AAO Exam-Quantitative Aptitude (Mock Test)

Calculation:

The series follows the following pattern:

⇒ 41 + 5 = 46

⇒ 46 + 7 = 53

⇒ 53 + 11 = 64

⇒ 64 + 13 = 77

⇒ 77 + 17 = 94 ∴ The missing term in the series is 77.

3. What will come in place of the question mark ‘?’ in the following question?

5, 6, 13, 40, 161, ?

987
806
876
888
654

4. What approximate value should come in place of the question mark (?) in the following question?

220.89 + 2.123 + 205.96 = 1.99 × ? + 214.98

90
100
120
80
110

Solution

Follow BODMAS rule to solve this question, as per the order given below.

Step - 1: Parts of an equation enclosed in 'Brackets' must be solved first, and following BODMAS rule in the bracket -

Step - 2: Any mathematical 'Of' or 'Exponent' must be solved next.

Step - 3: Next, the parts of the equation that contain 'Division' and 'Multiplication' are calculated.

Step - 4: Last but not the least, the parts of the equation that contain 'Addition' and 'Subtraction' should be calculated.

Since, we need to find out the approximate value, we can write these values to their nearest integers.

Given expression is

220.89 + 2.123 + 205.96 = 1.99 × ? + 214.98

⇒ 221 + 2³ + 206 = 2 × ? + 215

⇒ 221 + 8 + 206 = 2 × ? + 215

⇒ 435 = 2 × ? + 215

⇒ 220 = 2 × ?

⇒ ? = 110

5. What approximate value should come in the place of question mark (?) in the following question?

47.99% of 749.99 + 24.98 – 9.99 × 18.18 = ?

185
195
205
215
225

Solution

Follow BODMAS rule to solve this question, as per the order given below.

Step - 1: Parts of an equation enclosed in 'Brackets' must be solved first, and following BODMAS rule in the bracket -

Step - 2: Any mathematical 'Of' or 'Exponent' must be solved next.

Step - 3: Next, the parts of the equation that contain 'Division' and 'Multiplication' are calculated.

Step - 4: Last but not the least, the parts of the equation that contain 'Addition' and 'Subtraction' should be calculated.

Since, we need to find out the approximate value, we can write these values to their nearest integers.

Given expression is

47.99% of 749.99 + 24.98 – 9.99 × 18.18 = ?

⇒ 48% of 750 + 25 – 10 × 18 = ?

⇒ 360 + 25 – 10 × 18 = ?

⇒ 360 + 25 – 180 = ?

⇒ 385 – 180 = ?

∴ ? = 205

6. What approximate value should come in the place of question mark (?) in the following question?

49.9% of 499.9 × 19.9% of 99.9 ÷ ? = 49.9% of 1999.9

9
2
14
18
5

Solution

Follow the BODMAS rule to solve this question, as per the order given below.

Step - 1: Parts of an equation enclosed in 'Brackets' must be solved first, and following BODMAS rule in the bracket -

Step - 2: Any mathematical 'Of' or 'Exponent' must be solved next.

Step - 3: Next, the parts of the equation that contain 'Division' and 'Multiplication' are calculated.

Step - 4: Last but not the least, the parts of the equation that contain 'Addition' and 'Subtraction' should be calculated.

Since we need to find out the approximate value, we can write these values to their nearest integers.

Given Expression.

49.9% of 499.9 × 19.9% of 99.9 ÷ ? = 49.9% of 1999.9

⇒ 50% of 500 × 20% of 100 ÷ ? = 50% of 2000

⇒ 250 × 20 ÷ ? = 1000

⇒ ? = 5

7. Direction: Study the table carefully and answer the following questions.

The following table shows the total population in 6 different cities and the percentage of females in them.

Name of the City	Total number of Population	Percentage of female
A	32000	50%
B	64000	52%
C	48000	35%
D	50000	60%
E	40000	40%
F	45000	45%

Question:

The number of females in city A is approximately what percent more or less than the number of males in city E?

