NCERT Math Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables solutions (English Medium) || CBSE Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.3 Math Solution

Let cost of each bat \( = \) Rs. X
Let cost of each ball \( = \) Rs. \( Y \)
According to the question,
\(=7 x+6 y=3800 \text { Given }\)
\(=6 y=3800-7 x\)
\(=y=\frac{3800-7 x}{6} \ldots(i)\)
And
\(=3 x+5 y=1750 \ldots(ii) \) Given
Putting value of \(y\) from Equation (i) to equation (ii)
\(=3 x+5 \times \frac{3800-7 x}{6}=1750\)
\(=18 \mathrm{x}+1900-35 \mathrm{x}=1750 \times 6\)
\(=-17 x=10500-19000=-8500\)
\(=x=\frac{-8500}{-17}=500\)
Putting value of \( x \) in equation (i), we get
\(=y=\frac{3800-7 \times 500}{6}=\frac{300}{6}=50\)
Hence,
Cost of each bat \( = \) Rs. 500 and Cost of each ball \( = \) Rs. 50

Let the fixed charge for taxi \( = \) Rs. X
Let variable cost per \( \mathrm{km}= \) Rs. y
We know that,
Total cost \( = \) Fixed charge + Variable Charge
According to the question,
\(=x+10 y=105 \text { given }\)
\(=x=105-10 y \ldots(i)\)
And, For a journey of 15 km charge paid = Rs. 155
\(\text { so, } x+15 y=155 \ldots(ii)\)
Putting value of \( x \) from equation (i) to equation (ii). we get,
\(=105-10 y+15 y=155\)
\(=5 y=155-105=50\)
\(y=\frac{50}{5}=10\)
Putting value of \( y \) in equation (i). we get,
\(=x=105-10 \times 10=105-100=5\)
So, People have to pay for travelling a distance of 25 km
\(=x+25 y=5+25 \times 10=\text { Rs. } 255\)

Let numerator \( =x \)
Let denominator \( =\mathrm{y} \)
Fraction will be \( =\frac{x}{y} \)
According to the question,
Fraction become \( \frac{9}{11} \), if 2 is added in both. Numerator and denominator
\(=\frac{x+2}{y+2}=\frac{9}{11}\)
\(=11 \mathrm{x}+22=9 \mathrm{y}+18 \text { (by cross multiplication) }\)
\(=11 \mathrm{x}=9 \mathrm{y}-4\)
\(=x=\frac{9 y-4}{11} \ldots(i)\)
And, if 3 is added to both numerator and denominator it become \( \frac{5}{6} \)
\(=\frac{x+3}{y+3}=\frac{5}{6}\)
\( =6 x+18=5 y+15 \ldots(ii)\) By cross multiplication
Putting value of \( x \) from equation (i) to equation (ii)
\(=6 \times \frac{9 y-4}{11}+18=5 y+15\)
\(=54 y-24=55 y-33\)
\(=-y=-9\)
\(=y=9\)
Putting value of \( y \) in equation (i)
\(=x=\frac{9 y-4}{11}=\frac{9 \times 9-4}{11}=\frac{77}{11}=7\)
Hence, the fraction is \( \frac{7}{9} \).

Let present age of Jacob \( =X \) years
Let present age of his son \( = \) Y years
Five year hence,
Age of Jacob \( =X+5 \)
Age of son \( =Y+5 \)
And, age of Jacob is 3 times of his son Given
\(=x+5=3(y+5)\)
\(=x+5=3 y+15\)
\(=x=3 y+10 \ldots(i)\)
Five years ago,
Age of Jacob \( =\mathrm{X}-5 \)
Age of son \( = \) Y - 5
And, Jacob's age was 7 times of his son Given
\(=x-5=7(y-5)\)
\(=x-5=7 y-35\)
\(=x-7 y=-30 \ldots(ii)\)
Putting value of X from equation (i) to equation (ii)
\(=3 y+10-7 y=-30\)
\(=-4 y=-40\)
\(=y=\frac{-40}{-4}=10\)
Putting value of \( Y \) in (i)
\(=x=3 \times 10+10=30+10=40\)
Hence, present age of Jacob \( =40 \) years and his son's age \( =10 \) years.