Self Studies
Self Studies
Self Studies
Self Studies

Numerical Applications Test 20

Result Self Studies

Numerical Applications Test 20
  • Score

    -

    out of -
  • Rank

    -

    out of -
TIME Taken - -
Self Studies

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Self Studies Self Studies
Weekly Quiz Competition
  • Question 1
    1 / -0
    If 1st October is a Sunday, then 1st November will be
    Solution
    1st, 8th, 15th, 22nd and 29th October are Sundays.
    So, 31st October is Tuesday and 1st November will be a Wednesday.
  • Question 2
    1 / -0
    If the $${14}^{th}$$ day after $$5^{th}$$ March is a Wednesday, then what day of the week will fall on $${10}^{th}$$ December of the same year?
    Solution
    On the $${14}^{th} $$ day after $$ 5^{th} $$ March, will be $$ 19^{th} $$ March which will be Wednesday

    Now, no. of days from $$ 19^{th} $$ March to $$ 10^{th} $$ December.
     $$ = 12 + 30 + 31 + 30 + 31 + 31 + 30+ 31 + 30 + 10 = 266 = 266 / 7 = 38$$

    Since, it is divisible by $$ 7 $$, so, it will be a Wednesday on  $$ 10^{th} $$ December.
  • Question 3
    1 / -0
    At what time are the hands of a clock together between 5 and 6?

    Solution
    Minute hand moves $$6^0$$ in one minute.
    Hour hand moves $$\dfrac12^0$$ in one minute.
    Total degree moved by hour hand in $$x$$ minutes past $$5\>$$ O'clock is $$150^0 + \dfrac x2^0$$
    Total degree moved by minute hand in $$x$$ minute is $$6x^0$$
    Now both have to coincide, that only happens when,
    $$150^0+\dfrac x2^0 = 6x^0$$

    $$\Rightarrow x = \dfrac {300}{11} = 27\dfrac3{11}$$ min. past $$5\>$$ O'clock

    Hence, option D is the correct answer.
  • Question 4
    1 / -0
    The attendance of a class of $$45$$ boys for $$10$$ days is given as $$40, 42, 30, 35, 45, 44, 41, 38, 44$$ and $$41$$, then the mean attendance of a class is
    Solution
    Mean attendance of the class for 10 days $$= \cfrac{\text{Total attendance}}{\text{Number of days}}$$
    $$=\cfrac{40 + 42 + 30 + 35 + 45 + 44+ 41 + 38+ 44 +41}{10}$$
    $$=\cfrac{400}{10}$$
    $$= 10$$
  • Question 5
    1 / -0
    Kate was born on Saturday, 22nd March 1982. On what day of the week was she 14 years 7 month and 8 days of age ?
    Solution
    One normal year has $$365$$ Days i.e $$52$$ weeks and $$1$$ day extra.

    So, if we go $$14$$ years ahead from 1982, we will reach to 1996. In this duration, we will have $$4$$ leaps years and a leap year gives us $$2$$ Extra days.

    So, calculating the extra days:
    $$10 \times 1 + 4 \times 2 = 18 $$ and $$ 18$$ mod $$7$$ $$= 4$$

    So, on 22nd March 1986, we will get a (Saturday + 4 days) Wednesday, 

    Plus additional $$7$$ months will constitute:
    22 nd April $$\rightarrow$$ $$31$$ days 
    22 nd May  $$\rightarrow$$ $$30$$ days 
    22 nd June $$\rightarrow$$ $$31$$ days 
    22 nd July $$\rightarrow$$ $$30$$ days 
    22 nd August $$\rightarrow$$ $$31$$ days 
    22 nd September $$\rightarrow$$ $$31$$ days 
    22 nd October $$\rightarrow$$ $$30$$ days 
    And 8 days will result in $$222$$ days, 

