Download Clock and Calender Questions for SNAP <\/a><\/p>\n Enroll to SNAP 2022 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>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 5)\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 6:\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 6)\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 Enroll to SNAP & NMAT 2022 Crash Course<\/a><\/p>\n Question 7:\u00a0<\/b>If 29 January 2003 is a Wednesday, then what day of the week will be 26 February 2005?<\/p>\n a)\u00a0Thursday<\/p>\n b)\u00a0Saturday<\/p>\n c)\u00a0Friday<\/p>\n d)\u00a0Sunday<\/p>\n 7)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n Given,\u00a029 January 2003 is a Wednesday<\/p>\n Number of days between 29 January 2003 and January 29 2004 = 365 days<\/p>\n Number of days between 29 January 2004 and January 29 2005 = 366 days<\/p>\n Number of days between 29 January 2005 and 26 February 2005 = 28 days<\/p>\n Number of days between 29 January 2003 and 26 February 2005 = 365 + 366 + 28 = 759 days<\/p>\n Number of odd days between 29 January 2003 and 26 February 2005 = Remainder of $\\frac{759}{7}$ = 3<\/p>\n $\\therefore\\ $The day on 26 February 2005 = Wednesday + 3 = Saturday<\/p>\n Hence, the correct answer is Option B<\/p>\n Question 8:\u00a0<\/b>What day of the week was 5 February 2008?<\/p>\n a)\u00a0Thursday<\/p>\n b)\u00a0Monday<\/p>\n c)\u00a0Tuesday<\/p>\n d)\u00a0Wednesday<\/p>\n 8)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n Odd days in year 2007 = 6 +\u00a02 = 8 Question 9:\u00a0<\/b>What day of the week was 31 January 2007?<\/p>\n a)\u00a0Monday<\/p>\n b)\u00a0Tuesday<\/p>\n c)\u00a0Thursday<\/p>\n d)\u00a0Wednesday<\/p>\n 9)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n Number of odd days in 2006 = 7 Question 10:\u00a0<\/b>What day of the week was 29 June 2010 ?<\/p>\n a)\u00a0Sunday<\/p>\n b)\u00a0Monday<\/p>\n c)\u00a0Wednesday<\/p>\n d)\u00a0Tuesday<\/p>\n 10)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n Odd day till 2009 = 7 +\u00a02 + 2 = 11 Take\u00a0 SNAP mock tests here<\/strong><\/span><\/a><\/p>\n Enrol to 10 SNAP Latest Mocks For Just Rs. 499<\/a><\/strong><\/span><\/p>\n Question 11:\u00a0<\/b>A watch reads 4:30. If the minute hand points East in which direction will the hour hand point?<\/p>\n a)\u00a0North-East<\/p>\n b)\u00a0North<\/p>\n c)\u00a0South-West<\/p>\n d)\u00a0south<\/p>\n 11)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n According to the question, the given information is represented as shown below<\/p>\n $\\therefore\\ $The hour hand is pointing towards North-East<\/p>\n Hence, the correct answer is Option A<\/p>\n Question 12:\u00a0<\/b>If it was a Saturday on 10 November 2018, what was the day of the week on 15 August 2017?<\/p>\n a)\u00a0Monday<\/p>\n b)\u00a0Tuesday<\/p>\n c)\u00a0Sunday<\/p>\n d)\u00a0Friday<\/p>\n 12)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n Given,\u00a010 November 2018 was Saturday<\/p>\n Number of days between 15 August 2017 and 15 August 2018 = 365 days<\/p>\n Number of days between 15 August 2018 and 10 November 2018 = 16 + 30 + 31 + 10 = 87 days<\/p>\n Number of days between 15 August 2017 and 10 November 2018 = 452 days<\/p>\n Number of odd days between 29 January 2003 and 26 February 2005 = Remainder of\u00a0$\\frac{452}{7}$ = 4<\/p>\n $\\therefore\\ $The day on 15 August 2017 = Saturday – 4 = Tuesday<\/p>\n Hence, the correct answer is Option B<\/p>\n Question 13:\u00a0<\/b>What was the day of the week on 26 January 2012 ?