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Numerical Applications Test 21

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Numerical Applications Test 21
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Weekly Quiz Competition
  • Question 1
    1 / -0
    Find the exact time between $$7$$ am and $$8$$ am when the two hands of a watch meet?
    Solution
    The minute hand covers $$6^o$$ in one minute.
    The hour hand covers  $$(\dfrac{1}{2})^o$$ in one minute

    Suppose both hands meet after $$x$$  minutes 
    so angle made by both with the $$12$$ is same 
    so $$6x=\dfrac{x}{2}+210$$
    $$\dfrac{11x}{2}=210$$
    $$x=38.18  min.$$
  • Question 2
    1 / -0
    Smt. Indira Gandhi died on 31st October, 1984. The day of the week was
    Solution
    1600 years contain 0 odd day, 300 years contain 1 odd day. Also, 83 years contain 20 leap years and 63 ordinary years and therefore (40+0) odd days i.e., 5 odd days.
    $$\therefore$$ 1983 years contain (0+1+5) i.e., 6 odd days.
    Number of days from Jan. 1984 to 31st Oct. 1984
    $$=(31+29+31+30+31+30+31+31+30+31)$$
    $$=305 days = 4 odd$$ days
    $$\therefore$$ Total number of odd days $$=6+4=3 $$ odd days.
    So, 31st Oct, 1984 was Wednesday.

  • Question 3
    1 / -0
    At what time between 9 to 10 will the hands of a watch be together?
    Solution
    To be together between 9 and 10, the minute hand has to gain 45 minute spaces.
    Now, 55 min spaces are gained in 60 minutes.
    $$\therefore 45$$ min spaces are gained in $$\left(\dfrac {60}{55}\times 45\right)$$ min or $$49\dfrac {1}{11}$$ min
    So, the hands are together at $$49\dfrac {1}{11}$$ minutes past $$9$$
  • Question 4
    1 / -0
    If we follows today's calendar, then the day on which first Republic day was celebrated, is
    Solution
    1600 years have $$0$$ odd days and $$300$$ years have 1 odd day.
    $$49$$ years contain $$12$$ leap year and $$37$$ ordinary years
    $$\therefore (24+37)$$ odd days.
    5 odd days thus 1949 years contain $$(0+1+5) = 6$$ odd days
    26 days of January contain 5 odd days.
    $$\therefore$$ Total odd days $$ = (6+5) =11 $$ or  4 odd days.
    So, the day was $$Thursday$$
  • Question 5
    1 / -0
    If the seventh day of a month is three days earlier than Friday, what day will it be on the nineteenth day of the month?
    Solution
    As mentioned, the seventh day of the month is three days earlier that Friday, which is Tuesday.
    So, the fourteenth day is also Tuesday and thus, the nineteenth day is Sunday.
  • Question 6
    1 / -0
    Following the current calendar, what will be the day of the week of $$1$$st January, $$2010$$?
    Solution
    Till $$2000$$, we have $$0$$ odd days.
    Till $$2009$$ we have $$11$$ odd days
    when diving $$11$$ by $$7$$, we get $$4$$ as remainder
    So calculating $$4$$ days from Monday we get Friday on 1st January, $$2010$$.

    Hence, option A.
  • Question 7
    1 / -0
    How many times in a day, the two hands of a clock coincide?
    Solution
    The hands of a clock coincide $$11$$ times in every $$12$$ hours (Since between $$11$$ and $$1$$, they coincide only once, i.e., at $$12$$ o'clock).
    $$AM$$
    $$12:00,1:05, 2:11, 3:16, 4:22, 5:27, 6:33, 7:38, 8:44, 9:49,10:55$$
    $$PM$$
    $$12:00,1:05, 2:11, 3:16, 4:22, 5:27, 6:33, 7:38, 8:44, 9:49,10:55$$
    The hands overlap about every $$65$$ minutes, not every $$60$$ minutes. 
    The hands coincide $$22$$ times in a day.
  • Question 8
    1 / -0
    If Monday falls on 4th April, 1998, then what was the day of 3rd November, 1987?
    Solution
    Counting the number of days after $$3$$rd November, $$1987$$, we have :
    Nov. Dec. Jan. Feb. March April
    days $$27 + 31 + 31 + 29 + 31 + 4$$
    $$= 153$$ days containing $$6$$ odd days.
    $$\therefore$$ $$6$$ days earlier to Monday is Tuesday.
    Hence option C is the correct answer.
  • Question 9
    1 / -0
    Find the day of the week on $$16^{th}$$ July, $$1776$$
    Solution
    $$16^{th}$$ July, $$1776$$ mean ($$1775$$ years $$+$$ $$6$$ months $$+$$ $$16$$ days)
    Now, $$1600$$ years have $$0$$ odd days.
    $$100$$ years have $$5$$ odd days.
    $$75$$ years contain $$18$$ leap years and $$57$$ ordinary years and therefore $$(36+57)$$ or $$93$$ or $$2$$ odd days.
    $$\therefore$$ $$1775$$ years given $$0+5+2=7$$ and so $$0$$ odd days,
    Also number of days from $$1^{st}$$ Jan. $$1776$$ to $$16^{th}$$ July, $$1776$$
    Jan. Feb. March April May June July
    $$31+29+31+30+31+30+16$$
    $$=198$$ days $$=28$$ weeks$$+$$ $$2$$ days $$=2$$ odd days
    $$\therefore$$ Total number of odd days $$=0+2=2$$.
    Hence the day on $$16^{th}$$ July, $$1776$$ was 'Tuesday'.
  • Question 10
    1 / -0
    A fort had provisions for  $$450$$ soldiers for  $$40$$ days. After  $$10$$ days, $$90$$ more soldiers came to fort. Find for how many days will the remaining provisions last, if consumed at the same rate ?
    Solution
    since $$450$$ soldiers had provision for $$40$$ days in a fort.

    After $$10$$ days: $$450$$ soldiers will remain with $$30$$ days of provision

    therefore for $$540$$ soldiers provision will decrease to $$x$$ days

    therefore By inverse proportion,
    $$x= \dfrac{450 * 30}{540}$$

    $$\therefore x=25$$ days
    Answer is $$25$$ days
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