SSC GD Constable 6 Feb 2025 Shift 1 Paper

Number of yellow bikes $={\mathit{n}}_{\mathit{y}}$
Number of green bikes $={\mathit{n}}_{\mathit{g}}$
We know that the total number of bikes is 100 :
${\mathit{n}}_{\mathit{y}}+{\mathit{n}}_{\mathit{g}}=100$
The average speed of all bikes is given by the weighted average of the speeds of yellow and green bikes:
$({\mathit{n}}_{\mathit{y}}\times .$ Average speed of yellow bikes $+{\mathit{n}}_{\mathit{g}}\times$ Average speed of green bikes) / Total number of bikes $=$ Average speed of all bikes
$\left({\mathit{n}}_{\mathit{y}}\times 55+{\mathit{n}}_{\mathit{g}}\times 30\right)∕100=35$
Multiply both sides by 100 :
$55\times {\mathit{n}}_{\mathit{y}}+30\times {\mathit{n}}_{\mathit{g}}=3500$
Now we have a system of two equations:
${\mathit{n}}_{\mathit{y}}+{\mathit{n}}_{\mathit{g}}=100$
$55\times {\mathit{n}}_{\mathit{y}}+30\times {\mathit{n}}_{\mathit{g}}=3500$
From equation 1 , we can express ${\mathit{n}}_{\mathit{y}}$ as:
${\mathit{n}}_{\mathit{y}}=100-{\mathit{n}}_{\mathit{g}}$
Substitute this into equation 2 :
$55\times \left(100-{\mathit{n}}_{\mathit{g}}\right)+30\times {\mathit{n}}_{\mathit{g}}=3500$
$5500-55\times {\mathit{n}}_{\mathit{g}}+30\times {\mathit{n}}_{\mathit{g}}=3500$
$5500-25\times {\mathit{n}}_{\mathit{g}}=3500$
Rearrange to solve for ${\mathit{n}}_{\mathit{g}}$ :
$25\times {\mathit{n}}_{\mathit{g}}=5500-3500$
$25\times {\mathit{n}}_{\mathit{g}}=2000$
${\mathit{n}}_{\mathit{g}}=2000∕25$
${\mathit{n}}_{\mathit{g}}=80$