{"id":598130,"date":"2022-05-02T10:37:14","date_gmt":"2022-05-02T10:37:14","guid":{"rendered":"https:\/\/www.indcareer.com\/schools\/?p=598130"},"modified":"2022-05-09T06:28:07","modified_gmt":"2022-05-09T06:28:07","slug":"selina-class-7-icse-solutions-mathematics-chapter-1-integers","status":"publish","type":"post","link":"https:\/\/www.indcareer.com\/schools\/selina-class-7-icse-solutions-mathematics-chapter-1-integers\/","title":{"rendered":"Selina Class 7 ICSE Solutions Mathematics : Chapter 1-\u00a0Integers"},"content":{"rendered":"\n<p>Class 7: Maths Chapter 1 solutions. Complete Class 7 Maths Chapter 1 Notes.<\/p>\n\n\n\n<h2 class=\"wp-block-heading\" id=\"h-selina-class-7-icse-solutions-mathematics-chapter-1-integers\">Selina Class 7 ICSE Solutions Mathematics : Chapter 1-&nbsp;Integers<\/h2>\n\n\n\n<p>Selina 7th Maths Chapter 1, Class 7 Maths Chapter 1 solutions<\/p>\n\n\n\n<h4 class=\"wp-block-heading\">Exercise 1A page: 6<\/h4>\n\n\n\n<p><strong>1. Evaluate:<\/strong><\/p>\n\n\n\n<p><strong>(i) 427 \u00d7 8 + 2 \u00d7 427<\/strong><\/p>\n\n\n\n<p><strong>(ii) 394 \u00d7 12 + 394 \u00d7 (-2)<\/strong><\/p>\n\n\n\n<p><strong>(iii) 558 \u00d7 27 + 3 \u00d7 558<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>(i) 427 \u00d7 8 + 2 \u00d7 427<\/p>\n\n\n\n<p>Using Distributive property<\/p>\n\n\n\n<p>= 427 \u00d7 (8 + 2)<\/p>\n\n\n\n<p>By further calculation<\/p>\n\n\n\n<p>= 427 \u00d7 10<\/p>\n\n\n\n<p>= 4270<\/p>\n\n\n\n<p>(ii) 394 \u00d7 12 + 394 \u00d7 (-2)<\/p>\n\n\n\n<p>Using Distributive property<\/p>\n\n\n\n<p>= 394 \u00d7 (12 \u2013 2)<\/p>\n\n\n\n<p>By further calculation<\/p>\n\n\n\n<p>= 394 \u00d7 10<\/p>\n\n\n\n<p>= 3940<\/p>\n\n\n\n<p>(iii) 558 \u00d7 27 + 3 \u00d7 558<\/p>\n\n\n\n<p>Using Distributive property<\/p>\n\n\n\n<p>= 558 \u00d7 (27 + 3)<\/p>\n\n\n\n<p>By further calculation<\/p>\n\n\n\n<p>= 558 \u00d7 30<\/p>\n\n\n\n<p>= 16740<\/p>\n\n\n\n<p><strong>2. Evaluate:<\/strong><\/p>\n\n\n\n<p><strong>(i) 673 \u00d7 9 + 673<\/strong><\/p>\n\n\n\n<p><strong>(ii) 1925 \u00d7 101 \u2013 1925<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>(i) 673 \u00d7 9 + 673<\/p>\n\n\n\n<p>Using Distributive property<\/p>\n\n\n\n<p>= 673 \u00d7 (9 + 1)<\/p>\n\n\n\n<p>By further calculation<\/p>\n\n\n\n<p>= 673 \u00d7 10<\/p>\n\n\n\n<p>= 6730<\/p>\n\n\n\n<p>(ii) 1925 \u00d7 101 \u2013 1925<\/p>\n\n\n\n<p>Using Distributive property<\/p>\n\n\n\n<p>= 1925 \u00d7 (101 \u2013 1)<\/p>\n\n\n\n<p>By further calculation<\/p>\n\n\n\n<p>= 1925 \u00d7 100<\/p>\n\n\n\n<p>= 192500<\/p>\n\n\n\n<p><strong>3. Verify:<\/strong><\/p>\n\n\n\n<p><strong>(i) 37 \u00d7 {8 + (-3)} = 37 \u00d7 8 + 37 \u00d7 (-3)<\/strong><\/p>\n\n\n\n<p><strong>(ii) (-82) \u00d7 {(- 4) + 19} = (-82) \u00d7 (- 4) + (-82) \u00d7 19<\/strong><\/p>\n\n\n\n<p><strong>(iii) {7 \u2013 (-7)} \u00d7 7 = 7 \u00d7 7 \u2013 (-7) \u00d7 7<\/strong><\/p>\n\n\n\n<p><strong>(iv) {(-15) \u2013 8} \u00d7 -6 = (-15) \u00d7 (-6) \u2013 8 \u00d7 (-6)<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>(i) 37 \u00d7 {8 + (-3)} = 37 \u00d7 8 + 37 \u00d7 (-3)<\/p>\n\n\n\n<p>Consider LHS = 37 \u00d7 {8 + (-3)}<\/p>\n\n\n\n<p>It can be written as<\/p>\n\n\n\n<p>= 37 \u00d7 {8 \u2013 3}<\/p>\n\n\n\n<p>By further calculation<\/p>\n\n\n\n<p>= 37 \u00d7 {5}<\/p>\n\n\n\n<p>= 185<\/p>\n\n\n\n<p>Similarly RHS = 37 \u00d7 8 + 37 \u00d7 (-3)<\/p>\n\n\n\n<p>It can be written as<\/p>\n\n\n\n<p>= 37 \u00d7 (8 \u2013 3)<\/p>\n\n\n\n<p>By further calculation<\/p>\n\n\n\n<p>= 37 \u00d7 5<\/p>\n\n\n\n<p>= 185<\/p>\n\n\n\n<p>Therefore, LHS = RHS.<\/p>\n\n\n\n<p>(ii) (-82) \u00d7 {(- 4) + 19} = (-82) \u00d7 (- 4) + (-82) \u00d7 19<\/p>\n\n\n\n<p>Consider LHS = (-82) \u00d7 {(- 4) + 19}<\/p>\n\n\n\n<p>It can be written as<\/p>\n\n\n\n<p>= (-82) \u00d7 {-4 + 19}<\/p>\n\n\n\n<p>By further calculation<\/p>\n\n\n\n<p>= (-82) \u00d7 {15}<\/p>\n\n\n\n<p>= \u2013 1230<\/p>\n\n\n\n<p>Similarly RHS = (-82) \u00d7 (- 4) + (-82) \u00d7 19<\/p>\n\n\n\n<p>It can be written as<\/p>\n\n\n\n<p>= (-82) \u00d7 (-4 + 19)<\/p>\n\n\n\n<p>By further calculation<\/p>\n\n\n\n<p>= \u2013 82 \u00d7 15<\/p>\n\n\n\n<p>= \u2013 1230<\/p>\n\n\n\n<p>Therefore, LHS = RHS.<\/p>\n\n\n\n<p>(iii) {7 \u2013 (-7)} \u00d7 7 = 7 \u00d7 7 \u2013 (-7) \u00d7 7<\/p>\n\n\n\n<p>Consider LHS = {7 \u2013 (-7)} \u00d7 7<\/p>\n\n\n\n<p>It can be written as<\/p>\n\n\n\n<p>= {7 + 7} \u00d7 7<\/p>\n\n\n\n<p>By further calculation<\/p>\n\n\n\n<p>= {14} \u00d7 7<\/p>\n\n\n\n<p>= 98<\/p>\n\n\n\n<p>Similarly RHS = 7 \u00d7 7 \u2013 (-7) \u00d7 7<\/p>\n\n\n\n<p>It can be written as<\/p>\n\n\n\n<p>= 7 \u00d7 7 + 7 \u00d7 7<\/p>\n\n\n\n<p>By further calculation<\/p>\n\n\n\n<p>= 7 \u00d7 (7 + 7)<\/p>\n\n\n\n<p>= 7 \u00d7 14<\/p>\n\n\n\n<p>= 98<\/p>\n\n\n\n<p>Therefore, LHS = RHS.<\/p>\n\n\n\n<p>(iv) {(-15) \u2013 8} \u00d7 -6 = (-15) \u00d7 (-6) \u2013 8 \u00d7 (-6)<\/p>\n\n\n\n<p>Consider LHS = {(-15) \u2013 8} \u00d7 -6<\/p>\n\n\n\n<p>It can be written as<\/p>\n\n\n\n<p>= {- 15 \u2013 8} \u00d7 -6<\/p>\n\n\n\n<p>By further calculation<\/p>\n\n\n\n<p>= {-23} \u00d7 -6<\/p>\n\n\n\n<p>= 138<\/p>\n\n\n\n<p>Similarly RHS = (-15) \u00d7 (-6) \u2013 8 \u00d7 (-6)<\/p>\n\n\n\n<p>It can be written as<\/p>\n\n\n\n<p>= \u2013 6 \u00d7 (- 15 \u2013 8)<\/p>\n\n\n\n<p>By further calculation<\/p>\n\n\n\n<p>= \u2013 6 \u00d7 \u2013 23<\/p>\n\n\n\n<p>= 138<\/p>\n\n\n\n<p>Therefore, LHS = RHS.<\/p>\n\n\n\n<p><strong>4. Evaluate:<\/strong><\/p>\n\n\n\n<p><strong>(i) 15 \u00d7 8<\/strong><\/p>\n\n\n\n<p><strong>(ii) 15 \u00d7 (-8)<\/strong><\/p>\n\n\n\n<p><strong>(iii) (-15) \u00d7 8<\/strong><\/p>\n\n\n\n<p><strong>(iv) (-15) \u00d7 -8<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>(i) 15 \u00d7 8 = 120<\/p>\n\n\n\n<p>(ii) 15 \u00d7 (-8) = \u2013 120<\/p>\n\n\n\n<p>(iii) (-15) \u00d7 8 = \u2013 120<\/p>\n\n\n\n<p>(iv) (-15) \u00d7 -8 = 120<\/p>\n\n\n\n<p><strong>5. Evaluate:<\/strong><\/p>\n\n\n\n<p><strong>(i) 4 \u00d7 6 \u00d7 8<\/strong><\/p>\n\n\n\n<p><strong>(ii) 4 \u00d7 6 \u00d7 (-8)<\/strong><\/p>\n\n\n\n<p><strong>(iii) 4 \u00d7 (-6) \u00d7 8<\/strong><\/p>\n\n\n\n<p><strong>(iv) (-4) \u00d7 6 \u00d7 8<\/strong><\/p>\n\n\n\n<p><strong>(v) 4 \u00d7 (-6) \u00d7 (-8)<\/strong><\/p>\n\n\n\n<p><strong>(vi) (-4) \u00d7 (-6) \u00d7 8<\/strong><\/p>\n\n\n\n<p><strong>(vii) (-4) \u00d7 6 \u00d7 (-8)<\/strong><\/p>\n\n\n\n<p><strong>(viii) (-4) \u00d7 (-6) \u00d7 (-8)<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>(i) 4 \u00d7 6 \u00d7 8 = 192<\/p>\n\n\n\n<p>(ii) 4 \u00d7 6 \u00d7 (-8) = \u2013 192<\/p>\n\n\n\n<p>It has one negative factor<\/p>\n\n\n\n<p>(iii) 4 \u00d7 (-6) \u00d7 8 = \u2013 192<\/p>\n\n\n\n<p>It has one negative factor<\/p>\n\n\n\n<p>(iv) (-4) \u00d7 6 \u00d7 8 = \u2013 192<\/p>\n\n\n\n<p>It has one negative factor<\/p>\n\n\n\n<p>(v) 4 \u00d7 (-6) \u00d7 (-8) = 192<\/p>\n\n\n\n<p>It has two negative factors<\/p>\n\n\n\n<p>(vi) (-4) \u00d7 (-6) \u00d7 8 = 192<\/p>\n\n\n\n<p>It has two negative factors<\/p>\n\n\n\n<p>(vii) (-4) \u00d7 6 \u00d7 (-8) = 192<\/p>\n\n\n\n<p>It has two negative factors<\/p>\n\n\n\n<p>(viii) (-4) \u00d7 (-6) \u00d7 (-8) = \u2013 192<\/p>\n\n\n\n<p>It has three negative factors<\/p>\n\n\n\n<p><strong>6. Evaluate:<\/strong><\/p>\n\n\n\n<p><strong>(i) 2 \u00d7 4 \u00d7 6 \u00d7 8<\/strong><\/p>\n\n\n\n<p><strong>(ii) 2 \u00d7 (-4) \u00d7 6 \u00d7 8<\/strong><\/p>\n\n\n\n<p><strong>(iii) (-2) \u00d7 4 \u00d7 (-6) \u00d7 8<\/strong><\/p>\n\n\n\n<p><strong>(iv) (-2) \u00d7 (-4) \u00d7 6 \u00d7 (-8)<\/strong><\/p>\n\n\n\n<p><strong>(v) (-2) \u00d7 (-4) \u00d7 (-6) \u00d7 (-8)<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>(i) 2 \u00d7 4 \u00d7 6 \u00d7 8 = 384<\/p>\n\n\n\n<p>(ii) 2 \u00d7 (-4) \u00d7 6 \u00d7 8 = -384<\/p>\n\n\n\n<p>The number of negative integer in the product is odd<\/p>\n\n\n\n<p>(iii) (-2) \u00d7 4 \u00d7 (-6) \u00d7 8 = 384<\/p>\n\n\n\n<p>The number of negative integer in the product is even<\/p>\n\n\n\n<p>(iv) (-2) \u00d7 (-4) \u00d7 6 \u00d7 (-8) = -384<\/p>\n\n\n\n<p>The number of negative integer in the product is odd<\/p>\n\n\n\n<p>(v) (-2) \u00d7 (-4) \u00d7 (-6) \u00d7 (-8) = 384<\/p>\n\n\n\n<p>The number of negative integer in the product is even<\/p>\n\n\n\n<p><strong>7. Determine the integer whose product with \u2018-1\u2019 is:<\/strong><\/p>\n\n\n\n<p><strong>(i) -47<\/strong><\/p>\n\n\n\n<p><strong>(ii) 63<\/strong><\/p>\n\n\n\n<p><strong>(iii) -1<\/strong><\/p>\n\n\n\n<p><strong>(iv) 0<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>(i) -47 = \u2013 1 \u00d7 47<\/p>\n\n\n\n<p>Therefore, the integer is 47.<\/p>\n\n\n\n<p>(ii) 63 = \u2013 1 \u00d7 \u2013 63<\/p>\n\n\n\n<p>Therefore, the integer is \u2013 63.<\/p>\n\n\n\n<p>(iii) -1 = \u2013 1 \u00d7 1<\/p>\n\n\n\n<p>Therefore, the integer is 1.<\/p>\n\n\n\n<p>(iv) 0 = -1 \u00d7 0<\/p>\n\n\n\n<p>Therefore, the integer is 0.<\/p>\n\n\n\n<p><strong>8. Eighteen integers are multiplied together. What will be the sign of their product, if:<\/strong><\/p>\n\n\n\n<p><strong>(i) 15 of them are negative and 3 are positive?<\/strong><\/p>\n\n\n\n<p><strong>(ii) 12 of them are negative and 6 are positive?<\/strong><\/p>\n\n\n\n<p><strong>(iii) 9 of them are positive and the remaining are negative?<\/strong><\/p>\n\n\n\n<p><strong>(iv) all are negative?<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>(i) Out of 18 integers, 15 of them are negative which is odd number. Therefore, the sign of product is negative.<\/p>\n\n\n\n<p>(ii) Out of 18 integers, 12 of them are negative which is even number. Therefore, the sign of product is positive.<\/p>\n\n\n\n<p>(iii) Out of 18 integers, 9 of them are negative which is odd number. Therefore, the sign of product is negative.<\/p>\n\n\n\n<p>(iv) All are negative which is even number. Therefore, sign of product is positive.<\/p>\n\n\n\n<p><strong>9. Find which is greater?