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Numerical Applications Test 28

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Numerical Applications Test 28
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  • Question 1
    1 / -0
    A booster pump can be used for filling as well as emptying a tank. The capacity of the tank is 24002400 m3\displaystyle m^{3}. The emptying capacity of the tank is 10m3\displaystyle m^{3}/min higher than its filling capacity and the pump needs 8 min lesser to empty the tank than it need to fill it. What is the filling capacity of the pump?
    Solution
    Let the filled capacity of the tank x  m3/minutex \  m^3/minute
    Then emptied capacity of the tank=(x+10)  m3/minute=(x+10) \  m^3/minute
    2400x2400x+10=8\therefore \dfrac{2400}{x}-\dfrac{2400}{x+10}=8
     2400(10x(x+10))=8\Rightarrow  2400\left(\dfrac{10}{x(x+10)} \right)=8
    x(x+10)=3000\Rightarrow x(x+10)=3000
    x=50  m3/min\Rightarrow x=50 \  m^3/min
  • Question 2
    1 / -0
    Pipes A and B can fill a tank in 55 and 66 hours respectively. Pipe C can empty it in 1212 hours. If all the three pipes are opened together then the tank will be filled in
    Solution
    Part filled by AA in 11 hour =15=\dfrac{1}{5}
    Part filled by BB in 11 hour =16=\dfrac1{6}
    Part emptied by CC in 11 hour =112=\dfrac1{12};
    Part filled by (A+BC)(A + B-C) in 11 hour =15+16112=1760=\dfrac1{5}+\dfrac1{6}-\dfrac1{12}=\dfrac{17}{60}
    Hence, three pipes together will fill the tank in 39173\dfrac{9}{17} hours.
  • Question 3
    1 / -0
    AA takes twice as much time as BB or thrice as much time as CC to finish a piece of work. Working together, they can finish the work in 22 days. BB  can do the work alone in
    Solution
    Suppose A,BA, B and CC take x,x2  x\,,\,\dfrac{x}{2}\,\, and x3\dfrac{x}{3} days respectively to finish the work.
    Then, (1x+2x+3x)\left (\dfrac{1}{x}+\dfrac{2}{x}+\dfrac{3}{x}\right)=12=\dfrac{1}{2}
      6x=12\therefore\,\,\dfrac{6}{x}=\dfrac{1}{2}
      x=12\therefore\,\,x=12
    So, BB takes 122=6\dfrac{12}{2}=6 days to finish the work.
  • Question 4
    1 / -0
    How many times are the hands of a clocks coincide in a day?
    Solution
    The hands of the clock makes 00 degrees with each other 1111 times in 1212 hours. 
    Therefore, in 2424 hours it would coincide 2222 times.
    Hence, option D is correct.
  • Question 5
    1 / -0
    If February 1, 1996 is a Thusrday, then what day is March 10, 1996 ?
    Solution
    \displaystyle \because 1996 is a leap year. 
    \displaystyle \therefore Days from 1 Feb to 10 March == 39 days == 4 odd days. 
    \displaystyle \therefore Thusrday + 3 odd days (\displaystyle \because 1 Feb is Thursday) ==  Sunday
  • Question 6
    1 / -0
    On what dates of October, 1975 did Tuesday fall ?
    Solution
    For determining the dates, we find the day on 1st Oct, 1975. 
    16001600 years have '0' odd days ............(A). 
    300300 years have '01' odd days ............(B). 
    7474 years have (18 leap years + 56 ordinary years) 
    2×18+1×56=922 \times 18 + 1 \times 56 = 92 odd days = '01' odd days.....(C) 
    Days from 1st January to 1st Oct., 1975 
    1st Jan - 30 June + 1st July to 1st Oct. 
    181+31+31+30+1=274181 + 31 + 31 + 30 + 1 = 274 days $$
    $$=\cfrac {274}{7} = '01'$$ odd days................... (D) 
    Adding (A) + (B) +(C) +(D) =0+01+01+01= 0 + 01 + 01 + 01 
    =03= '03' odd days 
    Ans. Wednesday( 1st Oct), hence 7th,14th,21st,28th7^{th}, 14^{th}, 21^{st},28^{th} Oct. will fall on Tuesday.
  • Question 7
    1 / -0
    Robert is travelling on his cycle and has calculated to reach point AA at 22 p.m.p.m. If he travels at 1010 kmphkmph, he will reach there at 1212 noon, if he travels at 1515 kmphkmph, at what speed must he travel to reach AA at 11 p.m.p.m.
  • Question 8
    1 / -0
    Find the day of the week on 26 January, 1950.
    Solution
    1600 years have '0' odd days ................ (1) 
    300 years have 1 odd day..........................(2) 
    49 years have 12 1eap years and 37 ordinary years. 
     (12×2+37)=617=5\displaystyle \therefore \quad \left( 12\times 2+37 \right) =\frac { 61 }{ 7 } =5odd years...................(3)
    26 January has 5 odd days................... (4) 
    On adding (1), (2), (3) & (4), 
    11 odd days =04 odd days  
      \displaystyle \therefore  
    day is Thursday. 
  • Question 9
    1 / -0
    Find the day of the week on 18 July, 1776 (leap year). 
    Solution
    Here 16001600 years have '0' odd day.................. (A)
    100100 years have '5' odd days................. (B).
    7575 years = (18 leap years + 57 ordinary years)
     =(18×2+57×1)= (18 \times 2 + 57 \times 1) 
    =93= 93 odd days 
    =(7×13+2)=2= (7 \times 13 + 2) = 2 odd days ....................(C) 
    Now, the number of days from 1st January to 18 July, 1776 
    =182+18=200= 182 + 18 = 200 days 
    =(28×7+4)= (28 \times 7 + 4) days =4= 4 odd days ....................(D) 
    Adding (A) + (B) +(C) +(D), 
    We get, 0+5+2+4=40 + 5 + 2 + 4 = 4 odd days
    Hence, the day is Thursday.
  • Question 10
    1 / -0
    Find the day of the week on 16 January, 1969
    Solution
    16001600 years have '0' odd day..................... (A) 
    300300 years have '1' odd day .......................(B) 
    6868 years have 1717 leap years and 5151 ordinary years. 
    Thus (17×2+51×1)=85(17 \times 2 + 51 \times 1)= 85 odd days 
     \displaystyle \cong  '01' odd day.................... (C) 
    16 January has =02= 02 odd days .........................(D) 
    Adding (A) + (B) +(C) +(D), 
    We get, 0+01+01+02=040 + 01 + 01 +02 = 04 odd days 
    Ans : Thursday
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