1.
NUMBERS
TYPES OF NUMBERS
1.Natural Numbers :
Counting numbers 1, 2, 3, 4, 5,..... are called natural
numbers.
2.Whole
Numbers : All counting numbers together with zero form the set of wholenumbers.
Thus,
(i) 0
is the only whole number which is not a natural number.
(ii) Every
natural number is a whole number.
3.Integers
: All natural numbers, 0 and negatives of counting numbers i.e.,{…,
- 3 , - 2 , - 1 , 0, 1, 2, 3,…..} together form the set of integers.
(i) Positive
Integers : {1, 2, 3, 4, …..} is the set of all positive integers.
(ii) Negative
Integers : {- 1,
- 2, - 3,…..} is the set of all negative integers.
(iii) Non-Positive
and Non-Negative Integers : 0 is neither positive nor negative.
So, {0, 1, 2, 3,….} represents the set of non-negative integers, while{0, - 1 ,
- 2 , - 3 ,…..} represents the set of non-positive integers.
4. Even Numbers : A
number divisible by 2 is called an even number, e.g., 2, 4, 6, 8,10, etc.
5. Odd Numbers : A
number not divisible by 2 is called an odd number. e.g., 1, 3, 5, 7,9, 11, etc.
6. Prime Numbers : A
number greater than 1 is called a prime number, if it has exactly two factors,
namely 1 and the number itself.
Prime numbers upto 100 are : 2, 3, 5, 7, 11, 13, 17, 19, 23,
29, 31, 37, 41, 43,
47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97.
Prime numbers Greater than 100 : Let p be a given number
greater than 100. To find out whether it is prime or not, we use the following
method :
Find a whole number nearly greater than the square root of
p. Let k > *jp. Test whether p
is divisible by any prime number less than k. If yes, then p
is not prime. Otherwise, p is
prime.
e.g,,We have to find whether 191 is a prime number or not.
Now, 14 > V191.
Prime numbers less than 14 are 2, 3, 5, 7, 11, 13.
191 is not divisible by any of them. So, 191 is a prime
number.
7.Composite Numbers : Numbers
greater than 1 which are not prime, are known as
composite numbers, e.g., 4, 6, 8, 9, 10, 12.
Note : (i)
1 is neither prime nor composite.
(ii) 2 is the only even number which is prime.
(iii) There are 25 prime numbers between 1 and 100.
8. Co-primes : Two
numbers a and b are said to be co-primes, if their H.C.F. is 1. e.g.,
(2, 3), (4, 5), (7, 9), (8, 11), etc.
are co-primes.
TESTS OF DIVISIBILITY
1. Divisibility By 2 : A
number is divisible by 2, if its unit's digit is any of 0, 2, 4, 6, 8.
Ex. 84932 is divisible by 2, while 65935 is not.
2. Divisibility By 3 : A
number is divisible by 3, if the sum of its digits is divisible by 3.
Ex.592482
is divisible by 3, since sum of its digits = (5 + 9 + 2 + 4 + 8 + 2) = 30,
which
is divisible by 3.
But, 864329 is not divisible by 3, since sum of its digits
=(8 + 6 + 4 + 3 + 2 + 9) = 32,
which is not divisible by 3.
3. Divisibility By 4 : A
number is divisible by 4, if the number formed by the last two digits is
divisible by 4.
Ex. 892648 is divisible by 4, since the number formed by the
last two digits is 48, which is divisible by 4.
But, 749282 is not divisible by 4, since the number formed
by the last tv/o digits is 82,which is not divisible by 4.
4. Divisibility By 5 : A
number is divisible by 5, if its unit's digit is either 0 or 5. Thus,20820 and
50345 are divisible by 5, while 30934 and 40946 are not.
5. Divisibility By 6 : A
number is divisible by 6, if it is divisible by both 2 and 3. Ex. The number
35256 is clearly divisible by 2.
Sum of its digits = (3 + 5 + 2 + 5 + 6) = 21, which is
divisible by 3. Thus, 35256 is divisible by 2 as well as 3. Hence, 35256 is
divisible by 6.
6. Divisibility By 8 : A
number is divisible by 8, if the number formed by the last three digits of the
given number is divisible by 8. Ex. 953360 is divisible by 8, since the number
formed by last three digits is 360, which is divisible by 8.
But, 529418 is not divisible by 8, since the number formed
by last three digits is 418, which is not divisible by 8.
7. Divisibility By 9 :
A number is divisible by 9, if the sum of its digits is divisible by 9 Ex.
60732 is divisible by 9, since sum of digits * (6 + 0 + 7 + 3 + 2) = 18, which
is divisible by 9.
