{"id":598385,"date":"2022-05-03T05:12:23","date_gmt":"2022-05-03T05:12:23","guid":{"rendered":"https:\/\/www.indcareer.com\/schools\/?p=598385"},"modified":"2022-05-10T06:52:15","modified_gmt":"2022-05-10T06:52:15","slug":"selina-class-7-icse-solutions-mathematics-chapter-10-simple-interest","status":"publish","type":"post","link":"https:\/\/www.indcareer.com\/schools\/selina-class-7-icse-solutions-mathematics-chapter-10-simple-interest\/","title":{"rendered":"Selina Class 7 ICSE Solutions Mathematics : Chapter 10-\u00a0Simple Interest"},"content":{"rendered":"\n<p>Class 7: Maths Chapter 10 solutions. Complete Class 7 Maths Chapter 10 Notes.<\/p>\n\n\n\n<h2 class=\"wp-block-heading\" id=\"h-selina-class-7-icse-solutions-mathematics-chapter-10-simple-interest\">Selina Class 7 ICSE Solutions Mathematics : Chapter 10-&nbsp;Simple Interest<\/h2>\n\n\n\n<p>Selina 7th Maths Chapter 10, Class 7 Maths Chapter 10 solutions<\/p>\n\n\n\n<h4 class=\"wp-block-heading\">Exercise 10 page: 116<\/h4>\n\n\n\n<p><strong>1. Find the S.I. and the amount on:<\/strong><\/p>\n\n\n\n<p><strong>(i) \u20b9 150 for 4 years at 5% per year.<\/strong><\/p>\n\n\n\n<p><strong>(ii) \u20b9 350 for 3 \u00bd years at 8% p.a.<\/strong><\/p>\n\n\n\n<p><strong>(iii) \u20b9 620 for 4 months at 8 p per rupee per month.<\/strong><\/p>\n\n\n\n<p><strong>(iv) \u20b9 3,380 for 30 months at 4 \u00bd % p.a.<\/strong><\/p>\n\n\n\n<p><strong>(v) \u20b9 600 from July 12 to Dec. 5 at 10% p.a.<\/strong><\/p>\n\n\n\n<p><strong>(vi) \u20b9 850 from 10<sup>th<\/sup> March to 3<sup>rd<\/sup> August at 2 \u00bd % p.a.<\/strong><\/p>\n\n\n\n<p><strong>(vii) \u20b9 225 for 3 years 9 months at 16% p.a.<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>(i) \u20b9 150 for 4 years at 5% per year<\/p>\n\n\n\n<p>We know that<\/p>\n\n\n\n<p>P = \u20b9 150<\/p>\n\n\n\n<p>R = 5% per year<\/p>\n\n\n\n<p>T = 4 years<\/p>\n\n\n\n<p>Here<\/p>\n\n\n\n<p>S.I = (P \u00d7 R \u00d7 T)\/ 100<\/p>\n\n\n\n<p>Substituting the values<\/p>\n\n\n\n<p>= (150 \u00d7 5 \u00d7 4)\/ 100<\/p>\n\n\n\n<p>= \u20b9 30<\/p>\n\n\n\n<p>Amount = P + S.I<\/p>\n\n\n\n<p>Substituting the values<\/p>\n\n\n\n<p>= 150 + 30<\/p>\n\n\n\n<p>= \u20b9 180<\/p>\n\n\n\n<p>(ii) \u20b9 350 for 3 \u00bd years at 8% p.a.<\/p>\n\n\n\n<p>We know that<\/p>\n\n\n\n<p>P = \u20b9 350<\/p>\n\n\n\n<p>R = 8% p.a.