CBSE 2025 · Region 2 · Set 3 · Q35 · 5 marks
The numerator of a fraction is $\displaystyle 3$ less than its denominator. If $\displaystyle 2$ is added to both numerator and denominator, then the sum of the new fraction and the original fraction is $\displaystyle 1 \frac{9}{20}$. Find the original fraction.A train travelling at a uniform speed for $\displaystyle 360$ km would have taken $\displaystyle 48$ minutes less to travel the same distance if its speed were $\displaystyle 5$ km/h more. Find the original speed of the train.
The numerator of a fraction is $\displaystyle 3$ less than its denominator. If $\displaystyle 2$ is added to both numerator and denominator, then the sum of the new fraction and the original fraction is $\displaystyle 1 \frac{9}{20}$. Find the original fraction.
A train travelling at a uniform speed for $\displaystyle 360$ km would have taken $\displaystyle 48$ minutes less to travel the same distance if its speed were $\displaystyle 5$ km/h more. Find the original speed of the train.
Marking-scheme solution
Let denominator be x
∴ Numerator \(\displaystyle =(\mathrm{x}-3)\)
Therefore, fraction \(\displaystyle =\frac{\mathrm{x}-3}{\mathrm{x}}\)
ATQ
\[\frac{x-3}{x}+\frac{x-3+2}{x+2}=\frac{29}{20}
\]
\(\displaystyle \Rightarrow 11 \mathrm{x}^{2}-98 \mathrm{x}-120=0\)
\(\displaystyle \Rightarrow(\mathrm{x}-10)(11 \mathrm{x}+12)=0\)
So, \(\displaystyle \mathrm{x}=10\)
∴ Fraction \(\displaystyle =\frac{7}{10}\)
Let the original speed of train be 'x' km/h ATQ
\[\frac{360}{x}-\frac{360}{x+5}=\frac{48}{60}
\]
\[\begin{aligned}
& \Rightarrow x^{2}+5 x-2250=0 \\
& \Rightarrow(x+50)(x-45)=0
\end{aligned}
\]
So, \(\displaystyle \mathrm{x}=45\)
∴ Original speed of the train is $\displaystyle 45$ km/h.
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