{"id":125502,"date":"2023-09-12T19:28:00","date_gmt":"2023-09-12T13:58:00","guid":{"rendered":"https:\/\/www.kopykitab.com\/blog\/?p=125502"},"modified":"2023-12-06T10:35:43","modified_gmt":"2023-12-06T05:05:43","slug":"rd-sharma-class-10-solutions-chapter-3-exercise-3-10","status":"publish","type":"post","link":"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-10-solutions-chapter-3-exercise-3-10\/","title":{"rendered":"RD Sharma Class 10 Solutions Chapter 3 Exercise 3.10 (Updated for 2024)"},"content":{"rendered":"\n<p><img class=\"alignnone size-full wp-image-125561\" src=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Class-10-Solutions-Chapter-3-Exercise3.10.jpg\" alt=\"RD Sharma Class 10 Solutions Chapter 3 Exercise 3.10\" width=\"1200\" height=\"675\" srcset=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Class-10-Solutions-Chapter-3-Exercise3.10.jpg 1200w, https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Class-10-Solutions-Chapter-3-Exercise3.10-768x432.jpg 768w\" sizes=\"(max-width: 1200px) 100vw, 1200px\" \/><\/p>\n<p><strong>RD Sharma Class 10 Solutions Chapter 3 Exercise 3.10:\u00a0<\/strong>This exercise focuses on problems that are specific to this application. The <a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-10-solutions-for-maths\/\"><strong>RD Sharma Class 10 Solutions<\/strong><\/a> can be of great support to students in clarifying their questions and teaching the proper approaches to problem-solving. For any queries about the exercise problems, the <a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-10-solutions-chapter-3-pair-of-linear-equations-in-two-variables\/\"><strong>RD Sharma Solutions for Class 10 Maths Chapter 3<\/strong><\/a> Pair of Linear Equations In Two Variables Exercise 3.10 PDF is available below.<\/p>\n<div id=\"ez-toc-container\" class=\"ez-toc-v2_0_47_1 counter-hierarchy ez-toc-counter ez-toc-grey ez-toc-container-direction\">\n<div class=\"ez-toc-title-container\">\n<p class=\"ez-toc-title\">Table of Contents<\/p>\n<span class=\"ez-toc-title-toggle\"><a href=\"#\" class=\"ez-toc-pull-right ez-toc-btn ez-toc-btn-xs ez-toc-btn-default ez-toc-toggle\" aria-label=\"ez-toc-toggle-icon-1\"><label for=\"item-69d7a8b74fd9d\" aria-label=\"Table of Content\"><span style=\"display: flex;align-items: center;width: 35px;height: 30px;justify-content: center;direction:ltr;\"><svg style=\"fill: #000000;color:#000000\" xmlns=\"http:\/\/www.w3.org\/2000\/svg\" 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class=\"ez-toc-section\" id=\"download-rd-sharma-class-10-solutions-chapter-3-exercise-310-free-pdf\"><\/span>Download RD Sharma Class 10 Solutions Chapter 3 Exercise 3.10 Free PDF<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<div id=\"example1\" style=\"text-align: justify;\">\u00a0<\/div>\n<p style=\"text-align: justify;\"><style>\n.pdfobject-container { height: 800px;}<br \/>\n.pdfobject { border: 1px solid #666; }<br \/>\n<\/style><\/p>\n<p style=\"text-align: justify;\"><script src=\"https:\/\/www.kopykitab.com\/_utility\/js\/pdfobject.min.js\"><\/script><br \/><script>PDFObject.embed(\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD_Sharma_Class_10_Chapter_3_Ex_3.10.pdf\", \"#example1\");<\/script><\/p>\n<p><a href=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD_Sharma_Class_10_Chapter_3_Ex_3.10.pdf\">RD Sharma Class 10 Solutions Chapter 3 Exercise 3.10<\/a><\/p>\n<h3><span class=\"ez-toc-section\" id=\"access-answers-to-rd-sharma-solutions-class-10-maths-chapter-3-exercise-310-important-question-with-answers\"><\/span>Access answers to RD Sharma Solutions Class 10 Maths Chapter 3 Exercise 3.10- Important Question with Answers<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p><strong>1. Points A and B are 70km. apart on a highway. A car starts from A and another car starts from B simultaneously. If they travel in the same direction, they meet in 7hrs, but if they travel towards each other, they meet in one hour. Find the speed of two cars.