<\/p>\n\n\n\n<p>T = 3 \u00bd years = 7\/2 years<\/p>\n\n\n\n<p>Here<\/p>\n\n\n\n<p>S.I = (P \u00d7 R \u00d7 T)\/ 100<\/p>\n\n\n\n<p>Substituting the values<\/p>\n\n\n\n<p>= (350 \u00d7 8 \u00d7 7)\/ (100 \u00d7 2)<\/p>\n\n\n\n<p>= \u20b9 98<\/p>\n\n\n\n<p>Amount = P + S.I<\/p>\n\n\n\n<p>Substituting the values<\/p>\n\n\n\n<p>= 350 + 98<\/p>\n\n\n\n<p>= \u20b9 448<\/p>\n\n\n\n<p>(iii) \u20b9 620 for 4 months at 8 p per rupee per month<\/p>\n\n\n\n<p>We know that<\/p>\n\n\n\n<p>P = \u20b9 620<\/p>\n\n\n\n<p>R = 8 p per rupee per month = 8% p.m.<\/p>\n\n\n\n<p>T = 4 months<\/p>\n\n\n\n<p>Here<\/p>\n\n\n\n<p>S.I = (P \u00d7 R \u00d7 T)\/ 100<\/p>\n\n\n\n<p>Substituting the values<\/p>\n\n\n\n<p>= (620 \u00d7 8 \u00d7 4)\/ 100<\/p>\n\n\n\n<p>= \u20b9 198.40<\/p>\n\n\n\n<p>Amount = P + S.I<\/p>\n\n\n\n<p>Substituting the values<\/p>\n\n\n\n<p>= 620 + 198.40<\/p>\n\n\n\n<p>= \u20b9 818.40<\/p>\n\n\n\n<p>(iv) \u20b9 3,380 for 30 months at 4 \u00bd % p.a.<\/p>\n\n\n\n<p>We know that<\/p>\n\n\n\n<p>P = \u20b9 3,380<\/p>\n\n\n\n<p>R = 4 \u00bd % p.a. = 9\/2 %<\/p>\n\n\n\n<p>P = 30 months = 30\/12 years<\/p>\n\n\n\n<p>Here<\/p>\n\n\n\n<p>S.I = (P \u00d7 R \u00d7 T)\/ 100<\/p>\n\n\n\n<p>Substituting the values<\/p>\n\n\n\n<p>= (3380 \u00d7 9 \u00d7 30)\/ (100 \u00d7 2 \u00d7 12)<\/p>\n\n\n\n<p>= \u20b9 380.25<\/p>\n\n\n\n<p>Amount = P + S.I<\/p>\n\n\n\n<p>Substituting the values<\/p>\n\n\n\n<p>= 3380 + 380.25<\/p>\n\n\n\n<p>= \u20b9 3760.25<\/p>\n\n\n\n<p>(v) \u20b9 600 from July 12 to Dec. 5 at 10% p.a.<\/p>\n\n\n\n<p>We know that<\/p>\n\n\n\n<p>P = \u20b9 600<\/p>\n\n\n\n<p>R = 10% p.a.<\/p>\n\n\n\n<p>T = July 12 to Dec 5<\/p>\n\n\n\n<p>July = 19 days<\/p>\n\n\n\n<p>Aug = 31 days<\/p>\n\n\n\n<p>Sep = 30days<\/p>\n\n\n\n<p>Oct = 31 days<\/p>\n\n\n\n<p>Nov = 30 days<\/p>\n\n\n\n<p>Dec = 05 days<\/p>\n\n\n\n<p>Total = 146 days<\/p>\n\n\n\n<p>T = 146\/365 years = 2\/5 years<\/p>\n\n\n\n<p>Here<\/p>\n\n\n\n<p>S.I = (P \u00d7 R \u00d7 T)\/ 100<\/p>\n\n\n\n<p>Substituting the values<\/p>\n\n\n\n<p>= (600 \u00d7 10 \u00d7 2)\/ (100 \u00d7 5)<\/p>\n\n\n\n<p>= \u20b9 24<\/p>\n\n\n\n<p>Amount = P + S.I<\/p>\n\n\n\n<p>Substituting the values<\/p>\n\n\n\n<p>= 600 + 24<\/p>\n\n\n\n<p>= \u20b9 624<\/p>\n\n\n\n<p>(vi) \u20b9 850 from 10<sup>th<\/sup> March to 3<sup>rd<\/sup> August at 2 \u00bd % p.a.<\/p>\n\n\n\n<p>We know that<\/p>\n\n\n\n<p>P = \u20b9 850<\/p>\n\n\n\n<p>R = 2 \u00bd% = 5\/2% p.a.