Solved Examples

1. LCM of 27,314and53is${\displaystyle \frac{2}{7},{\displaystyle \frac{3}{14}\mathrm{a}\mathrm{n}\mathrm{d}{\displaystyle \frac{5}{3}\mathrm{i}\mathrm{s}}}}$

a. 45

b. 35

c. 30

d. 25

Correct Option: C

Explanation:

LCM of numeratorsHCF of denominators=LCM of 2,3,5HCF of 7,14,3=301=30${\displaystyle \frac{\text{LCM of numerators}}{\text{HCF of denominators}}=\frac{\text{LCM of 2,3,5}}{\text{HCF of 7,14,3}}=\frac{30}{1}=30}$

2. About the number of pairs which have 16 as their HCF and 136 as their LCM, the conclusion can be

a. only one such pair exists

b. only two such pairs exist

c. many such pairs exist

d. no such pair exists

Correct Option: D

Explanation:

HCF is always a factor of LCM. ie., HCF always divides LCM perfectly.

3. The HCF of two numbers is 12 and their difference is also 12. The numbers are

a. 66, 78

b. 94, 106

c. 70, 82

d. 84, 96

Correct Option: D

Explanation:

The difference of required numbers must be 12 and every number must be divisible by 12. Therefore, they are 84, 96.

4. The HCF of two numbers is 16 and their LCM is 160.  If one of the numbers is 32, then the other number is 

a. 48

b. 80

c. 96

d. 112

Correct Option:b

Explanation:

The number = HCF×LCMGiven number=16×16032=80

5. HCF of three numbers is 12. If they are in the ratio 1:2:3, then the numbers are

a. 12,24,36

b. 10,20,30

c. 5,10,15

d. 4,8,12

Correct Option: A

Explanation:

Let the numbers be a, 2a and 3a.

Then, their HCF = a  so a=12

The numbers are 12,24,36

Logarithms - Solved Examples

Solved Example 1: (Important model)
How many digits are contained in the number 2100$2^{100}$
Sol: log2100$\log 2^{100}$ = 100 x log 2 = 100 x 0.3010 = 30.10
Number of digits in 2100$2^{100}$ are 30 + 1 = 31

To determine the characteristic of the logarithm of a decimal fraction: (Numbers between 0 to 1)
Look at this example:
Find the total zeroes after he decimal point of the expression 26$2^6$
We know that 26$2^6$ = 164${\displaystyle \frac{1}{64}}$ = 0.015625
log 164${\displaystyle \frac{1}{64}}$  = -1.806
Now when you calculate Antilog of -1.806 using calculator, you will get 0.0156.
But if you want to use antilog tables, you have to follow this procedure.
Now log 164${\displaystyle \frac{1}{64}}$ = log 126${\displaystyle \frac{1}{2^6}}$ = log 2−6$2^{-6}$ = -6 log 2.
We know that log 2 = 0.301
Now log 164${\displaystyle \frac{1}{64}}$ = -6 × 0.301 = -1.806.
Important: Now if you look at the antilog table for 0.80 and 6, you will get wrong answer.  Why? Because -1.806 = -1 + (-0.806)
But mantissa is always positive.
-1.806 should be written as -2 + (1 - 0.806) = -2 + 0.194
Now when you look at the anti log table 0.194 gives you 1563. 
So characteristic is 2.

That is the characteristic of the logarithm of a decimal fraction is one more than than the number of zeroes immediately after the decimal point  and is negative.

Working Rule to find the number of zeroes in a decimal number: 
1. Calculate the logarithm (you will get some negative number)
2. Subtract the decimal part from one and increase the integer part by 1 to make mantissa positive
3. Number of zeroes of that number = Integer part - 1

Solved Example 2: (Important model)
How many zeroes are there between the decimal point and the first significant digit in (12)1000${\left({\displaystyle \frac{1}{2}}\right)}^{1000}$
log (12)1000${\left({\displaystyle \frac{1}{2}}\right)}^{1000}$ = 1000 × log (1/2) = 1000 × -0.30102 = -301.02
But in logarithms the decimal point should be positive. (By using
- 301.02 = - 301 + -0.02 = -302 + (1 - 0.02) = 302___$\stackrel{\mathrm{\_}\mathrm{\_}\mathrm{\_}}{302}$.98
So number of zeroes are 302 - 1 = 301

