#class 10 maths chapter 3 solution
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ch 3 maths class 10 ncert solutions
Class 10 Maths Chapter 3, Pair of Linear Equations in Two Variables," features the most important questions strictly aligned with CBSE board exam standards. All the questions are either directly from the NCERT textbook or have been asked in previous board exams. Emphasis is placed on concepts that are frequently tested, ensuring students focus on what is essential for the exams. This study material is perfect for quickly revising key topics and gaining a thorough understanding of the concepts that matter most for success in the CBSE board exams.
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Class 5 math EV chapter 5 creative questions with ans
Class 5 math EV chapter 5 creative questions with ans
Question 01 Four bells rang at once in the beginning and then rang after every 5, 7, 12, 15 seconds. (a) What is the LCM of the time of ringing of first two bells (b) What is general price factor of time of ringing of 3rd and 4th bell? (c) Which smallest number can be divided by the ringing times of the bells without remainder? (d) After what minimum time the bells will ring together again? Solution: (a) 5 = 1x5 and 7 =1 x 7 LCM of 5 and 7 is = 1 x 5 x 7 = 35 (b) 12=2x2x3 and 15=3 x 5 The common prime factor of 12 and 15 is 3 (c) The LCM of the ringing times is required smallest number 3) 5, 7, 12, 15 5) 5, 7, 4, 5 1, 7, 4, 1 LCM of the numbers = 3 x 5 x 7 x 4 = 420 Required smallest number is 420 (d) The LCM of 5, 7, 12, 15 is the required minimum time The minimum time is 420 sec and = minute = 7 minute After 7 minutes the bell will ring together again Question 02 In a class the teacher told the students to write the smallest number of 4 digits and biggest number of 3 digits, whose digit of ones place will be 8. (a) Write the 2 numbers (b) What is the sum and differences of the two numbers? (c) Write the prime factors of the biggest number. (d) What is the LCM of the numbers? (e) What is the GCF of the numbers? Solution: (a) Smallest number of four digits whose digit in the ones place is 8 is = 1008 And, biggest number of three digits whose digit in the ones place is 8 is = 998 (b) Sum of the numbers = (1008 + 998) = 2006 Difference of the numbers = (1008 – 998) = 10 (c) Biggest number is 1008 1008 = 2 x 2 x 2 x 2 x 3 x 3 x 7 The prime factors of 1008 are 2,2,2,2,3,3,7 (d) 998 = 2x499 Prime factors of 998 are 2, 499 The common prime factors of 998 and 1008 is 2 GCF of 1008 and 998 is 2 (e) LCM of 1008 and 998 is =2 x 2 x 2 x 2 x 3 x 3 x 7 x 499 =16 x 9 x 7 x 499 = 502992 LCM of 1008 and 998 is 502992 Creative math questions and solutions for Class 5 EV Chapter 5 to make learning easier and more fun. Question 03 Three values are given, 24, 48, 72. (a) Write down 2 multiples of 24, 48, 72 (b) Write down the common multiples of 24 and 72 (c) Find the L.C.M of 24, 72 (d) Find the L.C.M of 24, 48, 72. Ans. (a) 2 multiples of 24 are = 48, 72 2 multiples of 48 are = 96, 144 2 multiples of 72 are = 144, 216 (b) Multiples of 24 are = 48, 72, 96, 120, 144, 168, 192, 216, 240, 264, 288 Multiples of 72 are = 144, 216, 288, 360, 432 Common multiples of 24 and 72 are = 144, 216, 288, ….. (c) 2)24, 72 2)12, 36 2)6, 18 3)3, 9 1, 3 L.C.M of 24, 72 is = 2 x 2 x 2 x3 x 1 x 3 = 72 (d) 2)24, 48, 72 2)12, 24, 36 2)6, 12, 18 3) 3, 6, 9 1, 2, 3 L.C.M of 24, 48, 72 is = 2 x 2 x 2 x 3 x 2 x 3 x 1 = 144 Question 04 Mr. Amzad brought about 40 mangoes, 85 apples from the market. He distributed the fruits equally. The price of 4 mangoes is 40 taka. (a) What is the price of mangoes? (b) If per piece cost 5 taka how much price will the apples be? (c) He distributed the fruits how many members? (d) Each person gave how many fruit? Answer: (a) The price of 4 piece of mango is 40 taka The price of 1 piece of mango is (40÷4) = 10 taka The price of 40 pieces of mango is (40x10) = 400 taka (b) 1 piece apples cost 5 taka 85 pieces of apples cost (5x85) = 425 taka (c) 40 = 2 x 20 = 2 x 2 x 10 = 2 x 2 x 2 x 5 85 = 5 x17 GCF = 5 He distributed the fruits among 5 person. (d) Each person gave (40 ÷ 5) = 8 piece mango Each person gave (85 ÷ 5) 17 piece apple Easy and fun questions with answers for Class 5 EV Chapter 5 to help students improve their math skills. Question 05 A number of saplings is such that when 3, 5, 6, 8, 10 or 15 are planted in each row, every time two saplings are left out. (a) What is the GCF of last 3 numbers? (b) What is the LCM of the given numbers? (c) what is the minimum number of saplings? (d) If we plant 18 saplings in 16 rows then how many more saplings do we need? Solution: (a) 8 = 1 x 2 x 2 x 2 10 = 1 x 2 x 5 15 = 1 x 3 x 5 GCD of 8, 10 and 15 is =1 (b) 2) 3, 5, 6, 8, 10, 15 3) 3, 5, 2, 4, 5, 15 1, 5, 1, 4, 1, 5 LCM = 2 x 3 x 5 x 4 = 120 (c) The minimum number of sampling is = (120 + 2) =122 (d) Number of extra saplings ={(16 x 8) – 122} =(128 – 122) = 6 Question 06 Tushi set a few no. of bell in their drawing room. After ringing together they rand after every 6, 9, 12, 15 seconds respectively. (a) Find