CSS Middle Standard “Mathematics” 51 iii. 430, 516, 817 1 10 516 817 43 430 516 1 430 301 516 0 301 1 215 301 215 2 86 215 172 2 43 86 86 0 HCF = 43 iv. 632, 790, 869 Ans: 1 8 790 869 79 632 790 10 632 79 790 0 790 0 HCF = 79 Q 10. Find the L.C.M by division method. i. 16, 24, 40 Ans: LCM = 2 × 2 × 2 × 2 × 3 × 5 = 240 ii. 40, 56, 60 Ans: LCM = 2 × 2 × 2 × 5 × 3 × 7 = 840 iii. 207, 138 Ans: LCM = 2 × 3 × 3 × 23 = 414 iv. 72, 96, 120 Ans: LCM = 2 × 2 × 2 × 2 × 2 × 3 × 3 × 5 = 1440 2 16, 24, 40 2 8, 12, 20 2 4, 6, 10 2 2, 3, 5 3 1, 3, 5 5 1, 1, 5 1, 1, 1 2 40, 56, 60 2 20, 28, 30 2 10, 14, 15 5 5, 7, 15 3 1, 7, 3 7 1, 7, 1 1, 1, 1 2 207, 138 3 207, 69 3 69, 23 23 23, 23 1, 1 2 72, 96, 120 2 36, 48, 60 2 18, 24, 30 2 9, 12, 15 2 9, 6, 15 3 9, 3, 15 3 3, 1, 5 5 1, 1, 5 1, 1, 1
CSS Middle Standard “Mathematics” 52 v. 120, 150, 135 Ans: LCM = 2 × 2 × 2 × 3 × 3 × 3 × 5 × 5 = 5400 vi. 102, 170, 136 Ans: LCM = 2 × 2 × 2 × 3 × 5 × 17 = 2040 Q 11. Find the smallest number which on adding to 19, is exactly divisible by 28, 36, and 45. Ans: Given that LCM of 28, 36 and 45 = x + 19 x = LCM – 19 LCM: LCM = 2 × 2 × 3 × 3 × 5 × 7 = 1260 x = 1260 – 19 = 1241 Q 12. Find the number which divides 167 and 95 leaving 5 as remainder? 9 5 18 167 18 95 162 90 5 5 Hence 18 is required number Q 13. Solve and prove whether each number is a prime number or composite number. i. 33 Ans: 33 is composite because it is divisible by 3 and 11. 11 3 3 33 11 33 33 33 0 0 ii. 111 Ans: 111 is composite because it is divisible by 3 also. 37 3 111 9 21 21 0 2 28, 36, 45 2 14, 18, 45 3 7, 9, 45 3 7, 3, 15 5 7, 1, 5 7 7, 1, 1 1, 1, 1 2 120, 150, 135 2 6, 75, 135 2 30, 75, 135 3 15, 75, 135 3 5, 25, 45 3 5, 25, 45 5 5, 25, 15 5 5, 25, 5 5 1, 5, 1 1, 1, 1 2 102, 170, 136 2 51, 85, 68 2 51, 85, 34 3 51, 85, 17 5 17, 85, 17 17 17, 17, 17 1, 1, 1
CSS Middle Standard “Mathematics” 53 iii. 55 Ans: 55 is composite because it is divisible by 5 and 11. 11 5 5 55 11 55 50 55 5 0 5 0 iv. 70 Ans: 70 is a composite number because it is divisible by 2, 5, 7, 10, 14, 35, 70 also. For example 10 35 14 7 5 2 7 70 2 70 5 70 10 70 14 70 35 70 –70 –6 –5 –70 70 70 0 10 20 0 0 0 10 –20 0 0 v. 317 Prime number because it has no division besides 1 and 317. vi. 18 Ans: 18 is a composite number because it is divisible by 2, 3, 6, 9 also. For example 3 9 6 2 6 18 2 18 3 18 9 18 18 –18 –18 –18 0 0 0 0 vii. 83 is Prime number because it has only the division. viii. 300 Ans: 300 is a composite number because it is divisible by 2, 3, 5, 10, 12, 25, 30, 60, 100, 150, 300 etc. For example: 10 30 300 300 0 ix. 120 Ans: 120 is composite number because it is divisible by 2, 3, 4, 5, 6, 8, 10, 12, 15, 20, 24, 30, 40, 60, 120 x. 88 Ans: 88 is a composite number because it is divisible by 2, 4, 8, 11, 22, 44, 88. xi. 100 Ans: 100 is composite number because it is divisible by 2, 4, 5, 10, 20, 25, 50, 100 xii. 1000 Ans: 1000 is composite number because it is divisible by 2, 4, 5, 8, 10, 20, 25, 50, 100, 200, 250, 500, 1000. xiii. 167 Ans: 167 is a prime number because it has only two divisiors. Model Paper No. 1 Section A Time Allowed: 50 Minutes (Objective Type) Total Marks: 40 Q.1: Fill the circle of correct answer. (40 Marks) 1. {0,1,2,3,4…..} is the set of __________ numbers. Natural Odd Whole Even
CSS Middle Standard “Mathematics” 54 2. If A = {1,2,3} and B = {1,2,3,4,5} then A is __________ subset of B. Proper Improper Super None of these 3. A set having no element is called __________ set. Super Improper Singleter Null 4. Two sets A and B are said to be __________ if they have equal number of elements. Equal Subsets Equivalent None of these 5. If two sets are equal, then they are also: Subset Equivalent Superset None of these 6. If A = {1,2,3,5,7}, then there are __________ possible subsets of A. 16 5 32 8 7. __________ is not Notation of a set. ∪ ∩ < 8. {a} is __________ set: Empty Infinite set singleton None of these 9. The __________ brackets are sometimes called set brackets. Round Square Curly None of these 10. If 30 ÷ 5, then quotient is: 0 30 5 6 11. Name the property 5+7 = 7+5 Commutative property w.r.t addition Commutative property w.r.t multiplication Associative property w.r.t multiplication Associative property w.r.t addition 12. Any number divided by 1 is equal to: 1 0 number itself 7 13. Any number added by zero is equal to: 0 1 number itself 5 14. Whole number starts with: –1 1 0 2 15. 6 × (5–2) = (____ × ____) – 6 × 2 6,5 6,2 6,–2 5,–2 16. 1 0 ------------- 100 100 1 0 50 17. 20 × 50 = __________ 500 100 1000 250 18. 650 130 : 3 2 5 10 19. 37 × 12 = 460 444 424 464 20. 372 + (210 + 100) = 662 672 682 692 21. 35 × (13 + 20) = 3125 1525 1155 1375 22. 137 × (25 – 16) 1233 1163 1353 1237 23. For any z, z + 0 = 0, then __________ is called __________ identity. 0, multiplicative 1, additive 1, multiplicative 0, additive
CSS Middle Standard “Mathematics” 55 24. Two numbers are called __________ if their only common factor is 1: Prime Coprime Compsite None of these 25. The numbers which are multiple of 2 are called __________ number: Even Odd Prime Composite 26. The natural number which has only two divsor 1 and number itself is called _________ number: Prime Composite Even Odd 27. A number exactly divisible by 8 if last __________ digits are either zero or divisible by 8: 1 2 3 4 28. A __________ number can be written as the product of all of its prime factors. Prime Composite Even Odd 29. The HCF of two or more numbers is the __________ of common factors. sum difference product division 30. LCM of 352 and 216 is: 5984 9504 3015 9506 31. 83 is __________ number: negative even prime composite 32. 167 is a/an __________ number, but also a __________ number: even, composite composite, prime odd, composite odd, prime 33. The smallest number which is common __________ of two or more __________ is called LCM. facter, number factor, dividend multiple, factor multiple, number 34. The only even prime number is: 1 2 4 0 35. Which of the following number is divisible by 5: 3156 21352 1623 1535 36. The index notation of 2×2×3×3×3×5 is: 2 2×33×52 2 3×32×51 2 2×32×51 2 2×33×5 37. HCF of 45,75,180 is: 25 15 35 45 38. LCM of 60, 75: 150 200 250 300 39. 254 is a number: Composite Prime Negative Coprime Section B Time Allowed: 2:10 minutes (Subjective Type) Total Marks: 60 Attempt all questions. Each question carries equal marks. Q.1: Find proper and improper subsets of {1,2,3,5,8}? Q.2: Prove the distributive law of multiplication over addition for the following 890,345,100 Q.3: Multiply the following: 3574 × 4592 × 320 Q.4: Solve the following: 5867316 ÷ 9 Q.5: An art gallery has 83542 painting. 32516 painting were damaged due to weather disaster and 1300 were shifted to other branch of gallery. How many painting left in gallery. Q.6: Find the prime factor of 512 by factor tree? Q.7: Factorize the given numbers and express their factors in index notation. Q.8: Find HCF by long division method: 48, 132, 352 Q.9: Find LCM by prime factorization: 340, 720, 480
CSS Middle Standard “Mathematics” 56 Q.10: Find LCM by division method: 300, 175, 525, 25 Q.11: Find least number of students exactly in groups of 348, 64 and 84 to participate in a Seminar. Model Paper No. 2 Section A Time Allowed: 50 Minutes (Objective Type) Total Marks: 40 Q.1: Fill the circle of correct answer. (40 Marks) 1. {1, 2, 3, 4…..} is the set of __________ numbers. Whole Natural Even Odd 2. If S = {a, b, c, d} and T = {a, b, c} then S is __________ of T. Proper subset Improper subset Superset None of these 3. __________ is notation of Empty set. {1} ∪ < 4. If two sets A and B have same and equal number of elements, then they are called __________ sets. Equivalent Equal Subsets Supersets 5. If X = {a, b, c, d}, then there and __________ possible proper subsets of X. 4 8 16 15 6. Two sets A and B are improper subsets of each other if A __________ B. = ≠ 7. Any set has at most __________ improper sets. 3 2 0 1 8. Set is a collection of __________ and objects. Well defined, distinct well defined, similar singleton None of these 9. A set contain __________ number of element is called finite set. indefinite uncountable definite both and b 10. Name the property a × (b × c) = (a × b) × c associative property w.r.t multiplication distributive property of multiplication over addition associative property w.r.t addition distributive property of multiplication over subtraction 11. Any number divided by itself is equal to __________. 1 number itself 1 2 12. a + (–a) = –a + a = 1 2a 0 a 13. If 20 ÷ 10, then quotient is: 0 10 20 2 14. 131 × 25 = __________ 3255 2375 3275 3165 15. 1 70 35 70 35 2 4 16. (a + b) + c = a + (_____ + _____) b, a a, c a, b b, c
CSS Middle Standard “Mathematics” 57 17. 360 40 6 12 9 24 18. (25 + 12) × 3 = __________ 121 311 111 141 19. For any z, z × 1 = 1 × z = z then: 0, multiplicative 1, multiplicative 1, additive 0, additive 20. The numbers which is not multiple of 2 called __________ number. prime odd even composite 21. A natural number which has more than 2 factors is called __________ number: prime odd composite even 22. (63 – 40) × 10 = 260 630 400 230 23. 36 × (20 × 10) = 3600 7200 6200 5200 24. A number is exactly divisible by 3 if the __________ of digits is divisible by __________. product, 6 differences, 3 product, 3 sum, 3 25. Which of following number is divisible by 8. 9004 1114 5300 20712 26. Factor tree continue till we have a __________ of prime number. column product row none of these 27. L.C.M is number which is completely __________ by the given number. add subtract multiple divisible 28. __________ numbers are those numbers whose common factor is only 1. prime odd even co-prime 29. The index notation of 2 × 3 × 3 × 4 × 4 × 4 is: 2 2 × 32 ×42 2 2 × 33 × 44 2 × 32 × 43 2 3 × 31 × 42 30. LCM of 70, 98, 175 is: 2350 2050 2130 2450 31. 7 and 9 are __________ numbers: prime composite even co-prime 32. H.C.F of 300, 75, 45 is: 25 15 30 35 33. 5 and 7n are __________ number: even competitive negative twin prime 34. The factor par of 15 are: 3 and 3 3 and 4 3 and 5 5 and 5 35. __________ is the number which is neither positive nor negative: 1 2 – 1 0 36. HCF can find by __________ methods: 3 4 5 2 37. 13 is a __________ because it is divisible by itself and __________. odd number, 0 prime number, 1 even number, 1 none 38. 143252 _____ 143123: > < = none 39. In the number line, numbers always increase on __________ side: left upper lower right
