4, 6, 12}] – [{3, 6, 9, 12, 15} {1, 2, 3, 4, 6, 12}] = {1, 2, 3, 4, 6, 9, 12, 15} – {3, 6, 12} = {1, 2, 4, 9, 15} (vi) A set of elements containing not in F, but not in M O i.e. F – (M O) = (U – F) – (M O) = [{1, 2, 3, 4, ...., 14, 15} – {1, 2, 3, 4, 6, 12}] – {3, 6, 9, 12, 15} {1, 3, 5, 7, 9, 11, 13, 15}] = {5, 7, 8, 9, 10, 11, 13, 14, 15} – {1, 3, 5, 6, 7, 9, 11, 12, 13, 15} = {8, 10, 14}
16 P Q R U 1 5 4 2 3 8 11 7 10 6 9 15 13 14 12 19 20 16 18 (c) Shading each of the sets in the Venn diagrams, (i) at least one set (ii) At most two sets. (i.e. A B C) (iii) neither all sets. (iv) M, but not in O F (i.e. ( M O F )) (i.e. M – (O F)) (v) M F, but not in M F (vi) not in F, but not in M O i.e. (M F) – (M F) i.e. F – (M O) 9. Use the given Venn diagram and find the sets by listing the elements. (a) P (Q – R) (b) (P Q) R (c) P – (Q – R) (d) (Q – P) (Q – R) (e) P – Q R (f) R Q – P (g) P – (Q R) (h) (P R) – (P R) (i) (P Q) – R Solution: Here, P = {1, 2, 3, 4, 5, 6, 7, 9, 10} Q = {1, 2, 3, 4, 5, 8, 11, 13, 15} R = {2, 3, 7, 8, 11, 12, 14} (a) P (Q – R) = {1, 2, 3, 4, 5, 6, 7, 9, 10}[{1, 2, 3, 4, 5, 8, 11, 13, 15}–{2, 3, 7, 8, 11, 12, 14}] = {1, 2, 3, 4, 5, 6, 7, 9, 10} {1, 4, 5, 13, 15} = {1, 2, 3, 4, 5, 6, 7, 9, 10, 13, 15} M O F U 915 6 3 12 1 5 7 11 13 4 2 8 10 14 M O F U 915 6 3 12 1 5 7 11 13 4 2 8 10 14 M O F U 915 6 3 12 1 5 7 11 13 4 2 8 10 14 M O F U 915 6 3 12 1 5 7 11 13 4 2 8 10 14 M O F U 915 6 3 12 1 5 7 11 13 4 2 8 10 14 M O F U 915 6 3 12 1 5 7 11 13 4 2 8 10 14
17 (b) (P Q) R = [{1, 2, 3, 4, 5, 6, 7, 9, 10} {1, 2, 3, 4, 5, 8, 11, 13, 15}] {2, 3, 7, 8, 11, 12, 14} = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 13, 15} {2, 3, 7, 8, 11, 12, 14} = {2, 3, 7, 8, 11} (c) P – (Q – R) = {1, 2, 3, 4, 5, 6, 7, 9, 10} – [{1, 2, 3, 4, 5, 8, 11, 13, 15} – {2, 3, 7, 8, 11, 12, 14}] = {1, 2, 3, 4, 5, 6, 7, 9, 10} – {1, 4, 5, 13, 15} = {2, 3, 6, 7, 9, 10} (d) (Q – P) (Q – R) = [{1, 2, 3, 4, 5, 8, 11, 13, 15} – {1, 2, 3, 4, 5, 6, 7, 9, 10}] [{1, 2, 3, 4, 5, 8, 11, 13, 15} – {2, 3, 7, 8, 11, 12, 14}] = {8, 11, 13, 15} {1, 4, 5, 13, 15} = {13, 15} (e) (P – Q) R = [{1, 2, 3, 4, 5, 6, 7, 9, 10} – {1, 2, 3, 4, 5, 8, 11, 13, 15}] {2, 3, 7, 8, 11, 12, 14} = {6, 7, 9, 10} {2, 3, 7, 8, 11, 12, 14} = {7} (f) R ( Q – P ) Now, Q – P = U – (Q – P) = {1, 2, 3, ...., 19, 20} – [{1, 2, 3, 4, 5, 8, 11, 13, 15} – {1, 2, 3, 4, 5, 6, 7, 9, 10}] = {1, 2, 3, ..... , 19, 20} – {8, 11, 13, 15} = {1, 2, 3, 4, 5, 6, 7, 9, 10, 12, 14} R ( Q – P ) = {2, 3, 7, 8, 11, 12, 14} {1, 2, 3, 4, 5, 6, 7, 9, 10, 12, 14} = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 14} (g) P – (Q R) = {1, 2, 3, 4, 5, 6, 7, 9, 10} – [{1, 2, 3, 4, 5, 8, 11, 13, 15} {2, 3, 7, 8, 11, 12, 14}] = {1, 2, 3, 4, 5, 6, 7, 9, 10} – {1, 2, 3, 4, 5, 7, 8, 11, 12, 13, 14, 15} = {6, 9, 10} (h) (P R) – (P R) = [{1, 2, 3, 4, 5, 6, 7, 9, 10} {2, 3, 7, 8, 11, 12, 14}] – [{1, 2, 3, 4, 5, 6, 7, 9, 10} {2, 3, 7, 8, 11, 12, 14}] = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 14} – {2, 3, 7} = {1, 4, 5, 6, 8, 9, 10, 11, 12, 14} (i) (P Q) – R = (P Q) – R = [{1, 2, 3, 4, 5, 6, 7, 9, 10} {1, 2, 3, 4, 5, 8, 11, 13, 15}] – {2, 3, 7, 8, 11, 12, 14} = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 13, 15} – {2, 3, 7, 8, 11, 12, 14} = {1, 4, 5, 6, 9, 10, 13, 15}
18 10. Suppose, U = {x/x ≤ 20 and x ∈ N}, A = {x/x is a prime number}, B = {x/x is an odd number} and C = {x/x is an even number}. (a) Write the given information in the set notation. (b) List the elements containing each of the following. i. A ii. A C iii. A B iv. U − A C v. A B C vi. B C – A (c) Shade each of the above sets in the Venn diagrams. Solution: Given, U = {x : x 20 and x N} A = {x : x is a prime number} B = {x : x is an odd number} C = {x : x is an even number} (a) Writing above information in set notation: U = {1, 2, 3, ....., 19, 20} A = {2, 3, 5, 7, 11, 13, 17, 19} B = {1, 3, 5, 7, 9, 11, 13, 15, 17, 19} C = {2, 4, 6, 8, 10, 12, 14, 16, 18, 20} (b) (i) A = U – A = {1, 2, 3, ..., 19, 20} – {2, 3, 5, 7, 11, 13, 17, 19} = {1, 4, 6, 8, 9, 10, 12, 14, 15, 16, 18, 20} (ii) A C = U – (A C) = {1, 2, 3, ......, 19, 20} – [{2, 3, 5, 7, 11, 13, 17, 19} {2, 4, 6, 8, 10, 12, 14, 16, 18, 20}] = {1, 2, 3, ...., 19, 20} – {2} = {1, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20} (iii) A B = U – (A B) = {1, 2, 3, ......., 19, 20} – [{2, 3, 5, 7, 11, 13, 17, 19} {1, 3, 5, 7, 9, 11, 13, 15, 17, 19}] = {1, 2, 3, ...., 19, 20} – {1, 2, 3, 5, 7, 9, 11, 13, 15, 17, 19} = {4, 6, 8, 10, 12, 14, 16, 18, 20} (iv) U – ( A C ) = A C = A C = {2, 3, 5, 7, 11, 13, 17, 19} {2, 4, 6, 8, 10, 12, 14, 16, 18, 20} = {2} (v) A B C = U – (A B C) = {1, 2, 3, ......, 19, 20} – [{2, 3, 5, 7, 11, 13, 17, 19} {1, 3, 5, 7, 9, 11, 13, 15, 17, 19} {2, 4, 6, 8, 10, 12, 14, 16, 18, 20}] = {1, 2, 3, ..... , 19, 20} – {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20} = { }
19 (vi) B ( C – A ) = B [U – (C – A)] C – A = U – (C – A) = {1, 2, 3, ......, 19, 20} – [{2, 4, 6, 8, 10, 12, 14, 16, 18, 20} – {2, 3, 5, 7, 11, 13, 17, 19}] = {1, 2, 3, ....... , 19, 20} – {4, 6, 8, 10, 12, 14, 16, 18, 20} = {1, 2, 3, 5, 7, 9, 11, 13, 15, 17, 19} B ( C – A ) = {1, 3, 5, 7, 9, 11, 13, 15, 17, 19} {1, 2, 3, 5, 7, 9, 11, 13, 15, 17, 19} = {1, 3, 5, 7, 9, 11, 13, 15, 17, 19} (c) Shading each set in Venn diagrams. (i) A (ii) A C (iii) A B (iv) U – A C (v) A B C (vi) B ( C – A ) 11. Suppose, U = { a, b, c, .............., k}, A = {a, b, d, e, g}, B = {c, e, g, i} and C = {o, a, e, h, i} are the given sets. Verify the following relations: (a) A – B = A B = B – A (b) A B = A B (c) (A – B) (B – A) = (A B) – (A B) (d) A (B – C) = (A B) – (A – C) (e) A – (B C) = (A – B) (A – C) (f) A B C = A B C 3 5 7 11 13 17 19 2 1 9 15 4 6 8 10 12 14 16 18 20 BA C U 3 5 7 11 13 17 19 2 1 9 15 4 6 8 10 12 14 16 18 20 BA C U 3 5 7 11 13 17 19 2 1 9 15 4 6 8 10 12 14 16 18 20 BA C U 3 5 7 11 13 17 19 2 1 9 15 4 6 8 10 12 14 16 18 20 BA C U 3 5 7 11 13 17 19 2 1 9 15 4 6 8 10 12 14 16 18 20 BA C U 3 5 7 11 13 17 19 2 1 9 15 4 6 8 10 12 14 16 18 20 BA C U
