CBSE 10 Maths NCERT Class 10 Maths

 3. (iv) The taxi charges in a city consist of a fixed charge together with the charge for the distance covered. For a distance of 10 km, the charge paid is Rs.105 and for a journey of 15 km, the charge paid is Rs.155. What are the fixed charges and the charge per km? How much does a person have to pay for travelling a distance of 25 km? Solution: Let be fixed charges = Rs.x Let be per km charges = Rs.y     X+10y=115--------(1)     X+15y=155--------(2)  By solving equation (1)     X+10y=105     10y=105-x   Y=(105-x)/10--------(3)  Now substitute y=105-x/10 in equation (2)    X+15y=155    x/1+15(105-x)/10=155    10x+1575-15x/10=155    1575-5x/10=155    1575-5x=1550    1575-1550=5x    5x=25    X=25/5=5  X=5 substitute in equation (3)   Y=105-x/10   Y=105-5/10=100/10=10    Y=10  Fixed charges (x)=5   Per km charges (y...

 3. Form the pair of linear equations for the following problems and find their solution by substitution    method (iii) The coach of a cricket team buys 7 bats and 6 balls for Rs.3800. Later, she buys 3 bats and 5 balls for  Rs.1750. Find the cost of each bat and each ball. Solution:  let be one bat cost=Rs.x  let be one ball cost=Rs.y  7 bats and 6 balls cost-3800      7x+6Y=3800------(1)  3 bats and 5 balls cost = 1750      3x+5y=1750-------(2)   By solving equation (1)      7x+6y=3800      6y=3800-7x      Y=(3800-7x)/6--------(3)    Substitute y=(3800-7x)/6 in equation (2)     3x+5y=1750     3x/1+5(3800-7x)/6=1750      18x+19000-35x/6=1750     19000-10500=17x     x=8500/17=500     X=500 Substitute x=500 in equation (3)     y=(3800-7x)/6=3800-7(500)/6      =3800-3500/...

 3. Form the pair of linear equations for the following problems and find their solution by substitution    method. (ii) The larger of two supplementary angles exceeds the smaller by 18 degrees. Find them. Solution:         Supplementary angle total =180°        Let be one angle = x        Let be another angle = y     We know two supplementary angles total = 180°           x+y=180°------(1)      larger angle is exceeds 18° more than smaller angle           x=y+18----------(2)      substitute equation (2) in equation (1)           x+y=180           y+18+y=180           2y=180-18=162           Y=162/2           Y=81°      substitute y=81° in equation (2)       ...

3. Form the pair of linear equations for the following problems and find their solution by substitution    method. (i) The difference between two numbers is 26 and one number is three times the other. Find them Solution:  Let be numbers = x, y difference between two numbers is 26  x-y=26--------(1)  One number is three times the other  x=3y-------(2)  substitute x=3y in equation (1)   x-y=26  3y-y=26  2y=26  y=13  substitute y=13 in equation (2)  x=3y  x=3x13=39  x=39, y=13

 2. Solve 2x+3y=11 and 2x–4y=–24 and hence find the value of ‘m’ for which y=mx+3 Solution: 2x+3y=11---------(1) 2x-4y=-24--------(2) By solving equation (1) 2x+3y=11 3y=11-2x Y=(11-2x)/3--------(3) Now substitute y=(11-2x)/3 in equation (2) 2x-4y=-24 2x-4(11-2x)/3=-24 (6x-44+8x)/3=24 (14x-44)/3=-24 14x-44=-72 14x=-72+44 14x=-28 X=-2 Substitute x=-2 in equation (3) Y=(11-2x)/3 Y=11-2(2)/3=11+4/3=15/3=5 Y=5 Substitute x,y values in y=mx+3 Y=mx+3 5=m(-2)+3 5=-2m+3 -2m=5-3 -2m=2 m=-1

 1. Solve the following pair of linear equations by the substitution method. (vi) 3x/2-5y/3=-2, x/3+y/2=13/6 Solution (3x)/2-(5y)/3=2 Make lcm of 2/3 (9x-10y)/6=-2 9x-10y=-12 x/3+y/2=13/6 make lcm of 3,2 (2x+3y)/6=13/6 2x+3y=6x13/6 2x+3y=13--------(2) Let we take equation (1) and make x form 9x-10y=-12 9x=10y-12 X=(10y=12)/3--------(3) Substitute x value in equation (2) 2x+3y=13 2(10y-12)/9+3y=13 (20y-24)/9+3y=13 (20y-24+27y)/9+3y=13 47y-24=117 47y=117+24=141 Y=141/47=3 Substitute y value in equation (3) X=(10y-12)/9=10(3)-12/9=30-12=9=18/9=2 X=2, Y=3