CONCEPT 3
CHANGES IN THE NUMERATOR:
The changes in the numerator and the denominator will change (increase or decrease)the overall value of the ratio.
1. The changes in the numerator is directly proportional to the value of the ratio.
=> If the value of the numerator increases, then the value of the ratio will increase.
=> If the value of the numerator decreases, then the value of the ratio will decrease.
2. In fact, the percentage change in the value of the numerator is equal to the percentage change in the value of the ratio. An example is, if a value of an item increases from 20/10 to 30/10, the percentage change in the value of the numerator is (30-20/20 = 50%. Also, the percentage change in the value of the ratio is (3-2)/2 =50%
Example 1:
If there is threefold increase in all the sides of a cyclic quadrilateral, then the percentage increase in its areawill be? (IIFT 2009)
Solution:
A square is a cyclic quadrilateral. So, we can find the area as x^2 when the side is x
If the sides increase three folds, it means 3x then the area will be 9x^2
So, the percentage increase in area will be (9x^2 -x^2)/x^2 = 8x^2/x^2 = 800%
CONCEPT 4:
CHANGES IN THE DENOMINATOR:
1. The changes in the denominator is inversely proportional to the value of the ratio.
=> If the value of the denominator increases, the value of the ratio will decrease.
=> If the value of the denominator decreases, the value of the ratio will increase.
2. For example, consider the value of an item decreased from 20/10 to 20/12 (that is, denominator has increased by 20%). The value of the ratio decreased by 16.67%. It means that, a 20% increase in the denominator has resulted in 16.67% decrease in the value of the ratio.
3. A real-life example could be that if the price of a commodity increases by 20%, then in order for the expenditure( which is the product of the price and consumption) to remain constant, the percentage reduction in the consumption should be 16.67%
NOTE:
1. If the numerator increases by a%, then the value of the ratio will increase by a% and vice versa. But this is not the case in concern with the changes in denominator. Instead for the changes in denominator, the change in the value of the ratio is calculated as below.
2. If the price of the commodity increases by a%, then the percentage reduction in consumption, so that the expenditure remains same is (a/a+100)*100.
3. If the price of the commodity decreases by b%, then the percentage increase in consumption, so that the expenditure remains same is (b/b+100)*100.
Example 1:
Rice is now being sold at Rs.27 per kg. During last month it was sold at Rs.24 per kg. Find how much percent that a family should reduce its consumption to keep the expenditure fixed. (SNAP 2009)
Solution:
Assume the family consumes 1kg.
The reduction in consumption is 24/27 => 8/9
So, the percentage of reduction reduced is [1-(8/9)] /1* 100 = 11.11%
Example 2:
The price of a commodity decreases by 20%. By what percentage should the quantity increase so as to keep revenue constant?
Solution:
The formula for percentage increase is b/100-b => 20/100-20 => 25%
Example 3:
Sowmiya generally spent Rs.800 for buying a month's provision of potatoes. However, lower yield this year caused the cost of one kilogram of potatoes to be increased by 60%. Owing to this, Sowmiya had to buy 30 kg less potatoes than usual. What was the cost of potatoes this year?
Solution:
If the cost of one kilogram of potatoes increases, the decrease in the consumption so as to keep the expenditure constant should be (60/(100+60))*100 = 37.5%
Decrease in the consumption is what which is given (the decrease of 30 kg potatoes in whole consumption than usual) So, 37.5% decrease = 30 kg
Actual consumption after price increase= (30/37.5)*100 = 80kg
So, the usual consumption is 80 kg. This year's consumption = 80-30 = 50 kg
=> Cost of potatoes this year = 800/50 = Rs.16 per kg.
CONCEPT 5:
CHANGES IN BOTH NUMERATOR AND THE DENOMINATOR:
1. If the numerator increases and the denominator decreases, then the value of the ratio will surely increase.
2. If the numerator decreases and the denominator increases, then the value of the ratio will surely decrease.
3. However, if the numerator and the denominator simultaneously increase/decrease then it is not quote so apparent how the value of the ratio will change.
4. We already know how the individual changes in the numerator and the denominator affect the ratio. Now, we ll try to combine these concepts.
5. Consider that if the numerator increases by a% and the denominator increases by b%. Since the numerator increases by a%, the value of the ratio will also increase by a%. And since the denominator increase by b%, the value of the ratio will decrease by b/100+b% Thus by applying successive changes, we can find out the net percentage increase.
Example 1:
A student took a certain entrance exam and scored 180 out of 250 marks in the first attempt. However, during the second attempt, the total marks were increased to 330, out of which the student scored 250. Find the percentage change in the percentage marks of the student in the first two attempts.
Solution:
The numerator increases by (250-180)*180 = 38.89% (a%)
The denominator increases by (330-250)*250 =32%
So, b% is calculated as 32/(100+32) = 24.24% (b%) (value of ratio will decrease by 24.24%)
Net percentage change= a%+(-b%)+ (a*-b)/100 = 5.22% (approximately)
Example 2:
The numerator of the ratio M/N is increased by 20% and its denominator is increased by 25%. Then the numerator of the new ratio formed is increased 10% to get the fraction P/Q. What is net percentage change between the ratios M/N and P/Q?
Solution:
An increase of 25% in denominator results in 25/100+25 = 20% decrease in the value of the ratio.
