NCERT Intext Question Page No. 259<\/span><\/p>\nQuestion 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
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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
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NCERT Intext Question Page No. 251<\/span><\/p>\nQuestion 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
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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>\nQuestion 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>\nQuestion 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 \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 \nWe have 1221 \u00f7 37 = 33, remainder = 0<\/p>\n
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3. Chosen number =117 \nOther numbers are 711 and 171 \nSum of the three numbers \n \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 \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 \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n\t \n\t \n\t \n