But, 68956 is not divisible by 9, since sum of digits = (6 +
8 + 9 + 5 + 6) = 34, which is not divisible by 9.
8. Divisibility By 10 : A
number is divisible by 10, if it ends with 0.
9. Divisibility By 11 : A
number is divisible by 11, if the difference of the sum of its digits at odd
places and the sum of its digits at even places, is either 0 or a number
divisible by 11.
Ex. The
number 4832718 is divisible by 11, since :
(sum of digits at odd places) - (sum of digits at even
places) ie (8 + 7 + 3 + 4) - (1 + 2 + 8) = 11, which is divisible by 11.
10. Divisibility
By 12 ; A number is divisible by 12, if it is divisible by both 4 and 3.
Ex. Consider
the number 34632.
(i) The number formed by last two digits is 32, which is
divisible by 4,
(ii) Sum of digits = (3 + 4 + 6 + 3 + 2) = 18, which is
divisible by 3. Thus, 34632 is divisible by 4 as well as 3. Hence, 34632 is
divisible by 12.
11.
Divisibility By 14 : A number is divisible by 14, if it is divisible by 2 as
well as 7.
12. Divisibility
By 15 : A number is divisible by 15, if it is divisible by both 3 and 5.
13. Divisibility
By 16 : A number is divisible by 16, if the number formed by the last4 digits
is divisible by 16.
Ex.7957536
is divisible by 16, since the number formed by the last four digits is 7536,
which is divisible by 16.
14. Divisibility By 24 : A
given number is divisible by 24, if it is divisible by both3 and 8.
15. Divisibility By 40 : A
given number is divisible by 40, if it is divisible by both 5 and 8.
16. Divisibility By 80 : A
given number is divisible by 80, if it is divisible by both 5 and 16.
Note : If
a number is divisible by p as well as q, where p and q are co-primes, then the
given number is divisible by pq.
If p arid q are not co-primes, then the given number need
not be divisible by pq, even when it is divisible by both p and q.
Ex. 36
is divisible by both 4 and 6, but it is not divisible by (4x6) = 24, since 4
and 6 are not co-primes.
BASIC FORMULAE
1. (a + b)2
=
a2 +
b2 +
2ab
2. (a - b)2
=
a2 +
b2 -
2ab
3. (a + b)2
-
(a - b)2 =
4ab
4. (a + b)2
+
(a - b)2 =
2 (a2 +
b2)
5. (a2
-
b2) = (a + b) (a - b)
6. (a + b + c)2
=
a2 +
b2 +
c2 +
2 (ab + bc + ca)
7. (a3
+
b3) = (a +b) (a2
-
ab + b2)
8. (a3
-
b3) = (a - b) (a2
+
ab + b2)
9. (a3
+
b3 +
c3 -3abc)
= (a + b + c) (a2 +
b2 +
c2 -
ab - bc - ca)
10. If a + b + c = 0, then a3
+
b3 +
c3 =
3abc.
DIVISION ALGORITHM OR EUCLIDEAN
ALGORITHM
If we divide a given number by another number, then : Dividend
= (Divisor x Quotient) + Remainder
{i) (xn
-
an )
is divisible by (x - a) for all values of n.
(ii) (xn
-
an) is divisible by (x
+ a) for all even values of n.
(iii) (xn
+
an) is divisible by (x
+ a) for all odd values of n.
PROGRESSION
A succession of numbers formed and arranged in a definite
order according to certain definite rule, is called a progression.
1. Arithmetic Progression (A.P.) :
If each term of a progression differs from its preceding
term by a constant, then such a progression is called an arithmetical progression.
This constant difference is called the common difference of
the A.P.
An A.P. with first term a and common difference d is given
by a, (a + d), (a + 2d), (a +3d)...
The nth term of this A.P. is given by
Tn =a
(n - 1) d.
The sum of n terms of this A.P.
Sn = n/2 [2a + (n - 1) d] = n/2 (first
term + last term).
SOME IMPORTANT RESULTS :
(i) (1 + 2 + 3 +…. + n) =n(n+1)/2
(ii) (l2
+
22 +
32 +
... + n2) = n (n+1)(2n+1)/6
(iii) (13
+
23 +
33 +
... + n3) =n2(n+1)2
2. Geometrical Progression (G.P.)
: A progression of numbers in which every term bears a constant
ratio with its preceding term, is called a geometrical progression.
The constant ratio is called the common ratio of the G.P. A
G.P. with first term a and common ratio r is : a, ar, ar2,
In this G.P. Tn = arn-1
2 H.C.F. AND L.C.M. OF NUMBERS
I. Factors and Multiples : If
a number a divides another number b exactly, we say that a is a factor of b. In
this case, b is called a multiple of a.
