There are 4 semi-circular gardens on each side of a square-shaped pond with each side 21 m. The cost of fencing the entire plot at the rate of Rs. 12.50 per metre is :
A wire can be bent in the form of a circle of radius 56 cm. If it is bent in the form of a square, then its area will be :
(A) 3520 cm2
(B) 6400 cm2
(C) 7744 cm2
(D) 8800 cm2
Solution:
Length of wire : $$\eqalign{ & = 2\pi \times R \cr & = \left( {2 \times \frac{{22}}{7} \times 56} \right)cm \cr & = 352\,cm \cr} $$ Side of the square : $$\eqalign{ & = \frac{{352}}{4}\,cm \cr & = 88\,cm \cr} $$ Area of the square : $$\eqalign{ & = \left( {88 \times 88} \right)c{m^2} \cr & = 7744\,c{m^2} \cr} $$
83.
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?
The diagonal of a rectangular field is 15 metres and the difference between its length its length and width is 3 metres. The area of the rectangular field is :
(A) $$9{m^2}$$
(B) $$12{m^2}$$
(C) $$21{m^2}$$
(D) $$108{m^2}$$
Solution:
Let l and b be the length and breadth of the rectangle respectively. Then, $$\eqalign{ & \Rightarrow \sqrt {{l^2} + {b^2}} = 15 \cr & \Rightarrow \left( {{l^2} + {b^2}} \right) = {\left( {15} \right)^2} \cr & \Rightarrow {l^2} + {b^2} = 225 \cr} $$ And, $$\eqalign{ & \Rightarrow l + b = 3 \cr & \Rightarrow {\left( {l - b} \right)^2} = 9 \cr & \Rightarrow {l^2} + {b^2} - 2lb = 9 \cr & \Rightarrow 225 - 2lb = 9 \cr & \Rightarrow 2lb = 216 \cr & \Rightarrow lb = 108 \cr} $$ Hence, area of the field $$ = lb = 108\,{m^2}$$
85.
The area of a circle whose radius is the diagonal of a square whose area is 4 sq. units is :
(A) 16π sq. units
(B) 4π sq. units
(C) 6π sq. units
(D) 8π sq. units
Solution:
Area of square = 4 sq.units Side of square = $$\sqrt 4 $$ = 2 units Diagonal of square = $$2\sqrt 2 $$ units Radius of the circle = $$2\sqrt 2 $$ units Radius of the circle : $$\eqalign{ & = \pi {r^2} \cr & = \pi \times {\left( {2\sqrt 2 } \right)^2} \cr & = 8\pi \,\text{sq. units} \cr} $$
86.
A room 5m ÃÂÃÂÃÂÃÂÃÂÃÂ 8 m is to be carpeted leaving a margin of 10 cm from each wall. If the cost of the carpet is Rs. 18 per sq. meter, the cost of carpeting the room will be :
(A) Rs. 673.92
(B) Rs. 682.46
(C) Rs. 691.80
(D) Rs. 702.60
Solution:
Area of the carpet : = [(5 - 0.20) × (8 - 0.20)] m2 = (4.8 × 7.8) m2 = 37.44 m2 ∴ Cost of carpeting : = Rs. (37.44 × 18) = Rs. 673.92
87.
Three plots having areas 110, 130 and 190 square metres are to be subdivided into flower beds of equal size. If the breadth of a bed is 2 metres, the maximum length of a bed van be :
(A) 5 m
(B) 11 m
(C) 13 m
(D) 19 m
Solution:
Maximum possible size of a flower bed : = (H.C.F. of 110, 130, 190) sq. m = 10 sq. m ∴ Maximum possible length = $$\left( {\frac{{10}}{2}} \right)$$ m = 5 m
88.
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 :
(A) 10%
(B) 10.08%
(C) 20%
(D) 28%
Solution:
Let the original length = x and original breadth = y Decrease in area : $$\eqalign{ & = xy - \left( {\frac{{80}}{{100}}x \times \frac{{90}}{{100}}y} \right) \cr & = \left( {xy - \frac{{18}}{{25}}xy} \right) \cr & = \frac{7}{{25}}xy \cr} $$ ∴ Decrease% : $$\eqalign{ & = \left( {\frac{7}{{25}}xy \times \frac{1}{{xy}} \times 100} \right)\% \cr & = 28\% \cr} $$
89.
A one-rupee coin is placed on a plain paper. How many coins of the same size can be placed round it so that each one touches the centre and adjacent coins ?
(A) 3
(B) 4
(C) 6
(D) 7
Solution:
When 3 congruent circles touch each other externally, the triangle formed with their centers is an equilateral triangle. Hence when a circle is surrounded by identical circles, centers of two consecutive circles make an angle of 60° with the central circle. Thus, six identical circles a surround a circle of equal radius.
90.
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