Practice Simple average - quantitative aptitude Online Quiz (set-1) For All Competitive Exams
Q-1) The average weight of 15 oarsmen in a boat is increased by 1.6 kg when one of the crew, who weighs 42 kg is replaced by a new man. Find the weight of the new man (in kg).
(a)
(b)
(c)
(d)
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
If the average age decreases by ‘t’ years after entry of new person, then the age of the new person = T – N.t
Here, N = 15, T = 42, t = 1.6
Weight of new oarsman
= (42 + 15 × 1.6) kg.
= (42 + 24) kg. = 66 kg.
Q-2) The mean of 9 observations is 16. One more observation is included and the new mean becomes 17. The 10th observation is
(a)
(b)
(c)
(d)
Mean of Ten observations – Mean of nine observations
Tenth observation
= 10 × 17 – 16 × 9
= 170 – 144 = 26
Aliter : Using Rule 19,
Here, N = 9, T = 16
n = 1, t = 1
10th observation
= T+ $({N/n}+1)$t
= 16+$(9/1+1) ×1$
= 16 + 10 = 26
Q-3) The average of some natural numbers is 15. If 30 is added to first number and 5 is subtracted from the last number the average becomes 17.5 then the number of natural number is
(a)
(b)
(c)
(d)
Number of natural numbers = x
∴ Their sum = 15x
According to the question,
15x + 30 – 5 = x × 17.5
⇒ 17.5x – 15x = 25
⇒ 2.5x = 25
⇒ x = $25/2.5$ = 10
Q-4) The average of 30 results is 20 and the average of other 20 results is 30. What is the average of all the results ?
(a)
(b)
(c)
(d)
Using Rule 10,
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
$Average = {n_1a_1 + n_2a_2}/{n_1 + n_2}$
Required average
= ${20×30+20×30}/{30+20}$
= ${600+600}/50$
= $1200/50$ = 24
Q-5) The average of 20 numbers is 15 and the average of first five is 12. The average of the rest is
(a)
(b)
(c)
(d)
If the average of remaining numbers be x, then
20 × 15 = 5 × 12 + 15x
⇒ 300 = 60 + 15x
⇒ 15x = 300 – 60 = 240
⇒ x = $240/15$ = 16
Aliter : Using Rule 13,
Here, m = 20, x = 15
n = 5, y = 12
Average of remaining Numbers = $({mx-ny}/{m-n})$
= $({20×15-5×12}/{20-5})$
= $({300-60}/15)$ = $240/15$ = 16
Q-6) The average of marks obtained by 100 candidates in a certain examination is 30. If the average marks of passed candidates is 35 and that of the failed candidates is 10, what is the number of candidates who passed the examination?
(a)
(b)
(c)
(d)
Number of successful students in the exam = x
∴ Number of unsuccessful students = 100 – x
According to the question,
30 = ${35x+10(100-x)}/100$
⇒ 3000 = 35x + 1000 – 10x
⇒ 3000 = 25x + 1000
⇒ 25x = 3000 – 1000 = 2000
⇒ x = $2000/25$ = 80
Q-7) The average of 30 numbers is 15. The average of the first 18 numbers is 10 and that of the next 11 numbers is 20. The last number is
(a)
(b)
(c)
(d)
Let the last number be x.
According to the question,
18 × 10 + 11 × 20 + x = 30 × 15
⇒ 180 + 220 + x = 450
⇒ 400 +x = 450
⇒ x = 450 – 400 = 50
Q-8) A student was asked to find the arithmetic mean of the following 12 numbers : 3, 11, 7, 9, 15, 13, 8, 19, 17, 21, 14 and x He found the mean to be 12. The value of x will be :
(a)
(b)
(c)
(d)
Using Rule 1,
Average of two or more numbers/quantities is called the mean of these numbers, which is given by
$\text"Average(A)" = \text"Sumof observation / quantities"/\text"No of observation / quantities"$
∴ S = A × n
Mean = ${3 +11+ 9+7+15+ 13+ 8 +19 +17 +21 +14+x}/12$
According to question,
${137+x}/12$ = 12
∴ 137 + x = 144
∴ x = 144 – 137 = 7
Q-9) The average marks obtained by a class of 60 students is 65. The average marks of half of the students is found to be 85. The average marks of the remaining students is
(a)
(b)
(c)
(d)
Let average marks of remaining 30 students be x.
∴ 65 = ${30×85+30×x}/60$
⇒ 65 × 60 = 2550 + 30x
⇒ 3900 = 2550 + 30x
⇒ 30x = 3900 – 2550 = 1350
⇒ x = $1350/30$ = 45.
Q-10) A librarian purchased 50 story– books for his library. But he saw that he could get14 more books by spending Rs. 76 more and the average price per book would be reduced by Re. 1. The average price (in Rs.) of each book he bought, was :
(a)
(b)
(c)
(d)
Let the average cost of each book bought (of 64 books) be Rs.x.
According to the question,
64 × x – 50(x + 1) = 76
⇒ 64x – 50x – 50 = 76
⇒ 14x = 76 + 50 = 126
⇒ x = $126/14$ = 9
∴ Required average price
= 9 + 1 = Rs. 10