Basic & Commercial Arithmetic - Byju'sbyjus.com/.../ias_files/basic_and_commercial_arithmetic.pdfThen (c + d) = 1.2 (a + b) Also b = 30, d = 40, c = 80 80 + 40 = 1.2 (a + 30) => 120

CSAT-2012 Arithmetic

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Basic & Commercial Arithmetic

CONTENTS 1)Percentages 1.1) Concepts

1.2) Percentage Equivalence Table 1.3) Introduction to Constant Product Rule

2) Simple Interest & Compound Interest 2.1) Simple Interest 2.2) Compound Interest

2.3) Applications of Compound Interest 2.4) Introduction to Pascal’s Triangle

3) Profit & Loss 3.1) Concepts & Terminologies

Miscellaneous Arithmetic

CONTENTS 4) RATIO, PROPORTION, VARIATION & PARTNERSHIP

4.1) Introduction 4.2) Properties Of Ratios 4.3) Comparison Of Ratios 4.4) Standard Results on Ratios 4.5) Variation 4.6) Partnership

5) WEIGHTED AVERAGE, MIXTURES AND ALLIGATIONS

5.1) Classification Of Averages 5.2) Applications Of Averages 5.3) Weighted Average 5.4) Shortcut Techniques in Averages 5.5) Mixtures & Alligation



1) PERCENTAGES

1.1:- CONCEPTS “Per cent” means for every hundred. If you get 25 marks in an exam where the full marks is 50, your percentage marks will be what you will get in a total

of hundred marks. (25

50× 100)

Using unitary method: In 50 you get 25; So in 1 you get 25

50

And in 100 you get 25

50 × 100 = 50 percent

Thus, what you get out of hundred becomes the percentage. For a fraction, if the denominator is hundred, it is called a percentage and the numerator of the fraction is called rate per cent 1.1.1) PERCENTAGE FOR COMPARISON: Percentage helps us to compare between different fractions when the denominator or the total number is different in each case. It is one of the simplest tools for the comparison of data. Take for example, this table below which shows the marks obtained by a student in 3 different subjects From this data alone, we cannot compare the marks obtained for the various subjects.. Now suppose we have the data of the total marks obtained as follows

Subject Marks obtained Total Marks Marks obtained/ total marks x 100

History 60 100 60%

Math 25 25 100%

English 45 90 50%

Now as all the three subjects are represented on a scale of 100, it is easy to compare the marks for the three subjects, and decide which subject has the student scored the maximum in. 1.1.2) REPRESENTATIONS OF PERCENTAGES 1) a% of b.

a% of b ( a percent of b) = b% of a a% of b is represented mathematically as 𝑎×𝑏

100.

Eg. 24% of 25 = 24𝑥25

100 = 6 25% of 24 =

25×24

100 = 6

2) What percentage of a is b

This is represented as 𝑏

𝑎× 100

Eg. What percentage of 75 is 25? 25

75 × 100 = 33.33%

1.1.3) CONVERSIONS 1) To convert a fraction into a percentage

Subject Marks obtained

History 60

Math 25

English 45



Multiply and divide by 100. Keep the denominator as 100, the numerator you obtain is the required answer. Take for example

21

22=

21

22𝑥100

100=

2100

22

100 = 95

6

11%

2) To find the fraction equivalent of a percentage

Divide by 100(after removing the percentage sign) e.g. ) 11 3

8% =

91

8𝑥100=

91

800

eg) 25% =25

100=

1

4

ILLUSTRATION: 1. Recently I went to buy a laptop for myself. The dealer said that he has laptops of two companies: HP and Lenevo.

He was selling the HP laptop for Rs. 42000 and told me that he will offer me the same at 7

8 of that price while the

Lenevo laptop was for Rs. 46000 and he was offering it at 4

5

𝑡𝑕 of that price. I decided to buy the laptop on which I was

getting a better percentage discount. Which one should I buy? Solution:

HP: 7

8 means

7×100

8 = = 87.5% means a discount of 100 – 87.5 = 12.5%

LENEVO: 4

5 means

4𝑥100

5 =

400

5= 80 % means a discount of 100 – 80 = 20%. So I should buy the LENEVO laptop.

1.1.4) CONCEPT OF CHANGE There are two types of change 1. Absolute value change:- It is the actual change in the quantity. For example, if there are 10 rabbits in the first

year and 15 rabbits in the second year, the absolute change in the number of rabbits is 5 2. Percentage change:- This can be obtained by calculating the absolute change and dividing it by the initial

number of rabbits present

e.g. ) percentage change = 𝐴𝑏𝑠𝑜𝑙𝑢𝑡𝑒 𝑣𝑎𝑙𝑢𝑒 𝑐𝑕𝑎𝑛𝑔𝑒

𝑂𝑟𝑖𝑔𝑖𝑛𝑎𝑙 𝑞𝑢𝑎𝑛𝑡𝑖𝑡𝑦× 100=

5

10=

1

2× 100= 50%

1.1.5) PERCENTAGE CHANGE AND PERCENTAGE POINT CHANGE If the pass percentage of a class was 75 % in 1991 and 85% in 1992, we can calculate the percentage point change and the percentage change as follows Percentage point change = Final percentage – Initial percentage= 85 %– 75% = 10 percentage points

Percentage change= 𝐹𝑖𝑛𝑎𝑙 𝑝𝑒𝑟𝑐𝑒𝑛𝑡𝑎𝑔𝑒 – 𝐼𝑛𝑖𝑡𝑖𝑎𝑙 𝑝𝑒𝑟𝑐𝑒𝑛𝑡𝑎𝑔𝑒

𝐼𝑛𝑖𝑡𝑖𝑎𝑙 𝑝𝑒𝑟𝑐𝑒𝑛𝑡𝑎𝑔𝑒× 100 =

85 – 75

75𝑥 100 =

40

3%

1.1.6) TO INCREASE A NUMBER BY A GIVEN PERCENT

Use the formula: 100 + 𝑟𝑎𝑡𝑒

100

e.g.) Increase 40 by 20% Here the rate = 20 Multiply the number by the above formula to get the answer

