{"id":44869,"date":"2022-05-30T05:00:00","date_gmt":"2022-05-29T23:30:00","guid":{"rendered":"https:\/\/cbselibrary.com\/?p=44869"},"modified":"2023-01-25T12:16:38","modified_gmt":"2023-01-25T06:46:38","slug":"ml-aggarwal-class-8-solutions-for-icse-maths-chapter-5-ex-5-1","status":"publish","type":"post","link":"https:\/\/cbselibrary.com\/ml-aggarwal-class-8-solutions-for-icse-maths-chapter-5-ex-5-1\/","title":{"rendered":"ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 5 Playing with Numbers Ex 5.1"},"content":{"rendered":"

ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 5 Playing with Numbers Ex 5.1<\/h2>\n

Question 1.
\nWrite the following numbers in generalized form:
\n(i) 89
\n(ii) 207
\n(iii) 369
\nSolution:
\n(i) 89 = 8 \u00d7 10 + 9
\n(ii) 207 = 2 \u00d7 100 + 0 \u00d7 10 + 7 \u00d7 1
\n(iii) 369 = 3 \u00d7 100 + 6 \u00d7 10 + 9 \u00d7 1<\/p>\n

Question 2.
\nWrite the quotient, when the sum of a 2-digit number 34 and number obtained by reversing the digits is divided by
\n(i) 11
\n(ii) sum of digits
\nSolution:
\nSum of two-digit number 34 and the number
\nobtained by reversing the digit 43 = 34 + 43 = 77
\n(i) 77 \u00f7 11 = 7
\n(ii) 77 \u00f7 (Sum of digit) = 77 \u00f7 (4 + 3)
\n= 77 \u00f7 7 = 11<\/p>\n

Question 3.
\nWrite the quotient when the difference of a 2-digit number 73 and number obtained by reversing the digits is divided by
\n(i) 9
\n(ii) a difference of digits.
\nSolution:
\nDifference of two digit number 73 and the number
\nobtained by reversing the digits
\n= 73 – 37 = 36
\n(i) Now, 36 \u00f7 9 = 4
\n(ii) 36 \u00f7 (7 – 3) = 36 + 4 = 9<\/p>\n

Question 4.
\nWithout actual calculation, write the quotient when the sum of a 3-digit number abc and the number obtained by changing the order of digits cyclically i.e. bca and cab is divided by
\n(i) 111
\n(ii) (a + b + c)
\n(iii) 37
\n(iv) 3
\nSolution:
\nSum of 3-digit number abc and the number
\nobtained by changing the order of digits i.e. bca and cab.
\n\u2234 abc + bca + cab
\n= 100a + 10b + c + 1006 + 10c + a + 100c + 10a + b
\n= 111a + 111b + 111c = 111 (a + b + c)
\n(i) When divided by 111 =\u00a0111 (a + b + c)\u00a0\u00f7 111 =\u00a0a + b + c
\n(ii) When divided by (a + b + c) =\u00a0111 (a + b + c)\u00a0\u00f7 (a + b + c) = 111
\n(iii) When divided by 37 =\u00a0111 (a + b + c)\u00a0\u00f7 37 = 3(a + b + c)
\n(iv) When divided by 3 =\u00a0111 (a + b + c)\u00a0\u00f7 3 = 37(a + b + c)<\/p>\n

Question 5.
\nWrite the quotient when the difference of a 3-digit number 843 and number obtained by reversing the digits is divided by
\n(i) 99
\n(ii) 5
\nSolution:
\nDifference of 3-digit number 843 and the number
\nobtained by reversing the digit 348
\n= 843 – 348 = 495
\n(i) Divided by 99, then \\(\\frac{495}{99}\\) = 5
\n(ii) Divided by 5, then = \\(\\frac{495}{5}\\) = 99<\/p>\n

Question 6.
\nThe sum of digits of a 2-digit number is 11. If the number obtained by reversing the digits is 9 less than the original number, find the number.
\nSolution:
\nSum of two digit number = 11
\nLet unit’s digit = x
\nand tens digit = y,
\nthen x + y = 11 …(i)
\nand number = x + 10y
\nBy reversing the digits,
\nUnit digit = y
\nand tens digit = x
\nand number = y + 10x
\nNow, y + 10x + 9 = x + 10y
\n10x + y – 10y – x = -9
\n9x – 9y = -9
\nx – y = -1 …(ii)
\nAdding (i) and (ii),
\n2x= 10 \u21d2 x= \\(\\frac{10}{2}\\) = 5
\n\u2234 y = 1 + 5 = 6
\nand number = x + 10y = 5 + 10 \u00d7 6
\n= 5 + 60 = 65<\/p>\n

