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MOST IMPORTANT quantitative aptitude - 9 EXERCISES

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The following question based on average topic of quantitative aptitude

Questions : The average monthly income (in ₨) of certain agricultural workers is S and that of other workers is T. The number of agricultural workers is 11 times that of other workers. Then, the average monthly income (in ₨) of all the workers is :

(a) $\text"S+11T"/12$

(b) $\text"S+T"/12$

(c) $\text"11S+T"/12$

(d) $1/\text"11S"+T$

The correct answers to the above question in:

Answer: (c)

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Read more simple average Based Quantitative Aptitude Questions and Answers

Question : 1

If the average of x and $1/x (x ≠ 0)$ is M, then the average of $x^2$ and $1/x^2$is :

a) $1-M^2$

b) $1-2M$

c) $2M^2-1$

d) $2M^2+1$

Answer: (c)

Question : 2

If the average of m numbers is $n^2$ and that of n numbers is $m^2$, then average of (m + n) numbers is

a) $m/n$

b) m+n

c) mn

d) m-n

Answer: (c)

The average of m numbers = n2

Total number of ‘m’ numbers = m × n²

And the average of n numbers = m2

Total number of ‘n’ numbers = n × m²

∴ Average of (m + n) numbers

= ${mn² + m²n}/{m + n} = {mn(n + m)}/{m + n} = mn$

Question : 3

Numbers of boys and girls are ‘x’ and ‘y’ respectively. Ages of a girl and a boy are ‘a’ years and ‘b’ years respectively. The average age (in years) of all boys and girls is

a) $\text"x+y"/\text"bx+ay"$

b) $\text"x+y"/\text"bx+ay"$

c) $\text"bx+ay"/\text"x+y"$

d) $\text"x+y"/\text"ax+by"$

Answer: (c)

According to question,

Total age of boys = ap years

Total age of girls = bq years

∴ Required average =${ap + bq}/{p +q}$

Question : 4

If a, b, c, d, e are five consecutive odd numbers, their average is

a) $5(a+4)$

b) $\text"abcde"/5$

c) $5(a+b+c+d+e)$

d) $a+4$

Answer: (d) 

Question : 5

The average of 5 consecutive integers starting with ‘m’ is n. What is the average of 6 consecutive integers starting with (m + 2) ?

a) $\text"2n+5"/2$

b) $n+2$

c) $n+3$

d) $\text"2n+9"/2$

Answer: (a)

According to the question let m = 1

∴ 5 consecutive integers are = 1,2,3,4,5

$\text"1+2+3+4+5"/5=n$

$n = 15/5=3$

6 consecutive integers starting with (m + 2) are = 3,4,5,6,7,8

∴$\text"3+4+5+6+7+8"/6=33/6=11/2$

Now check from option to put n = 3

While in Option :(a) $\text"2n+5"/2 = 11/2 $

Question : 6

The average of nine consecutive numbers is n. If the next two numbers are also included the new average w

a) increase by 2

b) remain the same

c) increase by 1.5

d) increase by 1

Answer: (d)

Given,

$\text"(a + a + 1 + a + 2 + a + 3 + a + 4 + a + 5 + a + 6 + a + 7 + a + 8)"/9 = n$

$=> \text"(9a + 36)"/9 = n$

=> a + 4 = n -------------------- ( 1 )

If the next 2 numbers are included , let new average = k

$\text"(a + a + 1 + a + 2 + a + 3 + a + 4 + a + 5 + a + 6 + a + 7 + a + 8 + a + 9 + a + 10)"/11 = k$

$\text"(11a + 55)"/11 = k$

a + 5 = k -------------------- ( 2 )

Subtracting ( 1 ) from ( 2 ) , we get :

a + 5 - a - 4 = k - n

=> k - n = 1

=> k = n + 1

Therefore the new average is n + 1

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