Averages Part 5

Problem 1: Average of n numbers is‘a’. The first number is increased by 2, the second one is increased by 4, the third one is increased by 8 and so on. The average of the new numbers is?

Solution:

Average of n numbers is a

Total =an+2+4+8+16

a=2, r=2, n=n

S_{n}=\frac{a\times (r^{n}-1)}{r-1} , r>1

\frac{a^{n}+\frac{2\times (2^{n}-1)}{2-1}}{n}

a+2\times \frac{2^{n-1}}{n}

Problem 2: In a class of 50 boys, the average height of 30 boys is 160 cm. If the average height of remaining boys is 100 cm, then what will be the average height of boys(in cm)?

Solution:

\frac{30\times 160+20\times 100}{50}

\frac{30\times 160}{50}+\frac{20\times 100}{50}=136

Problem 3: The average of the first three numbers is thrice the fourth number. If the average of all the four numbers is 10, then find the fourth?

Solution:

\frac{a+b+c}{3}=3d

a+b+c=9d

\frac{a+b+c+d}{4}=10

9d+d=4\times 10

10d=40

d=4

Problem 4: The average of some natural numbers is 15. If 30 is added to the first number and 5 is subtracted from the last number, the average becomes 17.5. Then the numbers of natural numbers are?

Solution:

15n+30-5 = 17.5n

25 = 2.5n

n = 10

Shortcut:

\frac{30-5}{2.5}=\frac{25}{2.5}=\frac{250}{25}=10

 

Leave a Reply

Your email address will not be published. Required fields are marked *