{"id":6020,"date":"2025-06-09T14:14:10","date_gmt":"2025-06-09T14:14:10","guid":{"rendered":"https:\/\/thecompanyboy.com\/?p=6020"},"modified":"2025-06-09T14:14:10","modified_gmt":"2025-06-09T14:14:10","slug":"rs-aggarwal-quantitative-aptitude-pdf-clocks-download","status":"publish","type":"post","link":"https:\/\/www.reilsolar.com\/drive\/rs-aggarwal-quantitative-aptitude-pdf-clocks-download\/","title":{"rendered":"RS Aggarwal Quantitative Aptitude PDF Free Download: CLOCKS"},"content":{"rendered":"

CLOCKS<\/strong><\/h1>\n

IMPORTANT FACTS<\/h2>\n

The Face or dial of a watch is a circle whose circumference is divided into 60\u00a0 equal parts, called\u00a0 minute spaces<\/em><\/strong>.<\/p>\n

A clock\u00a0 has two hands, the smaller one is called the hour hand <\/strong><\/em>or short<\/em><\/strong> hand<\/strong> <\/em>while the larger one is called the minute hand <\/em>or long hand.<\/em>.<\/strong><\/p>\n

    \n
  1. i) In 60 minutes, the minute hand gains 55 minutes on the hour hand.<\/li>\n
  2. ii) In every hour, both the hands coincide once.<\/li>\n<\/ol>\n

    iii) The hands are in the same straight line when they are coincident or opposite to each other.<\/p>\n

      \n
    1. iv) When the two hands are at right angles, they are 15 minute spaces apart.<\/li>\n<\/ol>\n

      v)When the hand’s are in opposite directions, they are 30 minute spaces apart.<\/p>\n

      \u00a0vi)Angle traced by hour hand in 12 hrs = 360\u00b0.<\/p>\n

      vii)Angle traced by minute hand in 60 min. = 360\u00b0.<\/p>\n

      Too Fast <\/em><\/strong>and Too Slow: If <\/em>a watch or <\/em>a clock indicates <\/em>8.15, when the correct time <\/em>, 8 is said to be <\/em>15 minutes too fast.<\/em><\/strong><\/p>\n

      On the other hand, if it indicates <\/em><\/strong>7.45, when the correct time is <\/em>8, it is said to be 15 minutes too <\/em><\/strong>slow.<\/em><\/strong><\/p>\n

      \u00a0<\/em><\/strong>SOLVED EXAMPLES<\/strong><\/p>\n

      Ex 1<\/strong>:Find the angle between the hour hand and the minute hand of a clock when 3.25.<\/p>\n

      Solution<\/strong>:angle \u00a0traced by the hour hand in 12 hours = 360\u00b0<\/p>\n

      Angle traced by it in three hours 25 min (ie) 41\/12 hrs=(360*41\/12*12)\u00b0 =102*1\/2\u00b0<\/p>\n

      angle traced by minute hand in 60 min. = 360\u00b0.<\/p>\n

      Angle traced by it in 25 min. = (360 <\/u>X 25 )\/60= 150\u00b0<\/p>\n

      Required angle = 1500 \u2013 102*1\/2\u00b0= 47*1\/2\u00b0<\/p>\n

      \u00a0<\/em>Ex 2:<\/strong>At what time between 2 and 3 o’clock will the hands of a clock be together?<\/em><\/p>\n

      \u00a0<\/strong>Solution:<\/strong> At 2 o’clock, the hour hand is at 2 and the minute hand is at 12, i.e. <\/em>they are 10 min\u00a0 spaces apart.<\/p>\n

      To be together, the minute hand must gain 10 minutes over the hour hand.<\/p>\n

      Now, 55 minutes are gained by it in 60 min.<\/p>\n

      10 minutes will be\u00a0 gained in (60 <\/u>x 10)\/55\u00a0 min. = 120\/11 min.<\/p>\n

      The hands will coincide at 120\/11 min. past 2.<\/p>\n

      \u00a0<\/em>Ex. 3.<\/strong> At what time between 4 and 5 o’clock will the hands of a clock be at right angle?<\/p>\n

      \u00a0<\/em>\u00a0\u00a0\u00a0\u00a0 Sol:<\/strong> At 4 o’clock, the minute hand will be 20 min. spaces behind the hour hand, Now, when the two hands are at right angles, they are 15 min. spaces apart. So, they are at right angles in following two cases.<\/p>\n

      \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Case I. When minute hand is 15 min. spaces behind the hour hand:<\/em><\/h3>\n

      In this case min. hand will have to gain (20 – 15) = 5 minute spaces. 55 min. spaces are gained by it in 60 min.<\/p>\n

      \u00a05 min spaces will be gained by it in 60*5\/55 <\/u>\u00a0min=60\/11min.<\/p>\n

