# Aptitude Area Online Test 2020 | Aptitude Area Mock Test 2020 |

## Aptitude Area Online Test, Free Aptitude Quiz

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Aptitude Area Online Test, Free Aptitude Quiz, Online Aptitude Area Test. Aptitude Area Question and Answers 2020. Aptitude Area Quiz. Aptitude Area Free Mock Test 2020. **Aptitude Area Question and Answers in PDF. **The Aptitude Area online mock test paper is free for all students.The below Aptitude questions and answers can improve your skills in order to face the Interview, Competitive examination, Govt Exams and various entrance test with full confidence.** Aptitude online test** is very useful for exam preparation and getting for Rank. Aptitude Area Question and Answers in Hindi and English. Aptitude Area Mock test for topic via Online Mode. Here we are providing** Aptitude Area Mock Test in Hindi**. Now Test your self for “**Aptitude Area Online Test in Hindi**” Exam by using below quiz…

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- Question 1 of 20
##### 1. Question

The ratio between the length and the breadth of a rectangular park is 3 : 2. If a man cycling along the boundary of the park at the speed of 12 km/hr completes one round in 8 minutes, then the area of the park (in sq. m) is:

CorrectPerimeter = Distance covered in 8 min. = (12000/60 × 8) m = 1600 m.

Let length = 3x metres and breadth = 2x metres.

Then, 2(3x + 2x) = 1600 or x = 160.

∴ Length = 480 m and Breadth = 320 m.

∴ Area = (480 × 320) m^{2}= 153600 m^{2}.IncorrectPerimeter = Distance covered in 8 min. = (12000/60 × 8) m = 1600 m.

Let length = 3x metres and breadth = 2x metres.

Then, 2(3x + 2x) = 1600 or x = 160.

∴ Length = 480 m and Breadth = 320 m.

∴ Area = (480 × 320) m^{2}= 153600 m^{2}. - Question 2 of 20
##### 2. Question

An error 2% in excess is made while measuring the side of a square. The percentage of error in the calculated area of the square is:

Correct100 cm is read as 102 cm.

∴ A_{1}= (100 x 100) cm^{2}and A_{2}(102 x 102) cm^{2}.

(A_{2}– A_{1}) = [(102)^{2}– (100)^{2}]

= (102 + 100) x (102 – 100)

= 404 cm^{2}.

∴ Percentage error = (404/100×100 × 100)x % = 4.04%Incorrect100 cm is read as 102 cm.

∴ A_{1}= (100 x 100) cm^{2}and A_{2}(102 x 102) cm^{2}.

(A_{2}– A_{1}) = [(102)^{2}– (100)^{2}]

= (102 + 100) x (102 – 100)

= 404 cm^{2}.

∴ Percentage error = (404/100×100 × 100)x % = 4.04% - Question 3 of 20
##### 3. Question

The ratio between the perimeter and the breadth of a rectangle is 5 : 1. If the area of the rectangle is 216 sq. cm, what is the length of the rectangle?

CorrectIncorrect - Question 4 of 20
##### 4. Question

The percentage increase in the area of a rectangle, if each of its sides is increased by 20% is:

CorrectIncorrect - Question 5 of 20
##### 5. Question

A rectangular park 60 m long and 40 m wide has two concrete crossroads running in the middle of the park and rest of the park has been used as a lawn. If the area of the lawn is 2109 sq. m, then what is the width of the road?

CorrectArea of the park = (60 x 40) m

^{2}= 2400 m^{2}.

Area of the lawn = 2109 m^{2}.

∴ Area of the crossroads = (2400 – 2109) m^{2}= 291 m^{2}.

Let the width of the road be*x*metres. Then,

60x + 40x – x^{2}= 291

⇒ x^{2}– 100x + 291 = 0

⇒ (x – 97)(x – 3) = 0

⇒ x = 3.IncorrectArea of the park = (60 x 40) m

^{2}= 2400 m^{2}.

Area of the lawn = 2109 m^{2}.

∴ Area of the crossroads = (2400 – 2109) m^{2}= 291 m^{2}.

Let the width of the road be*x*metres. Then,

60x + 40x – x^{2}= 291

⇒ x^{2}– 100x + 291 = 0

⇒ (x – 97)(x – 3) = 0

⇒ x = 3. - Question 6 of 20
##### 6. Question

The diagonal of the floor of a rectangular closet is 7.5 feet. The shorter side of the closet is 4.5 feet. What is the area of the closet in square feet?

CorrectIncorrect - Question 7 of 20
##### 7. Question

A towel, when bleached, was found to have lost 20% of its length and 10% of its breadth. The percentage of decrease in area is:

CorrectIncorrect - Question 8 of 20
##### 8. Question

A man walked diagonally across a square lot. Approximately, what was the percent saved by not walking along the edges?

CorrectIncorrect - Question 9 of 20
##### 9. Question

The diagonal of a rectangle is √41 cm and its area is 20 sq. cm. The perimeter of the rectangle must be:

CorrectIncorrect - Question 10 of 20
##### 10. Question

What is the least number of squares tiles required to pave the floor of a room 15 m 17 cm long and 9 m 2 cm broad?

