By using property

The average of n consecutive numbers is always middle number (where n is odd number)

Therefore, Average = 31. Hence (a).

The average of n consecutive numbers is always middle number (where n is odd number)

Therefore, Average = 31. Hence (a).

Find the average of cubes of first 5 prime numbers.

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).

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

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).

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?

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).

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

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).

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.

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).

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).

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).

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).

= \({(52*75)+(55*65)+(58*90) \over {52+55+58}}\)

= \({3900+3575+5220 \over {165}}\) = \({12695 \over {165}}\)

= 76.93 Hence (c).

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.

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

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?

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).

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).

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).

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:

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).

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?

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).

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?

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).

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?

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).