business mathematics

Facts And Formulae:

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

i.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 HCF.

Example : Find HCF of 26 * 32*5*74 , 22 *35*52 * 76 ,
2*52 *72
Solution: The prime numbers given common numbers are 2,5,7
Therefore HCF is 22 * 5 *72 .

ii.Division Method : Divide the larger number by smaller one. Now divide the divisor by remainder. Repeat the process of dividing preceding number last obtained till zero is obtained as number. The last divisor is HCF.


Least common multiple[LCM] : The least number which is divisible by each one of given numbers is LCM.

There are two methods for this:
i.Factorization method : Resolve each one into product of prime factors. Then LCM is product of highest powers of all factors.

ii.Common division method.

Most of you have a question of “what is remainder theorem?”
Here it is………….

When a polynomial function f(x) is divided by (x-a), the remainder is f(a)
For example, when x^2 – 2x + 5 is divided by x-1, the remainder will be f(1),
i.e. 12 – 2(1) + 5 = 4

We can see that if f(x) is divided by (x+a), then the remainder will be f(-a).
For example, when x^3 + x^2 -5x – 4 is divided by (x+1), then the remainder will be f(-1).
i.e., (-1)^3 + (-1)^2 – 5(-1) – 4. i.e. 1

If f(a) is zero it means that the remainder is zero. Hence we can say that (x-a) is a factor of f(x).

In simplifying an expression of various operations must be performed as per the fallowing order.

V B O D M A S

V -------- Vinculum
B -------- Brackets - in the order (),{},[]
O -------- Of
D -------- Division
M -------- Multiplication
A -------- Addition
S -------- Subtraction

In the above order simplify the given expression.

Take the square root of the number.
Round of the square root to the next highest integer call this number as Z.
Check for divisibility of the number N by all prime numbers below Z. If
there is no numbers below the value of Z which divides N then the number
will be prime.

Example 239 is prime or not?
√239 lies between 15 or 16.Hence take the value of Z=16.
Prime numbers less than 16 are 2,3,5,7,11 and 13.
239 is not divisible by any of these. Hence we can conclude that 239
is a prime number.

Binary and Decimal systems:

What is decimal:
To describe the numbers we generally use the following ten digits.
1,2,3,4,5,6,7,8,9,0.
We know that the system in which we use these ten digits to describe numbers is called decimal system. The decimal system is also called as base ten system.

What is binary:
Only two digits 0 and 1 are used for computation in computers. This system is called BINARY

system. It is also called base two system.

0,1,10,11,100,101,110,111.......and so on are few numbers in binary system.

Binary - Decimal
0 - 0
1 - 1
10 - 2
11 - 3
100 - 4
101 - 5
110 - 6
111 - 7
1000 - 8
1001 - 9

Conversion of Binary to Decimal:
1. convert 1100(base 2) to decimal
1100(2) = 1*8 + 1*4 + 0*2 + 0*1
= 8 + 4 + 0 + 0
= 12


Conversion of Decimal to Binary:
We use division method to convert decimal to binary

Eg:
56 in the binary system:

2 | 56 unit's place
_______
2 | 28-0 two's place
_______
2 | 14-0 fourth's place
_______
2 | 7-0 eighth's place
_______
2 | 3-1 sixteen's place
_______
2 | 1-1 thirty two's place
_______
| 0-1 sixty four's place


Therefore 56 = 111000 (binary)

ROMAN numbers in maths and their rules:

some of the ROMAN numerals and their corresponding indo-arabic numerals:
I - 1
V - 5
X - 10
L - 50
C - 100
D - 500
M - 1000

While writing ROMAN numbers, the following rules are to be kept in mind.

1. If a digit is repeated a number of times, the value of the digit is added as many times as it occurs.
examples: II = 1+1 = 2
XXX = 10+10+10 = 30
MM = 1000+1000 = 2000

2. To write a number in which the smallest digit always comes to the right of the right of the greater digit, we add all the values of all the digits.
examples: VIII = 5+1+1+1 = 8
LXVI = 50+10+5+1 = 66
DCLXVII = 500+100+50+10+5+1+1 = 667

3. TO WRITE A NUMBER IN WHICH the smaller digit is placed before the greater digit. we subtract the value of the smaller digit from that of the greater digit.
examples: IV = 5-1 =4
XL = 50-10 = 40
LIX = 50+(10-1) = 59

Simple Problems on Numbers

Simple Problems on Numbers

Simple problems:

1.What least number must be added to 3000 to obtain a number
exactly divisible by 19?

Solution:
On dividing 3000 by 19 we get 17 as remainder
Therefore number to be added = 19-17=2.

2.Find the unit's digit n the product 2467 153 * 34172?

Solution:
Unit's digit in the given product=Unit's digit in 7 153 * 172
Now 7 4 gives unit digit 1
7 152 gives unit digit 1
7 153 gives 1*7=7.Also 172 gives 1
Hence unit's digit in the product =7*1=7.

3.Find the total number of prime factors in 411 *7 5 *112 ?

Solution:
411 7 5 112= (2*2) 11 *7 5 *112
= 222 *7 5 *112
Total number of prime factors=22+5+2=29

4.The least umber of five digits which is exactly
divisible by 12,15 and 18 is?
a.10010 b.10015 c.10020 d.10080

Solution:
Least number of five digits is 10000
L.C.Mof 12,15,18 s 180.
On dividing 10000 by 180,the remainder is 100.
Therefore required number=10000+(180-100)
=10080.
Ans (d).

5.The least number which is perfect square and is divisible
by each of the numbers 16,20 and 24 is?
a.1600 b.3600 c.6400 d.14400

Solution:
The least number divisible by 16,20,24 = L.C.M of 16,20,24=240
=2*2*2*2*3*5
To make it a perfect square it must be multiplied by 3*5.
Therefore required number =240*3*5=3600.
Ans (b).