20 Solution: Given, U = {a, b, c, ..., k}, A = {a, b, d, e, g}, B = {c, e, g, i}, C = {o, a, e, h, i} (a) A – B = A B = B – A A – B = {a, b, d, e, g} – {c, e, g, i} = {a, b, d} A B = A (U – B) = {a, b, d, e, g} [{a, b, c, ..., k} – {c, e, g, i}] = {a, b, d, e, g} {a, b, d, f, h, j, k} = {a, b, d} B – A = (U – B) – (U – A) = [{a, b, ..., k} – {c, e, g, i}] – [{a, b, ... , k} – {a, b, d, e, g}] = {a, b, d, f, h, j, k} – {c, f, h, i, j, k} = {a, b, d} Hence, A – B = A B = B – A , proved. (b) A B = A B A B = U – (A B) = {a, b, c, ...., k} – [{a, b, d, e, g} {c, e, g, i}] = {a, b, c, d, ..., k} – {a, b, c, d, e, g, i} = {f, h, j, k} A B = (U – A) (U – B) = [{a, b, c, ..., k} – {a, b, d, e, g}] [{a, b, c, ... , k} – {c, e, g, i}] = {c, f, h, i, j, k} {a, b, d, f, h, j, k} = {f, h, j, k} Hence, A B = A B , proved. (c) (A – B) (B – A) = (A B) – (A B) A – B = {a, b, d, e, g} – {c, e, g, i} = {a, b, d} B – A = {c, e, g, i} – {a, b, d, e, g} = {c, i} LHS = (A – B) (B – A) = {a, b, d} {c, i} = {a, b, c, d, i} A B = {a, b, d, e, g} {c, e, g, i} = {a, b, c, d, e, g, i} A B = {a, b, d, e, g} {c, e, g, i} = {e, g} RHS = (A B) – (A B) = {a, b, c, d, e, g, i} – {e, g} = {a, b, c, d, i} Hence, LHS = RHS, proved.
21 (d) A (B – C) = (A B) – (C – A) LHS = A (B – C) = {a, b, d, e, g} [{a, b, d, e, g} – {o, a, c, e, h, i}] = {a, b, d, e, g} {b, d} = {a, b, c, d, e, g} RHS = (A B) – (A C) = [{a, b, d, e, g} {c, e, g, i}] – [{a, b, d, e, g} – {o, a, e, h, i}] = {a, b, c, d, e, g, i} – {b, d, g} = {a, b, c, d, e, g} (e) A – (B C) = (A – B) (A – C) LHS = {a, b, d, e, g} – [{c, e, g, i} {a, e, h, i}] = {a, b, d, e, g} – {e, i} = {a, b, d, g} RHS = (A – B) (A – C) = [{a, b, d, e, g} – {c, e, g, i}] [{a, b, d, e, g} – {a, e, h, i}] = {a, b, d} {b, d, g} = {a, b, d, g} LHS = RHS, proved. (f) A B C = A B C LHS = U – (A B C) = {a, b, c, ...., j, k} – [{a, b, d, e, g} {c, e, g, i} {a, e, h, i}] = {a, b, c, ......, j, k} – {e} = {a, b, c, d, f, g, h, i, j, k} RHS = A B C = (U – A) (U – B) (U – C) = [{a, b, c, ..., k} – {a, b, d, e, g}] [{a, b, c, ..., k} – {c, e, g, i}] [{a, b, c, ..., j, k} – {a, e, h, i}] = {c, f, h, i, j, k} {a, b, d, f, h, j, k} {b, c, d, f, g, j, k} = {a, b, c, d, f, g, h, i, j, k} LHS = RHS, proved.
22 1.3 Cardinality of Sets PRACTICE 1.3 Read / Understand / Think / Do Keeping Skill Sharp 1. (a) Define cardinality of a set. (b) What is the cardinality of a null set ? (c) Write the two examples of singleton set. (d) What is the cardinality of E = {6, 7, 8, 9, ........... 23} ? (e) If P = {u, n, i, o} and Q = {i, n, t, e, r, s, c, t, i, o, n} the cardinality of n(P Q) = ....... Solution: (a) The cardinality of a set is the number of elements of the set. (b) The cardinality of a null set is 0. (c) Two examples of singleton sets are (i) a set of highest peak of the world, (ii) a set of capital of Nepal. (d) Cardinality of E, n(E) = 18. (e) P Q = {u, n, i, o} {i, n, t, e, r, s, c, t, i, o, n} = {n, i, o} n(P Q) = 3. 2. Circle (±) the correct answer. (a) What is the cardinality of singleton set? (i) 0 (ii) 1 (iii) 2 (iv) 3 (b) What is the cardinality of equivalent sets? (i) equal (ii) equivalent (iii) unequal (iv) non-equivalent (c) Which is the cardinality of union of the sets N = {1, 3, 5, 7} and P = {2, 4, 8} ? (i) 5 (ii) 6 (iii) 7 (iv) 8 (d) What is the cardinality of the intersection of the sets M = {m, a, n, g, o} and P = {p, o, m, e, g, r, a, n, t} ? (i) 9 (ii) 5 (iii) 14 (iv) 13 (e) If P and Q are overlapping sets in the universal set U, then ............. (i) n(U) = n(C) + n(P) + n( C P ) (ii) n(U) = n(C) + n(P) + n(C P) + n( C P ) (iii) n(U) = n(C) + n(P) – n(C P) + n( C P ) (iv) n(U) = no(C) + no(P) – n(C P) + n( C P ) Solution: (a) (ii) (b) (i) (c) (iii) (d) (ii) (e) (iii)
23 Check Your Performance Answer the given questions for each problem. 3. Think about the following sets and answer the given questions: C = Set of colours in rainbow, H = {Hands of a normal human being}, L = Set of the longest river of the world, Y = {Name of your parents}, P = {Pages of your the Leading Maths}, D = Set of the province of Nepal, N = {National parks of Nepal}, S = Set of the capitals of SAARC countries, A = {Sum of angles of first 5 regular polygons}, M = {World heritage sites in Nepal}, M = Set of the height of mountains more than 8000 m in Nepal} [If you do not know the name of different things mentioned above, you can use google search.] (a) Write the elements of the given sets in set notation. (b) Define cardinality of a set. (c) Find the cardinality of the above mentioned sets. (d) Which set has the greatest cardinality and that of the least cardinality ? Solution: (a) Here, C = {violet, indigo, blue, green, yellow, orange, red} H = {left hand, right hand}, L = {Nile} Y = {Ram, Sita}, P = {1, 2, 3, ....., 416} D = {Koshi, Madhesh, Bagmati, Gandaki, Lumbini, Karnali, Sudur Pachchhim} N = {Shey Phoksundo NP, Khaptad NP, Bardia NP, Langtang NP, Sagarmath NP, Rara NP, Banke NP, Koshi Tappu wildlife reserve, Dhorpatan Hunting Reserve, Annapurna Conservation Area, Shulklapanta NP} S = {Delhi, Kathmandu, Colombo, Islamabad, Kabul, Dhaka, Thimpu, Male} A = {180°, 360°, 540°, 720°, 900°} N = {Pashupatinath temple, Boudhanath stupa, Swayambhunath, Kathmandu Durbar Square, Lalitpur Durbar square, Bhaktapur Durbar square, Lumbini, Changunarayan, Chitwan National Park, Sagarmatha National Park} M = {Mount Everest, Kanchanjunga, Lhotse, Makalu, Cho Oyu, Dhoulagri I, Manaslu, Annapurna I} (b) The number of distinct elements of a finite set is called cardinality of the set. (c) n(C) = 7, n(H) = 2, n(L) = 1, n(Y) = 2, n(P) = 416, n(D) = 7, n(N) = 10, n(S) = 8, n(A) = 5, n(M) = 8 (d) Set P has the greatest cardinality and set L has the least cardinality. 4. Answer the given questions for the following numbers (a) and (b) both. (a) It is given that a universal set U = {1, 2, 3, ...., 12} and its subsets A = {1, 3, 9, 12} and B = {2, 3, 5, 7, 11}. (b) If A = {Even Prime numbers} and B = {Whole numbers less than 6}, then find cardinal numbers of A B, A B, A – B and B – A. Also, represent in a Venn diagram. (i) Write the cardinality of the sets U, A and B. (ii) Find the cardinal numbers of A B, A B, A – B and B – A. (iii) Represent the above results in a Venn diagram. (iv) Why is n(A B) not more than n(A B) ? Given suitable reason. Solution:
24 Given, (a) U = {1, 2, 3, ...., 12}, A = {1, 3, 9, 12} and B = {2, 3, 5, 7, 11} (i) n(U) = 12, n(A) = 4, n(B) = 5 (ii) A B = {1, 5, 9, 12} {2, 3, 5, 7, 11} = {1, 2, 3, 5, 7, 9, 11, 12} Hence, n(A B) = 8 A B = {1, 3, 9, 12} {2, 3, 5, 7, 11} = {3} n(A B) = 1 A – B = {1, 3, 9, 12} – {2, 3, 5, 7, 11} = {1, 9, 12} n(A – B) = 3 B – A = {2, 3, 5, 7, 11} – {1, 3, 9, 12} = {2, 5, 7, 11} n(B – A) = 4 (iii) Showing the above results in a Venn diagram, (iv) From above Venn diagram, no(A) = 3, no(B) = 4, n(A B) = 1 n(A B) = no(A) + no(B) + n(A B) = 3 + 4 + 1 = 8 > 1 which means n(A B) is greater than n(A B). Another reason, (A B) is a proper subset of (A B) hence n(A B) n(A B). So, n(A B) is not more than n(A B). (b) A = {Even prime numbers}, B = {whole numbers less than 6} A = {2}, B = {0, 1, 2, 3, 4, 5} Now, A B = {2} {0, 1, 2, 3, 4, 5} = {0, 1, 2, 3, 4, 5} n(A B) = 6 A B = {2} {0, 1, 2, 3, 4, 5} = {2} n(A B) = 1 A – B = {2} – {0, 1, 2, 3, 4, 5} = { } n(A – B) = 0 B – A = {0, 1, 2, 3, 4, 5} – {2} = {0, 1, 3, 4, 5} n(B – A) = 5 A B U 314 4 n(A B) A B U 314 4 n(A B) A B U 314 4 n(A – B) A B U 314 4 n(B – A)
25 M NU 1 2 6 5 7 10 4 3 8 9 5. If M = Set of multiples of 5 less than 40 and E = Set of factors of 24, then; (a) Write the elements of the given sets and their cardinalities in the set notation. (b) Show the above information in a Venn diagram. (c) Find n(M E), n(M E), n(M – E) and n(E – M). (d) Establish the relation n(M E) = n(M) + n(E) – n(M E). Solution: Given, M = Set of multiples of 5 less than 40 and E = set of factors of 24 (a) M = {5, 10, 15, 20, 25, 30, 35} and E = {1, 2, 3, 4, 6, 8, 12, 24} n(M) = 7 and n(E) = 8 (b) Showing above information in a Venn diagram alongside, (c) M E = {5, 10, 15, 20, 25, 30, 35} {1, 2, 3, 4, 6, 8, 12, 24} = {1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 20, 24, 30, 35} Again, M E = {5, 10, 15, 20, 25, 30, 35} {1, 2, 3, 4, 6, 8, 12, 24} = { } M – E = {5, 10, 15, 20, 25, 30, 35} – {1, 2, 3, 4, 6, 8, 12, 24} = {5, 10, 15, 20, 23, 30, 35} E – M = {1, 2, 3, 4, 6, 8, 12, 24} – {5, 10, 15, 20, 25, 30, 35} = {1, 2, 3, 4, 6, 8, 12, 24} n(M E) = 15 n(M E) = 0, n(M – E) = 7, n(E – M) = 8 (d) n(M E) = n(M) + n(E) – n(M E) or, 15 = 7 + 8 – 0 or, 15 = 15, proved. 6. Observe the given Venn diagram and answer the following questions; (a) Write the elements of the sets mentioned in the adjoining Venn diagram in roster and setbuilder forms. (b) Find the cardinalities of U, M, N, M N and M N . (c) Show the above cardinalities in another Venn diagram. (d) Find n( M ) and n( M N ). (e) Establish the relation between n(U), n(M), n(N), n(M N) and n( M N ). (f) Prove that: n(U) = no(M) + n(N) + n( M N ). B U A 2 1 n(A B) 0 3 4 5 B U A 2 n(A B) 0 1 3 4 5 B U A 2 n(A – B) 0 1 3 4 5 B U A 2 n(B – A) 0 1 3 4 5 M E U 5 15 25 35 10 20 30 2 4 6 8 12 1 3 24
26 Solution: (a) Writing the elements of sets in roster form: U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}, M = {1, 2, 3} and N = {2, 4, 6, 8} Set builder form: U = {x : x 10, x N} M = {x : x 3, x N} and N = {x : x is an ever number less than 10} (b) n(U) = 10, n(M) = 3, n(N) = 4 M N = {1, 2, 3} {2, 4, 6, 8} = {2} n(M N) = 1 M N = U – (M N) = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} – [{1, 2, 3} {2, 4, 6, 8}] = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} – {1, 2, 3, 4, 6, 8} = {5, 7, 9, 10} n( M N ) = 4 (c) Showing the above cardinalities in another Venn diagram, (d) n( M ) = ? Now, M = U – M = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} – {1, 2, 3} = {4, 5, 6, 7, 8, 9, 10} n( M ) = 7 n( M N ) = ? Again, M N = U – (M N) = {1, 2, 3, ......, 9, 10} – {2} = {1, 3, 4, 5, 6, 7, 8, 9, 10} n( M N ) = 9 (e) Here, n(U) = 10, n(M) = 3, n(N) = 4, n(M N) = 1 and ( M N ) = 4 From above figure, n(U) = 10 or, 10 = 4 + 3 + 1 + 4 ` i.e. n(U) = n(N) + n(M) – n(M N) + n( M N ) Hence, n(U) = n(M) + n(N) – n(M N) + n( M N ) (f) From above relation in (e), we have, n(U) = n(M) + n(N) – n(M N) + n( M N ) or, n(U) = n(M) – n(M N) + n(N) + n( M N ) or, n(U) = no(M) + n(M) + n( M N ) [ n(M) – n( M N ) = no(M)] proved. n(M N) = 1 M N U 2 4 1 3 n(M) = 2 + 1 = 3 n(N) = 1 + 3 = 4 n(M N) = 4 n(U) = 10
27 P Q R U s h n a o c k q r j m f d e p 7. Observe the given Venn diagram and answer the following questions; (a) Write the elements of the sets mentioned in the given Venn diagram in roster form. (b) Find the following cardinalities: (i) n(U), n(P), n(Q), n(R) (ii) n(P Q), n(Q R) (iii) no(P R), no(P Q) (v) no(P), no(R) (iv) n(P Q R), n(P Q R) (vi) n(P (Q R)), n(Q (P R)) (vii) n( P R ), n( P Q ) (viii) n( P Q R ), n( P Q R ) (c) Prove that: (i) n(P Q R) = n(P Q) + no(R) (ii) n(U) = n(P Q R) + n(P Q R) Solution: (a) Writing the elements of the sets mentioned in the given Venn diagram, In roster form: P = {c, e, h, k, o, s}, Q = {a, h, n, p, q, r, s} and R = {a, e, h, j, n, o}, U = {a, b, c, d, e, f, h, j, k, m, n, o, p, q, r, s} (b) (i) n(U) = 16, n(P) = 6, n(Q) = 7, n(R) = 6 (ii) n(P Q) = ? (P Q) = {c, e, h, k, o, s} {a, h, n, p, q, r, s} = {a, c, e, h, k, j, n, o, p, q, r, s} n(P Q) = 11 n(Q R) = ? (Q R) = {a, h, n, p, q, r, s} {a, e, h, j, n, o} = {a, h, n} n(Q R) = 3 (iii) P R = {c, e, h, k, o, s} {a, e, h, j, n, o} = {e, h, o} no(P R) = n(P R) – n(P Q R) = 3 – 1 = 2 no(P R) = 2 (P Q) = {c, e, h, k, o, s} {a, h, n, p, q, r, s} = {h, s} no(P Q) = n(P Q) – n(P Q R) = 2 – 1 no(P Q) = 1 (iv) n(P Q R) = {h} (P Q R) = {c, e, h, k, o, s} {a, h, n, p, a, r, s} {a, e, h, j, n, o} = {h} n(P Q R) = 1 P Q R = {c, e, h, k, o, s} {a, h, n, p, q, r, s} {a, e, h, j, n, o} = {a, c, e, h, j, k, o, n, p, q, r, s} n(P Q R) = 12 (v) Now, no(P) = n(P) – n(P Q) – n(P R) + n(P Q R) = 6 – 2 – 3 + 1 = 7 – 5 = 2. no(R) = n(R) – n(Q R) – n(P R) + n(P Q R) = 6 – 3 – 3 – 1 = 7 – 6 = 1
