NCERT Solutions for Class 10 Maths Chapter 3: Pair of Linear Equations in Two Variables || CBSE Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables solutions Ex 3.7 Math Solution

Let the speed of the train be \( x \mathrm{~km} / \mathrm{h} \) and the time taken by train to travel the given distance be \( t \) hours and the distance to travel was d km .
We know that,
Speed \( =\frac{\text { distance travelled }}{\text { timetaken to travel that distance }} \)
\(x=\frac{d}{t}\)
Or, \( \mathrm{d}=\mathrm{xt}\ldots(i) \)
If the train would have been \( 10 \mathrm{~km} / \mathrm{h} \) faster, it would have taken 2 hours less than the scheduled time
\(\Rightarrow(x+10)=\frac{d}{t-2}\)
\(\Rightarrow(x+10)(\mathrm{t}-2)=\mathrm{d}\)
\(\Rightarrow \mathrm{xt}+10 \mathrm{t}-2x-20=\mathrm{d}\)
From (i), we have
\(\Rightarrow \mathrm{d}+10 \mathrm{t}-2x-20=\mathrm{d}\)
\(\Rightarrow-2 x+10 \mathrm{t}=20\)
\(\Rightarrow x-5 \mathrm{t}=-10 \Rightarrow x=5 \mathrm{t}-10\ldots(ii)\)
Also,
if the train were slower by \( 10 \mathrm{~km} / \mathrm{h} \); it would have taken 3 hours more than the scheduled time \( \Rightarrow(x-10)=\frac{d}{t-3} \)
\(\Rightarrow(x-10)(\mathrm{t}+3)=\mathrm{d}\)
\(\Rightarrow x\mathrm{t}-10 \mathrm{t}+3 x-30=\mathrm{d}\)
By using equation (i),
\(
\Rightarrow \mathrm{d}-10 \mathrm{t}+3 x-30=\mathrm{d} \Rightarrow 3 x-10 \mathrm{t}=30 \ldots(iii)\)
Substituting the value of \( x \) from eq (ii) into eq (iii), we get
\(\Rightarrow 3(5 \mathrm{t}-10)-10 \mathrm{t}=30\)
\(\Rightarrow 15 \mathrm{t}-30-10 \mathrm{t}=30\)
\(\Rightarrow 5 \mathrm{t}=60 \Rightarrow \mathrm{t}=12 \text { hours }\)
Putting this in eq(ii) \( \Rightarrow x=5 \mathrm{t}-10 \)
\(=5(12)-10\)
\(=50 \mathrm{~km} / \mathrm{h}\)
From equation (i), we obtain
Distance to travel \( =\mathrm{d}=x\mathrm{t} \)
\(=50 \times 12\)
\(=600 \mathrm{~km}\)
Hence, the distance covered by the train is 600 km .

Let the number of rows be \(x\) and number of students in a row be \(y\).
Total students of the class \( = \) Number of rows \( x \) Number of students in a row \( =xy \)
Using the information given in the question,
Condition 1
Total number of students \( =(x-1)(y+3) \)
\(=(x-1)(y+3)\)
\(=x y-y+3 x-3\)
\(3 x-y-3=0\)
\(
3 x-y=3\ldots(i)\)
Condition 2
Total number of students \( =(x+2)(y-3) \)
\(=x y+2 y-3 x-6\)
\(\Rightarrow 3 x-2 y=-6 \ldots(ii)\)
Subtracting equation (ii) from (i),
\((3 x-y)-(3 x-2 y)=3-(-6)\)
\(\Rightarrow-y+2 y=9\)
\(\Rightarrow 3+6 y=9\)
By using equation (i), we obtain \( 3 x-9=3 \),
\(\Rightarrow 3 x=9+3\)
\(\Rightarrow 3 x=12\)
\(\Rightarrow x=4\)
from (i),
\(\Rightarrow 3(4)-y=3\)
\(\Rightarrow 12-y=3 \Rightarrow 9=y\)
Number of rows \( =x=4 \)
Number of students in a row \( =y=9 \)
Number of total students in a class \( = \) Number of students in 1 row \( \times \) Number of rows
\(=x y\)
\(=4 \times 9=36\)