ExamVeda
Login
Home

Average - Aptitude MCQ Questions and Solutions with Explanations

Average MCQ Questions and answers with easy and logical explanations.Arithmetic Ability provides you all type of quantitative and competitive aptitude mcq questions on Average with easy and logical explanations. Average MCQ is important for exams like Banking exams,IBPS,SCC,CAT,XAT,MAT etc. Page-6 section-3

Each section contains 100 MCQs on Average .
51
The total marks obtained by a student in Physics, Chemistry and Mathematics together is 120 more than the marks obtained by him in Chemistry. What is the average marks obtained by him in Physics and Mathematics together?
Discuss
Answer & Solution
Answer: Option B
Solution:
P + C + M = C + 120
⇒ P + M = 120
∴ Required average
= $$\frac{P + M}{2}$$
= $$\frac{120}{2}$$
= 60
52
A student was asked to find the arithmetic mean of the numbers 3, 11, 7, 9, 15, 13, 8, 19, 17, 21, 14 and x. He found mean to be 12. What should be the number in place of x?
Discuss
Answer & Solution
Answer: Option B
Solution:
Clearly, we have
$$ \Rightarrow \frac{{3 + 11 + 7 + 9 + 15 + 13 + 8 + 19 + 17 + 21 + 14 + x}}{{12}} = 12$$
$$\eqalign{ & \Rightarrow 137 + x = 144 \cr & \Rightarrow x = 144 - 137 \cr & \Rightarrow x = 7 \cr} $$
53
If the mean of a, b, c, is M and ab + bc + ca = 0, then the mean of a2, b2, c2 is -
Discuss
Answer & Solution
Answer: Option B
Solution:
We have :
$$\frac{a + b + c}{3}$$   = M
or (a + b + c) = 3M
Now, (a + b + c)2 = (3M)2 = 9M2
⇔ a2 + b2 + c2 + 2 (ab + bc + ca) = 9M2
⇔ a2 + b2 + c2 = 9M2
[$$\because $$ (ab + bc + ca) = 0]
∴ Required mean
$$\eqalign{ & = \left( {\frac{{{a^2} + {b^2} + {c^2}}}{3}} \right) \cr & = \frac{{9{M^2}}}{3} \cr & = 3{M^2} \cr} $$
54
The average of 2, 7, 6 and x is 5 and the average of 18, 1, 6, x and y is 10. What is the value of y?
Discuss
Answer & Solution
Answer: Option C
Solution:
We have:
$$\eqalign{ & \Rightarrow \left( {\frac{{2 + 7 + 6 + x}}{5}} \right) = 5 \cr & \Rightarrow 15 + x = 20 \cr & \Rightarrow x = 5 \cr} $$
Also,
$$\eqalign{ & \Rightarrow \left( {\frac{{18 + 1 + 6 + x + y}}{5}} \right) = 10 \cr & \Rightarrow 25 + 5 + y = 50 \cr & \Rightarrow y = 20 \cr} $$
55
The average of x1, x2, x3 and x4 is 16. Half the sum of x2, x3, x4 is 23. What is the value of x1?
Discuss
Answer & Solution
Answer: Option B
Solution:
x1 + x2 + x3 + x4 = 16 × 4 = 64
⇒ $$\frac{1}{2}$$ (x2 + x3 + x4) = 23
⇒ x2 + x3 + x4 = 46
∴ x1 = 64 - 46
        = 18
56
Company C sells a line of 25 products with an average retail price of Rs. 1200. If none of these products sells for less than Rs. 420 and exactly 10 of the products sell for less than Rs. 1000, then what is the greatest possible selling price of the most expensive product?
Discuss
Answer & Solution
Answer: Option D
Solution:
To find the greatest possible selling price of the most expensive product, we need to consider the minimum selling price of the remaining 24 products which is Rs. 420 each for 10 products and Rs. 1000 each for other 14 products.
Minimum selling price of 24 products
= Rs. (420 × 10 + 1000 × 14)
= Rs. (4200 + 14000)
= Rs. 18200
Total selling price of 25 products
= Rs. (1200 × 25)
= Rs. 30000
∴ Greatest possible selling price of the most expensive products
= Rs. (30000 - 18200)
= Rs. 11800
57
If the mean of 5 observations x, x + 2, x + 4, x + 6 and x + 8 is 11, then the mean of the last three observations is-
Discuss
Answer & Solution
Answer: Option B
Solution:
We have :
$$ \Rightarrow \frac{{x + \left( {x + 2} \right) + \left( {x + 4} \right) + \left( {x + 6} \right) + \left( {x + 8} \right)}}{4} = 11$$
$$\eqalign{ & \Rightarrow 5x + 20 = 55 \cr & \Rightarrow x = 7 \cr} $$
So, the numbers are 7, 9, 11, 13, 15
∴ Required mean
$$\eqalign{ & = \left( {\frac{{11 + 13 + 15}}{3}} \right) \cr & = \frac{{39}}{3} \cr & = 13 \cr} $$
58
The mean of 12, 22, 32, 42, 52, 62, 72 is -
Discuss
Answer & Solution
Answer: Option B
Solution:
12 + 22 + 32 + . . . . . + n2 = $$\frac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6}$$
∴ 12 + 22 + 32 + . . . . . + 72 = $$\left( {\frac{{7 \times 8 \times 15}}{6}} \right)$$   = 140
So, required average
$$ = \left( {\frac{{140}}{7}} \right)$$
= 20
59
Total of Arun’s marks in Sanskrit and Mathematics together are 80 more than his marks in Science. His average marks in the three subjects are 100. What are his marks in Science?
Discuss
Answer & Solution
Answer: Option B
Solution:
S + M + Sc = 100 × 3 = 300
And
S + M = Sc + 80
⇒ Sc + 80 + Sc = 300
⇒ 2Sc = 220
⇒ Sc = 110
60
If a, b, c, d, e are five consecutive odd numbers, their average is
Discuss
Answer & Solution
Answer: Option D
Solution:
Clearly,
b = a + 2
c = a + 4
d = a + 6
e = a + 8
∴ Average
$$\eqalign{ & = \frac{{a + \left( {a + 2} \right) + \left( {a + 4} \right) + \left( {a + 6} \right) + \left( {a + 8} \right)}}{5} \cr & = \left( {\frac{{5a + 20}}{5}} \right) \cr & = \left( {a + 4} \right) \cr} $$