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Syllabus

30. A boat sails a distance of 44 km in 4 hours with the current. It takes 4 hours 48minute longer to cover the same distance against the current. Find the speed of boat on still water and the speed of the current.

0.2x + 0.1y = 25

2 (x - 2) - 1.6y = 116

Plz solve this question by linear equations.

Take no. of customers as 'x' & no. of lines as 'y'.

Find x & y

Please tell question no.9

Q9. The ratio of two numbers is $\frac{2}{3}$. If 2 is subtracted from the first and 8 from the second, the ratio becomes the reciprocal of the original ratio. Find the numbers.

2/x - 1/y = 1

Solve by reducing them to pairs of linear equations

Please explain the steps...

(27/8)

^{2/3}- (1/4)^{-2}+ 5^{0}4x - 3y = -1

Find p, if y = px - 3

Linear Equations by substitutional method

33. A man travels 600km partly by train and partly by car. If he covers 120km by train and the next by car, it takes him 8 hours. But, if he travels 200km by train and the rest by car he takes 20 minutes longer. Find the speed of the car and that of the train.

Q. The sides of an equilateral triangle are (6x + 3y) cm, (8x + 9y - 5) cm and (10x + 12y - 8) respectively. Find the length of each side.

[

Hint.Take 6x + 3y = 8x + 9y - 5 and 8x + 9y - 5 = 10x + 12y - 8 as two equations.]Q. 2x + 3y = 17

3x - 2y = 6

$40.\mathrm{Solve}:\frac{15}{\mathrm{u}}+\frac{2}{\mathrm{v}}=17\mathrm{and}\frac{1}{\mathrm{u}}+\frac{1}{\mathrm{v}}=\frac{36}{5}$

11x+13y=74

Method of elimination by equating coefficient

3/x + 4y = 7

5/x + 6y = 13

3/x + 4y =7. 5/x + 6y =13

Q.48. 8 men and 12 boys can finish a work in 10 days while 6 men and 8 boys can finish the same work in 14 days. Find the time taken by 1 man and that 1 boy alone to finish the work.

Q. 8 The lengths of the sides of the triangle are given in centimetres as:

$\left(2x+\frac{y}{2}\right),\left(\frac{2x}{3}+2y+\frac{5}{2}\right)and\left(y+\frac{5x}{3}+\frac{1}{2}\right).$

If the triangle is equilateral, prove that the triangle whose sides are (2xy+1), (2x+1) and (6y+1) cm each is also equilateral.