<\/strong><\/p>\n\n\n\n<p><strong>(i) (8 + 10) \u00d7 15 or 8 + 10 \u00d7 15<\/strong><\/p>\n\n\n\n<p><strong>(ii) 12 \u00d7 (6 \u2013 8) or 12 \u00d7 6 \u2013 8<\/strong><\/p>\n\n\n\n<p><strong>(iii) {(-3) \u2013 4} \u00d7 (-5) or (-3) \u2013 4 \u00d7 (-5)<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>(i) (8 + 10) \u00d7 15 or 8 + 10 \u00d7 15<\/p>\n\n\n\n<p>We know that<\/p>\n\n\n\n<p>(8 + 10) \u00d7 15 = 18 \u00d7 15 = 270<\/p>\n\n\n\n<p>8 + 10 \u00d7 15 = 8 + 150 = 158<\/p>\n\n\n\n<p>Therefore, (8 + 10) \u00d7 15 &gt; 8 + 10 \u00d7 15.<\/p>\n\n\n\n<p>(ii) 12 \u00d7 (6 \u2013 8) or 12 \u00d7 6 \u2013 8<\/p>\n\n\n\n<p>We know that<\/p>\n\n\n\n<p>12 \u00d7 (6 \u2013 8) = 12 (-2) = \u2013 24<\/p>\n\n\n\n<p>12 \u00d7 6 \u2013 8 = 72 \u2013 8 = 64<\/p>\n\n\n\n<p>Therefore, 12 \u00d7 (6 \u2013 8) &lt; 12 \u00d7 6 \u2013 8.<\/p>\n\n\n\n<p>(iii) {(-3) \u2013 4} \u00d7 (-5) or (-3) \u2013 4 \u00d7 (-5)<\/p>\n\n\n\n<p>We know that<\/p>\n\n\n\n<p>{(-3) \u2013 4} \u00d7 (-5) = {- 3 \u2013 4} \u00d7 (-5)<\/p>\n\n\n\n<p>By further calculation<\/p>\n\n\n\n<p>= \u2013 7 \u00d7 -5 = 35<\/p>\n\n\n\n<p>Similarly<\/p>\n\n\n\n<p>(-3) \u2013 4 \u00d7 (-5) = \u2013 3 + 20 = 17<\/p>\n\n\n\n<p>Therefore, {(-3) \u2013 4} \u00d7 (-5) &gt; (-3) \u2013 4 \u00d7 (-5)<\/p>\n\n\n\n<p><strong>10. State, true or false:<\/strong><\/p>\n\n\n\n<p><strong>(i) product of two different integers can be zero.<\/strong><\/p>\n\n\n\n<p><strong>(ii) product of 120 negative integers and 121 positive integers is negative.<\/strong><\/p>\n\n\n\n<p><strong>(iii) a \u00d7 (b + c) = a \u00d7 b + c<\/strong><\/p>\n\n\n\n<p><strong>(iv) (b \u2013 c) \u00d7 a = b \u2013 c \u00d7 a.<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>(i) True.<\/p>\n\n\n\n<p>Example: 5 \u00d7 0 = 0, 0 \u00d7 \u2013 8 = 0<\/p>\n\n\n\n<p>(ii) False.<\/p>\n\n\n\n<p>The total number of negative integers is 120 which is an even number and we know that the product of even numbers of negative integers is always positive. Hence, the sign of the product will be positive.<\/p>\n\n\n\n<p>(iii) False.<\/p>\n\n\n\n<p>a \u00d7 (b + c) \u2260 a \u00d7 b + c<\/p>\n\n\n\n<p>ab + ac \u2260 ab + c<\/p>\n\n\n\n<p>(iv) False.<\/p>\n\n\n\n<p>(b \u2013 c) \u00d7 a \u2260 b \u2013 c \u00d7 a<\/p>\n\n\n\n<p>ab \u2013 ac \u2260 b \u2013 ca<\/p>\n\n\n\n<h4 class=\"wp-block-heading\">Exercise 1B page: 11<\/h4>\n\n\n\n<p><strong>1. Divide:<\/strong><\/p>\n\n\n\n<p><strong>(i) 117 by 9<\/strong><\/p>\n\n\n\n<p><strong>(ii) (-117) by 9<\/strong><\/p>\n\n\n\n<p><strong>(iii) 117 by (-9)<\/strong><\/p>\n\n\n\n<p><strong>(iv) (-117) by (-9)<\/strong><\/p>\n\n\n\n<p><strong>(v) 225 by (-15)<\/strong><\/p>\n\n\n\n<p><strong>(vi) (-552) \u00f7 24<\/strong><\/p>\n\n\n\n<p><strong>(vii) (-798) by (-21)<\/strong><\/p>\n\n\n\n<p><strong>(viii) (-910) \u00f7 26<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>(i) 117 by 9<\/p>\n\n\n\n<p>= 117\/9<\/p>\n\n\n\n<p>It can be written as<\/p>\n\n\n\n<p>= (13 \u00d7 9)\/ 9<\/p>\n\n\n\n<p>= 13<\/p>\n\n\n\n<p>(ii) (-117) by 9<\/p>\n\n\n\n<p>= \u2013 117\/ 9<\/p>\n\n\n\n<p>It can be written as<\/p>\n\n\n\n<p>= (-13 \u00d7 9)\/ 9<\/p>\n\n\n\n<p>= -13<\/p>\n\n\n\n<p>(iii) 117 by (-9)<\/p>\n\n\n\n<p>= 117\/ -9<\/p>\n\n\n\n<p>It can be written as<\/p>\n\n\n\n<p>= (13 \u00d7 9)\/ -9<\/p>\n\n\n\n<p>= \u2013 13<\/p>\n\n\n\n<p>(iv) (-117) by (-9)<\/p>\n\n\n\n<p>= \u2013 117\/-9 = 117\/9<\/p>\n\n\n\n<p>It can be written as<\/p>\n\n\n\n<p>= (13 \u00d7 9)\/ 9<\/p>\n\n\n\n<p>= 13<\/p>\n\n\n\n<p>(v) 225 by (-15)<\/p>\n\n\n\n<p>= 225\/ (-15)<\/p>\n\n\n\n<p>It can be written as<\/p>\n\n\n\n<p>= (15 \u00d7 15)\/ -15<\/p>\n\n\n\n<p>= \u2013 15<\/p>\n\n\n\n<p>(vi) (-552) \u00f7 24<\/p>\n\n\n\n<p>= \u2013 552\/ 24<\/p>\n\n\n\n<p>It can be written as<\/p>\n\n\n\n<p>= (- 23 \u00d7 24)\/ 24<\/p>\n\n\n\n<p>= \u2013 23<\/p>\n\n\n\n<p>(vii) (-798) by (-21)<\/p>\n\n\n\n<p>= \u2013 798\/ \u2013 21 = 798\/21<\/p>\n\n\n\n<p>It can be written as<\/p>\n\n\n\n<p>= (38 \u00d7 21)\/ 21<\/p>\n\n\n\n<p>= 38<\/p>\n\n\n\n<p>(viii) (-910) \u00f7 26<\/p>\n\n\n\n<p>= \u2013 910\/ 26<\/p>\n\n\n\n<p>It can be written as<\/p>\n\n\n\n<p>= (- 35 \u00d7 26)\/ 26<\/p>\n\n\n\n<p>= \u2013 35<\/p>\n\n\n\n<p><strong>2. Evaluate:<\/strong><\/p>\n\n\n\n<p><strong>(i) (-234) \u00f7 13<\/strong><\/p>\n\n\n\n<p><strong>(ii) 234 \u00f7 (-13)<\/strong><\/p>\n\n\n\n<p><strong>(iii) (-234) \u00f7 (-13)<\/strong><\/p>\n\n\n\n<p><strong>(iv) 374 \u00f7 (-17)<\/strong><\/p>\n\n\n\n<p><strong>(v) (-374) \u00f7 17<\/strong><\/p>\n\n\n\n<p><strong>(vi) (-374) \u00f7 (-17)<\/strong><\/p>\n\n\n\n<p><strong>(vii) (-728) \u00f7 14<\/strong><\/p>\n\n\n\n<p><strong>(viii) 272 \u00f7 (-17)<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>(i) (-234) \u00f7 13<\/p>\n\n\n\n<p>= \u2013 234\/ 13<\/p>\n\n\n\n<p>It can be written as<\/p>\n\n\n\n<p>= (-18 \u00d7 13)\/ 13<\/p>\n\n\n\n<p>= \u2013 18<\/p>\n\n\n\n<p>(ii) 234 \u00f7 (-13)<\/p>\n\n\n\n<p>= 234\/ -13<\/p>\n\n\n\n<p>It can be written as<\/p>\n\n\n\n<p>= (18 \u00d7 13)\/ \u2013 13<\/p>\n\n\n\n<p>= \u2013 18<\/p>\n\n\n\n<p>(iii) (-234) \u00f7 (-13)<\/p>\n\n\n\n<p>= \u2013 234\/ \u2013 13 = 234\/13<\/p>\n\n\n\n<p>It can be written as<\/p>\n\n\n\n<p>= (18 \u00d7 13)\/ 13<\/p>\n\n\n\n<p>= 18<\/p>\n\n\n\n<p>(iv) 374 \u00f7 (-17)<\/p>\n\n\n\n<p>= 374\/ -17<\/p>\n\n\n\n<p>It can be written as<\/p>\n\n\n\n<p>= (22 \u00d7 17)\/ (-17)<\/p>\n\n\n\n<p>= \u2013 22<\/p>\n\n\n\n<p>(v) (-374) \u00f7 17<\/p>\n\n\n\n<p>= \u2013 374\/ 17<\/p>\n\n\n\n<p>It can be written as<\/p>\n\n\n\n<p>= (-22 \u00d7 17)\/ (17)<\/p>\n\n\n\n<p>= \u2013 22<\/p>\n\n\n\n<p>(vi) (-374) \u00f7 (-17)<\/p>\n\n\n\n<p>= \u2013 374\/ -17 = 374\/17<\/p>\n\n\n\n<p>It can be written as<\/p>\n\n\n\n<p>= (22 \u00d7 17)\/ (17)<\/p>\n\n\n\n<p>= 22<\/p>\n\n\n\n<p>(vii) (-728) \u00f7 14<\/p>\n\n\n\n<p>= \u2013 728\/14<\/p>\n\n\n\n<p>It can be written as<\/p>\n\n\n\n<p>= (-52 \u00d7 14)\/ 14<\/p>\n\n\n\n<p>= -52<\/p>\n\n\n\n<p>(viii) 272 \u00f7 (-17)<\/p>\n\n\n\n<p>= 272\/ -17<\/p>\n\n\n\n<p>It can be written as<\/p>\n\n\n\n<p>= (16 \u00d7 17)\/ (-17)<\/p>\n\n\n\n<p>= -16<\/p>\n\n\n\n<p><strong>3. Find the quotient in each of the following divisions:<\/strong><\/p>\n\n\n\n<p><strong>(i) 299 \u00f7 23<\/strong><\/p>\n\n\n\n<p><strong>(ii) 299 \u00f7 (-23)<\/strong><\/p>\n\n\n\n<p><strong>(iii) (-384) \u00f7 16<\/strong><\/p>\n\n\n\n<p><strong>(iv) (-572) \u00f7 (-22)<\/strong><\/p>\n\n\n\n<p><strong>(v) 408 \u00f7 (-17)<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>(i) 299 \u00f7 23<\/p>\n\n\n\n<p>= 299\/23<\/p>\n\n\n\n<p>It can be written as<\/p>\n\n\n\n<p>= (23 \u00d7 13)\/ 23<\/p>\n\n\n\n<p>= 13<\/p>\n\n\n\n<p>(ii) 299 \u00f7 (-23)<\/p>\n\n\n\n<p>= 299\/ -23<\/p>\n\n\n\n<p>It can be written as<\/p>\n\n\n\n<p>= (23 \u00d7 13)\/ \u2013 23<\/p>\n\n\n\n<p>= \u2013 13<\/p>\n\n\n\n<p>(iii) (-384) \u00f7 16<\/p>\n\n\n\n<p>= \u2013 384\/16<\/p>\n\n\n\n<p>It can be written as<\/p>\n\n\n\n<p>= (- 24 \u00d7 16)\/ 16<\/p>\n\n\n\n<p>= \u2013 24<\/p>\n\n\n\n<p>(iv) (-572) \u00f7 (-22)<\/p>\n\n\n\n<p>= \u2013 572\/ \u2013 22 = 572\/22<\/p>\n\n\n\n<p>It can be written as<\/p>\n\n\n\n<p>= (26 \u00d7 22)\/ 22<\/p>\n\n\n\n<p>= 26<\/p>\n\n\n\n<p>(v) 408 \u00f7 (-17)<\/p>\n\n\n\n<p>= 408\/ -17<\/p>\n\n\n\n<p>It can be written as<\/p>\n\n\n\n<p>= (24 \u00d7 17)\/ (-17)<\/p>\n\n\n\n<p>= \u2013 24<\/p>\n\n\n\n<p><strong>4. Divide:<\/strong><\/p>\n\n\n\n<p><strong>(i) 204 by 17<\/strong><\/p>\n\n\n\n<p><strong>(ii) 152 by \u2013 19<\/strong><\/p>\n\n\n\n<p><strong>(iii) 0 by 35<\/strong><\/p>\n\n\n\n<p><strong>(iv) 0 by (-82)<\/strong><\/p>\n\n\n\n<p><strong>(v) 5490 by 10<\/strong><\/p>\n\n\n\n<p><strong>(vi) 762800 by 100<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>(i) 204 by 17<\/p>\n\n\n\n<p>= 204\/17<\/p>\n\n\n\n<p>It can be written as<\/p>\n\n\n\n<p>= (12 \u00d7 17)\/ 17<\/p>\n\n\n\n<p>= 12<\/p>\n\n\n\n<p>(ii) 152 by \u2013 19<\/p>\n\n\n\n<p>= 152\/ -19<\/p>\n\n\n\n<p>It can be written as<\/p>\n\n\n\n<p>= (8 \u00d7 19)\/ -19<\/p>\n\n\n\n<p>= \u2013 8<\/p>\n\n\n\n<p>(iii) 0 by 35<\/p>\n\n\n\n<p>= 0\/35<\/p>\n\n\n\n<p>= 0<\/p>\n\n\n\n<p>(iv) 0 by (-82)<\/p>\n\n\n\n<p>= 0\/ -82<\/p>\n\n\n\n<p>= 0<\/p>\n\n\n\n<p>(v) 5490 by 10<\/p>\n\n\n\n<p>= 5490\/10<\/p>\n\n\n\n<p>It can be written as<\/p>\n\n\n\n<p>= (549 \u00d7 10)\/ 10<\/p>\n\n\n\n<p>= 549<\/p>\n\n\n\n<p>(vi) 762800 by 100<\/p>\n\n\n\n<p>= 762800\/100<\/p>\n\n\n\n<p>It can be written as<\/p>\n\n\n\n<p>= (7628 x 100)\/ 100<\/p>\n\n\n\n<p>= 7628<\/p>\n\n\n\n<p><strong>5. State, true or false:<\/strong><\/p>\n\n\n\n<p><strong>(i) 0 \u00f7 32 = 0<\/strong><\/p>\n\n\n\n<p><strong>(ii) 0 \u00f7 (-9) = 0<\/strong><\/p>\n\n\n\n<p><strong>(iii) (-37) \u00f7 0 = 0<\/strong><\/p>\n\n\n\n<p><strong>(iv) 0 \u00f7 0 = 0<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>(i) True.<\/p>\n\n\n\n<p>(ii) True.<\/p>\n\n\n\n<p>(iii) False. It is not defined.<\/p>\n\n\n\n<p>(iv) False. It is not defined.