Math Soultion-3

Hello, guys here is the set of solutions ,to yesterday's questions 

1) let the present ages of mther and daughter  be M and D
    M-5=4(D-5)  =>(1)
    M+15=2(D+15) => (2)

  solving (1) and (2) ,we get
 D= 15 years

2) Portion of work completed in 4 days = 4*1/8 = 1/2

   Portion of work to be completed in next 4 days

   = 4*1/12 = 1/3

   portion of work to be completed =

   = 1-(1/2+ 1/3)
  
   = 1/6

   Portion of work that B and C can complete in 1 day = 1/12 + 1/24 = 1/8
   Time taken to do 1/6 work by B and C  = (1/6)/(1/8)
      =  4/3 days
 Time taken to complete the entire work is ( 4 + 4 + 4/3 ) days
           = 9.33 days

3) equations is 8x^2 -22x +15 =0
   roots are given by
   (-b + root of (b^2 - 4*a*c))/2*a and (-b - root of (b^2 - 4*a*c))/2*a

   hence from the above equations roots are
   (22 + root of (484 -480 ))/16 and   (22 - root of (484 -480 ))/16
   => roots are 3/2 ,5/4
    LCM (3/2 ,5/4) = LCM(3,5)/HCF(2,4) = 15/2
    HCF(3/2 ,5/4) = HCF(3,5) /LCM(2,4) =1/4
   => LCM:HCF = (15/2)/(1/4) = 30:1

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Quantitative Maths- 3

Hello, guys here is the set of  few questions in maths ,the solutions will be posted tomorrow,try out the solutions and you can even post your answer's.  

1) Five years ago , the age of a mother was four times  the age of her daughter.
   Fifteen years hence , the mother's age will be twice the age of her daughter.
   Find the present age of the daughter.
a) 24 years
b) 20 years
c) 15 years
d) 10 years


2) A,B,C can complete a peice of work in 8, 12,  24 days resp. A worked for 4 days
   and then quit the job. B worked for 4 more days and then C joined him. In how
   many days can the work be completed?
a) 1.3 days
b) 2.3 days
c) 9.3 days
d) 9 days

3) Find the ratio of L.C.M and the H.C.F of the roots of the equation 8x^2 -22x +15 = 0.
a) 30:1
b) 20:1
c) 1:1
d) 15:1

 

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Math Solution-2

Hello, guys here is the set of solutions ,to yesterday's questions

1) 5/2+9/4+17/8+33/16+.... to n terms

   = 2+ 1/2 + 2 + 1/4 + 2 + 1/8 + 2+ 1/16 + ...

   = (2+2+2+....n terms) + (1/2+1/4+1/8+....n terms)
 
   =  2n + 1/2 (1-1/2^n)/(1-1/2)
 
   = 2n+1 -1/2^n


short cut :-
                substitue 1 in option one and see if it is 5/2
                similarly substitue 2 and add 5/2 and 9/4
                if it matches then it is d correct option else go to option two.


2) let the 1st term and the common difference (c.d) be a and d
 

    now, [(a+(2-1)d) + (a + (9-1)d)+ (a+ (18-1)d)] {since nth term in A.P is a + (n-1)d}

         = [(a+7d)+ (a+15d)] ,{given that sum of 2nd 9th 18th is equal to sum of 8th and 16th }

       => 3a + 26d = 2a + 22d

       => a + 4d = 0
 
     Hence 5th term is zero.


3) Here 777k + 65 is the number. 777 is divisible by 37
 
    Therefore reminder when divided by 37 is same as the reminder when divied by
    37. therefore the reminder is rem(65/37) which is 28.




4) (A/(x-1)) + (B/(x+2))=1

   when  A = 0 ,

    B/(x+2)= 1

   => x= B-2
 
   => only one root


   when B=0,
 
   A/(x+2) =1
 
   => x=A+1

   =>only one root

  when both are not equal to zero

 we get 2 distinct roots.

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Quantitative Maths-2

Hello, guys here is the set of  few questions in maths ,the solutions will be posted tomorrow,try out the solutions and you can even post your answer's.  

1) Find the sum of the first n terms of the series 5/2,9/4,17/8,33/16,....
 a) 2n+1-1/2^n
 b) 2n+1+1/2^n
 c) n+2-1/2^n
 d) 2n+1/2^n

2) The sum of the 2nd, 9th, and 18th terms of an arthematic progression (A.P)
   is equal to the sum of the 8th and the 16th terms of the A.P . Which terms of the
   series should necessarily be equal to zero?
a) 1st
b) 2nd
c) 3rd
d) none

3) A number , when divided by 777, leaves a reminder of 65.What is the reminder ,
   if the same number id divided by 37?
a) 19
b)  9
c) 28
d) none

4) The number of real roots of the quation (A/x-1) + (B /x+2) =1 ,when A and B are real numbers which are not equal to zero simultaneously is
a) zero
b) 1
c) 2
d) 1 or 2   

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Math solution-1

Hello, guys here is the set of solutions ,to yesterday's questions  ..

1) Let the cost of each pen, each eraser and each refil be Rs p, Rs e and Rs r resp then
 
   5p+7e+9r=46-(1)
   2p+5e+8r=36-(2)
   - -
   __________________
   3  2     (take diff of any two of the 3 variabels using trial and error  )

Now multiply (1) by 3 and (2) by 2 and then subtract smaller value 4m larger value
we get
   11(p+e+r)=66 =>p+e+r=6

2)
   concept of men-days is used.
   work done= men* number of days ->(1)

   Therefore from (1)
   
   work done = 10 *8 ->(2)

3)The given word is "LASER" .Alphabetical order of the letters of the given word is:
   A,E,L,R,S.

    Number of words beggining with A=4!=24
    Number of words beggining with E=4!=24
    Number of words beggining with LAE=2!=2
    Number of words beggining with  LAR=2!=2
    Number of words beggining with LASER=1
    Required word is LASER

4) First thing to remember in speed and distance is don't be frigtened by seeing the size of the problem.
    
  At the end of the day only formula in T&d is (distance= speed * time)

 Let the speed of the motorcyclist be ukmph and lenght of the train be L km.

 L 
    __________ =    4    ->(1)  (Time taken  is given by Lenght / Speed)
      45+u      (Bike and train in opposite directions  so sum of speeds)

 L
    __________  =   6              ->(2) (Time taken  is given by Lenght / Speed)
       45-5          (Cycle  and train in same  diection so difference of speeds)


solving we get u = 15.
   
  Therefore,      Rank of the word 'LASAR' =24+24+2+2+1=53 

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