II. 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 divides each
of them exactly.
There are two methods of
finding the H.C.F. of a given set of numbers :
1. Factorization Method : Express
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 number 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 the 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 numbers. Similarly, the H.C.F. of more than three numbers may be
obtained.
III. 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.
1. Factorization Method of Finding
L.C.M.: 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. Common Division Method
{Short-cut Method) of Finding L.C.M.:
Arrange the given numbers in a row in any order. Divide by a
number which divides 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,
IV. Product of two numbers
=Product of their H.C.F. and L.C.M.
V. Co-primes: Two
numbers are said to be co-primes if their H.C.F. is 1.
VI. H.C.F. and L.C.M. of
Fractions:
1.H C F= H.C.F. of Numerators 2.L C M = L.C.M of Numerators
L.C.M. of Denominators H.C.F. of Denominators
VII. H.C.F. and L.C.M. of Decimal
Fractions: In 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.
VIII. 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.
3. DECIMAL FRACTIONS
I.
Decimal Fractions : Fractions in which denominators are
powers of 10 are known as decimal
fractions.
Thus,
1/10=1 tenth=.1;1/100=1 hundredth =.01;
99/100=99
hundreths=.99; 7/1000=7 thousandths=.007, etc
II.
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=25/100=1/4; 2.008=2008/1000=251/125.
III.
1. Annexing zeros to the extreme right of a decimal fraction
does not change its value
Thus,
0.8 = 0.80 = 0.800, etc.
2.
If numerator and denominator of a fraction contain the same number of decimal
places, then we remove the decimal sign.
Thus,
1.84/2.99 = 184/299 = 8/13; 0.365/0.584 = 365/584=5
IV.
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 = 0.0730 x 10000 = 730.
3.Multiplication
of Decimal Fractions : Multiply the given numbers
considering them without the 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 .02 x .002). Now, 2x2x2 = 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,
0.00066/0.11 = (0.00066*100)/(0.11*100) = (0.066/11) = 0.006V
V.
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. Suppose, we have to arrange the fractions 3/5, 6/7
and 7/9 in descending order. now, 3/5=0.6,6/7 = 0.857,7/9 = 0.777....
since
0.857>0.777...>0.6, so 6/7>7/9>3/5
VI.
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.
In
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
1/3 = 0.3333….= 0.3; 22 /7 = 3.142857142857.....= 3.142857
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 = 5/9; 0.53 = 53/59 ;0.067 = 67/999;etc...
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.
e.g.,
0.17333.= 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) / 90 = 15/19 = 1/6;
____
0.2273 = (2273 – 22)/9900 = 2251/9900
VII.
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-ca)
8. When a + b + c = 0, then a3
+
b3+ c3
= 3abc
4. SIMPLIFICATION
executed,
so as to find out the value of a given expression.
Here,
‘B’ stands for ’bracket’ ,’O’for ‘of’ , ‘D’ for’ division’ and ‘M’ for
‘multiplication’, ‘A’
for
‘addition’ and ‘S’ for ‘subtraction’.
Thus,
in simplifying an expression, first of all the brackets must be removed,
strictly in the order(), {} and [].
After
removing the brackets, we must use the following operations strictly in the
order:
(1)of
(2)division (3) multiplication (4)addition (5)subtraction.
II.
Modulus of a real number : Modulus of a real number a is defined
as
|a|
= {a, if a>0
-a, if a<0
Thus,
|5|=5 and |-5|=-(-5) =5.
III.
Virnaculum (or bar): When an expression contains Virnaculum, before applying
the ‘BODMAS’ rule, we simplify the expression under the Virnaculum.
5. SQUARE ROOTS AND CUBE ROOTS
Square
Root: If x2
=
y, we say that the square root of y
is x and we write, √y = x.
Thus,
√4 = 2, √9 = 3, √196 = 14.
Cube
Root: The cube root of a given number x is the number whose cube
is x. We denote the cube
root
of x by 3√x.
Thus,
3√8 = 3√2
x 2 x 2 = 2, 3√343 = 3√7
x 7 x 7 = 7 etc.
Note:
1.√xy
= √x * √y 2. √(x/y) = √x / √y = (√x / √y) * (√y / √y) = √xy / y
6. SURDS AND
INDICES
1.
LAWS OF INDICES:
(i) am x an = am + n
(ii) am / an = am-n
(iii) (am)n = amn
(iv) (ab)n = anbn
(v) ( a/ b )n = ( an / bn )
(vi) a0 = 1
2.
SURDS: Let a be a rational number and n be
a positive integer such that a1/n = nsqrt(a)
is
irrational. Then nsqrt(a)
is called a surd of order n.
3.
