A typist uses a paper 30 cm by 15 cm. He leaves a margin of 2.5 cm at the top and bottom and 1.25 cm on either side. What percentage of paper area is approximately available for typing ?
(A) 60%
(B) 65%
(C) 70%
(D) 80%
Solution:
Area of the sheet = (30 × 15) cm2 = 450 cm2 Area used for typing = [(30 - 5) × (15 - 2.5)] cm2 = 312.5 cm2 ∴ Required percentage : $$\eqalign{ & = \left( {\frac{{312.5}}{{450}} \times 100} \right)\% \cr & = 69.4\% \approx 70\% \cr} $$
62.
The area of a rectangle is thrice that of a square. If the length of the rectangle is 40 cm and its breadth is times that of the side pf the square, then the side if the square is :
(A) 15 cm
(B) 20 cm
(C) 30 cm
(D) 60 cm
Solution:
Let the side of the square be x cm Then, its area = x2 cm2 Area of the rectangle = 3x2 cm2 ∴ 40 × $$\frac{3}{2}$$ × x = 3x2 ⇔ x = 20 cm
63.
A cow is tethered in the middle of a field with a 14 feet long rope. If the cow grazes 100 sq.ft per day, they approximately what time will be taken by the cow to graze the whole field ?
(A) 2 days
(B) 6 days
(C) 18 days
(D) 24 days
Solution:
Area of the field grazed : $$\eqalign{ & = \left( {\frac{{22}}{7} \times 14 \times 14} \right)sq.ft \cr & = 616\,sq.ft \cr} $$ Number of days taken to graze the field : $$\eqalign{ & = \frac{{616}}{{100}}\text{days} \cr & = 6\,\text{days}(\text{approx}) \cr} $$
64.
A carpenter is designing a table. The table will be in the form of a rectangle whose length is 4 feet more than its width. How long should the table be if the carpenter wants the area of the table to be 45 sq. ft ?
(A) 6 ft
(B) 9 ft
(C) 11 ft
(D) 13 ft
Solution:
Let the width of the table be x feet. Then, length of the table = (x + 4) ft $$\eqalign{ & \therefore x\left( {x + 4} \right) = 45 \cr & \Rightarrow {x^2} + 4x - 45 = 0 \cr & \Rightarrow {x^2} + 9x - 5x - 45 = 0 \cr & \Rightarrow x\left( {x + 9} \right) - 5\left( {x + 9} \right) = 0 \cr & \Rightarrow \left( {x + 9} \right)\left( {x - 5} \right) = 0 \cr & \Rightarrow x = 5 \cr} $$ Hence, length of the table = (5 + 4) = 9 feet
65.
Which of the following figures has the longest perimeter ?
(A) A square of side 10 cm
(B) A rectangular of sides 12 cm and 9 cm
(C) A circle of radius 7 cm
(D) A rhombus of side 9 cm
Solution:
(a) Perimeter = (4 × 10) cm = 40 cm (b) Perimeter = 2(12 + 9) cm = 42 cm (c) Perimeter = $$\left( {2 \times \frac{{22}}{7} \times 7} \right)$$ cm = 44 cm (d) Perimeter = (4 × 9) cm = 36 cm
66.
Two equal circle are drawn in square in such a way that a side of the square forms diameter of each circle. If the remaining area of the square is 42 cm2, how much will the diameter of the circle measure ?
(A) 3.5 cm
(B) 4 cm
(C) 14 cm
(D) 7.5 cm
Solution:
Let length of each side of square = 2π According to the question, $$\eqalign{ & \frac{{\pi {r^2}}}{2} + \frac{{\pi {r^2}}}{2} + 42 = {\text{Area of square}} \cr & \Rightarrow \pi {r^2} + 42 = 4{r^2} \cr & \Rightarrow 4{r^2} - \pi {r^2} = 42 \cr & \Rightarrow {r^2}\left( {4 - \frac{{22}}{7}} \right) = 42 \cr & \Rightarrow {r^2}\left( {\frac{{28 - 22}}{7}} \right) = 42 \cr & \Rightarrow \frac{{6{r^2}}}{7} = 42 \cr & \Rightarrow {r^2} = \frac{{42 \times 7}}{6} \cr & \Rightarrow {r^2} = 7 \times 7 \cr & \Rightarrow r = 7 \cr & \therefore 2r = 14\,cm \cr} $$
67.
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
68.
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 fourth side fencing. The dimension of the garden is :
(A) 15 m × 6.67 m
(B) 20 m × 5 m
(C) 30 m × 3.33 m
(D) 40 m × 2.5 m
Solution:
We have : $$\eqalign{ & 2b + l = 30 \cr & \Rightarrow l = 30 - 2b \cr} $$ $$\eqalign{ & {\text{Area}} = {\text{100 }}{m^2} \cr & \Rightarrow l \times b = 100 \cr & \Rightarrow b\left( {30 - 2b} \right) = 100 \cr & \Rightarrow {b^2} - 15b + 50 = 0 \cr & \Rightarrow \left( {b - 10} \right)\left( {b - 5} \right) = 0 \cr & \Rightarrow b = 10{\text{ or }}b = 5 \cr} $$ When, b = 10, $$l$$ = 10 and when b = 5, $$l$$ = 20 Since the garden is rectangular, so its dimension is 20 m × 5 m
69.
The base of triangle is 15 cm and height is 12 cm the height of another triangle of double the area having base 20 cm is :
(A) 22 cm
(B) 20 cm
(C) 18 cm
(D) 10 cm
Solution:
Given base of triangle and its height is 15 cm and 12 cm respectively Area of first triangle : $$\eqalign{ & = \frac{1}{2} \times {\text{Base}} \times {\text{Height}} \cr & = \frac{1}{2} \times 15 \times 12 \cr & = 90{\text{ sq}}{\text{. cm}} \cr} $$ According to the question, Let height of triangle be h cm Area of new triangle = 180 sq. cm Base = 20 sq. cm $$\eqalign{ & \Rightarrow 180 = \frac{1}{2} \times 20 \times {\text{Height}} \cr & \Rightarrow {\text{h}} = \frac{{2 \times 180}}{{20}} \cr & \Rightarrow {\text{h}} = 18{\text{ cm}} \cr} $$
70.
The radius of a circle is 20% more than the height of a right-angled triangle. The base of the triangle is 36 cm. If the area of triangle and circle be equal, what will be area of circle ?
(A) 72 cm2
(B) 128 cm2
(C) 144 cm2
(D) 216 cm2
Solution:
Let the height of the triangle be x cm Then, radius of the circle = (120% of x) cm = $$\left( {\frac{{6x}}{5}} \right)$$ cm $$\eqalign{ & \therefore \frac{1}{2} \times 36 \times x = \frac{{22}}{7} \times \frac{{6x}}{5} \times \frac{{6x}}{5} \cr & \Rightarrow x = \left( {\frac{{18 \times 7 \times 5 \times 5}}{{22 \times 6 \times 6}}} \right)cm \cr} $$ So, radius of the circle : $$\eqalign{ & = \left[ {\frac{6}{5} \times \left( {\frac{{18 \times 7 \times 5 \times 5}}{{22 \times 6 \times 6}}} \right)} \right]cm \cr & = \left( {\frac{{105}}{{22}}} \right)cm \cr} $$ ∴ Area of the circle :$$\eqalign{ & = \left( {\frac{{22}}{7} \times \frac{{105}}{{22}} \times \frac{{105}}{{22}}} \right)c{m^2} \cr & = \left( {\frac{{1575}}{{22}}} \right)c{m^2} \cr & = 71.6\,c{m^2} \approx 72\,c{m^2} \cr} $$