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Numerical Applications Test 30

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Numerical Applications Test 30
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  • Question 1
    1 / -0
    If the day, two days after tomorrow be Thursday, what day would have been two days before yesterday ?
    Solution
    According to the question
    Tomorrow will be Tuesday
    Today is Monday
    Yesterday was Sunday
    Then Two days before yesterday was Friday.
  • Question 2
    1 / -0
    Find the day of the week on 26 January, 1950
    Solution
    Total number of odd days in first $$1600$$ years $$= 0$$
    Number of odd days in next $$300$$ years $$= 1$$
    Number of odd days in 49 years = $$12 \times 2 + 37 \times 1 = 61$$ days or $$5$$ odd days
    Thus total number of odd days in year $$1949 = 1 + 5 = 6$$ days
    Number of days in year $$1950 = 26 = 3 weeks + 5$$ days. 
    Thus, number of odd days in $$1950 = 5$$

    Hence, total number of odd days $$= 5 + 6 = 11 = 1$$ week + 4 days
    Now, 0 odd days then Sunday, 1 odd day then Monday and so on
    Since, 4 odd days hence it is Thursday.
  • Question 3
    1 / -0
    If $$5\,\text{January}\;1991$$ was a Saturday, what day of the week was $$3\,\text{March}\;1992$$ ?
    Solution
    Analysis:
    $$1991$$ is an ordinary year, and hence it has only one odd day. Thus, $$5\,\text{January}\;1992$$ was one day after Saturday, i.e., Sunday.
    Now, in January $$1992$$, there are $$26$$ days remaining $$(31-5=26)$$, i.e., $$5$$ odd days $$(26\div7=3\,\text{weeks and 5 odd days})$$. In February $$1992$$, there are $$29$$ days, i.e., $$1$$ odd day $$(29\div7=4\,\text{weeks and 1 odd day})$$.
    In March $$1992$$, there are $$3$$ odd days (we are calculating odd days only upto $$\text{3 March 1992}$$).
    Therefore, the total numbers of odd days after $$5\,\text{January}\;1992$$ till $$3\,\text{March}\;1992$$ is $$5+1+3=9 odd$$ days, which is equivalent to $$2$$ odd days$$(9\div 7= 1  week and 2  odd days)$$.
    Therefore, $$\text{3rd March 1992}$$ was $$2$$ days after Sunday, i.e., Tuesday.
  • Question 4
    1 / -0
    Ramesh travels $$760km$$ to his home partly by train and partly by car. He takes $$8hr$$, if he travels $$160km$$ by train and the rest by car. He takes $$12$$ minutes more, if he travels $$240km$$ by train and the rest by car. Find the speed of train and the car.
    Solution
    Let the speed of train be xkm/hr
    and the speed car be ykm/hr respectively.
    We know that $$Speed\quad =\dfrac { Distance }{ Time } \\ \Rightarrow Time\quad =\quad \dfrac { Distance }{ Speed }$$
    In Case 1 
    Distance travelled by train $$=$$ 160km
    Distance travelled by car $$=$$ (760-160)km $$=$$ 600km 
    Time taken by train+Time taken by car $$=$$ 8 hours
    Hence the equation becomes
    $$\dfrac { 160 }{ x } +\dfrac { 600 }{ y } \quad =\quad 8$$
    In case 2
    Distance travelled by train $$=$$ 240km
    Distance travelled by car $$=$$ (760-240)km = 520km
    Time taken by train+Time taken by car $$=$$ 8 hours+12 minutes
    Hence the equation becomes
    $$\dfrac { 240 }{ x } +\dfrac { 520 }{ y } \quad =\quad \dfrac { 41 }{ 5 } ......(2)$$
    Hence we get two equations
    $$\dfrac { 160 }{ x } +\dfrac { 600 }{ y } \quad =\quad 8.........(1)\\ \quad \dfrac { 240 }{ x } +\dfrac { 520 }{ y } \quad =\quad \dfrac { 41 }{ 5 } ......(2)$$
    $$Let\quad \dfrac { 1 }{ x\quad  } =p\quad and\quad \dfrac { 1 }{ y } =q$$
    Hence we get equations 
    160p + 600q $$=$$ 8........(3)
    240p + 520q $$=$$ 41/5...(4)
    Solving equations (3) and (4)we get p$$=$$1/80 and q $$=$$1/100
    $$\quad Hence\quad p=\dfrac { 1 }{ x } \\ \Rightarrow x\quad =\dfrac { 1 }{ p } \quad =80\\ Similary\quad y\quad =\quad \dfrac { 1 }{ q } \quad =\quad 100$$
    Solving we get x $$=$$ 80 and y $$=$$ 100
    hence, speed of train =80km/hr and speed of car =100km/hr.
  • Question 5
    1 / -0
    At what time are the hands of a clock together between 5 O'clock and 6 O'clock?
    Solution
    If the time is 5 hours and $$x$$ minutes after 5, when the two hands of the clock are same, then the angle formed by minute hand is: $$\cfrac{x}{60} \times 360 = 6x$$
    The hour hand moves $$\cfrac{1}{2}$$ degree each minute, then angle traced by hour hand is = $$150 + \cfrac{x}{2}$$
    Hence, $$6x = 150 + \cfrac{x}{2}$$
    $$150 = \cfrac{11x}{2}$$
    $$x = \cfrac{300}{11}$$
    $$x = 27 \cfrac{3}{11}$$ min. past 5
  • Question 6
    1 / -0
    $$A, B$$ and $$C$$ together can finish a piece of work in $$4$$ days. $$A$$ alone can do it in $$9$$ days and $$B$$ alone in $$18$$ days. How many days will be taken by $$C$$ to do it alone.
    Solution