CSS Middle Standard “Mathematics” 58 Section B Time Allowed: 2:10 minutes (Subjective Type) Total Marks: 60 Attempt all questions. Each question carries equal marks. Q.1: Verify the associative law w.r.t multiplication for following whole number 120, 170, 250. Q.2: Find the proper subsets of {a, b, c, d, e, f}. Q.3: Solve the following: 653892 ÷ 12 Q.4: Multiply the following: 324 × 512 × 132 Q.5: Which of following number are divisible by 15 and why? 352140, 8775, 843692, 732150, 43123, 654321 Q.6: Find the prime factorization of given numbers by repeated division method 7262, 36005, 4836? Q.7: Find the highest number which exactly divides these numbers? 575, 225, 600 Q.8: Find H.C.F by long division method? 145, 540, 675, 765 Q.9: Define the following: (i) Proper subsets (ii) Equal sets (iii) Singleton sets Q.10: Find the increase in the number of people die during 2001 and 2010. The number of people die in 2001 was 132352279 and in 2010 was 207774550? Model Paper No. 3 Section A Time Allowed: 50 Minutes (Objective Type) Total Marks: 40 Q.1: Fill the circle of correct answer. (40 Marks) 1. All the odd number in whole numbers are: empty set finite set infinite set singleton set 2. The set of prime numbers divisible by 2 is __________. empty set singleton set infinite set none of these 3. X is a set multiple of 2, Y is multiple of 4, and Z is a set multiple of 6, which on is true. X Y Y Z Z X 4. A = {a, b, c}, how many proper subset does A haves? 3 2 8 7 5. If A= {1, 2, 3, 4, 6, 8, 9} which one is superset of A: {1, 3, 4, 6, 8, 9} {1, 2, 3, 6, 8, 9, 10} {1, 2, 4, 6, 8, 9, 12} {1, 2, 3, 4, 6, 8, 9, 11} 6. A is a set of factor of 15, which one of following is not a number of A. 3 1 5 2 7. {a, b, c} __________ {1, 2, 3}. = 8. __________ is subset of every set: empty set singleton set empty set None of these 9. __________ is notation for membership. ^ 10. { } has __________ subsets. 2 0 1 3 11. Name the property a × (b + c) = a × b + a × c: associative property w.r.t addition associative property w.r.t multiplication distributive property of multiplication over addition distributive property of multiplication over subtraction 12. __________is the number which is neither negative nor positive. 1 0 2 None of these
CSS Middle Standard “Mathematics” 59 13. Any number multiply by 1 is equal to: 1 0 number itself 5 14. 1 1 a a a a 0 1 a –a 15. 131 ÷ 5 = __________ 25 35 37 27 16. If 120 ÷ 6, the quotient is: 6 120 20 10 17. 540 45 10 5 15 12 19. 0 4 40 1 3 40 1 10 0 20. 4 + 5 = _____ + _____ 4, 3 5, 4 3, 5 5, 3 21. 3 × (4 + 5) + _____ × 4 + 3 × _____ 3, 4 3, 5 4, 5 5, 3 22. a × (b × c) = (a × _____) × c b c a none of these 23. 35 × (4 + 8) = _____ 400 380 420 320 24. (850 + 300) + 250 = _____ 1200 4000 1400 1600 25. Which of following is divisible by 9: 35216 13724 45328 53721 26. If the common factor between any two number is only 1 they are called: composite Even prime co-prime 27. Sum of greatest three digit number and smallest digit number is: 199 1009 1091 1090 30. The factor 39 are __________ 1, 3, 9, 39 3, 13, 39 1, 3, 6, 9, 39 1, 3, 13, 39 31. Which of following is divisible by: 3040 15008 1836 25252 32. HCF of 5450 and 1000 is _____ 5 10 25 50 33. LCM of 120, 144 is __________ 1440 540 720 620 34. The index notation of 5 × 7 × 7 is 5 × 7 5 2 × 7 5 × 72 5 2 × 72 35. LCM of two or more numbers is the __________ of all their common __________: largest, divisor smallest, factor largest, multiple smallest, multiple 36. 0 is __________ number: whole number natural number negative number positive number 37. 125 × _____ = 125: 1 2 3 5
CSS Middle Standard “Mathematics” 60 38. Multiplicative Identity is __________ 0 1 2 none of these 39. If zero is multiplied by any number is equal to __________: number itself 1 0 none of these 40. The concept of sets was given by __________. George James Canter Mill Canter Jone George Canter Section B Time Allowed: 2:10 minutes (Subjective Type) Total Marks: 60 Attempt all questions. Each question carries equal marks. Q.1: Differentiate between Equal and Equivalent set with example? Q.2: Find all possible subsets of following set? X = {g, h, k, l, m, n} Q.3: Verify the distributive property of multiplication over subtraction for following numbers? 384, 560, 480 Q.4: If a = 340, b = 120, C = 50, then prove the associative law of multiplication. Q.5: Solve 6853896 ÷ 14 Q.6: Using divisibility test, find out which of the following is divisibly by 6, 8, 9, 10? And why 573282, 325460, 7146000 Q.7: Find the prime factorization of given number with factor three. 705642 Q.8: Find HCF of following numbers by long division method. 150, 315, 435, 675 Q.9: Find LCM of 120, 144, 160, 180 by prime factorization method? Q.10: The length of five strings are 105m, 125m, 145m, 175m, 225m. Find the distance which covers the five strong completely. Unit No. 4 Integers Lesson # 1 Teacher Objectives: To introduce base, exponent and value. To deduce law of exponents by using the rational numbers. ● Product law. ● Quotient law. ● Power law. ●For zero exponent. ● For exponent negative integer. To explain the concept of power of integer that is (–a)n when n is even or odd integers. To demonstrate the law of exponents to evaluate expression. Learning Outcomes: Student should be able to: Know that ● The natural numbers 1,2,3......, are also called positive integers and the corresponding negative numbers -1,-2,-3...., are called negative integers. ● 0” is an integer which is neither positive nor negative. Recognize integers. Represent integers on number line. Know that on the number line any number lying ● To the right of zero is positive ● To the left of zero is negative, ● To the right of another number is greater ● To the left of another number is smaller. ☻ Know that every positive integer is greater than a negative integer. Know that every negative integer is less than a positive integer. Arrange a given list of integers in ascending and descending order.
CSS Middle Standard “Mathematics” 61 Define absolute or numerical value of a number as its distance from zero on the number line and is always positive. Arrange the absolute or numerical values of the given integers in ascending and descending order. Use number line to display sum of two or more given negative integers, difference of two given positive integers, sum of two given integers. Add two integers (with like signs) in the following three steps: Take absolute values of given integers Add the absolute values Give the result the common sign Add two integers (with unlike signs) in the following three steps. Take absolute values of given integers Subtract the smaller absolute value from the larger. Give the result the sign of the integer with the larger absolute value. Recognize subtraction as the inverse process of addition. Subtract one integer from the other by changing the sign of the integer being subtracted and adding according to the rules for addition of integers. Recognize that the product of two integers of like signs is a positive integer. the product of two integers of unlike signs is a negative integer. Recognize that division is the inverse process of multiplication. Recognize that on dividing one integer by another. If both the integers have like signs, the quotient is positive. If both the integers have unlike signs, the quotient is negative. Know that division of an integer by “0” is not possible. Teacher materials: ● CSS Middle Standard Mathematics Book 6. ● Writing Board ●Marker ● Eraser. Procedure: Greet the students and begin by asking about the natural and whole numbers. Draw the circle given on page # 59 the book on the board and tell the students that the natural numbers 1, 2, 3 are also called positive integers and the correoponding negative numbers -1, -2, - 3 are called negative integers. Ask a student to draw a number line on the board. Extend the line and give it the shape given on page # 60 of the book to clarify the concept of representation of inegers on number line. Invite the class for book reading. Note for the teacher: Focus on all examples given in the book. Always solve few questions on the board by yourself. Then encourage students to solve other questions there. For notebooks work, begin by pair work and finally ending with individual work. Suppose you have an exercise consisting of 5 questions. Each question further consists of 5 parts then your strategy should be: Question No. Done by the teacher Board practice by the students Pair work Individual work Home work 1 Part i Part ii Part iii Part iv Part v Focus on the definitions and ask students to learn the definitions with understanding. Once in a week, homework can be assigned based on the definition given in the chapter / lesson / unit going on in the classroom. Exercise (4a) Q1. Separate positive and negative integers. Ans: Negative integers:
CSS Middle Standard “Mathematics” 62 –8, –339, –44, –91, –635, –435, –1100, –1432, –1072. Positive Integers: 11, 71, 88, 447, 834, 647, 839, 942, 1240, 1342, 4234. Q2. Draw on number line by selecting proper scale and show the integer on it. (i) 10 and 70 Ans: (ii) 45 and 55 Ans: (iii) 150 and 400 Ans: (iv) –200 and –600 Ans: (v) –300 and –200 Ans: (vi) –450 and +300 Ans: Q3. Write the absolute value of the given integers. |–732| = 732 |–4375| = 4375 |–432| = 432 |–5389| = 5389 |472| = 472 |3742| = 3742 |832| = 832 |–2314| = 2314 |941| = 941 |7100| = 7100 |–182| = 182 |–8432| = 8432 |1001| = 1001 |71105| = 71105 |2375| = 2375 |–83942| = 83942 Q4. For the following numbers write ascending and descending order and also arrange the absolute value of integers in descending order. (i) 430, 474, 894, –1140 Ans: Ascending order: –1140, 430, 474, 894 Descending order:894, 474, 430, –1140 Absolute value in descending order:1140, 894, 474, 430 (ii) –110, 730, –940, 940, 344, 7440 Ans: Ascending order: –940, –110, 344, 730, 940, 7440 Descending order: 7440, 940, 730, 344, –110, –940 Absolute value in descending order: 7440, 940, 940, 730, 344, 110 (iii) 8390, 11432, 37184, 74291, –3745, –4752 Ans: Ascending order: –4752, –3745, 8390, 11432, 37184, 74291 Descending order: 74291, 37484, 11432, 8390, –3745 –4752 Absolute value in descending order: 74291, 37184, 11432, 8390, 4752, 3745 Q5. Which one is greater? Ans: –41004, –41104, –4114
CSS Middle Standard “Mathematics” 63 –4114 is greater than other two numbers. Q6. Which one is smaller? Ans: –30001, –3001, –301 –30001 is smaller than other two numbers. Exercise (4b) Q1. Find the sum of following integers on the number line. (i) 7 + 8 Ans: 7+ 8 = 15 (ii) 50 + 30 Ans: 50 + 30 = 80 (iii) 6 + (–10) Ans: 6 – 10 = –4 (iv) 20 + (–10) Ans: = 20 – 10 = 10 (v) (–3) + (–5) Ans: = –3 – 5 = – 8 (vi) (–30) + (–60) Ans: = –30 –60 = –90 Q2. Find the difference of following integers on a number line. (i) (+10) – (+7) = +3 Ans: (ii) (+40) – (+20) = +20
CSS Middle Standard “Mathematics” 64 Ans: (iii) (+200) – (+100) = +100 Ans: Q3. Find the sum of the following integer on number line. (i) –15, –5 Ans:(–15) + (–5) = –20 (ii) –60, –10 Ans: (–60) + (–10) = –70 (iii) –12, –2 Ans: (–12) + (–2) = –14 Q4. Use the number line to find the sum. (i)
CSS Middle Standard “Mathematics” 65 Ans: 4 + (–5) = –1 (ii) Ans:–7 + 2 = –5 Q5. Add following. (i) 923, –713 Ans: 923 + (–713) = 210 (ii) –133, 115 Ans: (–133) + (115) = –18 (iii) –78434, –443 Ans: (–78434) + (–443) = –78877 (iv) –8434, –3474 Ans: (–8434) + (–3474) = –11908 (v) 4444, –474 Ans: 4444 + (–474) = 3970 (vi) –8434, –3474 Ans: (–8434) + (–3474) = –11908 Q6. Find the sum of the following by three step method. (i) +745, –392 Ans: Step - 1: Take the absolute values of the integers |745| = 745, |–392| = 392 Step - 2: Subtract the smaller absolute value from the larger absolute value. 745 – 392 = 353 Step - 3:Number 353 take the positive sign as 743 is larger number and it is positive. Hence (745) + (–392) = 353 (ii) 1496, – 1135 Ans: Step - 1: Take the absolute values of the integers |1496| = 1496, |–1135| = 1135 Step - 2: Subtract the smaller absolute value from the larger absolute value. 1496 – 1135 = 361 Step - 3:Number 361 will be positive sign as 1496 is a larger number and it is positive. Hence (1496) + (–1135) = 361 (iii) –87632, – 92523 Ans: Step - 1: Take the absolute value of the integers. |–87632| = 87632, |–92528| = 92528 Step - 2: Subtract the smaller absolute value from the larger absolute value. 87632 + 92528 = 180160 Step - 3:Number 180160 take the negative sign as both integers have negative signs. Hence (–87632) + (–92528) = –180160
CSS Middle Standard “Mathematics” 66 (iv) –8434, – 3474 Ans: Step - 1:Take the absolute value of the integers. |–8434| = 8434, |–3474| = 3474 Step - 2: Add the smaller absolute value and larger absolute value. 8434 + 3474 = 11908 Step - 3:Number 11908 will take negative sign as both integers have negative signs. Hence (–8434) + (–3474) = –11908 Q7: Ahmad borrows Rs. 300 from his mother. If he borrows another Rs. 900 from her. How much he own her altogether? Show his loan on numbers line. Ans: Rs. 300 + Rs. 900 = Rs. 1200 Exercise (4c) Q1. Subtract the following. (i) 64 from 89 Ans: 89 – 64 = 25 (ii) 1834 from 3397 Ans: 3397 – 1834 = 1563 (iii) –8643 from –9837 Ans:–9837 – (–8643) = –9837 + 8643 = –1194 (iv) –10937 from –35935 Ans:(–35935) – (–10937) = –35935 + 10937 = –24998 (v) –89325 from 93654 Ans: 93654 – (–89325) = 93654 + 89325 = 182979 (vi) –96384 from 118937 Ans:118937 – (–96384) = 118937 + 96384 = 215321 (vii) –65937 from –89563 Ans: –89563 – (–65937) = –89563 + 65937 = –23626 (viii) 80371 from –32596 Ans: –32596 – 80371 = –32596 – 80371 = –112967 Exercise (4d) Q1. Fill the missing integers. (i) Ans: 3 × –6 = –18 (ii) Ans: –15 × ( –2) = 30 (iii) Ans: (–5) × 15 = –75 (iv) Ans: 12 × 9 = 108 (v) Ans: 6 × 4 = 24 (vi) Ans: 7 × –13 = –91 (vii) Ans: –9 × (–1) = 9 (viii) Ans: (–2) × 5 = –10 Q2. Simply. (i) (–7) × (–8) × 1 Ans: Step - 1: |–7| = 7, |–8| = 8 Step - 2: 7 × 8 × 1 = 56 Step - 3: 56, we take positive sign due to rule no. 1 as both integers have same signs. Hence (–7) × (–8) × 1 = 56 (ii) (–8) × 9 × 7 Ans: Step - 1: |–8| = 8, |9| = 9, |7| = 7 Step - 2: 8 × 9 × 7 = 504 Step - 3: 504 take the negative signs as the sign of two integers are unlike Hence (–8) × 9 × 7 = –504 (iii) 3 × (–8) × 6 × 1 Ans: Step - 1: |3| = 3, |–8| = 8, |6| = 6, |1| = 1 Step - 2: 3 × 8 × 6 × 1 = 144
CSS Middle Standard “Mathematics” 67 Step - 3:144 will take negative sign as the signs of two integers are unlike. 3 × (–8) × 6 × 1 = – 144 (iv) (–16) × (–7) × (–9) Ans: Step - 1: |–16| = 16, |–7| = 7, |–9| = 9 Step - 2: 16 × 7 × 9 = 1008 Step - 3:1008 will take the negative sign. (–16) × (–7) × (–9) = –1008 (v) 4 × 0 × 7 Ans: Step - 1: 4 × 0 × 7 = 0 Step - 2: 0 has no sign. (vi) 2 × 3 × (–5) × (–16) × 2 Ans: Step - 1: |2| = 2, |3| = 3, |–5| = 5, |–16| = 16, |2| = 2 Step - 2: 2 × 3 × 5 × 16 × 2 = 960 Step - 3: 960 will take positive sign. Hence 2 × 3 × (–5) × (–16) × 2 = 960 (vii) (–16) × (–4) × (–9) 1 10 Ans: Step - 1: |–16| = 16, |–4| = 4, |–9| = 9, |1| = 1, |10| = 10 Step - 2: 16 4 9 1 10 = 5760 Step - 3: 5760 will take negative sign. Hence, (–16) × (–4) × (–9) 1 10 = –5760 (viii) 15 × (–3) × 14 × (–12) Ans: Step - 1: |15| = 15, |–3| = 3, |14| = 14, |–12| = 12 Step - 2: 15 × 3 × 14 × 12 = 7560 Step - 3: 7560 will take the positive sign Hence 15 × (–3) × 14 × (–12) = 7560 (ix) (–14) × (–5) × 3 × 8 × 10 Ans: Step - 1: |–14| = 14, |–5| = 5, |3| = 3, |8| = 8, |10| = 10 Step - 2: 14 × 5 × 3 × 8 × 10 = 16800 Step - 3: 16800 take positive sign. Hence (–14) × (–5) × 3 × 8 × 10 = 16800 (x) 7 × (–11) × (20) ×10 Ans: Step - 1: |7| = 7, |–11| = 11, |20| = 20, |10| = 10 Step - 2: 7 × 11 × 20 × 10 = 15400 Step - 3: 15400 will take the negative sign as. Hence, 7 × (–11) × 20 × 10 = –15400 Q3. Simply: (i) 15 × 15 Ans: Step - 1: 15 × 15 = 225 Step - 2: 225 will take positive sign. Hence 15 × 15 = 225 (ii) (14) × (–13) Ans: Step - 1: |14| = 14, |–13| = 13 Step - 2: 14 × 13 = 182 Step - 3:182 will take negative sign because both intege5rsw have unlike signs. Hence (14) × (–13) = –182 (iii) –150 × –14 Ans: Step - 1: |–150| = 150, |–14| = 14 Step - 2: 150 × 14 = 2100 Step - 3: 2100 will take the positive sign as both integers have like signs.
CSS Middle Standard “Mathematics” 68 Hence –150 × –14 = 2100 (iv) 130 × 130 Ans: Step - 1: 130 × 130 = 16900 Step - 2: 16900 has positive sign as both integers have like sign. Hence 130 × 130 = 16900 (v) (–70) × (25) Ans: Step - 1: |–70| = 70, |25| = 25 Step - 2: 70 × 25 = 1750 Step - 3: 1750 will take negative sign as both integers have unlike signs. Hence (–70) × 25 = –1750 (vi) –141 × –89 Ans: Step - 1: |–141| = 141, |89| = 89 Step - 2: 141 × 89 = 12549 Step - 3: 12549 has positive sign because both integers have like signs. Hence –141 × (–89) = 12549 (vii) –170 × 800 Ans: Step - 1: |–170| = 170, |800| = 800 Step - 2: 170 × 800 = 136000 Step - 3: 136000 has negative sign as both integers have unlike signs. Hence, –170 × 800 = –136000 (viii) 7539 × –189 Ans: Step - 1: |7539| = 7539, |–189| = 189 Step - 2: 7539 × 189 = 1424871 Step - 3: 1424871 has negative sign, as both integers have unlike signs. Hence 7539 × –189 = –14248171 Exercise (4e) Q.1 Solve it. (i) 55 ÷ 10 = (ii) 400 ÷ 2 = Ans Ans : : 55 ÷ 10 = – 5.5 400 2 200 (iii) 700 ÷ 14 = (iv) 75 + ( 15) = Ans Ans : 700 ÷ 14 = 50 75 15 0 : 9 (v) 10000 ÷ 5 = (vi) 175 ÷ 7 = Ans Ans : : 75 7 10000 ÷ 5 = 2000 1 25 (vii) 128 ÷ (8) = (viii) 128 ÷ 49 = Ans Ans : 128 ÷ (8) = 16 128 49 2.6 : 12 (ix) 364 ÷ 65 = (x) 209556 ÷ 5821 = Ans Ans : : 364 ÷ 65 = 5.6 209556 5821 36 Q.2 Fill the box with suitable integer. (i) –1800 ÷ = –18 (ii) –1400 ÷ –14 = –1800 ÷ = -18 Let unknown number is x, then 180 18 = x x = 100 –1400 –14 = 100
CSS Middle Standard “Mathematics” 69 (iii) ÷ 258 = 236 (iv) ÷ 3 = 888 Let x be the number Let unknown number is x. x ÷ 258 = 236 x ÷ 3 = 888 x = 888 – 3 = 2664 x × 1 236 258 x = 236 × 258 x = 60888 (v) ÷ –1 = 420 (vi) 12500 ÷ = –25 Let unknown number is x, then Let unknown number is x, x ÷ –1 = 420 12500 ÷ x = –25 x = –420 – 12500 25 =x x = –500 (vii) –2063635 ÷ = 58961 Let unknown number is x, then –2063635 ÷ x = 58961 x = 2063635 58961 = –35 Q.3 Find the quotient of the following. (i) –9504 ÷ 4 Ans: 2 3 7 6 4 9 5 0 4 –8 1 5 – 1 2 3 0 – 2 8 2 4 – 2 4 0 Hence –9504 ÷ 4 = –2376 (ii) 96768 ÷ –216 Ans: 4 4 8 216 9 6 7 6 8 –8 6 4 1 0 3 6 – 8 6 4 1 7 2 8 – 1 7 2 8 0 Hence 96768 ÷ –216 = –448 (iii) –243,892 ÷ –253 Ans: 9 6 4 253 2 4 3 8 9 2 – 2 2 7 7 1 6 1 9 – 1 5 1 8 1 0 1 2 – 1 0 1 2 0 Hence –243,892 ÷ –253 = 964 (iv) 335,750 ÷ 850 Ans: 3 9 5 850 3 3 5 7 5 0 – 2 5 5 0 8 0 7 5 – 7 6 5 0 4 2 5 0 – 4 2 5 0 0 Hence 335,750 ÷ 850 = 395 Q.4 440 ÷ 0 is this possible? Ans: No it is not possible because division by 0 does not work as any number divided by 0 is infinity. Q.5 Explain if any number divided by 1 remain the same. Ans: Since 1 is multiplicative identity, therefore any number when multiply by 1 remains the same. Review Exercise 4 Q 1: Choose the correct answer and fill the circle: i. On the left side of zero, on number line, integers are: negative zero whole number positive ii. Natural numbers are also called __________ numbers: whole number counting number integers real numbers iii. Negative numbers added to __________ become integers: natural number counting numbers whole numbers real numbers
