CTET 2 Math and Science 07 Jul 2024 Paper

Given :
- $\mathit{O}\mathit{P}=\left(\mathit{x}+7\right)\mathit{c}\mathit{m}$
- $\mathit{O}\mathit{Q}=\left(\mathit{x}+\mathit{y}\right)\mathit{c}\mathit{m}$
- $\mathit{O}\mathit{R}=20\mathit{c}\mathit{m}$
- $\mathit{O}\mathit{S}=16\mathit{c}\mathit{m}$
Concept used:
Properties of a Parallelogram:
- In a parallelogram, the diagonals bisect each other. This means that the diagonals PR and QS intersect at point O such that $\mathit{P}\mathit{O}=\mathit{O}\mathit{R}$ and $\mathit{Q}\mathit{O}=\mathit{O}\mathit{S}$.
Calculation:

Since the diagonals bisect each other at O , we have:
$\mathit{O}\mathit{P}=\mathit{O}\mathit{R}$, thus $\left(\mathit{x}+7\right)=20$
$\mathit{O}\mathit{Q}=\mathit{O}\mathit{S}$, thus $\left(\mathit{x}+\mathit{y}\right)=16$
From the equation $\left(\mathit{x}+7\right)=20$ :
We solve for $\mathit{x}:\mathit{x}=20-7=13$
Next, we use the value of $\mathit{x}$ to solve for $\mathit{y}$ from the equation $\left(\mathit{x}+\mathit{y}\right)=16$ :
Substitute $\mathit{x}=13$ into $\left(\mathit{x}+\mathit{y}\right)=16$ :
$13+\mathit{y}=16$
Solve for $\mathit{y}:\mathit{y}=16-13=3$
Finally, we calculate the value of $\left(7\mathit{y}+5\right)$ :
Substitute $\mathit{y}=3$ into $\left(7\mathit{y}+5\right)$ :
$\left(7\times 3\right)+5=21+5=26$
Thus, the value of $\left(7\mathit{y}+5\right)$ is 26 .