Download Clock Questions for XAT PDF \u2013 XAT Clock questions pdf by Cracku. Practice XAT solved Clock Questions paper tests, and these are the practice question to have a firm grasp on the Clocks topic in the XAT exam. Top 20 very Important Clock Questions for XAT based on asked questions in previous exam papers. \u00a0The XAT question papers contain actual questions asked with answers and solutions.<\/p>\n
Download Clocks Questions for XAT<\/a><\/p>\n Enroll to XAT 2023 Crash Course<\/a><\/p>\n Question 1:\u00a0<\/b>An alarm was set at 11 AM on Monday on a clock that was set correctly at 1 AM on Sunday, but the clock started gaining 20 seconds every 24 hours. What was the actual time when the alarm went off?<\/p>\n a)\u00a010:31:40 AM<\/p>\n b)\u00a010:51:20 AM<\/p>\n c)\u00a010:30:00 AM<\/p>\n d)\u00a011:30:00 AM<\/p>\n 1)\u00a0Answer\u00a0(E)<\/strong><\/p>\n Solution:<\/b><\/p>\n Total time between sunday at 1 AM to monday at 11 AM =24hours + 10hours = 34 hours<\/p>\n Now total gain in seconds in 34 hours = $\\frac{20}{24}$\u00d734= $\\frac{85}{3}$<\/p>\n The actual time at which alarm went off is<\/p>\n =10:59:(60-$\\frac{85}{3}$)<\/p>\n =10:59:$\\frac{95}{3}$<\/p>\n =10:59:32<\/p>\n Question 2:\u00a0<\/b>How much does a watch lose per day if hand coincides every 64 minutes?<\/p>\n a)\u00a0$17\\frac{5}{11}$<\/p>\n b)\u00a0$32\\frac{8}{11}$<\/p>\n c)\u00a0$\\frac{32}{11}$<\/p>\n d)\u00a0$\\frac{16}{11}$<\/p>\n 2)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n In 24 hours the hands of hour and minute coincide 22 times. 24 hours equals 24*60=1440 minutes.<\/p>\n 1440\/22 = 65 and 5\/11 minutes.=> every 65 and 5\/11 min, they coincide.<\/p>\n Now the clock in question coincide every 64 minutes; therefore 24 hours in your clock means 64*22 minutes or 1408 minutes.<\/p>\n So your clock should lose 1440-1408=32 minutes.<\/p>\n Otherwise, the clock gains 32 minutes as it is faster than usual.<\/p>\n Question 3:\u00a0<\/b>Find the angle between the hands at 3:30 p.m.<\/p>\n a)\u00a0$120^\\circ$<\/p>\n b)\u00a0$75^\\circ$<\/p>\n c)\u00a0$90^\\circ$<\/p>\n d)\u00a0$105^\\circ$<\/p>\n 3)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n At 3:30\u00a0, the minute hand is at number 6. At 3:00 the hour hand<\/b>\u00a0was at number 3.<\/p>\n total 360 degree which is divided by 12 parts<\/p>\n hence , the angle between two number is 30 degree<\/p>\n so,<\/p>\n The angle between 3 and 6 is 90 degree\u00a0but hour has moved 15 degree.(between 3 and 4 )<\/p>\n Hence the angle betweenthe two is ( 90 -15) = 75 degree.<\/p>\n Question 4:\u00a0<\/b>What is the obtuse angle formed by the hands of a clock when the time in the clock is 2:30?<\/p>\n a)\u00a0$95^\\circ$<\/p>\n b)\u00a0$120^\\circ$<\/p>\n c)\u00a0$105^\\circ$<\/p>\n d)\u00a0$165^\\circ$<\/p>\n 4)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n Angle between the hands of a clock is given by the formula $\\dfrac{11}{2}H – 30M$ or $30M – \\dfrac{11}{2}H$ where H is hours and M is minutes. Question 5:\u00a0<\/b>A watch loses 5 minutes every hour and was set right at 6 a.m. on a Monday. When will it a)\u00a06 a.m. on next Sunday<\/p>\n b)\u00a03 a.m. on next Monday<\/p>\n c)\u00a03 a.m. on next Sunday<\/p>\n d)\u00a06 a.m. on next Monday<\/p>\n 5)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n For the watch to show the correct time again, it should lose 12 hours. So , the answer would be option a)\u00a06 a.m. on next Sunday.