{"id":27995,"date":"2021-07-02T16:16:30","date_gmt":"2021-07-02T10:46:30","guid":{"rendered":"https:\/\/mcq-questions.com\/?p=27995"},"modified":"2022-03-02T10:30:56","modified_gmt":"2022-03-02T05:00:56","slug":"ncert-solutions-for-class-8-maths-chapter-16-intext-questions","status":"publish","type":"post","link":"https:\/\/mcq-questions.com\/ncert-solutions-for-class-8-maths-chapter-16-intext-questions\/","title":{"rendered":"NCERT Solutions for Class 8 Maths Chapter 16 Playing with Numbers InText Questions"},"content":{"rendered":"

These NCERT Solutions for Class 8 Maths<\/a> Chapter 16 Playing with Numbers InText Questions and Answers are prepared by our highly skilled subject experts.<\/p>\n

NCERT Solutions for Class 8 Maths Chapter 16 Playing with Numbers InText Questions<\/h2>\n

NCERT Intext Question Page No. 250<\/span><\/p>\n

Question 1.
\nWrite the following numbers in generalised form.;
\n(i) 25
\n(ii) 73
\n(iii) 129
\n(iv) 302
\nAnswer:
\n(i) 25 = 20 + 5 = 2 \u00d7 10 + 5 \u00d7 1 = 10 \u00d7 2 + 5
\n(ii) 73 = 70 + 3 = 7 \u00d7 10 + 3 \u00d7 1 = 10 \u00d7 7 + 3
\n(iii) 129 = 100 + 20 + 9
\n= 1 \u00d7 100 + 2 \u00d7 10 + 9 \u00d7 1
\n= 100 \u00d7 1 + 10 \u00d7 2 + 9
\n(iv) 302 = 3 \u00d7 100 + 0 \u00d7 10 + 2 \u00d7 1
\n= 100 \u00d7 3 + 10 \u00d7 0 + 2<\/p>\n

\"NCERT<\/p>\n

Question 2.
\nWrite the following in the usual form.
\n(i) 10 \u00d7 5 + 6
\n(ii) 100 \u00d7 7 + 10 \u00d7 1 + 8
\n(iii) 100 \u00d7 a + 10 \u00d7 c + b
\nAnswer:
\n(i) 10 \u00d7 5 + 6 = 50 + 6 = 56
\n(ii) 100 \u00d7 7 + 10 \u00d7 1 + 8
\n= 700 + 10 + 8 = 718
\n(iii) 100 \u00d7 a + 10 \u00d7 c + b
\n= 100a + 10c + b = a c b<\/p>\n

NCERT Intext Question Page No. 257<\/span><\/p>\n

Question 1.
\nIf the division N + 5 leave a remainder of 3, what might be the one’s digit of N?
\nAnswer:
\nThe unit digit when divided by 5 must leave a remainder of 3.50, the one\u2019s digit must be either 3 or 8.<\/p>\n

Question 2.
\nIf the division N + 5 leaves a remainder of 1, what might be the ones digit of N?
\nAnswer:
\nThe one\u2019s digit when divided by 5 must have a remainder of 1. So the one\u2019s digit must be either 1 or 6.<\/p>\n

Question 3.
\nIf the division N + 5 leaves a remainder of 4. What might be the one\u2019s digit of N?
\nAnswer:
\nThe one\u2019s digit, when divided by 5 must leave a remainder of 4. So the one\u2019s digit must be either 4 or 9.<\/p>\n

\"NCERT<\/p>\n

NCERT Intext Question Page No. 259<\/span><\/p>\n

Question 1.
\nCheck the divisiblity of the following number by 9.
\n(i) 108
\n(ii) 616
\n(iii) 294
\n(iv) 432
\n(v) 927
\nAnswer:
\n(i) 108
\nSum of the digits = 1 + 0 + 8 = 9 which is divisible by 9.
\n\u2234 108 is divisible by 9.<\/p>\n

(ii) 616
\nSum of the digits = 6 + 1 + 6
\n= 13 which is not divisible by 9.
\n\u2234 616 is not divisible by 9.<\/p>\n

(iii) 294
\nSum of the digits = 2 + 9 + 4 = 15
\nwhich is not divisible by 9.
\n\u2234 294 is not divisible by 9.<\/p>\n

(iv) 432
\nSum of the digit = 4 + 3 + 2 = 9
\nwhich is divisible by 9.
\n\u2234 432 is divisible by 9.<\/p>\n

(v) 927
\nSum of the digits = 9 + 2 + 7 = 18
\nwhich is divisible by 9.
\n\u2234 927 is divisible by 9.<\/p>\n

\"NCERT<\/p>\n

Question 9.
\nCheck the divisibility of the following number by 3.
\n(i) 108
\n(ii) 616
\n(iii) 294
\n(iv) 432
\n(v) 927
\nAnswer:
\n(i) 108
\nSum of the digits =1 + 0 + 8 = 9
\nwhich is divisible by 3.
\n\u2234 108 is divisible by 3.<\/p>\n

(ii) 616
\nSum of the digits = 6 + 1 + 6 = 13
\nwhich is not divisible by 3.
\n\u2234 613 is also not divisible by 3.<\/p>\n

(iii) 294
\nSum of the digits = 2 + 9 + 4 = 15
\nwhich is divisible by 3.
\n\u2234 294 is also divisible by 3.<\/p>\n

(iv) 432
\nSum of the digits = 4 + 3 + 2 = 9
\nwhich is divisible by 3.
\n\u2234 432 is also divisible by 3.<\/p>\n

(v) 927
\nSum of the digits = 9 + 2 + 7 = 18 which is divisible by 3.
\n\u2234 Thus 927 is also divisible by 3.<\/p>\n