CSS Middle Standard “Mathematics” 70 iv. Absolute value is the distance of a number from____________: number line total sum zero 1 v. Integers are denoted by: Z W I F vi. To add two positive numbers, add their absolute values, the result is _______: positive zero negative none of these vii. |–12| =_________: 12 –12 ± 12 –|12| viii. (–1) + (–3) = _________ + 4 –4 ± 4 3 ix. (60) × (–50) =_________: 300 3000 –3000 –300 x. (–60) × (–50) = _________ –185 150 3000 –3000 Q2. Fill the missing integers. (i) (–120) + ___133___ = 13 (ii) __67874__ + (–67894)= –20 (iii) 190 – ____5_____ = 185 (iv) ___–200___ – (–150)= –50 (v) 18 – ____4_____ = 14 (vi) ___334______ (–340)= –6 (vii) __560______ –220= +340 (viii) 9 × ___2_____ = 18 (ix) 4–__94___ = –90 (x) (–1) ×___10___ = –10 (xi) 3 + ____–19____ = –16 (xii) ____0____ + (–4)= –4 (xiii) 4 – ___13____ = –9 (xiv) ___16____ – 17= –1 Q3. Represent on number line to describe each situation. (i) Karachi king loss the match by 50 runs. Ans: (ii) Peshawar Zalmi won the match by 5 runs. Ans: (iii) Anoral ran 1 meter to right. Ans: (iv) Alishba ran 2 meters toward right. Ans: Q4. Draw the following numbers on number line. 0, ±10, ±30, ±50, ±60 Ans:
CSS Middle Standard “Mathematics” 71 Q5. Add –50 and 20 on the number line. Ans: –50 + 20 = –30 Q6. Display the sum of –9 and 8 on a number line. Ans: – 9 + 8 = –1 Q7. Given integers –18, 10, 19, –20, 400, –1, 0, 35, –15. Ans: Step 1:(i) Arrange integer in ascending order. –20, –18, –15, –1, 0, 10, 19, 35 400 Step 2: (ii) Find absolute values of integers. |–18| = 18, |10| = 10, |19| = 19, |–20| = 20, |400| = 400, |–1| = 1, |0| = 0, |35| = 35, |–15| = 15. 18, 10, 19, 20, 400, 1, 0, 35, 15 (iii)Arrange the absolute values of integers in descending order. 400, 35, 20, 19, 18, 15, 10, 1,0 Q8. Add the following. (i) +78945, –38493 Ans: 78945 + (–38493) = 78945 – 38493 = 40452 (ii) –937546, –105935 Ans: –937546 + (–105935) = –937546 – 105935 = –1043481 (iii) 345968, 698324 Ans: 345968 + 698324 = 1044292 (iv) –923540, 564239 Ans: –923540 + 564239 = –359301 Q9. Find the following products. (i) 9378 × 325 9378 325 46890 187560 2813400 3047850 (multiply by 5) (multiply by 20) (multiply by 300) (ii) 10056 × 136 Ans: 10056 136 60336 301680 1005600 1367616 (multiply by 6) (multiply by 30) (multiply by 100) Hence 10056 × 136 = 1367616 (iii) 1235 × 123 Ans: 1235 123 3705 24700 133500 151905 (multiply by 3) (multiply by 20) (multiply by 100) Hence 1235 × 123 = 151905
CSS Middle Standard “Mathematics” 72 Q9 Find the quotient in the following. (i) 98547 ÷ 307 Ans: 321 307 98547 921 644 614 307 307 0 Hence 98547 ÷ 307 = 321 Qoutient = 321 (ii) 338520 ÷ 546 Ans: 620 546 338520 3276 10920 10920 0 Hence 338520 ÷ 546 = 620 Qoutient = 620 (iii) 1032400 ÷ 145 Ans: 7120 145 1032400 1015 174 145 2900 2900 0 Hence 1032400 ÷ 145 = 7120 Qoutient = 7120 Unit No. 5 Simplification Lesson # 1 Teacher Objectives: To introduce perfect square. To explain the method to check whether a number is a perfect square or not. To introduce and explain the following properties of perfect square of a number. The square of an even number is even. The square of an odd number is odd. The square of a proper fraction is loss then itself. The square of a decimal less than 1 is smaller item decimal. To introduce the square root of natural number and its notation. Explain the method to find square root, by division method and factorization method of, natural number, fraction and decimal, which is perfect square. Explain the solution of real life problems involving square roots. Learning Outcomes: Know that the following four kinds of brackets –––– vinculum, –––– vinculum, ( ) parentheses or curved brackets or round brackets, { } braces or curly brackets,
CSS Middle Standard “Mathematics” 73 [ ] square brackets or box brackets, are used to group two or more numbers together with operations. Know the order of preference as, ––––, ( ), { } and [ ], to remove (simplify) them from an expression. Recognize BODMAS rule to follow the order in which the operations, to simplify mathematical expressions, are performed. Simplify mathematical expressions involving fractions and decimals grouped with brackets using BODMAS rule. Solve real life problems involving fractions and decimals. Teacher materials. ● CSS Middle Standard Mathematics Book 6. ● Writing Board ●Marker ● Eraser. Procedure: Intorduce the word BODMAS rule on the board and ask students what they know about it. Encourage them to write down the kinds of brackets they know on the board. Also talk about the order of brackets. Invite students for book reading and examples solving. Note for the teacher: Focus on all examples given in the book. Always solve few questions on the board by yourself. Then encourage students to solve other questions there. For notebooks work, begin by pair work and finally ending with individual work. Suppose you have an exercise consisting of 5 questions. Each question further consists of 5 parts then your strategy should be: Question No. Done by the teacher Board practice by the students Pair work Individual work Home work 1 Part I Part ii Part iii Part iv Part v Focus on the definitions and ask students to learn the definitions with understanding. Once in a week, homework can be assigned based on the definition given in the chapter / lesson / unit going on in the classroom. Exercise (5a) Q1. Simplify the following. (i) 157 – {170 ÷ 10 (144 + 850 – 707)} Ans: = 157 – {170 ÷ 10 (994 – 707)}= 157 – {170 ÷ 10 (287)} = 157 – {170 ÷ 2870} 170 157 2870 17 157 287 45059 17 287 45042 287 (ii) 234740 + {3750 + (570550 ÷ 5)} Ans: = 234740 + {3750 + 114110} = 234740 + 117860= 352600 (iii) 100 – {50 – (10 2 )} Ans: = 100 – {50 – 12} = 100 – 38= 62 (iv) 410000 – {2345 + 1745 (1050 ÷ 10)} Ans: = 410000 – {2345 + 1745(105)} = 410000 – {2345 + 183225}= 410000 – 185570 = 224430 (v) 147484 + [18760 × 3 –{1474 – (875 ÷ 5)}] Ans: = 147484 + [18760 × 3 – {1474 – 175}] = 147484 + [18760 × 3 – 1299}= 147484 + [56280 – 1299] = 147484 + 54981= 202465 (vi) 53474 + [843 × 5 –{1500 ÷ (3000 ÷ 100)}] Ans: = 53474 + [843 × 5 –{1500 ÷ 30}] = 53474 + [843 × 5 – 50}= 53474 + [4215 – 50] = 53474 + 4165= 57639 (vii) 3764564 + [14450 ÷ 100 + { 3987- 274 + (880 ÷ 10)}] Ans: = 3764564 + [14450 ÷ 100 + {3713 + (880 ÷ 10)}] = 3764564 + [14450 ÷ 100 + {3713 + 88}] = 3764564 + [14450 ÷ 100 + 3801] 14450 3764564 3801 100
CSS Middle Standard “Mathematics” 74 289 289 7602 3764564 3801 3764564 2 2 = 3764564 + 3945.5 3768509.5 OR 3768510 Exercise (5b) Q1. Simplify the following. 4 3 4 16 100 (i) + × ÷ 14 20 9 15 130 4 3 4 16 100 + × ÷ 14 20 9 15 130 Ans : 2 3 4 15 10 7 20 9 16 13 = 2 3 4 15 10 7 20 9 16 13 2 1 10 + 7 16 13 416 91 1120 1456 613 1456 10 13 9 25 (ii) 13 + 8 8 10 40 10 13 9 25 13 + 8 8 10 40 Ans : 114 13 9 5 + 8 8 10 8 114 13 36 + 25 8 8 40 114 13 61 8 8 40 114 65 61 8 40 114 4 8 40 57 1 4 10 285 2 20 283 20 3 14 20 1 11 100 1 14 (iii) 12 ÷ 15 2 7 4 3 181 11 100 1 14 15 2 7 4 3 Ans : 181 11 400 7 14 15 2 28 3 181 11 393 14 15 2 28 3 181 11 131 15 2 2 181 11 2 15 2 131 181 11 15 131 23711 - 165 1965 23546 1965 1931 11 1965 14 16 14 300 (iv) 8 150 23 15 500 Ans: 1214 16 14 150 23 15 3 00 5 5 00 1214 16 14 150 23 25 1214 400 322 150 575 1214 78 150 575 1214 575 150 78 13961 234 or 155 59 234 (v) 150 180 15 12 13 + 16 150 180 15 12 29 Ans : 150 180 15 17 150 180 15 17 150 180 32 150 148 2 1 1 18 50 3 (vi) 4 180 1 ÷ 8 8 7 9 40 4 33 1 18 50 7 180 8 8 7 9 40 4 Ans : 33 1 5 7 1 180 2 8 7 4 4 8 33 1 35 180 2 8 7 128 33 1 70 180 8 7 128 33 1 35 180 8 7 64
CSS Middle Standard “Mathematics” 75 33 64 245 180 8 448 33 309 180 8 448 33 80640 309 8 448 33 80331 8 448 1848 80331 448 82179 448 195 or183 448 7 1 3 1 2 7 (vii) 11+ + 180 15 10 3 8 14 3 2 7 541 123 1 2 7 11 10 3 8 14 3 2 Ans : 7 541 123 3 28 147 11 10 3 8 42 7 541 123 96 11 10 3 8 42 7 541 733 11 10 3 56 7 396553 11 10 168 9240 588 1982765 1992593 840 840 13 1 14 1 16 12 13 (viii) 15 + 1 5 6 15 13 12 17 10 7 103 1 14 13 16 12 13 5 6 15 13 12 17 10 7 Ans : 103 1 14 13 16 12 13 1 6 15 13 12 17 10 7 5 103 1 14 13 16 78 6 15 13 12 17 175 103 1 14 13 1248 6 15 13 12 2975 103 1 14 3875 14976 6 15 13 35700 103 1 14 23699 6 15 13 35700 103 1 331786 6 15 464100 103 30940 331786 6 464100 103 300846 21061 661 or16 6 464100 1275 1275 70 2 4 1 1 3 1 (ix) 15 + + 4 8 1 2 100 5 12 5 5 25 5 70 22 100 1 6 53 15 5 100 5 12 5 5 25 Ans : 7 22 25 1 6 53 15 10 5 3 5 5 5 7 22 25 1 6 53 15 10 5 3 5 7 22 25 60 15 10 5 3 5 7 22 15 100 10 5 7 22 500 15 10 5 7 522 15 10 5 150 7 1044 10 1 120 10 Exercise (5c) Q1. Simplify the following. (i) 130.75 × [4.22 + {7.14 × (1.96 ÷ 1.4 × 3.05)}] Ans: =130.75 × [4.22 + {7.14 × (1.4 × 3.05)}] =130.75 × [4.22 + {7.14 × 4.27}] =130.75 × [4.22 + 30.488] =130.75 × 34.7078=4538.045 (ii) 24 180.3 + 123.14 + 8.6 (19.6 ÷ 32.2) 18 144 Ans: 2616 = 180.3+ 123.14 + 8.6 0.609 144 109 = 180.3+ 123.14 + 7.991 6 =180.3+131.131 18.1667 = 293.26