the GCF (b) Find the LCM (c) What is the summation of 4 multiples of 9? (d) When the bells will ring together again? Ans. (a) 6 = 2 x 3 9 = 3 x 3 12 = 2 x 2 x 3 15 = 3 x 5 GCF of 6, 9, 12, 15 is = 3 (b) 3) 6, 9, 12, 15 2) 2, 3, 4, 5 1, 3, 2, 5 LCM of 6, 9, 12, 15 = 3 x 2 x 3 x 2 x 5 x 1 = 180 (c) 4 multiples of 9 are = 9, 18, 27, 36. Sum of 4 multiples of 9 is = 9 + 18 + 27 + 36=90 (d) 3)6, 9, 12, 15 2)1, 3, 2, 5 1, 3, 2, 5 LCM = 3 x 2 x 3 x 2 x 5 x 1=180 The bells of the drawing room will ring again together after 10 seconds. Question 07 16, 24, 2 and 40 are four even numbers. (a) Write the factors of first 2 numbers (b) Write 2 multiples of last wo numbers. (c) Which smallest number when divided by the four number gives 6 as remainder? (d) Find the next number after 4 & 6 which can be divided by 32 without any remainder. Solution: (a) Factors of 16 are = 1, 2, 4, 8, 16 Factors of 24 are = 1, 2, 3, 4, 8, 12, 24 (b) 32 x 1 = 32 32 x 2= 64 The multiples of 32 are 32, 64 and 40 x 2 = 80 (c) The multiples of 40 are 40, 80. 2) 16, 24, 32, 40 2) 8, 12, 16, 20 2) 4, 6, 8, 10 2) 2, 3, 4, 5 1, 3, 2, 5 LCM of 16, 24, 32, 40 is = 2x2x2x3x2x5 = 480 Required smallest number = (480 + 6) = 486 (d) Required number = (486 + 32) -6 = 518 – 6 = 512 Question 08 210 mangoes and lychees are divided among some boys and girls. If the number of mangoes is 60 then. (a) What is the number of lychees? (b) Write the number of mangoes and lychees in prime factors. (c) Among how many maximum number of boys and girls the fruits can be divided equally so that no fruit is left? (d) How many fruits will each one get? (e) How many mangoes and how many lychees will each get? Solution: (a) Number of lychees = (210-60) = 150 (b) Number of mangoes and lychees are 60 and 150 60 = 2 x 2 x 3 x 5 150 = 2 x 3 x 5 x 5 (c) If we divide 60 mangoes and 150 lychees among boys and girls with no fruit being left, then GDS of 60 and 150 is the required maximum number of boys and girls. The common prime factors of 60 and 150 are = 2,3 and 5 Required maximum number of boys and girls is 30 (d) Number of fruits each one will get is = (210 ÷ 30) =7 (e) Number of mangoes each one will get is = (60 ÷ 30)= 2 Number of lychees each one will get is = (150 ÷ 30) = 5 Master Class 5 Math EV Chapter 5 with creative questions and simple answers for better learning. Question 09 The length and breath of a rectangular house is 7.20 meters and 44 decimeters. The floor of the house has to be fitted with marbles so that no marble has to be broken. (a) Express the length of the house in decimeter (b) What is the area of the house in square decimeter? (c) Express the length and breath of the house in prime factors. (d) What is the maximum size of the marble stone needed? (e) How many marble stones are needed for the floor? Solution: (a) Length of the house = (7.20 x 10) decimeter = 72 decimeters (b) Area of the rectangular house =(length x breath) =(72 x 44) square decimeters = 3168 square decimeters (c) Length of the house is 72 decimeters and breath is 44 decimeters. 72 = 8 x 9 = 2 x 2 x 2 x 3 x 3 44 = 2 x 22 = 2 x 2 x 11 (d) The length of the required marble will be equal to the GCF of 72 and 44. GCF of 72 and 44 = 2 x 2 = 4 The maximum length of the required marble stone is 4 decimeters. (e) Are of the square marble stones is =(4 x 4) square decimeters = 16 square decimeters Numbers of marble stones needed for the floor is (3168 ÷ 16) = 198 Question 10 The capacity of holding water for 2 drums are 228 liters and 348 liters. (a) Express the capacity of first drum in prime factors. (b) What are the common prime factors of the capacity of 1st and 2nnd drum? (c) What will be the maximum capacity of a pitcher with which we can fill the drums by pouring water integer numbers of times? (d) How many pitcher of water each drum can hold? (e)How many pitcher of water is totally needed to fill 2 drums? Solution: (a) Water holding capacity of the 1st drum is 228 liter. 228= 2 x 2 x 3 x 19 (b) Water holding capacity of the 2nd drum is 348 liters. 348 = 2 x 2 x 3 x29 The common factors of 228 and 348 are 2, 2,3 (c) The GCF of the capacities of the two drums is the required maximum capacity of the pitcher. GCF of 228 and 348 is = 2 x 2 x 3 = 12 The maximum water holding capacity of the pitcher is 12 liters. (d) Amount of water needed to fill the first drum = (228 ÷ 12) pitchers = 19 pitchers Amount of water needed to fill the 2nd drum = (348 ÷ 12) pitchers = 29 pitchers (e) To fill the two drums total amount of water needed is = (29 + 19) pitchers = 48 pitchers More Questions: Q-11 Three bells having tolled together began to toll after every 9, 12 and 15 minutes. (a) What is done to find out after what minimum time will the bells toll together again? (b) After what time will the bells toll together again? (c) If the bells began tolling after every 6, 9 and 12 minutes, after what minimum times will the bells toll together again? Q-12
