CAT 2021 Set-3 | Quantitative Aptitude | Question: 16

1 vote
103 views
A park is shaped like a rhombus and has area $96 \; \text{sq m}.$ If $40 \; \text{m}$ of fencing is needed to enclose the park, the cost, in $\text{INR},$ of laying electric wires along its two diagonals, at the rate of $₹ \; 125 \; \text{per m},$ is

retagged

1 vote

Let the length of the diagonal of a rhombus be $2a$ meter, $2b$ meter.

Area of the rhombus $= \dfrac{\text{Product of diagonal}}{2}$

$\Rightarrow 96 = \frac{2a \times 2b}{2}$

$\Rightarrow \boxed{ab = 48}$

In $\triangle \text{AOB},$ we can apply the Pythagorean theorem.

$\text{Hypotenuse}^{2} = \text{Perpendicular}^{2} + \text{Base}^{2}$

$\Rightarrow \text{AB}^{2} = \text{OA}^{2} + \text{OB}^{2}$

$\Rightarrow 10^{2} = a^{2} + b^{2}$

$\Rightarrow \boxed{a^{2} + b^{2} = 100}$

Now, $(a+b)^{2} – 2ab =100$

$\Rightarrow (a+b)^{2} – 96 =100 \quad [\because ab=48]$

$\Rightarrow (a+b)^{2} =196$

$\Rightarrow \boxed{a+b =14} \; \longrightarrow (1)$

And $(a-b)^{2} + 2ab =100$

$\Rightarrow (a-b)^{2}+96 =100$

$\Rightarrow (a-b)^{2} =4$

$\Rightarrow \boxed{a-b =2} \; \longrightarrow (2)$

Adding the equation $(1)$, and equation $(2).$

$\qquad \require{cancel} \begin{array} {} a+\cancel{b} = 14 \\ a-\cancel{b} = 2 \\\hline 2a \quad = 16 \end{array}$

$\Rightarrow \boxed{a=8}$

From equation $(1),$

$\Rightarrow a+b=14$

$\Rightarrow 8+b=14$

$\Rightarrow \boxed{b=6}$

So,

• The length of the one diagonal $=2a=16$ meters.
• The length of the other diagonal $=2b=12$ meters.

$\therefore$ The cost in INR, of laying electric wires along its two diagonal at the rate of ₹$125$ per meter $=28 \times 125= ₹3500.$

Correct Answer $:3500$

10.1k points 4 8 30
edited

Related questions

1 vote
1
77 views
In a triangle $\text{ABC}, \angle \text{BCA} = 50^{\circ}. \text{D}$ and $\text{E}$ are points on $\text{AB}$ and $\text{AC},$ respectively, such that $\text{AD = DE}.$ If $\text{F}$ is a point on $\text{BC}$ such that $\text{BD = DF},$ then $\angle \text{FDE, in degrees},$ is equal to $96$ $72$ $80$ $100$
1 vote
2
59 views
Let $\text{ABCD}$ be a parallelogram. The lengths of the side $\text{AD}$ and the diagonal $\text{AC}$ are $10 \; \text{cm}$ and $20 \; \text{cm},$ respectively. If the angle $\angle \text{ADC}$ is equal to $30^{\circ}$ then the area of the parallelogram, in sq. cm, is $\frac{25(\sqrt{5} + \sqrt{15})}{2}$ $25 (\sqrt{5} + \sqrt{15})$ $\frac{25 (\sqrt{3} + \sqrt{15})}{2}$ $25 (\sqrt{3} + \sqrt{15})$
1 vote
The cost of fencing a rectangular plot is $₹ \; 200 \; \text{per ft}$ along one side, and $₹ \; 100 \; \text{per ft}$ along the three other sides. If the area of the rectangular plot is $60000 \; \text{sq. ft},$ then the lowest possible cost of fencing all four sides, in $\text{INR},$ is $160000$ $100000$ $120000$ $90000$
Anil can paint a house in $12 \; \text{days}$ while Barun can paint it in $16 \; \text{days}.$ Anil, Barun, and Chandu undertake to paint the house for $₹ \; 24000$ and the three of them together complete the painting in $6 \; \text{days}.$ If Chandu is paid in proportion to the work done by him, then the amount in $\text{INR}$ received by him is
The arithmetic mean of scores of $25$ students in an examination is $50.$ Five of these students top the examination with the same score. If the scores of the other students are distinct integers with the lowest being $30,$ then the maximum possible score of the toppers is