Averages and Mixtures - Formulae, Shortcuts

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.

Mixtures

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

‘Averages’ is the term used for living things and ‘Mixtures’ is the term used for non-living things.

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

Solved Examples

1.       What is the average of 10, 20, 30, 40, 50?

Soln:

Method 1:

No. of values: 5

Avg = (10+20+30+40+50)/5 = 30

Method 2:

As the given values are in A.P., we can use shortcut as follows:

Avg = (10+50)/2 = 30

2.       If the Arithmetic mean of 10 terms is 20, find the sum of the terms?

Soln: Sum = avg * no. of units = 20 * 10 = 200

3.       If the average marks of 2 students is 80 and the 3^rd student with 68 marks joined, then find the average of 3 students?

Soln:

Total marks of 2 students = 80*2=160

Total marks of 3 students = 160+68 = 228

Avg of 3 students = 228/3 = 73 marks

4.        If the average marks of 2 students is 80 and when the 3^rd student joined, the average increased to 82. Find the marks of 3^rd student?

Soln:

Method1:

Total marks of 2 students = 80*2=160

Total marks of 3 students = 82*3 = 246

Marks of 3^rd student = 246-160=86marks

Method2 (by glance):

As the 3^rd student joined, the avg increased by 2 marks for 3 students => total 3*2 = 6marks

So the marks of 3^rd student = 80+6=86 marks

5.       If the average of marks of first 10 students of a class is 50, and the average of next 20 students is 75, find the average marks of the class if the total strength is 30?

Soln:

Here there are 2 sections of people. So we need to take weighted average.

Avg = (mp + nq)/(m + n)

ð  Avg = (10*50 + 20*75)/(10+20) = (500+1500)/30 = 2000/30

6.       If the average of marks of first 10 students of a class is 50, and the average of next 20 students is 75, find the average marks of the class if the total strength is 40?

Soln:

The details are given for first 10+20 = 30 students, but the class strength is 40.

So the class average cannot be determined as we don’t know the details of last 10 students.

7.       If the 2 persons average age is 25, find the avg age after 5 years?

Soln:

After 5 years, age of both persons will increase by 5 years.

So avg age will be increased by 5 years = 25 + 5 = 30 years

8.       If the 2 persons average age is 25, find the avg age before 5 years?

Soln:

Cannot be determined for the following reason:

Let present age of 2 persons be 46 and 4 years.

Before 5 years, 1^st person age = 46 – 5 = 41 years

                         2^nd person age = 4-5 = -1 which is not possible, i.e. 2^nd person is not yet born before 5 years. So, in this particular scenario, the avg age before 5 years cannot be determined.

9.       Find the average of 79, 87, 93, 82?

Soln:

Method1:

Avg =( 79+87+93+82)/4 = 341/4 = 85.25

Method2:

As the above given numbers are large no.s, instead of adding directly, we can follow as below:

Assume the avg as 80, and add or subtract each no. accordingly as follows:

New sub avg = (-1+7+13+2)/4 = 21/4 = 5.25

Therefore, avg = 80+5.25 = 85.25

Same is the shortcut for weighted averages.

10.   If the average of first half of class is 20marks and the average of remaining half of the class is 24, find the average of the class?

Soln: Avg = (avg1+avg2)/2 = (20+24)/2 = 22 marks

Note: As the no. of students is same in first and next half of the class, avg of the class cab be determined as the mean of two averages.

11.   If the average of boys is 20 marks and the average of girls is 24, find the combined average of boys and girls?

Soln:

Cannot be determined as the no. of boys and girls is unknown.

12.   If the ratio of averages of 10 boys and 20 girls is 3:2, find the combined average?

Soln:

Cannot be determined because of the following reason:

Let average of 10 boys be 3x

Average of 20 girls be 2x

Combined average = (3x + 2x)/30 -> cannot be determined as x is unknown.

Note: If the ratio of averages is given, then the combined average cannot be determined.

Time and Work Solved Problems

Difficulty Level - Medium

      

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

                                    = 1/15^th 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, 1/4^th 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