So, consider the initial value of the ratio to be 100
100 => increase if 20% as per increase in num => 120 => decrease of 20% as per increase in den => 96
The net percent change will be |96-100| = 4% decrease
The numerator of the new ratio is increased by 10%
100=> decrease of overall 4% => 96 => increase of 10% as per increase in num => 105.6
Net percentage change between the two ratios = 105.6 - 100 = 5.6% increase
CONCEPT 6:
PERCENTAGE CHANGE IN A QUANTITY (A*B) WHEN BOTH A AND B CHANGE:
The difference between this method and the above mentioned is, earlier it is stated how to find the percentage change in one of the variables (say b), when the other changes (say a) and the product remains constant (a*b) 1. Now, we see how to calculate the percentage change in (a*b) when both a and b change.
2. An illustration : Let the original value of an item be A=a*b. This changes to B= x*y in the next year. So, to find the percentage change between A and B, we will find the percentage changes between a and x (p%) and b and y (q%). So,
p= (x-a)/a % and q = (y-b)/b % (Here p and q will be negative if there is a percentage decrease)
3. An example: Let A = 10*12 and B= 9* 16.=> p=-10% and q=33.33%
So, the net percentage change = 120-100 = 20%
Example 1:
Due to erosion of the soil from some parts of his field, a farmer considered increasing the length of his rectangular field by 25% and reducing the breadth by 12%. What will be the percentage change in the area of his plot?
Solution:
Let the area be 100. => 25% increase => 125 => 12% decrease => 110
The percentage change will be 110-100 = 10%
CONCEPT 7:
SUCCESSIVE PERCENTAGE CHANGES:
1. Two successive increases on a particular value of a% and b% would be equal to a net increase of
[a+b+(ab/100)] %
In case of decline in growth or a discount, the value of a,b or both is negative.
2. Three successive increases on a particular value of a%, b% and c% would be equal to a net increase of
([(100+a)/100]*[(100+b)/100]*[(100+c)/100] -1) %
Example 1:
Ravi's income has increased by 10% over the last year and will be 20% higher next year. If last year his salary was Rs.15000, what will it be next year?
Solution:
Successive increases of 20% and 10% = 20+10+ (20*10)/100 = 32%
Ravi's salary next year = 15000*1.32 = Rs.19800
Example 2:
At the end of year 1998, Shepherd bought nine dozen goats. Henceforth, every year he added p% of the goats at the beginning of the year and sold q% at the end of the year where p>0 and q>0. If Shepherd had nine goats at the end of the year 2002, after making the sales for that year, which of the following true?
1. p=q 2.p<q 3. p>q 4.p=q/2 (CAT 2003)
Solution:
Shepherd bought 9 dozen goats at the end of year 1998. Since p% and q% are same every year, and that at the end of year they should negate which means the number of goats added and sold are same every year,
So, consider he bought 1 dozen of goats to it, Percentage added = 1/9*100 = 11.11% (p)
He sold 1 dozen of goats to get back the same, Percentage sold = 1/10* 100 = 10%(q)
So, always P>Q
Example 3:
A student took five papers in an examination where the full marks were the same for each paper. His marks in these papers were in proportion of 6:7:8:9:10 In all papers together, the candidate obtained 60% of marks. Then the number of papers in which he got more than 50% marks are?
Solution:
Average scored in all the papers = (6+7+8+9+10)/5 = 40/5 = 8
Because he scored a total of 60% of marks, 50% of marks will be (8/60)*50 = 6.67
So, number of subjects he scored more than 6.67 is 4
Example 4:
A college raised 75% of the amount needed for a new building by receiving an average donation of Rs.600 from the people already solicited. The people already solicited represents 60% of the people the college will ask for the donations. If the college is to raise exactly the amount needed for the new building, what should be the average donation from the remaining people solicited? (CAT 2001)
Solution:
Let the population be p. Then the amount already received = 600* 0.6p
Remaining amount = 25% => (3/4) of the total amount = 360p
(1/4) of the total amount = 25 percent = 120p
Remaining people = 0.4p So, Amount per head (Average donation) = 120p/0/4p = Rs.300
Example 5:
Fresh grapes contain 90% of water by weight while dry grapes contain 20% of water by weight.
What will be the weight of dry grapes available from 20 kg of fresh grapes? (CAT 2001)
Solution:
Total weight of fresh grapes = 20 kg
As water has 90% by weight in this total weight, the weight of grape mass will be 10%
So, the actual weight of grape mass in total weight of fresh grapes = (10/100)*20 = 2 kg
(Note- In this question it is evident that both fresh and dry grapes has same grape mass, only the water content percentage is different)
Therefore, the weight of dry grapes available from 20 kg of fresh grapes is calculated by taking into account the percentage of grape mass that can be contained (which is 100 -20% of weight of water) => 80%
For 80% of grape mass, the actual weight can be 2 kg.
So, for 100% of grape mass, what can be the weight that can be made available from 20 kg of fresh grapes?
=> (2/80)*100 = 2.5 kg
Example 6:
The present value of an optical instrument is Rs.20000. If its value will depreciate 5% in the first year, 4% in second year and 2% in the third year, what will be its value after three years? (FMS 2009)
Solution:
Value after 3 years = 0.95*0.96*0.98*20000 = Rs. 17875.2
Example 7:
BSNL offers its share at a premium of Rs.40, whereas its par value is Rs.160. Parul Mehra invested Rs.50000 in this stock. After one year, BSNL declared a dividend of 19%. What rate of interest did Ms.Mehra receive on her investment? (FMS 2009)
Solution:
Premium is the excess of money that the company receives over its par value
So, cost of buying one share = 160+40 = Rs.200
No. of shares she bought = 50000/200 = 250 shares
Dividend = 0.19*250*160 = Rs. 7600 (Dividend is always calculated on par value)
Rate of interest of dividend on investment = (7600/50000)*100 = 15.2 %
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