APTITUDE MATERIALS - HCF and LCM

HCF and LCM

1. Factors and Multiples:

If number a divided another number b exactly, we say that a is a factor of b.

In this case, b is called a multiple of a.

2. Highest Common Factor (H.C.F.) or Greatest Common Measure (G.C.M.) or Greatest Common Divisor (G.C.D.):

The H.C.F. of two or more than two numbers is the greatest number that divided each of them exactly.

There are two methods of finding the H.C.F. of a given set of numbers:

1. Factorization Method: Express the each one of the given numbers as the product of prime factors. The product of least powers of common prime factors gives H.C.F.
2. Division Method: Suppose we have to find the H.C.F. of two given numbers, divide the larger by the smaller one. Now, divide the divisor by the remainder. Repeat the process of dividing the preceding number by the remainder last obtained till zero is obtained as remainder. The last divisor is required H.C.F.

Finding the H.C.F. of more than two numbers: Suppose we have to find the H.C.F. of three numbers, then, H.C.F. of [(H.C.F. of any two) and (the third number)] gives the H.C.F. of three given number.

Similarly, the H.C.F. of more than three numbers may be obtained.

3. Least Common Multiple (L.C.M.):

The least number which is exactly divisible by each one of the given numbers is called their L.C.M.

There are two methods of finding the L.C.M. of a given set of numbers:

1. Factorization Method: Resolve each one of the given numbers into a product of prime factors. Then, L.C.M. is the product of highest powers of all the factors.
2. Division Method (short-cut): Arrange the given numbers in a rwo in any order. Divide by a number which divided exactly at least two of the given numbers and carry forward the numbers which are not divisible. Repeat the above process till no two of the numbers are divisible by the same number except 1. The product of the divisors and the undivided numbers is the required L.C.M. of the given numbers.
1. Product of two numbers = Product of their H.C.F. and L.C.M.
2. Co-primes: Two numbers are said to be co-primes if their H.C.F. is 1.
3. H.C.F. and L.C.M. of Fractions:

   1. H.C.F. =	H.C.F. of Numerators
   	L.C.M. of Denominators

4. 2. L.C.M. =	L.C.M. of Numerators
   	H.C.F. of Denominators

5. H.C.F. and L.C.M. of Decimal Fractions:

In a given numbers, make the same number of decimal places by annexing zeros in some numbers, if necessary. Considering these numbers without decimal point, find H.C.F. or L.C.M. as the case may be. Now, in the result, mark off as many decimal places as are there in each of the given numbers.

9. Comparison of Fractions:

Find the L.C.M. of the denominators of the given fractions. Convert each of the fractions into an equivalent fraction with L.C.M as the denominator, by multiplying both the numerator and denominator by the same number. The resultant fraction with the greatest numerator is the greatest.

APTITUDE MATERIALS - DECIMAL FRACTIONS

DECIMAL FRACTIONS

1. Decimal Fractions:

Fractions in which denominators are powers of 10 are known as decimal fractions.

Thus,	1	= 1 tenth = .1;	1	= 1 hundredth = .01;
	10		100

99	= 99 hundredths = .99;	7	= 7 thousandths = .007, etc.;
100		1000

2. Conversion of a Decimal into Vulgar Fraction:

Put 1 in the denominator under the decimal point and annex with it as many zeros as is the number of digits after the decimal point. Now, remove the decimal point and reduce the fraction to its lowest terms.

Thus, 0.25 =	25	=	1	;       2.008 =	2008	=	251	.
	100		4		1000		125

3. Annexing Zeros and Removing Decimal Signs:

Annexing zeros to the extreme right of a decimal fraction does not change its value. Thus, 0.8 = 0.80 = 0.800, etc.

If numerator and denominator of a fraction contain the same number of decimal places, then we remove the decimal sign.

Thus,	1.84	=	184	=	8	.
	2.99		299		13

4. Operations on Decimal Fractions:
   1. Addition and Subtraction of Decimal Fractions: The given numbers are so placed under each other that the decimal points lie in one column. The numbers so arranged can now be added or subtracted in the usual way.
   2. Multiplication of a Decimal Fraction By a Power of 10: Shift the decimal point to the right by as many places as is the power of 10.

Thus, 5.9632 x 100 = 596.32;   0.073 x 10000 = 730.

3. Multiplication of Decimal Fractions: Multiply the given numbers considering them without decimal point. Now, in the product, the decimal point is marked off to obtain as many places of decimal as is the sum of the number of decimal places in the given numbers.

Suppose we have to find the product (.2 x 0.02 x .002).

Now, 2 x 2 x 2 = 8. Sum of decimal places = (1 + 2 + 3) = 6.

