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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 of x numbers is y and average of y numbers is x. Then the average of all the numbers taken together is

(a) ${x^2 +y^2}/ {x+y}$

(b) ${xy}/{x + y}$

(c) ${2xy}/{x + y}$

(d) ${x+y}/{2xy}$

The correct answers to the above question in:

Answer: (c)

Sum of x numbers = xy

Sum of y numbers = xy

∴ Required average = ${xy+xy}/{x+y}$ = ${2xy}/{x+y}$

Aliter : Using Rule 10,

Here, $n_1$ = x, $a_1$ = y

$n_2$ = y, $a_2$ = x

∴ Average = ${n_1a_1+n_2a_2}/{n_1+n_2}$

= ${xy+yx}/{x+y}$ = ${2xy}/{x+y}$

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Read more simple average Based Quantitative Aptitude Questions and Answers

Question : 1

A man bought 13 articles at Rs.70 each, 15 at Rs.60 each and 12 at Rs.65 each. The average price per article is

a) Rs.65.75

b) Rs.62.25

c) Rs.64.75

d) Rs.60.25

Answer: (c)

Using Rule 2,

If the given observations (x) are occuring with certain frequency (A) then,

$Average = {A_1x_1+A_2x_2+...............+A_nx_n}/{x_1+x_2+......+x_n}$

where, $A_1, A_2, A_3, .......... A_n$ are frequencies

Required average price

= ${13×70+15×60+12×65}/{13+15+12}$

= ${910+900+780}/40$ =$2590/40$

= Rs.64.75

Question : 2

The average weight of A, B and C is 45 kg. If the average weight of A and B be 40 kg and that of B and C be 43 kg, then the weight of B is :

a) 29.5 kg.

b) 35 kg.

c) 32 kg.

d) 31 kg.

Answer: (d)

B’s weight = (A + B)’s weight + (B + C)’s weight – (A + B + C)’s weight

= (40 × 2 + 2 × 43 – 45 × 3) kg.

= (80 + 86 – 135) kg.

= 31 kg.

Question : 3

The average weight of five persons sitting in a boat is 38 kg. The average weight of the boat and the persons sitting in the boat is 52kg. What is the weight of the boat ?

a) 232 kg

b) 242 kg

c) 122 kg

d) 228 kg

Answer: (c)

Weight of the boat

= 6 × 52 – 5 × 38

= 312 – 190 = 122 kg

Question : 4

There are 100 students in 3 sections A, B and C of a class. The average marks of all the 3 sections was 84. The average of B and C was 87.5 and the average marks of A is 70. The number of students in A was

a) 20

b) 25

c) 35

d) 30

Answer: (a)

Number of students in section A = x

∴ Number of students in sections B and C

= (100 – x)

∴ x × 70 + (100 – x) × 87.5

= 84 × 100

⇒ 70x + 87.5 × 100 – 87.5x = 8400

⇒ 8750 – 17.5x = 8400

⇒ 17.5x = 8750 – 8400 = 350

⇒ x = $350/{17.5}$ = 20

Question : 5

The average of the marks obtained in an examination by 8 students was 51 and by 9 other students was 68. The average marks of all 17 students was :

a) 60

b) 60.5

c) 59.5

d) 59

Answer: (a)

Sum of total number of 8 students in exam = 8 × 51 = 408

Sum of total number of 9 students in exam = 9 × 68 = 612

∴ Required average

= ${408+ 612}/17$ =${1020}/17$ = 60

Aliter : Using Rule 10,

Here, $n_1$ = 8, $a_1$ = 51

$n_2$ = 9, $a_2$ = 68

∴ Average = ${n_1a_1+ n_2a_2}/{n_1+ n_2}$

${8×51+9×68}/{8+9}$

${408+612}/17$ = $1020/17$

= 60 marks.

Question : 6

The average of x numbers is $y^2$ and the average of y numbers is $x^2$ . So the average of all the numbers taken together is

a) ${x^2 + y^2}/{x + y}$

b) x$y^2$ + y$x^2$

c) xy

d) ${x^3 + y^3}/{x + y}$

Answer: (c)

Total sum of x numbers = x$y^2$

Total sum of y numbers = y$x^2$

∴ Required average

= ${xy^2+yx^2}/{x+y}$

=${xy(y+x)}/{x+y}$ = xy

Aliter : Using Rule 10,

Here, $n_1$ = x, $a_1$ = $y^2$

$n_2$ = y, $a_2$ = $x^2$

∴ Average = ${n_1a_1+n_2a_2}/{n_1+n_2}$

= ${xy^2+yx^2}/{x+y}$

= xy$({x+y}/{x+y})$ = xy

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