1.3 Marks

Consider a sequence of seven consecutive integers. The average of thefirst five integers is n. The average of all the seven integers is

[CAT 2000]

1) n

2) n + 1

3) K × n, where K is a function of n

4) n + (2/7)

Solution:

Average of first five integers is n.

∴ The integers are (n – 2), (n – 1), n, (n +1), (n + 2).

∴ The last two numbers are (n + 3), (n + 4).

∴ The average of the seven numbers = (7n + 7)/7 = n + 1

Hence, option 2.

2.3 Marks

Three math classes; X, Y, and Z, take an algebra test.

The average score in class X is 83.

The average score in class Y is 76.

The average score in class Z is 85.

The average score of all students in classes X and Y together is 79.

The average score of all students in classes Y and Z together is 81.

What is the average for all three classes?

[CAT 2001]

1) 81

2) 81.5

3) 82

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Section I

4) 84.5

Solution:

Let there be x number of students in class X, y number of students in class

Y and z number of students in class Z.

∴ 83x + 76y = 79(x + y)

∴ 4x = 3y

Similarly,

76y + 85z = 81(y + z)

∴ 4z = 5y

∴ 20x = 15y = 12z

∴ x : y : z = 3 : 4 : 5

∴ The average of all the three classes

Hence, option 2.

3.3 Marks

A set of consecutive positive integers beginning with 1 is written on theblackboard. A student came along and erased one number.

What was the number erased?

[CAT 2001]

1) 7

2) 8

3) 9

4) None of these

Solution:

Let there were n consecutive integers starting with 1 in the original set.

∴ Average = 35, if n = 69

∴ Average = 35.5, if n = 70

However, as the new average has 17 in the denominator, we can say thatthe number of numbers in the new set (n − 1) is 68.

∴ n = 69

∴ 68 numbers that remained on the blackboard added up to 2408.

∴ The number that was erased was = 2415 − 2408 = 7

Hence, option 1.

4.3 Marks

A boy finds the average of 10 positive integers. Each integer contains twodigits. By mistake, the boy interchanges the digits of one number say ba

for ab. Due to this, the average becomes 1.8 less than the previous one.What was the difference of the two digits a and b?

[CAT 2002]

1) 4

2) 2

3) 6

4) 8

Solution:

Let the Arithmetic Mean of the 10 numbers be x and s be the sum of theremaining 9 numbers.

Interchanging the number ab with ba,

Subtracting (i) from (ii), we get,

10b + a − (10a + b) = 9(b − a) = 18

b − a = 2

Hence, option 2.

5.3 Marks

Ten years ago, the ages of the members of a joint family of eight peopleadded up to 231 years. Three years later, one member died at the age of60 years and a child was born during the same year. After another threeyears, one more member died, again at 60, and a child was born duringthe same year. The current average age of this eight member joint familyis nearest to:

[CAT 2007]

1) 23 years

2) 22 years

3) 21 years

4) 25 years

5) 24 years

Solution:

The sum of the ages of the members of the family ten years ago = 231

∴ The sum of the ages of the members of the family seven years ago =231 + (3 × 8) – 60 = 195

∴ The sum of the ages of the members of the family four years ago = 195+ (3 × 8) – 60 = 159

∴ The sum of the ages of the members of the family now = 159 + (4 × 8) =191

∴ Required average = 191/8 = 23.875 ≈ 24

Hence, option 5.