Percentage

1.1.14 expressed as a per cent of 1.9 is:

**Answer:**Option** 1**

Solution:

$$\begin{array}{rl}& \text{Required}\phantom{\rule{thinmathspace}{0ex}}\text{Percentage}\\ & =\frac{\left(1.14\times 100\right)}{1.9}\\ & =60\mathrm{\%}\end{array}$$

2.In an examination 80% candidates passed in English and 85% candidates passed in Mathematics. If 73% candidates passed in both these subjects, then what per cent of candidates failed in both the subjects?

**Answer:**Option** 1**

Solution:

Students passed in English = 80%Students passed in Math's = 85%

Students passed in both subjects = 73%

Then, number of students passed in at least one subject = (80+85)-73 = 92%.

Thus, students failed in both subjects = 100-92 = 8%.

3.Half percent, written as a decimal, is

**Answer:**Option** 1**

Solution:

$$\begin{array}{rl}& \text{As}\phantom{\rule{thinmathspace}{0ex}}\text{we}\phantom{\rule{thinmathspace}{0ex}}\text{know},\phantom{\rule{thinmathspace}{0ex}}1\mathrm{\%}=\frac{1}{100}\\ & \text{Hence},\phantom{\rule{thinmathspace}{0ex}}\\ & \frac{1}{2}\mathrm{\%}=\left(\frac{1}{2}\times \frac{1}{100}\right)\\ & \phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}=\frac{1}{200}\\ & \phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}=0.005\end{array}$$

4.The population of a town increases every year by 4%. If its present population is 50,000, then after 2 years it will be

**Answer:**Option** 1**

Solution:

$$\begin{array}{rl}& \text{Here we can use the compound interest based formula,}\\ & \text{Population after}n\text{years}\\ & =P\times {\left[1+\left(\frac{r}{100}\right)\right]}^{n}\\ & \text{Population after 2 years}\\ & =50000\times {\left[1+\left(\frac{4}{100}\right)\right]}^{2}\\ & \text{Population after 2 years}\\ & =54080\end{array}$$

we can use, net percentage change graphic as well,

Then, population after 2 years= 54,080.

In this calculation, we need to find 1% of 50,000 first, which is easily calculated by dividing 50,000 by 100.

5.A and B are two fixed points 5 cm apart and C is a point on AB such that AC is 3cm. if the length of AC is increased by 6%, the length of CB is decreased by

**Answer:**Option** 1**

Solution:

As A and B are fixed, C is any point on AB, so if AC is increases then CB decreases.Then, solution can be visualized as,

Increase in AC 6% = (106*3)/100=3.18 cm.

Decrease in CB = 0.18 cm

% decrease = (0.18/2)*100 = 9%.

AC = 3 Cm.

BC = 2 Cm.

Increase in AC by 6%, then

New, AC = 3 + 6% of 3 = 3 + 0.18 = 3.18 cm.

0.18 cm increase in AC means 0.18 cm decrease in BC as already mentioned AB as the fixed point.

So, % decrease in BC,

= (Actual Decrease in BC /Original BC)*100

= (0.18/2) *100 = 9%.

6.The cost of an article was Rs.75. The cost was first increased by 20% and later on it was reduced by 20%. The present cost of the article is:

**Answer:**Option** 1**

Solution:

Initial Cost = Rs. 75After 20% increase in the cost, it becomes,

(75+ 20% of 75) = Rs. 90

Now, Cost is decreased by 20%, So cost will become,

(90 - 20% of 90) = Rs. 72.

So, present cost is Rs. 72.

7.The price of the sugar rise by 25%. If a family wants to keep their expenses on sugar the same as earlier, the family will have to decrease its consumption of sugar by

**Answer:**Option** 1**

Solution:

Let the initial expenses on Sugar was Rs. 100.Now, Price of Sugar rises 25%. So, to buy same amount of Sugar, they need to expense,

= (100 + 25% of 100) = Rs. 125.

But, They want to keep expenses on Sugar, so they have to cut Rs. 25 in the expenses to keep it to Rs. 100.

Now, % decrease in Consumption,

(25/125)*100 = 20%.

100-----25%↑---→125------X%↓---→100.

Here, X = (25/125)*100 = 20%.

8.If the price of a commodity is decreased by 20% and its consumption is increased by 20%, what will be the increase or decrease in expenditure on the commodity?

**Answer:**Option** 1**

Solution:

Let the initial expenditure on the commodity be Rs. 100.Now, the price decreases by 20%,

Current Price = (100 - 20% of 100) = Rs. 80.

Same time due to decrement in price 20% consumption has been increased. So,

Current expenses on commodity = (80 + 20% of 80)= Rs. 96.

Here, the initial expenditure was Rs. 100 which became 96 at the end, it means there is 4% decrement in the expenditure of the commodity.

100===20%↓(Decrement in Price)===>80===20%↑(Increment in Consumption)===>96.

Thus, there is a decrement of 4%.

9.If A's salary is 25% more than B's salary, then B's salary is how much lower than A's salary?

**Answer:**Option** 1**

Solution:

Let B's Salary is Rs. 100. Then,A's Salary = (100 + 25% of 100) = Rs. 125.

Difference between A's Salary and B's Salary = 125 - 100 = Rs. 25.

% Difference (lower) = (25/125)*100 = 20%

100(B salary)----25%↑---→125(A salary)-----20%↓---→100 (B salary).

B's salary is 20% lower than A's.

10.Population of a town increase 2.5% annually but is decreased by 0.5 % every year due to migration. What will be the percentage increase in 2 years?

**Answer:**Option** 1**

Solution:

Net percentage increase in Population = (2.5-0.5) = 2% each year.Let the Original Population of the town be 100.

Population of Town after 1 year = (100 + 2% of 100) = 102.

Population of the town after 2nd year = (102 + 2% of 102 ) = 104.4 Now, % increase in population = (4.04 /100)*100 = 4.04%

100==2% Up(1

% population increase in 2 years = 4.04%.