General Questions
Practice and master this topic with our carefully crafted questions.
Radhika runs along the boundary of a rectangular park at the rate of 12 km/hr and completes one full round in 15 minutes. If the length of the park is 4 times its breadth, the area of the park is ?
Solution:
Radhika runs at the rate of 12 km/hr and completes one round of the park in 15 minutes.
Distance covered in one round = Perimeter of the rectangular park.
Time = 15 minutes = 15/60 hour = 1/4 hour
Perimeter = Speed × Time
= 12 × 1/4 km
= 3 km = 3000 m
Let the breadth of the park be x m.
Since the length is 4 times the breadth,
Length = 4x
The perimeter of the rectangle is:
2(Length + Breadth) = 3000
2(4x + x) = 3000
10x = 3000
x = 300 m
Therefore,
Breadth = 300 m
Length = 4 × 300 = 1200 m
Now, the area of the park is:
Area = Length × Breadth
= 1200 × 300
= 360000 m²
Therefore, the area of the park is 360000 m².
Correct Answer: (A) 360000 m² ✅
⚡ Quick Trick:
First find the perimeter using:
Perimeter = Speed × Time
= 12 × 1/4 km = 3 km = 3000 m
Length : Breadth = 4 : 1
Total ratio parts = 4 + 1 = 5
2 × 5 = 10 parts = 3000 m
1 part = 300 m
Area = 1200 × 300 = 360000 m²
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 :
Solution:
The ratio between the length and breadth of the rectangular park is 3 : 2.
Let the length and breadth be 3x m and 2x m respectively.
A man cycles at the speed of 12 km/hr and completes one round in 8 minutes.
Time = 8/60 hr = 2/15 hr
Perimeter = Speed × Time
= 12 × 2/15 km
= 1.6 km = 1600 m
Since one round of the park is its perimeter,
2(Length + Breadth) = 1600
2(3x + 2x) = 1600
10x = 1600
x = 160
Therefore,
Length = 3 × 160 = 480 m
Breadth = 2 × 160 = 320 m
Area of the park is:
Area = Length × Breadth
= 480 × 320
= 153600 sq. m
Therefore, the area of the park is 153600 sq. m.
Correct Answer: (B) 153600
⚡ Quick Trick:
Length : Breadth = 3 : 2, so let them be 3x and 2x.
Perimeter = 12 × (8/60) km
= 1.6 km = 1600 m
2(3x + 2x) = 1600
x = 160
Area = 480 × 320 = 153600 sq. m
A farmer wishes to start a 100 sq. m rectangular vegetable garden. Since he has only 30 m barbed wire, he fences three sides of the garden letting his house compound wall act as the forth side fencing . The dimension of the garden is :
Solution:
Let the length of the garden along the compound wall be l m and the breadth be b m.
The area of the rectangular garden is 100 sq. m.
Area = Length × Breadth
l × b = 100 ...(1)
Since one side is along the house compound wall, fencing is required only for the other three sides.
l + 2b = 30 ...(2)
From equation (2),
l = 30 − 2b
Substituting in equation (1),
(30 − 2b)b = 100
30b − 2b² = 100
b² − 15b + 50 = 0
(b − 5)(b − 10) = 0
b = 5 m or 10 m
If b = 5 m, then
l = 30 − 2 × 5 = 20 m
If b = 10 m, then
l = 30 − 2 × 10 = 10 m
Since the garden is rectangular and the fenced side is usually taken as the longer side, the required dimensions are:
Length = 20 m
Breadth = 5 m
Therefore, the dimensions of the garden are 20 m × 5 m.
Correct Answer: (B) 20 m × 5 m ✅
⚡ Quick Trick:
Fencing is required for only three sides, so:
Length + 2 × Breadth = 30
Check the options whose product is 100:
15 × 6.67 ≈ 100
20 × 5 = 100 ✓
30 × 3.33 ≈ 100
Only 20 + 2 × 5 = 30 satisfies the fencing condition.
Let each side = x cm
Then, (x + 4 )2 - x2 = 60
⇒ x 2 + 8x + 16 - x2 = 60
∴ x = 5.5 cm
Length = (40 x 10 ) dm = 400 dm.
Breadth = (15 x 10 ) dm = 150 dm.
Area of veranda = (400 x 150 ) dm2
Area of one stone = (6 x 5 ) dm2
∴ Required number of stones = (400 x 150) /(6 x 5) = 2000
Perimeter = 2 x (36 + 21 ) m = 144 m
∴ Number of flagstaffs = 144 / 3 = 38
Area = 1/2 x (Diagonal)2
= (1/2) x 5.2 x 5.2 cm2
= 13.52 cm2
Let breadth = b, length = 2b
∴ Area of rectangle = 2b x b
= 2b2
As per question.
∵ (2b - 5 ) (b + 5 ) = 2b2 + 75
⇒ 5b = 75 + 25
⇒ 5b = 100
∴ b = 100 / 5 = 20
Hence, length of the rectangle =2b
= 2 x 20
= 40 cm.
Original area = π(d/2)2
= (πd2) / 4
New area = π(2d/2)2
= πd2
Increase in area = (πd2 - πd2/4)
= 3πd2/4
∴ Required increase percent
= [(3πd2)/4 x 4/(πd2) x 100]%
= 300%
Let the diagonal of one square be (2d) cm
Then, diagonal of another square = d cm
∴ Area of first square = [ 1/2 x (2d)2] cm2
Area of second square = (1/2 x d2) cm2
∴ Ratio of area = (2d)2/ d2
= 4/1 = 4 : 1