{"id":5989,"date":"2025-06-09T09:52:42","date_gmt":"2025-06-09T09:52:42","guid":{"rendered":"https:\/\/thecompanyboy.com\/?p=5989"},"modified":"2025-06-09T09:52:42","modified_gmt":"2025-06-09T09:52:42","slug":"rs-aggarwal-quantitative-aptitude-pdf-download-time-distance-and","status":"publish","type":"post","link":"https:\/\/www.reilsolar.com\/drive\/rs-aggarwal-quantitative-aptitude-pdf-download-time-distance-and\/","title":{"rendered":"RS Aggarwal Quantitative Aptitude PDF Free Download: TIME AND DISTANCE"},"content":{"rendered":"<h1 style=\"text-align: center\"><strong>TIME AND DISTANCE\u00a0<\/strong><\/h1>\n<h2><strong>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 IMPORTANT FACTS AND FORMULAE<\/strong><\/h2>\n<p>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Distance\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Distance<\/p>\n<ol start=\"2\">\n<li>Speed = Time ,\u00a0 Time=\u00a0\u00a0\u00a0\u00a0\u00a0 Speed\u00a0\u00a0\u00a0\u00a0\u00a0 , Distance\u00a0 =\u00a0 (Speed *\u00a0 Time)<\/li>\n<li>x km \/ hr = x *\u00a0 <u>5<\/u><\/li>\n<\/ol>\n<p>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 18\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/p>\n<ol start=\"4\">\n<li>x m\/sec = (x * 18\/5) km \/hr<\/li>\n<li>If the ratio of the speeds of A and B is a:b , then the ratio of the times taken by them to cover the same distance is <u>1<\/u>: <u>1<\/u>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 a\u00a0\u00a0 b<\/li>\n<\/ol>\n<p>or b:a.<\/p>\n<ol start=\"5\">\n<li>Suppose a man covers a certain distance at x km\/ hr and an equal distance at y km \/ hr . Then , the average speed during the whole journey is <u>2xy <\/u>\u00a0\u00a0\u00a0km\/ hr.<\/li>\n<\/ol>\n<p>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0x+y<\/p>\n<h2>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 <strong>SOLVED EXAMPLES<\/strong><\/h2>\n<p><strong>\u00a0<\/strong><strong>Ex. 1.<\/strong> <strong>How many minutes does Aditya take to cover a distance of 400 m, if he runs at a speed of 20 km\/hr?<\/strong><\/p>\n<p><strong>Sol.<\/strong> Aditya\u2019s speed = 20 km\/hr\u00a0 = {20 * <u>5<\/u>} m\/sec\u00a0 =\u00a0\u00a0 <u>50<\/u> m\/sec<\/p>\n<ul>\n<li>9<\/li>\n<\/ul>\n<p>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \\Time taken to cover 400 m= { 400 * <u>9<\/u> } sec =72 sec = 1 <u>12<\/u>\u00a0 min 1 <u>1<\/u> min.<\/p>\n<p>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 50\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 60\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 5\u00a0<\/p>\n<p><strong>\u00a0<\/strong><strong>Ex. 2.<\/strong> <strong>A cyclist covers a distnce of 750 m in 2 min 30 sec. What is the speed in km\/hr of the cyclist?<\/strong><\/p>\n<p><strong>Sol.<\/strong> Speed = { <u>750<\/u> } m\/sec\u00a0 =5 m\/sec\u00a0 = { 5\u00a0 *\u00a0 <u>18<\/u> } km\/hr =18km\/hr<\/p>\n<ul>\n<li>5<\/li>\n<\/ul>\n<p><strong>Ex. 3.<\/strong> <strong>A dog takes 4 leaps for every 5 leaps of a hare but 3 leaps of a dog are equal to 4 leaps of the hare. Compare their speeds.<\/strong><\/p>\n<p><strong>Sol.<\/strong> Let the distance covered in 1 leap of the dog be x and that covered in 1 leap of the hare by y.<\/p>\n<p>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Then , 3x = 4y =&gt; x = <u>4<\/u> y\u00a0 =&gt;\u00a0 4x = <u>16<\/u>\u00a0 y.<\/p>\n<ul>\n<li>3<\/li>\n<\/ul>\n<p>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \\ Ratio of speeds of dog and hare = Ratio of distances covered by them\u00a0 in the same time<\/p>\n<p>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 = 4x : 5y = <u>16<\/u> y : 5y\u00a0 =<u>16 <\/u>\u00a0: 5\u00a0 = 16:15<\/p>\n<ul>\n<li>3<\/li>\n<\/ul>\n<p><strong>Ex. 4.While covering a distance of 24 km, a man noticed that after walking for 1 hour and 40 minutes, the distance covered by him was <u>5<\/u> of the remaining distance. What was his speed in metres per second?