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#### Information

**Name: **Mensuration aptitude questions and answers test – Level 1

**Subject: **Aptitude ( Quantitative Aptitude / Maths)

**Topic: **Mensuration : Length and Area

**Questions: **12 Objective type

**Time Allowed: **20 minutes

**Important for: **IBPS Clerk, SBI Clerk, SSC CHSL, SSC GD, SSC CGL, Railway, Police, Army, BSF, CRPF, CTET, UPTET, REET, HTET, UTET etc Competitions & Polytechnic, IIT JEE, UPTU, UTU, WBJEE, AIEEE, AIPMT, B Sc, M Sc Entrance examination.

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- Question 1 of 12
##### 1. Question

1 pointsIn ΔABC,∠B = 90°, AB = 18cm and BC = 24cm. Calculate the length of AC.

Correct**Explanation:**By Pythagoras Theorem, AC

^{2}= AB^{2}+ BC^{2}

= 18^{2}+ 24^{2}

= 324 + 576

= 900

AC = √900

= 30cmIncorrect**Explanation:**By Pythagoras Theorem, AC

^{2}= AB^{2}+ BC^{2}

= 18^{2}+ 24^{2}

= 324 + 576

= 900

AC = √900

= 30cm - Question 2 of 12
##### 2. Question

1 pointsA square has the perimeter 40cm. What is the sum of the diagonals?

CorrectLet ‘a’ be the length of the side of the square ∎ABCD and AC is a diagonal of the square.

Perimeter of square ∎ABCD = 4 a units

4a = 40cm

a = 4

40 = 10cm

We know that in a square each angle is 90° and the diagonals are equal.

In ΔABC, AC^{2}= AB^{2}+ BC^{2}= 10^{2}+ 10^{2}= 100 + 100 = 200

∴AC = √200= 10√2 = 10 × 1.414 = 14.14cm

Diagonal AC = Diagonal BD

Hence, Sum of the diagonals = 14.14 + 14.14 = 28.28cmIncorrectLet ‘a’ be the length of the side of the square ∎ABCD and AC is a diagonal of the square.

Perimeter of square ∎ABCD = 4 a units

4a = 40cm

a = 4

40 = 10cm

We know that in a square each angle is 90° and the diagonals are equal.

In ΔABC, AC^{2}= AB^{2}+ BC^{2}= 10^{2}+ 10^{2}= 100 + 100 = 200

∴AC = √200= 10√2 = 10 × 1.414 = 14.14cm

Diagonal AC = Diagonal BD

Hence, Sum of the diagonals = 14.14 + 14.14 = 28.28cm - Question 3 of 12
##### 3. Question

1 pointsThe two adjacent sides of a parallelogram are 5cm and 4 cm respectively, and if the respective diagonal is 7cm then find the area of the parallelogram?

CorrectRequired area = 2√(s(s-a)(s-b)(s-c))

Where, S = (a + b + D)/2; i.e.; (5 + 4 + 7)/2 = 8

⟹2√(8(8-5)(8-4)(8-7))

⟹2√(8×3×4)

⟹8√6

⟹19.6 cm^{2}IncorrectRequired area = 2√(s(s-a)(s-b)(s-c))

Where, S = (a + b + D)/2; i.e.; (5 + 4 + 7)/2 = 8

⟹2√(8(8-5)(8-4)(8-7))

⟹2√(8×3×4)

⟹8√6

⟹19.6 cm^{2} - Question 4 of 12
##### 4. Question

1 pointsIn a parallelogram, the lengths of adjacent sides are 12cm and 14cm respectively. If the length of one diagonal is 16cm, find the length of the other diagonal.

CorrectIn a parallelogram,

The sum of the squares of the diagonals = 2 × (Sum of the squares of the two adjacent sides)

D_{1}^{2}+ D_{2}^{2}= 2 (a^{2}+ b^{2})

16^{2}+ D_{2}^{2}= 2(12^{2}+ 14^{2})

256 + D_{2}^{2}= 2(144 + 196)

D_{2}^{2}= 680 – 256 = 424

∴ D_{2}= √424 = 20.6cmIncorrectIn a parallelogram,

The sum of the squares of the diagonals = 2 × (Sum of the squares of the two adjacent sides)

D_{1}^{2}+ D_{2}^{2}= 2 (a^{2}+ b^{2})

16^{2}+ D_{2}^{2}= 2(12^{2}+ 14^{2})

256 + D_{2}^{2}= 2(144 + 196)

D_{2}^{2}= 680 – 256 = 424

∴ D_{2}= √424 = 20.6cm - Question 5 of 12
##### 5. Question

1 pointsIn a trapezium, parallel sides are 60 and 90 centimeters respectively and non-parallel sides are 40 and 50 centimeters respectively. Find its area.

