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Olympiad Test: Elementary Mensuration-I -1 - Class 7 MCQ


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20 Questions MCQ Test Mathematics Olympiad Class 7 - Olympiad Test: Elementary Mensuration-I -1

Olympiad Test: Elementary Mensuration-I -1 for Class 7 2024 is part of Mathematics Olympiad Class 7 preparation. The Olympiad Test: Elementary Mensuration-I -1 questions and answers have been prepared according to the Class 7 exam syllabus.The Olympiad Test: Elementary Mensuration-I -1 MCQs are made for Class 7 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Olympiad Test: Elementary Mensuration-I -1 below.
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Olympiad Test: Elementary Mensuration-I -1 - Question 1

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:

Detailed Solution for Olympiad Test: Elementary Mensuration-I -1 - Question 1

Perimeter = Distance covered in 8 min.
= (12000/60 × 8) = 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) m2 = 153600 m2.

Olympiad Test: Elementary Mensuration-I -1 - Question 2

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:

Detailed Solution for Olympiad Test: Elementary Mensuration-I -1 - Question 2

100 cm is read as 102 cm.
∴ A1 = (100 × 100) cm2 and
A2 = (102 × 102) cm2.
(A2 – A1) = [(102)2 – (100)2]
 = (102 + 100) × (102 – 100)
 = 404 cm2
∴ Percentage error
= [404/(100 × 100) × 100]% = 4.04%

Olympiad Test: Elementary Mensuration-I -1 - Question 3

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?

Detailed Solution for Olympiad Test: Elementary Mensuration-I -1 - Question 3

Given
2(l + b)/b = 5/l
⇒ 2l + 2b = 5b
⇒ 3b = 2l
b = 2l/3
Then, Area = 216 cm2
⇒ l × b = 216
⇒ l × 2l/3 = 216
⇒ l2 = 324
⇒ l = 18 cm

Olympiad Test: Elementary Mensuration-I -1 - Question 4

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

Detailed Solution for Olympiad Test: Elementary Mensuration-I -1 - Question 4

Let original length = x metres and original breadth = y metres.
Original area = (xy) m2.
New length = (120/100) × m = (6/5) × m
New breadth = (120/100) y m = (6/5) y m
New area = (6/5) × m × (6/5) y m
= (36/25)xy m2
The difference between the original area
= xy  and new area 36/25 xy is
= (36/25)xy – xy
= xy( 36/25 – 1)
= xy( 11/25) or (11/25)xy
∴ Increase %
= [(11/25)xy × 1/xy × 100]%
= 44%

Olympiad Test: Elementary Mensuration-I -1 - Question 5

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?

Detailed Solution for Olympiad Test: Elementary Mensuration-I -1 - Question 5

Area of the park = (60 × 40) m2
= 2400 m2
Area of the lawn = 2109 m2.
∴ Area of the crossroads
= (2400 – 2109) m2 = 291 m2.
Let the width of the road be x metres. Then,
60x + 40x – x2 = 291
⇒ x2 – 100x + 291 = 0
⇒ (x – 97)(x – 3) = 0
⇒ x = 3

Olympiad Test: Elementary Mensuration-I -1 - Question 6

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?

Detailed Solution for Olympiad Test: Elementary Mensuration-I -1 - Question 6


∴ Area of closet = (6 × 4.5) sq. ft = 27 sq. ft.

Olympiad Test: Elementary Mensuration-I -1 - Question 7

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:

Detailed Solution for Olympiad Test: Elementary Mensuration-I -1 - Question 7

Let original length = x
and original breadth = y
Decrease in area
= xy – [(80/100)x × (90/100)y]
= (7/25)xy
∴ Decrease %  = [(7/25)xy × 1/xy × 100]%
= 28%

Olympiad Test: Elementary Mensuration-I -1 - Question 8

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

Detailed Solution for Olympiad Test: Elementary Mensuration-I -1 - Question 8

Let the side of the square (ABCD) be x metres.
Then, AB + BC = 2x metres.

AC = √2x = (1.41x) m.
Saving on 2x metres = (0.59x) m.
Saving % = (0.59x/2x) × 100%
= 30 %( approx)

Olympiad Test: Elementary Mensuration-I -1 - Question 9

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

Detailed Solution for Olympiad Test: Elementary Mensuration-I -1 - Question 9

If l and b are length and breadth then

Olympiad Test: Elementary Mensuration-I -1 - Question 10

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?