<\/p>\n\n\n\n<p>T = 10<sup>th<\/sup> March to 3<sup>rd<\/sup> August<\/p>\n\n\n\n<p>March = 21 days<\/p>\n\n\n\n<p>April = 30 days<\/p>\n\n\n\n<p>May = 31 days<\/p>\n\n\n\n<p>June = 30 days<\/p>\n\n\n\n<p>July = 31 days<\/p>\n\n\n\n<p>Aug = 3 days<\/p>\n\n\n\n<p>Total = 146 days<\/p>\n\n\n\n<p>T = 146\/365 = 2\/5 years<\/p>\n\n\n\n<p>Here<\/p>\n\n\n\n<p>S.I = (P \u00d7 R \u00d7 T)\/ 100<\/p>\n\n\n\n<p>Substituting the values<\/p>\n\n\n\n<p>= (850 \u00d7 5 \u00d7 2)\/ (100 \u00d7 2 \u00d7 5)<\/p>\n\n\n\n<p>= \u20b9 8.50<\/p>\n\n\n\n<p>Amount = P + S.I<\/p>\n\n\n\n<p>Substituting the values<\/p>\n\n\n\n<p>= 850 + 8.50<\/p>\n\n\n\n<p>= \u20b9 858.50<\/p>\n\n\n\n<p>(vii) \u20b9 225 for 3 years 9 months at 16% p.a.<\/p>\n\n\n\n<p>We know that<\/p>\n\n\n\n<p>P = \u20b9 225<\/p>\n\n\n\n<p>R = 16% p.a.<\/p>\n\n\n\n<p>T = 3 years 9 months = 3 and 9\/12 years = 3 \u00be years = 15\/4 years<\/p>\n\n\n\n<p>Here<\/p>\n\n\n\n<p>S.I = (P \u00d7 R \u00d7 T)\/ 100<\/p>\n\n\n\n<p>Substituting the values<\/p>\n\n\n\n<p>= (225 \u00d7 16 \u00d7 15)\/ (100 \u00d7 4)<\/p>\n\n\n\n<p>= \u20b9 135<\/p>\n\n\n\n<p>Amount = P + S.I<\/p>\n\n\n\n<p>Substituting the values<\/p>\n\n\n\n<p>= 225 + 135<\/p>\n\n\n\n<p>= \u20b9 360<\/p>\n\n\n\n<p><strong>2. On what sum of money does the S.I. for 10 years at 5% become \u20b9 1,600?<\/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>S.I = \u20b9 1,600<\/p>\n\n\n\n<p>R = 5% p.a.<\/p>\n\n\n\n<p>T = 10 years<\/p>\n\n\n\n<p>We know that<\/p>\n\n\n\n<p>P = (S.I \u00d7 100)\/ (R \u00d7 T)<\/p>\n\n\n\n<p>Substituting the values<\/p>\n\n\n\n<p>= (1600 \u00d7 100)\/ (5 \u00d7 10)<\/p>\n\n\n\n<p>So we get<\/p>\n\n\n\n<p>= \u20b9 3,200<\/p>\n\n\n\n<p><strong>3. Find the time in which \u20b9 2,000 will amount to \u20b9 2,330 at 11% p.a.<\/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>A = \u20b9 2,330<\/p>\n\n\n\n<p>P = \u20b9 2,000<\/p>\n\n\n\n<p>We know that<\/p>\n\n\n\n<p>S.I = A \u2013 P<\/p>\n\n\n\n<p>Substituting the values<\/p>\n\n\n\n<p>= 2330 \u2013 2000<\/p>\n\n\n\n<p>= \u20b9 330<\/p>\n\n\n\n<p>Here<\/p>\n\n\n\n<p>Time = (S.I \u00d7 100)\/ (P \u00d7 R)<\/p>\n\n\n\n<p>Substituting the values<\/p>\n\n\n\n<p>= (330 \u00d7 100)\/ (2000 \u00d7 11)<\/p>\n\n\n\n<p>So we get<\/p>\n\n\n\n<p>= 3\/2<\/p>\n\n\n\n<p>= 1 \u00bd years<\/p>\n\n\n\n<p><strong>4. In what time will a sum of money double itself at 8% p.a.