CSS Middle Standard “Mathematics” 76 (iii) 16.29 2.15 + 14.4 ÷ 3.44 (1.8×0.8 0.3) Ans: = 16.29 2.15 14.4 ÷ 3.44 1.44 0.3 = 16.29 - 2.15 + 14.4 ÷ 3.44 -1.14 = 16.29 - 2.15 + 14.4 ÷ 2.3 =16.29 2.15 6.261 =16.29 6.261 2.15 = 20.401 (iv) 14 21 3.21+ 349.682 2.59 + 840.321 5880 +1.203 66 Ans: 1400 = 3.21 349.682 + 2.59 840.321} 5889 1.203 66 = 21.21 3.21 349.682 + 2.59 840.321 5890.203 = 21.21 3.21 349.682 + 842.911 5890.203 = 21.21 3.21 349.682 + 4964939.659 = 21.21 3.21 4965289.3415 = 4965663.44= 4965291 (v) 850.17 14.2× 110.20 + 14.689 1000.1 981.2 Ans: = 850.17 14.2 110.20 + 14.689 1981.3 = 850.17 14.2 110.20 +14.689 1981.3 = 850.17 14.2 124.889 1981.3 = 850.17 14.2 1856.411 = 850.17 26361.0362 = 25510.866 (vi) 334.70 170.0 ÷ 11.1 + 18.0 14.1 110.2 110.1 75 Ans: = 4.463 170.0 11.1+ 18.0 14.1 220.3 = 4.463 170.0 11.1+ 3.9 220.3 = 4.463 170.0 11.1+ 859.17 = 4.463 170.0 870.27 = 4.463 0.1953 = 4.658 (vii) 2 130.5 62.52 20.5 2.92 × 5.6 × 3 14 Ans: 44 = 130.5 62.52 20.5 2.92 5.6 14 = 130.5 62.52 20.5 16.35 3.143 = 130.5 62.52 36.85 3.143 = 130.5 25.67 3.143 = 130.5 80.68 = 211.18 Exercise (5d) Q1. Three cousins shared a sum of money Rs. 10,000, They received on passing the annual exam. First received 1 6 thof it, the second devided 1 4 thof this money, remaining is given to the third one. Find the amount of each cousin. Ans: Total Money = Rs.10,000 Share of 1st cousin = 1 10,000 6 = 5000 3 = Rs. 1666 Share of 2nd cousin = 1 10,000 4 = Rs. 2500 Share of 3rd cousin = 10,000 – (1666 + 2500) = 10,000 – 4166 = Rs.5834 Q.2 Amna gets Rs. 20,000 per week. She saves 5 8 of it. Find (i) How much she saves. (ii) How much she spends. (iii) Her monthly saving. Ans: Total amount of Amna per week = Rs. 20,000
CSS Middle Standard “Mathematics” 77 i) She saves per week = 5 20,000 8 = 12,500 ii) Amna’s spends amount = 20,000 – 12500 = Rs. 7500 iii) Amna’s monthly saving = 4 × 12500 = Rs. 50,000 Q.3 The lawn of a house is 1 12 2 feet long and 1 11 6 feet wide. What is the total area of the lawn? Ans: Length of lawn = 1 12 2 25 feet 5 Width of lawn = 1 11 6 67 feet 6 Area of lawn = length width Area of lawn 25 67 × 2 6 1675 2 feet 12 2 139.583feet Q.4 Ali had to travel from Lahore to Gujranwala which is 70 km. He traveled 9 10 of the distance by train, 1 12 by local bus and remaining on auto-rickshaw. Find the distance traveled by train, bus and auto-rickshaw. Ans: Total distance from Lahore to Gujrawala = 70 km Distance traveled by train = 9 70 10 63km Distance traveled by local bus 1 ×70 = 5.83 km 12 Distance traveled by Rickshaw = (70 – 63 – 5.83) km = 1.17 km Q.5 Ibrahim has to pay zakat of Rs. 20,000. He decided to pay the Zakat to NGO, hospital and widow such as 1 5 of Zakat to NGO and 0.6667 to hospital. What amount of Zakat has been left for widow. Ans: Ibrahim has to pay Zakat = Rs. 20,000 Amount of Zakat to NGO = 1 20,000 5 Rs. 4,000 Amount of Zakat to hospital 0.6667 20,000 Distance traveled by Rickshaw = Rs. 13334 Amount of Zakat to widow = 20,000 – (4,000 + 13334) = Rs. 2666 Q.6 Sidra has Rs. 1300 balance in her mobile. She gave 3 5 toSamia. Samia gave 1 6 to her own share to her friend Mariam. How much balance Mariam got? Ans: Sidra has balance = Rs. 1300 She gave to Samia = 3 1300 5 Rs. 780 Samia gave to Mariam 1 780 6 = Rs. 130
CSS Middle Standard “Mathematics” 78 Review Exercise 5 Q.1 Choose the correct answer and fill the circle. i. 14 1 + is 7 16 3.124 2.1667 2.186 4.423 ii. 5 8 of 40 is: 26 29 25 30 iii. 0.5 ÷ 0.2 is: 3.5 2.5 4.1 1.3 iv. 13.2 – 8.32 is equal to: 11.2 6.88 4.88 none of these v. 1 6 of 10,000 is _______: 500 678.11 589.17 1666.667 vi. 116.5 – 8.91 is: 107.59 113.14 110.89 89.794 vii. 1300 10 is _________: 133.3 130 1300 none of these viii. BODMAS stand for: brackets observation division multiplication addition subtraction brackets order division multiplication addition subtraction all of them none of them ix. [ ] is called _______ brackets: vinculum square brackets braces parenthese x. 15.1 + (14.2 – 13.2) =: 18.2 16.1 15.1 14.2 Q.2 Simplify the followings. (i) 1300 – 24 [150 – {150 ÷ 30 × 40 – (300 + 530 630 )}] Ans: = 1300 – 24 [150 – {150 ÷ 30 × 40 – (300 + 530 630 )}] = 1300 – 24 [150 – {150 ÷ 30 × 40 – (300 + 1160)}] = 1300 – 24 [150 – {150 ÷ 30 × 40 – 1460}]= 1300 – 24 [150 – { –1260}] = 1300 – 24 [150 + 1260]= 1300 – 24 [1410] = 1300 – 33840 = –32540 1 50 250 210 (ii) 135 120 150 250 1 50 250 210 × 135 120 150 250 Ans : 1 5 25 21 × 135 12 15 25 1 5 5 21 × 135 12 3 25 1 5 125 63 × 135 12 75 1 5 62 × 135 12 75 1 31 × 135 90 31 12150 31 12150 = 0.0025
CSS Middle Standard “Mathematics” 79 1 2 1 3 480 (iii) 41 12 11 10 1.44 ÷ 22 890 15 16 80 903 10682 166 163 80 + × + 1.44× 22 890 15 16 480 Ans : 903 10682 166 163 6 + × + 22 890 15 16 25 903 10682 13529 6 + + 22 890 120 25 903 + 12.00 112.98 22 903 100.98 22 41.045 100.98 = – 59.935 11 1 1 340 (iv) 44 6 12 3.778 22 30 6 90 979 181 73 340 3.778 22 30 6 90 Ans : 979 181 73 3.778 3.778 22 30 6 979 181 73 0 22 30 6 979 181 73 22 30 6 979 13213 22 180 12935527 3960 3266.55 (v) 80.792 + [139.49 – {140.78 × 150.0 + (150.4 – 4 10 79 4 . . }] Ans: = 80.792 + [139.49 – {140.78 × 150.0 + (150.4 – 83.5)}] = 80.792 + [139.49 – {140.78 × 150.0 + 66.9}] = 80.792 + [139.49 – {21117 + 66.9}] = 80.792 + [139.49 – 21183.9] = 80.792 – 21044.41= – 20963.618 (vi) 1890 – {1432.10 × 100 + 71.09 ÷ (174 × 1.25)} Ans: =1890 – {1432.10 × 100 + 71.09 ÷ 217.5} =1890 – {1432.10 × 100 + 0.327} =1890 – {143210 + 0.327} =1890 – 143210.327= –141320.327 (vii) 1311.01 + [1410.1 × 22.00 + {11.87 – (180.89 + 17.89)}] Ans: =1311.01 + [1410.1 × 22.00 + {11.87 – 198.78}] =1311.01 + [1410.1 × 22.00 – 186.91] =1311.01 + [31022.2 – 186.91] =1311.01 + [30835.29]= 32146.3 (viii) 1768.142 ÷ [1182.1 + {18.732 – 4.893 + (43.25 ÷ 23.25)}] Ans: =1768.142 ÷ [1182.1 + {18.732 – 4.893 + (43.25 ÷ 23.25)}] =1768.142 ÷ [1182.1 + {18.732 – 4.893 + 1.860}] =1768.142 ÷ [1182.1 + {20.592 – 4.893}]=1768.142 ÷ [1182.1 + 15.699] =1768.142 ÷ 1197.799=1.476 Q.3 A telecom company has 18000 employees 0.5 are in HR, 0.4 are in marketing and 0.06 in quality control. The rest are in IT. Tell how many employees are in IT. Ans: Total employees = 18000 Employees in HR = 0.5 of total= 0.5 × 18000 = 9000 No. of employees in marketing = 0.4 of total = 0.4 × 18000 = 7200 No. of Employees in quality control = 0.06 of total= 0.06 × 18000 = 1080 No. of employess in IT = 18000 – (9000 + 7200 + 1080) = 18000 – 17280 = 720 Q.4 Alishaba bought 2 purses of Rs. 3500 having a discount of 17%. She also bought one suit of Rs. 2700 with discount of 13%. Tell what was the total amount that she paid? Ans: Total price of two purses = Rs. 3500/-
CSS Middle Standard “Mathematics” 80 Discount money in purses = 17 3500 100 = Rs. 595 Alishaba paid for two purses = 3500 – 595 = Rs. 2905 Price of one suit = Rs. 2700/- Discount amount = 13 2700 100 = Rs. 351 Alishaba paid for suit = 2700 – 351 = Rs. 2349 Total amount paid by Alishaba = Rs. 2905+2349 = Rs. 5254 Q.5 Sum of two fractions is 3 15 4 . If one fraction is 6 4 8 find the other fraction.? Ans:Sum of two fraction = 3 15 4 = 63 4 One fraction = 6 4 8 Other fraction = 38 8 Other fraction = 63 38 40 8 = 126 38 8 88 11 8 Q.6 Paid a ticket of Rs. 2100 while traveling from Lahore to Peshawar. The distance between Lahore and Peshawar is 550km. What will be the rate of 1 kilometer. Ans: Price of Ticket = Rs. 2100 Distance between Lahore and Peshawar = 550 km Rate of 550 killometers = Rs. 100 Rate of 1 killometer = 2100 550 = 42 11 = Rs. 3.82 Q.7 Alishaba ate 1 1 2 a pizza, Anoral ate 1 1 2 of pizza and Ibrahim ate 1 2 4 of pizza. How much pizza did they eat altogether? Ans: Alishaba eat pizza = 1 1 2 = 3 2 Anoral eat pizza = 1 1 2 = 3 2 Ibrahim eat pizza = 1 2 4 = 9 4 They eat pizza altogether = 3 3 9 2 2 4 = 21 4 = 1 5 4 Q.8 Ali has a cable wire of length 55.55m. He wants to distribute his wire into five houses equally. Find the length of wire distributed to each house. If the price of each part of wire is Rs. 15.25, find the cost of total wire? Ans: Length of wire = 55.55 m The length of wire distributed to each house = 55.55 m 5 = 11.11 m The cost of total wire = 15.25 × 5 = Rs. 76.25
CSS Middle Standard “Mathematics” 81 Unit No. 6 Ratio Teacher Objectives: To introduce continued ratio and revise the direct and inverse proportion. To explain the unitary and proportion method to solve the real life problems. To explain the solution of real life problems related to time and work using proportion. To explain relation between time and distance. To explain the conversion of units of speed kilometer per hour into meter per second and vice versa. To explain the method to solve variation related problems involving time and distance. Learning Outcomes: Define ratio as a relation which one quantity bears to another quantity of the same kind, with regard to their magnitudes. Know that of the two quantities forming a ratio, the first one is called antecedent and the second one consequent. Know that ratio has no units. Calculate ratio of two numbers. Reduce given ratio into lowest (equivalent) form. Describe the relationship between ratio and fraction. Know that an equality of two ratios constitutes a proportion, e.g., a : b :: c : d, where a, d are known as extremes and b, c are called the means. Find proportion (direct and inverse). Solve real life problems involving direct and inverse proportion. Teacher materials: CSS middle Standard Mathematics Book 6 Writing Board Marker Eraser Procedure: Greet the students and ask them what they know about ratio by using the bubble map. Tell them that the method of compering two queutitties of the same kind and in the same units by division is known as a ratio. Ask them about the quantities which can be compared. Invite them for book reading and focus on reading page # 91 and 92. Invite them for book work till page # 95. Before starting propostion, ask students about it. Tell them that equality of two ratios is called a proportion. Invite them for book reading and exercises. Note for the teacher: Focus on all examples given in the book. Always solve few questions on the board by yourself. Then encourage students to solve other questions there. For notebooks work, begin by pair work finally ending with individual work. Suppose you have an exercise consisting of 5 questions. Each question further consists of 5 parts then your strategy should be: Question No. Done by the teacher Board practice by the students Pair work Individual work Home work 1 Part I Part ii Part iii Part iv Part v Ratio
