{"id":8190,"date":"2024-01-03T12:46:31","date_gmt":"2024-01-03T07:16:31","guid":{"rendered":"https:\/\/wordpress-505192-1602719.cloudwaysapps.com\/?p=8190"},"modified":"2024-01-03T17:07:01","modified_gmt":"2024-01-03T11:37:01","slug":"samacheer-kalvi-8th-maths-solutions-term-3-chapter-1-ex-1-5","status":"publish","type":"post","link":"https:\/\/samacheerguru.com\/samacheer-kalvi-8th-maths-solutions-term-3-chapter-1-ex-1-5\/","title":{"rendered":"Samacheer Kalvi 8th Maths Solutions Term 3 Chapter 1 Numbers Ex 1.5"},"content":{"rendered":"

Students can Download Maths Chapter 1 Numbers Ex 1.5 Questions and Answers, Notes Pdf, Samacheer Kalvi 8th Maths Book Solutions<\/a> Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.<\/p>\n

Tamilnadu Samacheer Kalvi 8th Maths Solutions Term 3 Chapter 1 Numbers Ex 1.5<\/h2>\n

Miscellaneous and Practice Problems<\/span><\/p>\n

Question 1.
\nA square carpet covers an area of 1024 m\u00b2 of a big hall. It is placed in the middle of the hall. What is the length of a side of the carpet?
\nSolution:
\nArea of the carpet = 1024 m\u00b2
\nside \u00d7 side = 1024 m\u00b2
\n(side)\u00b2 = 1024 m\u00b2
\n\"Samacheer
\n(side)\u00b2 = 2 \u00d7 2 \u00d7 2 \u00d7 2 \u00d7 2 \u00d7 2 \u00d7 2 \u00d7 2 \u00d7 2 \u00d7 2
\n= 2\u00b2 \u00d7 2\u00b2 \u00d7 2\u00b2 \u00d7 2\u00b2 \u00d7 2\u00b2
\n= (2 \u00d7 2 \u00d7 2 \u00d7 2 \u00d7 2)\u00b2
\n(side)\u00b2 = 32\u00b2
\nside = 32
\nLength of a side of the carpet = 32 m<\/p>\n

\"Samacheer<\/p>\n

Question 2.
\nThere is a large square portrait of a leader that covers an area of 4489 cm\u00b2. If each side has a 2 cm liner, what would be its area?
\nSolution:
\nArea of the square = 4489 cm\u00b2
\n(side)\u00b2 = 4489 cm\u00b2
\n\"Samacheer
\n(side)\u00b2 = 67 x 67
\nside\u00b2 = 67\u00b2
\nLength of a side = 67
\n\"Samacheer
\nLength of a side Length of a side with liner = 67 + 2 + 2 cm = 71 cm
\nArea of the larger square = 71 x 71 cm\u00b2
\n= 5041 cm\u00b2
\nArea of the liner = Area of big square-Area of small square
\n= (5041 – 4489) cm\u00b2 = 552 cm\u00b2<\/p>\n

Question 3.
\n2401 plants are planted in a garden such that each contains as many plants as the number of rows. Find the number of rows and the number of plants in each row.
\nSolution:
\nGiven number of plants in a row = Number of rows.
\nNumber of rows \u00d7 number of plants in a row = Total plants
\nTotal plants = 2401
\n\"Samacheer
\n= 7 \u00d7 7 \u00d7 7 \u00d7 7 = 7\u00b2 \u00d7 7\u00b2
\n= 49 \u00d7 49
\n\u2234 number of rows = 49
\nnumber of plants in a row = 49<\/p>\n

Question 4.
\nIf \\(\\sqrt[3]{1906624} \\times \\sqrt{x}\\) = 3100, find x.
\nSolution:
\n\"Samacheer
\n\\(\\sqrt{x}\\) = 25
\nSquaring on both sides \\((\\sqrt{x})^{2}\\) = 25\u00b2
\nx = 625<\/p>\n

Question 5.
\nIf (625)x<\/sup> = 15625, find x\u00b2 and x\u00b3
\nSolution:
\n(625)x<\/sup> = 15625
\n\"Samacheer
\n(5 x 5 x 5 x 5)x<\/sup> = 5 x 5 x 5 x 5 x 5 x 5
\n(54<\/sup>)x<\/sup> = 56<\/sup>
\n54x<\/sup>\u00a0= 56<\/sup>
\n54x<\/sup> = 56<\/sup>
\nComparing the powers of 5 both sides
\n4x = 6
\nx = \\(\\frac{6}{4}\\)
\nx = \\(\\frac{3}{2}\\)
\n\"Samacheer<\/p>\n