<\/p>\n\n\n\n<p><strong>6. Evaluate:<\/strong><\/p>\n\n\n\n<p><strong>(i) 42 \u00f7 7 + 4<\/strong><\/p>\n\n\n\n<p><strong>(ii) 12 + 18 \u00f7 3<\/strong><\/p>\n\n\n\n<p><strong>(iii) 19 \u2013 20 \u00f7 4<\/strong><\/p>\n\n\n\n<p><strong>(iv) 16 \u2013 5 \u00d7 3 + 4<\/strong><\/p>\n\n\n\n<p><strong>(v) 6 \u2013 8 \u2013 (-6) \u00f7 2<\/strong><\/p>\n\n\n\n<p><strong>(vi) 13 \u2013 12 \u00f7 4 \u00d7 2<\/strong><\/p>\n\n\n\n<p><strong>(vii) 16 + 8 \u00f7 4 \u2013 2 \u00d7 3<\/strong><\/p>\n\n\n\n<p><strong>(viii) 16 \u00f7 8 + 4 \u2013 2 \u00d7 3<\/strong><\/p>\n\n\n\n<p><strong>(ix) 16 \u2013 8 + 4 \u00f7 2 \u00d7 3<\/strong><\/p>\n\n\n\n<p><strong>(x) (-4) + (-12) \u00f7 (-6)<\/strong><\/p>\n\n\n\n<p><strong>(xi) (-18) + 6 \u00f7 3 + 5<\/strong><\/p>\n\n\n\n<p><strong>(xii) (-20) \u00d7 (-1) + 14 \u00f7 7<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>(i) 42 \u00f7 7 + 4<\/p>\n\n\n\n<p>It can be written as<\/p>\n\n\n\n<p>= 42\/7 + 4<\/p>\n\n\n\n<p>By further calculation<\/p>\n\n\n\n<p>= 6 + 4<\/p>\n\n\n\n<p>= 10<\/p>\n\n\n\n<p>(ii) 12 + 18 \u00f7 3<\/p>\n\n\n\n<p>It can be written as<\/p>\n\n\n\n<p>= 12 + 18\/3<\/p>\n\n\n\n<p>By further calculation<\/p>\n\n\n\n<p>= 12 + 6<\/p>\n\n\n\n<p>= 18<\/p>\n\n\n\n<p>(iii) 19 \u2013 20 \u00f7 4<\/p>\n\n\n\n<p>It can be written as<\/p>\n\n\n\n<p>= 19 \u2013 20\/4<\/p>\n\n\n\n<p>By further calculation<\/p>\n\n\n\n<p>= 19 \u2013 5<\/p>\n\n\n\n<p>= 14<\/p>\n\n\n\n<p>(iv) 16 \u2013 5 \u00d7 3 + 4<\/p>\n\n\n\n<p>It can be written as<\/p>\n\n\n\n<p>= 16 \u2013 15 + 4<\/p>\n\n\n\n<p>By further calculation<\/p>\n\n\n\n<p>= 20 \u2013 15<\/p>\n\n\n\n<p>= 5<\/p>\n\n\n\n<p>(v) 6 \u2013 8 \u2013 (-6) \u00f7 2<\/p>\n\n\n\n<p>It can be written as<\/p>\n\n\n\n<p>= 6 \u2013 8 \u2013 (-6\/2)<\/p>\n\n\n\n<p>By further calculation<\/p>\n\n\n\n<p>= 6 \u2013 8 \u2013 (-3)<\/p>\n\n\n\n<p>So we get<\/p>\n\n\n\n<p>= 6 \u2013 8 + 3<\/p>\n\n\n\n<p>= 9 \u2013 8<\/p>\n\n\n\n<p>= 1<\/p>\n\n\n\n<p>(vi) 13 \u2013 12 \u00f7 4 \u00d7 2<\/p>\n\n\n\n<p>It can be written as<\/p>\n\n\n\n<p>= 13 \u2013 12\/4 \u00d7 2<\/p>\n\n\n\n<p>By further calculation<\/p>\n\n\n\n<p>= 13 \u2013 3 \u00d7 2<\/p>\n\n\n\n<p>= 13 \u2013 6<\/p>\n\n\n\n<p>= 7<\/p>\n\n\n\n<p>(vii) 16 + 8 \u00f7 4 \u2013 2 \u00d7 3<\/p>\n\n\n\n<p>It can be written as<\/p>\n\n\n\n<p>= 16 + 8\/4 \u2013 2 \u00d7 3<\/p>\n\n\n\n<p>By further calculation<\/p>\n\n\n\n<p>= 16 + 2 \u2013 2 \u00d7 3<\/p>\n\n\n\n<p>So we get<\/p>\n\n\n\n<p>= 16 + 2 \u2013 6<\/p>\n\n\n\n<p>= 18 \u2013 6<\/p>\n\n\n\n<p>= 12<\/p>\n\n\n\n<p>(viii) 16 \u00f7 8 + 4 \u2013 2 \u00d7 3<\/p>\n\n\n\n<p>It can be written as<\/p>\n\n\n\n<p>= 16\/8 + 4 \u2013 2 \u00d7 3<\/p>\n\n\n\n<p>By further calculation<\/p>\n\n\n\n<p>= 2 + 4 \u2013 6<\/p>\n\n\n\n<p>So we get<\/p>\n\n\n\n<p>= 6 \u2013 6<\/p>\n\n\n\n<p>= 0<\/p>\n\n\n\n<p>(ix) 16 \u2013 8 + 4 \u00f7 2 \u00d7 3<\/p>\n\n\n\n<p>It can be written as<\/p>\n\n\n\n<p>= 16 \u2013 8 + 4\/2 \u00d7 3<\/p>\n\n\n\n<p>By further calculation<\/p>\n\n\n\n<p>= 16 \u2013 8 + 2 \u00d7 3<\/p>\n\n\n\n<p>So we get<\/p>\n\n\n\n<p>= 16 \u2013 8 + 6<\/p>\n\n\n\n<p>Here<\/p>\n\n\n\n<p>= 22 \u2013 8<\/p>\n\n\n\n<p>= 14<\/p>\n\n\n\n<p>(x) (-4) + (-12) \u00f7 (-6)<\/p>\n\n\n\n<p>It can be written as<\/p>\n\n\n\n<p>= (-4) + (-12\/-6)<\/p>\n\n\n\n<p>By further calculation<\/p>\n\n\n\n<p>= \u2013 4 + 2<\/p>\n\n\n\n<p>= -2<\/p>\n\n\n\n<p>(xi) (-18) + 6 \u00f7 3 + 5<\/p>\n\n\n\n<p>It can be written as<\/p>\n\n\n\n<p>= (-18) + 6\/3 + 5<\/p>\n\n\n\n<p>By further calculation<\/p>\n\n\n\n<p>= (-18) + 2 + 5<\/p>\n\n\n\n<p>So we get<\/p>\n\n\n\n<p>= \u2013 18 + 7<\/p>\n\n\n\n<p>= \u2013 11<\/p>\n\n\n\n<p>(xii) (-20) \u00d7 (-1) + 14 \u00f7 7<\/p>\n\n\n\n<p>It can be written as<\/p>\n\n\n\n<p>= (-20) \u00d7 (-1) + 14\/7<\/p>\n\n\n\n<p>By further calculation<\/p>\n\n\n\n<p>= (-20) \u00d7 (-1) + 2<\/p>\n\n\n\n<p>So we get<\/p>\n\n\n\n<p>= 20 + 2<\/p>\n\n\n\n<p>= 22<\/p>\n\n\n\n<h4 class=\"wp-block-heading\">Exercise 1C page: 15<\/h4>\n\n\n\n<p><strong>Evaluate:<\/strong><\/p>\n\n\n\n<p><strong>1. 18 \u2013 (20 \u2013 15 \u00f7 3)<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>18 \u2013 (20 \u2013 15 \u00f7 3)<\/p>\n\n\n\n<p>It can be written as<\/p>\n\n\n\n<p>= 18 \u2013 (20 \u2013 15\/3)<\/p>\n\n\n\n<p>By further calculation<\/p>\n\n\n\n<p>= 18 \u2013 (20 \u2013 5)<\/p>\n\n\n\n<p>= 18 \u2013 20 + 5<\/p>\n\n\n\n<p>So we get<\/p>\n\n\n\n<p>= 18 + 5 \u2013 20<\/p>\n\n\n\n<p>= 23 \u2013 20<\/p>\n\n\n\n<p>= 3<\/p>\n\n\n\n<p><strong>2. \u2013 15 + 24 \u00f7 (15 \u2013 13)<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>\u2013 15 + 24 \u00f7 (15 \u2013 13)<\/p>\n\n\n\n<p>It can be written as<\/p>\n\n\n\n<p>= \u2013 15 + 24 \u00f7 2<\/p>\n\n\n\n<p>So we get<\/p>\n\n\n\n<p>= \u2013 15 + 12<\/p>\n\n\n\n<p>= \u2013 3<\/p>\n\n\n\n<p><strong>3. 35 \u2013 {15 + 14 \u2013 (13 + )}<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>35 \u2013 {15 + 14 \u2013 (13 +<br><img decoding=\"async\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2022\/05\/selina-solutions-concise-maths-class-7-chapter-1-ex-1c-image-2.gif\" alt=\"Selina Solutions Concise Maths Class 7 Chapter 1 Ex 1C Image 2\">)}<\/p>\n\n\n\n<p>It can be written as<\/p>\n\n\n\n<p>= 35 \u2013 [15 + 14 \u2013 (13 + 4)]<\/p>\n\n\n\n<p>By further calculation<\/p>\n\n\n\n<p>= 35 \u2013 [15 + 14 \u2013 17]<\/p>\n\n\n\n<p>Multiplying the negative sign<\/p>\n\n\n\n<p>= 35 \u2013 15 \u2013 14 + 17<\/p>\n\n\n\n<p>So we get<\/p>\n\n\n\n<p>= 52 \u2013 29<\/p>\n\n\n\n<p>= 23<\/p>\n\n\n\n<p><strong>4. 27 \u2013 {13 + 4 \u2013 (8 + 4 \u2013 )}<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>27 \u2013 {13 + 4 \u2013 (8 + 4 \u2013<br><img decoding=\"async\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2022\/05\/selina-solutions-concise-maths-class-7-chapter-1-ex-1c-image-4.gif\" alt=\"Selina Solutions Concise Maths Class 7 Chapter 1 Ex 1C Image 4\">)}<\/p>\n\n\n\n<p>It can be written as<\/p>\n\n\n\n<p>= 27 \u2013 {13 + 4 \u2013 (8 + 4 \u2013 4)}<\/p>\n\n\n\n<p>By further calculation<\/p>\n\n\n\n<p>= 27 \u2013 {13 + 4 \u2013 8}<\/p>\n\n\n\n<p>So we get<\/p>\n\n\n\n<p>= 27 \u2013 {13 + (-4)}<\/p>\n\n\n\n<p>= 27 \u2013 9<\/p>\n\n\n\n<p>= 18<\/p>\n\n\n\n<p><strong>5. 32 \u2013 [43 \u2013 {51 \u2013 (20 \u2013 )}]<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>32 \u2013 [43 \u2013 {51 \u2013 (20 \u2013<br><img decoding=\"async\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2022\/05\/selina-solutions-concise-maths-class-7-chapter-1-ex-1c-image-6.gif\" alt=\"Selina Solutions Concise Maths Class 7 Chapter 1 Ex 1C Image 6\">)}]<\/p>\n\n\n\n<p>It can be written as<\/p>\n\n\n\n<p>= 32 \u2013 [43 \u2013 {51 \u2013 (20 \u2013 11)}]<\/p>\n\n\n\n<p>By further calculation<\/p>\n\n\n\n<p>= 32 \u2013 [43 \u2013 {51 \u2013 9}]<\/p>\n\n\n\n<p>So we get<\/p>\n\n\n\n<p>= 32 \u2013 [43 \u2013 42]<\/p>\n\n\n\n<p>= 32 \u2013 1<\/p>\n\n\n\n<p>= 31<\/p>\n\n\n\n<p><strong>6. 46 \u2013 [26 \u2013 {14 \u2013 (15 \u2013 4 \u00f7 2 \u00d7 2)}]<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>46 \u2013 [26 \u2013 {14 \u2013 (15 \u2013 4 \u00f7 2 \u00d7 2)}]<\/p>\n\n\n\n<p>It can be written as<\/p>\n\n\n\n<p>= 46 \u2013 [26 \u2013 {14 \u2013 (15 \u2013 2 \u00d7 2)}]<\/p>\n\n\n\n<p>By further calculation<\/p>\n\n\n\n<p>= 46 \u2013 [26 \u2013 {14 \u2013 (15 \u2013 4)}]<\/p>\n\n\n\n<p>So we get<\/p>\n\n\n\n<p>= 46 \u2013 [26 \u2013 {14 \u2013 11}]<\/p>\n\n\n\n<p>= 46 \u2013 [26 \u2013 3]<\/p>\n\n\n\n<p>Here<\/p>\n\n\n\n<p>= 46 \u2013 23<\/p>\n\n\n\n<p>= 23<\/p>\n\n\n\n<p><strong>7. 45 \u2013 [38 \u2013 {60 \u00f7 3 \u2013 (6 \u2013 9 \u00f7 3) \u00f7 3}]<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>45 \u2013 [38 \u2013 {60 \u00f7 3 \u2013 (6 \u2013 9 \u00f7 3) \u00f7 3}]<\/p>\n\n\n\n<p>It can be written as<\/p>\n\n\n\n<p>= 45 \u2013 [38 \u2013 {60 \u00f7 3 \u2013 (6 \u2013 3) \u00f7 3}]<\/p>\n\n\n\n<p>By further calculation<\/p>\n\n\n\n<p>= 45 \u2013 [38 \u2013 {20 \u2013 3 \u00f7 3}]<\/p>\n\n\n\n<p>So we get<\/p>\n\n\n\n<p>= 45 \u2013 [38 \u2013 {20 \u2013 1}]<\/p>\n\n\n\n<p>By subtraction<\/p>\n\n\n\n<p>= 45 \u2013 [38 \u2013 19]<\/p>\n\n\n\n<p>= 45 \u2013 19<\/p>\n\n\n\n<p>= 26<\/p>\n\n\n\n<p><strong>8. 17 \u2013 [17 \u2013 {17 \u2013 (17 \u2013 )}]<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>17 \u2013 [17 \u2013 {17 \u2013 (17 \u2013<br><img decoding=\"async\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2022\/05\/selina-solutions-concise-maths-class-7-chapter-1-ex-1c-image-8.gif\" alt=\"Selina Solutions Concise Maths Class 7 Chapter 1 Ex 1C Image 8\">)}]<\/p>\n\n\n\n<p>It can be written as<\/p>\n\n\n\n<p>= 17 \u2013 [17 \u2013 {17 \u2013 (17 \u2013 0)}]<\/p>\n\n\n\n<p>By further calculation<\/p>\n\n\n\n<p>= 17 \u2013 [17 \u2013 {17 \u2013 17}]<\/p>\n\n\n\n<p>So we get<\/p>\n\n\n\n<p>= 17 \u2013 [17 \u2013 0]<\/p>\n\n\n\n<p>= 17 \u2013 17<\/p>\n\n\n\n<p>= 0<\/p>\n\n\n\n<p><strong>9. 2550 \u2013 [510 \u2013 {270 \u2013 (90 \u2013 )}]<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>2550 \u2013 [510 \u2013 {270 \u2013 (90 \u2013<br><img decoding=\"async\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2022\/05\/selina-solutions-concise-maths-class-7-chapter-1-ex-1c-image-10.gif\" alt=\"Selina Solutions Concise Maths Class 7 Chapter 1 Ex 1C Image 10\">)}]<\/p>\n\n\n\n<p>It can be written as<\/p>\n\n\n\n<p>= 2550 \u2013 [510 \u2013 {270 \u2013 (90 \u2013 87)}]<\/p>\n\n\n\n<p>By further calculation<\/p>\n\n\n\n<p>= 2550 \u2013 [510 \u2013 {270 \u2013 3}]<\/p>\n\n\n\n<p>So we get<\/p>\n\n\n\n<p>= 2550 \u2013 [510 \u2013 267]<\/p>\n\n\n\n<p>= 2550 \u2013 243<\/p>\n\n\n\n<p>= 2307<\/p>\n\n\n\n<p><strong>10. 30 + [{-2 \u00d7 (25 \u2013 )}]<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>30 + [{-2 \u00d7 (25 \u2013<br><img decoding=\"async\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2022\/05\/selina-solutions-concise-maths-class-7-chapter-1-ex-1c-image-12.gif\" alt=\"Selina Solutions Concise Maths Class 7 Chapter 1 Ex 1C Image 12\">)}]<\/p>\n\n\n\n<p>It can be written as<\/p>\n\n\n\n<p>= 30 + [{-2 \u00d7 (25 \u2013 10)}]<\/p>\n\n\n\n<p>By further calculation<\/p>\n\n\n\n<p>= 30 + [{-2 \u00d7 15}]<\/p>\n\n\n\n<p>So we get<\/p>\n\n\n\n<p>= 30 + [-30]<\/p>\n\n\n\n<p>= 30 \u2013 30<\/p>\n\n\n\n<p>= 0<\/p>\n\n\n\n<p><strong>11. 88 \u2013 {5 \u2013 (-48) \u00f7 (-16)}<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>88 \u2013 {5 \u2013 (-48) \u00f7 (-16)}<\/p>\n\n\n\n<p>It can be written as<\/p>\n\n\n\n<p>= 88 \u2013 {5 \u2013 (- 48\/ -16)}<\/p>\n\n\n\n<p>By further calculation<\/p>\n\n\n\n<p>= 88 \u2013 {5 \u2013 3}<\/p>\n\n\n\n<p>So we get<\/p>\n\n\n\n<p>= 88 \u2013 2<\/p>\n\n\n\n<p>= 86<\/p>\n\n\n\n<p><strong>12. 