3

Q1. Rekha's mother is five times as old as her daughter Rekha. Five years later, Rekha's mother will be three times as old as her daughter Rekha. Find the present age of Rekha and her mother's age.

Solution

Let Rekha's age be 'x' years And her mother's age be 'y' years y = 5x, as per given data ---(1) After 5 years, y + 5 = 3(x+5) y - 3x = 10 ----(2) Substituting (1) in (2), we get, 2x = 10 or x = 5 So, y = 5x = 25 Hence, Rekha's age is 5 years and her mother's age is 25 years.

Q2. A train covered a certain distance at a uniform speed. If the train would have been 10km/h faster, it would have taken 2 hours less than the schedule time. And, if the trains were slower by 10km/hr, it would have taken 3 hours more than the scheduled time. Find the distance covered by the train.  

Solution

Let the original speed of train be x km/hr and the schedule time of journey be y hours. Therefore, s = d/t i.e. d = xykm.   If speed of the train increases by 10km/hr, then increased speed = (x + 10)km/hr Time reduces by 2hrs, so reduced time = (y - 2)h Therefore, Distance = (x + 10)(y-2) = xy -2x + 10y - 20   In both the cases distances are equal   xy = xy - 2x + 10y - 20  x - 5y = -10    x = 5y -10  ----- (1)   If speed is reduced by 10km/hr, then the reduced speed = (x - 10)km/h If time is increased by 3hr, then the increased time = (y + 3)hr Then, distance = (x - 10)(y + 3) = xy + 3x - 10y - 30    xy = xy + 3x - 10y - 30  3x - 10y = 30 ----- (2)   Using substitution method, subs. (1) in (2), we have   3(5y - 10) - 10y = 30  15y - 30 - 10y = 30  5y = 60  y = 12   Subs. value of y in (1) x = 5(12) - 10 = 50 x = 50   Therefore, distance = xy = 50 x 12 = 600km Hence, distance covered by the train is 600km.  

Q3. A motor boat can cover 30 km upstream and 28 km downstream in 7 hrs. It can cover 21 km upstream and returns in 5 hrs. Find the speed of the boat and the stream.

Solution

Q4. 10 years ago, mother`s age was 12 times that of her daughter and 10 years hence, her age would be twice that of her daughter. Find their present ages. Let the present age mother be x years and daughter's present age be y years.  

Solution

Q5. Solve for x,ax+by= a²-2bbx+ay=ab-2a

Solution

ax+by= a²-2b...(1)bx+ay=ab-2a...(2)adding the two equations,(a+b)x+(a+b)y= a²+ab-2a-2b(x+y)(a+b)=a(a+b)-2(a+b) x+y=a-2...(3)Subtracting (2) from (1), we get,(a-b)x-(a-b)y= a²-ab-2b+2a(x-y)(a-b)=a(a-b)+2(a-b) x-y=a+2...(4)Now,Adding (3) and (4)2x=2a x=aSubstituting this value in (4) we get,y=-2So the solution to the given system isx=a, y=-2