6.A positive number which when added to 1000 gives a sum ,
which is greater than when it is multiplied by 1000.
The positive integer is?
a.1 b.3 c.5 d.7

Solution:
1000+N>1000N
clearly N=1.

7.How many numbers between 11 and 90 are divisible by 7?

Solution:
The required numbers are 14,21,28,...........,84.
This is an A.P with a=14,d=7.
Let it contain n terms
then T =84=a+(n-1)d
=14+(n-1)7
=7+7n
7n=77 =>n=11.

8.Find the sum of all odd numbers up to 100?

Solution:
The given numbers are 1,3,5.........99.
This is an A.P with a=1,d=2.
Let it contain n terms 1+(n-1)2=99
=>n=50
Then required sum =n/2(first term +last term)
=50/2(1+99)=2500.

9.How many terms are there in 2,4,6,8..........,1024?

Solution:
Clearly 2,4,6........1024 form a G.P with a=2,r=2
Let the number of terms be n
then 2*2 n-1=1024
2n-1 =512=29
n-1=9
n=10.

10.2+22+23+24+25..........+28=?

Solution:
Given series is a G.P with a=2,r=2 and n=8.
Sum Sn=a(1-r n)/1-r=Sn=2(1-28)/1-2.
=2*255=510.

11.Find the number of zeros in 27!?

Solution:
Short cut method :
number of zeros in 27!=27/5 + 27/25
=5+1=6zeros.

Medium Problems:

12.The difference between two numbers 1365.When the larger
number is divided by the smaller one the quotient is 6 and
the remainder is 15.The smaller number is?
a.240 b.270 c.295 d.360

Solution:
Let the smaller number be x, then larger number =1365+x
Therefore 1365+x=6x+15
5x=1350 => x=270
Required number is 270.

13.Find the remainder when 231 is divided by 5?

Solution:
210 =1024.
unit digit of 210 * 210 * 210 is 4
as 4*4*4 gives unit digit 4
unit digit of 231 is 8.
Now 8 when divided by 5 gives 3 as remainder.
231 when divided by 5 gives 3 as remainder.

14.The largest four digit number which when divided by 4,7
or 13 leaves a remainder of 3 in each case is?
a.8739 b.9831 c.9834 d.9893. Solution:

solution:
Greatest number of four digits is 9999
L.C.M of 4,7, and 13=364.
On dividing 9999 by 364 remainder obtained is 171.
Therefore greatest number of four digits divisible by 4,7,13
=9999-171=9828.
Hence required number=9828+3=9831.
Ans (b).

15.What least value must be assigned to * so that th number
197*5462 is divisible by 9?

Solution:
Let the missing digit be x
Sum of digits = (1+9+7+x+5+4+6+2)=34+x
For 34+x to be divisible by 9 , x must be replaced by 2
The digit in place of x must be 2.

16.Find the smallest number of 6 digits which is exactly
divisible by 111?

Solution:
Smallest number of 6 digits is 100000
On dividing 10000 by 111 we get 100 as remainder
Number to be added =111-100=11.
Hence,required number =10011.

17.A number when divided by 342 gives a remainder 47.When
the same number is divided by 19 what would be the remainder?

Solution:
Number=342 K + 47 = 19 * 18 K + 19 * 2 + 9=19 ( 18K + 2) + 9.
The given number when divided by 19 gives 18 K + 2 as quotient
and 9 as remainder.

18.In doing a division of a question with zero remainder,a
candidate took 12 as divisor instead of 21.The quotient
obtained by him was 35. The correct quotient is?
a.0 b.12 c.13 d.20

Solution:
Dividend=12*35=420.
Now dividend =420 and divisor =21.
Therefore correct quotient =420/21=20.

19.If a number is multiplied by 22 and the same number is
added to it then we get a number that is half the square
of that number. Find the number.
a.45 b.46 c.47 d. none

Solution:
Let the required number be x.
Given that x*22+x = 1/2 x2
23x = 1/2 x2
x = 2*23=46
Ans (b)

20.Find the number of zeros in the factorial of the number 18?

Solution:
18! contains 15 and 5,which combined with one even number
gives zeros. Also 10 is also contained in 18! which will
give additional zero .Hence 18! contains 3 zeros and the
last digit will always be zero.

21.The sum of three prime numbers is 100.If one of them
exceeds another by 36 then one of the numbers is?
a.7 b.29 c.41 d67.

Solution:
x+(x+36)+y=100
2x+y=64
Therefore y must be even prime which is 2
2x+2=64=>x=31.
Third prime number =x+36=31+36=67.

22.A number when divided by the sum of 555 and 445 gives
two times their difference as quotient and 30 as remainder .
The number is?
a.1220 b.1250 c.22030 d.220030.

Solution:
Number=(555+445)*(555-445)*2+30
=(555+445)*2*110+30
=220000+30=220030.

23.The difference of 1025-7 and 1024+x is divisible by 3 for x=?
a.3 b.2 c.4 d.6

Solution:
The difference of 1025-7 and 1024+x is
=(1025-7)-(1024-x)
=1025-7-1024-x
=10.1024-7 -1024-x
=1024(10-1)-(7-x)
=1024*9-(7+x)
The above expression is divisible by 3 so we have to
replace x with 2.
Ans (b).

Complex Problems:

24.Six bells commence tolling together and toll at intervals
of 2,4,6,8,10,12 seconds respectively. In 30 minutes how many
times do they toll together?

Solution:
To find the time that the bells will toll together we have
to take L.C.M of 2,4,6,8,10,12 is 120.
So,the bells will toll together after every 120 seconds
i e, 2 minutes
In 30 minutes they will toll together [30/2 +1]=16 times

25.The sum of two numbers is 15 and their geometric mean is
20% lower than their arithmetic mean. Find the numbers?
a.11,4 b.12,3 c.13,2 d.10,5

Solution:
Sum of the two numbers is a+b=15.
their A.M = a+b / 2 and G.M = (ab)1/2
Given G.M = 20% lower than A.M
=80/100 A.M
(ab)1/2=4/5 a+b/2 = 2*15/5= 6
(ab)1/2=6
ab=36 =>b=36/a
a+b=15
a+36/a=15
a2+36=15a
a2-15a+36=0
a2-3a-12a+36=0
a(a-3)-12(a-3)=0
a=12 or 3.
If a=3 and a+b=15 then b=12.
If a=12 and a+b=15 then b=3.
Ans (b).