40 𝑥 100 + 20

100=

120𝑥40

100 = 48. (We are effectively computing 40 +

20

100× 40

It is easier to remember that given the base value=x, the final value after increase can be found as 20% increase(x+0.2x= 1.2x) 30% increase(x+0.3x=1.3x) 5% increase(x+0.05x=1.05x)



1.1.7) TO DECREASE A NUMBER BY A GIVEN PERCENT

Use the formula 100 – 𝑅𝑎𝑡𝑒

100

e.g. ) Decrease 40 by 20% Here the rate = 20. Multiply the number by the above formula to get the answer

40 𝑥 100− 20

100=

80𝑥40

100= 32. (We are effectively computing 40 −

20

100× 40 )

It is easier to remember that given the base value=x, the final value after decrease can be found as 20% decrease 0.8 x (0.8= 1-0.2) 30% decrease 0.7x (0.7= 1-0.3) 5% decrease 0.95x (0.95= 1-0.05) TO FIND THE PERCENTAGE INCREASE OR DECREASE OF A GIVEN NUMBER, WE USE THE CONCEPT OF PERCENTAGE CHANGE AS EXPLAINED ABOVE

1) Percentage Increase = 𝐹𝑖𝑛𝑎𝑙 𝑉𝑎𝑙𝑢𝑒 – 𝐼𝑛𝑖𝑡𝑖𝑎𝑙 𝑉𝑎𝑙𝑢𝑒

𝐼𝑛𝑖𝑡𝑖𝑎𝑙 𝑉𝑎𝑙𝑢𝑒 =

𝑇𝑜𝑡𝑎𝑙 𝐼𝑛𝑐𝑟𝑒𝑎𝑠𝑒

𝐼𝑛𝑖𝑡𝑖𝑎𝑙 𝑉𝑎𝑙𝑢𝑒 𝑥 100%

2) Percentage Decrease = 𝐼𝑛𝑖𝑡𝑖𝑎𝑙 𝑉𝑎𝑙𝑢𝑒 – 𝐹𝑖𝑛𝑎𝑙 𝑉𝑎𝑙𝑢𝑒

𝐼𝑛𝑖𝑡𝑖𝑎𝑙 𝑉𝑎𝑙𝑢𝑒 =

𝑇𝑜𝑡𝑎𝑙 𝐷𝑒𝑐𝑟𝑒𝑎𝑠𝑒

𝐼𝑛𝑖𝑡𝑖𝑎𝑙 𝑉𝑎𝑙𝑢𝑒 𝑥 100%

ILLUSTRATION: 2. In the IPL match an analysis was done for the two openers of both teams: Delhi Daredevils (DD) and Deccan Chargers (DC). It was found that the two DC openers Gilchrist and Gibbs together scored 20% more than what the two DD openers Sehwag and Gambhir scored. Gambhir scored 30 and Gibbs scored 40. How much percentage less/more did Sehwag score than Gilchrist if Gilchrist scored 80? Solution: Say runs scored by Sehwag = a, Gambhir = b, Gilchrist = c and Gibbs = d. Then (c + d) = 1.2 (a + b) Also b = 30, d = 40, c = 80

80 + 40 = 1.2 (a + 30) => 120

1.2 = a + 30 => a = 70

So Sehwag = 70. So Sehwag scored 80 – 70

80 x 100 =

10

80𝑥 100 = 12.5 % less than Gilchrist.

Note: Here we took 80 as the denominator because we need to calculate how much Sehwag scored less “than” Gilchrist. So Gilchrist’s score has to be made the denominator as it is the base value we are comparing the difference to. 1.1.8) BASIC FORMULA a) If the price of a commodity increases by r%, then the reduction in consumption so as to not increase the

expenditure is 𝑟

100+𝑟𝑥 100 .This formula can also be used to compare incomes between two people.

e.g.) If P’s income is r% more than Q’s income, then Q’s income is less than P’s income by 𝑟

100+𝑟𝑥 100

b) If the price of a commodity decreases by r%, then the increase in consumption so as to not decrease the

expenditure is 𝑟

100−𝑟𝑥 100 . This formula can also be used to compare incomes between two people.

e.g.) If P’s income is r% less than Q’s income, then Q’s income is greater than P’s income by 𝑟

100−𝑟𝑥 100

c) If there is a consecutive percentage change of a% and b%, the net change can be calculate as



Net change= 𝑎 + 𝑏 + 𝑎𝑏

100

ILLUSTRATIONS: 3. Anil is taking an examination which has two sections A and B. He starts with section A and when he proceeds to section B he realizes that the marks carried for each question has decreased by 40% as compared to section A. How many more questions (in percentage) should he solve in section B as compared to section A if he wants to score same marks in section A and section B? Given that both sections carry equal marks. Solution: Here marks per question decreases by 40%, so using the formula (b), he should solve

40

100−40 × 100 % = 66.66% more questions in section B if he wants to score same marks.

4. The petrol prices shot up by 20% due to Iraq War. Amit travels 2000 kms every month and his car gives a mileage of 20 kms per liter. By how many kilometers should he reduce his travel to maintain his expenditure to the previous level. Solution:

At 20 kms per liter he bought 2000

20= 100 liters of petrol.

Now the price the increased by 20%; using formula (a), the consumption has to be reduced by (20

100+20) × 100) % =

16.66%.