9 \u00d7 (8 \u2013 ) \u2013 2 (2 + )<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>9 \u00d7 (8 \u2013<br><img decoding=\"async\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2022\/05\/selina-solutions-concise-maths-class-7-chapter-1-ex-1c-image-15.gif\" alt=\"Selina Solutions Concise Maths Class 7 Chapter 1 Ex 1C Image 15\">) \u2013 2 (2 +<br><img decoding=\"async\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2022\/05\/selina-solutions-concise-maths-class-7-chapter-1-ex-1c-image-16.gif\" alt=\"Selina Solutions Concise Maths Class 7 Chapter 1 Ex 1C Image 16\">)<\/p>\n\n\n\n<p>It can be written as<\/p>\n\n\n\n<p>= 9 \u00d7 (8 \u2013 5) \u2013 2 (2 + 6)<\/p>\n\n\n\n<p>By further calculation<\/p>\n\n\n\n<p>= 9 \u00d7 3 \u2013 2 \u00d7 8<\/p>\n\n\n\n<p>So we get<\/p>\n\n\n\n<p>= 27 \u2013 16<\/p>\n\n\n\n<p>= 11<\/p>\n\n\n\n<p><strong>13. 2 \u2013 [3 \u2013 {6 \u2013 (5 \u2013 )}]<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>2 \u2013 [3 \u2013 {6 \u2013 (5 \u2013<br><img decoding=\"async\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2022\/05\/selina-solutions-concise-maths-class-7-chapter-1-ex-1c-image-18.gif\" alt=\"Selina Solutions Concise Maths Class 7 Chapter 1 Ex 1C Image 18\">)}]<\/p>\n\n\n\n<p>It can be written as<\/p>\n\n\n\n<p>= 2 \u2013 [3 \u2013 {6 \u2013 (5 \u2013 1)}]<\/p>\n\n\n\n<p>By further calculation<\/p>\n\n\n\n<p>= 2 \u2013 [3 \u2013 {6 \u2013 4}]<\/p>\n\n\n\n<p>So we get<\/p>\n\n\n\n<p>= 2 \u2013 [3 \u2013 2]<\/p>\n\n\n\n<p>= 2 \u2013 1<\/p>\n\n\n\n<p>= 1<\/p>\n\n\n\n<h4 class=\"wp-block-heading\">Exercise 1D page: 15<\/h4>\n\n\n\n<p><strong>1. The sum of two integers is \u2013 15. If one of them is 9, find the other.<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>It is given that<\/p>\n\n\n\n<p>Sum of two integers = \u2013 15<\/p>\n\n\n\n<p>One integer = 9<\/p>\n\n\n\n<p>Other integer = \u2013 15 \u2013 9<\/p>\n\n\n\n<p>Taking negative sign as common<\/p>\n\n\n\n<p>= \u2013 (15 + 9)<\/p>\n\n\n\n<p>= \u2013 24<\/p>\n\n\n\n<p><strong>2. The difference between integers x and \u2013 6 is \u2013 5. Find the values of x.<\/strong><\/p>\n\n\n\n<p><strong>x \u2013 (-6) = \u2013 5 or \u2013 6 \u2013 x = \u2013 5<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>It is given that<\/p>\n\n\n\n<p>The difference between integers x and -6 is -5<\/p>\n\n\n\n<p>x \u2013 (-6) = \u2013 5 or \u2013 6 \u2013 x = \u2013 5<\/p>\n\n\n\n<p>So the value of x is<\/p>\n\n\n\n<p>x + 6 = \u2013 5 or \u2013 x = \u2013 5 + 6<\/p>\n\n\n\n<p>By further calculation<\/p>\n\n\n\n<p>x = \u2013 5 \u2013 6 or \u2013 x = 1<\/p>\n\n\n\n<p>x = \u2013 11 or x = \u2013 1<\/p>\n\n\n\n<p><strong>3. The sum of two integers is 28. If one integer is \u2013 45, find the other.<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>It is given that<\/p>\n\n\n\n<p>Sum of two integers = 28<\/p>\n\n\n\n<p>One integer = \u2013 45<\/p>\n\n\n\n<p>Other integer = 28 \u2013 (-45)<\/p>\n\n\n\n<p>By further calculation<\/p>\n\n\n\n<p>= 28 + 45<\/p>\n\n\n\n<p>= 73<\/p>\n\n\n\n<p><strong>4. The sum of two integers is \u2013 56. If one integer is \u2013 42, find the other.<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>It is given that<\/p>\n\n\n\n<p>Sum of two integers = \u2013 56<\/p>\n\n\n\n<p>One integer = \u2013 42<\/p>\n\n\n\n<p>Other integer = \u2013 56 \u2013 (- 42)<\/p>\n\n\n\n<p>By further calculation<\/p>\n\n\n\n<p>= \u2013 56 + 42<\/p>\n\n\n\n<p>= \u2013 14<\/p>\n\n\n\n<p><strong>5. The difference between an integer x and (-9) is 6. Find all possible values of x.<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>It is given that<\/p>\n\n\n\n<p>The difference between an integer x and (-9) is 6<\/p>\n\n\n\n<p>x \u2013 (-9) = 6 or \u2013 9 \u2013 x = 6<\/p>\n\n\n\n<p>So the value of x is<\/p>\n\n\n\n<p>x \u2013 (-9) = 6 or \u2013 9 \u2013 x = 6<\/p>\n\n\n\n<p>By further calculation<\/p>\n\n\n\n<p>x + 9 = 6 or \u2013 x = 6 + 9<\/p>\n\n\n\n<p>So we get<\/p>\n\n\n\n<p>x = 6 \u2013 9 or \u2013 x = 15<\/p>\n\n\n\n<p>Here<\/p>\n\n\n\n<p>x = \u2013 3 or x = \u2013 15<\/p>\n\n\n\n<p>Therefore, possible values of x are \u2013 3 and \u2013 15.<\/p>\n\n\n\n<p><strong>6. Evaluate:<\/strong><\/p>\n\n\n\n<p><strong>(i) (-1) \u00d7 (-1) \u00d7 (-1) \u00d7 \u2026\u2026\u2026 60 times.<\/strong><\/p>\n\n\n\n<p><strong>(ii) (-1) \u00d7 (-1) \u00d7 (-1) \u00d7 (-1) \u00d7 \u2026\u2026.. 75 times.<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>(i) (-1) \u00d7 (-1) \u00d7 (-1) \u00d7 \u2026\u2026\u2026 60 times = 1 because (-1) is multiplied even number of times.<\/p>\n\n\n\n<p>(ii) (-1) \u00d7 (-1) \u00d7 (-1) \u00d7 (-1) \u00d7 \u2026\u2026\u2026 75 times = -1 because (-1) is multiplied odd number of times.<\/p>\n\n\n\n<p><strong>7. Evaluate:<\/strong><\/p>\n\n\n\n<p><strong>(i) (-2) \u00d7 (-3) \u00d7 (-4) \u00d7 (-5) \u00d7 (-6)<\/strong><\/p>\n\n\n\n<p><strong>(ii) (-3) \u00d7 (-6) \u00d7 (-9) \u00d7 (-12)<\/strong><\/p>\n\n\n\n<p><strong>(iii) (-11) \u00d7 (-15) \u00d7 (-11) \u00d7 (-25)<\/strong><\/p>\n\n\n\n<p><strong>(iv) 10 \u00d7 (-12) + 5 \u00d7 (-12)<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>(i) (-2) \u00d7 (-3) \u00d7 (-4) \u00d7 (-5) \u00d7 (-6)<\/p>\n\n\n\n<p>= 6 \u00d7 20 \u00d7 (-6)<\/p>\n\n\n\n<p>By further calculation<\/p>\n\n\n\n<p>= 120 \u00d7 (-6)<\/p>\n\n\n\n<p>= \u2013 720<\/p>\n\n\n\n<p>(ii) (-3) \u00d7 (-6) \u00d7 (-9) \u00d7 (-12)<\/p>\n\n\n\n<p>= 18 \u00d7 108<\/p>\n\n\n\n<p>By further calculation<\/p>\n\n\n\n<p>= 1944<\/p>\n\n\n\n<p>(iii) (-11) \u00d7 (-15) + (-11) \u00d7 (-25)<\/p>\n\n\n\n<p>= 165 + 275<\/p>\n\n\n\n<p>By further calculation<\/p>\n\n\n\n<p>= 440<\/p>\n\n\n\n<p>(iv) 10 \u00d7 (-12) + 5 \u00d7 (-12)<\/p>\n\n\n\n<p>= \u2013 120 \u2013 60<\/p>\n\n\n\n<p>By further calculation<\/p>\n\n\n\n<p>= \u2013 180<\/p>\n\n\n\n<p><strong>8. (i) If x \u00d7 (-1) = \u2013 36, is x positive or negative?<\/strong><\/p>\n\n\n\n<p><strong>(ii) If x \u00d7 (-1) = 36, is x positive or negative?<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>(i) x \u00d7 (-1) = \u2013 36<\/p>\n\n\n\n<p>So we get<\/p>\n\n\n\n<p>\u2013 x = \u2013 36<\/p>\n\n\n\n<p>By further simplification<\/p>\n\n\n\n<p>x = 36<\/p>\n\n\n\n<p>Hence, it is a positive integer.<\/p>\n\n\n\n<p>(ii) x \u00d7 (-1) = 36<\/p>\n\n\n\n<p>So we get<\/p>\n\n\n\n<p>\u2013 x = 36<\/p>\n\n\n\n<p>By further simplification<\/p>\n\n\n\n<p>x = \u2013 36<\/p>\n\n\n\n<p>Hence, it is a negative integer.<\/p>\n\n\n\n<p><strong>9. Write all the integers between \u2013 15 and 15, which are divisible by 2 and 3.<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>Here the integers between \u2013 15 and 15 are<\/p>\n\n\n\n<p>\u2013 12, \u2013 6, 0, 6 and 12 which are divisible by 2 and 3.<\/p>\n\n\n\n<p><strong>10. Write all the integers between \u2013 5 and 5, which are divisible by 2 or 3.<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>Here the integers between \u2013 5 and 5 are<\/p>\n\n\n\n<p>\u2013 4, \u2013 3, \u2013 2, 0, 2, 3 and 4 which are divisible by 2 or 3.<\/p>\n\n\n\n<p><strong>11. Evaluate:<\/strong><\/p>\n\n\n\n<p><strong>(i) (-20) + (-8) \u00f7 (-2) \u00d7 3<\/strong><\/p>\n\n\n\n<p><strong>(ii) (-5) \u2013 (-48) \u00f7 (-16) + (-2) \u00d7 6<\/strong><\/p>\n\n\n\n<p><strong>(iii) 16 + 8 \u00f7 4 \u2013 2 \u00d7 3<\/strong><\/p>\n\n\n\n<p><strong>(iv) 16 \u00f7 8 \u00d7 4 \u2013 2 \u00d7 3<\/strong><\/p>\n\n\n\n<p><strong>(v) 27 \u2013 [5 + {28 \u2013 (29 \u2013 7)}]<\/strong><\/p>\n\n\n\n<p><strong>(vi) 48 \u2013 [18 \u2013 {16 \u2013 (5 \u2013 )}]<\/strong><\/p>\n\n\n\n<p><strong>(vii) \u2013 8 \u2013 {- 6 (9 \u2013 11) + 18 \u00f7 -3}<\/strong><\/p>\n\n\n\n<p><strong>(viii) (24 \u00f7 \u2013 12) \u2013 (3 \u00d7 8 \u00f7 4 + 1)<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>(i) (-20) + (-8) \u00f7 (-2) \u00d7 3<\/p>\n\n\n\n<p>It can be written as<\/p>\n\n\n\n<p>= \u2013 20 + 4 \u00d7 3<\/p>\n\n\n\n<p>By further calculation<\/p>\n\n\n\n<p>= \u2013 20 + 12<\/p>\n\n\n\n<p>= \u2013 8<\/p>\n\n\n\n<p>(ii) (-5) \u2013 (-48) \u00f7 (-16) + (-2) \u00d7 6<\/p>\n\n\n\n<p>It can be written as<\/p>\n\n\n\n<p>= (-5) \u2013 3 + (-2) \u00d7 6<\/p>\n\n\n\n<p>By further calculation<\/p>\n\n\n\n<p>= \u2013 5 \u2013 3 \u2013 12<\/p>\n\n\n\n<p>So we get<\/p>\n\n\n\n<p>= \u2013 8 \u2013 12<\/p>\n\n\n\n<p>= \u2013 20<\/p>\n\n\n\n<p>(iii) 16 + 8 \u00f7 4 \u2013 2 \u00d7 3<\/p>\n\n\n\n<p>It can be written as<\/p>\n\n\n\n<p>= 16 + 2 \u2013 2 \u00d7 3<\/p>\n\n\n\n<p>By further calculation<\/p>\n\n\n\n<p>= 16 + 2 \u2013 6<\/p>\n\n\n\n<p>So we get<\/p>\n\n\n\n<p>= 18 \u2013 6<\/p>\n\n\n\n<p>= 12<\/p>\n\n\n\n<p>(iv) 16 \u00f7 8 \u00d7 4 \u2013 2 \u00d7 3<\/p>\n\n\n\n<p>It can be written as<\/p>\n\n\n\n<p>= 2 \u00d7 4 \u2013 2 \u00d7 3<\/p>\n\n\n\n<p>By further calculation<\/p>\n\n\n\n<p>= 8 \u2013 6<\/p>\n\n\n\n<p>= 2<\/p>\n\n\n\n<p>(v) 27 \u2013 [5 + {28 \u2013 (29 \u2013 7)}]<\/p>\n\n\n\n<p>It can be written as<\/p>\n\n\n\n<p>= 27 \u2013 [5 + {28 \u2013 22}]<\/p>\n\n\n\n<p>By further calculation<\/p>\n\n\n\n<p>= 27 \u2013 [5 + 6]<\/p>\n\n\n\n<p>So we get<\/p>\n\n\n\n<p>= 27 \u2013 11<\/p>\n\n\n\n<p>= 16<\/p>\n\n\n\n<p>(vi) 48 \u2013 [18 \u2013 {16 \u2013 (5 \u2013&nbsp;<img decoding=\"async\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2022\/05\/selina-solutions-concise-maths-class-7-chapter-1-ex-1d-image-3.gif\" alt=\"Selina Solutions Concise Maths Class 7 Chapter 1 Ex 1D Image 3\">)}]<\/p>\n\n\n\n<p>It can be written as<\/p>\n\n\n\n<p>= 48 \u2013 [18 \u2013 {16 \u2013 (5 \u2013 5)}]<\/p>\n\n\n\n<p>By further calculation<\/p>\n\n\n\n<p>= 48 \u2013 [18 \u2013 {16 \u2013 0}]<\/p>\n\n\n\n<p>So we get<\/p>\n\n\n\n<p>= 48 \u2013 [18 \u2013 16]<\/p>\n\n\n\n<p>= 48 \u2013 2<\/p>\n\n\n\n<p>= 46<\/p>\n\n\n\n<p>(vii) \u2013 8 \u2013 {- 6 (9 \u2013 11) + 18 \u00f7 -3}<\/p>\n\n\n\n<p>It can be written as<\/p>\n\n\n\n<p>= \u2013 8 \u2013 {- 6 (-2) \u2013 6}<\/p>\n\n\n\n<p>By further calculation<\/p>\n\n\n\n<p>= \u2013 8 \u2013 {12 \u2013 6}<\/p>\n\n\n\n<p>So we get<\/p>\n\n\n\n<p>= \u2013 8 \u2013 6<\/p>\n\n\n\n<p>= \u2013 14<\/p>\n\n\n\n<p>(viii) (24 \u00f7&nbsp;<img decoding=\"async\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2022\/05\/selina-solutions-concise-maths-class-7-chapter-1-ex-1d-image-4.gif\" alt=\"Selina Solutions Concise Maths Class 7 Chapter 1 Ex 1D Image 4\">\u2013 12) \u2013 (3 \u00d7 8 \u00f7 4 + 1)<\/p>\n\n\n\n<p>It can be written as<\/p>\n\n\n\n<p>= (24 \u00f7 3 \u2013 12) \u2013 (3 \u00d7 2 + 1)<\/p>\n\n\n\n<p>By further calculation<\/p>\n\n\n\n<p>= (8 \u2013 12) \u2013 (6 + 1)<\/p>\n\n\n\n<p>So we get<\/p>\n\n\n\n<p>= \u2013 4 \u2013 7<\/p>\n\n\n\n<p>= \u2013 11<\/p>\n\n\n\n<p><strong>12. Find the result of subtracting the sum of all integers between 20 and 30 from the sum of all integers from 20 to 30.