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**Quiz Description** :

**Name: **Mensuration objective question answer – Level 2

**Subject: **Aptitude

**Topic: **Mensuration

**Questions: **12 Objective type

**Time Allowed: **20 min

**Important for: **IBPS Clerk, IBPS PO, IBPS RRB, SBI Clerk, SBI PO, UPSSSC, State Police, Army, BSF, CRPF, IIT JEE and 11th & 12th exam etc.

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

1 pointsA door-frame of dimensions 3 m × 2 m is fixed on the wall of dimension 10 m × 10 m. Find the total labor charges for painting the wall if the labor charges for painting 1 m

^{2}of the wall is Rs 2.50.CorrectPainting of the wall has to be done excluding the area of the door.

Area of the door = l × b = 3 × 2 m^{2}= 6 m^{2}

Area of wall including door = side × side = 10 m × 10 m = 100 m^{2}

Area of wall excluding door = (100 − 6) m^{2}= 94 m^{2}

Total labor charges for painting the wall = Rs 2.50 × 94 = Rs 235IncorrectPainting of the wall has to be done excluding the area of the door.

Area of the door = l × b = 3 × 2 m^{2}= 6 m^{2}

Area of wall including door = side × side = 10 m × 10 m = 100 m^{2}

Area of wall excluding door = (100 − 6) m^{2}= 94 m^{2}

Total labor charges for painting the wall = Rs 2.50 × 94 = Rs 235 - Question 2 of 12
##### 2. Question

1 pointsThe area of a rectangular sheet is 500 cm

^{2}. If the length of the sheet is 25 cm, what is the perimeter of the rectangular sheet?CorrectArea of the rectangular sheet = 500 cm

^{2}

Length (l) = 25 cm

Area of the rectangle = l × b (where b = width of the sheet)

Therefore, width b = Area/l = 500/25 = 20cm

Perimeter of sheet = 2 × (l + b) = 2 × (25 + 20) cm = 90 cm

So, the width of the rectangular sheet is 20 cm and its perimeter is 90 cm.IncorrectArea of the rectangular sheet = 500 cm

^{2}

Length (l) = 25 cm

Area of the rectangle = l × b (where b = width of the sheet)

Therefore, width b = Area/l = 500/25 = 20cm

Perimeter of sheet = 2 × (l + b) = 2 × (25 + 20) cm = 90 cm

So, the width of the rectangular sheet is 20 cm and its perimeter is 90 cm. - Question 3 of 12
##### 3. Question

1 pointsA wire is in the shape of a square of side 10 cm. If the wire is rebent into a rectangle of length 12 cm, find its breadth. Which encloses more area, the square or the rectangle?

CorrectSide of the square = 10 cm

Length of the wire = Perimeter of the square = 4 × side = 4 × 10 cm = 40 cm

Length of the rectangle, l = 12 cm.

Let b be the breadth of the rectangle.

Perimeter of rectangle = Length of wire = 40 cm

Perimeter of the rectangle = 2 (l + b)

Thus, 40 = 2 (12 + b)

or, 40/2 = 12 + b

Therefore, b = 20 – 12 = 8 cm

The breadth of the rectangle is 8 cm.

Area of the square = (side)^{2}= 10 cm × 10 cm = 100 cm^{2}

Area of the rectangle = l × b= 12 cm × 8 cm = 96 cm^{2}

So, the square encloses more area even though its perimeter is the same as that of the rectangle.IncorrectSide of the square = 10 cm

Length of the wire = Perimeter of the square = 4 × side = 4 × 10 cm = 40 cm

Length of the rectangle, l = 12 cm.

Let b be the breadth of the rectangle.

Perimeter of rectangle = Length of wire = 40 cm

Perimeter of the rectangle = 2 (l + b)

Thus, 40 = 2 (12 + b)

or, 40/2 = 12 + b

Therefore, b = 20 – 12 = 8 cm

The breadth of the rectangle is 8 cm.

Area of the square = (side)^{2}= 10 cm × 10 cm = 100 cm^{2}

Area of the rectangle = l × b= 12 cm × 8 cm = 96 cm^{2}

So, the square encloses more area even though its perimeter is the same as that of the rectangle. - Question 4 of 12
##### 4. Question

1 pointsThe area of a square and a rectangle are equal. If the side of the square is 40 cm and the breadth of the rectangle is 25 cm, find the perimeter of the rectangle.