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Let\u2019s consider the car starting from point A as X and its speed as x km\/hr.<\/p>\n<p>And the car starting from point B as Y and its speed as y km\/hr.<\/p>\n<p>It\u2019s seen that there are two cases in the question:<\/p>\n<p># Case 1: Car X and Y are moving in the same direction<\/p>\n<p># Case 2:\u00a0Car X and Y are moving in the opposite direction<\/p>\n<p>Let\u2019s assume that the meeting point in case 1 as P and in case 2 as Q.<\/p>\n<p><strong>Now, solving for case 1:<\/strong><\/p>\n<p>The distance travelled by car X = AP<\/p>\n<p>And, the distance travelled by car Y = BP<\/p>\n<p>As the time taken for both the cars to meet is 7 hours,<\/p>\n<p>The distance travelled by car X in 7 hours = 7x km [\u2235 distance = speed x time]<\/p>\n<p>\u21d2 AP = 7x<\/p>\n<p>Similarly,<\/p>\n<p>The distance travelled by car Y in 7 hours = 7y km<\/p>\n<p>\u21d2 BP = 7Y<\/p>\n<p>As the cars are moving in the same direction (i.e. away from each other), we can write<\/p>\n<p>AP \u2013 BP = AB<\/p>\n<p>So, 7x \u2013 7y = 70<\/p>\n<p>x \u2013 y = 10 \u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026. (i) [after taking 7 common out]<\/p>\n<p><strong>Now, solving for case 2:<\/strong><\/p>\n<p>In this case, as it\u2019s clearly seen,<\/p>\n<p>The distance travelled by car X = AQ<\/p>\n<p>And,<\/p>\n<p>The distance travelled by car Y = BQ<\/p>\n<p>As the time taken for both the cars to meet is 1 hour,<\/p>\n<p>The distance travelled by car x in 1 hour = 1x km<\/p>\n<p>\u21d2 AQ = 1x<\/p>\n<p>Similarly,<\/p>\n<p>The distance travelled by car y in 1 hour = 1y km<\/p>\n<p>\u21d2 BQ = 1y<\/p>\n<p>Now, since the cars are moving in the opposite direction (i.e. towards each other), we can write<\/p>\n<p>AQ + BQ = AB<\/p>\n<p>\u21d2 x + y = 70 \u2026\u2026\u2026\u2026\u2026 (ii)<\/p>\n<p>Hence, by solving (i) and (ii) we get the required solution<\/p>\n<p>From (i), we have x = 10 + y\u2026\u2026. (iii)<\/p>\n<p>Substituting this value of x in (ii).<\/p>\n<p>\u21d2 (10 + y) + y = 70<\/p>\n<p>\u21d2 y = 30<\/p>\n<p>Now, using y = 30 in (iii), we get<\/p>\n<p>\u21d2 x = 40<\/p>\n<p>Therefore,<\/p>\n<p>\u2013 Speed of car X = 40km\/hr.<\/p>\n<p>\u2013 Speed of car Y = 30 km\/hr.<\/p>\n<p><strong>2. A sailor goes 8 km downstream in 40 minutes and returns in 1 hour. Determine the speed of the sailor in still water and the speed of the current.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p><strong><br \/><\/strong><br \/>Let\u2019s assume,<\/p>\n<p>The speed of the sailor in still water as x km\/hr<\/p>\n<p>And,<\/p>\n<p>The speed of the current as y km\/hr<\/p>\n<p>We know that,<\/p>\n<p>Speed of the sailor in upstream = (x \u2013 y) km\/hr<\/p>\n<p>Speed of the sailor in downstream = (x + y) km\/hr<\/p>\n<p>So, time taken to cover 8 km upstream =\u00a08\/ (x \u2013 y) hr [\u2235 time = distance\/ speed]<\/p>\n<p>And, time taken to cover 8 km downstream =\u00a08\/ (x + y hr [\u2235 time = distance\/ speed]<\/p>\n<p>It\u2019s given that time taken to cover 8 km downstream in 40 minutes or,\u00a040\/ 60\u00a0hour or 2\/3\u00a0hr.