100 mangoes and 180 leychees are divided among some children. (a) What is the largest number of children among whom mangoes and lychees are divided without any reminder? (b) How many mangoes will each of them get? (c) How many lychees will each of them get? Q-13 There are 126 mangoes, 231 lychees and 357 jackfruit sapling are distributed for planting a village. (a) What is the largest number of villagers among whom sapling can be divided equally? (b) How many mangoes, how many lychees and how many jackfruit will each of them get? Q-14 There are two bells. One bell rings after each 12 minutes and the other bell rings after 5 minutes. (a) What is the L.C.M of the numbers representing ringing time of the two bells? (b) If a bell ringing after every 7 minutes is included with the given two bells, then the 3 bells will ring altogether. after how long time they once ring together? (c) If the two bells ring together at 3 p.m, then when will they ring again altogether? Q-15 There are two drums of capacity to contain water of 228 litres and 348 litres. (a) What is the G.C.F. of 228 and 348? (b) How many pitchers of water of the highest capacity will be required to fill both the drums separately? (c) What is the least number exactly divisible by 228 and 348? Q-16 The length and the breadth of a hall room of rectangular size are 12 metre and 7 metre respectively. (a) What is the biggest square size tile that can be used to cover the hall room without breaking any of them. (b) If the length of the hallroom is increased by 3 metre and the breadth is decreased by 2 metre, what will be the change of area of the hallroom? Q-17 12, 18, 24 and 30 are four even numbers with 2 digits. (a) Find out the first 4 multiple of 12. (b) Find out the LCM of the given numbers. (c) What is the greatest common number among the given numbers? Explore creative and easy math questions with answers for Class 5 EV Chapter 5 to help students learn effortlessly. Q-18 Four bells rang at the same time and then rang again at an interval of 5, 7, 12 and 15 seconds respectively. (a) What is a prime number? (b) Write the 1st 4 multiples of 15. (c) After what interval of time will the bell ring again together? (d) What is the G.C.F. of 5, 7, 12 and 15? Q-19 147 lichees were distributed among some boys and it was found that each boy got 11 lichees and 4 lichees were in excess. (a) Express the above information mathematically in the form of an open sentence? (b) Find out the number of boys among whom the lichees were distributed. (c) If of the liches are distributed among some boys and each boy gets 11 lichees leaving 5 as excess, express the above information in an open sentence. Q-20 A particular numbers is multiplied by and then the product is divided by 6 to get 21 as quotient. (a) Write the open sentence with the help of the information given above. (b) What is the value of the particular number according to the open sentence in (a)? (c) If 3 times x equals 3 more than 45, what is the value of x? Read the full article
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MP Board Class 12th Maths Book Solutions in English Medium
MP Board Class 12th Maths Chapter 1 Relations and Functions
Chapter 1 Relations and Functions Ex 1.1
Chapter 1 Relations and Functions Ex 1.2
Chapter 1 Relations and Functions Ex 1.3
Chapter 1 Relations and Functions Ex 1.4
Chapter 1 Relations and Functions Miscellaneous Exercise
MP Board Class 12th Maths Chapter 2 Inverse Trigonometric Functions
Chapter 2 Inverse Trigonometric Functions Ex 2.1
Chapter 2 Inverse Trigonometric Functions Ex 2.2
Chapter 2 Inverse Trigonometric Functions Miscellaneous Exercise
MP Board Class 12th Maths Chapter 3 Matrices
Chapter 3 Matrices Ex 3.1
Chapter 3 Matrices Ex 3.2
Chapter 3 Matrices Ex 3.3
Chapter 3 Matrices Ex 3.4
Chapter 3 Matrices Miscellaneous Exercise
MP Board Class 12th Maths Chapter 4 Determinants
Chapter 4 Determinants Ex 4.1
Chapter 4 Determinants Ex 4.2
Chapter 4 Determinants Ex 4.3
Chapter 4 Determinants Ex 4.4
Chapter 4 Determinants Ex 4.5
Chapter 4 Determinants Ex 4.6
Chapter 4 Determinants Miscellaneous Exercise
MP Board Class 12th Maths Chapter 5 Continuity and Differentiability
Chapter 5 Continuity and Differentiability Ex 5.1
Chapter 5 Continuity and Differentiability Ex 5.2
Chapter 5 Continuity and Differentiability Ex 5.3
Chapter 5 Continuity and Differentiability Ex 5.4
Chapter 5 Continuity and Differentiability Ex 5.5
Chapter 5 Continuity and Differentiability Ex 5.6
Chapter 5 Continuity and Differentiability Ex 5.7
Chapter 5 Continuity and Differentiability Ex 5.8
Chapter 5 Continuity and Differentiability Miscellaneous Exercise
MP Board Class 12th Maths Chapter 6 Application of Derivatives
Chapter 6 Application of Derivatives Ex 6.1
Chapter 6 Application of Derivatives Ex 6.2
Chapter 6 Application of Derivatives Ex 6.3
Chapter 6 Application of Derivatives Ex 6.4
Chapter 6 Application of Derivatives Ex 6.5
Chapter 6 Application of Derivatives Miscellaneous Exercise
MP Board Class 12th Maths Chapter 7 Integrals
Chapter 7 Integrals Ex 7.1
Chapter 7 Integrals Ex 7.2
Chapter 7 Integrals Ex 7.3
Chapter 7 Integrals Ex 7.4
Chapter 7 Integrals Ex 7.5
Chapter 7 Integrals Ex 7.6
Chapter 7 Integrals Ex 7.7
Chapter 7 Integrals Ex 7.8
Chapter 7 Integrals Ex 7.9
