# Aptitude Percentage Test Paper 6

26) If the length of a rectangle is increased by 37.5% and its breadth is decreased by 20%, find the change in its area.

1. 15% increase
2. 13% decrease
3. 10% increase
4. 10% decrease

Explanation:
Area of rectangle = length * breadth
Length is Increased by 37.5% = 37(1/2)* 1/100 = 3/8
Breath is decreased by 20% = 1/5
l * b = Area
Old:       8 * 5 = 40
New:    11 * 4 = 44
Difference = 44-40 = 4
Original = 40
The percentage change is equal to (difference / original) * 100.
(4/40) *100 = 10% increase.

27) The price of a product is reduced by 25%, but the daily sale of the article is increased by 30%. Find the net effect on the daily sale.

1. 2.5% increase
2. 1.5% decrease
3. 2% increase
4. 2.5% decrease

Explanation:
We can say that the price of one unit * no. of units sold = Total Sale
Let the initial price of product = 100, sale =100 items.
Then, Total Sale = 100 * 100 = 10000
ATQ, price reduces to 25%.
So, the new price = 100-25 = 75
Sale increase to 30%,
So, the new sale = 100+30= 130
Now, the new total sale = 75* 130 = 9750
Difference = 10000 - 9750 = 250
% net effect = (diff/original)* 100
= (250/10,000) * 100 = 2.5
So, the net effect on daily sale = 2.5% decrease

28) A mixture of 20 kg of milk and water contains 90% milk. How many kilograms of water should be added so that water becomes 40%?

1. 10 kg
2. 20 kg
3. 12 kg
4. 8 kg

Explanation:
Milk : water
Original Percentage: (90% : 10%)
Or                                       9 : 1
Total milk = 9 + 1 = 10 units of milk
We have 10 units = 20 kg
So, 1 unit = 20/10 = 2 kg.
New Percentage:60% : 40%
Or,                           (3 : 2)
On multiplying by 3 to make the ratio of milk equal, as the quality of milk remains the same, the ratio of water becomes 6.
(3 : 2) * 3
Or,   9 : 6 (ratio of water)
Now, the difference between the old and new ratio of water = 6 - 1 = 5 units.
1 unit = 2 kg
So, the required kg of water = 5 * 2 = 10 kg

29) The population of a town increases by 16(2/3) % in the first year, decreases by 37.5% in the second year, and increases by 57(1/7) % in the third year. The population of the town becomes 55000 after these 3 years. Find the initial population of the town 3 years ago.

1. 52000
2. 50000
3. 48000
4. 56250

Explanation:
Let the initial population = P
16(2/3) % = 16(2/3) * 1/100 = 1/6
Similarly, 37.5 % = 3/8
57(1/7) % = 4/7
1st-year          2nd-year         3rd-year
(+1/6)            (-3/8)               (+4/7)
Concept: 50% = The above concept shows that when old value is 2 units then a 50% increment means an increase of one unit, i.e. the old value becomes 3 after a 50% increment.
I.e. old value = 2 new value after 50% increment = 2+1 + 3
Now, let the initial population P.

As per question:

After an increment of 1/6, the new value will be 7/6.
And, after a decrease of -3/8, the new value will be 5/8.
And, after an increase of 4/7, the new value will be 11/7.
Now, as per question = [[[P * (7/6)] * (5/8)] * (11/7)] = 55000
So, the initial population, P = 48000

30) The salary of a man per hour is increased by 20%, and his working hours is decreased by 25%. If the original income is Rs. 4000. Find the new income.

1. 2000
2. 2500
3. 3000
4. 3600

Explanation:
Income = per hour salary * duration
Increase in per hour salary = 20% = 1/5
Working hours are reduced by 25% = - ¼
Per hour salary * duration = Income
Use Concept: 50% The above concept shows that when old value is 2 units then a 50% increment means an increase of one unit, i.e. the old value becomes 3 after a 50% increment.
I.e. old value = 2 new value after 50% increment = 2+1 + 3
AS per question:
An increase of 20 % in per hour salary = 6/5
An decrease of 25% = 3/4
Now, as per above two values:
Old income: 5 (per hour salary) * 4 (working hour) = 20 (but the original income is given 4000)
Similarly, new income: 6 * 3 = 18
Now, we multiply 20 with 200 to make it = 4000
Then, 18 is also multiplied with 200 to find the required new income. 18 * 200 = 3600

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