28 A B C U 8 3 6 2 5 4 7 4 (vi) P (Q R) = {c, e, h, k, o, s} {a, h, n} = {a, c, e, h, k, n, o, s} n(P (Q R)) = 8 Q (P R) = {a, h, n, p, q, r, s} [{c, e, h, k, o, s} {a, e, h, j, n, o}] = {a, h, n, p, q, r, s} {a, c, e, h, j, k, n, o, s} = {a, h, n, s} n(Q (P R)) = 4 (vii) n( P R ) = n(U) – n(P R) = 15 – 9 = 6 n( P Q ) = n(U) – n(P Q) = 15 – 2 = 13 (viii) n( P Q R ) = n(U) – n(P Q R) = 15 – 12 = 3 n( P Q R ) = n(U) – n(P Q R) = 15 – 1 = 14 (c) (i) To prove: n(P Q R) = n(P Q) + no(R) From above calculation, n(P Q R) = 12 n(P Q) + no(R) = n(P Q) + n(R) – n(P R) – n(Q R) + n(P Q R) = 11 + 6 – 3 – 3 + 1 = 18 – 6 = 12 Hence, n(P Q R) = n(P Q) + no(R), proved. (ii) From above, n(U) = 15 n(P Q R) + n( P Q R ) = 12 + 3 = 15 Hence, n(U) = n(P Q R) + n( P Q R ), Proved. 8. Observe the given Venn diagram of the cardinality and answer the following questions; (a) Find the cardinalities of n(U), n(A), n(B) and n(C). (b) Prove that: (i) n(A B C) = n(A B) + no(C) (ii) n(A B C) = n(A) + no(B) + no(C) + no(B C) (iii) n(A) = no(A) + no(A B) + no(A C) + n(A B C) (iv) n( A ) = no(B) + no(C) + no(B C) + n( A B C ) (v) n(U) = n(A B C) + n( A B C ) (vi) n(A B C) = n(A) + n(B) + n(C) – n(A B) – n(A C) – n(BC) + n(ABC) (vii) n(U) = no(A) + n(B C) + n( A B C ) Solution: (a) From the given Venn diagram n(A) = 8 + 3 + 2 + 5 = 18, n(B) = 3 + 6 + 4 + 2 = 15, n(C) = 2 + 4 + 7 + 5 = 18, n(U) = 8 + 3 + 6 + 7 + 4 + 2 + 5 + 4 = 39 (b) (i) To prove, n(A B C) = n(A B) + no(C) Here, n(A B C) = 8 + 3 + 6 + 4 + 7 + 5 + 2 = 35 n(A B) = 8 + 5 + 3 + 2 + 6 + 4 = 28 and no(C) = 7
29 n(A B) + no(C) = 28 + 7 = 35 Hence, n(A B C) = n(A B) + no(C), proved. (ii) n(A B C) = n(A) + no(B) + no(C) + no(B C) Here, n(A B C) = 35, n(A) = 18, no(B) = 6, no(C) = 7, no(B C) = 4 n(A) + no(B) + no(C) + no(B C) = 18 + 6 + 7 + 4 = 35 = n(A B C) Hence, n(A B C) = n(A) + no(B) + no(C) + no(B C), proved. (iii) To prove, n(A) = no(A) + no(A B) + no(A C) + n(A B C) Here, n(A) = 18, no(A) = 8, no(A B) = 3, no(A C) = 5 and n(A B C) = 2 no(A) + no(A B) + no(A C) + n(A B C) = 8 + 3 + 5 + 2 = 18 = n(A) Hence, n(A) = no(A) + no(A B) + no(A C) + n(A B C), proved. (iv) n( A ) = no(B) + no(C) + n(B C) + n( A B C ) Here, n( A ) = n(U) – n(A) = 39 – 18 = 21, no(B) = 6, no(C) = 7, no(B C) = 4, n( A B C ) = 4 no(B) + no(C) + n(B C) + n( A B C ) = 6 + 7 + 4 + 4 = 21 = n( A ) Hence, n( A ) = no(B) + no(C) + n(B C) + n( A B C ), proved. (v) n(U) = n(A B C) + n( A B C ) Here, n(U) = 39, n(A B C) = 35, n( A B C ) = 4 n(A B C) + n( A B C ) = 35 + 4 = 39 = n(U) Hence, n(U) = n(A B C) + n( A B C ), proved. (vi) n(A B C) = n(A) + n(B) + n(C) – n(A B) – n(A C) – n(B C) + n(A B C) Here, n(A B C) = 35, n(A) = 18, n(B) = 15, n(C) = 18, n(A B) = 3 + 2 = 5 n(A C) = 2 + 5 = 7, n(B C) = 2 + 4 = 6, n(A B C) = 2 n(A) + n(B) + n(C) – n(A B) – n(A C) – n(B C) + n(A B C) = 18 + 15 + 18 – 5 – 7 – 6 + 2 = 53 – 18 = 35 = n(A B C) Hence, n(ABC) = n(A) + n(B) + n(C) – n(AB) – n(AC) – n(B C) + n(ABC) (vii) n(U) = no(A) + n(B C) + n( A B C ) Here, n(U) = 39, no(A) = 8, no(B C) = 3 + 6 + 2 + 4 + 5 + 7 = 27, n( A B C ) = 4 no(A) + n(B C) + n( A B C ) = 8 + 27 + 4 = 39 = n(U) Hence, n(U) = no(A) + n(B C) + n( A B C ), proved. (viii) n(U) = no(A) + no(B) + no(C) + no(A B) + no( A C ) + no(B C) + n(A B C) + n( A B C ) Here, n(U) = 39, no(A) = 8, no(B) = 6, no(C) = 7, no(A B) = 3
9. O t ( ( ( ( Soluti ( ( ( ( 10. I ( ( ( ( ( Soluti G ( ( no no Hence, Observe the diff the set notation b (a) Write the e (b) Show the ab (c) Find n(S), n (d) Is n(S K) ion: (a) S = {footbal K = {tennis (b) Showing the (c) n(S) = 6, n(K (d) Yes, n(S K If F = {Set of fac (a) Write the e (b) Find n(F), n (c) Show the ab (d) Prove that: (e) Why is no(E ion: Given, F = Set of (a) F = {1, 2, 3, (b) n(F) = 6, n(O n(F F – O = n(F – O (O – F) n(O – F fo bas ha ten ba vol go bow S K o(A C) = 5, no( o(A) + no(B) + no + n( A B , n(U) = no(A) + + n(A ferent balls of S by supposing S f lements of the g bove informatio n(K), n(S K) a ) = n(K)? Why? ll, basketball, han ball, volleyball, g e above informati K) = 8, n(S K) K) = n(K) becaus ctors of 18} and O lements of the g n(O), n(F O), bove cardinalitie n(F O) = no(F E) not more than f factors of 18 & O , 6, 9, 18} & O = O) = 7, F O = O) = 2 = {1, 2, 3, 6, 9, 1 O) = 4 ) = {3, 5, 7, 9, 11 F) = 5 ootball sketball andball nnis ball aseball lleyball olfball wling ball U 30 (B C) = 4, n(A o(C) + no(A B) C ) = 8 + 6 + 7 + + no(B) + no(C) + B C) + n( A Saru and Karina for the set of Sar given sets and the n in a Venn diag and n(S K). ndball, tennis ball golf ball, base ba ion in a Venn dia = 8, n(S K) = se S K. O = {Set of odd given sets in the s n(F – O) and n( es in a Venn dia F) + no(O) + n(F n n(E) ? Give sui O = set of odd nu {3, 5, 7, 9, 11, 1 {1, 2, 3, 6, 9, 18} 8} – {3, 5, 7, 9, 1 1, 13, 15} – {1, 2 A B C) = 2, n + no(A C) + n + 3 + 5 + 4 + 2 + no(A B) + no(A A B C ), prov a in the adjoinin ru's balls and K eir cardinalities gram. l, baseball, volley all, bowling ball, agram alongside 6 numbers from 2 set notation. O – F). agram. O) itable reason. umbers from 2 to 3, 15} } {3, 5, 7, 9, 11 11, 13, 15} = {1, , 3, 6, 9, 18} = {5 n( A B C ) = no(B C) + n(A + 4 = 39 = n(U) A C) + no(B ved. ng pictures. Nam K for the set of K in set notation. y ball} handball, footbal 2 to 15} then; 15. 1, 13, 15} = {3, 9 2, 6, 18} 5, 7, 11, 13, 15} = 4 B C) C) me their balls in Karina's balls. ll, basketball} 9} n