LAWS OF SURDS:
(i) n√a = a1/2
(ii) n √ab
= n √a * n √b
(iii) n √a/b
= n √a / n √b
(iv) (n √a)n =
a
(v) m√(n√(a)) = mn√(a)
7. PERCENTAGE
1.
Concept of Percentage : By a certain percent ,we mean that many
hundredths. Thus x percent means x hundredths, written as x%.
To
express x% as a fraction : We have , x% = x/100.
Thus,
20% =20/100 =1/5; 48% =48/100 =12/25, etc.
To
express a/b as a percent : We have, a/b =((a/b)*100)%.
Thus,
¼ =[(1/4)*100] = 25%; 0.6 =6/10 =3/5 =[(3/5)*100]% =60%.
2.
If the price of a commodity increases by R%, then the
reduction in consumption so as not to increase the expenditure is
[R/(100+R))*100]%.
If
the price of the commodity decreases by R%, then the increase in consumption so
as to decrease the expenditure is
[(R/(100-R)*100]%.
3.
Results on Population : Let the population of the town be P now
and suppose it increases at the rate of R% per annum, then :
1.
Population after n years = P [1+(R/100)]^n.
2.
Population n years ago = P /[1+(R/100)]^n.
4.
Results on Depreciation : Let the present value of a machine be
P. Suppose it depreciates at the rate R% per annum. Then,
1.
Value of the machine after n years = P[1-(R/100)]n.
2.
Value of the machine n years ago = P/[1-(R/100)]n.
5.
If A is R% more than B, then B is less than A by
[(R/(100+R))*100]%.
If
A is R% less than B , then B is more than A by
[(R/(100-R))*100]%.
8. PROFIT AND LOSS
Cost Price: The price at which article is purchased.
Abbrevated as C.P.
Selling Price: The price at which article is sold.
Abbrevated as S.P.
Profit or Gain: if Sp is greater than Cp, the selling price
is said to have profit or gain.
Loss: if sp is
less than cp, the seller is said
to incured a loss.
formula
1.gain=(sp)-(cp).
2.loss=(cp)-(sp).
3.loss or gain is always reckoned on cp
4. gain %={gain*100}/cp.
5.loss%={loss*100}/cp.
6.sp={(100+gain%) /100}*cp.
7.sp={(100-loss%)/100}*cp.
8.CP={100/(100+gain%)}
*sp
9.cp=100/(100-loss%)}*sp
10.if the
article is sold at a gain of say 35%, then sp
=135% of cp
11.if a
article is sold at a loss of say 35%. Then sp=65%
of cp.
12.when a
person sells two items, One at a gain of x% and other at a loss of x%. then the seller always incures a loss
given:
{loss%=(comon loss and gain ) 2}/10. =(x/10) 2
13.if the
trader professes to sell his goods at cp
but uses false weights, then GAIN=[ERROR/(TRUE VALUE)-(ERROR)*100]%
9.
RATIO AND PROPORTION
I.
RATIO: The ratio of two quantities a and b in the same units, is the
fraction a/b and we write it as a:b.
In
the ratio a:b, we call a as the first term or antecedent and
b, the second term or consequent.
Ex.
The ratio 5: 9 represents 5/9 with antecedent = 5,
consequent = 9.
Rule:
The multiplication or division of each term of a ratio by
the same
non-zero
number does not affect the ratio.
Ex.
4: 5 = 8: 10 = 12: 15 etc. Also, 4: 6 = 2: 3.
2.
PROPORTION: The
equality of two ratios is called proportion.
If
a: b = c: d, we write, a:
b:: c : d and we say that a, b,
c, d are in proportion . Here a
and d are called extremes, while b and c are called mean
terms.
Product of means = Product of extremes.
Thus,
a: b:: c : d <=> (b x c) = (a x d).
3.
(i) Fourth Proportional: If a :
b = c: d, then d is called the fourth proportional to a,
b, c.
(ii)
Third Proportional: If a:
b = b: c, then c is called the third proportional to a and b.
(iii)
Mean Proportional: Mean proportional between a and
b is square
root of ab
4.
(i) COMPARISON OF RATIOS:
We
say that (a: b)
> (c: d) <=>
(a/b)>(c /d).
(ii)
COMPOUNDED RATIO:
The
compounded ratio of the ratios (a: b), (c:
d), (e : f) is (ace:
bdf)
5.
(i) Duplicate ratio of (a
: b) is (a2
: b2).
(ii)
Sub-duplicate ratio of (a : b)
is (√a
: √b).
(iii)Triplicate
ratio of (a : b) is
(a3 : b3).
(iv)
Sub-triplicate ratio of (a : b)
is (a ⅓ : b
⅓ ).