<\/p>\n<p>8\/ (x + y)\u00a0=\u00a02\/3<\/p>\n<p>8 \u00d7 3 = 2(x + y)<\/p>\n<p>24 = 2x + 2y<\/p>\n<p>x + y = 12 \u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026 (i) [After taking 2 common out and rearranging]<\/p>\n<p>Similarly, the time taken to cover 8 km upstream in 1hour can be written as,<\/p>\n<p>8\/ (x \u2013 y)\u00a0= 1<\/p>\n<p>8 = 1(x \u2013 y)<\/p>\n<p>\u21d2 x \u2013 y = 8 \u2026\u2026\u2026\u2026\u2026.. (ii)<\/p>\n<p>Hence, by solving (i) and (ii) we get the required solution<\/p>\n<p>On adding (i) and (ii) we get,<\/p>\n<p>2x = 20<\/p>\n<p>\u21d2 x = 10<\/p>\n<p>Now, putting the value of x in (i), we find y<\/p>\n<p>10 + y = 12<\/p>\n<p>\u21d2 y = 2<\/p>\n<p>Therefore, the speed of the sailor is 10km\/hr and the speed of the current is 2km\/hr.<\/p>\n<p><strong>3. The boat goes 30km upstream and 44km downstream in 10 hours. In 13 hours, it can go 40km upstream and 55km downstream. Determine the speed of stream and that of the boat in still water.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Let\u2019s assume,<\/p>\n<p>The speed of the boat in still water as x km\/hr<\/p>\n<p>And,<\/p>\n<p>The speed of the stream as y km\/hr<\/p>\n<p>We know that,<\/p>\n<p>Speed of the boat in upstream = (x \u2013 y) km\/hr<\/p>\n<p>Speed of the boat in downstream = (x + y) km\/hr<\/p>\n<p>So,<\/p>\n<p>Time taken to cover 30 km upstream =\u00a030\/ (x \u2212 y) hr [\u2235 time = distance\/ speed]<\/p>\n<p>Time taken to cover 44 km downstream =44\/ (x + y) hr [\u2235 time = distance\/ speed]<\/p>\n<p>It\u2019s given that the total time of journey is 10 hours. So, this can be expressed as<\/p>\n<p>30\/ (x \u2013 y)\u00a0+ 44\/ (x + y)\u00a0= 10 \u2026\u2026.. (i)<\/p>\n<p>Similarly,<\/p>\n<p>Time taken to cover 40 km upstream =\u00a040\/ (x \u2013 y) hr [\u2235 time = distance\/ speed]<\/p>\n<p>Time taken to cover 55 km downstream =\u00a055\/ (x + y) hr [\u2235 time = distance\/ speed]<\/p>\n<p>And for this case, the total time of the journey is given as 13 hours.<\/p>\n<p>Hence, we can write<\/p>\n<p>40\/ (x \u2013 y) +\u00a055\/ (x + y)\u00a0= 13 \u2026\u2026. (ii)<\/p>\n<p>Hence, by solving (i) and (ii) we get the required solution<\/p>\n<p>Taking 1\/ (x \u2013 y) = u and 1\/ (x + y) = v in equations (i) and (ii) we have<\/p>\n<p>30u + 44v = 10<\/p>\n<p>40u + 55v = 10<\/p>\n<p>Which may be re-written as,<\/p>\n<p>30u + 44v \u2013 10 = 0 \u2026\u2026. (iii)<\/p>\n<p>40u + 55v \u2013 13 = 0\u2026\u2026\u2026 (iv)<\/p>\n<p>Solving these equations by cross multiplication, we get,<\/p>\n<p><img title=\"R D Sharma Solutions For Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables ex 3.10 - 1\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-3-48.png\" alt=\"R D Sharma Solutions For Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables ex 3.10 - 1\" \/><\/p>\n<p>Now,<\/p>\n<p>1\/ (x \u2013 y)\u00a0=\u00a02\/10<\/p>\n<p>\u21d2 1 x 10 = 2(x \u2013 y)<\/p>\n<p>\u21d2 10 = 2x \u2013 2y<\/p>\n<p>\u21d2 x \u2013 y = 5 \u2026\u2026. (v)<\/p>\n<p>And,<\/p>\n<p>1\/ (x + y) = 1\/11<\/p>\n<p>\u21d2 x + y = 11 \u2026\u2026. (vi)<\/p>\n<p>Again, solving (v) and (vi)<\/p>\n<p>Adding (v) and (vi), we get<\/p>\n<p>2x = 16<\/p>\n<p>\u21d2 x = 8<\/p>\n<p>Using x in (v), we find y<\/p>\n<p>8 \u2013 y = 5<\/p>\n<p>\u21d2 y = 3<\/p>\n<p>Therefore, the speed of the boat in still water is 8 km\/hr and the speed of the stream is 3 km\/hr.