CSS Middle Standard “Mathematics” 82 Focus on the definitions and ask students to learn the definitions with understanding. Once in a week, homework can be assigned based on the definition given in the chapter / lesson / unit going on in the classroom. Exercise (6a) Q.1 Simplify the following. (i) : 150 70 160 100 Ans: =150 70 : 160 100 = 15 7 : 16 10 = 15 7 ×80 : 80 16 10 = 75:56 (ii) : 13 39 40 80 Ans: = 13 39 ×80 : 80 40 80 = 26 : 39 = 2:3 (iii) 0.450: 0.750 Ans: = 450 750 : 1000 1000 = 450 750 ×1000 : 1000 1000 1000 = 450:750 225 : 375 45:75 9:15 3: 5 (iv) 82:112:142 Ans: 41: 56: 71 (v) 1.4: 2 5 : 1 3 Ans:=14 2 1 : : 10 5 3 = 14 2 1 ×30 : 30 : 30 10 5 3 = 42: 12: 10= 21: 6: 5 (vi) 120 : 240 : 280 Ans: =120 : 240 : 280 60: 120 : 140 30: 60: 70 15: 30: 35 3: 6 : 7 Q.2 Find the ratio between the following and make them as the lowest form. (i) 250 cm, 1m Ans: 250 cm , 100 cm As 1m = 100 cm 250 : 100 25 : 10 = 5 : 2 (ii) 800g to 1kg Ans: 800 g: 1kg As 1kg = 1000g So required ratio is 800 : 1000 = 4 : 5 (iii) 364 maths books to 455 English books Ans: 364:455. 364 4 455 5 4:5 (iv) 2 kilometers, 8 meters and 672 meters. Ans: As 2kilometers = 2000 meters so the required ratio is. 2008: 672 1004: 336 502: 168 251: 84 (v) 4 hours, 8 minutes and 3 hours. Ans: As 1 hour = 60 minutes So 3 hours = 3 × 60 = 180 minutes Note: a:b
CSS Middle Standard “Mathematics” 83 And 4 hours = 4 × 60 = 240 minutes + 8 minutes = 248 minutes Now required ratio is. 248 : 180 124 : 90 62 : 45 Q.3 The ratio of monthly income to the savings in a family is 5:4. If the saving is Rs. 9000 find the income and the expenses. Ans: Income to saving = 5 : 4 Saving = 9000 Income : Saving = 5 : 4 Income 5 Saving 4 Income = 5 9000 4 Income = Rs. 11250 Expenses = Income – saving = 11250 – 9000= Rs. 2250 Q.4 A sum of Eidi is divided between Ibrahim and Hadi in the ratio 4:7. If Hadi’s share is Rs. 6160, find the total money and share of Ibrahim. Ans: Sum of Eidi divided between Ibrahim and Hadi = 4 : 7 Hadi’s Share = Rs. 6160 Ibrahim’s share : Hadi’s share = 4 : 7 Ibrahim's share 4 Hadi's share 7 Ibrahim’s share = 4 6160 7 Ibrahim’s share = Rs. 3520 Total money = 6160 + 3520= Rs. 9680 Q.5 Mosa has notes of Rs. 100, Rs. 50 and Rs. 10. The ratio of these notes is 2 : 3 : 5 respectively and total amount is Rs. 200,000. Find the numbers of notes of each kind. Ans: Total amount = 200000 Ratio of notes of Rs. 100, Rs. 50, Rs. 10 = 2 : 3 : 5 Sum of ratio = 2 + 3 + 5 = 10 Notes of Rs.100 = 2 200000 10 = 40000 Number of notes of Rs. 100 = 400 Notes of Rs. 50 = 3 200000 10 = 60000 Number of notes of Rs. 50 = 1200 Notes of Rs. 10 = 5 20000 10 0 10 0000 Number of the notes of Rs.10 = 10000 Q.6 A certain sum of money is divided among Amna, Fatima and Umer in the ratio 2 : 3 : 4. If Amna’s share is Rs. 25000 find the share of Fatima and Umer? Ans: Ratio among Amna, Fatima and Umer = 2 : 3 : 4 Amna share = Rs. 25000 Sum of share = 2 + 3 + 4 = 9
CSS Middle Standard “Mathematics” 84 Amna’s share = Rs. 25,000 2 9 of total 25000 Total = 9 25000 2 = 112,500 Fatima’s share = 3 112500 9 = Rs. 37,500 Umer’s share = 4 112500 9 = Rs. 50,000 Q.7 The ratio of number of male and female students in a university is 3 : 4 . If there are 16000 female students, find the number of male students? Ans: Ratio of male and female = 3 : 4 Sum of Ratio = 3 + 4 = 7 No. of female students = 16,000 4 7 of female students = 16,000 4 7 × Total = 16000 Total = 16000 × 7 4 = 28,000 No. of male students = 28000 × 3 7 = 12000 Q.8 In a library, the ratio of English books to Math books is 3 : 4. If there are 12000 books of English, what will be the number of Maths books? Ans: Ratio of English to Maths books = 3 : 4 Sum of Ratio = 3 + 4 = 7 No. of English books = 12000 3 7 of total = 12,000 3 7 × total = 12,000 Total = 7 12000 28000 3 No. of Maths books = 28000 4 7 =16,000 Exercise (6b) Q.1 Which of the following proportion are trues? (i) True (ii)False (iii) False (iv) False (v) False (vi) True Q.2 Solve the following. (i) 10 : 2 :: x : 4 Ans: 10 : 2 :: x : 4 10 × 4 = 2 × x 40 = 2x x = 40 2 x = 20 (ii) 15 : 6 :: 5 : x
CSS Middle Standard “Mathematics” 85 Ans: 15 : 6 :: 5 : x 15x = 6 × 5 = 30 x = 30 15 x = 2 (iii) 8 : x :: 2 : 6 Ans: 8 : x :: 2 : 6 2x = 8 × 6 = 48 2x = 48 x = 48 2 x = 24 (iv) x : 10 :: 12 : 2 Ans: x : 10 :: 12 : 2 2x = 10 × 12 2x = 120 x = 120 2 x = 60 Q.3 If 3 men earns Rs. 480 in a day, find how much will 7 men earn in a day? Ans: Men : Rs 3 : 480 7 : x 7 3 = x 480 x = 7 3 × 480 = 7 × 160 = Rs. 1120 Q.4 6 typists working 5 hours a day can type the manuscript of a book in 16 days. How many days will 4 typists take to do the same job, each working 5 hours a day? Ans: Typists: Days 6 : 16 4 : x 6 4 = 16 x x = 6 16 4 x = 24 days Q.5 6 oxen can graze a field in 28 days. How long would 9 oxen take to graze the same field? Ans: Oxen: Days 6 : 28 9 : x 6 9 = 28 x x = 28 9 6 = 18.7 days Q.6 The cost of 16 packets of besan, each weight 900 grams is Rs. 84. What will be the cost of 27 packets of besen, each weigh 900 grams?
CSS Middle Standard “Mathematics” 86 Ans: Packets : Rs 16 : 84 27 : x 27 16 = 84 x x = 84 16 27 = Rs. 142 Q.7 To improve her vocabulary, Sumera is asked to learn 39 words a day from dictionary. If she learn one word in 13 minute,how long will she take to complete her task? Ans: Words : Minut 1 : 13 39 : x 39 1 = 13 x x = 39 × 13 x = 507 minutes Q.8 12 men, working 8 hours a day, complete a piece of work in 10 days. Find the number of men required to complete the same work in 8 days, working 8 hours a day. Ans: Men : Days 12 : 10 x : 8 12 x = 8 10 x = 12 8 10 x = 15 men Q.9 39 persons can repair a road in 12 days, working 5 hours a day. In how many days will 30 persons, working 5 hours a day, complete the work? Ans: Days : Persons 12 : 39 x : 30 12 x = 30 39 x = 39 30 10 4 2 12 5 = 15.6 Q. 10 Food for 150 students in a boarding school is sufficient for 20 days. 100 more students join them, for how many days the food will be sufficient for students. Ans: Students : Days 150 : 20 250 : x 250 150 = 20 x x = 20 25 15 x = 12 days Review Exercise 6 Q.1 Choose the correct answer.
CSS Middle Standard “Mathematics” 87 i. If14kg of pulses cost Rs. 441, what is cost of 22 kg of pulses? Rs. 627Rs. 649 Rs. 671 Rs. 693 ii. If 36 men can do a piece of work in 25 days, in how many days will 15 men do it? 5056 60 72 iii. If 20 men can but a wall of 56 meters long in 6-days. What length of a similar wall be built by 35 man in 3 days. 49 meters 16 meters52 meters 42 meters iv. If a car covers 102km in 6.8 litres of petrol, how much distance will it cover in 242 liters of petrol? 363km 330km 375km 396km v. On a penticular day, 200 US dollars are worth Rs.9666. On that day, how many dollars could be bought for Rs. 5074.65? 115 US dollar 117 US dollar 172 US dollar 105 US dollar vi. If 5 men of 7 women earn Rs. 525 per day, how much would 7 men would earn per day? Rs. 1331 Rs. 735 Rs. 710 Rs. 1420 vii. The 16 bags of washing power, each weighing 1.5kg is Rs. 672. Find the cost of 18 bags of same kind and weight. Rs. 682 Rs. 1128 Rs. 756 Rs. 1000 viii. If 3 persons can weave 168 shows in 14 days, how many shawls will be woven by 8 person in the same number of day? 153 484 727 448 ix. If the cost f transporting 160kg of goods for 125km is Rs.60, what will be the cost of transporting 200kg of goods for the same distance. Rs. 85 Rs. 50 Rs. 75 Rs. 175 x. If 18 dolls cost Rs 630, how many dolls can be bought for Rs 455? 9 11 13 15 xi. If a man earns Rs 805 per week, in how many days he will earn Rs. 1840? 7 days 16 days 19 days 23 days Q.2 If 4A = 5B = 6C, find the ratio of A:B:C. Ans: 4A = 5B = 6C 4A = 5B, 5B = 6C A B = 5 4 , B C = 6 5 A:B = 5:4 B:C = 6:5 A:B:C 5:4 6:5 30:24:20 15:12:10 Q.3 Divid Rs. 43000 into 3 parts such that ratio of A,B, and C is 15:12:10? Ans: Total Money = Rs. 43000 Ratio of A, B, C = 15 : 12 : 10 Sum of ratio = 15 + 12 + 10= 37 Share of A = 43000 37 15 = 17432 Share of B = 43000 37 12 = 13946
CSS Middle Standard “Mathematics” 88 Share of C = 43000 37 10 = 11622 Q.4 A certain sum of money is divided among A,B,C in the ratio 2 : 3 : 4. If A’s share is Rs. 20000, find the share of B and C. Ans: A : B : C = 2 : 3 : 4 Sum of ratio = 2 + 3 + 4 = 9 A’s share = 20,000 9 2 of total money = 20, 000 Total money = 20,000 × 2 9 Total money = Rs. 90, 000 B’s share = 9 3 of total money = 90000 9 3 = Rs. 30,000 C’s share = 9 4 of total money = 9 4 90000 = Rs. 40,000 Q.5 Divid Rs. 150000 among A,B,C in the ratio 3 1 , 4 1 , 5 1 . Ans: A : B : C = 3 1 : 4 1 : 5 1 = 60 3 1 : 60 4 1 : 60 5 1 A:B:C = 20 : 15 : 12 Sum of ratio = 20 + 15 + 12 = 47 A’s share = 150000 47 20 = 63829.7 B’s share = 150,000 47872 47 15 C’s share = 150,000 38297.8 47 12 Q.6 The ratio of number of male and female teachers in a school is 5:4. If there are 16 female teachers, find the number of male teachers. Ans: Ratio of male and female teachers = 5:4 No. of male : No. of female = 5:4 5 = No.of female 4 No. of male No. of male = ×No.of female 4 5 = 16 4 5 = 20 Q.7 In Rs. 166.50 is the cost of 9kg of rice, how much rice can be purchased for Rs. 259. Ans: Rs : kg 166.50 : 9 259 : x 9 x = 166.50 259 x = 9 166.50 259 = 14kg Q.8 In a house of 10 persons food is sufficient for 30 days, 3 persons left the house. For how many days the food would be sufficient?