So, he should reduce his travel by 16.66×2000

100= 333.33 kms to maintain the same expenditure level.

5. After receiving two successive pay hikes if Ashish’s salary becomes 𝟏𝟓

𝟖 of his initial salary. By how much percent

was the salary raised the first time if the second raise was twice as high (in %) as the first? a) 15 % b) 20% c) 25% d) 30% Solution: We can solve this question using two approaches: Approach 1: (conventional) Suppose first raise was of a % then the second raise = 2a %

Now, using formula (c) net change: 𝑎 + 𝑏 +𝑎𝑏

100𝑎 + 2𝑎 +

2𝑎𝑥𝑎

100

Now if initial salary was X, then % change = 15 𝑋

8– 𝑋

𝑋× 100 =

7

8× 100

So 3a + 2𝑎2

100 =

700

8 solving we get a = 25%. So the first raise was of 25%.

Approach 2: (Recommended) In these types of questions it is quicker to go from the answer options.

First calculate the net change (15 𝑋

8– 𝑋 ) × 100 =

700

8% = 87.5 %

Start with option (b) 20% First year raise 20% next year raise= 40% Net change (use formula (c))

20 + 20 + 400

100 = 44%

With option (d)

Net change = 30 + 60 + 1800

100 = 108%

This way, you will get option (c) 25% is the correct option.



1.2 PERCENTAGE EQUIVALENCE TABLE It’s good to know the reciprocals of numbers till 20.Once the reciprocals are known, the value of several fractions can be easily obtained from the table shown Rows

1 2 3 4 5 6 7 8

1 100

2 50 100

3 33.33 66.67 100

4 25 50 75 100

5 20 40 60 80 100

6 16 33 50 67 83 100

7 14.28 28 43 57 72 86 100

8 12.5 25 37.5 50 62.5 75 87 100

9 11 22 33 44 55 66 77

10 10 20 30 40 50 60 70

11 9.09 18 27 36 45 54 63

The table is read as follows

First row, first column 1

1𝑥 100 = 100% Second row, first column

1

2𝑥 100 = 50%

Fourth row, Third column 3

4𝑥 100 = 75%

It is advantageous to know the reciprocals of prime numbers (the reciprocals below are multiplied by 100)

1

13= 7.69

1

17= 5.88

1

19= 5.26

1

23= 4.35

1

29= 3.45

Application of the table

e.g 1) Suppose you need to do the approximate multiplication 42 x 142

The faster way to do it is to take 42 x 142 as 42 x 14.2 x 10 = 42 × 1

7 × 100 × 10 (as 14.28 =

1

7× 100)

Thus 42 × 142= 6000

e.g 2) Similarly for the multiplication 114 x 8.37 = 114 × 5

6 × 10 ( 83 =

5

6× 100) = 950 (Approximately)

Initially, it may take time to get used to the conversion, but after repeated practice, it saves a lot of time for calculation, and can be one of your best time saving strategies for the exam.

1.3 Constant Product Rule-Introduction This rule can be used when there is a relationship or inverse proportionality between two parameters, i.e. when the product of two parameters is a constant. This concept can be applied to many arithmetic topics based on percentages like Time and Work, Time Speed Distance, Interest and Profit & Loss to just name a few For example,

Time x Speed = Distance Efficiency x Time= Work Done Price x Consumption = Expenditure

In the third example, if the price of a commodity is increased by a particular percentage, then to keep the “Expenditure”, which is the product of Price and Consumption constant, we need to reduce Consumption by a particular percentage.

Columns



This reduction in percentage depends entirely on the percentage by which Price has been increased. If the increase

in price is 50% 1

2 , then the consumption has to be decreased by 33%

1

3 to keep the expenditure constant.

So the rule states that,

A 𝟏

𝒙 increase in one of the parameters will result in a

𝟏

𝒙+𝟏 decrease in the other parameter if the parameters are inversely

proportional. Note that two inversely proportional quantities have a constant product.

Also

A 𝟏

𝒙 decrease in one of the parameters will result in a

𝟏

𝒙−𝟏 increase in the other parameter if the parameters are inversely

proportional.

Using this concept, we can draw a table for the change in component Q reflected due to a change in component P, when their product is constant, i.e when P and Q are inversely proportional.

SOME APPLICATIONS OF CONSTANT PRODUCT RULE: 6) Price of sugar is going up by 20%. Find the percentage reduction in consumption a family should adopt so that the expenditure remains a constant? Solution: Here price x quantity = expenditure (constant)

20 % 1

5 increase in price will result in

1

6 i.e 16.66% decrease in consumption

7) Normally A does a certain work. But A is absent so Ravi has two options: B or C. B can finish the work by taking 6 hrs more than A while C is 1.5 times more efficient than B and can finish the work 4 hrs earlier. Find the usual time taken by A. Solution: Again we have two cases Case 1 B efficiency = x 6 hrs more than A Case 2 C efficiency = 1.5x 4 hrs less than A Now we know efficiency and time are two inversely proportional quantities. Taking case 1 as original, efficiency increases by ½ so the time taken will decrease by 1/3 from case1 to case 2. 6 hrs more than A and 4 hrs less than A means a difference of 10 hrs.

As we have already calculated, time taken will decrease by 1

3 and the difference is of 10 hrs this means the original

time taken in case 1 will be 30 hrs, i.e. B takes 30 hours. (1

3 of 30 hrs = 10 hrs)

So time taken by A = 30-6=24 hrs.

