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Numerical Applications Test 8

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Numerical Applications Test 8
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  • Question 1
    1 / -0
    What is the arithmetic mean of the progression $$11, 22, 33, 44, 55, 66, 77?$$
    Solution
    Using the formula for required arithmetic mean $$=\dfrac{\text{sum of the terms}}{\text{number of terms}}$$

    After substituing the values we get: $$=\dfrac{11+22+33+44+55+66+77}{7}$$
                                                    $$\quad \quad \quad =\dfrac{11(1+2+3+4+5+6+7)}{5}=\dfrac{11\cdot 28}{7}=11\cdot 4=44$$
  • Question 2
    1 / -0
    If 6th March, 2005 is Monday, what was the day of the week on 6th March, 2004?
    Solution
    The year 2004 is a leap year. So, it has 2 odd days.
    But, Feb 2004 not included because we are calculating from March 2004 to March 2005. So it has 1 odd day only.
    $$\displaystyle \therefore $$     The day on 6th March, 2005 will be 1 day beyond the day on 6th March, 2004.
    Given that, 6th March, 2005 is Monday.
    $$\displaystyle \therefore $$ 6th March, 2004 is Sunday (1 day before to 6th March, 2005).
  • Question 3
    1 / -0
    Tara's three bowling scores in a tournament were $$167, 178$$, and $$186$$. What was her average score for the tournament?
    Solution
    Three bowling scores of tournament are $$167, 178$$ and $$186$$.
    Average score will be $$=\dfrac { 167+178+186 }{ 3 } =177$$
  • Question 4
    1 / -0
    It was Sunday on Jan 1, 2006. What was the day of the week Jan 1, 2010?
    Solution
    On 31st December, 2005 it was Saturday.
    Number of odd days from the year 2006 to the year 2009 = (1 + 1 + 2 + 1) = 5 days.
    ∴∴ On 31st December 2009, it was Thursday.
    Thus, on 1st Jan, 2010 it is Friday.
  • Question 5
    1 / -0
    What was the day of the week on 28th May, 2006?
    Solution
    28 May, 2006 = (2005 years + Period from 1.1.2006 to 28.5.2006)
    Odd days in 1600 years = 0
    Odd days in 400 years = 0
    5 years = (4 ordinary years + 1 leap year) = (4 x 1 + 1 x 2)  6 odd days
    Jan. Feb. March April May
    (31 +  28  +  31   +   30   +   28 ) = 148 days
    $$\displaystyle \therefore $$ 148 days = (21 weeks + 1 day)  1 odd day.
    Total number of odd days = (0 + 0 + 6 + 1) = 7  0 odd day.
    Given day is Sunday.
  • Question 6
    1 / -0
    An application was received by inward clerk in the afternoon of the week day. Next day he forwarded it to the table of the senior clerk, who was on leave that day. The senior clerk next day evening put up the application to the desk officer. Desk officer studied the application and disposed off the matter on the same day i.e. Friday. Which day was the application received by the inward clerk ?
    Solution
    Desk officer received the application on Friday. Clearly, the application was forwarded to the table of the senior clerk on Thursday. So, the application was received by the inward clerk on Wednesday.
  • Question 7
    1 / -0
    $$2, 10, m, 12, 4$$
    A group of $$5$$ integers is shown above. If the average (arithmetic mean) of the numbers is equal to $$m$$, find the value of $$m$$.
    Solution
    We know the average of a group of numbers is the sum of the numbers divided by the number of numbers, we can make an equation:
    $$\Rightarrow \dfrac { 2+10+m+12+4 }{ 5 } =m$$
    $$\Rightarrow 28+m=5m$$
    $$\Rightarrow m-5m=-28$$
    $$\Rightarrow -4m=-28$$
    $$\Rightarrow m=7$$
  • Question 8
    1 / -0
    Today is Monday. After 61 days, it will be:
    Solution
    Each day of the week is repeated after 7 days.
    So, after 63 days, it will be Monday.
    $$\displaystyle \therefore $$     After 61 days, it will be Saturday.
  • Question 9
    1 / -0
    What will be the day of the week 15th August, 2010?
    Solution
    15th August, 2010 = (2009 years + Period 1.1.2010 to 15.8.2010)
    Odd days in 1600 years = 0
    Odd days in 400 years = 0
    9 years = (2 leap years + 7 ordinary years) = (2 x 2 + 7 x 1) = 11 odd days  4 odd days.
    Jan. Feb. March April May June July Aug.
    (31+28+31+30+31+30+31+15) = 227 days.
    $$\displaystyle     \therefore $$ 227 days = (32 weeks + 3 days)  3 odd days.
    Total number of odd days = (0 + 0 + 4 + 3) = 7 $$\displaystyle \equiv $$ 0 odd days.
    Given day is Sunday.
  • Question 10
    1 / -0
    If 1st October is Sunday, then 1st November will be
    Solution
    Clearly, 1st, 8th, 15th, 22nd, and 29th October are Sundays.
    So, 31st October is Tuesday.
    $$\therefore$$ 1st November will be Wednesday.
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