## Results

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### #1. 1.14 expressed as a per cent of 1.9 is:

Solution: Required Percentage=(1.14×100)/1.9=60%

### #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?

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

Solution:
As we know,
1%=1/100
Hence,1/2%=(1/2×1/100)=1/200=0.005

### #4. If the price of the commodity is increased by 50% by what fraction must its consumption be reduced so as to keep the same expenditure on its consumption?

Solution:

Let the initial price of the commodity be 100.

After 50% increase in price, It will become,
100 ——50% increase—-> 150.
Now, we have to reduce the consumption to keep expenditure 100.

Increase in price= 150 – 100 = 50
We have to reduce the consumption,

=50/150×100=13 or 33.33%

Other Method:
Here, we use, Final product constant graphic.
100 ==50% up== 150===33.33% down===>100
Consumption Reduce = 33.33% = 1/3

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

Solution:

Here we can use the compound interest based formula,

Population after n years=P×[1 (r/100)]^n
Population after 2 years=50000×[1 (4/100)]^2
Population after n years=P×[1 (r/100)]^n
Population after 2 years=50000×[1 (4/100)]^2

Population after 2 years = 54080

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.

### #6. 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

Solution:

As A and B are fixed, C is any point on AB, so if AC is increases then CB decreases.
A________3 cm_________ _____2 cm____B
Then, solution can be visualized as,
Increase in AC 6% =

106×3/100=3.18cm.

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%

### #7. 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:

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

### #8. 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

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%

### #9. Each side of a rectangular field diminished by 40%. By how much per cent is the area of the field diminished?

Solution:

Let the Original length of the rectangle be 20 unit and breadth be 10 unit. Then
Original Area = length *breadth = 20*10 = 200 Square unit.
40% decrease in each side, then
Length = (20 – 40% of 20) = 12 unit.
Breadth = (10 – 40% of 10) = 6 unit.
Now, Area = 12 × 6 = 72 Square unit.
Decrease in area = 200 – 72 = 128 square unit.
% Decrease in Area =

128/200×100=64%

Mind Calculation Method:
Let the original area be 100 square unit.
100 === 40%↓(decrease in length) ==⇒ 60 === 40%↓ === (decrease in breadth) ===> 36
Diminished in area = 100 – 36 = 64%