Increase in P Decrease in Q

100% (1) 50% 1

2

50% 1

2 33.33%

1

3

33.33% 1

3 25%

1

4

25% 1

4 20%

1

5

20% 1

5 16.66%

1

6



8) An architect is planning to increase the length of a rectangular room by 10% without increasing the area of the room. By what percentage must he reduce the breadth of the room ? Solution: Length x breadth = Area ( constant)

Since there is a 10% (1

10) increase in length the breadth will be reduced by

1

11 = 9.09%.

9)Efficiency of A is 33.33% more than that of B. B takes 24 days complete a work. How many days A will take to complete the work? Solution: Efficiency x time = work( constant)

Efficiency of A is 33.33% 1

3 more. So time taken by A will be 25 % less than that by B => A will take 18 days.

10)24 men can finish a work in 25 days. In how many days will 30 men finish the work? Solution: Here (number of men) x (number of days) = man days( constant) By increasing the number of men by 25 %, time saved will be 20 %. So 30 men will take 20 days to finish the work.

Till now we have discussed questions where the % change can be written as 1

𝑥.

What if the change is not of the form 1

𝑥?

For eg = 30 % = 30

100=

3

10

IF there is a 3

10 increase in one parameter, the other parameter will get reduced by

3

10+3=

3

13

Looking at this in a table, let 𝐴 𝛼1

𝐵 i.e AB = K, a constant

Increase in A Decrease in B Change in K

40 % =2

5

2

7

0

30 % =3

10

3

13

0

60% = 3

5

3

8

0

2) SIMPLE INTEREST AND COMPOUND INTEREST The person who lends money is called the “Creditor/Lendor” and the one who borrows money is called the “Debtor”. The amount of money that is initially borrowed is called the “Capital or Principal money”. The duration of investment or loan period is the “Time” Amount received at the end of the “Time Period”= Principal + Interest There are two types of Interest

2.1) Simple Interest (SI) Here, throughout the period of the loan, the interest is paid on the principal or the original sum

SI = 𝑃𝑥𝑁𝑥𝑅

100

Where SI Simple Interest, P Principal , N Number of years, R Rate of interest



11) What is the interest to be paid on a principal of Rs 14000 borrowed at a rate of 15% for a period of 3 years and 6 months?

SI = 𝑃𝑥𝑁𝑥𝑅

100 =>SI=

14000𝑥15𝑥3.5

100= Rs 7350

2.2) Compound Interest (CI) It is called “Interest on Interest”. At the end of a year or another fixed period, the interest that is due is not paid to the creditor, but it is instead added to the sum lent and the amount obtained becomes the principal for the next year or period. This process is repeated till the amount for the last year is found. The amount which accrues after obtaining Compound Interest is calculated as follows

Amount = 𝑃 1 +𝑅

100 𝑁

P Principal , N Number of years, R Rate of interest Compound Interest= Amount - Principal 12) What will be the amount for a sum of Rs 1000 at 10% for 3 years compounded annually?


100 𝑁

= 1000 1 +10

100

3

=1000 110

100

3 = 1331

Note:- 1) In case interest is paid half yearly, N will become the number of half years and R will be the rate percent per half year Number of years x 2= Number of half years 𝑅𝑎𝑡𝑒 𝑝𝑒𝑟𝑐𝑒𝑛𝑡 𝑝𝑒𝑟 𝑦𝑒𝑎𝑟

2= Rate percent per 6 months

Extrapolating from this, If the interest is paid quarterly (once in 3 months) Number of years x 4= Number of quarters 𝑅𝑎𝑡𝑒 𝑝𝑒𝑟𝑐𝑒𝑛𝑡 𝑝𝑒𝑟 𝑦𝑒𝑎𝑟

4 = Rate percent per 3 months or rate percent per quarter

When rates are different for different years (say R1 and R2 for two consecutive years), we can calculate the total

amount as 𝑃 1 +𝑅1

100 𝑥 1 +

𝑅2

100

2.3 APPLICATIONS OF COMPOUND INTEREST-

POPULATION APPRECIATION /DEPRECIATION OVER TIME

If the original population of a town is P, and the annual increase is R%, the population of the town in “N” years will

be 𝑃 1 +𝑅

100 𝑁

If the original population of a town is P, and the annual decrease is R%, the population of the town in “N” years will

be 𝑃 1 −𝑅

100 𝑁

13) If the annual increase in the population of a village is 4% and the population at present is 17576, then what was the population 3 years ago?



From 𝑃 1 +𝑅

100 𝑁

, we get

Present population =Population 3 years ago x 104

100

3

Population 3 years ago x 263

253= 17576

population 3 years ago =17576 x 253

263= 15625

14) Population of a village is 10000 in year 2000 and is 14400 in 2002. Find the Compounded Annual Growth Rate? Solution:

CAGR = 14400

10000

1

2− 1 𝑥 100 = 20 %

This means an annual increase of 20 %. To verify 10000 x 1.2 = 12000& 12000 x 1.2 = 14400 We are using a factor of multiplication 1.2 because to increase a number by 20% and to get the final value, we need

to multiply by 120

100 = 1.2

CAGR is similar to rate percent, R in compound Interest calculations.

Final amount = 𝑃 1 +𝑅

100 𝑛

where P = initial amount, R = rate percent, n = number of years.

So, CAGR or R = [ 𝑓𝑖𝑛𝑎𝑙 𝑎𝑚𝑜𝑢𝑛𝑡

𝑖𝑛𝑖𝑡𝑖𝑎𝑙 𝑎𝑚𝑜𝑢𝑛𝑡

1

𝑛– 1] x100. CAGR implies the annual increase in growth.