<\/strong><\/p>\n<p><strong>\u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a07<\/strong><\/p>\n<p><strong>Sol.<\/strong> Let the speed be x km\/hr.<\/p>\n<p>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Then, distance covered in 1 hr. 40 min. i.e., 1 <u>\u00a02 <\/u>\u00a0hrs\u00a0 = <u>5x<\/u>\u00a0 km<\/p>\n<ul>\n<li>3<\/li>\n<\/ul>\n<p>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Remaining distance = { 24 \u2013 <u>5x<\/u> } km.<\/p>\n<p>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0 3\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/p>\n<ul>\n<li><u>5x <\/u>=\u00a0 <u>5<\/u> {\u00a0 24 &#8211;\u00a0 <u>5x<\/u>\u00a0 } \u00f3\u00a0 <u>5x<\/u>\u00a0 =\u00a0 <u>5<\/u> { <u>\u00a072-5x <\/u>\u00a0}\u00a0 \u00f3\u00a0 7x\u00a0 =72 \u20135x<\/li>\n<\/ul>\n<p>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 3\u00a0\u00a0\u00a0\u00a0 \u00a07\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 3\u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 3\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 7\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 3\u00a0\u00a0\u00a0\u00a0<\/p>\n<p>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \u00f3 12x = 72\u00a0 \u00f3\u00a0 x=6<\/p>\n<p>\u00a0 Hence speed = 6 km\/hr ={ 6 * <u>5<\/u> } m\/sec\u00a0 =\u00a0 <u>5<\/u>\u00a0 m\/sec = 1 <u>2<\/u><\/p>\n<p>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 18\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 3\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \u00a0\u00a03<\/p>\n<p><strong>\u00a0<\/strong><strong>Ex. 5.Peter can cover a certain distance in 1 hr. 24 min. by covering two-third of the distance at 4 kmph and the rest at 5 kmph. Find the total distance.<\/strong><\/p>\n<p>\u00a0<strong>Sol.<\/strong>\u00a0\u00a0 Let the total distance be x km . Then,<\/p>\n<p>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 <u>2<\/u> x\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 <u>1<\/u> x<\/p>\n<p>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 <u>3\u00a0\u00a0\u00a0 <\/u>\u00a0\u00a0+\u00a0 <u>\u00a03\u00a0\u00a0 <\/u>\u00a0\u00a0=\u00a0 <u>7\u00a0 <\/u>\u00a0\u00f3\u00a0 <u>x<\/u>\u00a0 +\u00a0 <u>x <\/u>\u00a0= <u>7<\/u>\u00a0\u00a0\u00a0 \u00f3\u00a0 7x\u00a0 = 42\u00a0 \u00f3\u00a0 x = 6<\/p>\n<p>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 4\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 5\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 5\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 6\u00a0\u00a0\u00a0\u00a0 15\u00a0\u00a0\u00a0 5<\/p>\n<p><strong>Ex. 6.A man traveled from the village to the post-office at the rate of 25 kmph and walked back at the rate of 4 kmph. If the whole journey took 5 hours 48 minutes, find the distance of the post-office from the village.<\/strong><\/p>\n<p>Sol.\u00a0\u00a0\u00a0 Average speed\u00a0\u00a0 = { <u>2xy\u00a0 <\/u>} km\/hr\u00a0 ={\u00a0 <u>2*25*4\u00a0 <\/u>} km\/hr\u00a0 = <u>200<\/u>\u00a0 km\/hr<\/p>\n<p>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 x+y\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 25+4\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 29<\/p>\n<p>\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0Distance traveled in 5 hours 48 minutes i.e., 5 <u>4<\/u>\u00a0 hrs.\u00a0 =\u00a0 { <u>200 <\/u>\u00a0*\u00a0 <u>29<\/u> } km\u00a0 = 40 km<\/p>\n<p>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 5\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 29\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 5<\/p>\n<p>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Distance of the post-office from the village ={\u00a0 <u>40<\/u>\u00a0 }\u00a0 = 20 km<\/p>\n<p>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 2<\/p>\n<p><strong>Ex. 7.An aeroplane files along the four sides of a square at the speeds of 200,400,600 and 800km\/hr.Find the average speed of the plane around the field.