\"Samacheer<\/p>\n

Question 6.
\nIf 2m-1<\/sup> + 2m+1<\/sup> = 640, then find \u2018m\u2019
\nSolution:
\nGiven 2m-1<\/sup> + 2m+1<\/sup> = 640
\n2m-1<\/sup> + 2m+1<\/sup> = 128 + 512 [consecutive powers of 2]
\n2m-1<\/sup> + 2m+1<\/sup> = 27<\/sup>+ 29<\/sup> [powers of 2: 2, 4, 8, 16, 32, 64, 128, 256, 512, …..]
\nm – 1 = 7
\nm = 7 + 1
\nm = 8<\/p>\n

Question 7.
\nSimplify \\(\\frac{16 \\times 10^{2} \\times 64}{4^{2} \\times 2^{4}}\\)
\nSolution:
\n\"Samacheer
\n\\(\\frac{16 \\times 10^{2} \\times 64}{4^{2} \\times 2^{4}}\\) = \\(\\frac{2^{4} \\times 10^{2} \\times 2^{6}}{\\left(2^{2}\\right)^{2} \\times 2^{4}}=\\frac{2^{4+6} \\times 10^{2}}{2^{4} \\times 2^{4}}=\\frac{2^{10} \\times 100}{2^{8}}\\)
\n= 210-8<\/sup> \u00d7 100= 22<\/sup> \u00d7 100 = 400<\/p>\n

Question 8.
\nGive the answer in scientific notation:
\nA human heart beats at an average of 80 beats per minute. How many times does it beat in
\n(i) an hour?
\n(ii) a day?
\n(iii) a year?
\n(iv) 100 years?
\nSolution:
\nHeart beat per minute = 80 beats
\n(i) an hour One hour = 60 minutes
\nHeart beat in an hour = 60 x 80 = 4800 = 4.8 x 103<\/sup><\/p>\n

(ii) In a day
\nOne day = 24 hours = 24 x 60 minutes
\n\u2234 Heart beat in one day = 24 x 60 x 80 = 24 x 4800 = 115200 = 1.152 x 105<\/sup><\/p>\n

(iii) a year
\nOne year = 365 days = 365 x 24 hours = 365 x 24 x 60 minutes
\n\u2234 Heart beats in a year = 365 x 24 x 60 x 80 = 42048000 = 4.2048 x 107<\/sup><\/p>\n

(iv) 100 years
\nHeart beats in one year = 4.2048 x 107<\/sup>
\nHeart beats in 100 years = 4.2048 x 107<\/sup> x 100 = 4.2048 x 107<\/sup> x 102<\/sup>
\n= 4.2048 x 109<\/sup><\/p>\n

Challenging Problems<\/span><\/p>\n

Question 9.
\nA greeting card has an area 90 cm2<\/sup>. Between what two whole numbers is the length of its side?
\nSolution:
\nArea of the greeting card = 90 cm2<\/sup>
\n(side)2<\/sup> = 90 cm2<\/sup>
\n\"Samacheer
\n(side)2<\/sup> = 2 x 5 x 3 x 3 = 2 x 5 x 32<\/sup>
\n\\(\\sqrt{({side})^{2}}=\\sqrt{2 \\times 5 \\times 3^{2}}\\)
\n\"Samacheer
\nside = 3 \\(\\sqrt{2 \u00d7 5}\\)
\nside = \\(\\sqrt{10}\\) cm
\nside = 3 \u00d7 3.2 cm
\nside = 9.6 cm
\n\u2234 Side lies between the whole numbers 9 and 10.<\/p>\n

\"Samacheer<\/p>\n

Question 10.
\n225 square shaped mosaic tiles, each of area 1 square decimetre exactly cover a square shaped verandah. How long is each side of the square shaped verandah?
\nSolution:
\nArea of one tile = 1 sq. decimeter
\nArea of 225 tiles = 225 sq.decimeter
\n225 square tiles exactly covers the square shaped verandah.
\n\u2234 Area of 225 tiles = Area of the verandah
\nArea of the verandah = 225 sq.decimeter
\nside x side = 15 x 15 sq.decimeter
\nside = 15 decimeters
\nLength of each side of verandah = 15 decimeters.<\/p>\n