CorrectLength of largest tile = H.C.F. of 1517 cm and 902 cm = 41 cm.

Area of each tile = (41 × 41) cm^{2}.

Required number of tiles = (1517 × 902/41 × 41) = 814.IncorrectLength of largest tile = H.C.F. of 1517 cm and 902 cm = 41 cm.

Area of each tile = (41 × 41) cm^{2}.

Required number of tiles = (1517 × 902/41 × 41) = 814. - Question 11 of 20
##### 11. Question

The difference between the length and breadth of a rectangle is 23 m. If its perimeter is 206 m, then its area is:

CorrectWe have: (l – b) = 23 and 2(l + b) = 206 or (l + b) = 103.

Solving the two equations, we get: l = 63 and b = 40.

∴ Area = (l x b) = (63 x 40) m^{2}= 2520 m^{2}.IncorrectWe have: (l – b) = 23 and 2(l + b) = 206 or (l + b) = 103.

Solving the two equations, we get: l = 63 and b = 40.

∴ Area = (l x b) = (63 x 40) m^{2}= 2520 m^{2}. - Question 12 of 20
##### 12. Question

The length of a rectangle is halved, while its breadth is tripled. What is the percentage change in area?

CorrectLet original length = x and original breadth = y.

Original area = xy.

New length = x/2 .

New breadth = 3y.

New area = (x/2 × 3y) = 3/2 xy.

∴ Increase % = (1/2 xy × 1/xy × 100) % = 50%.IncorrectLet original length = x and original breadth = y.

Original area = xy.

New length = x/2 .

New breadth = 3y.

New area = (x/2 × 3y) = 3/2 xy.

∴ Increase % = (1/2 xy × 1/xy × 100) % = 50%. - Question 13 of 20
##### 13. Question

The length of a rectangular plot is 20 metres more than its breadth. If the cost of fencing the plot @ 26.50 per metre is Rs. 5300, what is the length of the plot in metres?

CorrectLet breadth = x metres.

Then, length = (x + 20) metres.

Perimeter = (5300/26.50) m = 200 m.

∴ 2[(x + 20) + x] = 200

⇒ 2x + 20 = 100

⇒ 2x = 80

⇒ x = 40.

Hence, length = x + 20 = 60 m.IncorrectLet breadth = x metres.

Then, length = (x + 20) metres.

Perimeter = (5300/26.50) m = 200 m.

∴ 2[(x + 20) + x] = 200

⇒ 2x + 20 = 100

⇒ 2x = 80

⇒ x = 40.

Hence, length = x + 20 = 60 m. - Question 14 of 20
##### 14. Question

A rectangular field is to be fenced on three sides leaving a side of 20 feet uncovered. If the area of the field is 680 sq. feet, how many feet of fencing will be required?

CorrectWe have: l = 20 ft and lb = 680 sq. ft.

So, b = 34 ft.

∴ Length of fencing = (l + 2b) = (20 + 68) ft = 88 ft.IncorrectWe have: l = 20 ft and lb = 680 sq. ft.

So, b = 34 ft.

∴ Length of fencing = (l + 2b) = (20 + 68) ft = 88 ft. - Question 15 of 20
##### 15. Question

A tank is 25 m long, 12 m wide and 6 m deep. The cost of plastering its walls and bottom at 75 paise per sq. m, is:

CorrectArea to be plastered = [2(l + b) × h] + (l × b)

= {[2(25 + 12) × 6] + (25 × 12)} m^{2}

= (444 + 300) m^{2}

= 744 m^{2}.

∴ Cost of plastering = Rs. (744 × 75/100) = Rs. 558.IncorrectArea to be plastered = [2(l + b) × h] + (l × b)

= {[2(25 + 12) × 6] + (25 × 12)} m^{2}

= (444 + 300) m^{2}

= 744 m^{2}.

∴ Cost of plastering = Rs. (744 × 75/100) = Rs. 558. - Question 16 of 20
##### 16. Question

Each of the questions given below consists of a question followed by three statements. You have to study the question and the statements and decide which of the statement(s) is/are necessary to answer the question.

What is the area of the hall?

I. Material cost of flooring per square metre is Rs. 2.50

II. Labour cost of flooring the hall is Rs. 3500

III. Total cost of flooring the hall is Rs. 14,500.CorrectI. Material cost = Rs. 2.50 per m

^{2}

II. Labour cost = Rs. 3500.

III. Total cost = Rs. 14,500.

Let the area be A sq. metres.

∴ Material cost = Rs. (14500 – 3500) = Rs. 11,000.

∴ 5A/2 = 11000 ⇔ A = 11000 × 2/5 = 4400 m^{2}.

Thus, all I, II and III are needed to get the answer.

∴ Correct answer is (C).IncorrectI. Material cost = Rs. 2.50 per m

^{2}

II. Labour cost = Rs. 3500.

III. Total cost = Rs. 14,500.

Let the area be A sq. metres.

∴ Material cost = Rs. (14500 – 3500) = Rs. 11,000.

∴ 5A/2 = 11000 ⇔ A = 11000 × 2/5 = 4400 m^{2}.