26.When we multiply a certain two digit number by the
sum of its digits 405 is achieved. If we multiply the
number written in reverse order of the same digits
by the sum of the digits,we get 486.Find the number?
a.81 b.45 c.36 d. none

Solution:
Let the number be x y.
When we multiply the number by the sum of its digit
405 is achieved.
(10x+y)(x+y)=405....................1
If we multiply the number written in reverse order by its
sum of digits we get 486.
(10y+x)(x+y)=486......................2
dividing 1 and 2
(10x+y)(x+y)/(10y+x)(x+y) = 405/486.
10x+y / 10y+x = 5/6.
60x+6y = 50y+5x
55x=44y
5x = 4y.
From the above condition we conclude that the above
condition is satisfied by the second option i e b. 45.
Ans (b).

27.Find the HCF and LCM of the polynomials x2-5x+6 and x2-7x+10?
a.(x-2),(x-2)(x-3)(x-5)
b.(x-2),(x-2)(x-3)
c.(x-3),(x-2)(x-3)(x-5)
d. none

Solution:
The given polynomials are
x2-5x+6=0................1
x2-7x+10=0...............2
we have to find the factors of the polynomials
x2-5x+6 and x2-7x+10
x2-2x-3x+6 x2-5x-2x+10
x(x-2)-3(x-2) x(x-5)-2(x-5)
(x-3)(x-2) (x-2)(x-5)
From the above factors of the polynomials we can easily
find the HCF as (x-3)and LCM as (x-2)(x-3)(x-5).
Ans (c)

28.The sum of all possible two digit numbers formed from
three different one digit natural numbers when divided by
the sum of the original three numbers is equal to?
a.18 b.22 c.36 d. none

Solution:
Let the one digit numbers x,y,z
Sum of all possible two digit numbers
=(10x+y)+(10x+z)+(10y+x)+(10y+z)+(10z+x)+(10z+y) = 22(x+y+z)
Therefore sum of all possible two digit numbers when
divided by sum of one digit numbers gives 22.

29.A number being successively divided by 3,5,8 leaves
remainders 1,4,7 respectively. Find the respective
remainders if the order of divisors are reversed?

Solution:
Let the number be x.
3 - x
5 y - 1
8 z - 4
1 - 7

z=8*1+7=15
y=5z+4 = 5*15+4 = 79
x=3y+1 = 3*79+1=238
Now 8 238
5 29 - 6
3 5 - 4
1 - 2
Respective remainders are 6,4,2.

30.The arithmetic mean of two numbers is smaller by 24
than the larger of the two numbers and the GM of
the same numbers exceeds by 12 the smaller of the numbers.
Find the numbers?
a.6,54 b.8,56 c.12,60 d.7,55

Solution:
Let the numbers be a,b where a is smaller and b
is larger number.
The AM of two numbers is smaller by 24 than the
larger of the two numbers.
AM=b-24
AM of two numbers is a+b/2.
a+b/2 = b-24
a+b = 2b-48
a = b-48...................1
The GM of the two numbers exceeds by 12 the smaller
of the numbers
GM = a+12
GM of two numbers is (ab)1/2
(ab) 1/2= a+12
ab = a2+144+24a
from 1 b=a+48
a(a+48)= a2+144+24a
a2+48a = a2+144+24a
24a=144=>a=6
Therefore b=a+48=54.
Ans (a).

31.The sum of squares of the digits constituting a positive
two digit number is 13,If we subtract 9 from that number
we shall get a number written by the same digits in the
reverse order. Find the number?
a.12 b.32 c.42 d.52.

Solution:
Let the number be x y.
the sum of the squares of the digits of the number is 13
x2+y2=13
If we subtract 9 from the number we get the number
in reverse order
x y-9=y x.
10x+y-9=10y+x.
9x-9y=9
x-y=1
(x-y)2 =x2+y2-2x y
1 =13-2x y
2x y = 12
x y = 6 =>y=6/x
x-y=1
x-6/x=1
x2-6=x
x2-x-6=0
x+2x-3x-6=0
x(x+2)-3(x+2)=0
x=3,-2.
If x=3 and x-y=1 then y=2.
If x=-2 and x-y=1 then y=-3.
Therefore the number is 32.
Ans (b).

32.If we add the square of the digit in the tens place
of the positive two digit number to the product of the
digits of that number we get 52,and if we add the square
of the digit in the unit's place to the same product
of the digits we get 117.Find the two digit number?
a.18 b.39 c.49 d.28

Solution:
Let the digit number be x y
Given that if we add square of the digit in the tens place
of a number to the product of the digits we get 52.
x2+x y=52.
x(x+y)=52....................1
Given that if we add the square of the digit in the unit's plac
e to the product is 117.
y2+x y= 117
y(x+y)=117.........................2
dividing 1 and 2 x(x+y)/y(x+y) = 52/117=4/9
x/y=4/9
from the options we conclude that the two digit number is 49
because the condition is satisfied by the third option.
Ans (c)

33.The denominators of an irreducible fraction is greater
than the numerator by 2.If we reduce the numerator of the
reciprocal fraction by 3 and subtract the given fraction
from the resulting one,we get 1/15.Find the given fraction?

Solution:
Let the given fraction be x / (x+2) because given that
denominator of the fraction is greater than the numerator by 2
1 – [(x – 1/(x+2))/3] = 1/15.
1 – (x2+2x-1) /3(x+2) = 1/15
(3x+6-x2-2x+1)/3(x+2) = 1/15
(7-x2+2x)/(x+2) = 1/5
-5x2+5x+35 = x+2
5x2-4x-33 = 0
5x2-15x+11x-33 = 0
5x(x-3)+11(x-3) = 0
(5x+11)(x-3) = 0
Therefore x=-11/5 or 3
Therefore the fraction is x/(x+2) = 3/5.