  • Question 7
    1 / -0
    A and B can do a working days B and C can do the same work in 12 days. A, B and C together can finish it in 6 days. A and C will do it in -
    Solution
    $$\displaystyle \frac {1}{A} + \frac {1}{B} = \frac {1}{8}$$    ....(i)

    $$\displaystyle \frac {1}{B} +  \frac {1}{C}=\frac {1}{12}$$   ...(ii)

    $$\displaystyle \frac {1}{A}+\frac {1}{B}+\frac {1}{C}=\frac {1}{16}$$    ...(iii) 

    (iii) - (i)

    $$\displaystyle \frac {1}{C} = \frac {1}{6} - \frac {1}{8} = \frac {4-3}{24} \frac {1}{24}$$

    (iii) - (ii)

    $$\displaystyle \frac {1}{A} = \frac {1}{6} - \frac {1}{12} = \frac {2-1}{12} =\frac {1}{12}$$

    $$\displaystyle \frac {1}{A} + \frac {1}{C} = \frac {1}{24} + \frac {1}{12} = \frac {1+2}{24} = \frac {3}{24} = \frac {1}{8}  $$

    So A, C do in 8 days
  • Question 8
    1 / -0
    In U.P.on 17th Oct. 1996, the president rule was declared. Find the day of week on that date.
    Solution
    Formula to calculate day of week based on date given is
    $$=$$ (Year Code + Month Code + Century Code + Number Date -Leap Year Code) $$mod \ 7$$.
    Year code $$=$$ (YY+(YY$$\div4)) \ mod \ 7$$
                      $$=(96+24) \  mod \ 7$$
                      $$=1$$
    Month code $$=0$$
    Century code $$=0$$
    Day code $$=(1+0+0+17) \ mod \ 7$$
                     $$=18 (mod \ 7)$$
                     $$=4$$.
    Hence, the day is Thursday.
  • Question 9
    1 / -0
    Find the day of the week on 15 August, 1947
    Solution
    $$15$$th August $$1947$$ has $$1946$$ years and period from $$1$$st January $$1947$$ to $$15$$th August $$1947$$.
    Now, first $$1600$$ years have 0 days.
    Next $$300$$ years have $$1$$ odd day.
    $$46$$ years = $$11$$ leap years + $$35$$ non leap years = $$11\times 2 + 35 = 57$$days = $$1$$ odd day
    Number of days from $$1$$st Jan $$1947$$ to $$15$$th August $$1947$$ = $$227$$ days = $$3$$ odd days
    Hence, total number of odd days = $$ 0 + 1 + 1 + 3 = 5$$
    Hence, if $$0$$ odd day it is Sunday, $$1$$ odd day it is a Monday and so on.
    Hence $$5$$ odd days is a Friday.
  • Question 10
    1 / -0
    Find the day of the week on $$\text{15 August, 1947}$$
    Solution
    $$\Rightarrow\;1600$$ year has $$0$$ odd days .....(i)
    $$300$$ years have $$1$$ odd day .....(ii)
    $$46$$ years have $$11$$ ;eap years and $$35$$ ordinary years.
    $$\therefore\;(11\times2+35)=57\,\text{odd days}=1\,\text{odd day}$$
    Till $$\text{30 June}$$ there are $$6$$ odd days
    From $$\text{1st July to 15 August}$$ there are $$46$$ days, i.e., $$4$$ odd days
    $$\therefore\;0+1+1+3=5$$ odd days
    $$\therefore$$ the day will be Thursday.
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