## Quick & Easy Calculations to Save Money while Shopping

On my recent visit to a supermarket, I realized the struggle and decision making fatigue which shoppers go through while making their household purchases.

I decided to write this post to ease this struggle to the best of my capacity.

For ease of understanding I’ve made categories based on the type of decisions we need to make.

## Comparison Shopping

All the examples given below are from the items I picked up from the grocery section of the supermarket.

1st comparison

• 24 mangoes for Rs.699
• 72 mangoes for Rs.1899

To compare these options, first thing you do is round off the numbers.

• 24 mangoes for Rs.699 700
• 72 mangoes for Rs.1899 1900

After that, you need to compare the price by making the quantity same. Ideal way is to scale up the lower quantity.

In this case if you multiply smaller quantity, i.e. 24 by 3 you get 72.

Hence 24×3 mangoes in 1st case will be Rs.700×3 = Rs.2100

Where in 2nd case, 72 mangoes are available for Rs.1900

So you know, if you can consume 72 mangoes, it’s available at a bargain. However, we need to consider the perishable nature of mangoes and the quantity we can consume in a reasonable time till which mangoes can remain fresh.

2nd Comparison

• 85 gm of honey for Rs.149
• 200 gm of honey for Rs.299

Again, 1st thing is to round off the prices.

• 85 gm of honey for Rs.149 150
• 200 gm of honey for Rs.299 300

Next, easier thing to do in this case will be to scale it up and see.

If we multiply the smaller quantity by 2, we get

85×2 = 170 gm for Rs.150 x 2 = Rs.300

Comparing once again,

In 1st case, we get 170 gm for Rs.300

In 2nd case, we get 200 gm of Rs.300

Thus clearly 2nd case is more beneficial.

3rd comparison

• 5kg of detergent powder for Rs.153
• 4kg of detergent powder for Rs.418 plus a bucket free.

We need to scale up the smaller quantity to match the larger quantity. Then the price comparison will be apple to apple. That too of the same size.

In the above case you saw that just simple multiplication was good enough to do that. However, here you need to do some quick thinking.

You remember HCF taught in school? Don’t panic. No worries even if you don’t.

HCF or highest common factor means finding a number which divides both the numbers.

Let me explain this for the above example. If we have smaller packets of 500 gm each, for buying 1.5 kg you need 3 packets and for buying 4 kg you need 8 packets.

So now, if 1.5kg is at Rs.153, 1 packet of 0.5kg is for Rs.153/3 = Rs.51

Therefore, 8 packets (i.e. 4kgs) will be for Rs.51*8 = Rs.408.

So, if you ignore the free bucket available you’re better off buying the smaller packet i.e. of 1.5 kg compared to 4 kg pack.

However, getting a bucket by paying additional Rs.10 is not a bad deal. So you should go for 4 kg packet.

In an earlier article I’ve touched upon calculations of gross margin, cost and sale price. That article will also help you in calculations of discounts. Have a look.

I hope I helped you to overcome your dilemma during shopping when you are spoilt for choices and one offer seems to be better than the other.

Did you all like my tips? If yes, write in the comments below. Stay tuned for another set of situations on Discount Shopping.

## Mentally Solve Problem of Averages in Seconds

In all competitive examinations and otherwise also in our day to day calculations we encounter issues related to averages. Especially when the composition of the group changes it becomes difficult to do computations.  Let us discuss quick and easy solutions to such problems.

### When a person leaves a group and another person replaces him, then there can be 2 scenarios:

Scenario I: When the average age increases then,

Age of the new comer = age of person who left + (no. of persons in the group * increase in average age)

Scenario II: When the average decreases then,

Age of the new comer = age of the person who left – (no. of persons in the group * decrease in average age)

Let us try to understand these with a simple example –

Question: The average age of 45 persons is decreased by 1/9 year when one of them whose age is 60 years is replaced by a new comer. What is the age of the new comer.

Solution:

Age of the new comer = age of the person who left – (no. of persons in the group * decrease in average age)

Age of the new comer = 60 – (45 * 1/9) = 55 years

Isn’t that easy to calculate using the formula I just gave you. However, to intuitively answer such questions you need to keep following things in mind

Firstly, the average age is reducing, which means an older man is replaced by relatively younger person. Hence the answer has to be lesser than 60. Hence we subtract from 60 and not add to it.

Secondly, average reduction of 1/9 kg for a group of 45 people means, in total there is a wait reduction of 1/9 per person x 45 persons = 5 kg for the entire group. This is result of the replacement of an old person by a relatively younger person in the group. That’s the reason we’re subtracting 5 from 60, to get the age of the new comer.