 .2 x .02 x .002 = .000008

4. Dividing a Decimal Fraction By a Counting Number: Divide the given number without considering the decimal point, by the given counting number. Now, in the quotient, put the decimal point to give as many places of decimal as there are in the dividend.

Suppose we have to find the quotient (0.0204 ÷ 17). Now, 204 ÷ 17 = 12.

Dividend contains 4 places of decimal. So, 0.0204 ÷ 17 = 0.0012

5. Dividing a Decimal Fraction By a Decimal Fraction: Multiply both the dividend and the divisor by a suitable power of 10 to make divisor a whole number.

Now, proceed as above.

Thus,	0.00066	=	0.00066 x 100	=	0.066	= .006
	0.11		0.11 x 100		11

5. Comparison of Fractions:

Suppose some fractions are to be arranged in ascending or descending order of magnitude, then convert each one of the given fractions in the decimal form, and arrange them accordingly.

Let us to arrange the fractions	3	,	6	and	7	in descending order.
	5		7		9

Now,	3	= 0.6,	6	= 0.857,	7	= 0.777...
	5		7		9

Since, 0.857 > 0.777... > 0.6. So,	6	>	7	>	3	.
	7		9		5

6. Recurring Decimal:

If in a decimal fraction, a figure or a set of figures is repeated continuously, then such a number is called a recurring decimal.

n a recurring decimal, if a single figure is repeated, then it is expressed by putting a dot on it. If a set of figures is repeated, it is expressed by putting a bar on the set.

Thus,	1	= 0.333... = 0.3;	22	= 3.142857142857.... = 3.142857.
	3		7

Pure Recurring Decimal: A decimal fraction, in which all the figures after the decimal point are repeated, is called a pure recurring decimal.

Converting a Pure Recurring Decimal into Vulgar Fraction: Write the repeated figures only once in the numerator and take as many nines in the denominator as is the number of repeating figures.

Thus, 0.5 =	5	; 0.53 =	53	; 0.067 =	67	, etc.
	9		99		999

Mixed Recurring Decimal: A decimal fraction in which some figures do not repeat and some of them are repeated, is called a mixed recurring decimal.

Eg. 0.1733333.. = 0.173.

Converting a Mixed Recurring Decimal Into Vulgar Fraction: In the numerator, take the difference between the number formed by all the digits after decimal point (taking repeated digits only once) and that formed by the digits which are not repeated. In the denominator, take the number formed by as many nines as there are repeating digits followed by as many zeros as is the number of non-repeating digits.

Thus, 0.16 =	16 - 1	=	15	=	1	;   0.2273 =	2273 - 22	=	2251	.
	90		90		6		9900		9900

7. Some Basic Formulae:
   1. (a + b)(a - b) = (a² + b²)
   2. (a + b)² = (a² + b² + 2ab)
   3. (a - b)² = (a² + b² - 2ab)
   4. (a + b + c)² = a² + b² + c² + 2(ab + bc + ca)
   5. (a³ + b³) = (a + b)(a² - ab + b²)
   6. (a³ - b³) = (a - b)(a² + ab + b²)
   7. (a³ + b³ + c³ - 3abc) = (a + b + c)(a² + b² + c² - ab - bc - ac)
   8. When a + b + c = 0, then a³ + b³ + c³ = 3abc.

APTITUDE MATERIALS - PROBLEMS ON AGES

Problems on ages:

1. The present ages of three persons in proportions 4 : 7 : 9. Eight years ago, the sum of their ages was 56. Find their present ages (in years).
2. Ayesha's father was 38 years of age when she was born while her mother was 36 years old when her brother four years younger to her was born. What is the difference between the ages of her parents?
3. A person's present age is two-fifth of the age of his mother. After 8 years, he will be one-half of the age of his mother. How old is the mother at present?
4. Q is as much younger than R as he is older than T. If the sum of the ages of R and T is 50 years, what is definitely the difference between R and Q's age?
5. The age of father 10 years ago was thrice the age of his son. Ten years hence, father's age will be twice that of his son. The ratio of their present ages is:
6. A man is 24 years older than his son. In two years, his age will be twice the age of his son. The present age of his son is:
7. Six years ago, the ratio of the ages of Kunal and Sagar was 6 : 5. Four years hence, the ratio of their ages will be 11 : 10. What is Sagar's age at present?
8. The sum of the present ages of a father and his son is 60 years. Six years ago, father's age was five times the age of the son. After 6 years, son's age will be:
9. At present, the ratio between the ages of Arun and Deepak is 4 : 3. After 6 years, Arun's age will be 26 years. What is the age of Deepak at present ?
10. Sachin is younger than Rahul by 7 years. If their ages are in the respective ratio of 7 : 9, how old is Sachin?
11. Present ages of Sameer and Anand are in the ratio of 5 : 4 respectively. Three years hence, the ratio of their ages will become 11 : 9 respectively. What is Anand's present age in years?
12. A is two years older than B who is twice as old as C. If the total of the ages of A, B and C be 27, the how old is B?
13. A father said to his son, "I was as old as you are at the present at the time of your birth". If the father's age is 38 years now, the son's age five years back was:
14. The sum of ages of 5 children born at the intervals of 3 years each is 50 years. What is the age of the youngest child?
15. Father is aged three times more than his son Ronit. After 8 years, he would be two and a half times of Ronit's age. After further 8 years, how many times would he be of Ronit's age?