31 = 3 = 3 = 5 P F U n(P) n(F) n n(PF) o(P) no(F) (c) Showing the above cardinalities in a Venn diagram alongside. (d) n(F O) = no(F) + no(O) + n(A O) F O = {1, 2, 3, 6, 9, 18} {3, 5, 7, 9, 11, 13, 15} = {1, 2, 3, 5, 6, 7, 9, 11, 13, 15, 18} n(F O) = 11 no(F) = 4, no(O) = 5, n(P O) = 2 no(F) + no(O) + n(P O) = 4 + 5 + 2 = 11 = n(F O) Hence, n(F O) = no(F) + no(O) + n(P O), proved. (e) no(E) is not more than n(E) because only E is a proper subset set E. 11. Answer the given questions for the following sets in the numbers (a) and (b): (a) Suppose, P = {x: x is a prime number less than 15} and F = {x: x is a factor of 30}. (b) Suppose, P = {Square numbers up to 100} and F = {Multiples of 9 less than 100}. (i) Write the elements of the given sets in the set notation. (ii) Establish the relation between n(P F), n(P), n(F), and n(P F). (iii) Establish the relation n(P F), no(P), no(F) and n(P F). (iv) Show the above cardinalities in a Venn diagram. (v) Why is n(P F) not more than n(P) ? Give suitable reason. Solution: (a) P = {x : x is a prime number less than 15} and F = {x : x is a factor of 30} (i) P = {2, 3, 5, 7, 11, 13} and F = {1, 2, 3, 5, 6, 10, 15, 30} (ii) Here, n(P) = 6, n(F) = 8, P F = {2, 3, 5, 7, 11, 13} {1, 2, 3, 5, 6, 10, 15, 30} = {2, 3, 5} n(P F) = 3 P F = {2, 3, 5, 7, 11, 13} {1, 2, 3, 5, 6, 10, 15, 30} = {1, 2, 3, 5, 6, 7, 10, 11, 13, 15, 30} n(P F) = 11 Hence, n(P F) = n(P) + n(F) – n(P F) = 6 + 8 – 3 = 14 – 3 = 11 (iii) Here, no(P) = n(P) – n(P F) = 6 – 3 = 3 and no(F) = n(F) – n(P F) = 8 – 3 = 5 Again, no(P) + no(F) + n(P F) = 3 + 5 + 3 = 11 Hence, n(P F) = no(P) + no(F) + n(P F) (iv) Showing the above cardinalities in a Venn diagram, (v) n(P F) is not more than n(P) because P F is a proper subset of the set P. n(F – O) n(O – F) 425 F O U n(F) n(O) n(F O)
32 (b) P = {Square number up to 100} and F = {Multiples of 9 less than 100} (i) P = {1, 4, 9, 16, 25, 36, 49, 64, 81, 100} and F = {9, 18, 27, 36, 45, 54, 63, 72, 81, 90, 99} (ii) n(P) = 10, n(F) = 11, P F = {1, 4, 9, 16, 25, 36, 49, 64, 81, 100} (9, 18, 27, 36, 45, 54, 63, 72, 81, 90, 99} = {9, 36, 81} n(P F) = 3 n(P) + n(F) – n(P F) = 10 + 11 – 3 = 21 – 3 = 18 Hence, n(P F) = n(P) + n(F) – n(P F) (iii) no(P) = n(P) – n(P F) = 10 – 3 = 7, no(F) = n(F) – n(P F) = 11 – 3 = 8 no(P) + no(F) + n(P F) = 7 + 8 + 3 = 18 Hence, n(P F) = no(P) + no(F) + n(P F) (iv) Showing the above cardinalities in a Venn diagram, (v) n(P F) is not more than n(P) because P F is a proper subset of the set P. 12. If n(A B) = 25, n(A) = 12 and n(B) = 17 then, (a) Define the cardinality of the union of A and B. (b) Illustrate the above information in a Venn diagram. (c) Find n(A B) and n(A Δ B). (d) Write the relation between n(A B), n(A B) and n(A Δ B). Solution: Given, n(A B) = 5, n(A) = 12 and n(B) = 17 (a) n(A B) = n(A) + n(B) – n(A B) = 12 + 17 – 5 = 29 – 5 = 24 (b) Illustrating the above information in a Venn diagram, (c) n(A B) = ?, n(A B) = ? n(A B) = 5 n(A B) = 7 + 12 = 19 (d) Here, n(A B) = 24, n(A B) = 5, n(A B) = 19 Now, n(A B) – n(A B) = 24 – 5 = 19 = n(A B) Hence, the relation between n(A B), n(A B) and n(A B) is, n(A B) = n(A B) – n(A B) = 7 = 3 = 8 P F U n(P) n(F) n n(PF) o(P) no(F) = 12 – 5 = 7 = 5 = 17 – 5 = 12 A B U n(A) n(B) n(AB) no(A) no(B)
33 13. If n(U) = 42, n(A) = 15, n(B) = 25 and n( A B ) = 9 then, (a) Define the cardinality of the intersection of A and B. (b) Present the above information in a Venn diagram. (c) Find n(A B) and n(A Δ B). (d) Write the relation between n(A B), n(A B) and n(A Δ B). Solution: Given, n(U) = 42, n(A) = 15, n(B) = 25 and n( A B ) = 9 (a) The number of elements in the intersection of A and B is called the cardinality of the intersection of A and B. (b) Presenting the above information in Venn diagram, (c) We know, n(U) = n(A B) + n( A B ) or, 42 = n(A B) = 9 n(A B) = 42 – 9 = 33 Also, n(A B) = n(A) + n(B) – n(A B) or, 33 = 15 + 25 – n(A B) n(A B) = 40 – 33 = 7 Hence, n(A B) = 7 n(A B) = 33 – 7 = 26 (d) Here, n(A B) = 7, n(A B) = 33 and n(A B) = 26 n(A B) – n(A B) = 33 – 7 = 26 = n(A B) Hence, n(A B) = n(A B) – n(A B) 14. It is given that n(U) = 45, n(A) = 25, n(B) = 28 and n(A B) = 13. (a) What is the meaning of n( A B ) ? (b) Illustrate the above information in a Venn diagram. (c) Find n( A B ), n(at most one set) and n(only one set). (d) Write the relation between n(U), n(A B), n(A Δ B) and n( A B ). Solution: Given, n(U) = 45, n(A) = 25, n(B) = 28 and n(A B) = 13 (a) The meaning of n( A B ) is the cardinality of the set whose elements are other than that of sets A and B. 15 – 7 = 8 7 25 – 7 = 18 A B U 9
34 25 – 13 = 12 = 13 28 – 13 = 15 A B U n(AB) 5 (b) Presenting the above information in a Venn diagram, (c) n( A B ) = ? We know, n(A B) = n(A) + n(B) – n(A B) = 25 + 28 – 13 = 40 n(U) = n(A B) + n( A B ) or, 45 = 40 + n( A B ) n( A B ) = 45 – 40 = 5. n(at most one) = n( A B ) = 45 – 13 = 32, n(only one) = 12 + 15 = 27 (d) Here, n(U) = 45, n(A B) = 13, n(A B) = 12 + 15 = 27 n( A B ) = 5 Now, n(A B) + n(A B) + n( A B ) = 13 + 27 + 5 = 45 = n(U) Hence, n(U) = n(A B) + n( A B ) + n(A B). 15. In a school, there are 500 students. 60% of the students like mathematics and 50% of the students like science. If every students like one of the two subjects. (a) Represent this information in a Venn diagram. (b) Find the number of students who like both mathematics and science. (c) Find the number of students who like mathematics only. (d) Find the number of students who like science only. Solution: Let n(M) and n(S) denote the student who like mathematics and science then, n(U) = 500 n(M) = 60% of 500 = 300 n(S) = 50% of 500 = 250 (a) Representing the given information in a Venn diagram, (b) Since every student like one of two subjects. n(U) = n(M S) or, 500 = n(M) + n(S) – n(M S) or, n(M S) = 550 – 500 = 50 Hence, 50 students like both mathematics and science. (c) Mathematics only, no(M) = n(M) – n(M S) = 300 – 50 = 250 (d) Science only, no(S) = n(S) – n(M S) = 250 – 50 = 200 M S U 250 50 200