### B. When a person joins a group without any replacement, then there can be 2 scenarios:

Scenario I: When the average age increases then,

Age of the incoming person = previous average age of the group + no. of persons including the person who joined * increase in the average age value

Scenario II: When the average age decreases then,

Age of the incoming person = previous average age of the group – no. of present persons including the person who joined * decrease in the average age value

I will explain this further with an example,

Question: The average age of 20 teachers is 45 years which is decreased by 6/7 years when a student joins the group. Then what is the age of the student?

Solution:

Age of the outgoing person = previous average age of the group – new no. of persons including the person who joined * decrease in the average age value

Therefore, age of the student = 45 – 21 * 6/7 = 27 years

Here also you need to observe that since the average age is going down when the new person is joining, the age of new comer will be lesser than the average, hence subtraction.

Moreover, the decrease is equal to 6/7 per person for each person in the new group. That is 21 times 6/7 = 18 kgs, which needs to be subtracted from the overall average to get the age of the new comer.

### C. When a person leaves a group without any replacement, then there can be 2 scenarios:

Scenario I: When the average age increases then,

Age of the outgoing person = previous average age of the group – no. of persons excluding the person who left * increase in the average age value

Scenario II: When the average age decreases then,

Age of the outgoing person = previous average age of the group + no. of present persons excluding the person who left * decrease in the average age value

Hope we’ve covered all scenarios and with the help of the above logical formulas we will be able to do our calculations.

### Practice questions for you to solve (try to solve mentally)

Question 1: In a boat there are 8 men whose average weight is increased by 1 kg when a man of 60 kg is replaced by a new man. What is the weight of the new comer?

Question 2: In a class there are 30 boys whose average weight is decreased by 200 grams when one boy whose weight was 25 kgs leaves the class and a new comer is admitted. Find the weight of the new comer?

Question 3: A cricketer has a certain average for 9 innings. In the 10th innings, he scores 100 runs, thereby increasing his average by 8 runs. His new average is?

## 5 Quick Shortcuts to Solve Time and Distance Problems

Time, Speed and Distance (TSD) is one of the most frequently occurring topics in quantitative aptitude section of many competitive exams. Here I’m putting together 5 short tricks which might come handy while answering various type of TSD questions. Let us start with some absolutely basic concepts –

### Shortcut #1: Basic Concepts of Time, Speed and Distance

• Speed =Distance/Time
• Time =Distance/Speed
• Distance = Speed x Time
• To convert km/hr to m/sec
• x km/hr =>x  * (5/18) m/sec
• To convert m/sec to km/hr
• y m/sec = y * (18/5) km/hr
• If the ratios of the speeds of A and B is a :b, then the ratio of the times taken to cover the same distance is (1/a) :(1/b) or b:a

### Shortcut #2: Finding out the Average Speed when Equal Distances are covered at Different Speeds

Lot of us make mistakes in calculation of average speed when the same distance is covered at different speeds.  We simply take the average of the given speeds. However, that gives absolutely wrong answer. So now get ready to find out what will give you the correct solution.

Theorem: If a certain distance is covered at x km/hr and the same distance is covered at y km/hr, then

Average speed =[2xy/(x+y)] km/hr

This is basically harmonic mean of the two speeds, i.e. 2/(1/x+1/y)

Example – If a car travels at 40 km/hr on a trip and at 60 km/hr on return trip. What is its average speed for the entire trip?

First thing we should be careful is we shouldn’t just average the 2 speeds. Overall average speed is not (S1+S2)/2.  From the above direct formula the answer will be

Solution: (2x40x60)/(40+60) = 48 km/hr

If we’ve to find the average of more than 2 speeds, average speed will be the harmonic mean of all such speeds

N / (1/a + 1/b + 1/c + 1/d)

Here N = 4, i.e. the number of variables (speeds in this case)

### Shortcut #3: Finding out the Distance when Equal Distances Covered at Different Speeds and Total Journey Time is given

Theorem: A person goes to a destination at a speed of S1 km/hr and returns to his place at a speed of S2 km/hr, if he takes T hours, then

One way distance = Total time taken x (Product of two speeds) / (Addition of two speeds)

= T * {S1*S2/(S1+S2)}

Example: A boy goes to school at a speed of 3 km/hr and returns to village at 2 km/hr. If he takes 5 hours, what is the distance between the school and the village?

