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Percentage
1.1.14 expressed as a per cent of 1.9 is:

Explanation:

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

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?

Explanation:

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%. [The percentage of students passed in English and Maths individually, have already included the percentage of students passed in both subjects. So, We are subtracting percentage of students who have passed in both subjects to find out percentage of students at least passed in one subject.]

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

3.Half percent, written as a decimal, is

Explanation:

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

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

Explanation:

Solution:

Alternatively,
we can use, net percentage change graphic as well,
50,000------4%↑---→ 52,000---- 4%↑---→ 54,080.
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

Explanation:

Solution:
As A and B are fixed, C is any point on AB, so if AC is increases then CB decreases.
A________3 cm_________ C _____2 cm____B
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%.

Alternatively,

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:

Explanation:

Solution:
Initial Cost = Rs. 75
After 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.

Mind Calculation Method: 75-----20%↑--→90-----20%↓-----→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

Explanation:

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%.

Mind Calculation Method;

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?

Explanation:

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.

Mind Calculation Method:

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?

Explanation:

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%

Mind Calculation Method:

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?

Explanation:

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%

Mind Calculation Method:

100==2% Up(1st year)==>102==2%Up(2nd year)==>104.04
% population increase in 2 years = 4.04%.

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