<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>Here the required number = sum of all integers from 20 to 30 \u2013 sum of all integers between 20 and 30<\/p>\n\n\n\n<p>Substituting the values<\/p>\n\n\n\n<p>= (20 + 21 + 22 + 23 + 24 + 25 + 26 + 27 + 28 + 29 + 30) \u2013 (21 + 22 + 23 + 24 + 25 + 26 + 27 + 28 + 29)<\/p>\n\n\n\n<p>On further calculation<\/p>\n\n\n\n<p>= 20 + 30 = 50<\/p>\n\n\n\n<p>Hence, the required number is 50.<\/p>\n\n\n\n<p><strong>13. Add the product of (-13) and (-17) to the quotient of (-187) and 11.<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>It is given that<\/p>\n\n\n\n<p>(-13) \u00d7 (-17) + (-187&nbsp;\u00f7 11)<\/p>\n\n\n\n<p>By further calculation<\/p>\n\n\n\n<p>= (-13) \u00d7 (-17) + (-17)<\/p>\n\n\n\n<p>So we get<\/p>\n\n\n\n<p>= 221 \u2013 17<\/p>\n\n\n\n<p>= 204<\/p>\n\n\n\n<p><strong>14. The product of two integers is \u2013 180. If one of them is 12, find the other.<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>Product of two integers = \u2013 180<\/p>\n\n\n\n<p>One integer = 12<\/p>\n\n\n\n<p>Other integer = \u2013 180\/ 12<\/p>\n\n\n\n<p>By division we get<\/p>\n\n\n\n<p>= \u2013 15<\/p>\n\n\n\n<p><strong>15. (i) A number changes from \u2013 20 to 30. What is the increase or decrease in the number?<\/strong><\/p>\n\n\n\n<p><strong>(ii) A number changes from 40 to \u2013 30. What is the increase or decrease in the number?<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>(i) A number changes from \u2013 20 to 30<\/p>\n\n\n\n<p>It can be written as<\/p>\n\n\n\n<p>\u2013 20 \u2013 30 = \u2013 50<\/p>\n\n\n\n<p>Hence, \u2013 50 will be the increase in the number.<\/p>\n\n\n\n<p>(ii) A number changes from 40 to \u2013 30<\/p>\n\n\n\n<p>It can be written as<\/p>\n\n\n\n<p>40 \u2013 (-30) = 40 + 30 = 70<\/p>\n\n\n\n<p>Hence, 70 will be the decrease in the number.<\/p>\n\n\n\n<p><strong>1. Divide:<\/strong><\/p>\n\n\n\n<p><strong>(i) 117 by 9<\/strong><\/p>\n\n\n\n<p><strong>(ii) (-117) by 9<\/strong><\/p>\n\n\n\n<p><strong>(iii) 117 by (-9)<\/strong><\/p>\n\n\n\n<p><strong>(iv) (-117) by (-9)<\/strong><\/p>\n\n\n\n<p><strong>(v) 225 by (-15)<\/strong><\/p>\n\n\n\n<p><strong>(vi) (-552) \u00f7 24<\/strong><\/p>\n\n\n\n<p><strong>(vii) (-798) by (-21)<\/strong><\/p>\n\n\n\n<p><strong>(viii) (-910) \u00f7 26<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>(i) 117 by 9<\/p>\n\n\n\n<p>= 117\/9<\/p>\n\n\n\n<p>It can be written as<\/p>\n\n\n\n<p>= (13 \u00d7 9)\/ 9<\/p>\n\n\n\n<p>= 13<\/p>\n\n\n\n<p>(ii) (-117) by 9<\/p>\n\n\n\n<p>= \u2013 117\/ 9<\/p>\n\n\n\n<p>It can be written as<\/p>\n\n\n\n<p>= (-13 \u00d7 9)\/ 9<\/p>\n\n\n\n<p>= -13<\/p>\n\n\n\n<p>(iii) 117 by (-9)<\/p>\n\n\n\n<p>= 117\/ -9<\/p>\n\n\n\n<p>It can be written as<\/p>\n\n\n\n<p>= (13 \u00d7 9)\/ -9<\/p>\n\n\n\n<p>= \u2013 13<\/p>\n\n\n\n<p>(iv) (-117) by (-9)<\/p>\n\n\n\n<p>= \u2013 117\/-9 = 117\/9<\/p>\n\n\n\n<p>It can be written as<\/p>\n\n\n\n<p>= (13 \u00d7 9)\/ 9<\/p>\n\n\n\n<p>= 13<\/p>\n\n\n\n<p>(v) 225 by (-15)<\/p>\n\n\n\n<p>= 225\/ (-15)<\/p>\n\n\n\n<p>It can be written as<\/p>\n\n\n\n<p>= (15 \u00d7 15)\/ -15<\/p>\n\n\n\n<p>= \u2013 15<\/p>\n\n\n\n<p>(vi) (-552) \u00f7 24<\/p>\n\n\n\n<p>= \u2013 552\/ 24<\/p>\n\n\n\n<p>It can be written as<\/p>\n\n\n\n<p>= (- 23 \u00d7 24)\/ 24<\/p>\n\n\n\n<p>= \u2013 23<\/p>\n\n\n\n<p>(vii) (-798) by (-21)<\/p>\n\n\n\n<p>= \u2013 798\/ \u2013 21 = 798\/21<\/p>\n\n\n\n<p>It can be written as<\/p>\n\n\n\n<p>= (38 \u00d7 21)\/ 21<\/p>\n\n\n\n<p>= 38<\/p>\n\n\n\n<p>(viii) (-910) \u00f7 26<\/p>\n\n\n\n<p>= \u2013 910\/ 26<\/p>\n\n\n\n<p>It can be written as<\/p>\n\n\n\n<p>= (- 35 \u00d7 26)\/ 26<\/p>\n\n\n\n<p>= \u2013 35<\/p>\n\n\n\n<p><strong>2. Evaluate:<\/strong><\/p>\n\n\n\n<p><strong>(i) (-234) \u00f7 13<\/strong><\/p>\n\n\n\n<p><strong>(ii) 234 \u00f7 (-13)<\/strong><\/p>\n\n\n\n<p><strong>(iii) (-234) \u00f7 (-13)<\/strong><\/p>\n\n\n\n<p><strong>(iv) 374 \u00f7 (-17)<\/strong><\/p>\n\n\n\n<p><strong>(v) (-374) \u00f7 17<\/strong><\/p>\n\n\n\n<p><strong>(vi) (-374) \u00f7 (-17)<\/strong><\/p>\n\n\n\n<p><strong>(vii) (-728) \u00f7 14<\/strong><\/p>\n\n\n\n<p><strong>(viii) 272 \u00f7 (-17)<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>(i) (-234) \u00f7 13<\/p>\n\n\n\n<p>= \u2013 234\/ 13<\/p>\n\n\n\n<p>It can be written as<\/p>\n\n\n\n<p>= (-18 \u00d7 13)\/ 13<\/p>\n\n\n\n<p>= \u2013 18<\/p>\n\n\n\n<p>(ii) 234 \u00f7 (-13)<\/p>\n\n\n\n<p>= 234\/ -13<\/p>\n\n\n\n<p>It can be written as<\/p>\n\n\n\n<p>= (18 \u00d7 13)\/ \u2013 13<\/p>\n\n\n\n<p>= \u2013 18<\/p>\n\n\n\n<p>(iii) (-234) \u00f7 (-13)<\/p>\n\n\n\n<p>= \u2013 234\/ \u2013 13 = 234\/13<\/p>\n\n\n\n<p>It can be written as<\/p>\n\n\n\n<p>= (18 \u00d7 13)\/ 13<\/p>\n\n\n\n<p>= 18<\/p>\n\n\n\n<p>(iv) 374 \u00f7 (-17)<\/p>\n\n\n\n<p>= 374\/ -17<\/p>\n\n\n\n<p>It can be written as<\/p>\n\n\n\n<p>= (22 \u00d7 17)\/ (-17)<\/p>\n\n\n\n<p>= \u2013 22<\/p>\n\n\n\n<p>(v) (-374) \u00f7 17<\/p>\n\n\n\n<p>= \u2013 374\/ 17<\/p>\n\n\n\n<p>It can be written as<\/p>\n\n\n\n<p>= (-22 \u00d7 17)\/ (17)<\/p>\n\n\n\n<p>= \u2013 22<\/p>\n\n\n\n<p>(vi) (-374) \u00f7 (-17)<\/p>\n\n\n\n<p>= \u2013 374\/ -17 = 374\/17<\/p>\n\n\n\n<p>It can be written as<\/p>\n\n\n\n<p>= (22 \u00d7 17)\/ (17)<\/p>\n\n\n\n<p>= 22<\/p>\n\n\n\n<p>(vii) (-728) \u00f7 14<\/p>\n\n\n\n<p>= \u2013 728\/14<\/p>\n\n\n\n<p>It can be written as<\/p>\n\n\n\n<p>= (-52 \u00d7 14)\/ 14<\/p>\n\n\n\n<p>= -52<\/p>\n\n\n\n<p>(viii) 272 \u00f7 (-17)<\/p>\n\n\n\n<p>= 272\/ -17<\/p>\n\n\n\n<p>It can be written as<\/p>\n\n\n\n<p>= (16 \u00d7 17)\/ (-17)<\/p>\n\n\n\n<p>= -16<\/p>\n\n\n\n<p><strong>3. Find the quotient in each of the following divisions:<\/strong><\/p>\n\n\n\n<p><strong>(i) 299 \u00f7 23<\/strong><\/p>\n\n\n\n<p><strong>(ii) 299 \u00f7 (-23)<\/strong><\/p>\n\n\n\n<p><strong>(iii) (-384) \u00f7 16<\/strong><\/p>\n\n\n\n<p><strong>(iv) (-572) \u00f7 (-22)<\/strong><\/p>\n\n\n\n<p><strong>(v) 408 \u00f7 (-17)<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>(i) 299 \u00f7 23<\/p>\n\n\n\n<p>= 299\/23<\/p>\n\n\n\n<p>It can be written as<\/p>\n\n\n\n<p>= (23 \u00d7 13)\/ 23<\/p>\n\n\n\n<p>= 13<\/p>\n\n\n\n<p>(ii) 299 \u00f7 (-23)<\/p>\n\n\n\n<p>= 299\/ -23<\/p>\n\n\n\n<p>It can be written as<\/p>\n\n\n\n<p>= (23 \u00d7 13)\/ \u2013 23<\/p>\n\n\n\n<p>= \u2013 13<\/p>\n\n\n\n<p>(iii) (-384) \u00f7 16<\/p>\n\n\n\n<p>= \u2013 384\/16<\/p>\n\n\n\n<p>It can be written as<\/p>\n\n\n\n<p>= (- 24 \u00d7 16)\/ 16<\/p>\n\n\n\n<p>= \u2013 24<\/p>\n\n\n\n<p>(iv) (-572) \u00f7 (-22)<\/p>\n\n\n\n<p>= \u2013 572\/ \u2013 22 = 572\/22<\/p>\n\n\n\n<p>It can be written as<\/p>\n\n\n\n<p>= (26 \u00d7 22)\/ 22<\/p>\n\n\n\n<p>= 26<\/p>\n\n\n\n<p>(v) 408 \u00f7 (-17)<\/p>\n\n\n\n<p>= 408\/ -17<\/p>\n\n\n\n<p>It can be written as<\/p>\n\n\n\n<p>= (24 \u00d7 17)\/ (-17)<\/p>\n\n\n\n<p>= \u2013 24<\/p>\n\n\n\n<p><strong>4. Divide:<\/strong><\/p>\n\n\n\n<p><strong>(i) 204 by 17<\/strong><\/p>\n\n\n\n<p><strong>(ii) 152 by \u2013 19<\/strong><\/p>\n\n\n\n<p><strong>(iii) 0 by 35<\/strong><\/p>\n\n\n\n<p><strong>(iv) 0 by (-82)<\/strong><\/p>\n\n\n\n<p><strong>(v) 5490 by 10<\/strong><\/p>\n\n\n\n<p><strong>(vi) 762800 by 100<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>(i) 204 by 17<\/p>\n\n\n\n<p>= 204\/17<\/p>\n\n\n\n<p>It can be written as<\/p>\n\n\n\n<p>= (12 \u00d7 17)\/ 17<\/p>\n\n\n\n<p>= 12<\/p>\n\n\n\n<p>(ii) 152 by \u2013 19<\/p>\n\n\n\n<p>= 152\/ -19<\/p>\n\n\n\n<p>It can be written as<\/p>\n\n\n\n<p>= (8 \u00d7 19)\/ -19<\/p>\n\n\n\n<p>= \u2013 8<\/p>\n\n\n\n<p>(iii) 0 by 35<\/p>\n\n\n\n<p>= 0\/35<\/p>\n\n\n\n<p>= 0<\/p>\n\n\n\n<p>(iv) 0 by (-82)<\/p>\n\n\n\n<p>= 0\/ -82<\/p>\n\n\n\n<p>= 0<\/p>\n\n\n\n<p>(v) 5490 by 10<\/p>\n\n\n\n<p>= 5490\/10<\/p>\n\n\n\n<p>It can be written as<\/p>\n\n\n\n<p>= (549 \u00d7 10)\/ 10<\/p>\n\n\n\n<p>= 549<\/p>\n\n\n\n<p>(vi) 762800 by 100<\/p>\n\n\n\n<p>= 762800\/100<\/p>\n\n\n\n<p>It can be written as<\/p>\n\n\n\n<p>= (7628 x 100)\/ 100<\/p>\n\n\n\n<p>= 7628<\/p>\n\n\n\n<p><strong>5. State, true or false:<\/strong><\/p>\n\n\n\n<p><strong>(i) 0 \u00f7 32 = 0<\/strong><\/p>\n\n\n\n<p><strong>(ii) 0 \u00f7 (-9) = 0<\/strong><\/p>\n\n\n\n<p><strong>(iii) (-37) \u00f7 0 = 0<\/strong><\/p>\n\n\n\n<p><strong>(iv) 0 \u00f7 0 = 0<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>(i) True.<\/p>\n\n\n\n<p>(ii) True.<\/p>\n\n\n\n<p>(iii) False. It is not defined.<\/p>\n\n\n\n<p>(iv) False. It is not defined.