4. SIMPLIFICATION (SOLVED EXAMPLES 11 20.)

 Ex.11. If 2x + 3y = 34 and X + Y/y = 13/8, then find the value of 5y + 7x.

Sol. the given equations are:

2X + 3y =34 …..(I) and, X + y/y = 13/8 8X + 8y = 13y    6X – 5y = 0.(II)

Multiplying (I) by 5, (II) by 3 and adding, we get: 34 X = 170 or x= 5.

Putting x = 5 in (I), we get y =8.

5y + 7X = (5 x 8 +7 x 5) = 40 + 35 = 75.

Ex.12. If 2X +3y +z =55, X+ z - y =4 and y - x + z=12, then what are the values of x, y and z?

Sol. the given equations are:

2x + 3y + z =55 ..(I); X + z – y =4   ..(II); y – x + z= 12.

Subtracting (II) from (I), we get: X _ 4y = 51.

Subtracting (III) from (I), we get: 3x + 2y = 43.

Multiplying (v) by 2 and subtracting (iv) from it, we get: 5x = 35 or x=7.

Putting X =7, y = 11 in(I), we get: z =8.

Ex.13. find the value of (1-1/3) (1-1/4) (1-1/5) ….(1-1/100)

 Sol. given expression = 2/3 x 3/4 x 4/5 x …x 99/100 2/100 =1/50.

Ex.14. find the value of 1/2x3 + 1/3x4 + 1/4x5 + 1/5x6 +,…. 1/9x10.

Sol. given expression = (1/2 – 1/3) + (1/3 – 1/4) + (1/4 – 1/5) + (1/5 – 1/6) + … + (1/9 – 1/10)

= (1/2 – 1/10) = 4/10 = 2/5

Ex.15. simplify: 99 48/49 x 245.

Sol. given expression= (100-1/49) x 245 = 4899 = 4899 x 5 = 24495.

Ex. 16. A board 7 ft. 9 inches long is divided into 3 equal parts. What is the length of each part?

Sol. length of board = 7ft.9 inches = (7 x 12 +9) inches = 93 inches.

Length of each part – (93/3) inches = 3= 2 ft.7 inches.

Ex.17 . a man divides Rs. 8600 among 5 sons, 4 daughters and 2nephews. If each daughter receives 

four times as much as each nephew, and each son receives five times as much as each nephew, how much does each daughter receive?

Sol. let the share of each daughter= Rs.x.

The, share of each daughter = Rs.(4x); share of each son = Rs. (5x).

So, 5 x 5x + 4 x 4x +2 X x = 8600    25 X+16X + 2X =8600.

43X = 8600     X = 200.

Share of each daughter = Rs. (4 x 200) = Rs.800.

Ex.18. a man spends 2/5 of his salary on house rent, 3/10 of his salary on food and 1/8 of his salary on conveyance. If he has Rs. 1400 left with him, find his expenditure on food and conveyance.

Sol. part of the salary left = 1 – (2/5 + 3/10 + 1/8) = 1 – 33/40 = 7/40.

Let the monthly salary been Rs. X.

Then, 7/40 of X = 1400 x = (1400 x 40/7) = 800.

Expenditure on food = Rs. (1/8 x 8000) =Rs.1000.

Ex.19. a third of Arun’s marks in mathematics exceeds a half of his marks in English by 30. If he got 240 marks in the two subjects together, how many marks did he get in English?

Sol. let Arun’s marks in Mathematics and English be x and y respectively.

The, 1/3 X – 1/2   y = 30 2 X – 3y = 180..    (I) and X + y = 240..   (II)

Solving (I) and (II), we get: X=180 and y = 60.

Ex.20. a tin of oil was 4/5 full. When 6 bottles of oil were taken out and four bottles of oil wearer poured into it, it was 3/4 full. How many bottles of oil can the tin contain?

Sol. suppose x bottles can fill the tin completely.

The, 4/5 X = 3/4  (6 -4)    x/20=  X= 40.

Required number of bottles = 40.