Question 11.
\nA group of 1536 cadets wanted to have a parade forming a square design. Is it possible? If it is not possible how many more cadets would be required?
\nSolution:
\nNumber of cadets to form square design
\n\"Samacheer
\n1536 = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 3
\nThe numbers 2 and 3 are unpaired
\nIt is impossible to have the parade forming square design with 1536 cadets.
\n\"Samacheer
\n39 x 39 = 1521
\nAlso 40 x 40 = 1600
\n\u2234 We have to add (1600 – 1536) = 64 to make 1536 a perfect square.
\n\u2234 64 more cadets would be required to form the square design.<\/p>\n

Question 12.
\nFind the decimal fraction which when multiplied by itself gives 176.252176.
\nSolution:
\nWe will find the square root of 176.252176
\n\"Samacheer
\n176.252176 = 13.276 x 13.276
\n\u2234 The required number is 13.276<\/p>\n

Question 13.
\nEvaluate \\(\\sqrt{286225}\\) and use it to compute \\(\\sqrt{2862.25}\\) + \\(\\sqrt{28.6225}\\)
\nSolution:
\n\"Samacheer<\/p>\n

Question 14.
\nThe speed of light in glass is about 2 x 108<\/sup> m\/sec. Use the formula, time = \\(\\frac{distence}{speed}\\) to find the time (in hours) for a pulse of light to travel 7200 km in glass.
\nSolution:
\n\"Samacheer
\nRequired time = 10-5<\/sup> hours<\/p>\n

\"Samacheer<\/p>\n

Question 15.
\nSimplify : (3.769 x 105<\/sup>) + (4.21 x 105<\/sup>)
\nSolution:
\n\"Samacheer
\n(3.769 x 105<\/sup>) + (4.21 x 105<\/sup>) = 3,76,900 + 4,21,00
\n= 7,97,900 = 7.979 x 105<\/sup><\/p>\n

Question 16.
\nOrder the following from the least to the greatest: 1625, 8100, 3500, 4400, 2600
\nSolution:
\n1625<\/sup> = (24<\/sup>)25<\/sup> = 2100<\/sup>
\n8100<\/sup> = (23<\/sup>)100<\/sup> = 2300<\/sup>
\n4400<\/sup> = (22<\/sup>)400<\/sup> = 2800<\/sup>
\n2600<\/sup> = 2600<\/sup>
\nComparing the powers we have, 2100<\/sup> < 2300<\/sup> < 2600<\/sup>< 2800<\/sup>
\n\u2234 The required order : 1625<\/sup>, 8100<\/sup>, 2600<\/sup>, 3500<\/sup>, 4400<\/sup><\/p>\n","protected":false},"excerpt":{"rendered":"

Students can Download Maths Chapter 1 Numbers Ex 1.5 Questions and Answers, Notes Pdf, Samacheer Kalvi 8th Maths Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations. Tamilnadu Samacheer Kalvi 8th Maths Solutions Term 3 Chapter 1 Numbers Ex 1.5 Miscellaneous and […]<\/p>\n","protected":false},"author":2,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":"","footnotes":""},"categories":[4],"tags":[],"blocksy_meta":"","jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/samacheerguru.com\/wp-json\/wp\/v2\/posts\/8190"}],"collection":[{"href":"https:\/\/samacheerguru.com\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/samacheerguru.com\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/samacheerguru.com\/wp-json\/wp\/v2\/users\/2"}],"replies":[{"embeddable":true,"href":"https:\/\/samacheerguru.com\/wp-json\/wp\/v2\/comments?post=8190"}],"version-history":[{"count":1,"href":"https:\/\/samacheerguru.com\/wp-json\/wp\/v2\/posts\/8190\/revisions"}],"predecessor-version":[{"id":55406,"href":"https:\/\/samacheerguru.com\/wp-json\/wp\/v2\/posts\/8190\/revisions\/55406"}],"wp:attachment":[{"href":"https:\/\/samacheerguru.com\/wp-json\/wp\/v2\/media?parent=8190"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/samacheerguru.com\/wp-json\/wp\/v2\/categories?post=8190"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/samacheerguru.com\/wp-json\/wp\/v2\/tags?post=8190"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}