Sunday, January 26, 2014

Maths problem on Mixtures and easy solutions

Mixtures questions and its solutions

This is the extension of Averages which we have discussed earlier.

Formulae and shortcuts used to solve the following problems are discussed in the previous post.

Solved Problems
1. Let the cost of 2 quantities of rice be Rs.15 and Rs.19 per kg. Find the ratio of mixture which cost Rs.18 per kg?
Soln.:
          Method1:
          Let the quantity of rice of quality A (Rs.15/kg) be 'x'
          Let the quantity of rice of quality B (Rs.19/kg) be 'y'
          As per the problem, 15x+19y = 18(x+y)
          y = 3x => x/y = 1/3
          Therefore, the ratio of quantities mixed to make the quality of Rs.18 is 1:3.

          Method2: Using shortcut,
         
              Therefore, ratio is 1:3
 
2. Let the cost of 2 quantities of rice be Rs.15 (quality A) and Rs.19 (quality B) per kg. Find the quantity of rice of quality A that has to be mixed with 27kg of quality B to make the cost of rice as Rs.18?
Soln.:
          Method1:
          Let the quantity of rice of quality A (Rs.15/kg) be 'x'
          Let the quantity of rice of quality B (Rs.19/kg) be 'y'
          As per the problem, 15x+19y = 18(x+y)
          y = 3x => x/y = 1/3
          Therefore, the ratio of quantities mixed to make the quality of Rs.18 is 1:3.
          => 3 parts of quality B is to be mixed with 1 part of quality A 
          => 27kg of quality B is to be mixed with 9kg of quality A.
          Ans: 9kgs

          Method2: Using shortcut,
         
              Therefore, ratio is 1:3
               => 3 parts of quality B is to be mixed with 1 part of quality A 
               => 27kg of quality B is to be mixed with 9kg of quality A.
               Ans: 9kgs

3. Let the cost of 2 quantities of rice be Rs.15 and Rs.19 per kg. These 2 qualities of rice are mixed and sold at Rs.27 per kg of profit 50%. Find the ratio in which 2 qualities of rice mixed?
Soln.:
          Method1:
          Let the quantity of rice of quality A (Rs.15/kg) be 'x'
          Let the quantity of rice of quality B (Rs.19/kg) be 'y'
          Given Selling price, SP = 27 and the profit % is 50%
          We know that the profit%, p% = (SP-CP)/CP * 100
          => 50/100 = (27 - CP)/CP
          => CP = 18
          As per the problem, 15x+19y = 18(x+y)
          y = 3x => x/y = 1/3
          Therefore, the ratio of quantities mixed to make the quality of Rs.18 is 1:3.

          Method2: Using shortcut,
          Given Selling price, SP = 27 and the profit % is 50%
          We know that the profit%, p% = (SP-CP)/CP * 100
          => 50/100 = (27 - CP)/CP
          => CP = 18
         
              Therefore, the ratio of quantities mixed to make the quality of Rs.18 is 1:3.

4. A man purchased TV and Washing machine for Rs.30000. He sold the TV at 30% profit and washing machine at 60% profit. He makes overall profit of 50%. Then find for how much did he purchased TV and washing machine?
Soln:
          Method1:
          Let the cost price of TV be 'x'
          Let the cost price of washing machine be 'y'
          As per the problem,
          x + y = 30000 ------> Eq. 1
          SP of TV = 1.3x
          SP of washing machine = 1.6y
          1.3x + 1.6y = 30000 * 1.5
          1.3x + 1.6y = 45000 --------> Eq. 2
          On solving equations 1 and 2, x = 10000 and y = 20000
          Therefore, cost price of TV and washing machine are Rs.10000 and Rs.20000 respectively.

          Method2: Using shortcut,

         
              Therefore, ratio is 1:2. 3 parts is equivalent to 30000
              => Cost of TV is 10000 (1 part) and Cost of washing machine is 20000 (2 parts)

Averages and Mixtures problems | Formulae Shortcuts questions

Mixture and Alligation Shortcut Methods

 
Formulae:
1.       Average = sum of quantities/number of quantities
2.       Sum = avg * no. of quantities
3.       If the numbers are in A.P. (Arithmetic Progression), then the avg of numbers is given by
Avg = sum/n =[ (n/2)(a+l)]/n
                = (a+l)/2
Where  n – no. of quantities
                a – first term or value
                l – last term or value
4.       Weighted Average:
Section1:
                No. of quantities: m
                Avg of section1: p
Section2:
                No. of quantities: n
                Avg of section1: q
Avg of section1 and section2 = (m*p + n*q)/(m +n)

Points to remember/Shortcuts
1.       Average lies between minimum and maximum values.
a.       Avg > minimum value
b.      Avg < maximum value
2.       If every quantity is increased/decreased by ‘k’ value, then the avg also get increased/decreased by the same ‘k’ value.

time and work questions solutions

Time and work questions solutions 

Difficulty Level - Medium

      
      1.     A can work on 1km railway track in 1 day. In how many days, will he able to complete the work on 12km railway track?

Soln: no. of days = total work / work done in 1 day
Therefore, no. of days taken = 12/1 = 12 days


      2.     A can complete the work in 15 days. What fraction of work will be completed in 1 day?

Soln.:   Let the total work is 1 unit.
Work in 1day = total work/no. of days to complete
                                    = 1/15th of work

      3.     A can do a piece of work in 3 days and B can do a piece of work in 5 days. In how many days will the work be completed if both A and B work together?