Question 7.
\nIf the difference of two-digit number and the number obtained by reversing the digits is 36, find the difference between the digits of the 2-digit number.
\nSolution:
\nLet unit digit = x
\nand tens digit =y
\n\u2234 Number = x + 10y
\nand by reversing the digits
\nUnit digit =y
\nand tens digit = x
\nThen number = y + 10x
\n\u2234 x + 10y – y – 10x = 36
\n\u21d2 -9x + 9y = 36
\n(y – x) = \\(\\frac{36}{9}\\)
\n\u21d2 y – x = 4<\/p>\n

Question 8.
\nIf the sum of two-digit number and number obtained by reversing the digits is 55, find the siun of the digits of the 2-digit number.
\nSolution:
\nLet unit digit = x
\nand tens digit = y
\n\u2234 Number = x + 10y
\nand by reversing the digits
\nUnit digit =y
\nand tens digit = x
\nThen number = y + 10x
\n\u2234 x + 10y + y + 10x = 55
\n\u21d2 11x + 11y = 55
\n\u21d2 x + y = \\(\\frac{55}{11}\\)
\n\u21d2 x + y = 5
\n\u2234 Difference of the digits of the number = 4<\/p>\n

Question 9.
\nIn a 3-digit number, unit’s digit, ten’s digit and hundred’s digit are in the ratio 1 : 2 : 3. If the difference of original number and the number obtained by reversing the digits is 594, find the number.
\nSolution:
\nRatio in the digits of a three digit number
\n= 1 : 2 : 3
\nLet unit digit = x
\nThen tens digit = 2x
\nand hundreds digit = 3x
\nand number = x + 10 \u00d7 2x + 100 \u00d7 3x
\n= x + 20x + 300x = 321x
\nand reversing the digits,
\nUnit digit = 3x
\nTen’s digit = 2x
\nHundreds digit = x
\n\u2234 Number = 3x + 10 \u00d7 2x + 100 \u00d7 x
\n= 3x + 20x + 100 = 123x
\nAccording to the condition,
\n\u2234 321x – 123x = 594
\n\u21d2 198x = 594 \u21d2 x = \\(\\frac{594}{198}\\)= 3
\n\u2234 Number = 321x = 321 \u00d7 3 = 963<\/p>\n

Question 10.
\nIn a 3-digit number, unit’s digit is one more than the hundred’s digit and ten’s digit is one less than the hundred’s digit. If the sum of the original 3-digit number and numbers obtained by changing the order of digits cyclically is 2664, find the number.
\nSolution:
\nLet hundreds digit = x
\nThen unit digit = x + 1
\nand ten’s digit = x – 1
\nNumber = (x + 1) + 10(x – 1) + 100 \u00d7 x
\n= x + 1 + 10x – 10 + 100x
\n= 111x – 9
\nand by reversing the digits,
\nUnit digit = x – 1
\nTens digit = x
\nHundred digit = x + 1
\n\u2234 Number = x – 1 + 10x + 100x + 100
\n= 111x + 99
\nand number = x + 10(x + 1) + 100(x – 1)
\n= x + 10x + 10 + 100x – 100
\n= 111x – 90
\nNow according to the condition,
\n111x – 9 + 111x + 99 + 111x – 90 = 2664
\n\u21d2 333x + 99 – 99 = 2664
\n333x = 2664
\nx = \\(\\frac{2664}{333}\\) = 8
\n\u2234 Original number = 111x – 9
\n= 888 – 9 = 879<\/p>\n

ML Aggarwal Class 8 Solutions for ICSE Maths<\/a><\/h4>\n","protected":false},"excerpt":{"rendered":"

ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 5 Playing with Numbers Ex 5.1 Question 1. Write the following numbers in generalized form: (i) 89 (ii) 207 (iii) 369 Solution: (i) 89 = 8 \u00d7 10 + 9 (ii) 207 = 2 \u00d7 100 + 0 \u00d7 10 + 7 \u00d7 1 (iii) 369 … Read more<\/a><\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[3034],"tags":[],"yoast_head":"\nML Aggarwal Class 8 Solutions for ICSE Maths Chapter 5 Playing with Numbers Ex 5.1 - CBSE Library<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/cbselibrary.com\/ml-aggarwal-class-8-solutions-for-icse-maths-chapter-5-ex-5-1\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 5 Playing with Numbers Ex 5.1\" \/>\n<meta property=\"og:description\" content=\"ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 5 Playing with Numbers Ex 5.1 Question 1. 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