Hello Friends,
I am going to introduce you all to a method which can be used to solve various questions from different topics of Quantitative aptitude. I call this method as REVERSE RATIO method.
It can be applied to questions where the entities involved are inversely proportional to each other i.e. increase in one value results in decrease of other value and vice-versa.
Example: Men–work, speed–time, price–consumption etc are all inversely proportional to each other.
The method is quiet simple and straight forward once you get hang of it.
1. First we write the values of one of the given quantity in form of a ratio (say, a: b).
2. Then the ratio is reversed (b: a) to denote the other quantity which is inversely proportional to first one.
3. Now this ratio is multiplied by a constant so as to meet the given condition.
Thus, arriving at the solution.
We will discuss some questions to clear the concept. I will take some easy questions first. Please note that you may find these questions easy to approach by other methods as well but I will solve it using REVERSE RATIO just to make you familiar to the concept so that we can use them later to solve difficult questions.
Ques 2. Two cars Starts for a journey from A to B at the speed of 45km/h and 60km/h respectively. Second take 5hr less time to complete the journey. Find the distance.
Ques 5. If the price of an article is increased by 20%, by how much percentage should the consumption be decreased so as not to increase the expenditure.
Ques 4. Rajesh spends Rs.300 to buy sugar each month. Due to a reduction in 20% on the price of sugar he is able to buy 2kgs more. Find the new and original price of sugar.
Ques 5. When the price of rice was increased by 25%, a family reduced his consumption such that expenditure on rice was only 20% more than before. If previous consumption was 50kg, find new consumption.
Written by
Bipul Baibhav
It can be applied to questions where the entities involved are inversely proportional to each other i.e. increase in one value results in decrease of other value and vice-versa.
Example: Men–work, speed–time, price–consumption etc are all inversely proportional to each other.
The method is quiet simple and straight forward once you get hang of it.
1. First we write the values of one of the given quantity in form of a ratio (say, a: b).
2. Then the ratio is reversed (b: a) to denote the other quantity which is inversely proportional to first one.
3. Now this ratio is multiplied by a constant so as to meet the given condition.
Thus, arriving at the solution.
We will discuss some questions to clear the concept. I will take some easy questions first. Please note that you may find these questions easy to approach by other methods as well but I will solve it using REVERSE RATIO just to make you familiar to the concept so that we can use them later to solve difficult questions.
Ques 1. Few men decided to do a work in 40 days. 6 men were absent so the work was completed in 60 days. Find total men in beginning.
Sol. Here we have to deal with days and men which are inversely proportional to each other. If more men work then no. of days decrease.
Ratio of days – 40:60 or 4:6 (Previous: now)
Therefore, Ratio of men originally and now when 6 of them were absent will be – 6:4 since it is inversely proportional to days.
Here the difference of no. of men is 2 (6-4). But as per question, we require the difference of men working originally and now to be 6. So to meet this difference, we will multiply the ratio with 6(required difference) and divide by 2(the existing difference) to get the no. of men.
No. of Men = (6:4) x3 = 18:12 (Previous: now)
Answer – 18 i.e. number of men working previously or originally.
Ques 2. Two cars Starts for a journey from A to B at the speed of 45km/h and 60km/h respectively. Second take 5hr less time to complete the journey. Find the distance.
Sol. Speed Ratio – 45:60 or 3:4
Time Ratio – 4:3
Difference in time is given 5hr.
Time Taken = (4:3) x5 = 20:15 hrs
Distance = s x t = 45 x 20 OR 60 x 30 = 900km
Ques 3. Walking at 10km/h, a man is late for office by 15min. If he increases his speed by 2km/h, he is still late by 5min. Find distance of office.
Sol. Speed Ratio – 10:12 or 5:6
Time Ratio – 6:5
Difference in time is given 10min.
Time Taken = (6:5) x10 = 60:50 min.
Distance = s x t = 10x60/60 OR 12x50/60 = 10km
Ques 4. If a man goes from home to office at 60km/h, he is late by 20min. At 80km/h, he reaches 10min earlier. Find Distance.
Sol. Speed – 60:80 = 3:4
Time – (4:3) x30 = 120:90 (Multiply by 30 because the time gap between reaching 20 min late and 10min early is 30min which is the required time difference)
Distance = 60x120/60 = 120km
Ques 5. If the price of an article is increased by 20%, by how much percentage should the consumption be decreased so as not to increase the expenditure.
Sol. Price – 100:120 = 5:6
Consumption – 6:5
This means that if the consumption was 6 units earlier then now it became 5 units.
Percentage decrease – (1/6)*100 = 16 2/3 %
Ques 4. Rajesh spends Rs.300 to buy sugar each month. Due to a reduction in 20% on the price of sugar he is able to buy 2kgs more. Find the new and original price of sugar.
Sol. Price – 100:80 = 5:4
Consumption – 4:5 or 8:10 (Difference is 2)
Original Price – 300/8 = Rs.37.50
New Price – 300/10 = Rs.30
Ques 5. When the price of rice was increased by 25%, a family reduced his consumption such that expenditure on rice was only 20% more than before. If previous consumption was 50kg, find new consumption.
Sol. Price – 100:125 (Previous to new)
But as the family is ready to pay 20% more than before, so effective change in price or expenditure is 120:125 and consumption will reduce accordingly.
Consumption – 125:120 = 25:24 (Previous to new)
Written by
Bipul Baibhav