Soln.: Using formula:
                  Work done by A in 1 day = 1/3
                  Work done by B in 1 day = 1/5
                  Total work done by A and B in 1 day = 1/3 + 1/5 = 8/15
                  Therefore, no. of days to complete work by A and B together = 1/(Total work) = 1/(8/15) = 15/8 days which is less than 3 and 5

    Using shortcut/analysis/assumption
                  Let us consider the total work be 15 units (LCM of 3 and 5)
                  So work done by A in 1 day = 15/3 = 5 units
                  Similarly work done by B in 1 day = 15/5 = 3 units
                  So total work done by A and B in 1 day = 5 + 3 = 8 units
                  Therefore, no. of days to complete total work i.e. 15 units = total work/work done in 1 day = 15/8 days
       
      Note:
                a.     Work done by A and B in 1 day will always be greater than that of A and B individually
          b. No. of days taken by A and B together will always be less than that of A and B individually

      4.     A can do a piece of work in 6 days, B can do a piece of work in 4 days and C can do a piece of work in 12 days. Find the no. of days to complete the work if A, B and C work together?

Soln.: Using formula:
                  Work done by A in 1 day = 1/6
                  Work done by B in 1 day = ¼
                  Work done by C in 1 day = 1/12
                  Total work done by A, B and C in 1 day = 1/6 + ¼ + 1/12 = 12/24 = 1/2
                  Therefore, no. of days to complete work by A, B and C together = 1/(Total work) = 1/(1/2) = 2 days which is less than 4, 6, 12

      Using shortcut/analysis/assumption
                  Let us consider the total work be 24 units (LCM of 4, 6, 12)
                  So work done by A in 1 day = 24/4 = 6 units
                  work done by B in 1 day = 24/6 = 4 units
                  work done by C in 1 day = 24/12 = 2 units
                  So total work done by A, B and C in 1 day = 6 + 4 + 2 = 12 units
                  Therefore, no. of days to complete total work i.e. 24 units = total work/work done in 1 day = 24/12 = 2 days

The above Note is valid here as well.

      5.     A can do a piece of work in 6 days and B can do a piece of work in 12. Find the no. of days to complete the work if A and B work alternatively?

Soln.: Using formula:
                  Work done by A in 1 day = 1/6
                  Work done by B in 1 day = 1/12
                  Total work done by A and B working 1 day each = 1/6 + 1/12 = 3/12 = ¼
                  Therefore, 1/4th of work is done in 2days.
                  No. of days to complete total work if A and B work alternatively = 1/((1/4)/2) = 8 days

      Using shortcut/analysis/assumption
                  Let us consider the total work as 12 units (LCM of 6, 12)
                  So work done by A in 1 day = 12/6 = 2 units
                  work done by B in 1 day = 12/12 = 1 unit
                  Total work done by A and B working 1 day each = 2 + 1 = 3 units in 2 days
                  Therefore, work done in 1 day = work/no. of days = 3/2 units
                        No. of days to complete work = total work/work in 1 day = 12/(3/2) = 8 days

      6.     30 men can complete a job in 40 days. Then 25 men can complete the same job in how many days?

Soln.: As per M1D1 = M2D2
               30 * 40 = 25 * x  => x = 30 * 40/25 = 48 days

      7.   30 men can complete 1500 units in 24 days working 6hrs a day. In how many days can 18 men can complete 1800 units working 8 hrs a day?



Soln.: As per the formula  (from my earlier blog), M1D1h1/W1 = M2D2h2/W2
          => 30*24*6/1500 = 18*x*8/1800
          => x = 36 days


      8.     A and B can do a work in 10 and 15 days respectively. Then combinedly A & B, in how many days the work will be completed?

      Soln.: As per the formula  (from my earlier blog), x*y/(x + y)
               A and B together can complete the work in 10 * 15/(10 + 15) = 6 days

      9.   A can do a work in 10 and, A and B together can do a work in 6 days. In how many days B can complete the same work?


      Soln.: As per the formula  (from my earlier blog), x*y/(x - y)
               B alone can complete the work in 10 * 6/(10 - 6) = 15 days


      10.  A is twice faster than B and B can complete in 12 days alone. Find the number of days to complete if A and B together work?

            Soln.: Given B works in 12 days
      A is twice faster than B => A takes 2 times less time than B
      Therefore, A completes work in 12/2 = 6 days
            A and B together can complete in 12 * 6/(12 + 6) = 4 days

 

Time And Work questions formula for placement

Time And Work questions formula for placement


1.     Work from Days:
If A can do a piece of work in n days, then A's 1 day's work =
  1
.
n

2.     Days from Work:
If A's 1 day's work =
1
,
then A can finish the work in n days.
n

3.     Ratio:
If A is thrice as good a workman as B, then:
Ratio of work done by A and B = 3 : 1.
Ratio of times taken by A and B to finish a work = 1 : 3.

4.     No. of days = total work / work done in 1 day


5.     Relationship between Men and Work

More men         ------- can do ------->             More work

Less men         ------- can do ------->             Less work


6.     Relationship between Work and Time

More work        -------- takes------>               More Time

Less work         -------- takes------>               Less Time


7.     Relationship between Men and Time

More men        ------- can do in ------->          Less Time

Less men          ------- can do in ------->          More Time


8.     If M1 persons can do W1 work in D1 days and M2 persons can do W2 work in D2 days, then 
        

9.     If M1 persons can do W1 work in D1 days for h1 hours and M2 persons can do W2 work in D2 days for h2 hours, then

 Note:  If works are same, then M1D1h1 = M2D2h2 


10.  If A can do a work in ‘x’ days and B can do the same work in ‘y’ days, then the number of days required to complete the work if A and B work together is




11.  If A can do a work in ‘x’ days and A + B can do the same work in ‘y’ days, then the number of days required to complete the work if B works alone is