    So, $$ 222$$ mod $$7$$ $$=  5 $$
    which is $$31$$ weeks and $$5$$ Days (Wednesday +5 ) will result in $$Monday$$
  • Question 6
    1 / -0
    In U.P. on 17th October, 1996 President's rule was declared. Find the day of week on that date.
    Solution
    We use the concept of odd days, for such type of questions.
    The number of days more than the complete weeks are called odd days.
    $$1$$ ordinary year $$=365$$ days $$= ( 52$$ weeks $$+ 1$$ odd day$$)$$
    $$1$$ leap year $$= 366$$ days $$= (52$$ weeks $$+ 2$$ odd days$$)$$
    $$100$$ years $$= 76$$ ordinary years $$+ 24$$ leap years $$= 76(1)+24(2) = 124$$ odd days $$= 17$$ weeks $$+ 5$$ odd days $$= 5$$ odd days
    Number of odd days in $$200$$ years $$=5 \times 2=3$$ odd days
    Number of odd days in $$300$$ years $$= 5 \times 3 = 1$$  odd day
    Number of odd days in $$400$$ years $$= 5 \times 4+1= 0 $$ odd days 

    If number of odd days come as $$0$$, that means given date is on a Sunday, $$1$$ for Monday , $$2$$ for Tuesday, $$3$$ for Wednesday, $$4$$ for Thursday, $$5$$ for Friday and $$6$$ for Saturday.

    $$17$$ oct. $$1996$$ $$=$$ $$1995$$ years $$+ 9$$ months $$+ 17$$ days
    No. of odd days in $$1600$$ years $$ = 0$$
    No. of odd days in $$300$$ years $$= 1$$
    No. of odd days in $$95$$ years $$= 6$$
    $$\therefore $$ Total no. of odd days upto $$31$$st Dec$$'95$$ $$= 7 = 0$$ odd days
    Number of odd days in $$1996$$ till $$17$$ Oct $$= 31+29+31+30+31+30+31+31+30+17=291$$ days $$=41 $$ weeks $$+4$$ odd days $$=4$$ odd days
    $$\Rightarrow 17$$ Oct $$1996$$ is a Thursday.
  • Question 7
    1 / -0
    If February, 1, 1996 is Wednesday, what day is March 3, 1996?
    Solution
    1996 was a leap year. So we have 29 days in February, 1996. 
    So, February 29 will be a Wednesday if February 1 is a Wednesday. 
    Hence, March 3 will be a Saturday. 
  • Question 8
    1 / -0
    Find the Arithmetic Mean of the numbers 7, 6, 5, 9, 8, 0, 7
    Solution
    The Arithmetic Mean of numbers is calculated by using the below formula,

    A.M of numbers$$=\dfrac{\text{sum of the numbers}}{ \text{Total numbers}}$$

                                 $$=\dfrac{7+6+5+3+8+6+7}{7}$$

                                 $$=\dfrac{42}{7}$$
                  
                                 $$=6$$

    Option $$[B]$$ is correct.
  • Question 9
    1 / -0
    How many times in a day, the two hands of a clock coincide?
    Solution
    The hands coincide 11 times in every 12 hours (between 11 and 1 O'clock there is a common position at 12 O'clock). Hence, the hands coincide 22 times in a day.
  • Question 10
    1 / -0
    On January 12, 1980, it was a Saturday. The day of the week on January 12, 1979 was ____.
    Solution
    The year 1979 being an ordinary year, it has 1 odd day.
    So, the day on 12th January 1980 is one day beyond on the day on 12th January, 1979.
    But, January 12, 1980 being Saturday.
    $$\therefore$$ January 12, 1979 was Friday.
Self Studies
User
Question Analysis

  • Correct -

  • Wrong -

  • Skipped -

My Perfomance
  • Score

    -

    out of -
  • Rank

    -

    out of -
Re-Attempt Weekly Quiz Competition
Self Studies Get latest Exam Updates
& Study Material Alerts!
No, Thanks
Self Studies
Click on Allow to receive notifications
Allow Notification
Self Studies
Self Studies Self Studies
To enable notifications follow this 2 steps:
  • First Click on Secure Icon Self Studies
  • Second click on the toggle icon
Allow Notification
Get latest Exam Updates & FREE Study Material Alerts!
Self Studies ×
Open Now