<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>Consider the principal<\/p>\n\n\n\n<p>P = \u20b9 100<\/p>\n\n\n\n<p>It is given that<\/p>\n\n\n\n<p>A = 100 \u00d7 2 = \u20b9 200<\/p>\n\n\n\n<p>We know that<\/p>\n\n\n\n<p>S.I = A \u2013 P<\/p>\n\n\n\n<p>Substituting the values<\/p>\n\n\n\n<p>= 200 \u2013 100<\/p>\n\n\n\n<p>= \u20b9 100<\/p>\n\n\n\n<p>R = 8% p.a.<\/p>\n\n\n\n<p>Here<\/p>\n\n\n\n<p>Time = (S.I \u00d7 100)\/ (P \u00d7 R)<\/p>\n\n\n\n<p>Substituting the values<\/p>\n\n\n\n<p>= (100 \u00d7 100)\/ (100 \u00d7 8)<\/p>\n\n\n\n<p>So we get<\/p>\n\n\n\n<p>= 25\/2<\/p>\n\n\n\n<p>= 12 \u00bd years<\/p>\n\n\n\n<p><strong>5. In how many years will \u20b9 870 amount to \u20b9 1,044, the rate of interest being 2 \u00bd% p.a.?<\/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>P = \u20b9 870<\/p>\n\n\n\n<p>A = \u20b9 1044<\/p>\n\n\n\n<p>We know that<\/p>\n\n\n\n<p>S.I = A \u2013 P<\/p>\n\n\n\n<p>Substituting the values<\/p>\n\n\n\n<p>= 1044 \u2013 870<\/p>\n\n\n\n<p>= \u20b9 174<\/p>\n\n\n\n<p>R = 2 \u00bd = 5\/2 % p.a.<\/p>\n\n\n\n<p>We know that<\/p>\n\n\n\n<p>Time = (S.I \u00d7 100)\/ (P \u00d7 R)<\/p>\n\n\n\n<p>Substituting the values<\/p>\n\n\n\n<p>= (174 \u00d7 100 \u00d7 2)\/ (870 \u00d7 5)<\/p>\n\n\n\n<p>So we get<\/p>\n\n\n\n<p>= 8 years<\/p>\n\n\n\n<p><strong>6. Find the rate percent, if the S.I. on \u20b9 275 in 2 years is \u20b9 22.<\/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>P = \u20b9 275<\/p>\n\n\n\n<p>S.I = \u20b9 22<\/p>\n\n\n\n<p>T = 2 years<\/p>\n\n\n\n<p>We know that<\/p>\n\n\n\n<p>Rate = (S.I \u00d7 100)\/ (P \u00d7 T)<\/p>\n\n\n\n<p>Substituting the values<\/p>\n\n\n\n<p>= (22 \u00d7 100)\/ (275 \u00d7 2)<\/p>\n\n\n\n<p>So we get<\/p>\n\n\n\n<p>= 4% p.a.<\/p>\n\n\n\n<p><strong>7. Find the sum which will amount to \u20b9 700 in 5 years at 8% p.a.<\/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>Amount = \u20b9 700<\/p>\n\n\n\n<p>R = 8% p.a.<\/p>\n\n\n\n<p>T = 5 years<\/p>\n\n\n\n<p>Consider P = \u20b9 100<\/p>\n\n\n\n<p>We know that<\/p>\n\n\n\n<p>S.I = (P \u00d7 R \u00d7 T)\/ 100<\/p>\n\n\n\n<p>Substituting the values<\/p>\n\n\n\n<p>= (100 \u00d7 8 \u00d7 5)\/ 100<\/p>\n\n\n\n<p>= \u20b9 40<\/p>\n\n\n\n<p>Here<\/p>\n\n\n\n<p>A = P + S.I<\/p>\n\n\n\n<p>Substituting the values<\/p>\n\n\n\n<p>= 100 + 40<\/p>\n\n\n\n<p>= \u20b9 140<\/p>\n\n\n\n<p>If the amount is \u20b9 140 then the principal is \u20b9 100<\/p>\n\n\n\n<p>If the amount is \u20b9 700 then the principal = (100 \u00d7 700)\/ 140 = \u20b9 500<\/p>\n\n\n\n<p><strong>8. What is the rate of interest, if \u20b9 3,750 amounts to \u20b9 4,650 in 4 years?