(v)
If (a/b)=(c/d), then
((a+b)/(a-b))=((c+d)/(c-d)) (Componendo and dividendo)
6.
VARIATION:
(i)
We say that x is directly proportional to y, if
x = ky for some constant k and we
write, x ∞ y.
(ii)
We say that x is inversely proportional to y,
if xy = k for
some constant k and we write, x∞(1/y)
10. PARTNERSHIP
1.
Partnership: When two or more than two persons run a business jointly,
they are called
partners
and the deal is known as partnership.
2.
Ratio of Division of Gains:
i)
When investments of all the partners are for the same time,
the gain or loss is distributed a among the partners in the
ratio of their investments.
Suppose
A and B invest Rs. x and Rs. y respectively for a year in a business, then at
the end of the year:
(A’s
share of profit) : (B's share of profit) = x : y.
ii)
When investments are for
different time periods, then equivalent capitals are calculated
for a unit of time by taking (capital x number of units of time). Now, gain or
loss is divided in the ratio of these capitals.
Suppose
A invests Rs. x for p months and B invests Rs. y for q months, then
(A’s
share of profit) : (B's share of profit) = xp : yq.
3.
Working and Sleeping Partners: A partner who manages the business is
known. As a
working
partner and the one who simply invests the money is a sleeping
partner.
11. CHAIN RULE
1.
Direct Proportion: Two quantities are said to be directly
proportional, if on the increase (or decrease) of the one, the other increases
(or decreases) to the same
Ex.
1. Cost is directly proportional to the number of articles. (More Articles,
More Cost)
Ex.
2. Work done is directly proportional to the number of men working on it (More
Men, More Work)
2.
Indirect Proportion: Two quantities are said to be
indirectly proportional,if on the increase of the one, the other decreases to
the same extent and vice-versa.
Ex.
1. The time taken by a car in covering a certain distance is inversely
proportional to the speed of the car.(More speed, Less is the time taken to
cover a distance)
Ex.
2. Time taken to finish a work is inversely proportional to the num of persons
working at it.
(More
persons, Less is the time taken to finish a job)
Remark:
In solving questions by chain rule, we compare every item
with the term to be found out.
12.
TIME AND WORK
1.
If A can do a piece of work in n days, then A's 1 day's work = (1/n).
2.
If A’s 1 day's work = (1/n),then A can finish the work in n days.
3.
A is thrice as good a workman as B, then:
Ratio
of work done by A and B = 3 : 1.
Ratio
of times taken by A and B to finish a work = 1 : 3.
13.
PIPES AND CISTERNS
1. Inlet: A
pipe connected with a tank or a cistern or a reservoir, that fills it, is known
as an
inlet.
Outlet:
A pipe connected with a tank or a cistern or a reservoir,
emptying it, is known as an outlet.
2. (i) If a pipe can fill a tank in x hours,
then : part filled in 1 hour = 1/x
(ii) If a pipe can empty a full tank in y
hours, then : part emptied in 1 hour = 1/y
(iii) If a pipe can .fill a tank in x
hours and another pipe can empty the full tank in y hours
(where y>
x), then on opening both the pipes, the net part filled in 1
hour =
(1/x)-(1/y)
(iv) If a pipe
can fill a tank in x hours and another pipe can empty the full tank in y hours
(where
x
> y), then on opening both the pipes, the net
part emptied in 1 hour = (1/y)-(1/x)
14.
TIME AND DISTANCE
Distance Distance
1.
Speed = Time , Time= Speed , Distance = (Speed * Time)
2.
x km / hr = x * 5/18
3.
x m/sec = (x * 18/5) km /hr
4.
If the ratio of the speeds of A and B is a:b , then the ratio of the times
taken by them to cover the same distance is 1/a: 1/b or
b:a.
5.
Suppose a man covers a certain distance at x km/ hr and an equal distance at y
km / hr .
Then
, the average speed during the whole journey is 2xy/(x+y) km/ hr.
15. PROBLEMS ON TRAINS
1. a km/hr= (a* 5/18) m/s.
2. a m / s = (a*18/5) km/hr.
3 Time taken by
a train of length 1 metres to pass a pole or a standing man or a signal post is
equal to the time taken by the train to cover 1
metres.
4. Time taken by a train of length 1 metres to
pass a stationary object of length b
metres is the time taken by the train to cover (1 + b) metres.
5. Suppose two trains or two bodies are moving in the same
direction at u m / s and
v m/s, where u
> v, then their relatives speed = (u - v) m / s.
6. Suppose two trains or two bodies are moving in opposite
directions at u m / s
and v m/s, then their relative speed is = (u + v) m/s.