2.4) CALCULATION OF COMPOUND INTEREST USING PASCAL’S TRIANGLE Concept of Pascal’ triangle The Pascal’s triangle is a representation of numbers in the form of a triangle as follows

Each number in this triangle is obtained by adding the two numbers immediately above it. For eg) the 2 in the second line is obtained by adding the two 1s above it. Similarly the “6” in the fourth line is obtained by adding the two 3’s above it. This pattern of numbers has a unique application in the calculation of Compound Interest, and it simplifies the calculation tremendously. Let’s take the following example to calculate the Compound Interest using Pascal’s triangle 15) What is the Amount obtained on a principal amount of 8000 rupees with a rate of 5% for a period of 3 years compounded annually? Principal = 8000 Rate= 5%

For a period of 3 years, consider the 3rd line of the Pascal’s triangle

1 3 3 1 Amount is = 8000(1) + (5% of 8000=400)(3) + (5% of 400=20)(3) + (5% of 20 =1)(1)



Answer= 8000 + 1200 + 60 + 1 =9261 If we need to only find the interest, subtract the principal from the amount 9261- 8000 = 1261 ILLUSTRATIONS 16) At what rate of simple interest, will a sum of money double itself in 4 years? Solution: received by lender= double the amount given, this implies that interest = principal. If P is the principal, then P is the

simple interest P=𝑃𝑥4𝑥𝑅

100= R=

100

4 = 25%

17) Given that a certain sum of money amounts to Rs 108 is 2 years, Rs 112 in 3 years. Find the principal and

rate of Simple Interest. Solution: Amount after 2 years = 108. amount after 3 years = 112 In SI, interest amount remains same year after year This means that interest for each year = Rs 4 Principal = 108-2(4)= 100

Rate of Interest = 4𝑥100

100= 4%

18) What is the amount a person gets on investing Rs 80000 at 10% p.a., compounded semi-annually for 2

years? Solution: Here n=2x2= 4 time periods

R=10

2= 5% (half yearly)


100 𝑁

=80000 𝑥 105

100

4=Rs 97240.5

Alternative: Use Pascal’s triangle ( 4th line 1 4 6 4 1) 19) Out of Rs 80000, Anita invests Rs 26000 in shares, 30% of total in a plot and the remaining she deposited in 10% fixed deposit (calculated on compound interest) for 2 years. What is the gain from Fixed deposit after 2 years? Solution:

Amount in Fixed Deposit(FD)= 80000 − 26000 −30

100(80000) = 𝑅𝑠 30000

Amount after 2 years = 30000 110

100

2

=36300. Interest = 36300-30000= Rs 6300

3) PROFIT & LOSS

3.1) Terminologies involved 1) Cost Price (CP):- The amount paid to buy a product or the cost involved in manufacturing a product is called the Cost Price (CP) of that product. Types of cost a) Fixed Cost The kind of cost which is fixed in all cases b) Variable cost The cost which varies according to the number of units produced c) Semi-Variable cost Cost which is fixed for some entities and variable for some entities



2) Selling Price (SP):- The price at which an article is sold to the customer 3) Marked Price (MP):- The price which the shopkeeper fixes for the product ( the price on the label) in anticipation of some discount they may be asked by the customer for the product. 4) Discount:- the reduction made on the marked price is called the discount. When no discount is given, the Selling Price= Marked Price 5) Margin- When profit percentage is calculated as a percentage of SPM then it is known as margin Profit and Loss is always calculated on Cost Price.

%Profit = 𝑷𝒓𝒐𝒇𝒊𝒕

𝑪𝑷𝒙𝟏𝟎𝟎 =

𝑺𝑷−𝑪𝑷

𝑪𝑷𝒙 𝟏𝟎𝟎 =

𝑺𝑷

𝑪𝑷 – 𝟏 𝒙𝟏𝟎𝟎

% Loss= 𝑳𝒐𝒔𝒔

𝑪𝑷𝒙𝟏𝟎𝟎 =

𝑪𝑷−𝑺𝑷

𝑪𝑷𝒙𝟏𝟎𝟎 = 𝟏 −

𝑺𝑷

𝑪𝑷 𝒙𝟏𝟎𝟎

Discount is always calculated on Marked Price.

% Discount= 𝑫𝒊𝒔𝒄𝒐𝒖𝒏𝒕

𝑴𝑷𝒙𝟏𝟎𝟎

OTHER FORMULAE INVOLVED (a) Profit = SP-CP (SP>CP) (b) Loss= CP-SP (CP>SP) (c) SP= MP- discount

(d) 𝑀𝑃 =𝑆𝑃𝑥100

100 – 𝑑𝑖𝑠𝑐𝑜𝑢𝑛𝑡 %

ILLUSTRATIONS: 21) Monte Carlo is offering a summer discount of 20% on all the sweaters. Ram buys sweaters which cost him

Rs. 3000/- and then resold them in his shop at a discount of 10% on the market price. Find the total profit he earned.

Solution: When Ram bought sweaters from Monte Carlo he paid Rs. 3000.Thus, 3000 is Ram’s Cost Price

Using MP = 𝑆𝑃𝑥100

100 – 𝑑𝑖𝑠𝑐𝑜𝑢𝑛𝑡 %

MP = 3000 𝑥 100

100 – 20 = 3000 𝑥

100

80 = 3750. Rs 3750 is now, the new Marked Price

He sold all sweaters at a discount of 10%; discount = 10 𝑥3750

100 = Rs. 375.

So he sold the sweaters at Rs. 3750 – Rs. 375 = Rs. 3375. Profit = Rs. (3375 – 3000) = Rs. 375. 22) In the previous question, calculate the profit percentage earned by Ram. Solution: Profit earned = Rs. 375 Cost Price (CP) for Ram = Rs. 3000

So profit percentage = 375 ×100

3000 = 12.5%

23) The CP of 40 articles = SP of 50 articles. What is the profit or loss percentage? Solution: Since the Cost Price of 40 articles= Selling Price of 50 articles, we realize that the transaction is a loss for the shopkeeper.