Chapter 7 Integrals Ex 7.10
Chapter 7 Integrals Ex 7.11
Chapter 7 Integrals Miscellaneous Exercise
MP Board Class 12th Maths Chapter 8 Application of Integrals
Chapter 8 Application of Integrals Ex 8.1
Chapter 8 Application of Integrals Ex 8.2
Chapter 8 Application of Integrals Miscellaneous Exercise
MP Board Class 12th Maths Chapter 9 Differential Equations
Chapter 9 Differential Equations Ex 9.1
Chapter 9 Differential Equations Ex 9.2
Chapter 9 Differential Equations Ex 9.3
Chapter 9 Differential Equations Ex 9.4
Chapter 9 Differential Equations Ex 9.5
Chapter 9 Differential Equations Ex 9.6
Chapter 9 Differential Equations Miscellaneous Exercise
MP Board Class 12th Maths Chapter 10 Vector Algebra
Chapter 10 Vector Algebra Ex 10.1
Chapter 10 Vector Algebra Ex 10.2
Chapter 10 Vector Algebra Ex 10.3
Chapter 10 Vector Algebra Ex 10.4
Chapter 10 Vector Algebra Miscellaneous Exercise
MP Board Class 12th Maths Chapter 11 Three Dimensional Geometry
Chapter 11 Three Dimensional Geometry Ex 11.1
Chapter 11 Three Dimensional Geometry Ex 11.2
Chapter 11 Three Dimensional Geometry Ex 11.3
Chapter 11 Three Dimensional Geometry Miscellaneous Exercise
MP Board Class 12th Maths Chapter 12 Linear Programming
Chapter 12 Linear Programming Ex 12.1
Chapter 12 Linear Programming Ex 12.2
Chapter 12 Linear Programming Miscellaneous Exercise
MP Board Class 12th Maths Chapter 13 Probability
Chapter 13 Probability Ex 13.1
Chapter 13 Probability Ex 13.2
Chapter 13 Probability Ex 13.3
Chapter 13 Probability Ex 13.4
Chapter 13 Probability Ex 13.5
Chapter 13 Probability Miscellaneous Exercise
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WBBSE Solutions For Class 10 Maths Trigonometry Chapter 3 Trigonometric Ratios Of Complementary
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WBBSE Notes For Class 6 Maths Arithmetic Chapter 10 Highest Common Factor And Least Common Multiple Or 3 Numbers
#Class6maths#WBBSESolutions#Class6mathsarithmeticchapter10highestcommonfactorandleastcommonmultipleor3numbers
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#Jharkhandboardsolution#JACClass10MathsSolutions#JAC10thClassMathsChapter3#दोचरोंवालेरैखिकसमीकरणयुग्म
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Hots Question Class 10 Maths (2023 - 2024)
This document contains chapter-wise questions with solutions of the HOTS (Higher Order Thinking Skills) type, aligned with the 2018-19 curriculum and question paper design. This year, 17% of the exam, or 17 out of 80 marks, will consist of HOTS questions. These include four questions worth 3 marks each and one question worth 2 marks. Unlike regular questions, these are analytical and require a deep understanding of concepts combined with critical thinking skills to answer effectively.
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Class 8 math chapter 3 creative question with solution
Class 8 math chapter 3 creative question with solution
Creative Questions and Answers Question no : 01 The length of a rectangular garden is 60 metres and breadth is 40 metres. There is a road of width 2 metres inside around the garden. a. Find the area of the garden in square centimetre. b. Find out the area of the road. c. Find the perimeter of a rectangular region whose length is 6 times its breadth and area is equal to the rectangle given in the stem. Solution: (a) Here given for a rectangular garden, length = 60 m = 6000 cm and breadth = 40 m = 4000 cm ∴ Area of the garden = 6000 × 4000 sq.cm = 24000000 sq. cm (b) From (a) above the area of the rectangular garden,
ABCD, shown above is 24000000 sq. cm or 2400 sq. m including road. Again, the length of the garden excluding road = (60 - 4) m = 56 m and the breadth of the garden excluding road = (40 - 4) m = 36 m ∴ area of the garden excluding road = 56 x 36 sq. m = 2016 sq. m ∴ area of the road = (2400 – 2016 ) sq. m = 384 sq m. So, the required area of road is 384 sq. m. (c) Let the breadth of the proposed rectangular region be 'x' m. ∴ It’s length is 6x m ∴its area is 6x × x sq. m or 6x² sq. m Now according to the given information, 6x² = 2400, the area of the rectangle mentioned in (b) above or, x² = 400 or; x = 20 and 6x = 6 × 20 or, 120 That is, the length of the proposed rectangular region is 120 m and breadth is 20 m, ∴ The perimeter of the rectangular region = 2(length + breadth) = 2(120 + 20) m = 2(120 +20)m = 2(140) m = 280 m. So, The required perimeter of the proposed square region is 280 m. Question no : 02
