tougher problems in simple equation

A car did a journey in "t" hours. If the average speed had been "m" km/hr greater, the journey would have taken "y" hours less. What was the length of the journey?

tougher problems on simple equations

A fraction is such that if "c" is added to the numerator and "d" to the denominator, the value of the fraction becomes 1/x. If the numerator of the original fraction is doubled and the denominator is increased by "e", the fraction becomes 1/y. Find the original fraction.

tougher problems on simple equation

There are two mixtures of  acid and water, one of which contains twice as much water as acid, and  the other three times as much acid as water. How much must be taken from each in order to fill a liter measure, in which the water and the acid shall be equally mixed?

Simple Equation

A man bicycles half the distance from  one town to the another at 24 km/hr, and the other half at 16 km/hr. A second man bicycles all the way at 22.5 km/hr. If the difference in the time taken is 5 1/2 min, then what is the whole distance?

Factorization

Factorize: a³ + 2a²+ 2a + 1

Answer:

a³ + 2a²+ 2a + 1

= a³ + 3a²+ 3a + 1 - a² – a

= (a +1)³ – a(a + 1)

= (a + 1)[(a + 1)² – a]

= (a + 1)(a² + a + 1)

Factorize:  12a²- ab – b²

Answer:

12a²- ab – b²

= 12a²-4ab  + 3ab – b²

= 4a(3a – b) + b(3a – b)

= (4a + b)(3a – b)

 

Simultaneous Equation

A train running from A to B meets with an accident 75 km from A, after which it travels with 3/5th of its original velocity and arrives 3 hours late at B. If the accident had occurred 75 km further on, it would have been only 2 hours late. Find the distance from A to B and the original velocity of the train.

Simultaneous Equations

If 2 rabbits and 4 dogs cost $ 1750, three dogs and 2 cats cost $ 1725 and one cat and 3 rabbits cost $ 675. Find the price of each.
Let cost of 1 rabbit = R, cost of 1 dog = D, cost of 1 cat = C

simultaneous equation

A certain number of two digits is three times the sum of its digits and if 45 be added to it the digits will be reversed, find the number.

HCF and LCM

Find the HCF and LCM of 3x² + x -10,   6x² – x – 15 and 6x² -19x +15



Answer:

 3x² + x -10 =3x² +6 x- 5x -10 = 3x(x +2) - 5(x +2) = (3x - 5)(x + 2)

6x² – x – 15 = 6x² –10 x + 9x – 15  = 2x(3x - 5) + 3(3x - 5) = (2x + 3)(3x - 5)

 6x² -19x +15 = 6x² -10x - 9x +15 = 2x(3x - 5) -3(3x -5) = (2x - 3)(3x - 5)

The common factor is (3x -5). So, HCF = (3x - 5)

The LCM is (3x- 5)(x + 2)(2x + 3)(2x - 3)

rationalization

                                                   

rationalisation

                                                                   

proportion

if a/b = c/d = e/f
then show that these ratios are equal to 
(5a - 7c + 3e) / (5b -7d + 3f)



Answer:

let  a/b = c/d = e/f = k
therefore, a = bk, c = dk and e = fk
substituting in  (5a - 7c + 3e)
we get, 5bk - 7dk + 3fk  = k(5b - 7d + 3f)

Therefore,
(5a - 7c + 3e) / (5b -7d + 3f) = k(5b -7d + 3f) / (5b -7d + 3f) = k
Hence (5a - 7c + 3e) / (5b -7d + 3f)  has same value as the ratios.

If p/q = r/s = t/u, then prove that (p² – pr +t²)/(q² – qs + u²) = pt/qu

Answer

let,  p/q = r/s = t/u = k

so, p = qk, r = sk and t = uk

therefore, (p² – pr +t²) = q²k² - qsk² - u²k² = k²(q² - qs - u²)

therefore, (p² – pr +t²)/(q² – qs + u²) = k²(q² - qs - u²)/ (q² - qs - u²) = k²

= p/q x t/u

= pt/qu