Solution: Distance between school and village is 5 * (3*2) / (3+2)=6 km

### Shortcut #4.1: Finding out the Distance when Equal Distances Covered at Different Speeds

Theorem: A person goes to a destination at a speed of S1 km/hr and returns to his place at a speed of S2 km/hr, if he takes T1 and T2  hours respectively, then

One way distance = Total time taken x (Product of two speeds) / (Addition of two speeds)

= (T1 – T2) * {S1*S2/(S1-S2)}

### Shortcut #4.2: Shortcut for “Early and Late to Office” Type Problems

The same shortcut used above can be used in these type of problems. Here you go –

TheoremA person covers a certain distance having an average speed of x km/hr, he is late by x1 hours but with a speed of y km/hr, he reaches his destination yhours earlier, hence

Required distance = Product of two speeds x Difference between arrival times/Difference of two speeds

Example: A man covers a certain distance between his house and office on bike. Having an average speed of 30 km/hr, he is late by 10 minutes. However, with a speed of 40 km/hr, he reaches his office 5 minutes earlier. Find the distance between his house and office?

In the above case, the required distance = (30×40)x0.25/(40-30) =  30 km

Please note: 10+5 = 15 minutes = 15/60 hours = 0.25 hours

Now since you know few time, speed and distance shortcuts, let’s try this time and speed puzzle

### Shortcut #5: Finding Speed or Time Required after Crossing Each Other

Theorem: If two persons or trains A and B start their journey at the same time from two points P and Q towards each other and after crossing each other they take a and b hours in reaching Q and P respectively, then

Using this relationship you can find out the missing variables which can be either speed or time. Once these are known you can easily find the distance.

Example: Two, trains, one from Howrah to Patna and the other from Patna to Howrah, start simultaneously. After they meet, the trains reach their destinations after 9 hours and 16 hours respectively. The ratio of their speeds is:

Solution: using the above relationship, the ratio of their speed is √16/√9 = 4/3 or 4:3

Do you know more shortcuts for time, speed and distance problems?

## How to Solve Successive Percentage Problems?

This time I thought of giving you the crux of the whole article at the very beginning. See the formulae above and you will know what I am talking about. Read More

## Rule of 72 – Estimation of Compound Interest and Time

Effect of Compounding

The Rule of 72 is a good quick math shortcut to find out the following –

• Time required for an amount to double itself, at a given rate of interest
• Rate at which an amount should grow to double itself in given time

This formula can be applied for “Doubling Problems” related to money, population, etc. which grows at an annual compounded rate.

Formulae

1. To calculate the time; T = 72/R
2. To calculate the rate of interest; R= 72/T Read More

## EMI Calculation

In our daily life we face enormous application of mathematics. Calculation of equated monthly installments (EMI) for car or home loan is one such common application of mathematics.

EMI or equated monthly installments is the most popular form of loan payment.  It is a fixed amount of repayment made every month towards the loan, which includes payment towards both principal and interest. Most of us always believe the bank executives blindly on the figure which they quote as EMI.

## Sale Price, Gross Margin determination – Simple Trick

I am going to share with you one trick which one great teacher, Mr S.K. Jha taught me. Simple arithmetical problems of

Finding out selling price, when cost price and margin (percentage of profit on Sale) are given. Or,

Finding out cost price when Sale price is given and percentage of profit on cost is given.

Finding out the mark-up when cost price and Sale price are given.

We all have seen occurrence of such problems in our academic life and day-to-day life. All the time we need a paper – pen or a calculator. But now with the process given below I believe we can solve any such problem in our mind. Read More

## Time and Speed Puzzle

Raja’s office is 9 km from his home. One day, he decided to jog to work in order to increase his fitness. He estimates the journey time to be a little over 1 hour 45 minutes.

At first he feels fine, but before long his energy beings to fade and he realizes he will not be able to run the whole distance. A passing taxi is the answer to his prayer. Raja hails the cab and completes his journey, arriving at work one hour 10 minutes earlier than he had anticipated.

Assuming Raja could at 1/6 the speed of the cab, how far did he run before he hailed the taxi?

## How to Find the Average Speed?

Lot of students gets confused while finding out the average speed, when various distances are travelled with different speed. Say for example, trip to Agra from Delhi is made at an average speed of 40 km/hr and the trip back at an average speed of 60 km/hr. Find their average speed for the entire trip. (Hint: It’s Not 50 Km/hr)

Rule: If a person travels a distance at a speed of x km/hr and the same distance at a speed of y km/hr, then the average speed during the whole journey is given by 2xy/x + y km/hr.

Or, If a person travels half the distance at a speed of x km/hr and the other half at a speed of y km/hr, then the average speed during the whole journey is given by 2xy/x + y km/hr.

So answer to the above question:

(2*60*40)/(60+40)  = 48 Km/hr

If a person travels three equal distances at a speed of x km/hr, y km/hr and z km/hr respectively, then the average speed during the whole journey is 3xyz/xy + yz + xz km/hr.

This problem can also be dealt with assuming some hypothetical distance.

You can find thousands of such amazing faster calculation tricks in the MAGICAL BOOK on QUICKER MATHS – by M. Tyra

If you’ve a question related to time and speed or any other topic in maths, you can post it on QuickerMaths.com’s question answer platform