In figure, ABCD is a rectangular garden and there is a 1 metre wide road inside around the garden. Solution: a. Determine the area of rectangular garden by triangle. b. Determine the area of the road. c. There is a three metres wide road around the outside of a square land which perimeter is equal to the rectangular garden mentioned in the figure. How much money will be spent to planting grass at Tk. 7.00 per sq. metre? (a) According to the given stem, rectangle ABCD = 2 ×△ACD = 2 × × 40 x 30 sq. m = 1200 sq. m So, the area of the garden is 1200 sq. m. (b) From the stem, area of the garden with road = 40 x 30 sq. cm = 1200 sq. cm Again, the length of the garden without road = ( 40 - 1 × 2) m = 38m And the breadth of the garden without road = (30 – 1 × 2) m = 28 m. ∴area of the garden without road = 28 × 38 sq. m = 1064 sq. m ∴area of the road = (1200–1064) sq. m = 136 sq. m, (c) Here we have, perimeter of the rectangle = 2(40 + 30) m = 140 m Now according to the stem, perimeter of a square = 140 m ∴ length of each side of the square = = 35 m So, the area of the square = 35 x 35 sq. m = 1225 sq. m Again, length of each side of the square with 3m wide road around outside the square = (35 +3 + 3) m = 41 m. ∴area of the square with road = 41 x 41 = 1681 sq. m So, area of the road = (1681– 1225) = 456 sq. m. Now we have, cost of planting grass = Tk. 7 per sq. m ∴ ” ” ” ” in 456 sq. m = 456 x 7 = 3192 taka So, 3192 taka be spent for planting grass. Question no : 03 There are two crosswise roads of breadth 3 metres just in the middle of a field of length 50 metres and breadth 40 metres. a. Draw the figure with short description. b. Determine the sum of the area of the two roads. c. How many bricks of construct the two roads with a bricks of length 25 centimetres and breadth 10 centimetres. Solution: (a) A geometric figure based on the given information in the stem is drawn below :
(b) According to the figure drawn in (a) above, Area of the two roads = area of the rectangular region ABCD + area of the rectangular region PQRS + area of the rectangular region KLMN. = 50 × 3 sq. m + 08.5 × 3 sq. m + 18.5 × 3 sq. m = 150 sq. m + 55.5 sq. m + 55.5 sq. m = 261 sq. m The sum of the area of the two roads is 261 sq. m. (c) From (b) above, the total area of two roads = 261 sq. m The area of a brick = 25 x 10 sq. cm = 250 sq. cm = 0.025 sq. m ∴Number of bricks required for constructing the two roads = = 10,440 So, 10,440 piece of bricks are required. Question no : 04 The length of a rectangular plot is 90 meters and breadth is 70 meters. A pond was excavated in this plot with equal bank of width 4 metres around the plot. The depth of the pond is 2.5 meters. a. Find the perimeter of the plot. b. Determine the area of the bank of the pond. c. To dig the soil of the pond, it costs Tk 25 per cubic feet. How much money was spent to dig the soil of the pond? Solution: (a) The length and the breadth of a rectangular plot are 90 m and 70 m respectively. we know, perimeter of a rectangular region = 2(length + breadth) unit ∴The perimeter of the rectangular plot = 2(90+70) m = 2 x 160m = 320 m So, the required perimeter is 320 m. (b)
Here given that, length of the plot with pond and bank = 90 m breadth ” ” ” ” ” ” ” = 70 m area of the plot with pond and bank = 90 x 70 sq m = 6300 sq m Again, length of plot with pond only = (90 – 4×2)m = 82m breadth ” ” ” ” ” = (70 – 4×2)m = 62 m ∴area of plot with pond only = 82 x 62 sq. m = 5084 sq. m So, area of the bank = area of the plot with pond and bank area of the plot with pond only = 6300 sq. m – 5084 sq. m. = 1216 sq. m (c) Here we have, length of the pond = (90 – 4 x 2) m or, 82 m breadth ” ” ” = (70 – 4 x 2) m or, 62 m depth ” ” ” = 2.5 m ∴the volume of the pond = 82 × 62 × 2.5 cu m = 12,710 cu. m = 12,710 × 35.3 cu. ft = 4,48,663 cu. ft So, to excavate the pond, 4,48,663 cu. ft soilis to be dug. Now, if it costs 25 taka per cubic feet to dig the soil of the pond, total cost which will incur in this account = 4,48,663 × 25 = 1,12,16,575 taka ∴1,12,16,575 taka was spent to dig the soil of the pond. (Ans.) Question no : 05 The length of a rectangular garden is 60 metres and breadth is 40 metres. There is a 3 metres wide road outside around the garden. The road is metalled with bricks with 25 cm length and 12.5 cm breadth and price of each brick is Tk. 8. a. Determine the perimeter of garden. b. Determine the area of the road. c. How much money will be needed to metalise the road with brick? Solution: (a) We know, perimeter of a rectangle = 2(Length + breadth) Here in the case of the given rectangular garden, length = 60 m, breadth = 40 m. ∴Perimeter of the garden = 2(60 + 40) m = 200 m. (b) Given that, the length of the garden (without road) = 60 m the breadth of the garden without road = 40 m
∴ area of the garden 60 x 40 sq. m = 2400 sq, m Again, length of the garden with the road of width of 3 m = (60 + 3 + 3) m or 66 m breadth of the garden with road of with 3 m = (40 +3 + 3) m = 46 m ∴ area of the garden with 3 m wide road = 66 x 46 sq. m = 3036 sq. m ∴ area of the road = (3036 – 2400)sq. m = 636 sq. m So, the road is of area of 636 sq. m. (c) From (b) above, the area of the road = 636 sq. m = 6360000 sq. cm Again, the length of a brick =25 cm and the breadth of the brick =12.5 cm ∴area of a brick = 25 x 12.5 sq. cm = 312.5 sq. cm ∴the number of bricks required to cover 6360000 sq. cm road = = 20352 the price of a brick = 8 taka ∴the price of 20352 bricks = 20352 × 8 taka = 162816 taka So, to metalise the road with bricks 162816 taka will be needed. Question no : 06 The length of a rectangular field is 3 times its breadth. An amount of Tk. 1822.50 is spent to plant grass at Tk. 7.50 per sq. meters of that