7. If two trains of length a
metres and b metres
are moving in opposite directions at
u m / s and v m/s, then time taken by the trains to cross each other = (a + b)/(u+v)
sec.
8.If two trains of length a metres and b metres
are moving in the same direction
at u m / s and v
m / s, then the time taken by the
faster train to cross the slower
train = (a+b)/(u-v) sec.
9. If two trains (or bodies) start at the same time from
points A and B towards each
other and after crossing they take a and b sec in
reaching B and A respectively,
then
(A's speed) : (B’s speed) = (b1/2: a1/2).
16.BOATS AND STREAMS
1.In
water ,the direction along the stream is called downstream and ,the direction
against
the
stream is called upstream.
2.If
the speed of a boat in still water is u km/hr and the speed of the stream is v
km/hr, then:
speed
downstream=(u+v)km/hr.
speed
upstream=(u-v)km/hr.
3.If
the speed downstream is a km/hr and the speed upstream is b km/hr,then :
speed
in still water=1/2(a+b)km/hr, rate of stream=1/2(a-b)km/hr
17. SIMPLE INTEREST
1..
Principal: The money borrowed or lent out for a
certain period is called the principal or
the sum.
2.
Interest: Extra money paid for using other's money is called interest.
3.
Simple Interest (S.I.) :
If the interest on a sum borrowed for a certain period is reckoned
uniformly,
then it is called simple interest.
Let
Principal = P, Rate = R% per annum (p.a.) and Time = T years. Then,
(i)
S.I. = (P*R*T )/100
(ii)
P=(100*S.I)/(R*T) ;R=(100*S.I)/(P*T) and T=(100*S.I)/(P*R)
18.COMPOUND INTEREST
Compound
Interest: Sometimes it so happens that the borrower and the lender
agree to fix up a certain unit of time, say yearly or
half-yearly or quarterly
to settle the previous account. In such cases, the amount
after first unit of time becomes the principal for the second unit, the amount
after second unit becomes the principal for the third unit and so on.
After
a specified period, the difference between the
amount and the moneyborrowed is called the Compound
Interest (abbreviated as
C.I.) for that period.
Let
Principal = P, Rate = R% per annum, Time = n years.
I.
When interest is compound Annually:
Amount
= P(1+R/100)n
II.
When interest is compounded Half-yearly:
Amount = P[1+(R/2)/100]2n
III.
When interest is compounded Quarterly:
Amount = P[ 1+(R/4)/100]4n
IV.
When interest is compounded AnnuaI1y but time is in fraction, say 3(2/5) years.
Amount = P(1+R/100)3 x (1+(2R/5)/100)
V.
When Rates are different for different years, say Rl%, R2%, R3% for 1st, 2nd
and 3rd year respectively.
Then,
Amount = P(1+R1/100)(1+R2/100)(1+R3/100)
VI.
Present worth of Rs.x due n years
hence is given by :
Present
Worth = x/(1+(R/100))n
19 .AREA
FUNDEMENTAL CONCEPTS
I.RESULTS
ON TRIANGLES:
1.Sum
of the angles of a triangle is 180 degrees.
2.Sum
of any two sides of a triangle is greater than the third side.
3.Pythagoras
theorem:
In
a right angle triangle,
(Hypotenuse)
2 = (base)2 + (Height)2
4.The
line joining the midpoint of a side of a triangle to the opposite vertex is
called the
MEDIAN
5.The
point where the three medians of a triangle meet is called CENTROID.
Centroid
divides each of the medians in the ratio 2:1.
6.In
an isosceles triangle, the altitude from the vertex bi-sects the base
7.The
median of a triangle divides it into two triangles of the same area.
8.Area
of a triangle formed by joining the midpoints of the sides of a given triangle
is one-fourth of the area of the given triangle.
II.RESULTS
ON QUADRILATERALS:
1.
The diagonals of a parallelogram bisects each other .
2.
Each diagonal of a parallelogram divides it into two triangles of the same area
3.
The diagonals of a rectangle are equal and bisect each other.
4.
The diagonals of a square are equal and bisect each other at right angles.
5.
The diagonals of a rhombus are unequal and bisect each other at right angles.
6.
A parallelogram and a rectangle on the same base and between the same parallels
are equal
in
area.
7.
Of all the parallelograms of a given sides , the parallelogram which is a
rectangle has the
greatest
area.
IMPORTANT
FORMULAE
I.1.Area
of a rectangle=(length*breadth)
Therefore
length = (area/breadth) and breadth=(area/length)
2.Perimeter
of a rectangle = 2*(length+breadth)
II.Area
of a square = (side) 2 =1/2(diagonal) 2
III
Area of four walls of a room = 2*(length + breadth)*(height)
IV
1.Area of the triangle=1/2(base*height)
2.