<\/strong><\/p>\n<p><strong>Sol. :<\/strong><\/p>\n<p>Let each side of the square be x km and let the average speed of the plane around the field by y km per hour then ,<\/p>\n<p>\u00a0x\/200+x\/400+x\/600+x\/800=4x\/y\u00f325x\/2500\u00f34x\/y\u00f3y=(2400*4\/25)=384<\/p>\n<p>hence average speed =384 km\/hr<\/p>\n<p><strong>\u00a0<\/strong><strong>Ex. 8.Walking at <u>5<\/u> of its usual speed, a train is 10 minutes too late. Find its usual time to cover the journey.<\/strong><\/p>\n<p><strong>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 7 <\/strong><\/p>\n<p><strong>\u00a0<\/strong><strong>Sol. :<\/strong>New speed =5\/6 of the usual speed<\/p>\n<p>New time taken=6\/5 of the usual time<\/p>\n<p>So,( 6\/5 of the usual time )-( usual time)=10 minutes.<\/p>\n<p>=&gt;1\/5 of the usual time=10 minutes.<\/p>\n<ul>\n<li>usual time=10 minutes<\/li>\n<\/ul>\n<p><strong>\u00a0<\/strong><strong>Ex. 9.If a man walks at the rate of 5 kmph, he misses a train by 7 minutes. However, if he walks at the rate of 6 kmph, he reaches the station 5 minutes before the arrival of the train. Find the distance covered by him to reach the station.<\/strong><\/p>\n<p><strong>Sol. <\/strong>Let the required distance be x km<\/p>\n<p>Difference in the time taken at two speeds=1 min =1\/2 hr<\/p>\n<p>Hence x\/5-x\/6=1\/5&lt;=&gt;6x-5x=6<\/p>\n<p>\u00f3x=6<\/p>\n<p>Hence, the required distance is 6 km<\/p>\n<p><strong>Ex. 10. A and B are two stations 390 km apart. A train starts from A at 10 a.m. and travels towards B at 65 kmph. Another train starts from B at 11 a.m. and travels towards A at 35 kmph. At what time do they meet?<\/strong><\/p>\n<p><strong>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Sol. <\/strong>Suppose they meet x hours after 10 a.m. Then,<\/p>\n<p>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 (Distance moved by first in x hrs) + [Distance moved by second in (x-1) hrs]=390.\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 65x + 35(x-1) = 390\u00a0 =&gt; 100x = 425\u00a0 =&gt; x = 17\/4<\/p>\n<p>\u00a0So, they meet 4 hrs.15 min. after 10 a.m i.e., at 2.15 p.m.\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/p>\n<p><strong>Ex. 11. A goods train leaves a station at a certain time and at a fixed speed. After ^hours, an express train leaves the same station and moves in the same direction at a uniform speed of 90 kmph. This train catches up the goods train in 4 hours. Find the speed of the goods train.<\/strong><\/p>\n<p><strong>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Sol.<\/strong>\u00a0 Let the speed of the goods train be x kmph.<\/p>\n<p>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Distance covered by goods train in 10 hours= Distance covered by express train in 4 hours<\/p>\n<p>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 10x = 4 x 90 or x =36.<\/p>\n<p>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 So, speed of goods train = 36kmph.<\/p>\n<p><strong>Ex. 12. A thief is spotted by a policeman from a distance of 100 metres. When the policeman starts the chase, the thief also starts running. If the speed of the thief be 8km\/hr and that of the policeman 10 km\/hr, how far the thief will have run before he is overtaken?<\/strong><\/p>\n<p><strong>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Sol.<\/strong> Relative speed of the policeman = (10-8) km\/hr =2 km\/hr.<\/p>\n<p>Time taken by police man to cover 100m \u00a0\u00a0\u00a0\u00a0\u00a0 100\u00a0\u00a0 x\u00a0 1\u00a0 hr = 1\u00a0 hr.