field. a. If the breadth of field be x metre, find the area of the field. b. Find the length and breadth of the rectangular field. c. How many rocks of each 25 cm with sq. size rocks will be needed to construct a square room of which perimeter is equal to the perimeter of the rectangular field? Solution: (a) Let the breadth of the field = x metre ∴Length of the field = 3x metre So, area of the field = 3x × x sq. m = 3x² sq. m (b) Here breadth = x m ∴ length = 3x m ∴area of the rectangular field = 3x × x sq. m = 3x2 sq. m Again, total cost = 1822.50 taka Cost per sq. m = 7.50 taka ∴Total area = sq. m = 243 sq. m ∴3x² = 243 ⇒ x² = 81 ⇒ x = 9. That is, the breadth of the rectangular field is 9 m The length of the field = 3 × 9 m = 27 m (c) From (b), The length of the field = 27 m and the breadth of the field = 9 m. ∴ perimeter of the rectangular field = 2 (27 + 9) = 72 m. ∴ perimeter of the square room = 72 m. ∴ the length of 1 side of the square room=(72 4) m = 18m ∴ the area of the square room = 18 × 18 sq. m = 324 sq. m Again, area of the rock each of 25 cm with sq size = 25 × 25 sq. cm = 625 sq. cm = 0.0625 sq. m Now we have, area of the square room = 324 sq. m and area of each of square rock = 0.0625 sq. m ∴Number of sq rock = pieces = 5184 pieces. So, 5184 pieces of sq rock will be needed to construct the square room. Question no : 07 The area of a rectangular field is 100 acres and its length is three times the breadth. a. Find the area of the rectangular field in sq, metre. b. Find the length of the rectangular field. c. Find the area of a square field which perimeter is same to the rectangular field. Solution: (a) We know, 1 acre = 4046.86 sq. m ∴100 acre = 4046.86 × 100 sq. m = 404686 sq. m ∴the area of the rectangular field is 404686 sq. m. (b) Let the breadth of the rectangular field = x m ∴its length = 3xm ∴its area = 3x × x sq. m. Now according to the problem, 3x² = 404686 or, x² = 134895.33 or, x = 367.28, the breadth of the field ∴3x = 1101.84 That is, the length of the rectangular field is 1101.84 metre. (c) From (b), length of the rectangular field = 1101.84 m breadth of the rectangular field = 367.28 m ∴perimeter of the rectangular field = 2( 1101.84 + 367.28) m = 2938.24 m Now according to the problem, the perimeter of the square field = 2938.24 m ∴ the length of each side of the square = 734.56 rn ∴ area of the square field = 539578.37 sq. m Therefore, the desired area of the square field is 539578.39 sq. m. Question no : 08 The length of a rectangular tank is 5.5 metres and breadth is 4 metres. Breadth is 2 times of the height. It's four sides walls are mettelated with stones having size 1.5 x 1.5 m2. a. Find out the area of the base of the tank. b. What is the volume of water in litre if the tank is full of water and what is the weight of it in kilogram? c. How many stones are needed to mettelated to four sides walls of the tank? Solution: (a) Here the length of the base of a rectangular tank = 5.5 m the breadth of the base of the rectangular tank 4 m ∴ area of the base of the tank = 5.5 × 4 sq. m = 22.0 sq. m So, the desirous area of the base of the tank is 22 sq. m. (b) From (a), area of the base of the tank = 22 sq m height of the tank = × breadth = 2 m ∴ Volume of the tank = 22 × 2 cu. m = 44 cu. m = 44 × 1000000 cu. cm = 44000000 cu. cm Now if the tank is-full of water, the volume of water = 44000000 cu cm = litre = 44000 litre = 44000 kg, since the weight of 1litre of water = 1 kg Therefore, the volume of water of the tank is 44000 litre and its weight is 44000 kg. (c) Here the length, breadth of the base and height of the rectangular tank are 5.5 m, 4 m and 2 m from (a) and (b). ∴area of the 4 side /faces other than the base and the top of the tank = 2(5.5 × 2 + 4 × 2 ) sq. m = 2(11.0 + 8) sq. m =2(19) sq. m = 38 sq. m Again, area of a stone = 1.5 x 1.5 sq. m = 2.25 sq. m So, total area of 4 sides / faces is 38 sq. m and area of a stone is 2.25 sq. m ∴ number of stone = = 16.89 or 17 (approx) Therefore, stones needed to metal the four side wall of the tank is 17 pieces. Question no : 09
In figure, ABCD is a rectangular field, length of which is twice of it's breadth. The total cost of planting grass in the field is Tk. 12160 at the rate of Tk. 3.80 per sq. metre. a. Find the area of the field. b. Find the length of the diagonal AC? c. How much money will be spent at Tk. 7.25 per meter to erect a fence around that field? Solution: (a) Here total cost of planting grass = 12160 taka. Rate of cost of planting grass = 3.80 per sq. m. ∴Area of the field = sq. m = 3200 sq. m. (b) Let the breadth of the rectangular field be x metre. ∴The length of the rectangular field is 2x metre. ∴ its area = 2x × x sq. m = 2x² sq. m But from (a), area = 3200 sq, m. So, 2x² = 3200 or, x² = 1600 or, x = 40 That is, breadth of the rectangular fill is 40 m. ∴The length of the rectangular field= 40×2m= 80 m Now according to the given figure, △ABC is right triangle and is a half of the rectangle ABCD. ∴ AC² = AB² + BC² Here AB = 40 m and BC = 80 m. ∴ AC² = (40)²+ (80)² = 1600 + 6400 = 8000 ∴ AC = 89.44 that is, the length