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8 men and 12 boys can do a piece of work in 10 days while 6 men and 8 boys can do same work in 14 days

Posted by Prabhpreet Kuar 4 years ago

1 answers

Sia ? 4 years ago

Let us suppose that one man alone can finish the work in x days and one boy alone can finish it in y days. Then,
one man's one day's work {tex}= \frac { 1 } { x }{/tex}
One boy's one day's work {tex}= \frac { 1 } { y }{/tex}
{tex}\therefore{/tex} Eight men's one day's work = {tex}\frac { 8 } { x }{/tex}
{tex}12\ boy's{/tex} one day's work = {tex}\frac { 12 } { y }{/tex}
According to question it is given that {tex}8\ men{/tex} and {tex}12\ boys{/tex} can finish the work in {tex}10\ days{/tex}
{tex}10 \left( \frac { 8 } { x } + \frac { 12 } { y } \right) = 1 \Rightarrow \frac { 80 } { x } + \frac { 120 } { y } = 1{/tex} .................(i)
Again, {tex}6\ men{/tex} and {tex}8\ boys{/tex} can finish the work in {tex}14\ days{/tex}.
{tex}\therefore \quad 14 \left( \frac { 6 } { x } + \frac { 8 } { y } \right) = 1 \Rightarrow \frac { 84 } { x } + \frac { 112 } { y } = 1{/tex} ...........(ii)
Putting {tex}\frac { 1 } { x } = u{/tex} and {tex}\frac { 1 } { y } = v{/tex} in equations (i) and (ii), we get
{tex}80u + 120u - 1 = 0{/tex}
{tex}84u + 112v - 1 = 0{/tex}
By using cross-multiplication,
{tex}\Rightarrow \frac { u } { - 120 + 112 } = \frac { - v } { - 80 + 84 } = \frac { 1 } { 80 \times 112 - 120 \times 84 }{/tex}
{tex}\Rightarrow \quad \frac { u } { - 8 } = \frac { v } { - 4 } = \frac { 1 } { - 1120 }{/tex}
{tex}\Rightarrow \quad u = \frac { - 8 } { - 1120 } = \frac { 1 } { 140 } \text { and } v = \frac { - 4 } { - 1120 } = \frac { 1 } { 280 }{/tex}
{tex}u = \frac { 1 } { 140 } \Rightarrow \frac { 1 } { x } = \frac { 1 } { 140 } \Rightarrow x = 140{/tex}
{tex}v = \frac { 1 } { 280 } \Rightarrow \frac { 1 } { y } = \frac { 1 } { 280 } \Rightarrow y = 280{/tex}
One man alone can finish the work in {tex}140\ days{/tex} and one boy alone can finish the work in {tex}280\ days{/tex}.

1Thank You

ANSWER

Preeti Dabral 4 years ago

Similar as(side=14cm):

As per the given figure in question,
∠ABC = ∠BAC = ∠ACB = 60 ….[In an equilateral triangle all angles are of 60° each]
Now θ = 60°, r = {tex}{14 \over 2}{/tex} = 7 cm
Area of shaded region = ar(ΔABC) – 3{tex}\times{/tex} (area of minor sector)
{tex}{ = \frac{{\sqrt 3 }}{4}{{\left( {Side} \right)}^2}\; - \;3 \times \frac{\emptyset }{{360}}\;\pi {{\text{r}}^2}}{/tex}
{tex}{ = \;\;\frac{{1.73}}{4}\; \times 14 \times 14\; - \;3 \times \frac{{60}}{{360}}\; \times \;\frac{{22}}{7} \times 7 \times 7}{/tex}
{tex}{ = \;1.73\; \times 7 \times 7\; - \;3\; \times \frac{1}{6} \times 22 \times 7}{/tex}
{tex}{ = \;84.77\; - \;\frac{1}{2} \times 22 \times 7}{/tex}
{tex}{ = \;84.77\; - \;\;11 \times 7}{/tex}
= 84.77 - 77
= 7.77 cm²
Hence, the area of the shaded region = 7.77 cm²

0Thank You

Let r be the radius of the quadrant.
We know that, Perimeter of the quadrant {tex}= \frac { 25 r } { 7 } \mathrm { cm }{/tex}
Given that, perimeter of a quadrant {tex}= 25 cm{/tex}
{tex}\Rightarrow \frac { 25 r } { 7 } {/tex}{tex}= 25{/tex}
{tex}\Rightarrow r = 7 \mathrm { cm }{/tex}
{tex}\therefore {/tex} Area of the quadrant{tex}= \frac { 1 } { 4 } \pi r ^ { 2 }{/tex}
{tex}= \left( \frac { 1 } { 4 } \times \frac { 22 } { 7 } \times 7 \times 7 \right) \mathrm { cm } ^ { 2 }{/tex}
{tex}= \frac { 77 } { 2 } \mathrm { cm } ^ { 2 }{/tex}

2Thank You

Noni Jain 4 years ago

According to the question:- Let cosA= Hypotenuse sideadjacentA AB AC Similarly, cosB= Hypotenuse sideadjacentB = AB BC Given that cosA=cosB AB AC = AB BC AC=BC In triangle, angles opposite equal side are equal ∠B=∠A

Area of triangleAs shown in figure, largest triangle inscribed in a semi-circle is triangle with base as diameter of circle and height as radius of circle. Thus,the area of triangle is A=1/2×2r × r A=r^2

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