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

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.

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%

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

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%

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

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

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%

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)

If l and b are length and breadth then

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

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.

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%

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.

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

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

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).

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).

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).

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).

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).