Area of a triangle = √ (s*(s-a)(s-b)(s-c)), where a,b,c are
the sides of a triangle and s= ½(a+b+c)
3.Area
of the equilateral triangle =(√3/4)*(side) 2
4.Radius
of incircle of an equilateral triangle of side a=a/2√3
5.Radius
of circumcircle of an equilateral triangle of side a=a/√3
6.Radius
of incircle of a triangle of area del
V.1.Area
of the parellogram =(base *height)
2.Area
of the rhombus=1/2(product of the diagonals)
3.Area
of the trapezium=1/2(size of parallel sides)*distance between them
VI
1.Area of a circle =pi*r2,where r is the radius
2.
Circumference of a circle = 2ΠR.
3.
Length of an arc = 2ΠRθ/(360) where θ is the central angle
4.
Area of a sector = (1/2) (arc x R) = pi*R2*θ/360.
VII.
1. Area of a semi-circle = (pi)*R2.
2.
Circumference of a semi-circle =
(pi)*R.
20.VOLUME AND SURFACE AREA
I.
CUBOID
Let
length = 1, breadth = b and height = h units. Then,
1.
Volume = (1 x b x h) cubic units.
2.
Surface area= 2(lb + bh + lh) sq.units.
3.
Diagonal.= √(l2 +b2
+h2
) units
II.
CUBE
Let
each edge of a cube be of length a. Then,
1.
Volume = a3
cubic
units.
2.
Surface area = 6a2
sq.
units.
3.
Diagonal = √3
a units.
III.
CYLINDER
Let
radius of base = r and Height (or length) = h. Then,
1.
Volume = (π r2h) cubic units.
2.
Curved surface area = (2πrh).
units.
3.
Total surface area =2πr (h+r) sq. units
IV.
CONE
Let
radius of base = r and Height = h. Then,
1.
Slant height, l = √(h2+r2
)
2.
Volume = (1/3) πr2h
cubic units.
3.
Curved surface area = (πrl)
sq. units.
4.
Total surface area = (πrl
+ πr2
)
sq. units.
V.
SPHERE
Let
the radius of the sphere be r. Then,
1.
Volume = (4/3) πr3 cubic
units.
2.
Surface area = (4πr2)
sq. units.
VI.
HEMISPHERE
Let
the radius of a hemisphere be r. Then,
1.
Volume = (2/3) πr3
cubic
units.
2.
Curved surface area = (2πr2)
sq. units.
3.
Total surface area = (3πr2)
units.
Remember:
1 litre = 1000 cm3.
21. CALENDAR
Under this heading we mainly deal with finding the day of
the week on a particular
given date the process of finding it lies on obtaining the
number of odd days.
Odd Days : Number
of days more than the complete number of weeks in a given.
Period ., is the
number of odd days during that period.
LeapYear: Every
year which is divisible by 4 is called a leap year.
Thus each one of the years 1992, 1996, 2004, 2008, 2012,
etc. is a leap year. Every
4th century is a leap year but no other century is a leap
year. Thus each one of 400,
800, 1200,' 1600, 2000, etc. is a leap year.
None of 1900, 2010, 2020, 2100, etc. is a leap year.
An year which is not a leap year is called an
ordinary year.
(I )An ordinary year has 365 days. (II) A leap year has 366
days.
Counting of Odd Days:
i)1 ordinary year = 365 days = (52 weeks + 1 day).
:. An ordinary year has 1 odd day.
ii)1 leap year = 366 days = (52 weeks + 2 days).
:. A leap year has 2 odd days.
_ iii)100 years = 76 ordinary years + 24 leap years
= [(76 x 52) weeks + 76 days) + [(24 x 52) weeks + 48 days]
= 5200 weeks + 124 days = (5217 weeks + 5 days).
:. 100 years contain 5 odd days.
200 years contain 10 and therefore 3 odd days.
300 years contain 15 and therefore 1 odd day.
400 years contain (20 + 1) and therefore 0 odd day.
Similarly, each one of 800, 1200, 1600, 2000, etc. contains
0 odd days.
Remark: (7n + m) odd days, where m < 7 is equivalent to m
odd days.
Thus, 8 odd days ≡ 1 odd day etc.
22. CLOCKS
IMPORTANT FACTS
The Face or dial of a watch is a circle whose circumference
is divided into 60
equal parts, called minute
spaces.