\"NCERT<\/p>\n

NCERT Intext Question Page No. 251<\/span><\/p>\n

Question 1.
\nCheck what the result would have been if Sundaram had chosen the numbers shown below.
\n1. 27
\n2. 39
\n3. 64
\n4. 17
\nAnswer:
\n1. Chosen number = 27
\nNumber with reversed digits = 72
\nSum of the two numbers = 27 + 72 = 99
\nNow, 99 = 11 [9] = 11 [2 + 7]
\n= 11 [Sum of the digits of the chosen number]<\/p>\n

2. Chose number = 39
\nNumber with reversed digits = 93
\nSum of the two numbers = 39 + 93 = 132
\nNow, 132 \u00f7 11 = 12
\ni. e\u201e 132 = 11 [12] = 11 [3 + 9]
\n= 11 [Sum of the digits of the chose number]<\/p>\n

3. Chosen number = 17
\nNumber with reversed digits = 71
\nSum = 17 + 71 = 88
\nNow, 88 = 11 [8] = 11 [1 + 7]
\n= 11 [Sum of the digits of the chosen number]<\/p>\n

Question 2.
\nCheck what the result would have been if Sundaram had chosen the numbers shown.
\n1. 17
\n2. 21
\n3. 96
\n4. 37
\nAnswer:
\n1. Chosen number =17
\nNumber with reversed digits = 71
\nDifference = 71 – 17 = 54 = 9 \u00d7 [6]
\n= 9 x [Difference of the digits of the chosen number (7 – 1 = 6)]<\/p>\n

\"NCERT<\/p>\n

2. Chosen number = 21
\nNumber with reversed digits = 12
\nDifference = 21 – 12 = 9 = 9 \u00d7 [1]
\n= 9 x [Difference of the digits of the chosen number (12 -1 = 1)]<\/p>\n

3. Chosen number = 96
\nNumber with reversed digits = 69
\nDifference = 96 – 69 = 27 = 9 \u00d7 [3]
\n=9 x [Difference of the digits of the chosen number (9 – 1 = 3)]<\/p>\n

4. Chosen number = 37
\nNumber with reversed digits = 73
\nDifference = 73 – 37 = 36 = 9 \u00d7 [4]
\n=9 x [Difference of the digits of the chosen number (7-3 = 4)]<\/p>\n

NCERT Intext Question Page No. 252<\/span><\/p>\n

Question 1.
\nCheck what the result would have been if Minakshi had chosen the numbers shown below. In each case keep a record of the quotient obtained at the end.
\n1. 132
\n2. 469
\n3. 737
\n4. 901
\nAnswer:
\n1. 132 Chosen number = 132
\nReversed number = 231
\nDifference = 231 – 132 = 99
\nWe have 99 \u00f7 99 = 1, remainder = 0<\/p>\n

2. 469 Chosen number = 469
\nReversed number = 964
\nDifference = 964 – 469 = 495
\nWe have 99 \u00f7 99 = 5, remainder = 0<\/p>\n

3. Chosen number = 737
\nReversed number = 737
\nWe have Difference = 737 – 737 = 0
\n0 \u00f7 99 =0, remainder = 0<\/p>\n

4. Chosen number = 901
\nReversed number = 109
\nDifference = 901 – 109 = 792
\nWe have 792 \u00f7 99 = 8, remainder = 0
\nForming three-digit number with given three digits<\/p>\n

NCERT Intext Question Page No. 253<\/span><\/p>\n

Question 1.
\nCheck what the result would have been if Sundaram had chosen the numbers shown below.
\n1. 417
\n2. 632
\n3. 117
\n4. 937
\nAnswer:
\n1. Chosen number = 417
\nTwo other numbers with the same digits are 741 and 174
\nSum of the three numbers
\n\"NCERT
\nWe have 1332 \u00f7 37 = 36, remainder = 0<\/p>\n

2. Chosen number = 632
\nTwo other numbers are 263 and 326
\nSum of the three numbers
\n\"NCERT
\nWe have 1221 \u00f7 37 = 33, remainder = 0<\/p>\n

\"NCERT<\/p>\n

3. Chosen number =117
\nOther numbers are 711 and 171
\nSum of the three numbers
\n\"NCERT
\nWe have 1221 \u00f7 37 = 33, remainder = 0<\/p>\n

4. Chosen number = 937
\nOther two numbers are 793 and 379
\nSum of the three numbers
\n\"NCERT
\nWe have 2109 + 37 = 57, remainder = 0<\/p>\n","protected":false},"excerpt":{"rendered":"

These NCERT Solutions for Class 8 Maths Chapter 16 Playing with Numbers InText Questions and Answers are prepared by our highly skilled subject experts. NCERT Solutions for Class 8 Maths Chapter 16 Playing with Numbers InText Questions NCERT Intext Question Page No. 250 Question 1. Write the following numbers in generalised form.; (i) 25 (ii) …<\/p>\n

NCERT Solutions for Class 8 Maths Chapter 16 Playing with Numbers InText Questions<\/span> Read More »<\/a><\/p>\n","protected":false},"author":9,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"default","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","theme-transparent-header-meta":"default","adv-header-id-meta":"","stick-header-meta":"default","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","spay_email":""},"categories":[7],"tags":[],"yoast_head":"\nNCERT Solutions for Class 8 Maths Chapter 16 Playing with Numbers InText Questions - MCQ Questions<\/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:\/\/mcq-questions.com\/ncert-solutions-for-class-8-maths-chapter-16-intext-questions\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"NCERT Solutions for Class 8 Maths Chapter 16 Playing with Numbers InText Questions - MCQ Questions\" \/>\n<meta property=\"og:description\" content=\"These NCERT Solutions for Class 8 Maths Chapter 16 Playing with Numbers InText Questions and Answers are prepared by our highly skilled subject experts. 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