<\/p>\n<p><strong>4. A boat goes 24km upstream and 28km downstream in 6hrs. It goes 30km upstream and 21km downstream in 6.5 hours. Find the speed of the boat in still water and also speed of the stream.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Let\u2019s assume,<\/p>\n<p>The speed of the boat in still water as x km\/hr<\/p>\n<p>And,<\/p>\n<p>The speed of the stream as y km\/hr<\/p>\n<p>We know that,<\/p>\n<p>Speed of the boat in upstream = (x \u2013 y) km\/hr<\/p>\n<p>Speed of the boat in downstream = (x + y) km\/hr<\/p>\n<p>So, time taken to cover 28 km downstream =\u00a028\/ (x+y) hr [\u2235 time = distance\/ speed]<\/p>\n<p>Time taken to cover 24 km upstream =24\/ (x \u2013 y) hr [\u2235 time = distance\/ speed]<\/p>\n<p>It\u2019s given that the total time of journey is 6 hours. So, this can be expressed as<\/p>\n<p>24\/ (x \u2013 y)\u00a0+\u00a028\/ (x + y)\u00a0= 6\u2026\u2026 (i)<\/p>\n<p>Similarly,<\/p>\n<p>Time taken to cover 30 km upstream = 30\/ (x \u2212 y) [\u2235 time = distance\/ speed]<\/p>\n<p>Time taken to cover 21km downstream =21\/ (x + y) [\u2235 time = distance\/ speed]<\/p>\n<p>And for this case the total time of the journey is given as 6.5 i.e 13\/2 hours.<\/p>\n<p>Hence, we can write<\/p>\n<p>30\/ (x \u2013 y)\u00a0+\u00a021\/ (x + y)\u00a0=\u00a013\/2 \u2026.. (ii)<\/p>\n<p>Hence, by solving (i) and (ii) we get the required solution<\/p>\n<p>Taking 1\/ (x \u2013 y) = u and 1\/ (x + y) = v in equations (i) and (ii) we have (after rearranging)<\/p>\n<p>24u + 28v \u2013 6 = 0\u00a0\u2026\u2026 (iii)<\/p>\n<p>30u + 21v \u2013 13\/2 = 0\u00a0\u2026\u2026. (iv)<\/p>\n<p>Solving these equations by cross multiplication, we get,<\/p>\n<p><img title=\"R D Sharma Solutions For Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables ex 3. - \" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-3-49.png\" alt=\"R D Sharma Solutions For Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables ex 3.10 - 2\" \/><\/p>\n<p>u =\u00a01\/6 and v =\u00a01\/14<\/p>\n<p>Now,<\/p>\n<p>u =\u00a01\/ (x \u2212 y)\u00a0=\u00a01\/ 6<\/p>\n<p>x \u2013 y = 6 \u2026. (v)<\/p>\n<p>v =\u00a01\/ (x + y) = 1\/ 14<\/p>\n<p>x + y = 14\u2026\u2026. (vi)<\/p>\n<p>On Solving (v) and (vi)<\/p>\n<p>Adding (v) and (vi), we get<\/p>\n<p>2x = 20<\/p>\n<p>\u21d2 x = 10<\/p>\n<p>Using x = 10 in (v), we find y<\/p>\n<p>10 + y = 14<\/p>\n<p>\u21d2 y = 4<\/p>\n<p>Therefore,<\/p>\n<p>Speed of the stream = 4km\/hr.<\/p>\n<p>Speed of boat = 10km\/hr.<\/p>\n<p><strong>5. A man walks a certain distance with a certain speed. If he walks 1\/2 km an hour faster, he takes 1 hour less. But, if he walks 1km an hour slower, he takes 3 more hours. Find the distance covered by the man and his original rate of walking.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Let the actual speed of the man be x km\/hr and y be the actual time taken by him in hours.<\/p>\n<p>So, we know that<\/p>\n<p>Distance covered = speed \u00d7 distance<\/p>\n<p>\u21d2 Distance = x \u00d7 y = xy \u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026. (i)<\/p>\n<p>First condition from the question says that,<\/p>\n<p>If the speed of the man increases by 1\/2 km\/hr, the journey time will reduce by 1 hour.<\/p>\n<p>Showing this using variables, we have<\/p>\n<p>\u21d2 When speed is (x + 1\/2) km\/hr, time of journey = y \u2013 1 hours<\/p>\n<p>Now,<\/p>\n<p>Distance covered = (x + 1\/2) x (y \u2013 1) km<\/p>\n<p>Since the distance is the same, i.e. xy we can equate it, [from (i)]<\/p>\n<p>xy = (x + 1\/2) x (y \u2013 1)<\/p>\n<p>And we finally get,<\/p>\n<p>-2x + y \u2013 1 = 0 \u2026\u2026\u2026\u2026\u2026\u2026\u2026.. (ii)<\/p>\n<p>From the second condition of the question, we have<\/p>\n<p>If the speed reduces by 1 km\/hr, then the time of journey increases by 3 hours.