CP will be greater than SP 𝐶𝑃

𝑆𝑃=

50

40=

5

4

Loss percentage = 1 −𝑆𝑃

𝐶𝑃 𝑥100

Loss Percentage = 1 −4

5 =

1

5𝑥 100 = 20%

e) If two items are sold at the same price, one with a profit of x% and another

with a loss of x%, there will be an overall loss given by 𝑥2

100.

f) If the CP of two items is the same, and one is sold at x% gain, other at x % loss, then the net loss or net profit=0 ILLUSTRATION: 24) A shopkeeper sells two shirts at the same price; one at a discount of 20% and another at a gain of

20%.Find the overall loss or the gain he earned. As both shirts are sold at the same price one at 20% loss and other at 20% profit,

he will make a net loss of 202

100= 4%

g) On an offer of buy “a” goods and get “b” goods free, i.e., if “a+b” articles are

sold at the rate of “a” articles, then the percentage discount = 𝑏

𝑎+𝑏 × 100

h) If there are successive discounts of x% and y%, the effective discount can be

calculated as 𝑥 + 𝑦 – 𝑥𝑦

100 %

i) There are many shopkeepers who cheat by using false weights, then the gain to the shopkeeper on using the false weight can be calculated as follows

100 + 𝐺𝑎𝑖𝑛%

100=

𝑇𝑟𝑢𝑒 𝑊𝑒𝑖𝑔𝑕𝑡

𝐹𝑎𝑙𝑠𝑒 𝑊𝑒𝑖𝑔𝑕𝑡

Explanation: Suppose the shopkeeper is using a false weight in such a way that when his meter shows X kg (False weight) he is actually selling Y kg (True weight). Also say that the price he is selling the article is Rs. Z per kg. Thus, when his meter is showing X kg, he is actually selling Y kg. So for the shopkeeper, the cost price (CP) = YZ But the selling price (SP) = XZ

So, his profit = 𝑆𝑃 – 𝐶𝑃

𝐶𝑃=

𝑋𝑍 – 𝑌𝑍

𝑌𝑍 =

𝑋

𝑌– 1or

𝑋

𝑌– 1 𝑥100 %

𝑃𝑟𝑜𝑓𝑖𝑡 %

100=

𝑋

𝑌– 1

Note:-Although formulas are given for different cases, this is just for your reference. It is always better to approach every question by assuming values or going from answer options, as will be illustrated in the questions which follow.



Miscellaneous Arithmetic

4) RATIO, PROPORTION, VARIATION & PARTNERSHIP

4.1) INTRODUCTION What is a Ratio? A ratio is a fraction which represents the relationship between two similar quantities (ones with the same units). A ratio depicts how many times one number is of another. If we have two terms p and q, with values 4 and 5 respectively; the ratio is depicted as p:q, i.e. 4:5 and is measured

as 𝑝

𝑞

4

5. ‘p’ is called the antecedent and q is called the consequent.

There are two types of ratios Proper Ratio When the antecedent is less than the consequent (numerator < denominator)

Eg) 3

7,

1

2

Improper Ratio When the antecedent is more than the consequent (numerator > denominator)

eg) 7

3,

5

2

4.2) PROPERTIES OF RATIO A ratio is a comparison of two similar quantities: hence it will not have any units It will remain unchanged if both the antecedent and the consequent are multiplied and divided by the same non-zero multiplier Ratios of the squares of p & q p2 : q2 (Duplicate Ratio)

Ratios of the square roots of p &q 𝑝: 𝑞 (Sub Duplicate Ratio)

4.3) COMPARISON OF RATIOS Methods of comparison of ratios 1) By cross-multiplication

Eg 25) Find out which of 𝟏𝟐

𝟏𝟕 and

𝟕

𝟏𝟏 is greater?

Cross multiply the numerator of 1st fraction with denominator of 2nd fraction and numerator of 2nd fraction with denominator of 1st fraction 12 x 11 =132 17 x 7 = 119 Since 132> 119 The first fraction > second fraction 2) Using Decimals Take the following example 9

11= 0. 8181

6

10= 0.6

Here 0.8181 > 0.6. Hence the 1st fraction is greater than the second one



4.4) STANDARD RESULTS ON RATIOS

1) Componendo

𝐼𝑓𝑎

𝑏=

𝑐

𝑑𝑡𝑕𝑒𝑛

𝑎+𝑏

𝑏=

𝑐+𝑑

𝑑

5) Alterando

𝐼𝑓𝑎

𝑏=

𝑐

𝑑, 𝑡𝑕𝑒𝑛

𝑎

𝑐=

𝑏

𝑑

2) Dividendo

𝐼𝑓𝑎

𝑏=

𝑐

𝑑𝑡𝑕𝑒𝑛

𝑎−𝑏

𝑏=

𝑐−𝑑

𝑑

6) Continued Proportion a,b and c are said to be in continued proportion if 𝑎

𝑏=

𝑏

𝑐𝑜𝑟 𝑏2 = 𝑎𝑐. Also 𝑏 = 𝑎𝑐. Which means that

b is the geometric mean of a and c

3) Componendo and Dividendo

𝐼𝑓𝑎

𝑏=

𝑐


𝑎+𝑏

𝑎−𝑏=

𝑐+𝑑

𝑐−𝑑

7) Sum rule 𝑎

𝑏=

𝑐

𝑑=

𝑎+𝑐

𝑏+𝑑

In general, if 𝑎

𝑏=

𝑐

𝑑=

𝑒

𝑓= ⋯ . . 𝐾 (Some constant)

Then 𝑎

𝑏=

𝑐

𝑑=

𝑒

𝑓= ……… . . 𝐾 =

𝑎+𝑐+𝑒+⋯

𝑏+𝑑+𝑓+⋯..

