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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 : 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) $2M^2+1$

The correct answers to the above question in:

Answer: (c)

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Question : 1

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 : 2

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 : 3

Six friends have an average height of 167 cms. A boy with height 162 cm leaves the group. Find the new average height.

a) 169 cm

b) 167 cm

c) 166 cm

d) 168 cm

Answer: (d)

Total height of 5 friends

= (6 × 167 – 162) cm.

= (1002 – 162) cm.

= 840 cm.

∴ Required average = $840/ 5$

= 168 cm.

Question : 4

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

Answer: (c)

Question : 5

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 : 6

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 $

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