<\/p>\n<p>\u21d2 When speed is (x-1) km\/hr, time of journey is (y+3) hours<\/p>\n<p>Since, the distance covered = xy [from (i)]<\/p>\n<p>xy = (x-1)(y+3)<\/p>\n<p>\u21d2 xy = xy \u2013 1y + 3x \u2013 3<\/p>\n<p>\u21d2 xy = xy + 3x \u2013 1y \u2013 3<\/p>\n<p>\u21d2 3x \u2013 y \u2013 3 = 0 \u2026\u2026\u2026\u2026\u2026\u2026 (iii)<\/p>\n<p>From (ii) and (iii), the value of x can be calculated by<\/p>\n<p>(ii) + (iii) \u21d2<\/p>\n<p>x \u2013 4 = 0<\/p>\n<p>x = 4<\/p>\n<p>Now, y can be obtained by using x = 4 in (ii)<\/p>\n<p>-2(4) + y \u2013 1 = 0<\/p>\n<p>\u21d2 y = 1 + 8 = 9<\/p>\n<p>Hence, putting the value of x and y in equation (i), we find the distance<\/p>\n<p>Distance covered = xy<\/p>\n<p>= 4 \u00d7 9<\/p>\n<p>= 36 km<\/p>\n<p>Thus, the distance is 36 km and the speed of walking is 4 km\/hr.<\/p>\n<p><strong>6. A person rowing at the rate of 5km\/h in still water, takes thrice as much as time in going 40 km upstream as in going 40km downstream. Find the speed of the stream.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Let\u2019s assume x to be the speed of the stream.<\/p>\n<p>So, we know that<\/p>\n<p>Speed of boat in downstream = (5 + x) and,<\/p>\n<p>Speed of boat in upstream = (5 \u2013 x)<\/p>\n<p>It is given that,<\/p>\n<p>The distance in one way is 40km.<\/p>\n<p>And,<\/p>\n<p>Time taken during upstream = 3 \u00d7 time taken during the downstream<\/p>\n<p>Expressing it by equations, we have<\/p>\n<p>40\/ (5 \u2013 x) = 3 x 40\/ (5 + x) [\u2235 time = distance\/ speed]<\/p>\n<p>By cross-multiplication, we get<\/p>\n<p>(5+x) = 3(5-x)<\/p>\n<p>\u21d2 5 + x = 3(5 \u2013 x)<\/p>\n<p>\u21d2 x + 3x = 15 \u2013 5<\/p>\n<p>\u21d2 x = 10\/4 = 2.5<\/p>\n<p>Therefore, the speed of the stream is\u00a02.5 km\/hr.<\/p>\n<p><strong>7. Ramesh travels 760km to his home partly by train and partly by car. He takes 8 hours if he travels 160km by train and the rest by car. He takes 12 minutes more if he travels 240km by train and the rest by car. Find the speed of the train and car respectively.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Let\u2019s assume,<\/p>\n<p>The speed of the train be x km\/hr<\/p>\n<p>The speed of the car = y km\/hr<\/p>\n<p>From the question, it\u2019s understood that there are two parts<\/p>\n<p># Part 1:\u00a0When Ramesh travels 160 Km by train and the rest by car.<\/p>\n<p># Part 2:\u00a0When Ramesh travels 240Km by train and the rest by car.<\/p>\n<p>Part 1,<\/p>\n<p>Time taken by Ramesh to travel 160 km by train =\u00a0160\/x hrs [\u2235 time = distance\/ speed]<\/p>\n<p>Time taken by Ramesh to travel the remaining (760 \u2013 160) km i.e., 600 km by car =600\/y hrs<\/p>\n<p>So, the total time taken by Ramesh to cover 760Km = 160\/x hrs\u00a0+ 600\/y hrs<\/p>\n<p>It\u2019s given that,<\/p>\n<p>Total time taken for this journey = 8 hours<\/p>\n<p>So, by equations its<\/p>\n<p>160\/x\u00a0+\u00a0600\/y\u00a0= 8<\/p>\n<p>20\/x\u00a0+\u00a075\/y = 1 [on dividing previous equation by 8] \u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026 (i)<\/p>\n<p>Part 2,<\/p>\n<p>Time taken by Ramesh to travel 240 km by train = 240\/x hrs<\/p>\n<p>Time taken by Ramesh to travel (760 \u2013 240) = 520km by car =\u00a0520\/y hrs<\/p>\n<p>For this journey, it\u2019s given that Ramesh will take a total of is 8 hours and 12 minutes to finish.