<\/p>\n a)\u00a0Sunday<\/p>\n b)\u00a0Saturday<\/p>\n c)\u00a0Thursday<\/p>\n d)\u00a0Tuesday<\/p>\n 13)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n 26 January 2012 = (2011 years + period between 1-1-2012 to 26-01-2012)<\/p>\n odd days in 2000 = 0<\/p>\n in remaining 11 years = (2 leap years + 9 normal years) = (2 * 2 + 9 * 1) = 13 – 7 = 6 odd days<\/p>\n Total number of days in period between 1-1-2012 to 26-01-2012 = 26\u00a0days = 3 weeks + 5 days = 5 odd days<\/p>\n Total number of odd days = (6 + 5) = 11 – 7 =\u00a04 days<\/p>\n Hence, given day is Thursday (0 odd day means sunday, 1 odd day means mon and so on).<\/p>\n Question 14:\u00a0<\/b>What was the day of the week on 15 August 2013?<\/p>\n a)\u00a0Thursday<\/p>\n b)\u00a0Monday<\/p>\n c)\u00a0Wednesday<\/p>\n d)\u00a0Tuesday<\/p>\n 14)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n 15th August 2013 = (2013 years + period between 1-1-2013 to 15-8-2013)<\/p>\n odd days in 2000 = 0<\/p>\n in remaining 13 years = (3 leap years + 9 normal years) = (3 * 2 + 9 * 1) = 15 – 14 = 1 odd days<\/p>\n Total number of days in\u00a0period between 1-1-2013 to 15-8-2013 = 227 days = 32 weeks + 3 days = 3 odd days<\/p>\n Total number of odd days = (1 + 3) = 4 days<\/p>\n Hence, given day is thursday (0 odd day means sunday, 1 odd day means mon and so on )<\/p>\n Question 15:\u00a0<\/b>If it was a Friday on 1 January 2016, what was the day of the week on 31 December 2016?<\/p>\n a)\u00a0Sunday<\/p>\n b)\u00a0Saturday<\/p>\n c)\u00a0Monday<\/p>\n d)\u00a0Friday<\/p>\n 15)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n Year 2016 was a leap year<\/p>\n Therefore, number of days in 2016 = 366 days = 52 weeks + 2 odd days<\/p>\n Now 1st january 2016 was Friday so, 30th december 2016 was also Friday<\/p>\n Hence, 31th december 2016 was Saturday.<\/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
\nthe minute hand of a clock when it is 6:44 p.m.?<\/p>\n
\n($\\because$ 2004 is the leap year and odd days till 2000 is 0.)
\nOdd days in 5 February = odd days in January\u00a0+ 5 = 3 +\u00a05 = 8
\nTotal odd days = 8 +\u00a08 = 16
\nOdd days\u00a0= 16 = 2 weeks + 2 odd days
\nSo day\u00a0of the week was Tuesday.
\n$\\therefore$ The correct answer is option C.<\/p>\n
\n($\\because$ number of\u00a0Leap year from 2000 to 2007 = 1(2004))
\nOdd days in 31 January = 31
\nTotal odd days = (31 + 7) = 38
\nOdd day = 38\/7 = 3 remaining
\nThe day of the week was Wednesday.
\n$\\therefore$ The correct answer is option D.<\/p>\n
\n($\\because$ Year 2004 and 2008 are leap year and odd days till 2000 = 0)
\nOdd days in 29 June = Odd days in (January + February +\u00a0March +\u00a0April +\u00a0May + June) =\u00a0\u00a03 + 0 +\u00a03 +\u00a02 +\u00a03 + 1 = 12
\nTotal odd days = 11 +\u00a012 = 23
\nOdd days = 23\/7 = 2 remaining
\nThe days\u00a0of the week was Tuesday.
\n$\\therefore$ The correct answer is option D.<\/p>\n![](\"https:\/\/cracku.in\/media\/uploads\/269871.png\")
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