30%
45%
66.66%
35%
33.33%

Solution

Alternate Method

Given:

Name of the City	Total number of Population	Percentage of female
A	32000	50%
E	40000	40%

Formula Used:

Decreased percentage = {(Decreased value)/(Base value)} × 100

Calculation:

The number of females in city A = 50% of 32000

⇒ (50/100) × 32000

⇒ 32000/2

⇒ 16000

The percentage of males in city E = (100 - 40)%

⇒ 60%

Total number of males in city E = 60% of 40000

⇒ (60/100) × 40000

⇒ 60 × 400

⇒ 24000

The required decreased value = 24000 - 16000

⇒ 8000

The required decreased percentage = (8000/24000) × 100

⇒ 100/3

⇒ 33.33%

∴ The number of females in city A is approximately 33.33% less than the number of males in city E.

8. Direction: Study the table carefully and answer the following questions.

The following table shows the total population in 6 different cities and the percentage of females in them.

Name of the City	Total number of Population	Percentage of female
A	32000	50%
B	64000	52%
C	48000	35%
D	50000	60%
E	40000	40%
F	45000	45%

Question:

What is the difference between the number of males in cities A and D together and the number of females in cities B and F together?

16530
17000
17500
17530
18400

Solution

Given:

Name of the City	Total number of Population	Percentage of female
A	32000	50%
B	64000	52%
D	50000	60%
F	45000	45%

Calculation:

The percentage of males in city A = (100 - 50)%

⇒ 50%

Total number of males in city A = 50% of 32000

⇒ (50/100) × 32000

⇒ 32000/2

⇒ 16000

The percentage of males in city D = (100 - 60)%

⇒ 40%

Total number of males in city D = 40% of 50000

⇒ (40/100) × 50000

⇒ 40 × 500

⇒ 20000

Total number of males in city A and D together = 16000 + 20000

⇒ 36000

Total number of females in city B = 52% of 64000

⇒ (52/100) × 64000

⇒ 52 × 640

⇒ 33280

Total number of females in city F = 45% of 45000

⇒ (45/100) × 45000

⇒ 45 × 450

⇒ 20250

Total number of females in city B and F together = 33280 + 20250

⇒ 53530

The required difference = 53530 - 36000

⇒ 17530

∴ The difference between the number of males in cities A and D together and the number of females in cities B and F together will be 17530.

9. Direction: Study the table carefully and answer the following questions.

The following table shows the total population in 6 different cities and the percentage of females in them.

Name of the City	Total number of Population	Percentage of female
A	32000	50%
B	64000	52%
C	48000	35%
D	50000	60%
E	40000	40%
F	45000	45%

Question:

What is the average number of males in cities B, E and F together?

19000
19725
26490
28950
26000

Solution

Since population are in thousands we can ignore last three zeros to simplify the calculation

⇒ Average = (64 × (100 - 52)% + 40 × (100 - 40)% + 45 × (100 - 45)%)/3

⇒ Average = (64 × 48% + 40 × 60% + 45 × 55%)/3

⇒ Average = (64 × 16% + 40 × 20% + 15 × 55%)

⇒ Average = (64 × 16 + 40 × 20 + 15 × 55)/100 (%→ 1/100)

⇒ Average = (1024 + 800 + 825)/100

⇒ Average = (2649)/100

Since we ignored zeros at last we need to multiply with 3 zeros

∴ Required Average = 26490

Alternate Method

Given:

Name of the City	Total number of Population	Percentage of female
B	64000	52%
E	40000	40%
F	45000	45%

Calculation:

The percentage of males in city B = (100 - 52)%

⇒ 48%

Total number of males in city B = 48% of 64000

⇒ (48/100) × 64000

⇒ 48 × 640

⇒ 30720

The percentage of males in city E = (100 - 40)%

⇒ 60%

Total number of males in city E = 60% of 40000

⇒ (60/100) × 40000

⇒ 60 × 400

⇒ 24000

The percentage of males in city F = (100 - 45)%

⇒ 55%

Total number of males in city F = 55% of 45000

⇒ (55/100) × 45000

⇒ 55 × 450

⇒ 24750

Total number of male in cities B, E and F together = 30720 + 24000 + 24750

⇒ 79470

The required average = 79470/3

⇒ 26490

∴ The average number of males in cities B, E, and F together will be 26490.

