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Per-thousand

June 13, 2020
Category: Per-thousand

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☼ Per-thousand :

l = Per-thousand = $\frac{1}{1000}$

i.e. x l = $\frac{x}{1000}$ .

Two Important Results (Short Cut Method) :

Let, X and Y are closely associated with each other , then

$\large \mapsto$ If X is r % more than Y, then

Y is less than that of X, by : $\left [\frac{r}{\left (100+r \right )} \times 100 \right ]$ %.

$\large \mapsto$ If X is r % less than Y, then :

Y is more than that of X, by : $\left [\frac{r}{\left (100-r \right )} \times 100 \right ]$ %.

e.g. (i) If A’s height is 25 % more than that of B, how much percent is B’s height less than that of A?

Soln. Required Percentage = $\left [\frac{25}{\left (100-25 \right )} \times 100 \right ]$ % = 20 %

(ii) If A’s monthly income is 60 % less than that of B, how much percent is B, how much percent is B’s income more than that of A?

Soln. Required Percentage = $\left [\frac{60}{\left (100-60 \right )} \times 100 \right ]$ % = 150 %

(iii) If the price of oil is increased by 50 %, find by how much percent a householder must reduce her consumption of oil so as not to increase the expenditure.

Soln. Required Percentage = $\left [\frac{50}{\left (100-50 \right )} \times 100 \right ]$ % = 33 $\frac{1}{3}$ %

Results on Proportion : Let the percent population of a town be P and let there be an increase of R % per annum. Then,

(i) Population after n years = P $\left (1+\frac{R}{100} \right ) ^{n}$

(ii) Population n years ago = P $\left (1+\frac{R}{100} \right ) ^{-n}$ = . $\frac{1}{\left (1+\frac{R}{100} \right ) ^{n}}$

Where, P = Principal ( or Sum ) = Initial Population

A = Amount ( = P + I ) = Population after N yrs

R = Rate of Interest = Rate of growth in population

I = Interest = Increment/decrement in population

N = No. of years

To understand above formula, remember PARIN sir.

e.g. The population of a town is 193600. If it increases annually at 10 %, what will be its population 2 years hence ? What was it 2 years ago ?

Soln.

Population after 2 years

= 193600 $\times$ $\left (1+\frac{10}{100} \right )^{2}$

= 193600 $\times \frac{110}{100}\times\frac{110}{100}$

= 234256

Population 2 years ago =

= $\frac{193600}{\left (1+\frac{10}{100} \right )^{2}}$

= 193600 $\times \frac{100}{110}\times\frac{100}{110}$

= 160000

Results On Depreciation : Let the present value of a machine be P and let it depreciate at R % per annum. Then :

(i) Value of machine after n years = P $\left (1-\frac{R}{100} \right ) ^{n}$

(ii) Value of machine n years ago = P $\left (1-\frac{R}{100} \right ) ^{-n}$ = $\frac{1}{\left (1-\frac{R}{100} \right ) ^{n}}$ .

e.g. The value of a machine depreciates at the rate of 20 % per annum. If its present value is Rs.14400, what will be its worth after 2 years ? What was the value of the machine 2 years ago ?

Soln.

Value of the machine after 2 years

= 14400 $\times \left (1-\frac{20}{100} \right ) ^{2}$