35 C P U 30 17 20 7 16. In a group of 90 students, 54 talk English, 51 talk Newari and 9 talk neither. How many students talk both subjects. Draw Venn diagram to help you answer the questions. Solution: Let n(E) and n(N) denotes the number of students who talk English and Newari. Then, n(U) = 90, n(E) = 54, n(N) = 51,n( E N ) = 9 Now, n(U) = n(E N) + n( E N ) or, 90 = n(E N) + 9 or, n(E N) = 90 – 9 or, n(E) + n(N) – n(E N) = 81 or, 54 + 51 – n(E N) = 81 or, 105 – n(E N) = 81 or, – n(E N) = 81 –105 n(E N) = 24 Hence, 24 students talk both English and Newari. 17. In a group of 74 people, if the ratio of children who like Coke only and Pepsi only is 3:2 and the number of people, who like both drinks is 17 and who do not like both is 7, find the number of people who like Coke and represent the fact in a Venn diagram. Solution: Let n(C) and n(P) denotes the people who like Coke and Pepsi. Then by question, n(U) = 74 no(C) : no(P) = 3 : 2 n(C P) = 17 n( C P ) = 7 Let no(C) = 3x and no(P) = 2x Now, n(U) = n(C P) + n( C P ) or, 74 = n(C P) + 7 or, n(C P) = 74 – 7 or, no(C) + no(P) + n(C P) = 67 or, 3x + 2x + 17 = 67 or, 5x + 67 – 17 or, x = 50 5 = 10 no(C) = 3x = 3 × 10 = 30 people no(P) = 2x = 2 × 10 = 20 people Now, n(C) = no(C) + n(C P) = 30 + 17 = 47 Hence, 47 people like coke. Representing the above fact in the Venn diagram,
36 Additional Practice – I 1. Sets A = {multiples of 3}, B = {factors of 6} and C = {odd numbers} are the subsets of a universal set U = {natural numbers from 1 to 10}. (a) Write the cardinality of U. (b) Show the above information in a Venn diagram. (c) Prove that: A – (B C) = (A B) – (B C) (d) What are the type of the sets A B C and B C? Write with reason. Solution: Here, U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} A = {3, 6, 9}, B = {1, 2, 3, 6}, C = {1, 3, 5, 7, 9} (a) n(U) = 10 (b) Showing above information in a Venn diagram, (c) Here, A – (B C) = (A B) – (B C) LHS = A – (B C) = {3, 6, 9} – ({1, 2, 3, 6} {1, 3, 5, 7, 9}) = {3, 6, 9} – {1, 2, 3, 5, 6, 7, 9} = RHS = (A B) – (B C) = ({3, 6, 9} {1, 2, 3, 6}) – ({1, 2, 3, 6} {1, 3, 5, 7, 9}) = {1, 2, 3, 6, 9} – {1, 2, 3, 5, 6, 7, 9} = Hence, LHS = RHS, proved. (d) A B C is the union of the sets A, B and C. B C is the intersection of the sets B and C. 2. Sets U, A and B are given below. U = {1, 2, 3, 4, 5, 6, 7, 8, 9}, A = {2, 3, 5, 7} and B = {2, 3, 4, 8} (a) Write the cardinality of A. (b) Represent the above information in a Venn diagram. (c) Prove that: ( A B ) = A B (d) Is n(A – B) and n(B – A) always equal? Justify it. Solution: Here, U = {1, 2, 3, 4, 5, 6, 7, 8, 9}, A = {2, 3, 5, 7} and B = {2, 3, 4, 8} (a) n(A) = 4 (b) Representing the above information in a Venn diagram, (c) Here, ( A B ) = A B LHS = A B = U – (A B) = {1, 2, 3, ..., 9} – ({2, 3, 5, 7} {2, 3, 4, 8}) = {1, 2, 3, ..., 9} – {2, 3} = {1, 4, 5, 6, 7, 8, 9} RHS = A B = (U – A) (U – B) = ({1, 2, 3, ..., 9} – {2, 3, 5, 7}) ({1, 2, 3, ..., 9} – {2, 3, 4, 8}) = {1, 4, 6, 8, 9} {1, 5, 6, 7, 9} = {1, 4, 5, 6, 7, 8, 9} Hence, LHS = RHS, proved. A B U 2 3 8 4 5 7 9 1 6 A B C U 6 2 3 9 1 5 10 7 8 4
37 (d) No. n(A – B) and n(B – A) are not always equal. To justify, let A = {1, 3, 5, 7, 9} and B = {2, 3, 5, 7} Here, A – B = {1, 3, 5, 7, 9} – {2, 3, 5, 7} = {1, 9} n(A – B) = 2 and B – A = {2, 3, 5, 7} – {1, 3, 5, 7, 9} = {2} n(B – A) = 1 Hence, they are not always equal. 3. Sets U, E and F are given below. U = {natural number less than 15}, E = {even number less than 12} and F = {factors of 12}. (a) Define cardinality of the set. (b) Find the value of E F. (c) Prove that: n( E F ) = 11 (d) Are n( E F ) = n( E F )? Justify. Solution: Here, U = {1, 2, 3, ...., 14}, E = {2, 4, 6, 8, 10}, F = {1, 2, 3, 4, 6, 12} (a) Cardinality of a set is the number of elements or members of the set. (b) E F = {2, 4, 6, 8, 10} {1, 2, 3, 4, 6, 12} = {2, 4, 6} n(E F) = 3 (c) Here, n( E F ) = 11 LHS = n( E F ) = n(U) – n(E F) = 14 – 3 = 11, RHS, proved. (d) Yes, they are equal. LHS = n( E F ) = n(U) – n(E F) = 14 – 3 = 11 RHS = n( E F ) Now, E F = (U – E) (U – F) = ({1, 2, 3, ..., 14} – {2, 4, 6, 8, 10}) ({1, 2, 3, ..., 14} – {1, 2, 3, 4, 6, 12}) = {1, 3, 5, 7, 9, 11, 12, 13, 14} {5, 7, 8, 9, 10, 11, 13, 14} = {1, 3, 5, 7, 8, 9, 10, 11, 12, 13, 14} n( E F ) = 11, proved. 4. The relation of sets U, A, B and C are shown below in a Venn diagram. (a) Write the cardinality of A B C. (b) Find the value of A – B. (c) Prove that: A (B C) = (A B) (A C) (d) Is n(A B C ) + n( A B C ) = n(U)? Justify with reason. Solution: (a) n(A B C) = 1 (b) A – B = {2, 3, 5, 7} – {1, 3, 5} = {2, 7} (c) LHS = A (B C) = {2, 3, 5, 7} ({1, 3, 5} {2, 3, 4}) = {2, 3, 5, 7} {3} = {2, 3, 5, 7} RHS = (A B) (A C)
38 O E U 2 8 4 10 1 5 6 11 9 3 7 = ({2, 3, 5, 7} {1, 3, 5}) ({2, 3, 5, 7} {2, 3, 4}) = {1, 2, 3, 5, 7} {2, 3, 4, 5, 7} = {2, 3, 5, 7} Hence, LHS = RHS, proved. (d) Yes, they are equal. From Venn diagram. n( A B C ) + n(A B C) = n() or, 1 + 6 = 7 or, 7 = 7 Because, all means like and dislike. 5. The relation of the sets U, A and B are shown in the given Venn diagram. (a) Write the cardinality of the set A. (b) Find the value of n(A B). (c) Prove that: n( A B ) = n( A B ) (d) In what condition n(A B) = n(U)? Justify with reason. Solution: (a) From the given Venn diagram, A = {2, 3, 5, 7} n(A) = 4 (b) A B = {1, 2, 3, 5, 7} n(A B) = 5 (c) A B = {3, 5} A B = {1, 2, 6, 7} n( A B ) = 4 A B = {6} n( A B ) = 1 A = {1, 6}, B = {2, 6, 7} A B = {1, 2, 6, 7} n( A B ) = 4 Hence, n( A B ) = n( A B ), proved. (d) When n( A B ) = 0, then n(A B) = n(U). 6. Two sets O and E are as given below in the description method. O = {Set of odd numbers less than 11}, E = {Set of even numbers less than 11} (a) Write the cardinality of set O. (b) Show the above information in a Venn diagram. (c) Prove that: O – E ≠ E – O but n(O – E) = n(E – O) (d) Here, O – E = O? Is it always true? Why it so? Solution: Here, O = {1, 3, 5, 7, 9}, E = {2, 4, 6, 8, 10} (a) n(O) = 5 (b) Showing the above information in a Venn diagram, (c) O – E = {1, 3, 5, 7, 9} – {2, 4, 6, 8, 10} = {1, 3, 5, 7, 9} E – O = {2, 4, 6, 8, 10} – {1, 3, 5, 7, 9} = {2, 4, 6, 8, 10} Hence, O – E E – O. But, n(O – E) = 5 and n(E – O) = 5 Hence, n(O – E) = n(E – O), proved. (d) Yes, O – E = O. Yes, it is always true. When they are disjoint sets.
