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Saturday, July 20, 2013
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 questions on simple equation
A man rides one third of the distance from A to B at the rate of "a" km/hr and the remainder at "2b" km/hr. if he had traveled at the uniform rate of "3c" km/hr, he could have ridden from A to B and back again in the same time.
Prove that 2/c = (1/a) + (1/b)
Prove that 2/c = (1/a) + (1/b)
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)
Friday, July 19, 2013
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
Let cost of 1 rabbit = R, cost of 1 dog = D, cost of 1 cat = C
Simultaneous Equations
Two numbers are formed by the same two digits, and if the smaller number is divided by the greater the quotient is 4/7, and if the smaller is subtracted from the greater the remainder is 27. Find the number.
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)
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)
framing of equation
A man buys oranges at the rate of "p" for a $, and by selling them at "q" cents a dozen makes a profit of "r" %.
Show that pq -12r = 1200
Answer:
Cost of 12 oranges = 12/p
Selling price of 12 oranges = q/100
Gain = q/100 - 12/p = (pq - 1200)/100p
Gain % = r = 100 [(pq - 1200)/100p] / 12/p
or, 12r = 100p [(pq - 1200)/100p]
or, 12r = pq - 1200
or, pq - 12r = 1200
Show that pq -12r = 1200
Answer:
Cost of 12 oranges = 12/p
Selling price of 12 oranges = q/100
Gain = q/100 - 12/p = (pq - 1200)/100p
Gain % = r = 100 [(pq - 1200)/100p] / 12/p
or, 12r = 100p [(pq - 1200)/100p]
or, 12r = pq - 1200
or, pq - 12r = 1200
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