34.Three numbers are such that the second is as much
lesser than the third as the first is lesser than
the second. If the product of the two smaller numbers
is 85 and the product of two larger numbers is 115.
Find the middle number?

Solution:
Let the three numbers be x,y,z
Given that z – y = y – x
2y = x+z.....................1
Given that the product of two smaller numbers is 85
x y = 85................2
Given that the product of two larger numbers is 115
y z = 115...............3
Dividing 2 and 3 x y /y z = 85/115
x / z = 17 / 23
From 1
2y = x+z
2y = 85/y + 115/y
2y2 = 200
y2 = 100
y = 10

35.If we divide a two digit number by the sum of its digits
we get 4 as a quotient and 3 as a remainder. Now if we
divide that two digit number by the product of its digits
we get 3 as a quotient and 5 as a remainder .
Find the two digit number?

Solution:
Let the two digit number is x y.
Given that x y / (x+y)
quotient=4 and remainder = 3
we can write the number as
x y = 4(x+y) +3...........1
Given that x y /(x*y) quotient = 3 and remainder = 5
we can write the number as
x y = 3 x*y +5...............2
By trail and error method
For example take x=1,y=2
1............12=4(2+3)+3
=4*3+3
! =15
let us take x=2 y=3
1..............23=4(2+3)+3
=20+3
=23
2.............23=3*2*3+5
=18+5
=23
the above two equations are satisfied by x=2 and y=3
Therefore the required number is 23.

36.First we increased the denominator of a positive
fraction by 3 and then it by 5.The sum of the
resulting fractions proves to be equal to 2/3.
Find the denominator of the fraction if its numerator is 2.

Solution:
Let us assume the fraction is x/y
First we increasing the denominator by 3 we get x/(y-3)
Then decrease it by 5 we get the fraction as x/(y-5)
Given that the sum of the resulting fraction is 2/3
x/(y+3) + x/(y-5) = 2/3
Given numerator equal to 2
2*[ 1/y+3 + 1/y-5] =2/3
(y-5+y+3) / (y-3)(y+5) =1/3
6y – 6 = y2-5y+3y-15
y2-8y-9 = 0
y2-9y+y-9 = 0
y(y-9)+1(y-9) = 0
Therefore y =-1 or 9.

37.If we divide a two digit number by a number consisting
of the same digits written in the reverse order,we get 4
as quotient and 15 as a remainder. If we subtract 1 from
the given number we get the sum of the squares of the
digits constituting that number. Find the number?
a.71 b.83 c.99 d. none

Solution:
Let the number be x y.
If we divide 10x+y by a number in reverse order
i e,10y+x we get 4 as quotient and 15 as remainder.
We can write as
10x+y = 4(10y+x)+15......................1
If we subtract 1 from the given number we get square of the digits
10x+y = x2+y2.....................................2
By using above two equations and trail and error method
we get the required number. From the options also we can
solve the problem. In this no option is satisfied so answer is d.
Ans (d)

Square and Cube Roots problems

Square and Cube Roots problems

Formula:

The Product of two same numbers in easiest way as follow.
Example:let us calculate the product of 96*96
Solution: Here every number must be compare with the 100.
See here the given number 96 which is 4 difference with the 100.
so subtract 4 from the 96 we get 92 ,then the square of the
number 4 it is 16 place the 16 beside the 92 we get answer
as 9216.

9 6
- 4
--------------
9 2
--------------
4*4=16
9 2 1 6

therefore square of the two numbers 96*96=9216.

Example: Calculate product for 98*98
Solution: Here the number 98 is having 2 difference when compare
to 100 subtract 2 from the number then we get 96 square the
number 2 it is 4 now place beside the 96 as 9604

9 8
- 2
-------------
9 6
-------------
2*2=4
9 6 0 4.
so, we get the product of 98*98=9604.

Example: Calculate product for 88*88
Solution: Here the number 88 is having 12 difference when compare
to 100 subtract 12 from the 88 then we get 76 the square of the
number 12 is 144 (which is three digit number but should place
only two digit beside the 76) therefore in such case add one to
6 then it becomes 77 now place 44 beside the number 77 we will get
7744.
88
-12
------------
76
-----------
12*12=144

76
+ 144
--------------------
7744
--------------------

Example: Find the product of the numbers 46 *46?
Solution:consider the number 50=100/2. Now again go comparision with
the number which gets when division with 100.here consider the number
50 which is nearer to the number given. 46 when compared with the
number 50 we get the difference of 4. Now subtract the number 4 from
the 46, we get 42. As 50 got when 100 get divided by 2.
so, divided the number by 2 after subtraction.
42/2=21 now square the the number 4 i.e, 4*4=16
just place the number 16 beside the number 21
we get 2116.
4 6
4
----------------
4 2 as 50 = 100/2

42/2=21
now place 4*4=16 beside 21
2 1 1 6

Example: Find the product of the numbers 37*37
Solution:
consider the number 50=100/2
now again go comparision with the number which gets when
division with 100.
here consider the number 50 which is nearer to the number given.
37 when compared with the number 50 we get the difference of 13.
now subtract the number 13 from the 37, we get 24.
as 50 got when 100 get divided by 2.
so, divided the number by 2 after subtraction.
24/2=12
now square the the number 13 i.e, 13*13=169
just place the number 169 beside the number 21
now as 169 is three digit number then add 1 to 2 we get
1t as 13 then place 69 beside the 13
we get 1369.