CSS Middle Standard “Mathematics” 89 Ans: Days : Persons 30 : 10 x : 7 30 x = 7 10 x = 30 7 10 = 42.86 =43 days Q.9 9 hens can lay 9 eggs in 10 days. How many eggs can 4 hens lay in 10 days. Ans: Hens : Eggs 9 : 9 4 : x 9 4 = 9 x x = 9 9 4 = 4, x = 4 eggs Unit No. 5 Financial Arithmetic Teachers Objectives: To explain property tax and general sale tax. To explain the solution the tax-related problems. To introduce profit and markup. To explain the method to find rate of profit and markup per annum. To explain solution of real life problems involving profit markup. To introduce zakat and ushr. To explain solution of problems related to zakat and ushr. Learning Outcomes: Recognize percentage as a fraction with denominator of 100. Convert a percentage to a fraction by expressing it as a fraction with denominator 100 and then simplify. Convert a fraction to a percentage by multiplying it with 100%. Convert a percentage to a decimal by expressing it as a fraction with denominator. 100 and then as a decimal. Convert a decimal to a percentage by expressing it as a fraction with denominator 100 then as a percentage. Solve real life problems involving percentage. Define: selling price and cost price profit, loss and discount profit and loss percentage. Solve real life problem involving profit, loss and discount. Teacher materials. ● CSS Middle Standard Mathematics Book 6 ● Writing Board ● Marker ● Eraser Procedure: Greet the students and introduce the word percentage’ on the board. Ask the students what they know about it. Tell them that a percent is a ratio whose second term is 100. Percent means parts per hundred. Ask the students about the facilitites given by the government to the people. Ask them to think from where the government gets money to make expenses. Talk about the word taxes. Invite them for book reading and exercises. Note for the teacher: Focus on all examples given in the book. Always solve few questions on the board by yourself. Then encourage students to solve other questions there. For notebooks work, begin by pair work finally ending with individual work. Suppose you have an exercise consisting of 5 questions. Each question further consists of 5 parts then your strategy should be: Question No. Done by the teacher Board practice by Pair work Individual work Home work
CSS Middle Standard “Mathematics” 90 the students 1 Part i Part ii Part iii Part iv Part v Focus on the definitions and ask students to learn the definitions with understanding. Once in a week, homework can be assigned based on the definition given in the chapter / lesson / unit going on in the classroom. Exercise (7a) Q1. Convert the following fraction into percentage. (i) 17 20 Ans: Step-1 17 20 100 5 =17×5 = 85% Step-2 85% (ii) 440 100 Ans: 440 100 100 = 440% (iii) 15 8 900 Ans: 15 8 900 = 100 7215 900 = 801.66% (iv) 13 8 Ans: = 100 13 8 = 162.5% (v) 14 4 49 Ans: 14 210 4 49 49 210 100 428 5 9 % 4 . 7 (vi) 155 75 Ans: 100 175 75 = 206.67% (vii) 54 200 Ans: 54 2 00 100 = 27% (viii) 18 4 77 Ans: 18 326 4 77 77 = 100 326 77 =423.38% (ix) 4 50 250 Ans: 4 12504 50 250 250 100 12504 250 = 5001.6% (x) 7 19 Ans: 7 100 19 =36.84% (xi) 2 11 25 Ans: 2 277 11 25 25 = 100 277 25 =1108% (xii) 3 50 Ans: 100 3 50 = 6% Q2. Convert the following percentage to fractions. (i) 7% = 7 100 (ii) 83% = 83 100 (iii) 450% = 450 100 = 9 2 = 1 2 5 1 4 4 10 2 (iv) 47% = 47 100 (v) 33% = 33 100 (vi) 750% = 750 100 = 15 2 1 7 2 4 2 9 8 1 7 2 15 14 5 1 5077
CSS Middle Standard “Mathematics” 91 (vii) 154% =154 100 = 77 50 = 27 1 50 (viii) 210% = 210 100 = 21 10 = 1 2 10 (viii) 574% = 574 100 = 287 50 = 37 5 50 Exercise (7b) Q1. Convert the following from percentage to decimal. (i) 65%= 65 100 = 0.65 (ii) 165%= 165 100 = 1.65 (iii) 390%= 390 100 = 3.9 (iv) 86%= 86 100 = 0.86 (v) 145%= 145 100 = 1.45 (vi) 870%= 870 100 = 8.7 (vii) 114%= 114 100 = 1.14 (viii) 750%= 750 100 = 7.5 (ix) 5455%= 5455 100 = 54.55 (x) 6745%= 6745 100 = 67.45 Q2. Convert the following decimal fractions into percentage. (i) 0.006%= .006 × 100= 0.6% (ii) 8.76%= 8.76 × 100= 876% (iii) 0.76%= 0.76 × 100= 76% (iv) 876.40%= 876.40 × 100= 87640% (v) 0.04961= 0.0491 × 100= 4.961% (vi) 0.74896= 0.74896 × 100= 74.896% (vii) 0.7476= 0.7476 × 100= 74.76% (viii) 0.005= 0.005 × 100= 0.5% Exercise (7c) Q1. Express the following into fractions. (i) 15% of 240 15 240 100 36 (ii) 25% of 500 = 500 100 25 = 125 (iii) 4% of 150 4 150 100 = 6 (iv) 350% of 1000 = 350 100 10 00 = 3500 (v) 17% of 105 17 105 100 = 17.85 (vi) 10% of 1000 = 10 100 10 00 = 100 Q2. Convert the following into percentage. (i) 24 10 Ans: = 100 24 10 = 240% (ii) 2 11 25 Ans: = 2 277 11 25 75 100 277 75 = 1108% 5 50 287 250 37 2 10 20 20 1 21 20 1
CSS Middle Standard “Mathematics” 92 (iii) 3 50 Ans: = 100 3 50 = 6% (iv) 2 8 10 Ans: = 100 2 82 8 10 10 = 820% (v) 6 7 8 Ans: = 100 6 62 7 8 8 = 775% (vi) 7 50 Ans: = 100 7 50 = 14% Q3. Find the rate of percent in all the below questions. (i) 8 out of 32. 8 100 32 = 25% (ii) 44kg of sugar out of 66kg of sugar. = 44 100 66 = 66.67% (iii) Rs. 70 out of Rs. 150. 70 100 150 140 3 = 46.67% (iv) 40cm out of 60 cm 40 100 60 200 3 = 66.67% (v) 90 students out of 150. 90 100 150 = 60% Q4. What percent of 1 day is 46 minutes? Solution: 1 day = 1440 minutes Let the required percent is x% Then, x% percent of 1 day = 46 minutes x% = 100 46 1440 x = 3.2% Q5. What percent of 1 year is 1 day? Solution: 1 year = 364 days Let the required percent is x% Then according to given condition x% percent of 1 year = 1 day x% × 364 days = 1 day x% = 1 364 100 = 0.274 Q6. 5 members were absent out of 22 members in a meeting. Find percentage of absent members? Solution: Total members = 22 No. of absent members = 5 According to given condition 5 members absent out of 22 = 5 22 Percentage of absent member = 50 5 22 100 11 = 250 11 = 22.73% Q7. Ibrahim saves 27% of his pocket money, if he gets Rs.3100 per month. How much he save every month? Solution:
CSS Middle Standard “Mathematics” 93 Ibrahim got pocket money = Rs. 3100 27% of his pocket money = 27% of 3100 = 27 100 3100 = 837 Ibrahim save every month = Rs. 837 Q8. Due to crises in banking sectors, a bank retrenched 97 out of its 700 employees. What percentage of employees was retrenched? Solution: Bank retrenched 97 out of 700 employs = 97 700 Percentage of retrenched employs 97 100 700 = 13.86% Q9. A survey was conducted in which 120 persons were asked which cellular company provides good internet speed. (i) 24 said Jazz (ii)36 said Ufone (iii)54 said Telenor (iv)Rest said they are satisfied with services of Zong. Write these results in percentages. Solution: Persons who said Zong provide good internet speed = 120 – (24 + 36 + 54) = 6 Percentage of Jazz customers 24 100 120 = 20% Percentage of Ufone customers 36 100 120 = 30% Percentage of Telenor customers 54 100 120 = 45% Percentage of Zong customers 6 100 120 = 5% Q10. A company finds that % 1 4 4 of their cars made in the year 2012 to 2016 has some problem in ABS. The company made 28000 cars in this time duration. How many cars were defective? Solution: Cars having problem = 1 4 4 % = 17 4 % Company made the cars = 28000 17 4 % of 28000 = 17 4 % × 28000 17 28000 4 100 = 1190 No. of defected cars = 1190 Exercise (7d) Q1. A book seller purchased 20 dozens of pens at the rate of Rs. 375 per dozen. He sold each one of them at the rate of Rs. 33. What was his percentage profit? Solution: Cost price of 1 dozen pens = Rs. 375 Selling price of 1 pen = Rs. 33 Selling price of 1dozen pens = 12 × 33 = Rs. 396 Profit = 396 – 375 = Rs. 21
CSS Middle Standard “Mathematics” 94 Profit % = Profit 100 Cost price %= 21 100 375 = 5.6% Q2. 100 bananas are bought at the rate of Rs. 350 and sold at the rate of Rs. 48 per dozen. What is percentage profit or lose? Solution: Cost price of 100 bananas = Rs. 350 Cost price of 1 banana = Rs. 350 100 = Rs. 3.5 Selling price of 1 dozen (12) bananas = Rs. 48 Selling price of 1 banana = Rs. 48 12 = Rs. 4 Profit = selling price – cost price = 4 – 3.5 = Rs 0.5 Profit % = Profit 100 Cost price %= 0.5 100 3.5 % = 14.29% Q3. Anoral lost Rs. 28 by selling a hockey for Rs. 322. What was the loss percentage? Solution: Sale price = Rs. 322 Lost = Rs. 28 Cost price = 322 + 28= Rs. 350 Loss % = Loss 100 Cost price %= 28 100 350 % = 8% Q4. A used car was bought for Rs. 150,000 and sold for Rs. 120,000. Find loss percentage? Solution: Loss = cost price – sale price = 150,000 – 120,000= 30,000 Loss % = Loss 100 Lost price % 30,000 100 % 20% 150,000 Q5. At Ramdan bazar the price of dates was reduce by 12% and costs Rs. 440 per kg. What was the original price of dates? Solution: After 12% reduction the price of 1 kg dates 440 Let original price is 10% After reduction price is (100 – 12)%= Rs. 88% 88% of price of dates = Rs. 440 % of price of dates = 440 88 100% of the price of dates = 440 100 88 % Original price of dates = Rs. 500 Q6. During summer mega sale on electricity brand reduce the prices of all its goods by 15%. Calculate the original selling price of the LCD, which was sold for Rs. 23800 during sale? Solution: Let original selling price = 100% During sale 85% of selling price = Rs. 23800 1% of selling price = 23800 85 100% of selling price = 23800 100 85 = Rs. 28000 Original selling price = Rs. 28000
CSS Middle Standard “Mathematics” 95 Q7. Find the marked price of jeans and t-shirt, given the amount of discount. Solution: (i) 17% discount which is Rs. 49 (ii) 33 3 2 % discount which is Rs. 270 (i) Discount = 17% of total = 49 = 17 100 × total = 49 Total = 49 17 × 100 = 288.33 (ii) Discount = 2 33 3 %= 101 3 % 101 3 % of total = Rs. 270 101 300 × total = Rs. 270 Total = 270 300 3 = 801.98 Review Exercise 7 Q 1: Choose the correct answer and fill the circle: i. Which of the following is equal to 4 25 ? 106 8 to 100 16% none ii. 21.8% is equal to: 218 to 100 0.218 0.218 1000 none iii. In a class, Ms. Shazia gave an A grade to 15 out of every 100 students and Ms. Nazia gave an A grade to 3 out of 20 students. What percent of each teacher’s student received an A? 15% 20% 70% none iv. What is percentage of 3 to 20? 35% 15% 30% 40% v. What is the percent of 7 8 ? 78.5 87.5 81.5 86.5 vi. Which of the following is equal to 36%: 9 25 5 8 6 25 none vii. Maria sold a bicycle for Rs. 24900 at a profit of Rs. 600. Find the price at which she bought it. 24300 34000 24250 none viii. Danyal sold a book worth Rs. 850 at a loss of Rs. 180. Find the selling price of book? Rs. 700 Rs. 650 Rs. 670 none ix. If a chair was sold for Rs. 990 at the loss of Rs. 110. Find cost price of chair. Rs. 1200 Rs. 1100 Rs. 1250 Rs. 1150 Q2. If a 20 pound birthday cake of 3 flavours divide into 125 equal pieces. 30 are of chocolate, 20 are of strawberry, the remaining pieces are of coffee in the cake. What is the percentage of coffee flavour in cake? Solution: The pieces of coffee flavor = 125 – (30 + 20)= 75
CSS Middle Standard “Mathematics” 96 Percentage of coffee flavor = 15 4 75 125 100 25 % 60% Q3. In census of a village, there are 9500 people are resident, 7% of them were failed to provide the complete data. Calculate the number of people who give complete data. Solution: People are resident = 9500 Number of people failed to provide data = 7% of total = 7 100 9500 = 665 Number of people give complete data = 9500 – 665 = 8835 Q4. The cost price of 20 Articles is same as the selling price of x articles. If the profit is 25% then what is value of x? Solution: Let cost of 1 article = Rs. 1 Cost of x articles = x Sale price of x articles = 20 Profit = Sale price – Cost price = 20 – x Profit% = Profit 100 Cost price % 25 = 20 – x 100 x 25x = 2000 – 100x 125x = 2000 x = 2000 125 x = 16 Q5. In a medical store the profit is 320% of cost. If the cost increases by 25% but the selling price remain constant, approximately what percentage of selling price is the profit? Solution: Let Cost price be 100% Profit = 320% Profit = Sale price – Cost price Sale price = Profit + Cost price = 320 + 100 Sale price 420 If cost price increased by 25% The cost price = 100 + 25 = 125 Now profit = Sale price – Cost price = 420 – 125= 295 Profit % of selling price = 295 100 420 = 70.23% Q6. What decimal of an hour is a second? Solution: 1 hour = 60 minutes 1 minute = 60 seconds So 1 hour = 3600 seconds Decimal of an hour in second = 1 3600 = 0.000278 Q7. Fill in the blanks? (i) Cost price = Rs. 2400 Profit = Rs. 400, Sale price = ?, Loss = ?