A clock has two hands, the smaller one is called the hour
hand or short hand
while the larger one is called the minute hand or long
hand..
i) In 60 minutes, the minute hand gains 55 minutes on the
hour hand.
ii) In every hour, both the hands coincide once.
iii) The hands are in the same straight line when they are
coincident or opposite to
each other.
iv) When the two hands are at right angles, they are 15
minute spaces apart.
v)When the hand's are in opposite directions, they are 30
minute spaces apart.
vi)Angle traced by hour hand in 12 hrs = 360°.
vii)Angle traced by minute hand in 60 min. = 360°.
Too Fast and Too
Slow: If a watch or a clock
indicates 8.15, when the correct
time ,
8 is said to be 15 minutes
too fast.
On the other hand, if it indicates 7.45, when
the correct time is 8, it is said to be
15 minutes too slow
23. TRUE DISCOUNT
Suppose a man has to pay Rs. 156 after 4 years and the rate
of
interest is 14% per annum. Clearly, Rs. 100 at 14% will
amount to
Rs. 156 in 4 years. So, the payment of Rs. 100 now will
clear off
the debt of Rs. 156 due 4 years hence. We say that:
Sum due = Rs. 156 due 4 years hence;
Present Worth (P.W.) = Rs. 100;
True Discount (T.D.) = Rs. (156 - 100) = Rs. 56
(Sum due) - (P.W.).
We define : T.D. = Interest on P.W.
Amount = (P.W.) + (T.D.).
Interest is reckoned on P.W. and true discount is reckoned
on the amount.
IMPORTANT FORMULAE
Let rate = R% per annum and Time = T years. Then,
1. P.W.=[100 x Amount /100 + (R x T)
=100 x T.D./ RxT
2. T.D.=[(P.W.) x R x T /100]
= [ Amount x RxT/100 + (R x T)]
3.(S.I.)*(T.D.) /(S.I.)-(T.D.)
4. (S.I.) - (T.D.) - S.I. on T.D.
5. When the sum is put at compound interest, then
P.W. = Amount/[1
+R/100]^T.
Number Prefix Symbol
10 1 deka- da
10 2 hecto- h
10 3 kilo- k
10 6 mega- M
10 9 giga- G
10 12 tera- T
10 15 peta- P
10 18 exa- E
10 21 zeta- Z
10 24 yotta- Y
Number Prefix Symbol
10 -1 deci- d
10 -2 centi- c
10 -3 milli- m
10 -6 micro- u (greek mu)
10 -9 nano- n
10 -12 pico- p
10 -15 femto- f
10 -18 atto- a
10 -21 zepto- z
10 -24 yocto- y
Roman Numerals
I=1 V=5 X=10 L=50 C=100 D=500 M=1 000
_
V=5 000
_
X=10 000
_
L=50 000
_
C = 100 000
_
D=500 000
__
M=1 000 000
Fraction = Decimal
1/1 = 1
1/2 = 0.5
1/3 = 0.3, 2/3 = 0.6
1/4 = 0.25, 3/4 = 0.75
1/5 = 0.2, 2/5 = 0.4, 3/5 = 0.6, 4/5 = 0.8
1/6 = 0.16, 5/6 = 0.83
1/7 = 0.142857 , 2/7
= 0.285714 , 3/7 = 0.428571, 4/7 = 0.571428 5/7 = 0.714285, 6/7 = 0.857142
1/8 = 0.125 , 3/8 =
0.375, 5/8 = 0.625, 7/8 = 0.875
1/9 = 0.1, 2/9
= 0.2 , 4/9 = 0.4, 5/9 = 0.5
7/9 = 0.7,
8/9 = 0.8
1/10 = 0.1, 3/10
= 0.3, 7/10 = 0.7, 9/10 = 0.9
1/11 = 0.09, 2/11
= 0.18, 3/11 = 0.27, 4/11 = 0.36 5/11 = 0.45, 6/11 = 0.54, 7/11 = 0.63
8/11 = 0.72, 9/11 = 0.81, 10/11 = 0.90
1/12 = 0.083, 5/12 =
0.416, 7/12 = 0.583, 11/12 = 0.916
1/16 = 0.0625, 3/16 =
0.1875, 5/16 = 0.3125, 7/16 = 0.4375
11/16 = 0.6875, 13/16 = 0.8125, 15/16 = 0.9375
1/32 = 0.03125, 3/32 =
0.09375, 5/32 = 0.15625, 7/32 = 0.21875
9/32 = 0.28125,
11/32 = 0.34375, 13/32 = 0.40625
15/32 = 0.46875,
17/32 = 0.53125, 19/32 = 0.59375
21/32 = 0.65625,
23/32 = 0.71875, 25/32 = 0.78125
27/32 = 0.84375,
29/32 = 0.90625, 31/32 = 0.96875
_______________________________________________________________________________
Courtesy: Quantitative Aptitude by R.S. Agarwal
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