CorrectArea of square = (side)2 = = 40 cm × 40 cm = 1600 cm

^{2}

It is given that,

The area of the rectangle = The area of the square

Area of the rectangle = 1600 cm^{2}, breadth of the rectangle = 25 cm.

Area of the rectangle = l × b

or 1600 = l × 25

or 1600/25 = l or l= 64 cm

So, the length of rectangle is 64 cm.

Perimeter of the rectangle = 2 (l + b) = 2 (64 + 25) cm = 2 × 89 cm = 178 cm

So, the perimeter of the rectangle is 178 cm even though its area is the same as that of the square.IncorrectArea of square = (side)2 = = 40 cm × 40 cm = 1600 cm

^{2}

It is given that,

The area of the rectangle = The area of the square

Area of the rectangle = 1600 cm^{2}, breadth of the rectangle = 25 cm.

Area of the rectangle = l × b

or 1600 = l × 25

or 1600/25 = l or l= 64 cm

So, the length of rectangle is 64 cm.

Perimeter of the rectangle = 2 (l + b) = 2 (64 + 25) cm = 2 × 89 cm = 178 cm

So, the perimeter of the rectangle is 178 cm even though its area is the same as that of the square. - Question 5 of 12
##### 5. Question

1 pointsThe length and the breadth of a rectangular piece of land are 500 m and 300 m respectively. What will be its area?

CorrectArea of the rectangle = l × b = 500 × 300 = 150000 m

^{2}.IncorrectArea of the rectangle = l × b = 500 × 300 = 150000 m

^{2}. - Question 6 of 12
##### 6. Question

1 pointsA square of side 5 cm, divided into 4 equal triangles. What is the area of each triangle?

CorrectThe area of each triangle = 1/4× (Area of the square)

= 1/4× (side)^{2}= 1/4× (5)^{2}cm^{2}= 6.25 cm^{2}IncorrectThe area of each triangle = 1/4× (Area of the square)

= 1/4× (side)^{2}= 1/4× (5)^{2}cm^{2}= 6.25 cm^{2} - Question 7 of 12
##### 7. Question

1 pointsOne of the sides and the corresponding height of a parallelogram are 4 cm and 3 cm respectively. Find the area of the parallelogram.

CorrectGiven that;

The length of base (b) = 4 cm, height (h) = 3 cm

Area of the parallelogram = b × h = 4 cm × 3 cm = 12 cm^{2}IncorrectGiven that;

The length of base (b) = 4 cm, height (h) = 3 cm

Area of the parallelogram = b × h = 4 cm × 3 cm = 12 cm^{2} - Question 8 of 12
##### 8. Question

1 pointsWhat is the circumference of a circle of diameter 10 cm?

CorrectDiameter of the circle (d) = 10 cm

Circumference of circle = π d = 3.14 × 10 cm = 31.4 cm

So, the circumference of the circle of diameter 10 cm is 31.4 cm.IncorrectDiameter of the circle (d) = 10 cm

Circumference of circle = π d = 3.14 × 10 cm = 31.4 cm

So, the circumference of the circle of diameter 10 cm is 31.4 cm. - Question 9 of 12
##### 9. Question

1 pointsA gardener wants to fence a circular garden of diameter 21m. Find the length of the rope he needs to purchase, if he makes 2 rounds of fence. Also find the costs of the rope, if it cost Rs 4 per meter.

CorrectGiven that the diameter of the circular garden = 21 m

The radius (r) of circle = 10.5 m.

We know that the circumference of circle = 2 × π × r

∴ The circumference of the circular garden = 2 × π × 10.5 = 65.94 ≈ 66 m

∴ The length of rope needed to makes 2 rounds of fence = 2 × 66 = 132 m

Given the cost of rope (per meter) = Rs. 4 per meter

∴The total cost of rope = Rs.132 × 4 = Rs. 528IncorrectGiven that the diameter of the circular garden = 21 m

The radius (r) of circle = 10.5 m.

We know that the circumference of circle = 2 × π × r

∴ The circumference of the circular garden = 2 × π × 10.5 = 65.94 ≈ 66 m

∴ The length of rope needed to makes 2 rounds of fence = 2 × 66 = 132 m

Given the cost of rope (per meter) = Rs. 4 per meter

∴The total cost of rope = Rs.132 × 4 = Rs. 528 - Question 10 of 12
##### 10. Question

1 pointsFrom a circular sheet of radius 4 cm, a circle of radius 3 cm is removed. Find the area of the remaining sheet.

