{"id":121674,"date":"2021-02-15T12:14:55","date_gmt":"2021-02-15T12:14:55","guid":{"rendered":"https:\/\/www.indcareer.com\/schools\/?p=121674"},"modified":"2023-09-16T01:23:47","modified_gmt":"2023-09-16T01:23:47","slug":"ncert-solutions-for-8th-class-maths-chapter-16-playing-with-numbers","status":"publish","type":"post","link":"https:\/\/www.indcareer.com\/schools\/ncert-solutions-for-8th-class-maths-chapter-16-playing-with-numbers\/","title":{"rendered":"NCERT Solutions for 8th Class Maths: Chapter 16-Playing with Numbers"},"content":{"rendered":"\n

NCERT Solutions for 8th Class Maths: Chapter 16-Playing with Numbers<\/strong><\/h2>\n\n\n\n

Page No: 220<\/p>\n\n\n\n

Exercise 16.1<\/h2>\n\n\n\n

Find the values of the letters in each of the following and give reasons for the steps involved.<\/strong><\/p>\n\n\n\n

1. <\/strong><\/p>\n\n\n\n

\"\"\/<\/figure>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

On putting A = 1, 2, 3, 4, 5, 6, 7 and so on and we get, 7 + 5 = 12 in which ones place is 2.
\u2234 A = 7
And putting 2 and carry over 1, we get
B = 6
Hence A = 7 and B = 6<\/p>\n\n\n\n

2<\/strong>.<\/p>\n\n\n\n

\"\"\/<\/figure>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

On putting A = 1, 2, 3, 4, 5, 6, 7 and so on and we get,8 + 5 = 13 in which ones place is 3.
\u2234 A = 5
And putting 3 and carry over 1, we get
B = 4 and C = 1
Hence A = 5, B = 4 and C = 1<\/p>\n\n\n\n

3.<\/strong><\/p>\n\n\n\n

\"\"\/<\/figure>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

On putting A = 1, 2, 3, 4, 5, 6, 7 and so on and we get, A x A = 6 x 6 = 36 in which ones place is 6.
\u2234 A = 6
Hence A = 6<\/p>\n\n\n\n

4.<\/strong><\/p>\n\n\n\n

\"\"\/<\/figure>\n\n\n\n

Answer<\/strong>
Here, we observe that B = 5
So that 7 + 5 = 12.
Putting 2 at ones place and carry over 1 and A = 2, we get
2 + 3 + 1 = 6
Hence A = 2 and B = 5<\/p>\n\n\n\n

5.<\/strong><\/p>\n\n\n\n

\"\"\/<\/figure>\n\n\n\n

Answer<\/strong>
Here on putting B = 0,
we get 0\u00d73 = 0.
And A = 5, then 5\u00d73 = 15
\u21d2 A = 5 and C = 1
Hence A = 5, B = 0 and C = 1<\/p>\n\n\n\n

6.<\/strong><\/p>\n\n\n\n

\"\"\/<\/figure>\n\n\n\n

Answer<\/strong>
On putting B = 0, we get 0 , and A = 5, then 5\u00d7 5 = 25
\u21d2 A = 5, C = 2
Hence A = 5, B = 0 and C = 2<\/p>\n\n\n\n

7.<\/strong><\/p>\n\n\n\n

\"\"\/<\/figure>\n\n\n\n

Answer<\/strong>
Here product of B and 6 must be same as ones place digit as B.
6 \u00d7 1 = 6, 6 \u00d7 2 = 12, 6 \u00d7 3 = 18,
6 \u00d7 4 = 24
On putting B = 4, we get the ones digit 4 and remaining two B\u2019s value should be 44.
\u2234 For 6 \u00d7 7 = 42 and carry over 2 = 44
Hence A = 7 and B = 4<\/p>\n\n\n\n

8.<\/strong><\/p>\n\n\n\n

\"\"\/<\/figure>\n\n\n\n

Answer<\/strong>
On putting B = 9, we get 9 + 1 = 10
Putting 0 at ones place and carry over 1, we get
For A = 7 \u21d2 7 + 1 + 1 = 9
Hence A = 7 and B = 9<\/p>\n\n\n\n

9.<\/strong><\/p>\n\n\n\n

\"\"\/<\/figure>\n\n\n\n

Answer<\/strong>
On putting B = 7,
\u21d2 7 + 1 = 8
Now A = 4, then 4 + 7 = 11
Putting 1 at tens place and carry over 1, we get
2 + 4 + 1 = 7
Hence A = 4 and B = 7<\/p>\n\n\n\n

10.<\/strong><\/p>\n\n\n\n

\"\"\/<\/figure>\n\n\n\n

Answer<\/strong>
Putting A = 8 and B = 1, we get
8 + 1 = 9
Now again we add 2 + 8 = 10
Tens place digit is \u20180\u2019 and carry over 1.
Now 1 + 6 + 1 = 8 = A
Hence A = 8 and B = 1<\/p>\n\n\n\n

Page No. 260<\/p>\n\n\n\n

Exercise 16.2<\/h2>\n\n\n\n

1. If 21y5 is a multiple of 9, where y is a digit, what is the value of y?<\/strong><\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