<\/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>P = \u20b9 3,750<\/p>\n\n\n\n<p>A = \u20b9 4,650<\/p>\n\n\n\n<p>We know that<\/p>\n\n\n\n<p>S.I = A \u2013 P<\/p>\n\n\n\n<p>Substituting the values<\/p>\n\n\n\n<p>= 4650 \u2013 3750<\/p>\n\n\n\n<p>= \u20b9 900<\/p>\n\n\n\n<p>T = 4 years<\/p>\n\n\n\n<p>Here<\/p>\n\n\n\n<p>Rate = (S.I \u00d7 100)\/ (P \u00d7 T)<\/p>\n\n\n\n<p>Substituting the values<\/p>\n\n\n\n<p>= (900 \u00d7 100)\/ (3750 \u00d7 4)<\/p>\n\n\n\n<p>So we get<\/p>\n\n\n\n<p>= 6% p.a.<\/p>\n\n\n\n<p><strong>9. In 4 years, \u20b9 6,000 amounts to \u20b9 8,000. In what time will \u20b9 525 amount to \u20b9 700 at the same rate?<\/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>P = \u20b9 6,000<\/p>\n\n\n\n<p>A = \u20b9 8,000<\/p>\n\n\n\n<p>We know that<\/p>\n\n\n\n<p>S.I = A \u2013 P<\/p>\n\n\n\n<p>Substituting the values<\/p>\n\n\n\n<p>= 8000 \u2013 6000<\/p>\n\n\n\n<p>= \u20b9 2000<\/p>\n\n\n\n<p>T = 4 years<\/p>\n\n\n\n<p>Here<\/p>\n\n\n\n<p>Rate = (S.I \u00d7 100)\/ (P \u00d7 T)<\/p>\n\n\n\n<p>Substituting the values<\/p>\n\n\n\n<p>= (2000 \u00d7 100)\/ (6000 \u00d7 4)<\/p>\n\n\n\n<p>So we get<\/p>\n\n\n\n<p>= 25\/3%<\/p>\n\n\n\n<p>= 8 1\/3% p.a.<\/p>\n\n\n\n<p>It is given that<\/p>\n\n\n\n<p>P = \u20b9 525<\/p>\n\n\n\n<p>A = \u20b9 700<\/p>\n\n\n\n<p>We know that<\/p>\n\n\n\n<p>S.I = A \u2013 P<\/p>\n\n\n\n<p>Substituting the values<\/p>\n\n\n\n<p>= 700 \u2013 525<\/p>\n\n\n\n<p>= \u20b9 175<\/p>\n\n\n\n<p>R = 25\/3% p.a.<\/p>\n\n\n\n<p>Here<\/p>\n\n\n\n<p>Time = (S.I \u00d7 100)\/ (P \u00d7 R)<\/p>\n\n\n\n<p>Substituting the values<\/p>\n\n\n\n<p>= (175 \u00d7 100 \u00d7 3)\/ (525 \u00d7 25)<\/p>\n\n\n\n<p>So we get<\/p>\n\n\n\n<p>= 4 years<\/p>\n\n\n\n<p><strong>10. The interest on a sum of money at the end of 2 \u00bd years is 4\/5 of the sum. What is the rate percent?<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>Consider the sum P = \u20b9 100<\/p>\n\n\n\n<p>S.I = 100 \u00d7 4\/5 = \u20b9 80<\/p>\n\n\n\n<p>T = 2 \u00bd = 5\/2 years<\/p>\n\n\n\n<p>We know that<\/p>\n\n\n\n<p>Rate = (S.I \u00d7 100)\/ (P \u00d7 T)<\/p>\n\n\n\n<p>Substituting the values<\/p>\n\n\n\n<p>= (80 \u00d7 100 \u00d7 2)\/ (100 \u00d7 5)<\/p>\n\n\n\n<p>So we get<\/p>\n\n\n\n<p>= 32% p.a.