<\/p>\n\n\n\n<p><strong>6. Evaluate:<\/strong><\/p>\n\n\n\n<p><strong>(i) 42 \u00f7 7 + 4<\/strong><\/p>\n\n\n\n<p><strong>(ii) 12 + 18 \u00f7 3<\/strong><\/p>\n\n\n\n<p><strong>(iii) 19 \u2013 20 \u00f7 4<\/strong><\/p>\n\n\n\n<p><strong>(iv) 16 \u2013 5 \u00d7 3 + 4<\/strong><\/p>\n\n\n\n<p><strong>(v) 6 \u2013 8 \u2013 (-6) \u00f7 2<\/strong><\/p>\n\n\n\n<p><strong>(vi) 13 \u2013 12 \u00f7 4 \u00d7 2<\/strong><\/p>\n\n\n\n<p><strong>(vii) 16 + 8 \u00f7 4 \u2013 2 \u00d7 3<\/strong><\/p>\n\n\n\n<p><strong>(viii) 16 \u00f7 8 + 4 \u2013 2 \u00d7 3<\/strong><\/p>\n\n\n\n<p><strong>(ix) 16 \u2013 8 + 4 \u00f7 2 \u00d7 3<\/strong><\/p>\n\n\n\n<p><strong>(x) (-4) + (-12) \u00f7 (-6)<\/strong><\/p>\n\n\n\n<p><strong>(xi) (-18) + 6 \u00f7 3 + 5<\/strong><\/p>\n\n\n\n<p><strong>(xii) (-20) \u00d7 (-1) + 14 \u00f7 7<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>(i) 42 \u00f7 7 + 4<\/p>\n\n\n\n<p>It can be written as<\/p>\n\n\n\n<p>= 42\/7 + 4<\/p>\n\n\n\n<p>By further calculation<\/p>\n\n\n\n<p>= 6 + 4<\/p>\n\n\n\n<p>= 10<\/p>\n\n\n\n<p>(ii) 12 + 18 \u00f7 3<\/p>\n\n\n\n<p>It can be written as<\/p>\n\n\n\n<p>= 12 + 18\/3<\/p>\n\n\n\n<p>By further calculation<\/p>\n\n\n\n<p>= 12 + 6<\/p>\n\n\n\n<p>= 18<\/p>\n\n\n\n<p>(iii) 19 \u2013 20 \u00f7 4<\/p>\n\n\n\n<p>It can be written as<\/p>\n\n\n\n<p>= 19 \u2013 20\/4<\/p>\n\n\n\n<p>By further calculation<\/p>\n\n\n\n<p>= 19 \u2013 5<\/p>\n\n\n\n<p>= 14<\/p>\n\n\n\n<p>(iv) 16 \u2013 5 \u00d7 3 + 4<\/p>\n\n\n\n<p>It can be written as<\/p>\n\n\n\n<p>= 16 \u2013 15 + 4<\/p>\n\n\n\n<p>By further calculation<\/p>\n\n\n\n<p>= 20 \u2013 15<\/p>\n\n\n\n<p>= 5<\/p>\n\n\n\n<p>(v) 6 \u2013 8 \u2013 (-6) \u00f7 2<\/p>\n\n\n\n<p>It can be written as<\/p>\n\n\n\n<p>= 6 \u2013 8 \u2013 (-6\/2)<\/p>\n\n\n\n<p>By further calculation<\/p>\n\n\n\n<p>= 6 \u2013 8 \u2013 (-3)<\/p>\n\n\n\n<p>So we get<\/p>\n\n\n\n<p>= 6 \u2013 8 + 3<\/p>\n\n\n\n<p>= 9 \u2013 8<\/p>\n\n\n\n<p>= 1<\/p>\n\n\n\n<p>(vi) 13 \u2013 12 \u00f7 4 \u00d7 2<\/p>\n\n\n\n<p>It can be written as<\/p>\n\n\n\n<p>= 13 \u2013 12\/4 \u00d7 2<\/p>\n\n\n\n<p>By further calculation<\/p>\n\n\n\n<p>= 13 \u2013 3 \u00d7 2<\/p>\n\n\n\n<p>= 13 \u2013 6<\/p>\n\n\n\n<p>= 7<\/p>\n\n\n\n<p>(vii) 16 + 8 \u00f7 4 \u2013 2 \u00d7 3<\/p>\n\n\n\n<p>It can be written as<\/p>\n\n\n\n<p>= 16 + 8\/4 \u2013 2 \u00d7 3<\/p>\n\n\n\n<p>By further calculation<\/p>\n\n\n\n<p>= 16 + 2 \u2013 2 \u00d7 3<\/p>\n\n\n\n<p>So we get<\/p>\n\n\n\n<p>= 16 + 2 \u2013 6<\/p>\n\n\n\n<p>= 18 \u2013 6<\/p>\n\n\n\n<p>= 12<\/p>\n\n\n\n<p>(viii) 16 \u00f7 8 + 4 \u2013 2 \u00d7 3<\/p>\n\n\n\n<p>It can be written as<\/p>\n\n\n\n<p>= 16\/8 + 4 \u2013 2 \u00d7 3<\/p>\n\n\n\n<p>By further calculation<\/p>\n\n\n\n<p>= 2 + 4 \u2013 6<\/p>\n\n\n\n<p>So we get<\/p>\n\n\n\n<p>= 6 \u2013 6<\/p>\n\n\n\n<p>= 0<\/p>\n\n\n\n<p>(ix) 16 \u2013 8 + 4 \u00f7 2 \u00d7 3<\/p>\n\n\n\n<p>It can be written as<\/p>\n\n\n\n<p>= 16 \u2013 8 + 4\/2 \u00d7 3<\/p>\n\n\n\n<p>By further calculation<\/p>\n\n\n\n<p>= 16 \u2013 8 + 2 \u00d7 3<\/p>\n\n\n\n<p>So we get<\/p>\n\n\n\n<p>= 16 \u2013 8 + 6<\/p>\n\n\n\n<p>Here<\/p>\n\n\n\n<p>= 22 \u2013 8<\/p>\n\n\n\n<p>= 14<\/p>\n\n\n\n<p>(x) (-4) + (-12) \u00f7 (-6)<\/p>\n\n\n\n<p>It can be written as<\/p>\n\n\n\n<p>= (-4) + (-12\/-6)<\/p>\n\n\n\n<p>By further calculation<\/p>\n\n\n\n<p>= \u2013 4 + 2<\/p>\n\n\n\n<p>= -2<\/p>\n\n\n\n<p>(xi) (-18) + 6 \u00f7 3 + 5<\/p>\n\n\n\n<p>It can be written as<\/p>\n\n\n\n<p>= (-18) + 6\/3 + 5<\/p>\n\n\n\n<p>By further calculation<\/p>\n\n\n\n<p>= (-18) + 2 + 5<\/p>\n\n\n\n<p>So we get<\/p>\n\n\n\n<p>= \u2013 18 + 7<\/p>\n\n\n\n<p>= \u2013 11<\/p>\n\n\n\n<p>(xii) (-20) \u00d7 (-1) + 14 \u00f7 7<\/p>\n\n\n\n<p>It can be written as<\/p>\n\n\n\n<p>= (-20) \u00d7 (-1) + 14\/7<\/p>\n\n\n\n<p>By further calculation<\/p>\n\n\n\n<p>= (-20) \u00d7 (-1) + 2<\/p>\n\n\n\n<p>So we get<\/p>\n\n\n\n<p>= 20 + 2<\/p>\n\n\n\n<p>= 22<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Exercise 1C<\/h3>\n\n\n\n<p><strong>Evaluate:<\/strong><\/p>\n\n\n\n<p><strong>1. 18 \u2013 (20 \u2013 15 \u00f7 3)<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>18 \u2013 (20 \u2013 15 \u00f7 3)<\/p>\n\n\n\n<p>It can be written as<\/p>\n\n\n\n<p>= 18 \u2013 (20 \u2013 15\/3)<\/p>\n\n\n\n<p>By further calculation<\/p>\n\n\n\n<p>= 18 \u2013 (20 \u2013 5)<\/p>\n\n\n\n<p>= 18 \u2013 20 + 5<\/p>\n\n\n\n<p>So we get<\/p>\n\n\n\n<p>= 18 + 5 \u2013 20<\/p>\n\n\n\n<p>= 23 \u2013 20<\/p>\n\n\n\n<p>= 3<\/p>\n\n\n\n<p><strong>2. \u2013 15 + 24 \u00f7 (15 \u2013 13)<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>\u2013 15 + 24 \u00f7 (15 \u2013 13)<\/p>\n\n\n\n<p>It can be written as<\/p>\n\n\n\n<p>= \u2013 15 + 24 \u00f7 2<\/p>\n\n\n\n<p>So we get<\/p>\n\n\n\n<p>= \u2013 15 + 12<\/p>\n\n\n\n<p>= \u2013 3<\/p>\n\n\n\n<p><strong>3. 35 \u2013 {15 + 14 \u2013 (13 + )}<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>35 \u2013 {15 + 14 \u2013 (13 +<br><img decoding=\"async\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2022\/05\/selina-solutions-concise-maths-class-7-chapter-1-ex-1c-image-2.gif\" alt=\"Selina Solutions Concise Maths Class 7 Chapter 1 Ex 1C Image 2\">)}<\/p>\n\n\n\n<p>It can be written as<\/p>\n\n\n\n<p>= 35 \u2013 [15 + 14 \u2013 (13 + 4)]<\/p>\n\n\n\n<p>By further calculation<\/p>\n\n\n\n<p>= 35 \u2013 [15 + 14 \u2013 17]<\/p>\n\n\n\n<p>Multiplying the negative sign<\/p>\n\n\n\n<p>= 35 \u2013 15 \u2013 14 + 17<\/p>\n\n\n\n<p>So we get<\/p>\n\n\n\n<p>= 52 \u2013 29<\/p>\n\n\n\n<p>= 23<\/p>\n\n\n\n<p><strong>4. 27 \u2013 {13 + 4 \u2013 (8 + 4 \u2013 )}<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>27 \u2013 {13 + 4 \u2013 (8 + 4 \u2013<br><img decoding=\"async\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2022\/05\/selina-solutions-concise-maths-class-7-chapter-1-ex-1c-image-4.gif\" alt=\"Selina Solutions Concise Maths Class 7 Chapter 1 Ex 1C Image 4\">)}<\/p>\n\n\n\n<p>It can be written as<\/p>\n\n\n\n<p>= 27 \u2013 {13 + 4 \u2013 (8 + 4 \u2013 4)}<\/p>\n\n\n\n<p>By further calculation<\/p>\n\n\n\n<p>= 27 \u2013 {13 + 4 \u2013 8}<\/p>\n\n\n\n<p>So we get<\/p>\n\n\n\n<p>= 27 \u2013 {13 + (-4)}<\/p>\n\n\n\n<p>= 27 \u2013 9<\/p>\n\n\n\n<p>= 18<\/p>\n\n\n\n<p><strong>5. 32 \u2013 [43 \u2013 {51 \u2013 (20 \u2013 )}]<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>32 \u2013 [43 \u2013 {51 \u2013 (20 \u2013<br><img decoding=\"async\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2022\/05\/selina-solutions-concise-maths-class-7-chapter-1-ex-1c-image-6.gif\" alt=\"Selina Solutions Concise Maths Class 7 Chapter 1 Ex 1C Image 6\">)}]<\/p>\n\n\n\n<p>It can be written as<\/p>\n\n\n\n<p>= 32 \u2013 [43 \u2013 {51 \u2013 (20 \u2013 11)}]<\/p>\n\n\n\n<p>By further calculation<\/p>\n\n\n\n<p>= 32 \u2013 [43 \u2013 {51 \u2013 9}]<\/p>\n\n\n\n<p>So we get<\/p>\n\n\n\n<p>= 32 \u2013 [43 \u2013 42]<\/p>\n\n\n\n<p>= 32 \u2013 1<\/p>\n\n\n\n<p>= 31<\/p>\n\n\n\n<p><strong>6. 46 \u2013 [26 \u2013 {14 \u2013 (15 \u2013 4 \u00f7 2 \u00d7 2)}]<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>46 \u2013 [26 \u2013 {14 \u2013 (15 \u2013 4 \u00f7 2 \u00d7 2)}]<\/p>\n\n\n\n<p>It can be written as<\/p>\n\n\n\n<p>= 46 \u2013 [26 \u2013 {14 \u2013 (15 \u2013 2 \u00d7 2)}]<\/p>\n\n\n\n<p>By further calculation<\/p>\n\n\n\n<p>= 46 \u2013 [26 \u2013 {14 \u2013 (15 \u2013 4)}]<\/p>\n\n\n\n<p>So we get<\/p>\n\n\n\n<p>= 46 \u2013 [26 \u2013 {14 \u2013 11}]<\/p>\n\n\n\n<p>= 46 \u2013 [26 \u2013 3]<\/p>\n\n\n\n<p>Here<\/p>\n\n\n\n<p>= 46 \u2013 23<\/p>\n\n\n\n<p>= 23<\/p>\n\n\n\n<p><strong>7. 45 \u2013 [38 \u2013 {60 \u00f7 3 \u2013 (6 \u2013 9 \u00f7 3) \u00f7 3}]<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>45 \u2013 [38 \u2013 {60 \u00f7 3 \u2013 (6 \u2013 9 \u00f7 3) \u00f7 3}]<\/p>\n\n\n\n<p>It can be written as<\/p>\n\n\n\n<p>= 45 \u2013 [38 \u2013 {60 \u00f7 3 \u2013 (6 \u2013 3) \u00f7 3}]<\/p>\n\n\n\n<p>By further calculation<\/p>\n\n\n\n<p>= 45 \u2013 [38 \u2013 {20 \u2013 3 \u00f7 3}]<\/p>\n\n\n\n<p>So we get<\/p>\n\n\n\n<p>= 45 \u2013 [38 \u2013 {20 \u2013 1}]<\/p>\n\n\n\n<p>By subtraction<\/p>\n\n\n\n<p>= 45 \u2013 [38 \u2013 19]<\/p>\n\n\n\n<p>= 45 \u2013 19<\/p>\n\n\n\n<p>= 26<\/p>\n\n\n\n<p><strong>8. 17 \u2013 [17 \u2013 {17 \u2013 (17 \u2013 )}]<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>17 \u2013 [17 \u2013 {17 \u2013 (17 \u2013<br><img decoding=\"async\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2022\/05\/selina-solutions-concise-maths-class-7-chapter-1-ex-1c-image-8.gif\" alt=\"Selina Solutions Concise Maths Class 7 Chapter 1 Ex 1C Image 8\">)}]<\/p>\n\n\n\n<p>It can be written as<\/p>\n\n\n\n<p>= 17 \u2013 [17 \u2013 {17 \u2013 (17 \u2013 0)}]<\/p>\n\n\n\n<p>By further calculation<\/p>\n\n\n\n<p>= 17 \u2013 [17 \u2013 {17 \u2013 17}]<\/p>\n\n\n\n<p>So we get<\/p>\n\n\n\n<p>= 17 \u2013 [17 \u2013 0]<\/p>\n\n\n\n<p>= 17 \u2013 17<\/p>\n\n\n\n<p>= 0<\/p>\n\n\n\n<p><strong>9. 2550 \u2013 [510 \u2013 {270 \u2013 (90 \u2013 )}]<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>2550 \u2013 [510 \u2013 {270 \u2013 (90 \u2013<br><img decoding=\"async\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2022\/05\/selina-solutions-concise-maths-class-7-chapter-1-ex-1c-image-10.gif\" alt=\"Selina Solutions Concise Maths Class 7 Chapter 1 Ex 1C Image 10\">)}]<\/p>\n\n\n\n<p>It can be written as<\/p>\n\n\n\n<p>= 2550 \u2013 [510 \u2013 {270 \u2013 (90 \u2013 87)}]<\/p>\n\n\n\n<p>By further calculation<\/p>\n\n\n\n<p>= 2550 \u2013 [510 \u2013 {270 \u2013 3}]<\/p>\n\n\n\n<p>So we get<\/p>\n\n\n\n<p>= 2550 \u2013 [510 \u2013 267]<\/p>\n\n\n\n<p>= 2550 \u2013 243<\/p>\n\n\n\n<p>= 2307<\/p>\n\n\n\n<p><strong>10. 30 + [{-2 \u00d7 (25 \u2013 )}]<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>30 + [{-2 \u00d7 (25 \u2013<br><img decoding=\"async\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2022\/05\/selina-solutions-concise-maths-class-7-chapter-1-ex-1c-image-12.gif\" alt=\"Selina Solutions Concise Maths Class 7 Chapter 1 Ex 1C Image 12\">)}]<\/p>\n\n\n\n<p>It can be written as<\/p>\n\n\n\n<p>= 30 + [{-2 \u00d7 (25 \u2013 10)}]<\/p>\n\n\n\n<p>By further calculation<\/p>\n\n\n\n<p>= 30 + [{-2 \u00d7 15}]<\/p>\n\n\n\n<p>So we get<\/p>\n\n\n\n<p>= 30 + [-30]<\/p>\n\n\n\n<p>= 30 \u2013 30<\/p>\n\n\n\n<p>= 0<\/p>\n\n\n\n<p><strong>11. 88 \u2013 {5 \u2013 (-48) \u00f7 (-16)}<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>88 \u2013 {5 \u2013 (-48) \u00f7 (-16)}<\/p>\n\n\n\n<p>It can be written as<\/p>\n\n\n\n<p>= 88 \u2013 {5 \u2013 (- 48\/ -16)}<\/p>\n\n\n\n<p>By further calculation<\/p>\n\n\n\n<p>= 88 \u2013 {5 \u2013 3}<\/p>\n\n\n\n<p>So we get<\/p>\n\n\n\n<p>= 88 \u2013 2<\/p>\n\n\n\n<p>= 86<\/p>\n\n\n\n<p><strong>12. 