i.e. Each ratio = 𝑆𝑢𝑚 𝑜𝑓 𝑛𝑢𝑚𝑒𝑟𝑎𝑡𝑜𝑟𝑠

𝑆𝑢𝑚 𝑜𝑓 𝐷𝑒𝑛𝑜𝑚𝑖𝑛𝑎𝑡𝑜𝑟𝑠

Given that 𝑎

𝑏,𝑐

𝑑,𝑒

𝑓are all unequal ratios, the value

of 𝑎+𝑐+𝑒+⋯

𝑏+𝑑+𝑓+⋯ lies between the minimum and

maximum of all these ratios

4) Invertendo

𝐼𝑓𝑎

𝑏=

𝑐


𝑏

𝑎=

𝑑

𝑐

4) Power Rule

Given that 𝑎

𝑏=

𝑐

𝑑=

𝑒

𝑓= ⋯, then each ratio is equal

to 𝑚𝑎𝑝 + 𝑛𝑐𝑝 + …..

𝑚𝑏𝑝 + 𝑛𝑑𝑝 + ….

1

𝑝 where m,n,p all non zeroes

4.5) VARIATION Two quantities are said to vary with each other, if there exists a relationship between them such that a change is A entails a change in B and vice-versa. Eg) Speed varies inversely with time; that is if the Speed is more, then the time taken to cover a constant distance will be less

𝑆𝑝𝑒𝑒𝑑 𝛼1

𝑇𝑖𝑚𝑒



4.5.1) DIRECT PROPORTION

If Quantity 1 varies directly with Quantity 2; i.e. Quantity 2 increases if Quantity 1 increases and Quantity 2 decreases if Quantity 1 decreases; then they are said to be in direct proportion For example, if Time is constant, the Speed will be directly proportional to the distance. That is, if the Speed is more, then the distance covered will also be more, and vice-versa Speed α Distance

4.5.2) INVERSE PROPORTION

If Quantity 1 varies inversely with Quantity 2, then they are said to be in inverse proportion. For example, Speed varies inversely with time; that is if the Speed is more, then the time taken to cover a constant distance will be less

𝑆𝑝𝑒𝑒𝑑 𝛼1

𝑇𝑖𝑚𝑒

This means that when speed is minimum, then time is maximum; and when speed is maximum, then time is minimum

4.6) PARTNERSHIP In a partnership, two or more people or organizations invest their money and time and set up a business together. A partner who only invests money is called a Sleeping Partner and a partner who invests money and mainly manages the business is called the working partner. There are broadly two types of Partnerships Simple Partnership :- Where the money is invested for the same time period by all the investors Compound Partnership :- Where the money is invested for different periods of time by different investors. When the periods of investment are equal, the profits or losses are in the ratio of the corresponding investments 𝐼𝑛𝑣𝑒𝑠𝑡𝑚𝑒𝑛𝑡 𝑜𝑓 𝑃

𝐼𝑛𝑣𝑒𝑠𝑡𝑚𝑒𝑛𝑡 𝑜𝑓 𝑄 =

𝑃𝑟𝑜𝑓𝑖𝑡 𝑜𝑓 𝑃

𝑃𝑟𝑜𝑓𝑖𝑡 𝑜𝑓 𝑄 =

𝐿𝑜𝑠𝑠 𝑜𝑓 𝑃

𝐿𝑜𝑠𝑠 𝑜𝑓 𝑄

i.e. Investment of P :Investment of Q : Investment of R = Profit of P : Profit of Q : Profit of R (In other words, they are in direct proportion) In a compound partnership, the profits will depend on the period of investment as follows 𝐼𝑛𝑣𝑒𝑠𝑡𝑚𝑒𝑛𝑡 𝑜𝑓 𝑃𝑥 Period of investment by P

𝐼𝑛𝑣𝑒𝑠𝑡𝑚𝑒𝑛𝑡 𝑜𝑓 𝑄𝑥 Period of investment by Q =

𝑃𝑟𝑜𝑓𝑖𝑡 𝑜𝑓 𝑃

𝑃𝑟𝑜𝑓𝑖𝑡 𝑜𝑓 𝑄 =

𝐿𝑜𝑠𝑠 𝑜𝑓 𝑃

𝐿𝑜𝑠𝑠 𝑜𝑓 𝑄

𝑃’𝑠 𝑠𝑕𝑎𝑟𝑒 𝑖𝑛 𝑡𝑕𝑒 𝑝𝑟𝑜𝑓𝑖𝑡 =𝐼𝑛𝑣𝑒𝑠𝑡𝑚𝑒𝑛𝑡 𝑜𝑓 𝑃

𝑇𝑜𝑡𝑎𝑙 𝐼𝑛𝑣𝑒𝑠𝑡𝑚𝑒𝑛𝑡 𝑏𝑦 𝑎𝑙𝑙 𝑖𝑛𝑣𝑒𝑠𝑡𝑜𝑟𝑠𝑥 𝑃𝑟𝑜𝑓𝑖𝑡



5) WEIGHTED AVERAGE, MIXTURES AND ALLIGATIONS Average can be said to be the central value of a group of values, or the value around which a group of values concentrate. Average indicates a group of values, or is representative of a group of values. Average is traditionally calculated by dividing the sum of all the entities by the number of entities. For example, Eg 26) The average of the 5 numbers 100, 150, 200, 250 and 350 will be ? 100 +150 + 200 + 250 + 350

5=

1050

5 = 210

Average can also be calculated by

Assumption of Central Value Method

Take an example, with the numbers 300, 310, 320, 313, 315. Now let us assume the central value to be 313

Conventionally average = 300 + 310 + 320 + 313 + 315

5 = 311.6

Find the deviation of all the numbers from 313 313-313 = 0 300-313 = -13 310-313 = -3 320-313 = 7 315-313 = 2

Average of the deviations =0−13−3+7+2

5= −

7

5 = −1.4

Average = Assumed Central Value + Average of all deviations = 313 -1.4 =311.6

5.1) Classification of Averages Averages can be classified into the following 3 types 5.1.1) Mean: - Arithmetic Mean, Geometric Mean and Harmonic Mean : A) Arithmetic Mean (AM):

Arithmetic mean of n numbers a1,a2,……an is given by 𝑎1+𝑎2+𝑎3+⋯…𝑎𝑛

𝑛

It is nothing but the normal average.