<\/p>\n\n\n\n<p><strong>11. What sum of money lent out at 5% for 3 years will produce the same interest as \u20b9 900 lent out at 4% for 5 years?<\/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>P = \u20b9 900<\/p>\n\n\n\n<p>R = 4%<\/p>\n\n\n\n<p>T = 5 years<\/p>\n\n\n\n<p>We know that<\/p>\n\n\n\n<p>S.I = (P \u00d7 R \u00d7 T)\/ 100<\/p>\n\n\n\n<p>Substituting the values<\/p>\n\n\n\n<p>= (900 \u00d7 4 \u00d7 5)\/ 100<\/p>\n\n\n\n<p>= \u20b9 180<\/p>\n\n\n\n<p>It is given that<\/p>\n\n\n\n<p>S.I = \u20b9 180<\/p>\n\n\n\n<p>R = 5%<\/p>\n\n\n\n<p>T = 3 years<\/p>\n\n\n\n<p>We know that<\/p>\n\n\n\n<p>Sum P = (S.I \u00d7 100)\/ (R \u00d7 T)<\/p>\n\n\n\n<p>Substituting the values<\/p>\n\n\n\n<p>= (180 \u00d7 100)\/ (5 \u00d7 3)<\/p>\n\n\n\n<p>So we get<\/p>\n\n\n\n<p>= \u20b9 1200<\/p>\n\n\n\n<p><strong>12. A sum of \u20b9 1,780 becomes \u20b9 2,136 in 4 years. Find:<\/strong><\/p>\n\n\n\n<p><strong>(i) the rate of interest.<\/strong><\/p>\n\n\n\n<p><strong>(ii) the sum that will become \u20b9 810 in 7 years at the same rate of interest.<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>(i) It is given that<\/p>\n\n\n\n<p>P = \u20b9 1780<\/p>\n\n\n\n<p>A = \u20b9 2136<\/p>\n\n\n\n<p>We know that<\/p>\n\n\n\n<p>S.I = A \u2013 P<\/p>\n\n\n\n<p>Substituting the values<\/p>\n\n\n\n<p>= 2136 \u2013 1780<\/p>\n\n\n\n<p>= \u20b9 356<\/p>\n\n\n\n<p>T = 4 years<\/p>\n\n\n\n<p>Here<\/p>\n\n\n\n<p>Rate = (S.I \u00d7 100)\/ (P \u00d7 T)<\/p>\n\n\n\n<p>Substituting the values<\/p>\n\n\n\n<p>= (356 \u00d7 100)\/ (1780 \u00d7 4)<\/p>\n\n\n\n<p>So we get<\/p>\n\n\n\n<p>= 5% p.a.<\/p>\n\n\n\n<p>(ii) Consider P = \u20b9 100<\/p>\n\n\n\n<p>R = 5% p.a.<\/p>\n\n\n\n<p>T = 7 years<\/p>\n\n\n\n<p>We know that<\/p>\n\n\n\n<p>S.I = (P \u00d7 R \u00d7 T)\/ 100<\/p>\n\n\n\n<p>Substituting the values<\/p>\n\n\n\n<p>= (100 \u00d7 5 \u00d7 7)\/ 100<\/p>\n\n\n\n<p>= \u20b9 35<\/p>\n\n\n\n<p>Here amount = P + S.I<\/p>\n\n\n\n<p>Substituting the values<\/p>\n\n\n\n<p>= 100 + 35<\/p>\n\n\n\n<p>= \u20b9 135<\/p>\n\n\n\n<p>If the amount is \u20b9 135 then the principal is \u20b9 100<\/p>\n\n\n\n<p>If the amount is \u20b9 810 then principal = (100 \u00d7 810)\/ 135 = \u20b9 600<\/p>\n\n\n\n<p><strong>13. A sum amounts to \u20b9 2,652 in 6 years at 5% p.a. simple interest. Find:<\/strong><\/p>\n\n\n\n<p><strong>(i) the sum<\/strong><\/p>\n\n\n\n<p><strong>(ii) the time in which the same sum will double itself at the same rate of interest.