3 7
1 3
-----------------
2 4 as 50 = 100/2

24/2=12
square 13* 13=169

1 2
+ 1 6 9
-----------------------
1 3 6 9
-------------------------

Example: Find the product of 106*106
Solution: now compare it with 100 ,
The given number is more then 100
then add the extra number to the given number.
That is 106+6=112
then square the number 6 that is 6*6=36
just place beside the number 36 beside the 112,then
we get 11236.
1 0 6
+ 6
---------------------
1 1 2
--------------------
now 6* 6=36 place this beside the number 112, we get
1 1 2 3 6

Square root: If x2=y ,we say that the square root of y
is x and we write ,√y=x.

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 x1/3 .

Examples:

1.Evaluate 60841/2 by factorization method.

Solution: Express the given number as the product of prime
factors. Now, take the product of these prime factors choosing
one out of every pair of the same primes. This product gives the
square root of the given number.

Thus resolving 6084 in the prime factors ,we get 6084
2 6024
2 3042
3 1521
3 507
13 169
13
6084=21/2 *31/2 *131/2
60841/2=2*3*13=78.
Answer is 78.

2.what will come in place of question mark in each of the following
questions?

i)(32.4/?)1/2 = 2
ii)86.491/2 + (5+?1/2)2 =12.3

Solution: 1) (32.4/x)1/2=2
Squaring on both sides we get
32.4/x=4
=>4x=32.4
=>x=8.1

Answer is 8.1

ii)86.491/2 + √(5+x2)=12.3

solutin:86.491/2 + (5+x1/2 )=12.3
9.3+ √(5+x1/2 )=12.3
=> √(5+x1/2 ) =12.3-9.3
=> √(5+x1/2 )=3
Squaring on both sides we get
(5+x1/2 )=9
x1/2 =9-5
x1/2 =4
x=2.
Answer is 2.

3.√ 0.00004761 equals:

Solution: √ (4761/108)
√4761/√ 108
. 69/10000
0.0069.
Answer is 0.0069

4.If √18225=135,then the value of
√182.25 + √1.8225 + √ 0.018225 + √0.00018225.

Solution: √(18225/100) +√(18225/10000) +
√(18225/1000000) +√(18225/100000000)
=√(18225)/10 + (18225)1/2/100 +
√(18225)/1000 + √(18225)/10000
=135/10 + 135/100 + 135/1000 + 135/10000
=13.5+1.35+0.135+0.0135=14.9985.
Answer is 14.9985.

5.what should come in place of both the question
marks in the equation (?/ 1281/2= (162)1/2/?) ?

Solution: x/ 1281/2= (162)1/2/x
=>x1/2= (128*162)1/2
=> x1/2= (64*2*18*9)1/2
=>x2= (82*62*32)
=>x2=8*6*3
=>x2=144
=>x=12.

6.If 0.13 / p1/2=13 then p equals

Solution: 0.13/p2=13
=>p2=0.13/13
=1/100
p2=√(1/100)
=>p=1/10
therefore p=0.1
Answer is 0.1

7.If 13691/2+(0.0615+x)1/2=37.25 then x is equals to:

Solution
37+(0.0615+x)1/2=37.25(since 37*37=1369)
=>(0.0615+x)1/2=0.25
Squaring on both sides
(0.0615+x)=0.0625
x=0.001
x=10-3.
Answer is 10-3.

8.If √(x-1)(y+2)=7 x& y being positive whole numbers then
values of x& y are?

Solution: √(x-1)(y+2)=7
Squaring on both sides we get
(x-1)(y+2)=72
x-1=7 and y+2=7
therefore x=8 , y=5.
Answer x=8 ,y=5.

9.If 3*51/2+1251/2=17.88.then what will be the
value of 801/2+6*51/2?

Solution: 3*51/2+1251/2=17.88
3*51/2+(25*5)1/2=17.88
3*51/2+5*51/2=17.88
8*51/2=17.88
51/2=2.235
therefore 801/2+6 51/2=(16*51/2)+6*1/25
=4 51/2+6 51/2
=10*2.235
=22.35
Answer is 22.35

10.If 3a=4b=6c and a+b+c=27*√29 then Find c value is:

Solution: 4b=6c
=>b=3/2*c
3a=4b
=>a=4/3b
=>a=4/3(3/2c)=2c
therefore a+b+c=27*291/2
2c+3/2c+c=27*291/2
=>4c+3c+2c/2=27*291/2
=>9/2c=27*291/2
c=27*291/2*2/9
c=6*291/2

11.If 2*3=131/2 and 3*4=5 then value of 5*12 is

Solution:
Here a*b=(a2+b2)1/2
therefore 5*12=(52+122)1/2
=(25+144)1/2
=1691/2
=13
Answer is 13.

12.The smallest number added to 680621 to make
the sum a perfect square is

Solution: Find the square root number which
is nearest to this number
8 680621 824
64
162 406
324
1644 8221
6576
1645
therefore 824 is the number ,to get the nearest
square root number take (825*825)-680621
therefore 680625-680621=4
hence 4 is the number added to 680621 to make it
perfect square.

13.The greatest four digit perfect square number is

Solution: The greatest four digit number is 9999.
now find the square root of 9999.
9 9999 99
81
189 1819
1701
198
therefore 9999-198=9801 which is required number.
Answer is 9801.

14.A man plants 15376 apples trees in his garden and arranges
them so, that there are as many rows as there are apples trees
in each row .The number of rows is.

Solution: Here find the square root of 15376.
1 15376 124
1
22 53
44
244 976
976
0
therefore the number of rows are 124.

15.A group of students decided to collect as many paise from
each member of the group as is the number of members. If the
total collection amounts to Rs 59.29.The number of members
in the group is:

Solution: Here convert Money into paise.
59.29*100=5929 paise.
To know the number of member ,calculate the square root of 5929.
7 5929 77
49
147 1029
1029
0
Therefore number of members are 77.

16.A general wishes to draw up his 36581 soldiers in the form
of a solid square ,after arranging them ,he found that some of
them are left over .How many are left?
Solution: Here he asked about the left man ,So find the
square root of given number the remainder will be the left man
1 36581 191
1
29 265
261
381 481
381
100(since remaining)
Therefore the left men are 100.