<\/p>\n<p>240\/x\u00a0+\u00a0520\/y\u00a0= 8hrs 12mins = 8 + (12\/60) = 41\/5 hr<\/p>\n<p>240\/x\u00a0+\u00a0520\/y\u00a0=\u00a041\/5<\/p>\n<p>6\/x\u00a0+\u00a013\/y =\u00a041\/200 \u2026\u2026\u2026. (ii)<\/p>\n<p>Solving (i) and (ii), we get the required solution<\/p>\n<p>Let\u2019s take 1\/x = u and 1\/y = v,<\/p>\n<p>So, (i) and (ii) becomes,<\/p>\n<p>20u + 75v = 1 \u2026\u2026\u2026.. (iii)<\/p>\n<p>6u + 13v = 41\/200 \u2026\u2026. (iv)<\/p>\n<p>On multiplying (iii) by 3 and (iv) by 10,<\/p>\n<p>60u + 225v = 3<\/p>\n<p>60u + 130v = 41\/20<\/p>\n<p>Subtracting the above two equations, we get<\/p>\n<p>(225 \u2013 130)v = 3 \u2013 41\/20<\/p>\n<p>95v = 19\/ 20<\/p>\n<p>\u21d2 v = 19\/ (20 x 95) = 1\/100<\/p>\n<p>\u21d2 y = 1\/v = 100<\/p>\n<p>Using v = 1\/100 in (iii) to find v,<\/p>\n<p>20u + 75(1\/100) = 1<\/p>\n<p>20u = 1 \u2013 75\/100<\/p>\n<p>\u21d2 20u = 25\/100 = 1\/4<\/p>\n<p>\u21d2 u = 1\/80<\/p>\n<p>\u21d2 x = 1\/u = 80<\/p>\n<p>So, the speed of the train is 80km\/hr and the speed of car is100km\/hr.<\/p>\n<p><strong>8) A man travels 600 km partly by train and partly by car. If he covers 400km by train and the rest by car, it takes him 6 hours and 30 minutes. But, if he travels 200km by train and the rest by car, he takes half an hour longer. Find the speed of the train and the speed of the car.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Let\u2019s assume,<\/p>\n<p>The speed of the train be x km\/hr<\/p>\n<p>The speed of the car = y km\/hr<\/p>\n<p>From the question, it\u2019s understood that there are two parts<\/p>\n<p># Part 1:\u00a0When the man travels 400 km by train and the rest by car.<\/p>\n<p># Part 2:\u00a0When Ramesh travels 200 km by train and the rest by car.<\/p>\n<p>Part 1,<\/p>\n<p>Time taken by the man to travel 400 km by train = 400\/x hrs [\u2235 time = distance\/ speed]<\/p>\n<p>Time taken by the man to travel (600 \u2013 400) = 200 km by car = 200\/y hrs<\/p>\n<p>Time taken by a man to cover 600 km = 400\/x hrs + 200\/y hrs<\/p>\n<p>Total time taken for this journey = 6 hours + 30 mins = 6 + 1\/2 = 13\/2<\/p>\n<p>So, by equations its<\/p>\n<p>400\/x\u00a0+\u00a0200\/y\u00a0= 13\/2<\/p>\n<p>400\/x\u00a0+\u00a0200\/y\u00a0= 13\/2<\/p>\n<p>400\/x\u00a0+\u00a0200\/y\u00a0=\u00a013\/2<\/p>\n<p>200 (2\/x\u00a0+\u00a01\/y) =\u00a013\/2<\/p>\n<p>2\/x\u00a0+\u00a01\/y\u00a0=\u00a013\/400 .\u2026(i)<\/p>\n<p>Part 2,<\/p>\n<p>Time taken by the man to travel 200 km by train =\u00a0200\/x hrs. [\u2235 time = distance\/ speed]<\/p>\n<p>Time taken by the man to travel (600 \u2013 200) = 400km by car = 200\/y hrs<\/p>\n<p>For the part, the total time of the journey is given as 6hours 30 mins + 30 mins that is 7hrs,<\/p>\n<p>200\/x\u00a0+\u00a0400\/y\u00a0= 7<\/p>\n<p>200 (1\/x\u00a0+\u00a02\/y) = 7<\/p>\n<p>1\/x\u00a0+\u00a02\/y\u00a0=\u00a07\/200\u00a0\u2026..(ii)<\/p>\n<p>Taking\u00a01\/x\u00a0= u, and\u00a01\/y\u00a0= v,<\/p>\n<p>So, the equations (i) and (ii) become,<\/p>\n<p>2u + v =\u00a013\/400\u00a0\u2026.. (iii)<\/p>\n<p>u + 2v =\u00a07\/200\u00a0\u2026\u2026. (iv)<\/p>\n<p>Solving (iii) and (iv), by<\/p>\n<p>(iv) x 2 \u2013 (iii) \u21d2<\/p>\n<p>3v = 14\/200 \u2013 13\/400<\/p>\n<p>3v = 1\/400 x (28 \u2013 13)<\/p>\n<p>3v = 15\/400<\/p>\n<p>v = 1\/80<\/p>\n<p>\u21d2 y = 1\/v = 80<\/p>\n<p>Now, using v in (iii) we find u,<\/p>\n<p>2u + (1\/80) =\u00a013\/400<\/p>\n<p>2u = 13\/400 \u2013 1\/80<\/p>\n<p>2u = 8\/400<\/p>\n<p>u = 1\/100<\/p>\n<p>\u21d2 x = 1\/u = 100<\/p>\n<p>Hence, the speed of the train is 100 km\/hr and the speed of the car is 80 km\/hr.