      :. They are at right angles at 60\/11min. past 4.<\/p>\n

      Case II. When the minute hand is 15 min. spaces ahead <\/em>of the hour <\/em>hand:<\/h3>\n

      To be in this position, the minute hand will have to gain (20 + 15) = 35 minute spa’ 55 min. spaces are gained in 60 min.<\/p>\n

      35 min spaces are\u00a0 gained in (60 <\/u>x 35)\/55 min =40\/11<\/p>\n

      \u00a0 :. They are at right angles at 40\/11 min. past 4.<\/p>\n

      \u00a0<\/strong>Ex. 4. Find at what time between 8 and 9 o’clock will the hands of a <\/strong>clock being\u00a0 <\/strong>the same straight line but not together.<\/em><\/strong><\/p>\n

      \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Sol:<\/strong> At 8 o’clock, the hour hand is at 8 and the minute hand is at 12, i.e. the two hands_ are 20 min. spaces apart.<\/p>\n

      To be in the same straight line but not together they will be 30 minute spaces apart. So, the minute hand will have to gain (30 – 20) = 10 minute spaces over the hour hand.<\/p>\n

      55 minute spaces are gained. in 60 min.<\/p>\n

      10 minute spaces will be gained in (60 <\/u>x 10)\/55 min. = 120\/11min.<\/p>\n

      :. The hands will be in the same straight line but not together at 120\/11 min.<\/p>\n

      \u00a0<\/strong>Ex. 5. At what time between 5 and 6 o’clock are the hands of a clock 3minapart?<\/strong><\/p>\n

      . Sol<\/strong>. At 5 o’clock, the minute hand is 25 min. spaces behind the hour hand.<\/p>\n

      \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Case I. Minute hand is 3 min. spaces behind the hour hand.<\/em><\/p>\n

      In this case, the minute hand has to gain’ (25 – 3) = 22 minute spaces. 55 min. are gained in 60 min.<\/p>\n

      22 min. are gaineg in (60*22)\/55<\/u>min. = 24 min.<\/p>\n

      :. The hands will be 3 min. apart at 24 min. past 5.<\/p>\n

      \u00a0\u00a0\u00a0\u00a0 Case II. Minute hand is 3 min. spaces ahead <\/em>of the hour hand.<\/em><\/p>\n

      In this case, the minute hand has to gain (25 + 3) = 28 minute spaces. 55 min. are gained in 60 min.<\/p>\n

      \u00a028 min. are gained in\u00a0 (60 <\/u>x 28_)\/55=346\/11<\/p>\n

      The hands will be 3 min. apart at 346\/11 min. past 5.<\/p>\n

      Ex 6. Tbe minute hand of a clock overtakes the hour hand at intervals of 65 minutes of the correct time. How much a day does the clock gain or lose?<\/p>\n

      \u00a0 \u00a0<\/strong>\u00a0 \u00a0Sol:<\/strong> In a correct clock, the minute hand gains 55 min. spaces over the hour hand in 60 minutes.<\/p>\n

      To be together again, the minute hand must gain 60 minutes over the hour hand. 55 min. are gained in 60 min.<\/p>\n

      60 min are gained in\u00a0 60 <\/u>x 60 min =720\/11 min.<\/p>\n

      \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 55<\/p>\n

      But, they are together after 65 min.<\/p>\n

      Gain in 65 min \u00ad=720\/11-65 =5\/11min.<\/p>\n

      Gain in 24 hours =(5\/11 * (60*24)\/65)min =440\/43<\/p>\n

      The clock gains 440\/43 \u00a0<\/u>minutes in 24 hours.<\/p>\n

      Ex. 7. A watch which gains uniformly, is 6 min. slow at 8 o’clock in the morning Sunday <\/strong>and it is 6 min. 48 sec. fast at 8 p.m. on following Sunday. When was it correct<\/strong>?<\/strong><\/p>\n

      Sol.<\/strong> Time from 8 a.m. on Sunday to 8 p.m. on following Sunday = 7 days 12 hours = 180 hours<\/p>\n

      The watch gains (5 + 29\/5) min. or 54\/5 <\/u>min. in 180 hrs.<\/p>\n

      Now 54\/5\u00a0 min. are gained in 180 hrs.<\/p>\n

      5 min. are gained in (180 x 5\/54 <\/u>x 5) hrs. = 83 hrs 20 min. = 3 days 11 hrs 20 min.<\/p>\n

      Watch is correct 3 days 11 hrs 20 min. after 8 a.m. of Sunday.<\/p>\n

      It will be correct at 20 min. past 7 p.m. on Wednesday.\u00a0<\/strong><\/p>\n

      Ex 8. A clock is set right at 6 a.m. The clock loses 16 minutes in 24 hours. What will be the true time when the clock indicates 10 p.m. on 4th day?<\/strong><\/p>\n

      \u00a0<\/strong>Sol.<\/strong> Time from 5 a.m. on a day to 10 p.m. on 4th day = 89 hours.<\/p>\n

