{"id":95455,"date":"2019-10-04T11:32:05","date_gmt":"2019-10-04T06:02:05","guid":{"rendered":"https:\/\/www.cbselabs.com\/?p=95455"},"modified":"2021-09-18T15:16:26","modified_gmt":"2021-09-18T09:46:26","slug":"ncert-solutions-for-class-8-maths-chapter-16-playing-with-numbers","status":"publish","type":"post","link":"https:\/\/www.cbselabs.com\/ncert-solutions-for-class-8-maths-chapter-16-playing-with-numbers\/","title":{"rendered":"NCERT Solutions for Class 8 Maths Chapter 16 Playing with Numbers"},"content":{"rendered":"
NCERT Solutions for Class 8 Maths Chapter 16 Playing with Numbers Exercise 16.1<\/strong><\/p>\n Find the values of the letters in each of the following and give reasons for the steps involved. Ex 16.1 Class 8 Maths\u00a0Question 2. <\/p>\n Ex 16.1 Class 8 Maths\u00a0Question 3. Ex 16.1 Class 8 Maths\u00a0Question 4. <\/p>\n Ex 16.1 Class 8 Maths\u00a0Question 5. Ex 16.1 Class 8 Maths\u00a0Question 6. Ex 16.1 Class 8 Maths\u00a0Question 7. Ex 16.1 Class 8 Maths\u00a0Question 8. Ex 16.1 Class 8 Maths\u00a0Question 9. Ex 16.1 Class 8 Maths\u00a0Question 10. <\/p>\n <\/p>\n <\/p>\n <\/p>\n <\/p>\n <\/p>\n <\/p>\n <\/p>\n NCERT Solutions for Class 8 Maths Chapter 16 Playing with Numbers Class 8 Maths Playing with Numbers Exercise 16.1 Class 8 Maths Playing with Numbers Exercise 16.2 Playing with Numbers Class 8 Extra Questions NCERT Solutions for Class 8 Maths Chapter 16 Playing with Numbers Exercise 16.1 Find the values of the letters in each …<\/p>\n","protected":false},"author":27,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[2],"tags":[],"yoast_head":"\n
\nEx 16.1 Class 8 Maths Question 1.
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\nSolution:
\n<\/p>\n
\n
\nSolution:
\n<\/p>\n
\n
\nSolution:
\n<\/p>\n
\n
\nSolution:
\n<\/p>\n
\n
\nSolution:
\n3 \u00d7 B = B
\n\u21d2 B = D
\n3 \u00d7 A = CA
\n\u21d2 3 \u00d7 5 = 15
\nThus A = 5 and C = 1
\nHence A = 5, B = 0 and C = 1<\/p>\n
\n
\nSolution:
\n5 \u00d7 B = B
\n\u21d2 B = 0 or 5
\n5 \u00d7 A = CA
\n5 \u00d7 5 = 25
\nOnly possible when B = 0
\nThus A = 5 and C = 2
\nHence A = 5, B = 0 and C = 2<\/p>\n
\n
\nSolution:
\nB \u00d7 6 = B
\n6 \u00d7 4 = 24 \u2192 B = 4 and 2 is carried to
\n6 \u00d7 A = BB
\n\u21d2 6 \u00d7 7 = 42 + 2 (carried on) = 44
\nThus B = 7
\nHence A = 7 and B = 4<\/p>\n
\n
\nSolution:
\n1 + B = 0
\n1 + 9 = 10 \u2192 unit digit is 0 and 1 is carried to A
\n+ 1 +1 (carried on) = B = 9
\nA + 2 = 9 \u21d2 A = 9 – 2 = 7
\nHence A = 7 and B = 9<\/p>\n
\n
\nSolution:
\nB + 1 = 8 \u21d2 B = 8 – 1 = 7
\nA + B = a number with unit digit 1
\nA + B = 11
\n\u21d2 A + 7 = 11
\n\u21d2 A = 11 – 7 = 4 (1 Carried to)
\nNow 1 carried on + 2 + A = B
\n3 + A = 7
\n\u21d2 A = 7 – 3 = 4
\nHence A = 4, B = 7<\/p>\n
\n
\nSolution:
\n9 = A + B
\n9 = 1 + 8 or 2 + 7 or 3 + 6 or 4 + 5 or 8 + 1 or 7 + 2 or 6 + 3 or 5 + 4 or 0 + 9 or 9 + 0
\nNow 0 is required at unit place
\n2 + A = 10
\n\u21d2 A = 10 – 2 = 8
\nB = 9 – 8 = 1
\n1 + 6 + 1 (carried on) = A = 8
\nHence A = 8 and B = 1<\/p>\nMore CBSE Class 8 Study Material<\/h4>\n
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