of AC = 89.44 m. (c) From (b), we get, the length of the field 80 m and the breadth of the field = 40 m ∴perimeter of the rectangular field = 2(80 + 40) m = 2(120) m = 240m Now the cost for erecting fence around the field with perimeter of 240 m at the rate of 7.25 taka per metre = 240 × 7.25 taka = 1740 taka. ∴1740 taka will be spent for fencing the field. Question no : 10 Mr Nayeem has a rectangular garden of which length is one and half times its breadth and its area is 2400 sq. metres. There is a 3 metres wide path around the outside of the garden. There is planting grass at Tk. 3.25 per sq. meter of the path. a. In view of the above stem, draw a proportional figure of the rectangular garden with path. b. Find the length and breadth of the garden. c. How much money will be spent in total to the planting grass of the path? Solution: (a) Based on the given information, a proportional figure of the rectangular garden is drawn below :
A rectangular garden ABCD with 3 m path outside around. (b) Let us suppose, the breadth of the garden = x m ∴according to the condition of the problem, the length of the garden = 1.5x m or x m. ∴ area of the garden = x × x sq. m But according to the problem, area of the garden = 2400 sq. m ∴ = 2400 ⇒ 3x² = 2 × 2400 ⇒ x² = ⇒ x² = 1600 ⇒ x = 40 That is the breadth of the garden is 40 m. ∴The length of the garden = x m = = 60 m So, the length is 60 m and the breadth is 40 m of the garden. (c) According to the geometric figure drawn in (a) based on the stem,. area of the garden with 3 m path around outside = (60 +3 + 3) (40 +3 + 3) sq. m = 66 x 46 sq. m = 3036 sq. m Again, area of the garden = 2400 sq. m (given) ∴area of the path = (3036 - 2400) sq, m = 636 sq. m Now amount of money which will be required for planting grass in the path of 636 sq m at the rate of 3.25 taka per sq. m = 636 x 3.25 taka = 2067 taka So, total amount will be spent for planting grass in the path is 2067 taka. Question no : 11 The breadth of a rectangular field is half of its length. An amount of Tk. 12100 is spent to plant grass in the field at the rate of Tk. 2 per sq. metre. Around inside the field there is a road of breadth 4 metres. a. What is the area of the rectangular field. in square metre? b. The expenditure per square metre along the length is Tk. 15 and that of along the breadth is Tk. 10. How much money will be spent to erect a fence around that field? c. How much money will be spent to plant grass in the road at Tk. 12.50 per square metre ? Solution: (a) Here total cost of planting grass = 12100 taka the cost of planting grass per sq. m = 2 taka ∴area of the rectangular field = sq. m = 6050 sq. m. So, the required area of the field is 6050 sq. m. (b) Let the length of the rectangular field = 2x m ∴the breadth of the rectangular field = x m ∴area of the rectangular field = 2x × x sq. m = 2x2 sq. m. But according to (a) above, area is 6050 sq. Read the full article
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NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables
NCERT Solutions for Class 10 Maths Chapter 3. NCERT Solutions for Class 10 Math Chapter 3 Pair Of Linear Equations In Two Variables are provided here with simple step-by-step explanations.
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MP Board Class 12th Maths Book Solutions in Hindi Medium
MP Board Class 12th Maths Chapter 1 संबंध एवं फलन
Chapter 1 संबंध एवं फलन Ex 1.1
Chapter 1 संबंध एवं फलन Ex 1.2
Chapter 1 संबंध एवं फलन Ex 1.3
Chapter 1 संबंध एवं फलन Ex 1.4
Chapter 1 संबंध एवं फलन विविध प्रश्नावली
MP Board Class 12th Maths Chapter 2 प्रतिलोम त्रिकोणमितीय फलन
Chapter 2 प्रतिलोम त्रिकोणमितीय फलन Ex 2.1
Chapter 2 प्रतिलोम त्रिकोणमितीय फलन Ex 2.2
Chapter 2 प्रतिलोम त्रिकोणमितीय फलन विविध प्रश्नावली
MP Board Class 12th Maths Chapter 3 आव्यूह
Chapter 3 आव्यूह Ex 3.1
Chapter 3 आव्यूह Ex 3.2
Chapter 3 आव्यूह Ex 3.3
Chapter 3 आव्यूह Ex 3.4
Chapter 3 आव्यूह विविध प्रश्नावली
MP Board Class 12th Maths Chapter 4 सारणिक
Chapter 4 सारणिक Ex 4.1
Chapter 4 सारणिक Ex 4.2
Chapter 4 सारणिक Ex 4.3
Chapter 4 सारणिक Ex 4.4
Chapter 4 सारणिक Ex 4.5
Chapter 4 सारणिक Ex 4.6
Chapter 4 सारणिक विविध प्रश्नावली
MP Board Class 12th Maths Chapter 5 सांतत्य तथा अवकलनीयता
Chapter 5 सांतत्य तथा अवकलनीयता Ex 5.1
Chapter 5 सांतत्य तथा अवकलनीयता Ex 5.2
Chapter 5 सांतत्य तथा अवकलनीयता Ex 5.3
Chapter 5 सांतत्य तथा अवकलनीयता Ex 5.4
Chapter 5 सांतत्य तथा अवकलनीयता Ex 5.5
Chapter 5 सांतत्य तथा अवकलनीयता Ex 5.6
Chapter 5 सांतत्य तथा अवकलनीयता Ex 5.7
Chapter 5 सांतत्य तथा अवकलनीयता Ex 5.8
Chapter 5 सांतत्य तथा अवकलनीयता विविध प्रश्नावली
MP Board Class 12th Maths Chapter 6 अवकलज के अनुप्रयोग
Chapter 6 अवकलज के अनुप्रयोग Ex 6.1
Chapter 6 अवकलज के अनुप्रयोग Ex 6.2
Chapter 6 अवकलज के अनुप्रयोग Ex 6.3
Chapter 6 अवकलज के अनुप्रयोग Ex 6.4
Chapter 6 अवकलज के अनुप्रयोग Ex 6.5
Chapter 6 अवकलज के अनुप्रयोग विविध प्रश्नावली
MP Board Class 12th Maths Chapter 7 समाकलन
Chapter 7 समाकलन Ex 7.1
Chapter 7 समाकलन Ex 7.2
Chapter 7 समाकलन Ex 7.3
Chapter 7 समाकलन Ex 7.4
Chapter 7 समाकलन Ex 7.5
Chapter 7 समाकलन Ex 7.6
Chapter 7 समाकलन Ex 7.7
Chapter 7 समाकलन Ex 7.8
Chapter 7 समाकलन Ex 7.9
