Practice Average MCQ Questions with Details Solution
Q1. What will be the average of even numbers Between 11 to 63?
(a) 37.5
(b) 47
(c) 42
(d) 37
Answer: (d) 37 Solution: even number Between 11 to 63 are 12, 14, 16, … 60, 62 ∴ Required Average = $\frac{\mathrm{first} \mathrm{number}+\mathrm{last} \mathrm{number}}{2}=\,\,\frac{12+62}{2}=37$ Rule: The average of consecutive even natural numbers. A = $\frac{\mathrm{first} \mathrm{number}+\mathrm{last} \mathrm{number}}{2}$ |
Q2. Calculate the average of the cubes of first five natural numbers.
(a) 55
(b) 65
(c) 45
(d) 35
(e) None of the above
Answer: (c) 45 Solution: Required average = $\frac{5\left( 5+1 \right) ^2}{4}=45$ Rule: The average of cubes of first ‘n’ consecutive natural numbers A = $\frac{n\left( n+1 \right) ^2}{4}$ |
Q3. The average of 8 numbers is 14. The average of 6 of these numbers is 16. What is the average of the remaining two numbers?
(a) 12
(b) 6
(c) 8
(d) 10
Answer: (c) 8 Solution: Required average = $\frac{8\times 14-6\times 16}{2}=8$ |
Q4. The average of 40 consecutive even numbers is 59. What is the greatest number? [SSC SAS 2010]
(a) 98
(b) 59
(c) 112
(d) 20
Answer: (a) 98 Solution: Let 40 consecutive even numbers are a, a + 2, …., a+78 ( t_{40} = a + (n – 1)d, where n= 40, d = 2) Average = $\frac{\mathrm{first} \mathrm{number}+\mathrm{last} \mathrm{number}}{2}=\,\,\frac{\mathrm{a}+\mathrm{a}+78}{2}=\mathrm{a}+39$ ∴ a + 39 = 59 ⇒ a = 20 Greatest number = a + 78 = 20 + 78 = 98 |
Q5. Out of 40 boys in a class, average weight of 30 is 60 kg and the average weight of the remaining is 56 kg. The average weight (in kilogram) of the whole class is
(a) 58.5
(b) 58
(c) 57
(d) 59
Answer: (d) 59 Solution: Required average = $\frac{30\times 60+10\times 56}{30+10}=59$ Rule: If the average of ‘n_{1}’ numbers is a_{1} and the average of ‘n_{2}’ numbers is a_{2}, then average of total numbers n_{1} and n_{2} is Average (A) = $\frac{n_1a_1+n_2a_2}{n_1+n_2}$ |
Q6. A school has 4 section of Chemistry in Class X having 40, 35, 45 and 42 students. The mean marks obtained in Chemistry test are 50, 60, 55 and 45 respectively for the 4 sections. Determine the overall average of marks per student
(a) 50.25
(b) 52.25
(c) 51.25
(d) 53.25
Answer: (b) 52.25 Solution: Required average = $\frac{40\times 50+35\times 60+45\times 55+42\times 45}{40+35+45+42}$ = $\frac{2000+2100+2475+1890}{162}=52.2531$ Rules: If the given observations (x) are occurring with certain frequency (A) then, Average = $\frac{A_1x_1+A_2x_2+…….+A_nx_n}{x_1+x_2+…..+x_n}$ , where, A_{1}, A_{2}, A_{3}. ………. An are frequencies |
Q7. 3 years ago the average age of a family of 5 members was 17 years. With the birth of a new baby, the average age of six members remains the same even today. Find the age of the new baby.
(a) 1 year
(b) 2 years
(c) $1\frac{1}{2}$years
(d) Cannot be determined
Answer: (b) 2 years Solution: Sum of present ages of the six members = (17 × 6) years = 102 years. Sum of present ages of the 5 members (excluding baby) = 5 × (17 + 3) years = 100 years. ∴ Age of the baby = 102 – 100 = 2 years |
Q8. The average age of a group of person going for picnic is 16 years. Twenty new persons with an average age of 15 years join the group on the spot due to which their average becomes 15.5 years. Find the number of persons initially going for picnic.
(a) 20
(b) 18
(c) 22
(d) None of these
Answer: (a) 20 Solution: Let the number of persons, initially going for Picnic = x ∴ Sum of their ages = 16x Now new average = $\frac{16\mathrm{x}+15\times 20}{\mathrm{x}+20}$ ∴$\frac{16\mathrm{x}+15\times 20}{\mathrm{x}+20}=15.5$ ⇒ 16x +300 = 15.5x + 310 ⇒ 0.5x = 10 ⇒x = 20 |
Q9. The average age of 36 students in a group is 14 years. When teacher’s age is included to it, the average increases by one. What is the teacher’s age in years?
(a) 31
(b) 36
(c) 51
(d) Cannot be determined
Answer: (c) 51 Solution: Total age before teacher include = 36 × 14 = 504 Total age after teacher include = 37 × 15 = 555 Age of the teacher = (555 – 504) years = 51 years Using Trick: Age of the teacher = 14 + 37 = 51 Trick: The average of n numbers is ‘a’. After adding a new number average increase by x then The new numbers = a + x (n + 1) |
Q10. A pupil’s marks were wrongly entered as 83 instead of 63. Due to that the average marks for the class got increased by half. The number of pupils in the class is:
(a) 10
(b) 20
(c) 40
(d) 73
Answer: (c) 40 Solution: Let there be x pupils in the class. Total increase in marks = $\mathrm{x}\times \frac{1}{2}=\frac{\mathrm{x}}{2}$ $\frac{\mathrm{x}}{2}=\,\,83-63=20$ ⇒ x = 40 Using Trick: The number of pupils = $\frac{83-63}{\frac{1}{2}}=40$ Trick: The average of n numbers is ‘a’. If a number ‘x’ is replaced with another new number then average increase by ‘y’. n = $\frac{\mathrm{x}-\mathrm{new} \mathrm{number}}{\mathrm{y}}$ |