Practice New average from error - quantitative aptitude Online Quiz (set-1) For All Competitive Exams

Q-1)   The average marks obtained by 22 candidates in an examination are 45. The average marks of the first 10 candidates are 55 and those of the last eleven are 40. The number of marks obtained by the eleventh candidate is

(a)

(b)

(c)

(d)

Explanation:

Marks obtained by eleventh candidate

= 22 × 45 – (10 × 55 + 11 × 40)

= 990 – (550 + 440)

= 990 – 990 = 0


Q-2)   A student finds the average of ten 2-digit numbers. While copying numbers, by mistake, he writes one number with its digits interchanged. As a result his answer is 1.8 less than the correct answer. The difference of the digits of the number, in which he made mistake, is

(a)

(b)

(c)

(d)

Explanation:

Difference in average = 1.8

∴ Difference between the number and the number formed by interchanging the digits

= 1.8 × 10 = 18

(Since 53 – 35 = 18)

∴ Number = 35

∴ Difference of digits = 5 – 3 = 2


Q-3)   The average of 9 integers is found to be 11. But after the calculation, it was detected that, by mistake, the integer 23 was copied as 32, while calculating the average. After the due correction is made, the new average will be

(a)

(b)

(c)

(d)

Explanation:

Sum of 9 integers = 9 × 11 = 99.

New average = ${90+ 23- 32}/9$ = $90/9$ =10

Aliter : Using Rule 26,

The correct average will be = m +${\text"(a-b)"}/ \text"n"$.

Here, n = 9, m = 11

a = 23, b = 32

Correct mean = m + ${\text"(a - b)"}/\text"n"$

= 11 +${(23 -32)}/9$

= 11 +${(-9)}/9$

= 11 – 1 = 10


Q-4)   The average of six numbers is 20. If one number is removed, the average becomes 15. What is the number removed ?

(a)

(b)

(c)

(d)

Explanation:

Required number = sum of six numbers – sum of five numbers

= 6 × 20 – 15 × 5

= 120 – 75 = 45


Q-5)   Mean of 10 numbers is 30. Later on it was observed that numbers 15, 23 are wrongly taken as 51, 32. The correct mean is

(a)

(b)

(c)

(d)

Explanation:

Difference = 15 + 23 – 51 – 32 = –45

∴ Correct average = 30 -$45/10$ = 25.5

Aliter : Using Rule 27,

The correct average = m +${\text"(a+b-p-q)"}/ \text"n"$.

Here, n = 10, m = 30

a = 15, b = 23

p = 51, q = 32

Correct Average

= m + ${\text"(a + b - p - q)"}/ \text"n"n$

= 30 + ${(15 + 23 - 51 – 32)}/10$

= 30 + $({38-83}/10)$

= 30 - $45/10$

= 30 – 4.5 = 25.5


Q-6)   The average weight of 20 students in a class is increased by 0.75 kg when one of the students weighing 30 kg is replaced by a new student. Weight of the new student (in kg) is :

(a)

(b)

(c)

(d)

Explanation:

Required answer = 30 + 20 × 0.75

= 30 kg + 15 kg = 45 kg

Aliter : Using Rule 18,

If in the group of N persons, a new person comes at the place of a person of 'T’ years, so that average age,

increases by 't’ years

Then, the age of the new person = T + N.t.

Here, N = 20, T = 30, t = 0.75

Weight of New student = T + Nt

= 30 + 20 × 0.75

= 30 + 15 = 45 kg


Q-7)   The average of 9 observations was found to be 35. Later on, it was detected that an observation 81 was misread as 18. The correct average of the observations is :

(a)

(b)

(c)

(d)

Explanation:

Correct sum of 9 observations

= 9 × 35 – 18 + 81

= 315 + 63 = 378

∴ Required correct average = $378/9$ = 42


Q-8)   The average of 25 observations is 13. It was later found that an observation 73 was wrongly entered as 48. The new average is

(a)

(b)

(c)

(d)

Explanation:

Difference of two observations = 73 – 48 = 25

∴ New average = 13 + $25/25$=14

Aliter : Using Rule 26,

If average of n numbers is m later on it was found that a number 'a’ was misread as 'b’.

The correct average will be = m +${\text"(a-b)"}/ \text"n"$.

Here, n = 25, m = 13

a = 73, b = 48

Correct Average = m + ${\text"(a-b)"}/ \text"n"n$

= 13 +${(73- 48)}/25$

= 13 + 1 = 14


Q-9)   The average of 20 numbers is calculated as 35. It is dicovered later, that while calculating the average, one number, namely 85, was read as 45. The correct average is

(a)

(b)

(c)

(d)

Explanation:

Correct sum of 20 numbers

= 20 × 35 – 45 + 85

= 700 + 40 = 740

∴ Correct average = $740/20$ = 37

Aliter : Using Rule 26,

If average of n numbers is m later on it was found that a number 'a’ was misread as 'b’.

The correct average will be = m +${\text"(a-b)"}/ \text"n"$.

Here, n = 20, m = 35

a = 85, b = 45

Correct mean = m + ${\text"(a - b)"}/\text"n"$

= 35 +${(85 - 45)}/20$

= 35 + 2 = 37


Q-10)   A student, by mistake, wrote 64 in place of 46 as a number at the time of finding the average of 10 given numbers and got the average as 50. The correct average of the numbers is :

(a)

(b)

(c)

(d)

Explanation:

Correct sum of numbers

= 10 × 50 – 64 + 46

= 500 – 18 = 482

∴ Correct average = $482/10$ = 48.2