Area - Study Mode

[#226] A toothed wheel of diameter 50 cm is attached to a smaller wheel of diameter 30 cm. How many revolutions will the smaller wheel make the larger one makes 15 revolutions ?
Correct Answer

(C) 25

Explanation

Solution: Distance covered by smaller wheel in 1 revolution : $$eqalign{
& = left( {2pi imes 15}
ight)cm cr
& = left( {30pi }
ight)cm cr} $$ Distance covered by larger wheel in 1 revolution : $$eqalign{
& = left( {2pi imes 25}
ight)cm cr
& = left( {50pi }
ight)cm cr} $$ Let, $$k imes 30pi = 15 imes 50pi $$ Then, $$k = left( {frac{{15 imes 50pi }}{{30pi }}}
ight) = 25$$ ∴ Required number of revolution = 25

[#227] The perimeter of a square is equal to twice the perimeter of a rectangle of length 8 cm and breadth 7 cm. What is the circumference of a semi-circle whose diameter is equal to the side of the square ? (rounded off to two decimal places)
Correct Answer

(A) 23.57 cm

Explanation

Solution: Perimeter of rectangle : $$eqalign{
& = left[ {2left( {8 + 7}
ight)cm}
ight] cr
& = 30,cm cr} $$ Perimeter of square : $$eqalign{
& = left( {2 imes 30}
ight)cm cr
& = 60,cm cr} $$ Side of the square : $$eqalign{
& = left( {frac{{60}}{4}}
ight)cm cr
& = 15,cm cr} $$ Radius of the semi-circle $$ = left( {frac{{15}}{2}}
ight)cm$$ ∴ Circumference : $$eqalign{
& = left( {frac{{22}}{7} imes frac{{15}}{2}}
ight)cm cr
& = left( {frac{{165}}{7}}
ight)cm cr
& = 23.57,cm cr} $$

[#228] 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 :
Correct Answer

(B) Rs. 1650

Explanation

Solution: Length of the fence =$$4pi R$$ Where, $$R = frac{{21}}{2}m$$ $$eqalign{
& 4pi R cr
& = left( {4 imes frac{{22}}{7} imes frac{{21}}{2}}
ight)m cr
& = 132,m cr} $$ Cost of fencing : $$eqalign{
& = { ext{Rs}}{ ext{.}}left( {132 imes frac{{25}}{2}}
ight) cr
& = { ext{Rs}}{ ext{.1650}} cr} $$

[#229] 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 :
Correct Answer

(B) 20 m × 5 m

Explanation

Solution: We have : $$eqalign{
& 2b + l = 30 cr
& Rightarrow l = 30 - 2b cr} $$ $$eqalign{
& { ext{Area}} = { ext{100 }}{m^2} cr
& Rightarrow l imes b = 100 cr
& Rightarrow bleft( {30 - 2b}
ight) = 100 cr
& Rightarrow {b^2} - 15b + 50 = 0 cr
& Rightarrow left( {b - 10}
ight)left( {b - 5}
ight) = 0 cr
& Rightarrow b = 10{ ext{ or }}b = 5 cr} $$ When, b = 10, $$l$$ = 10 and when b = 5, $$l$$ = 20 Since the garden is rectangular, so its dimension is 20 m × 5 m

[#230] Two sides of a rectangle were measured. One of the sides (length) was measured 10% more than its actual length and the other side (width) was measured 5% less than its actual length. The percentage error in measure obtained for the area of the rectangle is :
Correct Answer

(A) 4.5%

Explanation

Solution: Let the actual length and width of the rectangle be $$l$$ and b respectively. Then, measured length : $$ = 100\% { ext{ of }}l = frac{{11l}}{{10}}$$ Measured width : $$ = 95\% { ext{ of }}b = frac{{19b}}{{20}}$$ Actual area = $$lb$$ Measured area : $$eqalign{
& = left( {frac{{11l}}{{10}} imes frac{{19b}}{{20}}}
ight) cr
& = frac{{209lb}}{{200}} cr} $$ Error in measurement : $$eqalign{
& = left( {frac{{209lb}}{{200}} - lb}
ight) cr
& = frac{{9lb}}{{200}} cr} $$ ∴ Error % : $$eqalign{
& = left( {frac{{9lb}}{{200}} imes frac{1}{{lb}} imes 100}
ight)\% cr
& = 4.5\% cr} $$