Since 21y5 is a multiple of 9.
Therefore according to the divisibility rule of 9, the sum of all the digits should be a multiple of 9.
\u2234 2+2+y+5= 8+y
\u21d2 8+y = 9
\u21d2 y = 1
Since 21y5 is a multiple of 9.<\/p>\n\n\n\n

2. If 31z5 is a multiple of 9, where z is a digit, what is the value of z? You will find that there are two answers for the last problem. Why is this so?<\/strong><\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

Since 31z5 is a multiple of 9.
Therefore according to the divisibility rule of 9, the sum of all the digits should be a multiple of 9.
\u2234 3+1+z+5 = 9+z
\u21d2 9+z = 9
\u21d2 z= 0
If 3 +1+z+5 = 9+z
\u21d2 9+a = 18
\u21d2 z= 9
Hence, 0 and 9 are two possible answers.<\/p>\n\n\n\n

3. If 24x is a multiple of 3, where x is a digit, what is the value of x?
(Since 24x is a multiple of 3, its sum of digits 6 + x is a multiple of 3; so 6 + x is one of these numbers: 0, 3, 6, 9, 12, 15, 18, … But since x is a digit, it can only be that 6 + x = 6 or 9 or 12 or 15. Therefore, x = 0 or 3 or 6 or 9. Thus, x can have any of four different values.)<\/strong><\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

Since, 24x is a multiple of 3.
Therefore according to the divisibility rule of 3, the sum of all the digits should be a multiple of 3.
\u2234 2+4+x = 6 +x
Since x is a digit.
\u21d26+x= 6 \u21d2 x= 0
\u21d2 6+x = 9 \u21d2 x= 3
\u21d2 6+ x = 12 \u21d2x= 6
\u21d2 6+x = 15 \u21d2 x = 9
Thus, x can have any of four different values.<\/p>\n\n\n\n

4. If 31z5 is a multiple of 3, where z is a digit, what might be the values of z?<\/strong><\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

Since 31z5 is a multiple of 3.
Therefore according to the divisibility rule of 3, the sum of all the digits should be a multiple of 3.
Since z is a digit.
\u2234 3+1+z+5 = 9+z
\u21d2 9+z=9 \u21d2 z= 0
If 3+1+z+5 = 9+z
\u21d2 9 +z = 12 \u21d2 z= 3
If 3+1+z+5 = 9+z
\u21d2 9 + z= 15 \u21d2 z= 6
If 3+1+z+5 = 9+z
\u21d2 9 +z = 18 \u21d2 z= 9
Hence 0, 3, 6 and 9 are four possible answers.<\/p>\n\n\n\n

Chapterwise NCERT Solutions for 8th Class Maths<\/h2>\n\n\n\n

Chapter 1 \u2013 Rational Numbers<\/a>
Chapter 2 \u2013 Linear Equations in One Variable<\/a>
Chapter 3 \u2013 Understanding Quadrilaterals<\/a>
Chapter 4 \u2013 Practical Geometry<\/a>
Chapter 5 \u2013 Data Handling<\/a>
Chapter 6 \u2013 Squares and Square Roots<\/a>
Chapter 7 \u2013 Cubes and Cube Roots<\/a>
Chapter 8 \u2013 Comparing Quantities<\/a>
Chapter 9 \u2013 Algebraic Expressions and Identities<\/a>
Chapter 10 \u2013 Visualizing Solid Shapes<\/a>
Chapter 11 \u2013 Mensuration<\/a>
Chapter 12 \u2013 Exponents and Powers<\/a>
Chapter 13 \u2013 Direct and Inverse Proportions<\/a>
Chapter 14 \u2013 Factorization<\/a>
Chapter 15 \u2013 Introduction to Graphs<\/a>
Chapter 16 \u2013 Playing with Numbers<\/a><\/p>\n\n\n\n

About<\/h2>\n\n\n\n

The National Council of Educational Research and Training is an autonomous organization of the Government of India which was established in 1961 as a literary, scientific, and charitable Society under the Societies Registration Act. Its headquarters are located at Sri Aurbindo Marg in New Delhi. Visit the Official NCERT website<\/a> to learn more. <\/p>\n\n\n\n

\n
NCERT Solutions<\/a><\/div>\n\n\n\n
NCERT Solutions for Class 8<\/a><\/div>\n\n\n\n
NCERT Solutions for Class 8 Maths<\/a><\/div>\n<\/div>\n","protected":false},"excerpt":{"rendered":"

NCERT Solutions for 8th Class Maths: Chapter 16-Playing with Numbers Page No: 220 Exercise 16.1 Find the values of the letters in each of the following and give reasons for the steps involved. 1.  Answer On putting A = 1, 2, 3, 4, 5, 6, 7 and so on and we get, 7 + 5 […]<\/p>\n","protected":false},"author":302,"featured_media":627569,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"newspack_featured_image_position":"","newspack_post_subtitle":"","newspack_article_summary_title":"Overview:","newspack_article_summary":"","newspack_hide_updated_date":false,"newspack_show_updated_date":false,"footnotes":""},"categories":[1411,58],"tags":[1506],"boards":[1180],"class_list":["post-121674","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-book-solutions","category-class-8","tag-ncert-maths-class-8","boards-ncert","entry"],"yoast_head":"\nNCERT Solutions for 8th Class Maths: Chapter 16-Playing 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