<\/p>\n<p><strong>9. Places A and B are 80 km apart from each other on a highway. One car starts from A and other from B at the same time. If they move in the same direction, they meet in 8 hours and if they move in the opposite direction, they meet in 1 hour and 20 minutes. Find the speeds of the cars.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Let\u2019s consider the car starting from point A as X and its speed as x km\/hr.<\/p>\n<p>And, the car starts from point B as Y and its speed as y km\/hr.<\/p>\n<p>It\u2019s seen that there are two cases in the question:<\/p>\n<p># Case 1: Car X and Y are moving in the same direction<\/p>\n<p># Case 2:\u00a0Car X and Y are moving in the opposite direction<\/p>\n<p>Let\u2019s assume that the meeting point in case 1 as P and in case 2 as Q.<\/p>\n<p><strong>Now, solving for case 1:<\/strong><\/p>\n<p>The distance travelled by car X = AP<\/p>\n<p>And, the distance travelled by car Y = BP<\/p>\n<p>As the time taken for both the cars to meet is 8 hours,<\/p>\n<p>The distance travelled by car X in 7 hours = 8x km [\u2235 distance = speed x time]<\/p>\n<p>\u21d2 AP = 8x<\/p>\n<p>Similarly,<\/p>\n<p>The distance travelled by car Y in 8 hours = 8y km<\/p>\n<p>\u21d2 BP = 8Y<\/p>\n<p>As the cars are moving in the same direction (i.e. away from each other), we can write<\/p>\n<p>AP \u2013 BP = AB<\/p>\n<p>So, 8x \u2013 8y = 80<\/p>\n<p>\u21d2 x \u2013 y = 10 \u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026. (i) [After taking 8 common out]<\/p>\n<p><strong>Now, solving for case 2:<\/strong><\/p>\n<p>In this case, as it\u2019s clearly seen,<\/p>\n<p>The distance travelled by car X = AQ<\/p>\n<p>And,<\/p>\n<p>The distance travelled by car Y = BQ<\/p>\n<p>As the time taken for both the cars to meet is 1 hour and 20 min, \u21d21 + (20\/60) = 4\/3 hr<\/p>\n<p>The distance travelled by car x in 4\/3 hour = 4x\/3 km<\/p>\n<p>\u21d2 AQ = 4x\/3<\/p>\n<p>Similarly,<\/p>\n<p>The distance travelled by car y in 4\/3 hour = 4y\/3 km<\/p>\n<p>\u21d2 BQ = 4y\/3<\/p>\n<p>Now, since the cars are moving in the opposite direction (i.e., towards each other), we can write<\/p>\n<p>AQ + BQ = AB<\/p>\n<p>\u21d2 4x\/3 + 4y\/3 = 80<\/p>\n<p>\u21d2 4x + 4y = 240<\/p>\n<p>\u21d2 x + y = 60 \u2026\u2026\u2026\u2026\u2026 (ii) [After taking LCM]<\/p>\n<p>Hence, by solving (i) and (ii) we get the required solution<\/p>\n<p>From (i), we have x = 10 + y\u2026\u2026. (iii)<\/p>\n<p>Substituting this value of x in (ii).<\/p>\n<p>\u21d2 (10 + y) + y = 60<\/p>\n<p>\u21d2 2y = 50<\/p>\n<p>\u21d2 y = 25<\/p>\n<p>Now, using y = 30 in (iii), we get<\/p>\n<p>\u21d2 x = 35<\/p>\n<p>Therefore,<\/p>\n<p>\u2013 Speed of car X = 35 km\/hr.<\/p>\n<p>\u2013 Speed of car Y = 25 km\/hr.<\/p>\n<p>We have provided complete details of RD Sharma Class 10 Solutions Chapter 3 Exercise 3.10. If you have any queries related to <a href=\"https:\/\/www.cbse.gov.in\/\" target=\"_blank\" rel=\"noopener\"><strong>CBSE<\/strong><\/a>\u00a0Class 10, feel free to ask us in the comment section below.<\/p>\n<h2><span class=\"ez-toc-section\" id=\"faqs-on-rd-sharma-class-10-solutions-chapter-3-exercise-310\"><\/span>FAQs on RD Sharma Class 10 Solutions Chapter 3 Exercise 3.10<span class=\"ez-toc-section-end\"><\/span><\/h2>\n\n\n<div id=\"rank-math-faq\" class=\"rank-math-block\">\n<div class=\"rank-math-list \">\n<div id=\"faq-question-1631107915673\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"where-can-i-download-rd-sharma-class-10-solutions-chapter-3-exercise-310-free-pdf\"><\/span>Where can I download RD Sharma Class 10 Solutions Chapter 3 Exercise 3.10 free PDF?