17.By what least number 4320 be multiplied to obtain number
which is a perfect cube?

Solution: find l.c.m for 4320.
2 4320
2 2160
2 1080
2 540
2 270
3 135
3 45
3 15
5
4320=25 * 33 * 5
=23 * 33 * 22 *5
so make it a perfect cube ,it should be multiplied by 2*5*5=50
Answer is 50.

18.3(4*12/125)1/2=?

Solution: 3(512/125)1/2
3(8*8*8)1/2/(5*5*5)
3(83)1/2/(53)
((83)/(53))1/3
=>8/5 or 1 3/5.

Simplifications problems

Simplifications problems

Introduction:

'BODMAS' rule: This rule depicts the correct sequence in which
the operations are to be executed, so as to find out the value of
a given expression.

Here B stands for Bracket, O for Of, D for Division, M for
Multiplication, A for Addition and S for Subtraction.

First of all the brackets must be removed, strictly in the
order () , {} , [].

After removing the brackets, we want use the following operations:
1.Of 2. Division 3. Multiplication 4. Addition 5. Subtraction

Modulus of a real number:
Modulus of a real number is a defined as
|a| = a, if a>0 or -a, if a < 0;

Problems:

1.(5004 /139) – 6= ?

Sol: Expression = 5004/ 139 – 6 = 36 – 6 = 30;


2.What mathematical operations should come at the place of ? in the
equation : (2 ? 6 – 12 / 4 + 2 = 11) ?

Sol: 2 ? 6 = 11 + 12 / 4 – 2
= 11 + 3 – 2
= 12
2 * 6 = 12


3.( 8 / 88) * 8888088 = ?

Sol : (1/11) * 8888088 = 808008


4.How many 1/8's are there in 371/2 ?

Sol: (371/2) /(1/8)= (75/2) /(1/8) = 300


5.Find the values of 1/2*3 +1/3*4 +1/4*5+ .................+1/9*10 ?

Sol: 1/2*3 +1/3*4+1/4*5+ ..................+1/9*10
= [½ -1/3] +[ 1/3 – ¼] + [¼- 1/5] +...............+[1/9-1/10]
= [ ½ – 1/10]
= 4/15 = 2/5


6.The value of 999 of 995/999* 999 is:

Sol: [1000- 4/1000]*999 = 999000-4
= 998996

7.Along a yard 225m long, 26 trees are planted at equal distance, one
tree being at each end of the yard. what is the distance between two
consecutive trees ?

Sol: 26 trees have 25 gaps between them.
Hence , required distance = 225/ 25 m= 9m


8.In a garden , there are 10 rows and 12 columns of mango trees. the
distance between the two trees is 2 m and a distance of one meter is
left from all sides of the boundary of the length of the garden is :

Sol: Each row contains 12 plants.
leaving 2 corner plants, 10 plants in between have 10 * 2 meters and
1 meter on each side is left.
length = (20 + 2) m = 22m


9.Eight people are planning to share equally the cost of a rental car,
if one person with draws from the arrangement and the others share
equally the entire cost of the car, then the share of each of the
remaining persons increased by?

Sol: Original share of one person = 1/8
new share of one person = 1/7
increase = 1/7 – 1/8 = 1/56
required fractions = (1/56)/(1/8) = 1/7


10.A piece of cloth cost Rs 35. if the length of the piece would
have been 4m longer and each meter cost Re 1 less , the cost
would have remained unchanged. how long is the piece?

Sol: Left the length of the piece be x m.
then, cost of 1m of piece = Rs [35 / x]
35/ x – 35 /x+4 = 1
x + 4 – x = x(x+ 4)/35
x2 + 4x – 140 = 0
x= 10


11.A man divides Rs 8600 among 5sons, 4 daughters and 2 nephews.
If each daughter receives four times as much as each nephew, and
each son receives five as much as each nephew. how much does each
daughter receive ?

Sol:
Let the share of each nephew be Rs x.
then, share of each daughter Rs 4x.
share of each son = 5x Rs
so, 5 *5x+ 4 * 4x + 2x =8600
2x + 16x + 25x= 8600
43x = 8600
x = 200
share of each daughter = 4 * 200 = Rs 800


12.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+1/9]
= 1- 33/40
=7/40
Let the monthly salary be rs x
then, 7/40 of x = 1400
x= [1400*40]/7
x= 8000
Expenditure on food = 3/10*8000 =Rs 2400
Expenditure on conveyance= 1/8*8000 =Rs 1000

Decimal Fractions problems

Decimal Fractions problems

1.Decimal fractions: Fractionin which denominations are powers
of 10 are decimal fractions.

Example:1 /10 = 0.1, 1 / 100 = 0.01

2.Convertion of Decimal into fraction:-

Example: 0.25 = 25/100 = 1/4

3.i) If numerator and denominator contain same number of decimal
places, then we remove decimal sign. Thus, 1.84/2.99 =184/299


Problems:

1.0.75 =75/100 =3/4


2.Find porducts= 6.3204*100
= 632.04


3.2.61*1.3=261*13=3393 some of decimal places 2 +1 =3

sol: 3.393

4.If 1/3.718 =0.2689,then find value of 1/0.0003718 ?

