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Average Practice Set 3

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Average Practice Set 3

shape Introduction

Average is a straight- forward concept and it can be solved easily by equal distribution method. Average is the sum of all the elements in a given data set divided by the total number of elements in the data set. The most commonly denoted term for average is Arithmetic Mean, simply termed as mean. The article Average Practice Set 3 very useful for for different competitive examinations like RRB ALP/Technical Exams/Junior Engineer Recruitment Exams, SSC, IBPS PO Exams and etc.

The Average (arithmetic mean) of a group or set of N numbers is defined as the sum of those numbers divided by N. Here N is the number of values or observations in a set

Average = \(\frac {Sum of Numbers / Values}{Numbers of Values / Observation}\)

A = \(\frac {Sum}{N}\) Or Sum = N * Average


shape Quiz

Q1. The average of five number is 42 ,if one number is excluded the average become 35.The excluded number is

    A. 7
    B. 40
    C. 70
    D. 20

Answer: C

Explanation:
\(\frac {x}{5} = 42 \Rightarrow 42 \times 5\) = 210

\(\frac {x}{4} = 35 \Rightarrow 35 \times 4\) = 140

210 – 140 = 70


Q2. The average age of 30 students in a class is 20 years. The average age of 25 students is 15. What is the average age of remaining students

    A. 42
    B. 54
    C. 34
    D. 45

Answer: D

Explanation:

Sum of age of 15 Students = \( (40 \times 20) – (25 \times 15) \) = 600 – 375 = 225

Average = \( \frac {225}{5}\) = 45


Q3. The average of nine number is x and the average of three of these is y if the average of the remaining three is z, then

    A. \(3 x = y + z\)
    B. \(2 x = y + z\)
    C. \( x = 3 y + 3 z\)
    D. None of these

Answer: A

Explanation:
\( x = \frac {3 y + 3 z}{9} \Rightarrow 3 x = y + z\)


Q4. If a, b, c, d, e, f, g are seven consecutive odd number,their average is

    A. (a+6)
    B. \( \frac {abcdefg}{7} \)
    C. \( \frac {(a+b+c+d+e+f+g)}{7} \)
    D. None of these

Answer: A

Explanation:
\( \frac {a + (a + 2) + (a + 4) + (a + 6) + (a + 8) + (a + 10) + (A + 12)}{7}\)

\( \frac {7 a + 42}{7}\) = a + 6


Q5. The mean of 1, 8, 27, 64, 125 ……… 1728

    A. 650
    B. 560
    C. 600
    D. 605

Answer: A

Explanation:
Sum of cube numbers = \( \frac {n (n + 1) (2 n + 1)}{6}\)

= \(\frac {12 \times 13 \times 25}{6}\)

= \(\frac {3900}{6}\) = 650

Q1. 4 years ago ,the average age of a family of 6member was 20 years.A baby having been born,the average age of the family is same today.The present age of the baby

    A. 1
    B. 3
    C. 2
    D. 4

Answer: D

Explanation:
Total age of 6 members (4 years ago) = \(20 \times 6\) = 120

Now total age of the family = 120 + \((4 \times 6) \Rightarrow 120 + 24\) = 144

Total age of 7 members (now) = \(20 \times 7\) = 140

Age of the baby = 144 – 140 = 4


Q2. Average of ten positive numbers is x.If each number is increased by 12% then x is incresed by

    A. 5 %
    B. 12 %
    C. 10 %
    D. 25 %

Answer: B

Explanation:
\(\frac {\frac {120}{100} {x}_{1} + \frac {120}{100} {x}_{2} + … + \frac {120}{100} {x}_{10}}{10}\) = \(\frac {12}{10} x \)

Average is increased by 12 %


Q3. The average age of husband and wife was 30yr, 4 yr ago. What will be their average age at present ?

    A. 30
    B. 34
    C. Cannot be determined
    D. None of these

Answer: B

Explanation:
Avg of H and W age 4yr ago = 30

Present avg age of H and W = 30 + 4 = 34


Q4. The average weight of 20 students is 60kg.If the weight of the teacher is added, average is increased by 2kg.What was the teacher’s weight ?

    A. 100kg
    B. 101kg
    C. 102kg
    D. 103kg

Answer: C

Explanation:
\( \frac {x}{20} =60; x \)= 1200

\( \frac {x}{21} = 62; x \)= 1302

1302 – 1200 = 102


Q5. The average mark in 2 subjects is 35 and in three other subjects is 40. Then find the average mark in all the five matches ?

    A. 37
    B. 37. 5
    C. 36
    D. 38

Answer: D

Explanation:
\( \frac {x}{2} = 35; x \)= 70

\( \frac {x}{3} = 40; x \)= 120

5 sub avg = \( \frac {(70 + 120)}{5} \)= 38

Q1. The average height of 15 students is calculated as 75. But later it was found that the height of 1 student wrongly entered as 35 instead of 38 and another as 46 instead of 63.The correct average is

    A. 71
    B. 73
    C. 75
    D. 76

Answer: D

Explanation:
\( \frac {(15 \times 75) – (35 + 46) + (38 + 63)} {15} \) = \( \frac {(1125 – 81 + 101)} {15} \)

= \( \frac {1145}{15} \) = 76.33 = 76


Q2. Among the three number, the first is thrice the third number and the second number is half of the first number. If the average of the three number is 65.8 then find the third number

    A. 35.56
    B. 35.85
    C. 35.89
    D. 35.69

Answer: C

Explanation:
Let third num = \( x, {1}^{st} number = 3x, \) second no = \( \frac {3 x}{2} \)

\( \frac {[3 x + ( \frac {3 x}{2}) + x]} {3} \) = 65.8

\( \frac {11 x}{2} \) = 197.4

\( X = \frac {(2 × 197.4)} {11} \) = 35.89


Q3. The average weight of 3 students P, Q, and R are 84kg. Another student S joins the group and the average becomes 80kg. If another man T whose weight is 3kg more than that of S, replaces P, then the average weight of Q, R, S, and T becomes 79 kg then the weight of P is

    A. 75
    B. 82
    C. 45
    D. 98

Answer: A

Explanation:
P + Q + R = \( 84 \times 3 \) = 252

P + Q + R + S = \( 4 \times 80 \) = 320

S = 320 – 252 = 68

Q + R + S + T = \( 79 \times 4 \) = 316

Q + R + 2S + 3 = 316

S = 68, Q + R = 177

P = 252 – 177 = 75


Q4. The average of 5 consecutive number is 58.Find the first number ?

    A. 55
    B. 56
    C. 57
    D. 58

Answer: B

Explanation:
\( x + x + 1 + x + 2 + x + 3 + x + 4 = 58 \times 5 \) = 290

\( 5x + 10 \)= 290

\( X = 290 – \frac {10}{5} = \frac {280}{5} \) = 56


Q5. The average weight of 8 staff is increased by 3kg when one of them whose weight is 50kg is replaced by a new staff. The weight of the new staff is

    A. 50
    B. 64
    C. 76
    D. 74

Answer:

Explanation:
50 + \( (8 \times 3) \)= 50 + 24 = 74.


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