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>You can download RD Sharma Class 10 Solutions Chapter 3 Exercise 3.10 free PDF from the above article.<\/p>\n\n<\/div>\n<\/div>\n<div id=\"faq-question-1631108112288\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"is-it-required-to-practice-all-of-the-questions-in-chapter-3-exercise-310-of-rd-sharma-solutions-for-class-10-maths\"><\/span>Is it required to practice all of the questions in Chapter 3 Exercise 3.10 of RD Sharma Solutions for Class 10 Maths?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>Yes, all of the questions in RD Sharma Solutions for Class 10 Maths Chapter 3 Exercise 3.10 must be learned. These questions may appear on both board exams and class tests. Students will be prepared for their board exams if they learn these questions.<\/p>\n\n<\/div>\n<\/div>\n<div id=\"faq-question-1631108671952\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"what-are-the-benefits-of-using-rd-sharma-class-10-solutions-chapter-3-exercise-310\"><\/span>What are the benefits of using RD Sharma Class 10 Solutions Chapter 3 Exercise 3.10?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>1. Correct answers according to the last CBSE guidelines and syllabus.<br \/>2. The RD Sharma Class 10 Solutions Chapter 3 Exercise 3.10 is written in simple language to assist students in their board examination, &amp; competitive examination preparation.<\/p>\n\n<\/div>\n<\/div>\n<\/div>\n<\/div>","protected":false},"excerpt":{"rendered":"<p>RD Sharma Class 10 Solutions Chapter 3 Exercise 3.10:\u00a0This exercise focuses on problems that are specific to this application. The RD Sharma Class 10 Solutions can be of great support to students in clarifying their questions and teaching the proper approaches to problem-solving. For any queries about the exercise problems, the RD Sharma Solutions for &#8230; <a title=\"RD Sharma Class 10 Solutions Chapter 3 Exercise 3.10 (Updated for 2024)\" class=\"read-more\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-10-solutions-chapter-3-exercise-3-10\/\" aria-label=\"More on RD Sharma Class 10 Solutions Chapter 3 Exercise 3.10 (Updated for 2024)\">Read more<\/a><\/p>\n","protected":false},"author":238,"featured_media":125561,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"fifu_image_url":"","fifu_image_alt":""},"categories":[73411,2985,73410],"tags":[3243,9206,73520,4388],"amp_enabled":true,"_links":{"self":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/125502"}],"collection":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/users\/238"}],"replies":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/comments?post=125502"}],"version-history":[{"count":4,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/125502\/revisions"}],"predecessor-version":[{"id":517682,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/125502\/revisions\/517682"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media\/125561"}],"wp:attachment":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media?parent=125502"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/categories?post=125502"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/tags?post=125502"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}