<\/p>\n Question 6:\u00a0<\/b>What will be the measure of the acute angle formed between the hour hand and the minute hand at 6:43 a.m.?<\/p>\n a)\u00a0$\u00a021.5^\\circ $<\/p>\n b)\u00a0$ 78^\\circ $<\/p>\n c)\u00a0$ 56^\\circ $<\/p>\n d)\u00a0$ 56.5^\\circ $<\/p>\n 6)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n A clock is a circle, and a circle always contains 360 degrees. Since there are 60 minutes on a clock, each minute mark is 6 degrees.<\/p>\n $\\frac{360^\\circ total}{60 minutes total}=6^\\circ\u00a0per minute$<\/p>\n The minute hand on the clock will point at 43 minutes, allowing us to calculate it’s position on the circle.<\/p>\n (43 min)(6)=$258^\\circ$<\/p>\n Since there are 12 hours on the clock, each hour mark is 30 degrees.<\/p>\n $\\frac{360^\\circ total}{12 hours total}=30^\\circ per hour$<\/p>\n We can calculate where the hour hand will be at 6:00.<\/p>\n $(6 hr)(30)=180^\\circ$<\/p>\n However, the hour hand will actually be between<\/em> the 6 and 7, since we are looking at 6:43 rather than an absolute hour mark. 43 minutes is equal to $\\frac{43}{60}$th of an hour. Use the same equation to find the additional position of the hour hand.<\/p>\n $180^\\circ + \\frac{43}{60} \\times 30 = 201.5^\\circ$<\/p>\n We are looking for the acute angle between the two hands of the clock. The will be equal to the difference between the two angle measures.<\/p>\n Required answer = $258^\\circ – 201.5^\\circ = 56.5^\\circ$<\/p>\n So, the answer would be option b)$ 56.5^\\circ $.<\/p>\n Take XAT 2023 Mock Tests<\/a><\/p>\n Question 7:\u00a0<\/b>What is the measure of the smaller of the two angles formed between the hour hand and a)\u00a0$\u00a062.5^\\circ$<\/p>\n b)\u00a0$\u00a062^\\circ $<\/p>\n c)\u00a0$ 84^\\circ $<\/p>\n d)\u00a0$ 83.5^\\circ $<\/p>\n 7)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n A clock is a circle, and a circle always contains 360 degrees. Since there are 60 minutes on a clock, each minute mark is 6 degrees.<\/p>\n $\\frac{360^\\circ total}{60 minutes total}=6^\\circ per\u00a0 minute$<\/p>\n The minute hand on the clock will point at 44 minutes, allowing us to calculate it’s position on the circle.<\/p>\n (44 min)(6)=$264^\\circ$<\/p>\n Since there are 12 hours on the clock, each hour mark is 30 degrees.<\/p>\n $\\frac{360^\\circ total}{12 hours total}=30^\\circ per\u00a0 hour$<\/p>\n We can calculate where the hour hand will be at 6:00.<\/p>\n $(6 hr)(30)=180^\\circ$<\/p>\n However, the hour hand will actually be between<\/em> the 6 and 7, since we are looking at 6:44 rather than an absolute hour mark. 44 minutes is equal to $\\frac{44}{60}$th of an hour. Use the same equation to find the additional position of the hour hand.<\/p>\n $180^\\circ + \\frac{44}{60} \\times 30 = 202^\\circ$<\/p>\n We are looking for the smaller angle between the two hands of the clock. The will be equal to the difference between the two angle measures.<\/p>\n Required answer = $264^\\circ – 202^\\circ = 62^\\circ$<\/p>\n So, the answer would be option b)$ 62^\\circ $.<\/p>\n Question 8:\u00a0<\/b>What would be the smaller of the two angles formed by the hour hand and the minute hand at 3:47 p.m.?