#### Averages

1) Find the average of 25, 27, 29, 31, 33, 35, 37.

Show Solution
By using property
The average of n consecutive numbers is always middle number (where n is odd number)
Therefore, Average = 31. Hence (a).

2)

Find the average of cubes of first 5 prime numbers.

Show Solution

Cubes of first 5 prime numbers are 8, 27, 125, 343 and 1331
Average = $${Sum of values \over {Number of values}}$$ =  $${8+27+125+343+1331 \over {5}}$$ =  $${1834 \over {5}}$$ = 366.8 .  Hence (b).

3)

The average age of 7 members of a committee is 32. What was the average age of the committee before 9 years?

Show Solution

By using property
If the average of m quantities is x and if a is added or subt racted from each given quantity , then average of m quantities becomes (x + a) or (x – a) respectively.
Therefore, Average = 32 – 9 = 23 .  Hence (d).

4)

The average temperature from Tuesday to Thursday is 38˚c and that from Wednesday to Friday is 34˚c. If the temperature on Tuesday is 27˚c, what is the temperature in ˚c on Friday?

Show Solution

Sum of the temperature from Tuesday to Thursday = 3 * 38 = 114
Sum of the temperature from Wednesday to Friday = 3 * 34 = 102
Tuesday – Friday = 12
Temperature on Tuesday = 27 ˚c
Friday’s  temperature = 15 ˚c .  Hence (c).

5)

A batsman makes a score of 73 runs in the 14th inning and thus increases his average by 4. Find his average after 14th inning.

Show Solution

Let the average after 14 th inning  be x.
Average after 13 th inning = x – 4
Sum = (x – 4) * 13
(x – 4) * 13 + 73 = 14x
x = 21 .  Hence (b).

6)

The average weight of Leena and Caroline is 63.5. The average weight of Kapoor and Leena is 57.5. The overall average weight of the 3 members is 70. Find the average weight of Kapoor and Caroline.

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L + C = 63.5 * 2 = 127 …..(i) K + L = 57.5 * 2 = 115 …..(ii) L + C + K = 70 * 3 = 210 …..(iii) Substitute equation (i) in (iii) 127 + k = 210 à k = 83 From (ii) 83 + L = 115 à L = 32 From (i) C = 95 Average weight of Kapoor and Caroline = $${K+C \over {2}}$$= $${83+95 \over {2}}$$ = 89. Hence (c).

7) Average cost of 8 Bananas, 3 Mangoes and 9 Apples is Rs.68. The average cost of 6 Bananas, 11 Grapes and 5 Apples is Rs.76. Find the total cost of 42 Bananas, 42 Grapes and 42 Apples.

Show Solution
Ave rage cost of 8 Bananas, 3 Mangoes and 9 Apples = Rs.68
Sum = 68 * 20 = 1360
A ver age cost of 6 Bananas, 11 Mangoes and 5 A pples =  Rs.76
Sum = 76 * 22 = 1672
Cost of 42 Bananas, 42 Grapes and 42 Apples = 3 * (1360 + 1672)
= 3 * 3032
= 9096 . Hence (b).

8) If the average marks of 3 classes of 52, 55 and 58 students are 75, 65 and 90 respectively. Find the average marks of all the students.

Show Solution
Average marks of all the students = $${Sum of marks \over {Number of students}}$$ =
= $${(52*75)+(55*65)+(58*90) \over {52+55+58}}$$
= $${3900+3575+5220 \over {165}}$$ =  $${12695 \over {165}}$$
= 76.93  Hence (c).

9)

In a Club, the average of a group of members is increased by 3 when 15 year old boy is replaced by a 45 year old lady. Find the total number of members in the group.

Show Solution

Average age = x, Number of members in group = n Sum = nx $${nx-15+45 \over {n}}$$ = x + 3 à n = 10. Hence (d).

10)

The average age of 9 members in a family is 72. If the age of the youngest member is 4, what Was the average age of the family at the time of birth of the youngest member?

Show Solution

Sum of the ages of 9 members = 72 *9 = 648
Sum of the ages at the time of birth of youngest member = 648 – (9*4) = 612
Average age of  family at the time of birth of youngest member =  $${612 \over {8}}$$ = 76.5 .  Hence (a).

11)There are 6 cans in a shop. The weight of the first can is 800kg. The weight of second can is 62.5% of the fourth can whose weight 25% more than the first can. The third can weight is 630kg and the sixth can weight is average of the second and fifth can. The fifth can weight is 20% of the sum of the first, second, third and fourth can weight. Find the difference in the average weight of the first five heaviestcans and the last five lightest cans.

Show Solution
Weight of the first can = 800 kg
Weight of the fourth can is  25% more than the first can = 1000 kg
Weight of the second can is  62.5% of the fourth can = 625 kg
Weight of the third can = 630 kg
Weight of the fifth can = 20% of (800 + 625 + 630 + 1000) = 611 kg
Weight of the sixth can = $${625+611 \over {2}}$$ = 618 kg
The average weight of five heaviest cans =  $${1000+800+630+625+618 \over {5}}$$
= $${3673 \over {5}}$$ = 734.6
The average weight of five lightest cans =  $${800+630+625+618+611 \over {5}}$$
= $${3284 \over {5}}$$ = 656.8
Difference = 734.6 – 656.8 = 77.8 .  Hence (d).

12)

Of the four numbers, the first is thrice the second and the second is thrice the third and fourth is two-third of the first number. The average of the reciprocal of the numbers is$${29 \over {144}}$$. The numbers are:

Show Solution

Let four numbers be a, b, c and d. a = 3b, b = 3c, d = $${2 \over {3}}$$a => a = 9c, d = 6c $${({1 \over {a}})+({1 \over {b}})+({1 \over {c}})+({1 \over {d}}) \over {4}}$$ = $${29 \over {144}}$$ $${bcd+acd+abd+abc \over {4abcd}}$$ = $${29 \over {144}}$$ Substitute a =9c, b=3c, d = 6c $${261 \over {648c}}$$ = $${29 \over {144}}$$à c= 2 Therefore, the four numbers are: 18, 6, 2 and 12. Hence (c).

13)

The height of a tree varies as the cube root of its age. When the age of the tree is 64 years and its height 70 feet, what will be the height of the tree at the age of 125?

Show Solution

Let height and age of tree be h and x respectively. h h =awhere a is constant 70 = a * à a = 17.5 Height at the age of 125 years = 17.5 * = 17.5 * 5 = 87.5Hence (a).

14)

Amul was asked to find the average of 10 two digit natural numbers. By mistake, during his calculations the digits of one of the numbers get reversed. Due to this, the average of the given set of numbers decreased by 5.4. What is the difference between the digits of the number?

Show Solution

Let the original number be 10x + y When the digits are reversed, the new number à 10y + x Decrease in average = 5.4 Sum of the 10 original numbers – sum of the 10 numbers with the reversed digits = 5.4 * 10 = 54 10x + y = 10y + x = 54 9x – 9y = 54 x – y = 6. Hence (b).

15)

The average price of 20 transistors is Rs.14 while the average price of 18 of these transistors is Rs.13. Of the remaining two transistors, if the price of one transistor is55.55% more than the price of the other, what is the price of each of these two transistors?

Show Solution

Total cost of 20 transistors = 20 * 14 = Rs.280
Total cost of 18 transistors = 18 * 13 = Rs.234
The cost of 2 transistors = 280 – 234 = Rs.46
The price of one transistor is 55.55 %  more than the price of the other transistor
a + b = 46
$$(a+{5 \over {9}}$$ a ) +a = 46
a = 18 , b = 28 .  Hence (d).

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