Sol: 10000/3.718 =10000*1/3.718
=10000*0.2689
= 2689

5.Find fractions :

i) 0.37 = 37/99
ii)3.142857 =3+0.142857
=3 +142857/999999
= 3 142857/ 999999
iii) 0.17=17-1/90 =16/90=8/45
iv)0.1254 =1254 -12/9900 =1242/9900=69/550

6.Fraction 101 27/100000

Sol: 101+27/100000
=101+0.00027
=101.00027

7.If 47.2506 =4A + 7/B +2C + 5/D + 6E then 40+7+0.2+0.05+0.0006

Sol: compairing terms
4A= 40 => A=10
7/B = 7 => B=1
2C= 0.2=> C=0.1
5/D= 0.05=>D=5/0.05 =>5*100/5 =100
6E= 0.0006=> E= 0.0001
5A + 3B+6C+ D+ 3E = 5*10+ 3*1+ 6*0.1 + 100+ 3*0.0001
=50+3+0.6+100+0.0003
=153.6003

8.4.036 divided by 0.04

Sol: 4.036/0.04 =4036/4 =100.9

9.[ 0.05/0.25 + 0.25/ 0.05]3

Sol: =>[5/25 + 25/5]
= [1/5+ 5]3
=26/53
=5.23
= 140.603

10.The least among the following :-
a. 0.2 b.1/0.2 c. 0.2 d. 0.22

sol:10/2 =5 0.2222 0.04 0.04 < 0.2 < 0.22 --------<5
Since 0.04 is least (0.2)2 is least.

11.Let F= 0.84181

Sol: when F is written as a fraction in lowest terms, denominator
exceeds numerator by
84181 -841 /99000 = 83340/99000 =463/550
Required distence = (550 – 463) = 87

12.2 .75 + 3.78

Sol: [-2+0.75]+[-3+0.78]
=-5+[0.75+0.78]
= -5+1.53
=-5+1+0.53
= -4+0.53
= 4.53

13.the sum of first 20 terms of series is 1/5*6 +1/6*7+1/7*8-----

Sol: [1/5 -1/6]+[1/6-1/7]+[1/7-1/8]+------------------------
= [1/5-1/25]
=4/25=0.16

14.13 +23+ ------------+93 =2025

Sol: value of (0.11) 3+ (0.22) 3+---------(0.99)3 =>
(0.11) [1+2+--------+9]
=0.001331*2025
=2.695275

15.(0.96)3 – (0.1)3/ (0.96)2 +0.096 +(0.1)2

Sol: formula => a3 -b3/a2 +ab +b2 =a -b
(0.96-0.1)=0.86

16.3.6*0.48*2.50 / 0.12*0.09*0.5

Sol: 36*48*250/12*9*5=800

17.find x/y = 0.04/1.5
= 4/150 =2/75
find y-x/y+x
(1- x/y) / (1+ x/y)
1 - 2/75 /1 +2/75 =73/77

18.0.3467+0.1333

Sol: 3467 -34/9900 + 1333-13/9900
= 3433 +1320/9900
= 4753/9900
= 4801 -48/9900 =0.4301

Problems on numbers

Problems on numbers


1.Simplify
a.8888+888+88+8
b.11992-7823-456

Solution: a.8888
888
88
8
9872
b.11992-7823-456=11992-(7823+456)
=11992-8279=3713

2.What could be the maximum value of Q in the following equation?
5PQ+3R7+2Q8=1114

Solution: 5 P Q
3 R 7
2 Q 8
11 1 4
2+P+Q+R=11
Maximum value of Q =11-2=9 (P=0,R=0)

3.Simplify: a.5793405*9999 b.839478*625

Solution:
a. 5793405*9999=5793405*(10000-1)
57934050000-5793405=57928256595
b. 839478*625=839478*54=8394780000/16=524673750.

4.Evaluate 313*313+287*287

Solution:
a²+b²=1/2((a+b)²+(a-b)²)
1/2(313+287)² +(313-287)²=1/2(600 ² +26 ² )
½(360000+676)=180338

5.Which of the following is a prime number?
a.241 b.337 c.391

Solution:
a.241
16>√241.Hence take the value of Z=16.
Prime numbers less than 16 are 2,3,5,7,11 and 13.
241 is not divisible by any of these. Hence we can
conclude that 241 is a prime number.
b. 337
19>√337.Hence take the value of Z=19.
Prime numbers less than 16 are 2,3,5,7,11,13 and 17.
337 is not divisible by any of these. Hence we can conclude
that 337 is a prime number.
c. 391
20>√391.Hence take the value of Z=20.
Prime numbers less than 16 are 2,3,5,7,11,13,17 and 19.
391 is divisible by 17. Hence we can conclude
that 391 is not a prime number.

6.Find the unit's digit n the product 2467 153 * 34172?

Solution: Unit's digit in the given product=Unit's digit in 7 153 * 172
Now 7 4 gives unit digit 1
7 152 gives unit digit 1
7 153 gives 1*7=7.Also 172 gives 1
Hence unit's digit in the product =7*1=7.

7.Find the total number of prime factors in 411 *7 5 *112 ?

Solution: 411 7 5 112= (2*2) 11 *7 5 *112
= 222 *7 5 *112
Total number of prime factors=22+5+2=29

8.Which of the following numbers s divisible by 3?
a.541326
b.5967013

Solution: a. Sum of digits in 541326=5+4+1+3+2+6=21 divisible by 3.
b. Sum of digits in 5967013=5+9+6+7+0+1+3=31 not divisible by 3.

9.What least value must be assigned to * so that th number 197*5462 is
divisible by 9?

Solution: Let the missing digit be x
Sum of digits = (1+9+7+x+5+4+6+2)=34+x
For 34+x to be divisible by 9 , x must be replaced by 2
The digit in place of x must be 2.

10.What least number must be added to 3000 to obtain a number exactly
divisible by 19?

Solution:On dividing 3000 by 19 we get 17 as remainder
Therefore number to be added = 19-17=2.

11.Find the smallest number of 6 digits which is exactly divisible by 111?

Solution:Smallest number of 6 digits is 100000
On dividing 10000 by 111 we get 100 as remainder
Number to be added =111-100=11.
Hence,required number =10011.

12.On dividing 15968 by a certain number the quotient is 89 and the remainder
is 37.Find the divisor?

Solution:Divisor = (Dividend-Remainder)/Quotient
=(15968-37) / 89
=179.

13.A number when divided by 342 gives a remainder 47.When the same number
is divided by 19 what would be the remainder?