___________________________

Solution:

1. Let their present ages be 4x, 7x and 9x years respectively.

Then, (4x - 8) + (7x - 8) + (9x - 8) = 56

 20x = 80

 x = 4.

 Their present ages are 4x = 16 years, 7x = 28 years and 9x = 36 years respectively.

2) Mother's age when Ayesha's brother was born = 36 years.

Father's age when Ayesha's brother was born = (38 + 4) years = 42 years.

 Required difference = (42 - 36) years = 6 years.

3) Let the mother's present age be x years.

Then, the person's present age =		2	x		years.
		5

		2	x + 8		=	1	(x + 8)
		5				2

 2(2x + 40) = 5(x + 8)

 x = 40.

4) Given that:
1. The difference of age b/w R and Q = The difference of age b/w Q and T.
2. Sum of age of R and T is 50 i.e. (R + T) = 50.
Question: R - Q = ?.
Explanation:
R - Q = Q - T
(R + T) = 2Q
Now given that, (R + T) = 50
So, 50 = 2Q and therefore Q = 25.
Question is (R - Q) = ?
Here we know the value(age) of Q (25), but we don't know the age of R.
Therefore, (R-Q) cannot be determined.

5) Let the ages of father and son 10 years ago be 3x and x years respectively.

Then, (3x + 10) + 10 = 2[(x + 10) + 10]

 3x + 20 = 2x + 40

 x = 20.

 Required ratio = (3x + 10) : (x + 10) = 70 : 30 = 7 : 3.

6) Let the son's present age be x years. Then, man's present age = (x + 24) years.

 (x + 24) + 2 = 2(x + 2)

 x + 26 = 2x + 4

 x = 22.

7) Let the ages of Kunal and Sagar 6 years ago be 6x and 5x years respectively.

Then,	(6x + 6) + 4	=	11
	(5x + 6) + 4		10

 10(6x + 10) = 11(5x + 10)

 5x = 10

 x = 2.

• Sagar's present age = (5x + 6) = 16 years.

8) Let the present ages of son and father be x and (60 -x) years respectively.

Then, (60 - x) - 6 = 5(x - 6)

 54 - x = 5x - 30

 6x = 84

 x = 14.

 Son's age after 6 years = (x+ 6) = 20 years..

9) Let the present ages of Arun and Deepak be 4x years and 3x years respectively. Then,

4x + 6 = 26        4x = 20

x = 5.

 Deepak's age = 3x = 15 years.

10) Let Rahul's age be x years.

Then, Sachin's age = (x - 7) years.

	x - 7	=	7
	x		9

 9x - 63 = 7x

 2x = 63

 x = 31.5

Hence, Sachin's age =(x - 7) = 24.5 years.

11) Let the present ages of Sameer and Anand be 5x years and 4x years respectively.

Then,	5x + 3	=	11
	4x + 3		9

 9(5x + 3) = 11(4x + 3)

 45x + 27 = 44x + 33

 45x - 44x = 33 - 27

 x = 6.

 Anand's present age = 4x = 24 years.

12) Let C's age be x years. Then, B's age = 2x years. A's age = (2x + 2) years.

 (2x + 2) + 2x + x = 27

 5x = 25

 x = 5.

Hence, B's age = 2x = 10 years.

13) Let the son's present age be x years. Then, (38 - x) = x

 2x = 38.

 x = 19.

 Son's age 5 years back (19 - 5) = 14 years.

14) Let the ages of children be x, (x + 3), (x + 6), (x + 9) and (x + 12) years.

Then, x + (x + 3) + (x + 6) + (x + 9) + (x + 12) = 50

 5x = 20

 x = 4.

 Age of the youngest child = x = 4 years.

15) Let Ronit's present age be x years. Then, father's present age =(x + 3x) years = 4x years.

	(4x + 8) =	5	(x + 8)
		2

 8x + 16 = 5x + 40

 3x = 24

 x = 8.

Hence, required ratio =	(4x + 16)	=	48	= 2.
	(x + 16)		24