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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 : Three Science classes A, B and C take a Life Science test. The average score of class A is 83. The average score of class B is 76. The average score of class C is 85. The average score of class A and B is 79 and average score of class B and C is 81. Then the average score of classes A, B and C is

(a) 80.5

(b) 80

(c) 81

(d) 81.5

The correct answers to the above question in:

Answer: (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

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

Students in class A ⇒ x

Students in class B ⇒ y

Students in class C ⇒ z

For classes A and B,

${83x + 76y}/{x+y}$ = 79

⇒ 83x + 76y = 79x + 79y

⇒ 83x – 79x = 79y – 76y

⇒ 4x = 3y

For classes B and C

${76y+85z}/{y+z}$= 81

⇒ 76y + 85z = 81y + 81z

⇒ 5y = 4z

∴ 20x = 15y = 12z

⇒ ${20x}/60$ = ${15y}/60$ = ${12z}/60$

⇒ $x/3$ = $y/4$ = $z/5$

∴ Required average

${83×3+76×4+85×5}/{3+4+5}$

${249+304+425}/12$ = $978/12$

= 81.5

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Read more simple average Based Quantitative Aptitude Questions and Answers

Question : 1

Out of 20 boys, 6 are each of 1 m 15 cm height, 8 are of 1 m 10 cm and rest of 1 m 12 cm. The average height of all of them is

a) 1 m 21 cm

b) 1 m 12 cm

c) 1 m 21.1 cm

d) 1 m 12.1 cm

Answer: (d)

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

Average height

= ${6×1.15+8×1.1+6×1.12}/20$

= ${6.9+8.8+6.72}/20$ = ${22.42}/20$

= 1 metre 12.1 cm

Question : 2

A library has an average number of 510 visitors on Sunday and 240 on other days. The average number of visitors per day in a month of 30 days beginning with Sunday is :

a) 300

b) 290

c) 295

d) 285

Answer: (d)

That month will have 5 sundays.

∴ Required average

= ${5×510+25×240}/30$

= ${2550+6000}/30$

= $8550/30$ = 285

Aliter : Using Rule 10,

Here, $n_1$ = 5, $a_1$ = 510

$n_2$ = 25, $a_2$ = 240

Question : 3

If the average weight of 6 students is 50 kg; that of 2 students is 51 kg; and that of other 2 students is 55 kg; then the average weight of all students is

a) 52 kg

b) 51.2 kg

c) 51.5 kg

d) 61 kg

Answer: (b)

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 weight

$(50×6+51×2+55×2)/10$

= ${300+102+110}/10$ = $512/10$

= 51.2 kg

Question : 4

The average of 100 observations was calculated as 35. It was found later, that one of the observations was misread as 83 instead of 53. The correct average is :

a) 35.7

b) 36.7

c) 34.7

d) 32.7

Answer: (c)

Difference = 83 – 53 = 30

Incorrect observation > Correct observation

∴ Required average

= 35 - $30/100$ = 35 – 0.3 = 34.7

Question : 5

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) 17

b) 31

c) 7

d) 3

Answer: (c)

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

Question : 6

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) 60

b) 50

c) 52

d) 56

Answer: (b)

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

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