Detailed Solution for Olympiad Test: Elementary Mensuration-I -1 - Question 10

Length of largest tile
= H.C.F. of 1517 cm and 902 cm = 41 cm.
Area of each tile = (41 × 41) cm2.
∴ Required number of tiles
= (1517 × 902)/(41 × 41) = 814

Olympiad Test: Elementary Mensuration-I -1 - Question 11

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

Detailed Solution for Olympiad Test: Elementary Mensuration-I -1 - Question 11

We have: (l – b) = 23 and 2(l + b) = 206 or
(l + b) = 103.
where, l =  length , b = breath
Solving these two equations, we get:
l = 63 and b = 40.
∴ Area = (l × b) = (63 × 40) m2
= 2520 m2.

Olympiad Test: Elementary Mensuration-I -1 - Question 12

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

Detailed Solution for Olympiad Test: Elementary Mensuration-I -1 - Question 12

Let 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%

Olympiad Test: Elementary Mensuration-I -1 - Question 13

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?

Detailed Solution for Olympiad Test: Elementary Mensuration-I -1 - Question 13

Let 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.

Olympiad Test: Elementary Mensuration-I -1 - Question 14

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?

Detailed Solution for Olympiad Test: Elementary Mensuration-I -1 - Question 14

We have: l = 20 ft and lb = 680 sq. ft.
So, b = 34 ft.
∴ Length of fencing = (l + 2b)
= (20 + 68) ft = 88 ft.

Olympiad Test: Elementary Mensuration-I -1 - Question 15

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:

Detailed Solution for Olympiad Test: Elementary Mensuration-I -1 - Question 15

Area to be plastered
= [2(l + b) × h] + (l × b)
= {[2(25 + 12) × 6] + (25 × 12)} m2
= (444 + 300) m2
= 744 m2.
∴ Cost of plastering = Rs (744 x 75/100)
= Rs 558

Olympiad Test: Elementary Mensuration-I -1 - Question 16

The area of playground is 1600 m2. What is the perimeter?
I. It is a perfect square playground.
II. It costs Rs 3200 to put a fence around the playground at the rate of Rs 20 per metre.

Detailed Solution for Olympiad Test: Elementary Mensuration-I -1 - Question 16

Area = 1600 m2.
I. Side = 1600 m = 40 m. So, perimeter
= (40 × 4) m = 160 m.
∴ I alone gives the answer.
II. Perimeter = Total cost = 3200 m
= 160 m
Cost per metre = 20
∴ II alone gives the answer.
∴ Correct answer is (c).

Olympiad Test: Elementary Mensuration-I -1 - Question 17

The area of a rectangle is equal to the area of right-angles triangle. What is the length of the rectangle?
I. The base of the triangle is 40 cm.
II. The height of the triangle is 50 cm.

Detailed Solution for Olympiad Test: Elementary Mensuration-I -1 - Question 17

Given: Area of rectangle
= Area of a right-angle triangle.
⇒ l x b = ½ x B x H
I gives, B = 40 cm.
II gives, H = 50 cm.
Thus, to find l, we need b also, which is not given.
∴ Given data is not sufficient to give the answer.
∴ Correct answer is (d).

Olympiad Test: Elementary Mensuration-I -1 - Question 18

What is the height of the triangle?
I. The area of the triangle is 20 times its base.
II. The perimeter of the triangle is equal to the perimeter of a square of side 10 cm.

Detailed Solution for Olympiad Test: Elementary Mensuration-I -1 - Question 18

I. A = 20 x B ½ x B x H = 20 x B
⇒ H = 40.
∴ I alone gives the answer.
II gives the perimeter of the triangle = 40 cm.
This does not give the height of the triangle.
∴ Correct answer is (a).

Olympiad Test: Elementary Mensuration-I -1 - Question 19

What will be the cost of painting the inner walls of a room if the rate of painting is Rs 20 per square foot?
I. Circumference of the floor is 44 feet.
II. The height of the wall of the room is 12 feet.

Detailed Solution for Olympiad Test: Elementary Mensuration-I -1 - Question 19

I gives, 2 πR = 44
II gives, H = 12.
∴ A = 2 πRH = (4 × 12)
Cost of painting = Rs (44 × 12 × 20)
Thus, I and II together give the answer.
∴ Correct answer is (d).

Olympiad Test: Elementary Mensuration-I -1 - Question 20

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.

Detailed Solution for Olympiad Test: Elementary Mensuration-I -1 - Question 20

I. Material cost = Rs 2.50 per m2
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 m2
Thus, all I, II and III are needed to get the answer.
∴ Correct answer is (c).

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