CorrectRadius of the outer circle = 4 cm

So, area of the outer circle = π × r^{2}= 3.14 × 4 × 4 = 50.24 cm^{2}

Radius of the inner circle = 3 cm

Area of the smaller circle = π × r^{2}= 3.14 × 3 × 3 = 28.2 cm^{2}

Area of the Remaining circular sheet = (50.24 – 28.26) cm2 = 21.98 cm^{2}≈ 22 cm^{2}IncorrectRadius of the outer circle = 4 cm

So, area of the outer circle = π × r^{2}= 3.14 × 4 × 4 = 50.24 cm^{2}

Radius of the inner circle = 3 cm

Area of the smaller circle = π × r^{2}= 3.14 × 3 × 3 = 28.2 cm^{2}

Area of the Remaining circular sheet = (50.24 – 28.26) cm2 = 21.98 cm^{2}≈ 22 cm^{2} - Question 11 of 12
##### 11. Question

1 pointsA rectangular park is 45 m long and 30 m wide. A path 2.5 m wide is constructed outside the park. Find the area of the path.

CorrectLet ABCD represent the rectangular park and the shaded region represent the path 2.5 m wide. To find the area of the path, we need to find (Area of rectangle PQRS – Area of rectangle ABCD).

We have, PQ = (45 + 2.5 + 2.5) m = 50 m

PS = (30 + 2.5 + 2.5) m = 35 m

Area of the rectangle ABCD = l × b = 45 × 30 m^{2}= 1350 m^{2}

Area of the rectangle PQRS = l × b = 50 × 35 m^{2}= 1750 m^{2}

Area of the path = Area of the rectangle PQRS − Area of the rectangle ABCD

= (1750 − 1350) m^{2}= 400 m^{2}IncorrectLet ABCD represent the rectangular park and the shaded region represent the path 2.5 m wide. To find the area of the path, we need to find (Area of rectangle PQRS – Area of rectangle ABCD).

We have, PQ = (45 + 2.5 + 2.5) m = 50 m

PS = (30 + 2.5 + 2.5) m = 35 m

Area of the rectangle ABCD = l × b = 45 × 30 m^{2}= 1350 m^{2}

Area of the rectangle PQRS = l × b = 50 × 35 m^{2}= 1750 m^{2}

Area of the path = Area of the rectangle PQRS − Area of the rectangle ABCD

= (1750 − 1350) m^{2}= 400 m^{2} - Question 12 of 12
##### 12. Question

1 pointsTwo cross roads, each of width 5 m, run at right angles through the centre of a rectangular park of length 70 m and breadth 45 m and parallel to its sides. Find the cost of constructing the roads at the rate of Rs 105 per m

^{2}.CorrectArea of the cross roads is the area of shaded portion, i.e., the area of the rectangle PQRS and the area of the rectangle EFGH. But while doing this, the area of the square KLMN is taken twice, which is to be subtracted.

Now, PQ = 5 m and PS = 45 m

EH = 5 m and EF = 70 m

KL = 5 m and KN = 5 mArea of the path = Area of the rectangle PQRS area of the rectangle EFGH – Area of the square KLMN

= PS × PQ + EF × EH – KL × KN

= (45 × 5 + 70 × 5 − 5 × 5) m^{2}

= (225 + 350 − 25) m2 = 550 m^{2}

Cost of constructing the path = Rs 105 × 550 = Rs 57,750IncorrectArea of the cross roads is the area of shaded portion, i.e., the area of the rectangle PQRS and the area of the rectangle EFGH. But while doing this, the area of the square KLMN is taken twice, which is to be subtracted.

Now, PQ = 5 m and PS = 45 m

EH = 5 m and EF = 70 m

KL = 5 m and KN = 5 mArea of the path = Area of the rectangle PQRS area of the rectangle EFGH – Area of the square KLMN

= PS × PQ + EF × EH – KL × KN

= (45 × 5 + 70 × 5 − 5 × 5) m^{2}

= (225 + 350 − 25) m2 = 550 m^{2}

Cost of constructing the path = Rs 105 × 550 = Rs 57,750