CBSE 2025 · Region 2 · Set 1 · Q33 · 5 marks
The sum of the areas of two squares is $\displaystyle 52 \mathrm{~cm}^{2}$ and difference of their perimeters is $\displaystyle 8$ cm . Find the lengths of the sides of the two squares.The time taken by a person to travel an upward distance of $\displaystyle 150$ km was $\displaystyle 2 \frac{1}{2}$ hours more than the time taken in the downward return journey. If he returned at a speed of $\displaystyle 10$ km/h more than the speed while going up, find the speeds in each direction.
The sum of the areas of two squares is $\displaystyle 52 \mathrm{~cm}^{2}$ and difference of their perimeters is $\displaystyle 8$ cm . Find the lengths of the sides of the two squares.
The time taken by a person to travel an upward distance of $\displaystyle 150$ km was $\displaystyle 2 \frac{1}{2}$ hours more than the time taken in the downward return journey. If he returned at a speed of $\displaystyle 10$ km/h more than the speed while going up, find the speeds in each direction.
Marking-scheme solution
Let the lengths of the sides of two squares be 'x' cm and 'y' cm such that \(\displaystyle \mathrm{x}>\mathrm{y}\). ATQ
\[x^{2}+y^{2}=52 \quad \text {--- ① }
\]
\[4 x-4 y=8 \text { or } x-y=2 \quad \text {--- ② }
\]
From ① and ②, we have
\[y^{2}+2 y-24=0
\]
\[\Longrightarrow(y+6)(y-4)=0
\]
\[\therefore y=4
\]
\[\text { So, } x=2+4=6
\]
\(\displaystyle \therefore\) Lengths of the sides of two squares are $\displaystyle 6$ cm and $\displaystyle 4$ cm respectively.
Let the speed in upward direction be 'x' km/h and the speed in downward direction \(\displaystyle =(\mathrm{x}+10) \mathrm{km} / \mathrm{h}\) ATQ
\[\frac{150}{x}-\frac{150}{x+10}=\frac{5}{2}
\]
\[\Rightarrow \mathrm{x}^{2}+10 \mathrm{x}-600=0
\]
\[\Rightarrow(x+30)(x-20)=0
\]
\[\therefore x=20
\]
and \(\displaystyle \mathrm{x}+10=20+10=30\)
Therefore, speeds in upward and downward direction are $\displaystyle 20$ km/h and $\displaystyle 30$ km/h respectively.
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