      \u00a0\u00a0\u00a0\u00a0\u00a0 Now 23 hrs 44 min. of this clock = 24 hours of correct clock.<\/p>\n

      356\/15 hrs of this clock = 24 hours of correct clock.<\/p>\n

      \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 89 hrs of this clock = (24 x 31556 <\/u>x 89) hrs of correct clock.<\/p>\n

      \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 = 90 hrs of correct clock.<\/p>\n

      \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 So, the correct time is 11 p.m.<\/p>\n

      Ex. 9. A clock is set right at 8 a.m. The clock gains 10 minutes in 24 hours will be the true time when the clock indicates 1 p.m. on the following day?<\/p>\n

      \u00a0<\/strong>Sol<\/strong>. <\/em>Time from 8 a.m. on a day\u00a0 1 p.m. on the following day = 29 hours.<\/p>\n

      \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 24 hours 10 min. of this clock = 24 hours of the correct clock.<\/p>\n

      145 \/6 \u00a0hrs of this clock = 24 hrs of the correct clock<\/p>\n

      29 hrs of this clock = (24 x\u00a0 6\/145 <\/u>x 29) hrs of the correct clock<\/p>\n

      = 28 hrs 48 min. of correct clock<\/p>\n

      The correct time is 28 hrs 48 min. after 8 a.m.<\/p>\n

      This is 48 min. past 12.<\/p>\n

      I can’t provide a free PDF download<\/strong> of RS Aggarwal\u2019s Quantitative Aptitude<\/strong> book as it is copyrighted material. However, I can help you with Clocks<\/strong> concepts from the book and provide key shortcuts and formulas for free.<\/p>\n

      \n

      \u00a0Important Formulas & Concepts for Clocks<\/strong><\/h3>\n

      Minute Spaces:<\/strong> A clock is divided into 60-minute spaces<\/strong>, and the hour hand moves 5-minute spaces per hour<\/strong>.
      \nAngle Between Hands Formula:<\/strong><\/p>\n

      \u03b8=\u2223(30H\u22125.5M)\u2223\\theta = \\left| (30H – 5.5M) \\right|<\/span>\u03b8<\/span>=<\/span><\/span>\u2223<\/span>(<\/span>30<\/span>H<\/span>\u2212<\/span>5.5<\/span>M<\/span>)<\/span>\u2223<\/span><\/span><\/span><\/span><\/span><\/span><\/p>\n

      where H = Hour<\/strong>, M = Minutes<\/strong>, and \u03b8 = Angle between hands<\/strong>.
      \nHands Coincide Every 65 5\/11 Minutes.<\/strong>
      \nHands Opposite (180\u00b0 Apart) Every 65 5\/11 Minutes.<\/strong>
      \nHands Perpendicular (90\u00b0 Apart) Twice Every Hour.<\/strong>
      \nGain\/Loss of Time by a Fast or Slow Clock<\/strong>:<\/p>\n

      Time\u00a0gained\u00a0or\u00a0lost=Total\u00a0minutes\u00a0in\u00a0a\u00a0day\u00d7Error\u00a0per\u00a0hour60\\text{Time gained or lost} = \\frac{\\text{Total minutes in a day} \\times \\text{Error per hour}}{60}<\/span>Time\u00a0gained\u00a0or\u00a0lost<\/span><\/span>=<\/span><\/span>60Total\u00a0minutes\u00a0in\u00a0a\u00a0day<\/span>\u00d7<\/span>Error\u00a0per\u00a0hour<\/span><\/span>\u200b<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/p>\n

      \n

      \u00a0Sample Problems & Solutions<\/strong><\/h3>\n

      Q1. At what time between 4 and 5 o’clock will the hands of the clock be at right angles?<\/strong>
      \nQ2. Find the angle between the hands of a clock at 3:20.<\/strong><\/p>\n

      Would you like detailed solutions<\/strong> or more practice questions<\/strong>? Let me know!<\/p>\n

      RS Aggarwal Quantitative Aptitude PDF Free Download: CLOCKS<\/h3>\n","protected":false},"excerpt":{"rendered":"

      CLOCKS IMPORTANT FACTS The Face or dial of a watch is a circle whose circumference is divided into 60\u00a0 equal parts, called\u00a0 minute spaces. A clock\u00a0 has two hands, the smaller one is called the hour hand or short hand while the larger one is called the minute hand or long hand.. i) In 60 […]<\/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-6020","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\/6020","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=6020"}],"version-history":[{"count":0,"href":"https:\/\/www.reilsolar.com\/drive\/wp-json\/wp\/v2\/posts\/6020\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.reilsolar.com\/drive\/wp-json\/wp\/v2\/media?parent=6020"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.reilsolar.com\/drive\/wp-json\/wp\/v2\/categories?post=6020"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.reilsolar.com\/drive\/wp-json\/wp\/v2\/tags?post=6020"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}