CSS Middle Standard “Mathematics” 97 As Profit = Sale price – Cost prie Sale price = Cost price + Profit = 2400 + 400 Sale price = 2800, Loss = 0 (ii) Cost price = Rs. 1900 Loss = Rs. 300, Sale price = ? Loss = Cost price – Sale price Sale price = Cost price – Loss= 1900 – 300 Sale price = 1600, Profit = 0 (iii) Sale price = Rs. 2900 Profit = Rs. 100, Cost price = ? Profit = Sale price – Cost price Cost price = Sale price – Profit = 2900 – 160 Cost price = 2800, Loss = 0 (iv)Sale price = Rs. 1590 Loss = Rs. 60 Loss = Cost price – Sale price 60 = Cost price – 1590 Cost price = 60 + 1590 Cost price = 1650, Profit = 0 (v) Cost price = Rs. 4100 Profit = Rs. 300, Sale price = ? Profit = Sale price – Cost price Selling price = Profit + Cost price = 300 + 4100 = Rs. 4400 Loss = 0 (vi) Cost price = Rs. 1200 Loss = Rs. 180, Sale price = ? Loss = Cost price – Sale price Sale price = Cost price – Loss = 1200 – 180 = Rs. 1020 Profit = 0 Unit No. 8 Introduction of Algebra Teacher Objectives: To introduce a constant as a symbol having a fixed numerical value. To variable as a quantity which can take various numerical values. To literal as an unknown number represented by an alphabet. To algebraic expression as a combination of constants and variable connected by the sign of fundamental operations. To introduce polynomial as an algebraic expression in which the powers of variables are all whole numbers. To recognize a monomial, a binomial and a trinomial as a polynomial having one term, two terms and three terms respectively. To explain the addition of two or more polynomials. To explain the subtraction of a polynomial from other polynomial. To explain the product of monomial with monomial, monomial with binomial/trinomial and binomial with binomial/trinomial. To explain the simplification of algebraic expression involving addition, subractions and multiplication, subtraction and multiplication. To introduce and explain the verification of algebraic identities such as: (x + a) (x + b) = x2 + (a + b) x + ab, (a + b)2 = (a + b) (a + b) = a2 + 2ab + b2 ,
CSS Middle Standard “Mathematics” 98 (a – b)2= (a – b) (a –b) = a2 – 2ab + b2 , a2 – b2 = (a – b) (a + b). To explain the factorization of an algebraic expression by algebraic identities. To explain the factorization of an algebraic expression by making groups. Learning Outcomes: Explain the term algebra as an extension of arithmetic in which letters replace the numbers. Know that: a sentence is a set of words making a complete grammatical structure and conveying full meaning. sentences that are true or false are known as statements. a statement must be either true or false but not both. a sentence that does not include enough information required to decide whether it is true or false is known as open statement (e.g., D + 2 = 9) a number that makes an open statement true is said to satisfy statement (e.g. D = 7 makes the statement D + 2 = 9 true). use English alphabet x in the open statement D + 2 = 9 to modify it to x + 2 = 9. Define variables as letters used to denote numbers in algebra. Know that any numeral, variable or combination of numerals and variables connected by one or more of the symbols “+” and “ –” is know as an algebraic expression (e.g. x+2y). Know that x, 2y and 5 are called the terms of the expression x+ 2y +5. Know that the symbol or number appearing as multiple of a variable used in algebraic term is called its coefficient (e.g. in 2y, 2 is the coefficient of y). Know that the number, appearing in algebraic expression, independent of a variable is called a constant term (e.g. in x + 2y + 5, number 5 is a constant terms). Differentiate between like and unlike terms. Know that: like terms can be combined to give a single term. addition or subtraction cannot be performed with unlike terms. Add and subtract given algebraic expressions. Simplify algebraic expressions grouped with brackets. Evaluate and simplify an algebraic expression when the values of variables involved are given. Teacher materials. ● CSS Middle Standard Mathematics Book 6 ● Writing Board ● Marker ● Eraser Procedure: Greet the students and encourage them to enlist the advantages of learning mamthematics in the daily life. Ask them about branches of mathematics they know. Tell them that today they will read about that branch of mathematics in which symbols, usually letters of the alphabet, represent numbers or quantitites and express general relationships that hold for all numbers of a specified set. Invite the students for book reading. Note for the teacher: Focus on all examples given in the book. Always solve few questions on the board by yourself. Then encourage students to solve other questions there. For notebooks work, begin by pair work finally ending with individual work. Suppose you have an exercise consisting of 5 questions. Each question further consists of 5 parts then your strategy should be Question No. Done by the teacher Board practice by the students Pair work Individual work Home work 1 Part i Part ii Part iii Part iv Part v Focus on the definitions and ask students to learn the definitions with understanding. Once in a week,
CSS Middle Standard “Mathematics” 99 homework can be assigned based on the definition given in the chapter / lesson / unit going on in the classroom. Exercise (8a) Q1. Tell whether these are sentence or words simply written together. (i) Red rose love. Ans: Simply word together (ii) Thanks God! Ans: Sentence (iii) Six is an odd number Ans: Sentence (iv) Baby sleep Ans: Word together (v) Urdu is our national language Ans: Sentence (vi) Water flows from higher level to lower level Ans: Sentence Q2. Find out whether the following statements are “true” “false” or “an open statement”. (i) 16 × 2 = 11 False (ii) 17 15 False (iii) 17x – 10 = 23 Open (iv) 6 × 4 = 24 True (v) 5 × 5 = 25 True (vi) 8 + 2 = y Open (vii) 4 2 False (viii) x + 14 = 17 Open (ix) (12 13) < 6 True Q3. Put “>” “<”or “=” in the blanks to make sentence true. (i) 7x – 2x > 4x (ii) 13xy + 10xy > 22xy (iii) – 30x2 < 30x2 (iv) 13 × 10 = 130 (v) 18xyz + 23xyz < 173xyz (vi) 180x2y – 140x2y = 40x2 y (vii) 112xy + 114xy > 203xy Exercise (8b) Q1. Write down all the variables, constants and coefficient from the given algebraic expression. (i) x + 2y – 3z. Variables x,y,z, Coefficients 1,2,–3, Constant (ii) 4x + 5y – 4. Variables x,y, Coefficients 4,5, Constant–4 (iii) a + 4b – 3b2 . Variables a,b, Coefficients 1,4, –3, Constant (iv) 3x + 4x2 – 16. Variables x, Coefficients 3,4, Constant –16 (v) 17xy – 14x2+ 17x. Variables x,y, Coefficients 17,–14,17, Constant (vi) 14x2 + 15z – 4x2 . Variables x,z, Coefficients 14,15,–4, Constant (vii) 27 + 16x3 – 14x. Variables x, Coefficients 16,–14, Constant 27 (viii) 15xyz + 3x3 + 1. Variables x,y,z, Coefficients 15,3, Constant 1 Q2. Write down separately the terms of following algebraic expression. (i) 3a + b – 2c Ans:3a, b, –2c (ii) –4xyz – 3xy + z Ans: –4xyz, –3xy, z (iii) 16x3 + 4xy –fgk Ans: 16x3 , 4xy, –fgk (iv) 4x – 1 x + 3x2 Ans: 4x, –1 x , 3x2 (v) abc + 13fgh – af2 – bg2 – ch2 Ans: abc, 13fgh, –af2 , –bg2 , –ch2 (vi) 18 1 x – 14x2 y 2 –kgk2 Ans: 18 1 x , 14x2 y 2 , –kgk2 (vi) –5abc + 7bcd + 3abd Ans: –5abc, 7bcd, 3abd Q3. Find the like terms in the following algebraic expression. (i) x + 3xy + 4x2 – 6xy Ans: 3xy, –6xy (ii) a 2b + ab + b2 a + 2ab42 ab Ans: No like term (iii) a 2 – 2a2b + 4a2b – 5a2 Ans: a2 , 5a2 and –2a2b, 4a2b
CSS Middle Standard “Mathematics” 100 (iv) xyz + x2y + 2xyz + 8x2 y Ans: xyz, 2xyz and x2y, 8x2 y (v) x + y + 4x – 3y Ans: x, 4x and y, 3y (vi) x 2y + 2xy2 + 3x2y + 5xy2 Ans: x 2y, 3x2y and 2xy2 , 5xy2 Q4. Translate each of the following word expressions into an algebraic expression. (i) Five times the number x increased by six. Ans: 5x + 6 (ii) Subtract 29 from half of 1. Ans: 1 2 b – 29 (iii) Sum of number 1 and square of x. Ans: 7 +x2 (iv) Five times a number which is 7 more than p. Ans: 5x = 7 + p (v) Total number of ranges in x crates where each crate contains y orange. Ans: xy Exercise (8c) Q1. Simplify the following by addition. (i) 3x + 4x + 10x + 6y. Ans: = 17x + 6y (ii) 18x2 – 10x2 + 13x – 4x. Ans: = 8x2 + 9x (iii) 3a2y + 4a2y + 10a2b. Ans: = 7a2y + 10a2b (iv) 13x2y – 18x2y + 14x2y – 26x2 y. Ans: = –5x2y – 12x2y= – 17x2 y (v) 103y3 + 10y2 – 100y. Ans: = 103y3 + 10y2 – 100y (vi) ax + bx + cx + dx. Ans: = (a + b + c + d)x (vii) 88xyz + 10xyz – 10xy + 20xy Ans: = 98xyz + 10xy Q2. Find the sum of the following expressions. (i) 3x2 + 4x + 2, 4x2 – 5x + 4, 7x2 + 8x – 1. 3x2 + 4x + 2 4x2 – 5x + 4 7x2 + 8x – 1 14x2 + 7x + 5 (ii) –3abc, +12abc, –7abc. –3abc +12abc –7abc 2abc (iii) 2m2 + mn + n2 , 3m2 – 3mn + 4n2 , –m2 + mn – 2n2 . 2m2 + mn + n2 3m2 – 3mn + 4n2 –m2 + mn – 2n2 4m2 – mn + 3n2 (iv) p 2 + 10pq + 3q2 , 2p2 – 3pq + q2 , 3p2 + pq – 2q2 . p 2 + 10pq + 3q2 2p2 – 3pq + q2 3p2 + pq – 2q2 6p2+ 8pq + 2q2 (v) 17a2 – ab + b2 , 8a2 – 3ab +2b2 , 4a2 + 8ab + 4b2 . 17a2 – ab + b2 8a2 – 3ab +2b2 4a2 + 8ab + 4b2 29a2+ 4ab + 7b2 (vi) 3x4 – 4x2 + 5x + 1, x4+ 2x2– 3x + 4, 4x4– 3x2 + 4x – 5. 3x4 – 4x2 + 5x + 1 x 4+ 2x2–3x + 4 4x4– 3x2 + 4x – 5 8x4 – 5x2 + 6x + 0 Q3. Subtract the following expressions. (i) –4x from x + 1. x + 1 4x 5x + 1 (ii) 5x2 – 4x3 + 4x2 – 3x from 1 (5x2 – 4x2 ) – 4x3 – 3x from 1 9x2 – 4x3 – 3x from 1 _9x2 4x3 3x –9x2+ 4x3 + 3x + 1 4x3 – 9x2 + 3x + 1 (iii) a 2b + 2b2 c + ac2 from 4a2b – 2b2 c + 6ac2 4a2b – 2b2 c +6ac2 _a2b 2b2 c ac2 3a2b – 4b2 c + 5ac2 (iv) x 2 + 4xy from 3x2 + 4xy 3x2 + 4xy _x 2 4xy 2x2 (v) 2x + 3 from –x (vi) 3x4 + 15xy – 13xyz from 7x4 – 10xy +