<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>(i) Consider P = \u20b9 100<\/p>\n\n\n\n<p>R = 5% p.a.<\/p>\n\n\n\n<p>T = 6 years<\/p>\n\n\n\n<p>We know that<\/p>\n\n\n\n<p>S.I = (P \u00d7 R \u00d7 T)\/ 100<\/p>\n\n\n\n<p>Substituting the values<\/p>\n\n\n\n<p>= (100 \u00d7 5 \u00d7 6)\/ 100<\/p>\n\n\n\n<p>= \u20b9 30<\/p>\n\n\n\n<p>Here amount = 100 + 30 = \u20b9 130<\/p>\n\n\n\n<p>If the amount is \u20b9 130 then principal is \u20b9 100<\/p>\n\n\n\n<p>If the amount is \u20b9 2652 then principal = (100 \u00d7 2652)\/ 130 = \u20b9 2040<\/p>\n\n\n\n<p>Consider sum P = \u20b9 100<\/p>\n\n\n\n<p>Amount = 100 \u00d7 2 = \u20b9 200<\/p>\n\n\n\n<p>We know that<\/p>\n\n\n\n<p>S.I = A \u2013 P<\/p>\n\n\n\n<p>Substituting the values<\/p>\n\n\n\n<p>= 200 \u2013 100<\/p>\n\n\n\n<p>= \u20b9 100<\/p>\n\n\n\n<p>R = 5% p.a.<\/p>\n\n\n\n<p>Here<\/p>\n\n\n\n<p>T = (S.I \u00d7 100)\/ (P \u00d7 R)<\/p>\n\n\n\n<p>Substituting the values<\/p>\n\n\n\n<p>= (100 \u00d7 100)\/ (100 \u00d7 5)<\/p>\n\n\n\n<p>So we get<\/p>\n\n\n\n<p>= 20 years<\/p>\n\n\n\n<p><strong>14. P and Q invest \u20b9 36,000 and \u20b9 25,000 respectively at the same rate of interest per year. If at the end of 4 years, P gets \u20b9 3,080 more interest than Q, find the rate of interest.<\/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>P\u2019s investment (P<sub>1<\/sub>) = \u20b9 36,000<\/p>\n\n\n\n<p>Q\u2019s investment (P<sub>2<\/sub>) = \u20b9 25,000<\/p>\n\n\n\n<p>T = 4 years<\/p>\n\n\n\n<p>Consider the rate of interest = x%<\/p>\n\n\n\n<p>So we get<\/p>\n\n\n\n<p>P\u2019s interest (S.I) = (P \u00d7 R \u00d7 T)\/ 100<\/p>\n\n\n\n<p>Substituting the values<\/p>\n\n\n\n<p>= (36000 \u00d7 x \u00d7 4)\/ 100<\/p>\n\n\n\n<p>= \u20b9 1440x<\/p>\n\n\n\n<p>Q\u2019s interest = (P \u00d7 R \u00d7 T)\/ 100<\/p>\n\n\n\n<p>Substituting the values<\/p>\n\n\n\n<p>= (25000 \u00d7 x \u00d7 4)\/ 100<\/p>\n\n\n\n<p>= \u20b9 1000x<\/p>\n\n\n\n<p>Here the difference in their interest = 1440x \u2013 1000x = \u20b9 440x<\/p>\n\n\n\n<p>The difference given = \u20b9 3080<\/p>\n\n\n\n<p>So we get<\/p>\n\n\n\n<p>440x = 3080<\/p>\n\n\n\n<p>x = 3080\/440<\/p>\n\n\n\n<p>x = 7%<\/p>\n\n\n\n<p>So the rate of interest = 7% p.a.