<\/p>\n a)\u00a0$168.5^{0}$<\/p>\n b)\u00a0$162^{0}$<\/p>\n c)\u00a0$166.5^{0}$<\/p>\n d)\u00a0$165^{0}$<\/p>\n 8)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n The minute hand moves 6 degrees per minute. (That\u2019s 360 degrees divided by 60 minutes)<\/p>\n The hour hand moves one-twelfth that, or $\\frac{1}{2}$ degree per minute.<\/p>\n So the angle between the minute and hour hand increases by 5.5 degrees every minute.<\/p>\n The minute and hour hand start together at noon.<\/p>\n It takes 227 minutes(60+60+60+47) to get to 3:47 . So in this time, the angle between the hour and minute hand has increased $227\\times 5.5$ degrees.<\/p>\n $227\\times 5.5$= 1248.5 degrees.<\/p>\n Which is the same as 168.5 degrees.(1 cicrle is 360 so in 1248.5 deg\u00a0 there are 3 circle and 168.5 deg left)<\/p>\n Question 9:\u00a0<\/b>In a week, how many times are the hands of clock at right angles with each other?<\/p>\n a)\u00a0308<\/p>\n b)\u00a044<\/p>\n c)\u00a024<\/p>\n d)\u00a0154<\/p>\n 9)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n If you switch to a rotating coordinate system in which the hour hand stands still, then the minute hand makes only 11 revolutions, and so it is at right angles with the hour hand 22 times<\/strong>. In a 24 hour day you get $2\\times 22= 44$<\/p>\n so in weekthere are 7 days<\/p>\n so,<\/p>\n $7\\times 44= 308 times$<\/p>\n Question 10:\u00a0<\/b>What will be the acute angle between the hour-hand and the minute-hand at 2:13 p.m?<\/p>\n a)\u00a0$16.5^\\circ$<\/p>\n b)\u00a0$18^\\circ$<\/p>\n c)\u00a0$13.5^\\circ$<\/p>\n d)\u00a0$12^\\circ$<\/p>\n 10)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n angle covered by minute hand in 1 minute is 6degree and angle covered by hour hand in 1minute is $\\frac {1}{2}$ degree.<\/p>\n So, angle covered by minute hand at 13minutes = $13\\times 6degree$ = 78degree.<\/p>\n Hence, at 2hours 13minutes\u00a0 = 133 x $\\frac {1}{2}$degree = 66degree.<\/p>\n Hence, the angle between the hour and minute hand at 2.13PM = 78degree – 66degree = $12^\\circ$<\/p>\n Question 11:\u00a0<\/b>What is the measure of the smaller of the two angles formed between the hour hand and the minute hand of a clock when it is 5:49 p.m.?<\/p>\n a)\u00a0$119^\\circ$<\/p>\n b)\u00a0$119.5^\\circ$<\/p>\n c)\u00a0$120^\\circ$<\/p>\n d)\u00a0$120.5^\\circ$<\/p>\n 11)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n The hour hand moves 360 degrees every 12 hours. At 5:49, its angle is $(5 + \\frac{49}{60}) \\times\\frac{ 360}{12} = 174.5 degrees$<\/p>\n The minute hand moves 360 degrees each 60 minutes, so at 15 minutes past the hour it has moved $\\frac{49}{60} \\times 360 = 294 degrees.$<\/p>\n Thus, the difference between the two hands is 294 – 174.5 = 119.5 degrees.<\/p>\n
\nHere, Given time = 02 : 30, H = 2 and M = 30.
\nAngle = $\\dfrac{11}{2} \\times 30 – 30 \\times 2 = 165 – 60 – 105^\\circ$<\/p>\n
\nshow the correct time again?<\/p>\n
\nIt loses 5 minutes in 1 hour.
\n\u21d2 It loses 1 minute in 12 minutes.
\n\u21d2 It will lose 12 hours (or 720 minutes) in 720 \u00d7 12 minutes = 8640 minutes = 144 hours = 6 days.
\n\u21d2Thus, the clock will show the correct time again at 6am on next Sunday.<\/p>\n
\nthe minute hand of a clock when it is 6:44 p.m.?<\/p>\n