9 \u00d7 (8 \u2013 ) \u2013 2 (2 + )<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>9 \u00d7 (8 \u2013<br><img decoding=\"async\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2022\/05\/selina-solutions-concise-maths-class-7-chapter-1-ex-1c-image-15.gif\" alt=\"Selina Solutions Concise Maths Class 7 Chapter 1 Ex 1C Image 15\">) \u2013 2 (2 +<br><img decoding=\"async\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2022\/05\/selina-solutions-concise-maths-class-7-chapter-1-ex-1c-image-16.gif\" alt=\"Selina Solutions Concise Maths Class 7 Chapter 1 Ex 1C Image 16\">)<\/p>\n\n\n\n<p>It can be written as<\/p>\n\n\n\n<p>= 9 \u00d7 (8 \u2013 5) \u2013 2 (2 + 6)<\/p>\n\n\n\n<p>By further calculation<\/p>\n\n\n\n<p>= 9 \u00d7 3 \u2013 2 \u00d7 8<\/p>\n\n\n\n<p>So we get<\/p>\n\n\n\n<p>= 27 \u2013 16<\/p>\n\n\n\n<p>= 11<\/p>\n\n\n\n<p><strong>13. 2 \u2013 [3 \u2013 {6 \u2013 (5 \u2013 )}]<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>2 \u2013 [3 \u2013 {6 \u2013 (5 \u2013<br><img decoding=\"async\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2022\/05\/selina-solutions-concise-maths-class-7-chapter-1-ex-1c-image-18.gif\" alt=\"Selina Solutions Concise Maths Class 7 Chapter 1 Ex 1C Image 18\">)}]<\/p>\n\n\n\n<p>It can be written as<\/p>\n\n\n\n<p>= 2 \u2013 [3 \u2013 {6 \u2013 (5 \u2013 1)}]<\/p>\n\n\n\n<p>By further calculation<\/p>\n\n\n\n<p>= 2 \u2013 [3 \u2013 {6 \u2013 4}]<\/p>\n\n\n\n<p>So we get<\/p>\n\n\n\n<p>= 2 \u2013 [3 \u2013 2]<\/p>\n\n\n\n<p>= 2 \u2013 1<\/p>\n\n\n\n<p>= 1<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Exercise 1D<\/h3>\n\n\n\n<p><strong>1. The sum of two integers is \u2013 15. If one of them is 9, find the other.<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>It is given that<\/p>\n\n\n\n<p>Sum of two integers = \u2013 15<\/p>\n\n\n\n<p>One integer = 9<\/p>\n\n\n\n<p>Other integer = \u2013 15 \u2013 9<\/p>\n\n\n\n<p>Taking negative sign as common<\/p>\n\n\n\n<p>= \u2013 (15 + 9)<\/p>\n\n\n\n<p>= \u2013 24<\/p>\n\n\n\n<p><strong>2. The difference between integers x and \u2013 6 is \u2013 5. Find the values of x.<\/strong><\/p>\n\n\n\n<p><strong>x \u2013 (-6) = \u2013 5 or \u2013 6 \u2013 x = \u2013 5<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>It is given that<\/p>\n\n\n\n<p>The difference between integers x and -6 is -5<\/p>\n\n\n\n<p>\u21d2 x \u2013 (-6) = \u2013 5 or \u2013 6 \u2013 x = \u2013 5<\/p>\n\n\n\n<p>So the value of x is<\/p>\n\n\n\n<p>x + 6 = \u2013 5 or \u2013 x = \u2013 5 + 6<\/p>\n\n\n\n<p>By further calculation<\/p>\n\n\n\n<p>x = \u2013 5 \u2013 6 or \u2013 x = 1<\/p>\n\n\n\n<p>x = \u2013 11 or x = \u2013 1<\/p>\n\n\n\n<p><strong>3. The sum of two integers is 28. If one integer is \u2013 45, find the other.<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>It is given that<\/p>\n\n\n\n<p>Sum of two integers = 28<\/p>\n\n\n\n<p>One integer = \u2013 45<\/p>\n\n\n\n<p>Other integer = 28 \u2013 (-45)<\/p>\n\n\n\n<p>By further calculation<\/p>\n\n\n\n<p>= 28 + 45<\/p>\n\n\n\n<p>= 73<\/p>\n\n\n\n<p><strong>4. The sum of two integers is \u2013 56. If one integer is \u2013 42, find the other.<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>It is given that<\/p>\n\n\n\n<p>Sum of two integers = \u2013 56<\/p>\n\n\n\n<p>One integer = \u2013 42<\/p>\n\n\n\n<p>Other integer = \u2013 56 \u2013 (- 42)<\/p>\n\n\n\n<p>By further calculation<\/p>\n\n\n\n<p>= \u2013 56 + 42<\/p>\n\n\n\n<p>= \u2013 14<\/p>\n\n\n\n<p><strong>5. The difference between an integer x and (-9) is 6. Find all possible values of x.<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>It is given that<\/p>\n\n\n\n<p>The difference between an integer x and (-9) is 6<\/p>\n\n\n\n<p>x \u2013 (-9) = 6 or \u2013 9 \u2013 x = 6<\/p>\n\n\n\n<p>So the value of x is<\/p>\n\n\n\n<p>x \u2013 (-9) = 6 or \u2013 9 \u2013 x = 6<\/p>\n\n\n\n<p>By further calculation<\/p>\n\n\n\n<p>x + 9 = 6 or \u2013 x = 6 + 9<\/p>\n\n\n\n<p>So we get<\/p>\n\n\n\n<p>x = 6 \u2013 9 or \u2013 x = 15<\/p>\n\n\n\n<p>Here<\/p>\n\n\n\n<p>x = \u2013 3 or x = \u2013 15<\/p>\n\n\n\n<p>Therefore, possible values of x are \u2013 3 and \u2013 15.<\/p>\n\n\n\n<p><strong>6. Evaluate:<\/strong><\/p>\n\n\n\n<p><strong>(i) (-1) \u00d7 (-1) \u00d7 (-1) \u00d7 \u2026\u2026\u2026 60 times.<\/strong><\/p>\n\n\n\n<p><strong>(ii) (-1) \u00d7 (-1) \u00d7 (-1) \u00d7 (-1) \u00d7 \u2026\u2026.. 75 times.<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>(i) (-1) \u00d7 (-1) \u00d7 (-1) \u00d7 \u2026\u2026. 60 times = 1 because (-1) is multiplied even number of times.<\/p>\n\n\n\n<p>(ii) (-1) \u00d7 (-1) \u00d7 (-1) \u00d7 (-1) \u00d7 \u2026\u2026.. 75 times = -1 because (-1) is multiplied odd number of times.<\/p>\n\n\n\n<p><strong>7. Evaluate:<\/strong><\/p>\n\n\n\n<p><strong>(i) (-2) \u00d7 (-3) \u00d7 (-4) \u00d7 (-5) \u00d7 (-6)<\/strong><\/p>\n\n\n\n<p><strong>(ii) (-3) \u00d7 (-6) \u00d7 (-9) \u00d7 (-12)<\/strong><\/p>\n\n\n\n<p><strong>(iii) (-11) \u00d7 (-15) \u00d7 (-11) \u00d7 (-25)<\/strong><\/p>\n\n\n\n<p><strong>(iv) 10 \u00d7 (-12) + 5 \u00d7 (-12)<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>(i) (-2) \u00d7 (-3) \u00d7 (-4) \u00d7 (-5) \u00d7 (-6)<\/p>\n\n\n\n<p>= 6 \u00d7 20 \u00d7 (-6)<\/p>\n\n\n\n<p>By further calculation<\/p>\n\n\n\n<p>= 120 \u00d7 (-6)<\/p>\n\n\n\n<p>= \u2013 720<\/p>\n\n\n\n<p>(ii) (-3) \u00d7 (-6) \u00d7 (-9) \u00d7 (-12)<\/p>\n\n\n\n<p>= 18 \u00d7 108<\/p>\n\n\n\n<p>By further calculation<\/p>\n\n\n\n<p>= 1944<\/p>\n\n\n\n<p>(iii) (-11) \u00d7 (-15) + (-11) \u00d7 (-25)<\/p>\n\n\n\n<p>= 165 + 275<\/p>\n\n\n\n<p>By further calculation<\/p>\n\n\n\n<p>= 440<\/p>\n\n\n\n<p>(iv) 10 \u00d7 (-12) + 5 \u00d7 (-12)<\/p>\n\n\n\n<p>= \u2013 120 \u2013 60<\/p>\n\n\n\n<p>By further calculation<\/p>\n\n\n\n<p>= \u2013 180<\/p>\n\n\n\n<p><strong>8. (i) If x \u00d7 (-1) = \u2013 36, is x positive or negative?<\/strong><\/p>\n\n\n\n<p><strong>(ii) If x \u00d7 (-1) = 36, is x positive or negative?<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>(i) x \u00d7 (-1) = \u2013 36<\/p>\n\n\n\n<p>So we get<\/p>\n\n\n\n<p>\u2013 x = \u2013 36<\/p>\n\n\n\n<p>By further simplification<\/p>\n\n\n\n<p>x = 36<\/p>\n\n\n\n<p>Hence, it is a positive integer.<\/p>\n\n\n\n<p>(ii) x \u00d7 (-1) = 36<\/p>\n\n\n\n<p>So we get<\/p>\n\n\n\n<p>\u2013 x = 36<\/p>\n\n\n\n<p>By further simplification<\/p>\n\n\n\n<p>x = \u2013 36<\/p>\n\n\n\n<p>Hence, it is a negative integer.<\/p>\n\n\n\n<p><strong>9. Write all the integers between \u2013 15 and 15, which are divisible by 2 and 3.<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>Here the integers between \u2013 15 and 15 are<\/p>\n\n\n\n<p>\u2013 12, \u2013 6, 0, 6 and 12 which are divisible by 2 and 3.<\/p>\n\n\n\n<p><strong>10. Write all the integers between \u2013 5 and 5, which are divisible by 2 or 3.<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>Here the integers between \u2013 5 and 5 are<\/p>\n\n\n\n<p>\u2013 4, \u2013 3, \u2013 2, 0, 2, 3 and 4 which are divisible by 2 or 3.<\/p>\n\n\n\n<p><strong>11. Evaluate:<\/strong><\/p>\n\n\n\n<p><strong>(i) (-20) + (-8) \u00f7 (-2) \u00d7 3<\/strong><\/p>\n\n\n\n<p><strong>(ii) (-5) \u2013 (-48) \u00f7 (-16) + (-2) \u00d7 6<\/strong><\/p>\n\n\n\n<p><strong>(iii) 16 + 8 \u00f7 4 \u2013 2 \u00d7 3<\/strong><\/p>\n\n\n\n<p><strong>(iv) 16 \u00f7 8 \u00d7 4 \u2013 2 \u00d7 3<\/strong><\/p>\n\n\n\n<p><strong>(v) 27 \u2013 [5 + {28 \u2013 (29 \u2013 7)}]<\/strong><\/p>\n\n\n\n<p><strong>(vi) 48 \u2013 [18 \u2013 {16 \u2013 (5 \u2013 )}]<\/strong><\/p>\n\n\n\n<p><strong>(vii) \u2013 8 \u2013 {- 6 (9 \u2013 11) + 18 \u00f7 -3}<\/strong><\/p>\n\n\n\n<p><strong>(viii) (24 \u00f7 \u2013 12) \u2013 (3 \u00d7 8 \u00f7 4 + 1)<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>(i) (-20) + (-8) \u00f7 (-2) \u00d7 3<\/p>\n\n\n\n<p>It can be written as<\/p>\n\n\n\n<p>= \u2013 20 + 4 \u00d7 3<\/p>\n\n\n\n<p>By further calculation<\/p>\n\n\n\n<p>= \u2013 20 + 12<\/p>\n\n\n\n<p>= \u2013 8<\/p>\n\n\n\n<p>(ii) (-5) \u2013 (-48) \u00f7 (-16) + (-2) \u00d7 6<\/p>\n\n\n\n<p>It can be written as<\/p>\n\n\n\n<p>= (-5) \u2013 3 + (-2) \u00d7 6<\/p>\n\n\n\n<p>By further calculation<\/p>\n\n\n\n<p>= \u2013 5 \u2013 3 \u2013 12<\/p>\n\n\n\n<p>So we get<\/p>\n\n\n\n<p>= \u2013 8 \u2013 12<\/p>\n\n\n\n<p>= \u2013 20<\/p>\n\n\n\n<p>(iii) 16 + 8 \u00f7 4 \u2013 2 \u00d7 3<\/p>\n\n\n\n<p>It can be written as<\/p>\n\n\n\n<p>= 16 + 2 \u2013 2 \u00d7 3<\/p>\n\n\n\n<p>By further calculation<\/p>\n\n\n\n<p>= 16 + 2 \u2013 6<\/p>\n\n\n\n<p>So we get<\/p>\n\n\n\n<p>= 18 \u2013 6<\/p>\n\n\n\n<p>= 12<\/p>\n\n\n\n<p>(iv) 16 \u00f7 8 \u00d7 4 \u2013 2 \u00d7 3<\/p>\n\n\n\n<p>It can be written as<\/p>\n\n\n\n<p>= 2 \u00d7 4 \u2013 2 \u00d7 3<\/p>\n\n\n\n<p>By further calculation<\/p>\n\n\n\n<p>= 8 \u2013 6<\/p>\n\n\n\n<p>= 2<\/p>\n\n\n\n<p>(v) 27 \u2013 [5 + {28 \u2013 (29 \u2013 7)}]<\/p>\n\n\n\n<p>It can be written as<\/p>\n\n\n\n<p>= 27 \u2013 [5 + {28 \u2013 22}]<\/p>\n\n\n\n<p>By further calculation<\/p>\n\n\n\n<p>= 27 \u2013 [5 + 6]<\/p>\n\n\n\n<p>So we get<\/p>\n\n\n\n<p>= 27 \u2013 11<\/p>\n\n\n\n<p>= 16<\/p>\n\n\n\n<p>(vi) 48 \u2013 [18 \u2013 {16 \u2013 (5 \u2013&nbsp;<img decoding=\"async\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2022\/05\/selina-solutions-concise-maths-class-7-chapter-1-ex-1d-image-3.gif\" alt=\"Selina Solutions Concise Maths Class 7 Chapter 1 Ex 1D Image 3\">)}]<\/p>\n\n\n\n<p>It can be written as<\/p>\n\n\n\n<p>= 48 \u2013 [18 \u2013 {16 \u2013 (5 \u2013 5)}]<\/p>\n\n\n\n<p>By further calculation<\/p>\n\n\n\n<p>= 48 \u2013 [18 \u2013 {16 \u2013 0}]<\/p>\n\n\n\n<p>So we get<\/p>\n\n\n\n<p>= 48 \u2013 [18 \u2013 16]<\/p>\n\n\n\n<p>= 48 \u2013 2<\/p>\n\n\n\n<p>= 46<\/p>\n\n\n\n<p>(vii) \u2013 8 \u2013 {- 6 (9 \u2013 11) + 18 \u00f7 -3}<\/p>\n\n\n\n<p>It can be written as<\/p>\n\n\n\n<p>= \u2013 8 \u2013 {- 6 (-2) \u2013 6}<\/p>\n\n\n\n<p>By further calculation<\/p>\n\n\n\n<p>= \u2013 8 \u2013 {12 \u2013 6}<\/p>\n\n\n\n<p>So we get<\/p>\n\n\n\n<p>= \u2013 8 \u2013 6<\/p>\n\n\n\n<p>= \u2013 14<\/p>\n\n\n\n<p>(viii) (24 \u00f7&nbsp;<img decoding=\"async\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2022\/05\/selina-solutions-concise-maths-class-7-chapter-1-ex-1d-image-4.gif\" alt=\"Selina Solutions Concise Maths Class 7 Chapter 1 Ex 1D Image 4\">\u2013 12) \u2013 (3 \u00d7 8 \u00f7 4 + 1)<\/p>\n\n\n\n<p>It can be written as<\/p>\n\n\n\n<p>= (24 \u00f7 3 \u2013 12) \u2013 (3 \u00d7 2 + 1)<\/p>\n\n\n\n<p>By further calculation<\/p>\n\n\n\n<p>= (8 \u2013 12) \u2013 (6 + 1)<\/p>\n\n\n\n<p>So we get<\/p>\n\n\n\n<p>= \u2013 4 \u2013 7<\/p>\n\n\n\n<p>= \u2013 11<\/p>\n\n\n\n<p><strong>12. Find the result of subtracting the sum of all integers between 20 and 30 from the sum of all integers from 20 to 30.