39 UNIT II ARITHMETIC CHAPTER TAX 2 2.1 Income TAX PRACTICE 2.1 Read / Understand / Think / Do Keeping Skill Sharp 1. (a) Write the formula to calculate the taxable income when total income and allowance are given. (b) Write the relation between the net income, income tax and total income. Solution: (a) Taxable income = Total income – allowance (b) Net income = Total income – income tax 2. (a) The annual income of a girl is Rs. 5,50,000. If she pays Rs. 55,000 annually in Citizen Investment Trust, what is her taxable income ? (b) The annual income of the staff of a commercial bank is Rs. 8,35,600. If he/she pays 1% of social security tax, how much tax should he/she pay? (c) A man earns Rs. 5,24,000 and tax allowance is Rs. 4,00,000. Find his income tax at 10% p.a. Solution: (a) Given, Annual income = Rs. 5,50,000 Annual payment in citizen investment trust = Rs. 55,000 Taxable income = Annual income – Annual payment in CIT = Rs. 5,50,000 – Rs. 55,000 = Rs. 4,95,000 (b) Here, Amount income of the staff = Rs. 8,35,600 Rate of social security tax = 1% Tax amount he should pay = 1% of Rs. 8,35,600 = 1 100 × Rs. 8,35,600 = Rs. 8,356 (c) Here, Income = Rs. 5,24,000 Tax allowance = Rs. 4,00,000
40 Tax rate = 10% p.a. Taxable income = Income – Tax allowance = Rs. 5,24,000 – Rs. 4,00,000 = Rs. 1,24,000 His income tax = 10% of taxable income = 10 100 × Rs. 1,24,000 = Rs. 12,400 3. (a) The royalty of a book author is Rs. 354000. If he pays 10% of income tax, how much tax should he pay? (b) A man pays Rs. 39810 as a income tax to the government. If his income is Rs. 265400, what is the rate of income tax? (c) The monthly salary of a married civil servant is Rs. 38,900. If 15% tax is levied on the yearly income more than Rs. 4,50,000, how much tax should he pay ? Solution: (a) Here, Royalty of book author = Rs. 3,54,000 Tax rate = 10% Tax amount = 10% of Royalty = 10 100 × Rs. 3,54,000 = Rs. 35,400 (b) Here, Tax amount = Rs. 39,810 Income = Rs. 2,65,400 Tax rate = ? We know, Tax rate = Tax amount Income or Taxable Amount × 100% = Rs. 39810 Rs. 265400 × 100% = 15% (c) Given, Monthly salary = Rs. 38,900 Yearly salary = Rs. 38,900 × 12 = Rs. 4,66,800 Tax free income = Rs. 4,50,000 Taxable income = Yearly income (salary) – Tax free income = Rs. 4,66,800 – Rs. 4,50,000 = Rs. 16,800 Tax rate = 15% Tax amount = Tax rate of taxable income = 15% of Rs. 16,800 = 15 100 × Rs. 16,800 = Rs. 2,520 Check Your Performance 4. (i) The monthly income of an unmarried man is Rs. 55,980. Then, (a) Define tax. (b) Find his yearly income. (c) How much does he pay income tax if the rates are used from the table? Band Tax for residential individual Rates 1st First Rs. 5,00,000 (Social Security tax, SST) 1% *
41 2nd From Rs. 5,00,001 to Rs. 7,00,000 10% 3rd From Rs.7,00,001 to Rs.10,00,000 20% 4th From Rs.10,00,001 to Rs.20,00,000 30% 5th Remaining above Rs. 20,00,000 36% ** (d) Find his net income after paying tax. (ii) The monthly salary of a girl is Rs. 72,082. 10% rebates in the income tax for female. Answer the questions of the equation 4. i). (a) What is income tax ? Define it. (b) Find his assessable income. (c) How much does he pay income tax if the rates are used from the table of Q. No. 4. i)? (d) Find his net income after paying tax. Solution: (i) (a) A compulsory contribution to state revenue, levied by the government on all types of employee's income is called tax. (b) Given, Monthly income = Rs. 55,980 Yearly income = Rs. 55,980 × 12 = Rs. 6,71,760 (c) For 1st Rs. 5,00,000 Tax rate = 1% Hence for 1st 5,00,000 Tax amount = 1% of Rs. 5,00,000 = 1 100 × Rs. 5,00,000 = Rs. 5,000 Remaining income = Rs. 6,71,760 – Rs. 5,00,000 = Rs. 1,71,760 Now, Tax rate for second 2 lakh = 10% Tax amount for Rs. 1,71,7600 = 10% of Rs. 1,71,760 = 10 100 × Rs. 1,71,760 = Rs. 17,176 Hence total tax = Rs. 5,000 + Rs. 17,176 = Rs. 22,176 Alternative Rs. 6,71,760 = Rs. 5,00,000 + Rs. 1,71,760 = 1% of Rs. 5,00,000 + 10% of Rs. 1,71,760 = Rs. 5000 + 17,176 = Rs. 22,176 (d) His net income after paying tax = Total income – Tax amount = Rs. 6,71,760 – Rs. 22,176 = Rs. 6,49,584 (ii) Monthly salary of girl = Rs. 72,082 Rebate = 10% (a) The tax levied directly on personal income above a certain minimum amount fixed by the government is called income tax. (b) Here, her assessable income = Yearly income = Rs. 72,082 × 12 = Rs. 8,64,984 (c) For the first Rs. 5,00,000 Tax amount = 1% of Rs. 5,00,000 = 1 100 × Rs. 5,00,000 = Rs. 5,000
42 For next 2 lakh, Tax amount = 10% of Rs. 2,00,000 = 10 100 × Rs. 2,00,000 = Rs. 20,000 For remaining (Rs. 8,64,984 – Rs. 7,00,000) i.e. Rs. 1,64,984 Tax amount = 20% of Rs. 1,64,984 = 20 100 × Rs. 1,64,984 = Rs. 32,996.80 Total tax amount = Rs. 5,000 + Rs. 20,000 + Rs. 32,996.88 = Rs. 57,996.80 Since she gets 10% rebates in the income tax being female, Tax amount with 10% rebate = Rs. 57,996.80 – 10% of Rs. 57,996.80 = Rs. 57,996.80 – 10 100 × Rs. 57,996.80 = Rs. 57,996.80 – Rs. 5,799.68 = Rs. 52,197.12 She pays Rs. 52,197.12. (d) Her net income = Assessable income – Tax amount with rebate = Rs. 8,64,984 – Rs. 52,197.12 = Rs. 8,12,786.88 5. (i) Mrs. Arpana is a section officer of Nepal Electricity Authority. Her monthly salary is Rs. 43,689. She gets Rs. 2,000 for dearness allowance in 12 months only and 24 months salary in a year. She deposits 15% of his yearly income in Citizen Investment Fund (CIF). Female rebates 10% on the tax liability. (a) Define assessable income. (b) Calculate her assessable income. (c) How much does she pay net income tax by using the tax slab table? Slab Tax for residential Couple Rates 1st First Rs. 6,00,000 (SST) 1% * 2nd Next Rs. 2,00,000 10% 3rd Next Rs. 3,00,000 20% 4th Next Rs. 9,00,000 30% 5th Remaining above Rs. 20,00,000 36% ** (d) How much percent tax does she pay in total ? Find it. (ii) Mr. Sharma is a deputy secretary of Kathmandu Upatyaka Khanepani Limited (KUKL). His monthly salary is Rs. 48,737. He gets Rs. 2,000 for dearness allowance in each month and 18 months net salary in a year. He deposits 10% of his assessable income in Citizen Investment Fund (CIF). He is married person. (a) Define taxable income. (b) Calculate his taxable income. (c) How much does he pay net income tax by using the above tax slab table? (d) How much percent tax does she pay in total ? Find it. Solution: (i) (a) All of the income earned each year that is subjected to taxation is called assessable or income.