Solved Example 3:
11+logabc+11+logbca+11+logcab=${\displaystyle \frac{1}{1+{\log }_{a}bc}+{\displaystyle \frac{1}{1+{\log }_{b}ca}+{\displaystyle \frac{1}{1+{\log }_{c}ab}=}}}$

a. 0	b. 3
c. 2	d. 1

Answer: d
Explanation:
11+logabc${\displaystyle \frac{1}{1+{\log }_{a}bc}}$ + 11+logbca${\displaystyle \frac{1}{1+{\log }_{b}ca}}$ + 11+logcab${\displaystyle \frac{1}{1+{\log }_{c}ab}}$= 
1logaa+logabc${\displaystyle \frac{1}{{\log }_{a}a+{\log }_{a}bc}}$+ 1logbb+logbca${\displaystyle \frac{1}{{\log }_{b}b+{\log }_{b}ca}}$ + 1logcc+logcab${\displaystyle \frac{1}{{\log }_{c}c+{\log }_{c}ab}}$
= 1logaabc${\displaystyle \frac{1}{{\log }_{a}abc}}$ + 1logbabc${\displaystyle \frac{1}{{\log }_{b}abc}}$ + 1logcabc${\displaystyle \frac{1}{{\log }_{c}abc}}$ = 
= logabca+logabcb+logabcc=logabcabc=1${\log }_{abc}a+{\log }_{abc}b+{\log }_{abc}c={\log }_{abc}abc=1$

Solved Example 4:
The value of  (yz)logy−logz×(zx)logy−logx×(xy)logx−logy$(yz{)}^{\log y-\log z}\times (zx{)}^{\log y-\log x}\times (xy{)}^{\log x-\log y}$

a. 2	b. 1
c. 0	d. 3

Answer: b
Explanation:
Assume K = (yz)logy−logz$(yz{)}^{\log y-\log z}$ × (zx)logy−logx$(zx{)}^{\log y-\log x}$ × (xy)logx−logy$(xy{)}^{\log x-\log y}$
Taking log on both sides
Log K = log ((yz)logy−logz$(yz{)}^{\log y-\log z}$ × (zx)logy−logx$(zx{)}^{\log y-\log x}$ × (xy)logx−logy$(xy{)}^{\log x-\log y}$)
= log(yz)logy−logz$\log (yz{)}^{\log y-\log z}$ + log(zx)logy−logx$\log (zx{)}^{\log y-\log x}$ + log(xy)logx−logy$\log (xy{)}^{\log x-\log y}$
= (logy−logz)log(yz)$(\log y-\log z)\log (yz)$ + (logz−logx)log(zx)$(\log z-\log x)\log (zx)$ + (logx−logy)log(xy)$(\log x-\log y)\log (xy)$
= (logy−logz)(logy+logz)$(\log y-\log z)(\log y+\log z)$ + (logz−logx)(logz+logx)$(\log z-\log x)(\log z+\log x)$ + (logx−logy)(logx+logy)$(\log x-\log y)(\log x+\log y)$ =0
log K = 0⇒$\Rightarrow$ K = 1

Solved Example 5:
log(x+y3)$\log \left({\displaystyle \frac{x+y}{3}}\right)$ = 12${\displaystyle \frac{1}{2}}$ (logx + logy)  then (xy+yx${\displaystyle \frac{x}{y}}+{\displaystyle \frac{y}{x}}$) is

a. 5	b. 7
c. 9	d. 0

Answer: b
Explanation: 
As the answer does not have any log, first we try to remove log from the given equation by simplifying it.
log(x+y3)$\left({\displaystyle {\displaystyle \frac{x+y}{3}}}\right)$ =  12${\displaystyle \frac{1}{2}}$(logx + logy)
⇒2log(x+y3)$\Rightarrow 2log\left({\displaystyle \frac{x+y}{3}}\right)$= log xy
⇒$\Rightarrow$ log(x+y3)2=logxy$\log {\left({\displaystyle \frac{x+y}{3}}\right)}^2=\log xy$
⇒$\Rightarrow$x2+y2+2xy=9xy$x^2+y^2+2xy=9xy$
⇒$\Rightarrow$x2+y2=7xy$x^2+y^2=7xy$
⇒$\Rightarrow$ xy+yx=7