<\/p>\n\n\n\n<p><strong>15. A sum of money is lent for 5 years at R% simple interest per annum. If the interest earned be one-fourth of the money lent, find the value of R.<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>Consider the sum P = \u20b9 100<\/p>\n\n\n\n<p>We know that<\/p>\n\n\n\n<p>S.I = 1\/4 \u00d7 100 = \u20b9 25<\/p>\n\n\n\n<p>T = 5 years<\/p>\n\n\n\n<p>Here<\/p>\n\n\n\n<p>Rate = (S.I \u00d7 100)\/ (P \u00d7 T)<\/p>\n\n\n\n<p>Substituting the values<\/p>\n\n\n\n<p>= (25 \u00d7 100)\/ (100 \u00d7 5)<\/p>\n\n\n\n<p>So we get<\/p>\n\n\n\n<p>= 5%<\/p>\n\n\n\n<p><strong>16. The simple interest earned on a certain sum in 5 years is 30% of the sum. Find the rate of interest.<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>Consider sum P = \u20b9 100<\/p>\n\n\n\n<p>We know that<\/p>\n\n\n\n<p>SI = 30\/100 \u00d7 100 = \u20b9 30<\/p>\n\n\n\n<p>T = 5 years<\/p>\n\n\n\n<p>Here<\/p>\n\n\n\n<p>Rate = (S.I \u00d7 100)\/ (P \u00d7 T)<\/p>\n\n\n\n<p>Substituting the values<\/p>\n\n\n\n<p>= (30 \u00d7 100)\/ (100 \u00d7 5)<\/p>\n\n\n\n<p>So we get<\/p>\n\n\n\n<p>= 6%<\/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 10-&nbsp;Simple Interest<\/p>\n\n\n\n<p><a href=\"https:\/\/www.indcareer.com\/docs\/1999b4b5-0eda-4662-bd00-6237921b6669\" target=\"_blank\" rel=\"noreferrer noopener\"><strong>Download PDF<\/strong>: Selina Class 7 ICSE Solutions Mathematics : Chapter 10-\u00a0Simple Interest 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 10 solutions. Complete Class 7 Maths Chapter 10 Notes. Selina Class 7 ICSE Solutions Mathematics : Chapter 10-&nbsp;Simple Interest Selina 7th Maths Chapter 10, Class 7 Maths Chapter 10 solutions Exercise 10 page: 116 1. Find the S.I. and the amount on: (i) \u20b9 150 for 4 years at 5% per [&hellip;]<\/p>\n","protected":false},"author":302,"featured_media":598387,"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-598385","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 10 - IndCareer Schools<\/title>\n<meta name=\"description\" content=\"Selina Class 7 ICSE Solutions Mathematics : Chapter 10-\u00a0Simple Interest | 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-10-simple-interest\/\" \/>\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 10-\u00a0Simple Interest\" \/>\n<meta property=\"og:description\" content=\"Class 7: Maths Chapter 10 solutions. Complete Class 7 Maths Chapter 10 Notes. 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