10. The radius of the circle is 13 cm more than the breadth of a rectangle. The area of the circle is 3850 sq.cm. The circumference of the circle is equal to the perimeter of a rectangle. The area of the rectangle is equal to the area of a square. Find the perimeter of a square.

352 cm
164 cm
176 cm
188 cm
196 cm

Solution

(Radius of circle)² = 3850 × (1/ π)

Radius of a circle = 35 cm

Breadth of a rectangle = 35 – 13 = 22 cm

Circumference of a circle = 2π × 35 = 220 cm

Perimeter of a rectangle = 2(Length + Breadth)

Length of a rectangle = (220/2) – 22 = 88 cm

Area of square = length × breadth = 88 × 22 = 1936 sq.cm

Side of a square = (1936)^1/2

Side of a square = 44 cm Perimeter of a square = 4 × 44 = 176 cm

11. Total age of 12 students in a class is 120 years. Two students of average age 8 years left the class and 5 new students of average age 11 years joined the same class. Find the new approximate average age of class.

13 years
12 years
10 years
8 years
7 years

Solution

Given:

Total age of 12 students = 120 years

Total age of two students who left class = 8 × 2 = 16 years

Total age of 5 students who joined class = 5 × 11 = 55 years

Total age of (12 - 2 + 5 = 15) students of a class = 120 - 16 + 55 = 159 years

∴ Required average = 159/15 = 10.6 years ≈ 10 years

12. A goods train started from P to Q at a speed of 54 kmph. After 10 hours, a passenger train started from Q to P at a speed of 63 kmph. Two trains will meet when passenger train travels 252 km from point Q. Find the distance between the city P and Q.

128 km
768 km
1008 km
960 km
458 km

Solution

GIVEN :

Speed of goods train = 54 kmph

Speed of passenger train = 63 kmph

Distance covered by goods trains in 10 hours = 54 × 10 = 540 km

CALCULATION :

Let the distance between the P and Q be M km.

Remaining distance to cover for goods trains = M - 540

Relative speed = 54 + 63 = 117 kmph

Two trains meet = 252 km from Q

Then,

⇒ (M - 540)/117 × 63 = 252

⇒ M - 540 = 468

⇒ M = 1008 The distance between the P and Q is 1008 km.

13. In the given question, two equations numbered l and II are given. Solve both the equations and mark the appropriate answer.

I. 2x² - 39x + 189 = 0 II. y² - 16y + 63 = 0

x > y
x < y
x ≥ y
x ≤ y
x = y or relationship between x and y cannot be established.

Solution

Calculation∶

I. 2x² - 39x + 189 = 0

⇒ 2x² - 18x - 21x + 189 = 0

⇒ 2x(x - 9) - 21(x - 9) = 0

⇒ (x - 9) (2x - 21) = 0

⇒ x = 9, 21/2

II. y² - 16y + 63 = 0

⇒ y² - 7y - 9y + 63 = 0

⇒ y(y - 7) - 9(y - 7) = 0

⇒ (y - 7) (y - 9) = 0

⇒ y = 7, 9

Value of x	Value of y	Relation
9	7	x > y
9	9	x = y
21/2	7	x > y
21/2	9	x > y

∴ x ≥ y

14. In the given question, two equations numbered l and II are given. Solve both the equations and mark the appropriate answer

I. 9x² - 3x - 2 = 0

II. y² - 8y + 16 = 0

x > y
x < y
x ≤ y
x ≥ y
x = y or the relationship between x and y cannot be established.