43 (b) Her monthly salary = Rs. 43,689 Her yearly income for 24 months = Rs. 43,689 × 24 = Rs. 10,48,536 Dearness allowance for 12 months = Rs. 2000 × 12 = Rs. 24,000 Her assessable income = Yearly income + dearness allowance = Rs. 10,48,536 + Rs. 24,000 = Rs. 10,72,536 Money deposited on CIF = 15% of yearly income = 15 100 × Rs. 10,72,536 = Rs. 1,60,880.40 Tax free income = Money deposited on CIT = Rs. 1,60,880.40 Taxable income = Assessable income – Amount deposited on CIF = Rs. 10,72,536 – Rs. 1,60,880.40 = Rs. 9,11,655.60 (c) Tax amount for 1st Rs. 6,00,000 = 1% of Rs. 6,00,000 = 1 100 × Rs. 6,00,000 = Rs. 6,000 Tax amount for next Rs. 2,00,000 = 10% of Rs. 2,00,000 = 10 100 × Rs. 2,00,000 = Rs. 20,000 Tax rate for remaining (Rs. 9,11,655.6 – Rs. 8,00,000) i.e. Rs. 1,11,655.60 = 20% of Rs. 1,11,655.60 = 20 100 × Rs. 1,11,655.60 = Rs. 22,331.12 Total tax amount = Rs. 6,000 + Rs. 20,000 + Rs. 22,331.12 = Rs. 48,331.12 Rebate amount = 10% of total tax amount = 10 100 × Rs. 48,331.12 = Rs. 4,833.11 Hence net income tax paid = Total tax amount – Rebate = Rs. 48,331.12 – Rs. 4,833.11 = Rs. 43,498 (d) Paid percent tax = Net income tax Taxable income × 100% = Rs. 43498 Rs. 911655.60 × 100% = 4.77% (ii) (a) Taxable income is the amount of income subject to tax, after deductions and exemptions. (b) Rs. Sharma's monthly salary = Rs. 48,737 His yearly income for 18 months = Rs. 48,737 × 18 = Rs. 8,77,266 Dearness allowance per month = Rs. 2,000 Total dearness allowance in 1 year = Rs. 2,000 × 12 = Rs. 24,000 His assessable income = Yearly income + Yearly dearness allowance = Rs. 8,77,266 + Rs. 24,000 = Rs. 9,01,266 Amount deposited on CIF = 10% of assessable income = 10 100 × Rs. 9,01,266 = Rs. 90,126.60 Tax free income = Amount deposited on CIF = Rs. 90,126.60
44 Here, Taxable income = Assessable income – Tax free income = Rs. 9,01,266 – Rs. 90,126.60 = Rs. 8,11,139.40 From above table, Tax amount for 1st Rs. 6,00,000 = 1% of Rs. 6,00,000 = 1 100 × Rs. 6,00,000 = Rs. 6,000 Tax amount for next Rs. 2,00,000 = 10% of Rs. 2,00,000 = 10 100 × Rs. 2,00,000 = Rs. 20,000 Tax amount for remaining (Rs. 8,11,139.40 – Rs. 8,00,000) i.e. Rs. 11,139.40 = 20% of Rs. 11,139.40 = 20 100 × Rs. 11,139.40 = Rs. 2,227.88 (c) Total tax amount paid = Rs. 6,000 + Rs. 20,000 + Rs. 2,227.88 = Rs. 28,227.88 (d) Total paid % of tax = Total tax amount Assessable income × 100% = Rs. 28227.88 Rs. 901266 × 100% = 3.13% 6. (i) Amar is a unmarried newly secondary class joint secretary of Ministry of Finance. His monthly salary with dearness allowance is Rs. 58,787. He gets one month salary for expense of festival at once. 10% of his monthly salary is deposited in Employees' Provident Fund (EPF) and Rs. 3,300 in life insurance in each month. The government deposits the same EPF amount in the fund. Dearness allowance is Rs. 2,000. (Use the tax rates in this book.) (a) What is assessable income ? Define it on the basis of Amar's income. (b) Find his yearly total assessable income. (c) Find taxable income of Amar. (d) How much income tax does he pay in total ? Find it. (ii) Sunil is a married fifth class heavy driver of Ministry of Home Affairs. His monthly salary with dearness allowance is Rs. 43,561. He gets one month salary for expense of festival at once. 10% of his monthly salary is deposited in Employees' Provident Fund (EPF) and Rs. 1,000 in life insurance in each month. The government deposits the same EPF amount in the fund. (Use the tax rates of this book.) Answer the similar questions in Q. No. 6. (i). Solution: (i) (a) The total income earned during a year with salary and deposited amount in EPF is called an assessable income. (b) Given, monthly salary of Amar with dearness allowance = Rs. 58,787 Monthly salary without dearness allowance = Rs. 58,787 – Rs. 2,000 = Rs. 56,787 Yearly income = Rs. 56,787 × 12 = Rs. 6,81,444 Amount of dearness allowance in 1 year = Rs. 2,000 × 12 = Rs. 24,000 Yearly deposited amount of EPF by government = 10% of Rs. 6,81,444 = 10 100 × Rs. 6,81,444 = Rs. 68,144.40 Yearly amount for life insurance = Rs. 3,300 × 12 = Rs. 39600
45 His yearly total assessable income = Yearly salary + Yearly allowance + Festive expense amount + Amount for EPF + Yearly life insurance = Rs. 6,81,444 + Rs. 24,000 + Rs. 56,787 + Rs. 68144.40 + Rs. 39,600 = Rs. 8,69,975.40 (c) Total EPF amount = Rs. 68144.40 × 2 = Rs. 136288.80 Total tax free income = Amount for life insurance + Total EPF amount = Rs. 39600 + Rs. 136288.80 = Rs. 1,75,888.80 (Now 1 3 of assessable income = 1 3 × Rs. 8,69,975.40 = Rs. 2,89,991.80, which is less than Rs. 3,00,000 Here, the sum of life insurance amount and total EPF amount is less than 1 3 of the total assessable income.) Hence the actual tax free amount = Rs. 1,75,888.80. Taxable income = Yearly total assessable income – actual tax free amount = Rs. 8,69,975.40 – Rs. 1,75,888.80 = Rs. 6,94,086.60 (d) Amar is unmarried person, From tax band table, Total tax amount = 1% of Rs. 5,00,000 + 10% of Rs. (6,94,086.60 – 5,00,000) = 1 100 × Rs. 5,00,000 + 10 100 × Rs. 1,94,086.60 = Rs. 5,000 + Rs. 19,408.66 = Rs. 24,408.66 He pays Rs. 24,408.66 tax in total. (ii) (b) Given, Monthly salary of Sunil with dearness allowance = Rs. 43,561 Monthly salary without dearness allowance = Rs. 43,561 – Rs. 2,000 = Rs. 41,561 Yearly salary = Rs. 41,561 × 12 = Rs. 4,98,732. Total amount of dearness allowance in 1 yr = Rs. 2000 × 12 = Rs. 24,000. Yearly deposited amount of EPF by government = 10% of Rs. 4,98,732 = 10 100 × Rs. 498732 = Rs. 49873.20 Yearly amount for life insurance = Rs. 1,000 × 12 = Rs. 12,000 Her yearly assessable income = Yearly salary + Yearly allowance + Festive expense + Amount for EPF + Yearly life insurance = Rs. (4,98,732 + 24,000 + 41,561 + 49,873.20 + Rs. 12,000) = Rs. 6,26,166.20 (c) Total EPF amount = Rs. 49873.20 × 2 = Rs. 99746.40 Total tax free income = Amount for life insurance + Total EPF amount = Rs. 12,000 + Rs. 99,746.40 = Rs. 1,11,746.40 Now, 1 3 of assessable income = 1 3 × Rs. 6,26,166.20 = 2,08,722.07 which is less than 3,00,000 Here the sum of life insurance amount and total EPF amount is less than 1 3 of the total
46 assessable amount income. Hence actual tax free income = Rs. 1,11,746.40. Taxable income = Yearly total assessable income – actual tax amount = Rs. 6,26,166.20 – Rs. 1,11,746.40 = Rs. 5,14,419.80 (d) Sunil is married, From the tax slab, total tax amount = 1% of Rs. 5,14,419.80 = 1 100 × Rs. 5,14,419.80 = Rs. 5,144.198 ~– Rs. 5,144.20 7. (i) The monthly net salary rate of a married secondary level teacher of 4 grades is Rs. 43,689. S/he gets Rs. 1,456 for one grade, Rs. 2,000 for dearness allowance in every month and one month salary for festival allowance at once. 10% of his/her monthly salary is deposited in Employees' Provident Fund (EPF), 10% of monthly salary in Citizen Investment Fund (CIF) and Rs. 400 in life insurance in each month. The government deposits the same EPF and insurance premium amounts in the related offices. (Use the tax rates in this book.) (a) What is tax allowance ? (b) Find his/her assessable income. (c) Find his/her total income tax. (d) How much amount does s/he get in each month without expense of festival ? (ii) The monthly net salary rate of a married Nayab Subba of 10 grades is Rs. 34,730. S/ he gets Rs. 1,158 for one grade and Rs. 2,000 for dearness allowance in every month. Also, he/she gets 13 months salary with one month salary for festival allowance at once. 10% of his/her monthly salary is deposited in Employees' Provident Fund (EPF), 5% in Citizen Investment Fund (CIF) and Rs. 400 in life insurance in each month. The government deposits the same EPF and insurance premium amounts in the related offices. (Use the tax rates in this book.) (a) Write the relation among assessable income, tax allowance and taxable income. (b) Find his/her total income. (c) Find his/her total income tax. (d) How much percent does s/he pay income tax in total ? Solution: (i) (a) Tax allowance is the amount of something that is permitted, especially within a set of regulations or for a specified purpose. (b) Given, monthly income of secondary level teacher = Rs. 43,689 Amount for 4 grades = Rs. 1456 × 4 = Rs. 5824 Monthly salary with grades amount = Rs. 43,689 + Rs. 5,824 = Rs. 49,513. Yearly salary = Rs. 49513 × 12 = Rs. 5,94,156 Total dearness allowance amount in a year = Rs. 2,000 × 12 = Rs. 24,000 EPF amount deposited = 10% of Rs. 5,94,156 = 10 100 × Rs. 5,94,156 = Rs. 59415.60