<\/p>\n<p>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 1000\u00a0\u00a0\u00a0\u00a0 2\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 20\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/p>\n<p>In 1\u00a0 hrs, the thief covers a distance of 8\u00a0 x\u00a0 1\u00a0 km = 2\u00a0 km\u00a0 = 400 m<\/p>\n<p>\u00a0\u00a0 20\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 20\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 5\u00a0\u00a0<\/p>\n<p><strong>\u00a0<\/strong><strong>Ex.13. I walk a certain distance and ride back taking a total time of 37 minutes. I could walk both ways in 55 minutes. How long would it take me to ride both ways?<\/strong><\/p>\n<p><strong>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Sol.<\/strong> Let the distance be x km. Then,<\/p>\n<p>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 ( Time taken to walk x km) + (time taken to ride x km) =37 min.<\/p>\n<p>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 ( Time taken to walk 2x km ) + ( time taken to ride 2x km )= 74 min.<\/p>\n<p>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 But, the time taken to walk 2x km = 55 min.<\/p>\n<p>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Time taken to ride 2x km = (74-55)min =19 min.<\/p>\n<p>I can&#8217;t provide a free PDF of <em>RS Aggarwal&#8217;s Quantitative Aptitude<\/em> due to copyright restrictions. However, you can check official sources like:<\/p>\n<p>If you&#8217;re looking for specific topics, I can help explain concepts or provide practice questions. Let me know how I can assist!\u00a0<\/p>\n<h3><a href=\"https:\/\/eltsindia.com\/StudyMaterialFiles\/ea007acc-bc55-4e17-8470-0d9e85313252quantitative-aptitude-for-competitive-examinations-by-rs-aggarwal-reprint-2017.pdf\" target=\"_blank\" rel=\"noopener\">RS Aggarwal Quantitative Aptitude PDF Free Download: TIME AND DISTANCE<\/a><\/h3>\n","protected":false},"excerpt":{"rendered":"<p>TIME AND DISTANCE\u00a0 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 IMPORTANT FACTS AND FORMULAE \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Distance\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Distance Speed = Time ,\u00a0 Time=\u00a0\u00a0\u00a0\u00a0\u00a0 Speed\u00a0\u00a0\u00a0\u00a0\u00a0 , Distance\u00a0 =\u00a0 (Speed *\u00a0 Time) x km \/ hr = x *\u00a0 5 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 18\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 x m\/sec = (x * 18\/5) km \/hr If the ratio of the speeds of A and B is a:b , then [&hellip;]<\/p>\n","protected":false},"author":41,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[126,127],"tags":[],"class_list":["post-5989","post","type-post","status-publish","format-standard","hentry","category-rs-aggarwal-quantitative-aptitude","category-rs-aggarwal-quantitative-aptitude-pdf"],"_links":{"self":[{"href":"https:\/\/www.reilsolar.com\/drive\/wp-json\/wp\/v2\/posts\/5989","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.reilsolar.com\/drive\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.reilsolar.com\/drive\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.reilsolar.com\/drive\/wp-json\/wp\/v2\/users\/41"}],"replies":[{"embeddable":true,"href":"https:\/\/www.reilsolar.com\/drive\/wp-json\/wp\/v2\/comments?post=5989"}],"version-history":[{"count":0,"href":"https:\/\/www.reilsolar.com\/drive\/wp-json\/wp\/v2\/posts\/5989\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.reilsolar.com\/drive\/wp-json\/wp\/v2\/media?parent=5989"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.reilsolar.com\/drive\/wp-json\/wp\/v2\/categories?post=5989"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.reilsolar.com\/drive\/wp-json\/wp\/v2\/tags?post=5989"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}