Solution

Given:

I) 9x2 - 3x - 2 = 0

II) y2 - 8y + 16 = 0

Concept used:

Using quadratic equation

Calculation:

From I,

9x2 - 3x - 2 = 0

⇒ 9x² - 6x + 3x - 2 = 0

⇒ 3x(3x - 2) + 1(3x - 2) = 0

⇒ (3x - 2) (3x + 1) = 0

⇒ x = 2/3, - 1/3

From II,

y2 - 8y + 16 = 0

⇒ y² - 4y - 4y + 16 = 0

⇒ y(y - 4) - 4(y - 4) = 0

⇒ (y - 4) (y - 4) = 0

⇒ y = 4, 4

Comparison between x and y (via Tabulation):

Value of x	Value of y	Relation
2/3	4	x < y
2/3	4	x < y
- 1/3	4	x < y
- 1/3	4	x < y

∴ x < y.

15. In the given question, two equations are numbered l and II as given. Solve both the equations and mark the appropriate answer.

I. x² + 20x + 64 = 0 II. y² + 27y + 110 = 0

x > y
x < y
x ≥ y
x ≤ y
x = y or relationship between x and y cannot be established.

Solution

Calculation∶

I. x² + 20x + 64 = 0

⇒ x² + 4x + 16x + 64 = 0

⇒ x(x + 4) + 16(x + 4) = 0

⇒ (x + 16)(x + 4) = 0

⇒ x = -16, -4

II. y² + 27y + 110 = 0

⇒ y² + 5y + 22y + 110 = 0

⇒ y(y + 5) + 22(y + 5) = 0

⇒ (y + 5)(y + 22) = 0

⇒ y = -5, -22

Value of x	Value of y	Relation
-16	-22	x > y
-16	-5	x < y
-4	-22	x > y
-4	-5	x > y

∴ Relationship between x and y cannot be established

16. Direction: Read the following information carefully and answer the questions that follow.

A survey is conducted on 540 people of society about their choice on Red and Green colors. The people have to select only Red or only Green or both as their choice. The number of males and females is in the ratio 5 ∶ 4 and 35% of male like only Red color. 25% of people select both colors as their choice. The number of males who selects only Green is double to that of who selects both as their choice. 30% of female like only Green color.

Question:

How many males like only Green color?

130
120
105
110
95

17. Direction: Read the following information carefully and answer the questions that follow.

A survey is conducted on 540 people of society about their choice on Red and Green colors. The people have to select only Red or only Green or both as their choice. The number of males and females is in the ratio 5 ∶ 4 and 35% of male like only Red color. 25% of people select both colors as their choice. The number of males who selects only Green is double to that of who selects both as their choice. 30% of female like only Green color.

Question:

The number of female like both colors is approximately what percentage of the number of female like only Green?

100%
103%
105%
97%
110%

18. Direction: Read the following information carefully and answer the questions that follow.

A survey is conducted on 540 people of society about their choice on Red and Green colors. The people have to select only Red or only Green or both as their choice. The number of males and females is in the ratio 5 ∶ 4 and 35% of male like only Red color. 25% of people select both colors as their choice. The number of males who selects only Green is double to that of who selects both as their choice. 30% of female like only Green color.

Question:

What is the difference of number of male and female who like both colors as their choice?

5
10
15
8
7

19. Direction: Read the following information carefully and answer the questions that follow.

A survey is conducted on 540 people of society about their choice on Red and Green colors. The people have to select only Red or only Green or both as their choice. The number of males and females is in the ratio 5 ∶ 4 and 35% of male like only Red color. 25% of people select both colors as their choice. The number of males who selects only Green is double to that of who selects both as their choice. 30% of female like only Green color.

Question:

The number of males who like Only Red color is what percentage more than the number of female like Only Green color?

46%
48%
42%
52%
56%

20. Direction: Read the following information carefully and answer the questions that follow.

A survey is conducted on 540 people of society about their choice on Red and Green colors. The people have to select only Red or only Green or both as their choice. The number of males and females is in the ratio 5 ∶ 4 and 35% of male like only Red color. 25% of people select both colors as their choice. The number of males who selects only Green is double to that of who selects both as their choice. 30% of female like only Green color.

Question:

What will be the average of number of males who like only Green color and number of female who like only Red color?

120
114
108
102
116

AAO Exam-Quantitative Aptitude (Mock Test)

AAO Exam-Quantitative Aptitude (Mock Test)

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