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Numerical Applications Test 1

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Numerical Applications Test 1
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Weekly Quiz Competition
  • Question 1
    1 / -0
    In how many ways can $$4$$ people be seated on a square table, one on each side? 
    Solution
    The number of circular arrangement for $$n$$ objects is $$(n-1)!.$$
     Similar to arrangements in a circle, there would be $$3!$$ ways possible of making $$4$$ people sit on a square table.
  • Question 2
    1 / -0
    The number of ways four boys can be seated around a round table in four chairs of different colours is:
    Solution
    Since the chair are of different colours, so you can treat it as linear permutation
    So number of ways will be $$4!=24$$
  • Question 3
    1 / -0
    If $$9^{th}$$ of the month falls on the day preceding Sunday then what day will be $$1^{st}$$ of the month fall?
    Solution
    $$9^{th}$$ of the month $$=$$ Saturday.
    Since day name repeats after $$7$$ days.
    $$\therefore 9-7=2^{nd}$$ of the given month is also Saturday.
    Then, $$1^{st}$$ of the given month would be Friday, option A.
  • Question 4
    1 / -0
    If the day before yesterday was Thursday, when will Sunday be?
    Solution
    If the day before yesterday was Thursday, then yesterday was Friday.
    Hence, today is Saturday.
    Then, tomorrow will be a Sunday.
  • Question 5
    1 / -0
    Seven people are seated in a circle. How many relative arrangements are possible?
    Solution
    The number of circular arrangement for $$n$$ objects is $$(n-1)!.$$
    There would be $$(7 - 1)! = 6!$$ arrangements possible.
  • Question 6
    1 / -0
    A work can be completed by $$40$$ workers in $$40$$ days. If $$5$$ workers leave every $$10$$ days, in how many days work will be completed?
    Solution
    A work can be completed by $$40$$ workers in $$40$$ days.
    Number of days required to complete the work by one worker $$=40\times 40=1600 $$ mandays(days per worker)
    Day----mandays----total mandays
    10             400              400
    20            350              750
    30            300              1050
    40            250              1300
    50            200              1500
    :
    After 50 days:
    1500 man-hrs have been accumulated, 15 men are on the job.
    Then, 100 more man-days required
    $$15 d = 100$$
    $$\therefore$$ $$d = 100/15=6.66$$
    $$\therefore$$ Total no. of days required $$=50+6.66=56.66$$ days.
    Hence, option C is correct.
  • Question 7
    1 / -0
    On what day of the week India will celebrate its Republic Day on 26th January, 2015 ?
    Solution
    Total number of odd days in first 2000 yeas = 0
    Number of odd days in 14 years = 3 x 2 + 11 x 1 = 17 days or  3 odd days
    Thus total number of odd days in 2015 = 26 odd days or 5 odd days

    Hence, total number of odd days = 5 + 3 = 8 days or 1 odd day
    Now, 0 odd days then Sunday, 1 odd day then Monday and so on
    Since, 1 odd days hence it is Monday.
  • Question 8
    1 / -0
    Monday, Tuesday etc. are numbered as 1,2 and so on. If 8th December 1994 was Monday, what would be the number of the day 8th January 1995 ? 
    Solution
    There are 31 days between 8th January 1995 and 8th December 1994.
    Thus, it has 4 weeks and 3 odd days.
    Since, 8th December is a Monday, hence, 8th January is a Thursday. or day 4.

  • Question 9
    1 / -0
    Mean of $$41, 39, 48, 52, 46, 62, 54, 40, 96, 52, 98, 49, 42, 52, 60 \ is\ 54.8$$
    Solution
    $$Mean$$ $$=$$ $$\dfrac{sum of all numbers}{no. of numbers}$$
    $$\therefore$$ $$mean$$ $$=$$ $$\dfrac{41+ 39 + 48 + 52 + 46 + 62 + 54 + 40 + 96 + 52 + 98 + 49 + 42 + 52 + 60}{15}$$ $$=$$ $$55.4$$
  • Question 10
    1 / -0
    Value of 0! is always 1.
    Solution
    we know $$1!=1$$
    Also 
    $$n!=n\times (n-1)\times (n-2)........3\times 2\times 1\\ n!=n\times (n-1)!\\ 1!=1(1-1)!\\ 1=1(0)!\\ 0!=1$$
    $$0! $$ is always $$1$$
    Hence its true that $$0!$$ is always $$1$$
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