CorrectLet k = difference between the parallel sides = 90 – 60 = 30 cm

Now, s= (k + c + d)/2 = (30 + 40 + 50)/2 = 120/2 = 60 cm

∴Area = (a + b )/k √(s(s-k)(s-c) (s-d))

= (60 + 90 )/30 √(60(60-30)(60-40) (60-50))

= 5√(60 × 30 ×20 ×10)

= 5 × 600

= 3000 cm^{2}IncorrectLet k = difference between the parallel sides = 90 – 60 = 30 cm

Now, s= (k + c + d)/2 = (30 + 40 + 50)/2 = 120/2 = 60 cm

∴Area = (a + b )/k √(s(s-k)(s-c) (s-d))

= (60 + 90 )/30 √(60(60-30)(60-40) (60-50))

= 5√(60 × 30 ×20 ×10)

= 5 × 600

= 3000 cm^{2} - Question 6 of 12
##### 6. Question

1 pointsIn a trapezium, parallel sides are 60 and 90 centimeters respectively and non-parallel sides are 40 and 50 centimeters respectively. Find the perpendicular distance between the two parallel sides of the trapezium.

CorrectLet k = difference between the parallel sides = 90 – 60 = 30 cm

Now, s= (k + c + d)/2 = (30 + 40 + 50)/2 = 120/2 = 60 cm

h= 2/k √(s(s-k)(s-c)(s-d))

=2/30 √(60(60-30)(60-40) (60-50))

= 2/30 √(60 × 30× 20× 10)

= 1/15×600

= 40 cmIncorrectLet k = difference between the parallel sides = 90 – 60 = 30 cm

Now, s= (k + c + d)/2 = (30 + 40 + 50)/2 = 120/2 = 60 cm

h= 2/k √(s(s-k)(s-c)(s-d))

=2/30 √(60(60-30)(60-40) (60-50))

= 2/30 √(60 × 30× 20× 10)

= 1/15×600

= 40 cm - Question 7 of 12
##### 7. Question

1 pointsA 5100 cm

^{2}trapezium has the perpendicular distance between the two parallel sides 60 m. If one of the parallel sides be 40 m then find the length of the other parallel side.CorrectA = 1/2 (a + b)h

5100 = 1/2 (40 + b)×60

∴ b= 130 mIncorrectA = 1/2 (a + b)h

5100 = 1/2 (40 + b)×60

∴ b= 130 m - Question 8 of 12
##### 8. Question

1 pointsWhat is the area of a rectangle if it is 23 meters 7 decimeters long and 14 meters and 4 decimeters 8 centimeters wide?

CorrectExplanation:

Length of the rectangle (l) = 23.7 meters

Breadth of the rectangle (b) = 14.48 meters

∴ Area of the rectangle (A) = (l × b) = (23.7 × 14.48) meter^{2}

= 343.18 meter^{2}IncorrectExplanation:

Length of the rectangle (l) = 23.7 meters

Breadth of the rectangle (b) = 14.48 meters

∴ Area of the rectangle (A) = (l × b) = (23.7 × 14.48) meter^{2}

= 343.18 meter^{2} - Question 9 of 12
##### 9. Question

1 pointsWhat amount needs to be spent in carpeting the floor if the carpet is available at Rs. 20 per meter while the carpet is 75 centimeter wide that used to cover the floor of a room which is 20 meters long and 12 meters broad?

CorrectAmount Required = Rate per meter × ((Length of room × Breadth of room)/(Width of carpet ))

= 20 × (20 × 12)/0.75

= Rs. 6400IncorrectAmount Required = Rate per meter × ((Length of room × Breadth of room)/(Width of carpet ))

= 20 × (20 × 12)/0.75

= Rs. 6400 - Question 10 of 12
##### 10. Question

1 pointsHow many paving stones each measuring 2.5 meter × 2 meter are required to pave a rectangular courtyard 30 meter long and 16.5 meter wide?

CorrectNumber of tiles required = (length × breadth of courtyard)/(length × breadth of each tile)

= (30 × 16.5 )/(2.5 × 2)

= 99IncorrectNumber of tiles required = (length × breadth of courtyard)/(length × breadth of each tile)

= (30 × 16.5 )/(2.5 × 2)

= 99 - Question 11 of 12
##### 11. Question

1 pointsIf the length of rectangle increases by 9% and the breadth of that rectangle decrease by 7%, then find the % change in area.

Correct% change in area = (% Change in length)+ (% Change in breadth)+ ((% change in length)× (% change in breadth))/100%

∴ % change in area = 9 – 7 + (9 × (-7))/90 = 2 – (63/100) = 1.37%

Since there is +ve sign, the area increased by 1.37%.

Incorrect% change in area = (% Change in length)+ (% Change in breadth)+ ((% change in length)× (% change in breadth))/100%

∴ % change in area = 9 – 7 + (9 × (-7))/90 = 2 – (63/100) = 1.37%

Since there is +ve sign, the area increased by 1.37%.

- Question 12 of 12
##### 12. Question

1 pointsIf sides of a square are increased by 28%, then its area is increased by what percentage?

Correct%increase in area = 2× (% Change in breadth)+ [ (% change in breadth) ]

^{2}/100%

∴ % increase in area = 2× 28 + [28]^{2}/100 = 56 + 7.84 = 63.84%≈64.0%Incorrect%increase in area = 2× (% Change in breadth)+ [ (% change in breadth) ]

^{2}/100%

∴ % increase in area = 2× 28 + [28]^{2}/100 = 56 + 7.84 = 63.84%≈64.0%