<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>Here the required number = sum of all integers from 20 to 30 \u2013 sum of all integers between 20 and 30<\/p>\n\n\n\n<p>Substituting the values<\/p>\n\n\n\n<p>= (20 + 21 + 22 + 23 + 24 + 25 + 26 + 27 + 28 + 29 + 30) \u2013 (21 + 22 + 23 + 24 + 25 + 26 + 27 + 28 + 29)<\/p>\n\n\n\n<p>On further calculation<\/p>\n\n\n\n<p>= 20 + 30 = 50<\/p>\n\n\n\n<p>Hence, the required number is 50.<\/p>\n\n\n\n<p><strong>13. Add the product of (-13) and (-17) to the quotient of (-187) and 11.<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>It is given that<\/p>\n\n\n\n<p>(-13) \u00d7 (-17) + (-187&nbsp;\u00f7 11)<\/p>\n\n\n\n<p>By further calculation<\/p>\n\n\n\n<p>= (-13) \u00d7 (-17) + (-17)<\/p>\n\n\n\n<p>So we get<\/p>\n\n\n\n<p>= 221 \u2013 17<\/p>\n\n\n\n<p>= 204<\/p>\n\n\n\n<p><strong>14. The product of two integers is \u2013 180. If one of them is 12, find the other.<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>Product of two integers = \u2013 180<\/p>\n\n\n\n<p>One integer = 12<\/p>\n\n\n\n<p>Other integer = \u2013 180\/ 12<\/p>\n\n\n\n<p>By division we get<\/p>\n\n\n\n<p>= \u2013 15<\/p>\n\n\n\n<p><strong>15. (i) A number changes from \u2013 20 to 30. What is the increase or decrease in the number?<\/strong><\/p>\n\n\n\n<p><strong>(ii) A number changes from 40 to \u2013 30. What is the increase or decrease in the number?<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>(i) A number changes from \u2013 20 to 30<\/p>\n\n\n\n<p>It can be written as<\/p>\n\n\n\n<p>\u2013 20 \u2013 30 = \u2013 50<\/p>\n\n\n\n<p>Hence, \u2013 50 will be the increase in the number.<\/p>\n\n\n\n<p>(ii) A number changes from 40 to \u2013 30<\/p>\n\n\n\n<p>It can be written as<\/p>\n\n\n\n<p>40 \u2013 (-30) = 40 + 30 = 70<\/p>\n\n\n\n<p>Hence, 70 will be the decrease in the number.<\/p>\n\n\n\n<h2 class=\"wp-block-heading\" id=\"h-download-pdf\"><strong>Download PDF<\/strong><\/h2>\n\n\n\n<p>Selina Class 7 ICSE Solutions Mathematics : Chapter 1-&nbsp;Integers<\/p>\n\n\n\n<p><a href=\"https:\/\/www.indcareer.com\/docs\/b913a35b-0a0e-4d28-bae0-1f17a4cfbc8b\" target=\"_blank\" rel=\"noreferrer noopener\"><strong>Download PDF<\/strong>: Selina Class 7 ICSE Solutions Mathematics : Chapter 1-\u00a0Integers PDF<\/a><\/p>\n\n\n\n<h2 class=\"wp-block-heading\"><strong>Chapterwise Selina Publishers&nbsp;ICSE Solutions for Class 7&nbsp;Mathematics :<\/strong><\/h2>\n\n\n\n<ul class=\"wp-block-list\"><li><a href=\"https:\/\/www.indcareer.com\/schools\/selina-class-7-icse-solutions-mathematics-chapter-1-integers\/\">Chapter 1- Integers<\/a><\/li><li><a href=\"https:\/\/www.indcareer.com\/schools\/selina-class-7-icse-solutions-mathematics-chapter-2-rational-numbers\/\">Chapter 2- Rational Numbers<\/a><\/li><li><a href=\"https:\/\/www.indcareer.com\/schools\/selina-class-7-icse-solutions-mathematics-chapter-3-fraction-including-problems\/\">Chapter 3- Fraction (Including Problems)<\/a><\/li><li><a href=\"https:\/\/www.indcareer.com\/schools\/selina-class-7-icse-solutions-mathematics-chapter-4-decimal-fractions-decimals\/\">Chapter 4- Decimal Fractions (Decimals)<\/a><\/li><li><a href=\"https:\/\/www.indcareer.com\/schools\/selina-class-7-icse-solutions-mathematics-chapter-5-exponents-including-laws-of-exponents\/\">Chapter 5- Exponents (Including Laws of Exponents)<\/a><\/li><li><a href=\"https:\/\/www.indcareer.com\/schools\/selina-class-7-icse-solutions-mathematics-chapter-6-ratio-and-proportion-including-sharing-in-a-ratio\/\">Chapter 6- Ratio and Proportion (Including Sharing in a Ratio)<\/a><\/li><li><a href=\"https:\/\/www.indcareer.com\/schools\/selina-class-7-icse-solutions-mathematics-chapter-7-unitary-method-including-time-and-work\/\">Chapter 7- Unitary Method (Including Time and Work)<\/a><\/li><li><a href=\"https:\/\/www.indcareer.com\/schools\/selina-class-7-icse-solutions-mathematics-chapter-8-percent-and-percentage\/\">Chapter 8- Percent and Percentage<\/a><\/li><li><a href=\"https:\/\/www.indcareer.com\/schools\/selina-class-7-icse-solutions-mathematics-chapter-9-profit-loss-and-discount\/\">Chapter 9- Profit, Loss and Discount<\/a><\/li><li><a href=\"https:\/\/www.indcareer.com\/schools\/selina-class-7-icse-solutions-mathematics-chapter-10-simple-interest\/\">Chapter 10- Simple Interest<\/a><\/li><li><a href=\"https:\/\/www.indcareer.com\/schools\/selina-class-7-icse-solutions-mathematics-chapter-11-fundamental-concepts-including-fundamental-operations\/\">Chapter 11- Fundamental Concepts (Including Fundamental Operations)<\/a><\/li><li><a href=\"https:\/\/www.indcareer.com\/schools\/selina-class-7-icse-solutions-mathematics-chapter-12-simple-linear-equations-including-word-problems\/\">Chapter 12- Simple Linear Equations (Including Word Problems)<\/a><\/li><li><a href=\"https:\/\/www.indcareer.com\/schools\/selina-class-7-icse-solutions-mathematics-chapter-13-set-concepts\/\">Chapter 13- Set Concepts<\/a><\/li><li><a href=\"https:\/\/www.indcareer.com\/schools\/selina-class-7-icse-solutions-mathematics-chapter-14-lines-and-angles-including-construction-of-angles\/\">Chapter 14- Lines and Angles (Including Construction of Angles)<\/a><\/li><li><a href=\"https:\/\/www.indcareer.com\/schools\/selina-class-7-icse-solutions-mathematics-chapter-15-triangles\/\">Chapter 15- Triangles<\/a><\/li><li><a href=\"https:\/\/www.indcareer.com\/schools\/selina-class-7-icse-solutions-mathematics-chapter-16-pythagoras-theorem\/\">Chapter 16- Pythagoras Theorem<\/a><\/li><li><a href=\"https:\/\/www.indcareer.com\/schools\/selina-class-7-icse-solutions-mathematics-chapter-17-symmetry-including-reflection-and-rotation\/\">Chapter 17- Symmetry (Including Reflection and Rotation)<\/a><\/li><li><a href=\"https:\/\/www.indcareer.com\/schools\/selina-class-7-icse-solutions-mathematics-chapter-18-recognition-of-solids-representing-3-d-in-2-d\/\">Chapter 18- Recognition of Solids (Representing 3-D in 2-D)<\/a><\/li><li><a href=\"https:\/\/www.indcareer.com\/schools\/selina-class-7-icse-solutions-mathematics-chapter-19-congruency-congruent-triangles\/\">Chapter 19- Congruency: Congruent Triangles<\/a><\/li><li><a href=\"https:\/\/www.indcareer.com\/schools\/selina-class-7-icse-solutions-mathematics-chapter-20-mensuration-perimeter-and-area-of-plane-figures\/\">Chapter 20- Mensuration (Perimeter and Area of Plane Figures)<\/a><\/li><li><a href=\"https:\/\/www.indcareer.com\/schools\/selina-class-7-icse-solutions-mathematics-chapter-21-data-handling\/\">Chapter 21- Data Handling<\/a><\/li><li><a href=\"https:\/\/www.indcareer.com\/schools\/selina-class-7-icse-solutions-mathematics-chapter-22-probability\/\">Chapter 22- Probability<\/a><\/li><\/ul>\n\n\n\n<h2 class=\"wp-block-heading\">About Selina Publishers&nbsp;ICSE<\/h2>\n\n\n\n<p>Selina Publishers has been serving the students since 1976 and is one of the quality ICSE school textbooks publication houses. Mathematics and Science books for classes 6-10 form the core of our business, apart from certain English and Hindi literature as well as a few primary books. All these books are based upon the syllabus published by the Council for the I.C.S.E. Examinations, New Delhi. The textbooks are composed by a panel of subject experts and vetted by teachers practising in ICSE schools all over the country. Continuous efforts are made in complying with the standards and ensuring lucidity and clarity in content, which makes them stand tall in the industry.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>Class 7: Maths Chapter 1 solutions. Complete Class 7 Maths Chapter 1 Notes. Selina Class 7 ICSE Solutions Mathematics : Chapter 1-&nbsp;Integers Selina 7th Maths Chapter 1, Class 7 Maths Chapter 1 solutions Exercise 1A page: 6 1. Evaluate: (i) 427 \u00d7 8 + 2 \u00d7 427 (ii) 394 \u00d7 12 + 394 \u00d7 (-2) [&hellip;]<\/p>\n","protected":false},"author":302,"featured_media":598132,"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":[1411,907],"tags":[2261],"boards":[],"class_list":["post-598130","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-book-solutions","category-class-7","tag-icse-solutions","entry"],"yoast_head":"<!-- This site is optimized with the Yoast SEO Premium plugin v27.0 (Yoast SEO v27.1.1) - https:\/\/yoast.com\/product\/yoast-seo-premium-wordpress\/ -->\n<title>Selina Solutions for Class 7, maths Chapter 1 - IndCareer Schools<\/title>\n<meta name=\"description\" content=\"Selina Class 7 ICSE Solutions Mathematics : Chapter 1-\u00a0Integers | Browse all Class 7 maths - IndCareer Schools\" \/>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/www.indcareer.com\/schools\/selina-class-7-icse-solutions-mathematics-chapter-1-integers\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"Selina Class 7 ICSE Solutions Mathematics : Chapter 1-\u00a0Integers\" \/>\n<meta property=\"og:description\" content=\"Class 7: Maths Chapter 1 solutions. 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