Solution:Number=342 K + 47 = 19 * 18 K + 19 * 2 + 9=19 ( 18K + 2) + 9.
The given number when divided by 19 gives 18 K + 2 as quotient and 9 as
remainder.


14.A number being successively divided by 3,5,8 leaves remainders 1,4,7
respectively. Find the respective remainders if the order of
divisors are reversed?

Solution:Let the number be x.

3 x 5 y - 1 8 z - 4 1 - 7 z=8*1+7=15
y=5z+4 = 5*15+4 = 79
x=3y+1 = 3*79+1=238
Now
8 238
5 29 - 6
3 5 - 4
1 - 2
Respective remainders are 6,4,2.

15.Find the remainder when 231 is divided by 5?

Solution:210 =1024.unit digit of 210 * 210 * 210 is 4 as
4*4*4 gives unit digit 4
unit digit of 231 is 8.
Now 8 when divided by 5 gives 3 as remainder.
231 when divided by 5 gives 3 as remainder.

16.How many numbers between 11 and 90 are divisible by 7?

Solution:The required numbers are 14,21,28,...........,84
This is an A.P with a=14,d=7.
Let it contain n terms
then T =84=a+(n-1)d
=14+(n-1)7
=7+7n
7n=77 =>n=11.

17.Find the sum of all odd numbers up to 100?

Solution:The given numbers are 1,3,5.........99.
This is an A.P with a=1,d=2.
Let it contain n terms 1+(n-1)2=99
=>n=50
Then required sum =n/2(first term +last term)
=50/2(1+99)=2500.

18.How many terms are there in 2,4,6,8..........,1024?

Solution:Clearly 2,4,6........1024 form a G.P with a=2,r=2
Let the number of terms be n
then 2*2 n-1=1024
2n-1 =512=29
n-1=9
n=10.

19.2+22+23+24+25..........+28=?

Solution:Given series is a G.P with a=2,r=2 and n=8.
Sum Sn=a(1-r n)/1-r=Sn=2(1-28)/1-2.
=2*255=510.

20.A positive number which when added to 1000 gives a sum ,
which is greater than when it is multiplied by 1000.The positive integer is?
a.1 b.3 c.5 d.7

Solution:1000+N>1000N
clearly N=1.

21.The sum of all possible two digit numbers formed from three
different one digit natural numbers when divided by the sum of the
original three numbers is equal to?
a.18 b.22 c.36 d. none

Solution:Let the one digit numbers x,y,z
Sum of all possible two digit numbers=
=(10x+y)+(10x+z)+(10y+x)+(10y+z)+(10z+x)+(10z+y)
= 22(x+y+z)
Therefore sum of all possible two digit numbers when divided by sum of
one digit numbers gives 22.

22.The sum of three prime numbers is 100.If one of them exceeds another by
36 then one of the numbers is?
a.7 b.29 c.41 d67.

Solution:x+(x+36)+y=100
2x+y=64
Therefore y must be even prime which is 2
2x+2=64=>x=31.
Third prime number =x+36=31+36=67.

23.A number when divided by the sum of 555 and 445 gives two times
their difference as quotient and 30 as remainder .The number is?
a.1220 b.1250 c.22030 d.220030.

Solution:Number=(555+445)*(555-445)*2+30
=(555+445)*2*110+30
=220000+30=220030.

24.The difference between two numbers s 1365.When the larger number is
divided by the smaller one the quotient is 6 and the remainder is 15.
The smaller number is?
a.240 b.270 c.295 d.360

Solution:Let the smaller number be x, then larger number =1365+x
Therefore 1365+x=6x+15
5x=1350 => x=270
Required number is 270.

25.In doing a division of a question with zero remainder,a candidate
took 12 as divisor instead of 21.The quotient obtained by him was 35.
The correct quotient is?
a.0 b.12 c.13 d.20

Solution:Dividend=12*35=420.
Now dividend =420 and divisor =21.
Therefore correct quotient =420/21=20.

Introduction:

Natural Numbers:

All positive integers are natural numbers.
Ex 1,2,3,4,8,......

There are infinite natural numbers and number 1 is the least natural number.
Based on divisibility there would be two types of natural numbers. They are

Prime and composite.

Prime Numbers:

A natural number larger than unity is a prime number if it
does not have other divisors except for itself and unity.
Note:-Unity i e,1 is not a prime number.

Properties Of Prime Numbers:

->The lowest prime number is 2.
->2 is also the only even prime number.
->The lowest odd prime number is 3.
->The remainder when a prime number p>=5 s divided by 6 is 1 or 5.However,
if a number on being divided by 6 gives a remainder 1 or 5 need not be
prime.
->The remainder of division of the square of a prime number p>=5 divide by
24 is 1.
->For prime numbers p>3, p²-1 is divided by 24.
->If a and b are any 2 odd primes then a²-b² is composite. Also a²+b²
is composite.
->The remainder of the division of the square of a prime number p>=5
divided by 12 is 1.

Process to Check A Number s Prime or not:

Take the square root of the number.
Round of the square root to the next highest integer call this number as Z.
Check for divisibility of the number N by all prime numbers below Z. If
there is no numbers below the value of Z which divides N then the number
will be prime.

Example 239 is prime or not?
√239 lies between 15 or 16.Hence take the value of Z=16.
Prime numbers less than 16 are 2,3,5,7,11 and 13.
239 is not divisible by any of these. Hence we can conclude that 239
is a prime number.


Composite Numbers:

The numbers which are not prime are known as composite numbers.

Co-Primes:

Two numbers a an b are said to be co-primes,if their H.C.F is 1.
Example (2,3),(4,5),(7,9),(8,11).....
Place value or Local value of a digit in a Number:

place value:

Example 689745132
Place value of 2 is (2*1)=2
Place value of 3 is (3*10)=30 and so on.
Face value:-It is the value of the digit itself at whatever
place it may be.

Example 689745132
Face value of 2 is 2.
Face value of 3 is 3 and so on.