Chapter 7 समाकलन Ex 7.10
Chapter 7 समाकलन Ex 7.11
Chapter 7 समाकलन विविध प्रश्नावली
MP Board Class 12th Maths Chapter 8 समाकलनों के अनुप्रयोग
Chapter 8 समाकलनों के अनुप्रयोग Ex 8.1
Chapter 8 समाकलनों के अनुप्रयोग Ex 8.2
Chapter 8 समाकलनों के अनुप्रयोग विविध प्रश्नावली
MP Board Class 12th Maths Chapter 9 अवकल समीकरण
Chapter 9 अवकल समीकरण Ex 9.1
Chapter 9 अवकल समीकरण Ex 9.2
Chapter 9 अवकल समीकरण Ex 9.3
Chapter 9 अवकल समीकरण Ex 9.4
Chapter 9 अवकल समीकरण Ex 9.5
Chapter 9 अवकल समीकरण Ex 9.6
Chapter 9 अवकल समीकरण विविध प्रश्नावली
MP Board Class 12th Maths Chapter 10 सदिश बीजगणित
Chapter 10 सदिश बीजगणित Ex 10.1
Chapter 10 सदिश बीजगणित Ex 10.2
Chapter 10 सदिश बीजगणित Ex 10.3
Chapter 10 सदिश बीजगणित Ex 10.4
Chapter 10 सदिश बीजगणित विविध प्रश्नावली
MP Board Class 12th Maths Chapter 11 त्रि-विमीय ज्यामिति
Chapter 11 त्रिविमीय ज्यामिति Ex 11.1
Chapter 11 त्रिविमीय ज्यामिति Ex 11.2
Chapter 11 त्रिविमीय ज्यामिति Ex 11.3
Chapter 11 त्रिविमीय ज्यामिति विविध प्रश्नावली
MP Board Class 12th Maths Chapter 12 रैखिक प्रोग्रामन
Chapter 12 रैखिक प्रोग्रामन Ex 12.1
Chapter 12 रैखिक प्रोग्रामन Ex 12.2
Chapter 12 रैखिक प्रोग्रामन विविध प्रश्नावली
MP Board Class 12th Maths Chapter 13 प्रायिकता
Chapter 13 प्रायिकता Ex 13.1
Chapter 13 प्रायिकता Ex 13.2
Chapter 13 प्रायिकता Ex 13.3
Chapter 13 प्रायिकता Ex 13.4
Chapter 13 प्रायिकता Ex 13.5
Chapter 13 प्रायिकता विविध प्रश्नावली
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WBBSE Solutions For Class 10 Maths Mensuration Chapter 3 Right Circular Cone
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WBBSE notes for class 6 maths arithmetic chapter 3 logical approximation of number
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How To Get Best Score In Class 10 Maths Paper
Mathematics is often regarded as one of the most scoring subjects. But, many students do not like the subject and get nightmares with the thought of it. However, it is essential to attain good marks in Maths to pass Class 10 with an appealing percentage. So what is the solution? Students can follow a few tips to get better marks in Maths. But before we detail the tips, let’s learn about the advantages of NCERT solutions in Class 10 Mathematics.
Advantages Of NCERT Solutions In Class 10 Mathematics
Over the years, NCERT solutions have been one of the top Class 10 Maths solutions and continue to be so. But, what is the reason? It helps candidates with well-researched study materials to improve their grades and help them grasp concepts better. The NCERT solutions for class 10 Maths mainly focus on questions and answers according to the NCERT pattern. These study materials provide answers to different problems and help the candidates get accurate answers.
Tips To Get Amazing Scores In Class 10 Mathematics Examination
So, let’s move forward and discuss some tips that you can adhere to get a 90+ in your final examination.
● Start With Easy Problems
The logic behind this is simple. You should be able to answer the easy questions atleast quickly, if not the difficult ones. If you know the easy subject by hard, you can also save more time to solve the complex sums.
● Understand Your Weak Areas
While preparing from NCERT Solutions for Class 10 Maths Chapter 3 or any other chapter, understand which topics you need to brush upon. Once you know which are they, solve more sample papers and write answers for the complex chapters to grasp concepts quicker.
● Find Solutions On Your Own
There are theories to every mathematical problem. Just by learning them, you won’t excel in the examination. Instead, try to apply those theories while you solve the questions. You can take the guidance from NCERT solutions for class 10 Maths if you feel stuck in the midway. Try to practice as much as possible to get better results.
● Maintain A Separate Sheet For Theories, Formulae, And Methods
The theories, formulas, and methods are endless in NCERT Solutions for Class 10 Maths Chapter 3 examination. Yet, you need to memorise them all to solve problems quickly in the test. The best way to do that is by listing all these on a piece of paper. Then, you can promptly revise it whenever you forget the formulas. Besides, it is the best way for last moment revision.
● Learn The Graphs And Figures
Graphs and figures are significant for mathematics examination. Thus, learn them well from the Class 10 Maths Solutions and implement them during the test. This way, you can get more marks than you expect.
● Focus On Neatness
While giving the examination, do not compromise on the handwriting. Keep the answer clean, neat, and clean, and avoid striking off excessively. This creates a wrong impression on the head examiner. Keep the answer sheet tidy because you can get one or two marks for handwriting.
Conclusion
Thus, these are how you can get the best score in the Class 10 Maths paper. So, what are you waiting for? Read the Class 10 Maths Solutions thoroughly to achieve an A grade in this subject.
#NCERT solutions for class 10 Maths#NCERT Solutions for Class 10 Maths Chapter 3#Class 10 Maths Solutions
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