Area of a rectangle is 150 sq. metre. When the breadth of the same rectangle is increased by 2 metres and the length decreased by 5 metres the area of the rectangle decreases by 30 square metres. What is the perimeter of the square whose sides are equal to the length of the rectangle ?
(A) 76 m
(B) 72 m
(C) 120 m
(D) 60 m
Solution:
Let the length of rectangle be $$l$$ metre and the breadth of the rectangle be $$b$$ metre Then area of the rectangle = $$l$$ × $$b$$ $$lb$$ = 150 m2 . . . . . (i) According to the question, $$\eqalign{ & \left( {l - 5} \right) \times \left( {b + 2} \right) = 150 - 30 \cr & \Rightarrow \left( {l - 5} \right) \times \left( {b + 2} \right) = 120 \cr & \Rightarrow lb - 5b + 2l - 10 = 120 \cr & \Rightarrow 150 - 5b + 2l - 10 = 120 \cr & \Rightarrow 5b - 2l = 20 \cr & \Rightarrow \frac{{5 \times 150}}{l} - 2{l^2} = 20l \cr & \Rightarrow 2{l^2} + 20l - 750 = 0 \cr & \Rightarrow {l^2} + 10l - 375 = 0 \cr & \Rightarrow l\left( {l + 25} \right) - 15\left( {l + 25} \right) = 0 \cr & \Rightarrow \left( {l + 15} \right)\left( {l + 25} \right) = 0 \cr} $$ On solving both equations we get, $$l$$ = 15 m and $$b$$ = 10 m Side of square = length of rectangle (given) So, the perimeter of the square : = 4 × $$l$$ = 4 × 15 = 60 m
22.
Two boys are running on two different circular paths with same centre. If their radii are 5 m and 10 m, the maximum possible distance between them is :
(A) 5 m
(B) 10 m
(C) 15 m
(D) 20 m
Solution:
Since the diameter is the longest chord of a circle So, maximum possible distance = (10 + 5) m = 15 m
23.
A rectangular garden (60 m ÃÂÃÂÃÂÃÂÃÂÃÂ 40 m) is surrounded by a road of width 2 m, the road is covered by tiles and the garden is fenced. If the total expenditure is Rs. 51600 and rate of fencing is Rs. 50 per metre, then the cost of covering 1 sq. m of road by tiles is :
(A) Rs. 10
(B) Rs. 50
(C) Rs. 100
(D) Rs. 150
Solution:
Length of the fence : = 2(60 + 40) m = 200 m Cost of fencing : = Rs. (200 × 50) = Rs. 10000 Area of the road : = [(64 × 44) - (60 × 40)] m2 = (2816 - 2400) m2 = 416 m2 Let the cost of tiling the road be Rs. x per sq.m ∴ 416x + 10000 = 51600 ⇒ 416x = 41600 ⇒ x = Rs. 100
24.
The radius of the wheel of a vehicle is 70 cm. The wheel makes 10 revolutions in 5 seconds. The speed of the vehicle is :
The area of a rectangle is 252 cm2 and its length and breadth are in the ratio of 9 : 7 respectively. What is its perimeter ?
(A) 64 cm
(B) 68 cm
(C) 96 cm
(D) 128 cm
Solution:
Let the length and breadth of the rectangle be (9x) cm and (7x) cm respectively. Then, $$\eqalign{ & 9x \times 7x = 252 \cr & \Rightarrow 63{x^2} = 252 \cr & \Rightarrow {x^2} = 4 \cr & \Rightarrow x = 2 \cr} $$ So, length = 18 cm, breadth = 14 cm ∴ Perimeter : = 2(18 + 14) cm = 64 cm
26.
The length of a room is 5.5 m and width is 3.75 m. Find the cost of paving the floor by slabs at the rate of Rs. 800 per square metre.
(A) Rs. 15000
(B) Rs. 15500
(C) Rs. 15600
(D) Rs. 16500
Solution:
Area of the floor = (5.5 × 3.75)m2 = 20.625m2 ∴ Cost of paying = Rs. (800 × 20.625) = Rs. 16500
27.
The circumference of the front wheel of a cart is 40 ft long and that of the back wheel is 48 ft long. What is the distance travelled by the cart, when the front wheel has done five more revolutions than the rear wheel ?
(A) 850 ft
(B) 950 ft
(C) 1200 ft
(D) 1450 ft
Solution:
Let the rear wheel make x revolutions Then, the front wheel makes (x + 5) revolutions (x + 5) × 40 = 48x ⇒ 8x = 200 ⇒ x = 25 Distance travelled by the cart : = (48 × 25) ft = 1200 ft
28.
The sides of a triangle are consecutive integers. The perimeter of the triangle is 120 cm. Find the length of the greatest side :
(A) 39 cm
(B) 40 cm
(C) 41 cm
(D) 42 cm
Solution:
Let the sides of the triangles be x cm, (x + 1) cm and (x + 2) cm respectively. Then, $$\eqalign{ & x + \left( {x + 1} \right) + \left( {x + 2} \right) = 120 \cr & \Rightarrow 3x + 3 = 120 \cr & \Rightarrow 3x = 117 \cr & \Rightarrow x = 39 \cr} $$ ∴ Length of greatest side : = (39 + 2) cm = 41 cm
29.
An equilateral triangle is described on the diagonal of a square. What is the ratio of the area of the triangle to that of the square ?
(A) 2 : $$\sqrt 3 $$
(B) 4 : $$\sqrt 3 $$
(C) $$\sqrt 3 $$ : 2
(D) $$\sqrt 3 $$ : 4
Solution:
Let the side of the square be a cm Then, the length of its diagonal = $$\sqrt 2 $$ a cm Area of equilateral triangle with side : $$\eqalign{ & = \sqrt 2 a \cr & = \frac{{\sqrt 3 }}{4} \times {\left( {\sqrt 2 a} \right)^2} \cr & = \frac{{\sqrt 3 {a^2}}}{2} \cr} $$ ∴ Required ratio : $$\eqalign{ & = \frac{{\sqrt 3 {a^2}}}{2}:{a^2} \cr & = \sqrt 3 :2 \cr} $$
30.
How many squares with side $$\frac{1}{2}$$ inch long are needed to cover a rectangle that is 4 feet long and 6 feet wide ?
(A) 24
(B) 96
(C) 3456
(D) 13824
Solution:
length of rectangle = 4ft = (4 ×12) inch = 48 inch length of rectangle = 6ft = (6 ×12) inch = 72 inch ∴ Number of squares : $$\eqalign{ & = \frac{{48 \times 72}}{{\frac{1}{2} \times \frac{1}{2}}} \cr & = 13824 \cr} $$