Thus, all I, II and III are needed to get the answer.

∴ Correct answer is (C). - Question 17 of 20
##### 17. Question

Each of the questions given below consists of a question followed by three statements. You have to study the question and the statements and decide which of the statement(s) is/are necessary to answer the question.

What is the area of a right-angled triangle?

I. The perimeter of the triangle is 30 cm.

II. The ratio between the base and the height of the triangle is 5 : 12.

III. The area of the triangle is equal to the area of a rectangle of length 10 cm.CorrectFrom II, base : height = 5 : 12.

Let base = 5x and height = 12x.

Then, hypotenuse = √(5x)^{2 }+ (12x)^{2 }= 13x.

From I, perimeter of the triangle = 30 cm.

∴ 5x + 12x + 13x = 30 ⇔ x = 1.

So, base = 5x = 5 cm, height = 12x = 12 cm.

∴ Area = (1/2 x 5 x 12)cm^{2}= 30 cm^{2}.

Thus, I and II together give the answer.

Clearly III is redundant, since the breadth of the rectangle is not given.

∴ Correct answer is (A).IncorrectFrom II, base : height = 5 : 12.

Let base = 5x and height = 12x.

Then, hypotenuse = √(5x)^{2 }+ (12x)^{2 }= 13x.

From I, perimeter of the triangle = 30 cm.

∴ 5x + 12x + 13x = 30 ⇔ x = 1.

So, base = 5x = 5 cm, height = 12x = 12 cm.

∴ Area = (1/2 x 5 x 12)cm^{2}= 30 cm^{2}.

Thus, I and II together give the answer.

Clearly III is redundant, since the breadth of the rectangle is not given.

∴ Correct answer is (A). - Question 18 of 20
##### 18. Question

Each of the questions given below consists of a question followed by three statements. You have to study the question and the statements and decide which of the statement(s) is/are necessary to answer the question.

What is the area of rectangular field?

I. The perimeter of the field is 110 metres.

II. The length is 5 metres more than the width.

III. The ratio between length and width is 6 : 5 respectively.CorrectI. 2(l + b) = 110 ⇒ l + b = 55.

II. l = (b + 5) ⇒ l – b = 5.

III. l/b = 6/5 ⇒ 5l – 6b = 0.

These are three equations in l and b. We may solve them pairwise.

∴ Any two of the three will give the answer.

∴ Correct answer is (B).IncorrectI. 2(l + b) = 110 ⇒ l + b = 55.

II. l = (b + 5) ⇒ l – b = 5.

III. l/b = 6/5 ⇒ 5l – 6b = 0.

These are three equations in l and b. We may solve them pairwise.

∴ Any two of the three will give the answer.

∴ Correct answer is (B). - Question 19 of 20
##### 19. Question

Each of these questions is followed by three statements. You have to study the question and all the three statements given to decide whether any information provided in the statement(s) is redundant and can be dispensed with while answering the given question.

What is the area of the given rectangle?

I. Perimeter of the rectangle is 60 cm.

II. Breadth of the rectangle is 12 cm.

III. Sum of two adjacent sides is 30 cm.CorrectFrom I and II, we can find the length and breadth of the rectangle and therefore the area can be obtained.

So, III is redundant.

Also, from II and III, we can find the length and breadth and therefore the area can be obtained.

So, I is redundant.

∴ Correct answer is “II and either I or III”.IncorrectFrom I and II, we can find the length and breadth of the rectangle and therefore the area can be obtained.

So, III is redundant.

Also, from II and III, we can find the length and breadth and therefore the area can be obtained.

So, I is redundant.

∴ Correct answer is “II and either I or III”. - Question 20 of 20
##### 20. Question

Each of these questions is followed by three statements. You have to study the question and all the three statements given to decide whether any information provided in the statement(s) is redundant and can be dispensed with while answering the given question.

What is the cost painting the two adjacent walls of a hall at Rs. 5 per m2 which has no windows or doors?

I. The area of the hall is 24 sq. m.

II. The breadth, length and height of the hall are in the ratio of 4 : 6 : 5 respectively.

III. Area of one wall is 30 sq. m.CorrectFrom II, let l = 4x, b = 6x and h = 5x.

Then, area of the hall = (24x^{2}) m^{2}.

From I. Area of the hall = 24 m^{2}.

From II and I, we get 24x^{2}= 24 ⇔ x = 1.

∴ l = 4 m, b = 6 and h = 5 m.

Thus, area of two adjacent walls = [(l x h) + (b x h)] m^{2}can be found out and so the cost of painting two adjacent walls may be found out.

Thus, III is redundant.

∴ Correct answer is (C).IncorrectFrom II, let l = 4x, b = 6x and h = 5x.

Then, area of the hall = (24x^{2}) m^{2}.

From I. Area of the hall = 24 m^{2}.

From II and I, we get 24x^{2}= 24 ⇔ x = 1.

∴ l = 4 m, b = 6 and h = 5 m.

Thus, area of two adjacent walls = [(l x h) + (b x h)] m^{2}can be found out and so the cost of painting two adjacent walls may be found out.

Thus, III is redundant.

∴ Correct answer is (C).