If a,b,c are in AP , then Arithmetic Mean, 𝑏 =𝑎+𝑐

2

In an AP the middle term will be the arithmetic mean of the previous term and the next term.

If you have three numbers in AP the middle term will be the arithmetic mean of the first term and the last term. And if you have 5 numbers in AP then the middle term i.e. third term will be the arithmetic mean. Also if you have an AP then sum of first term and last term divided by 2 will give the arithmetic mean of that AP which will also be the middle term. Point to Note: This is why while calculating the sum of ‘n’ terms we take the sum of first term and last term divided by 2 to get the average and then multiply by the number of terms to get the sum. Inserting n Arithmetic Means (AMs) between two numbers A and B Before explaining the concept of inserting AMs try to solve this question.



Eg 27) Find the sum of 500 arithmetic means between 2 and 3? When you insert n AMs between A and B, we will get an AP with n+2 terms, and the common difference is given

by 𝑩−𝑨

𝒏+𝟏

Let us see a similar question. For eg: Inserting 4 AMs between 10 and 20 implies that we will get an AP with 6 terms.

We can form the AP , by calculating the common difference as 20−10

5= 2.

Thus the 4 Arithmetic Means will be 12,14,16 and 18. If you observe, AM ( average) of 12,14,16 and 18 will be the AM of 10 and 20 = 15.

i.e 12+14+16+18

4=

10 +20

2= 15

So if the question is to find the sum of 4 AMs between 10 and 20 , answer is easy, it will be nothing but = 4 x (average of the 4AMs between 10 and 20)

= 4 𝑥 12+14+16+18

4

= 4 x 15 = 60. Extending the same concept to a larger number of terms, say “N” AMs, we can find out

The Sum of n AM’s between a and b = 𝑛 ×𝑎+𝑏

2

a= first term, b = last term 𝑎 + 𝑏

2 = average of terms. So for ‘n’ terms inserted in between ‘a’ and ‘b’ sum =

𝑛× 𝑎 + 𝑏

2

Thus the solution for question, Find the sum of 500 arithmetic means between 2 and 3 is 500× 2+3

2

= 500 × 2.5 = 1250 28. The first and the last term of an AP are 107 and 253. If there are five terms in this sequence, find the sum of the sequence. Solution: First term = 107 last term = 253

Average or mean = 107 + 253

2 = 180

Number of terms in the sequence = 5. So, sum of sequence = 180 * 5 = 900. B) Geometric Mean (GM):

Geometric mean of n numbers a1, a2,…..an is given by 𝑎1 × 𝑎2 × 𝑎3 × ……𝑎𝑛 1

𝑛

e.g. geometric mean four numbers a,b,c,d = 𝑎𝑏𝑐𝑑 1

4

e.g. geometric mean of 2, 4, 8 = 2 × 4 × 8 1

3 = (64)1/3 = 4

If a,b,c are in GP, geometric mean, 𝑏 = 𝑎𝑐 i.e In a GP if you take three consecutive terms middle term will be geometric mean of first term and the third term. And if you take a GP with 5 terms, them the middle term or 3rd term will be the GM of the series.



Application of GM : The Concept is GM is used in finding out the compounded annual growth rate (CAGR)

CAGR = (𝑓𝑖𝑛𝑎𝑙 𝑣𝑎𝑙𝑢𝑒

𝑖𝑛𝑖𝑡𝑖𝑎𝑙 𝑣𝑎𝑙𝑢𝑒)

1

𝑛 − 1 × 100 where n is the number of years.

29) Population of a village is 10000 in year 2000 and is 14400 in 2002. Find Compounded Annual Growth Rate?

CAGR = 14400

10000

1

2 − 1 × 100 = 20 %

This means an annual increase of 20 %. To verify 10000 × 1.2 = 12000 12000× 1.2 = 14400 We are using a factor of multiplication 1.2 because to increase a number by 20% and to get the final value, we need

multiply by 120

100= 1.2

CAGR is similar to rate percent, R in compound Interest calculations.

Final amount = 𝑃 1 +𝑅

100 𝑛

where P = initial amount, R = rate percent, n = number of years.

So, CAGR or 𝑅 = (𝑓𝑖𝑛𝑎𝑙 𝑎𝑚𝑜𝑢𝑛𝑡

𝑖𝑛𝑖𝑡𝑖𝑎𝑙 𝑎𝑚𝑜𝑢𝑛𝑡)

1

𝑛– 1 × 100

CAGR implies the annual increase in growth. Inserting n GM’s between A and B When we insert n GMs between A and B, we will get a GP with n+2 terms, the Common Ratio is given by

𝑟 = 𝑏

𝑎

1

𝑛+1 .

GM1 = A.r GM2 = A r2

………… For example 2 GMs between 1 and 27 will be 3,9. We can find out the GMs by finding the common ratio

= 27

1

1

3 = 3

We will get a GP with 4 terms 1,3,9,27 3 GMs between 1 and 16 will be 2,4 & 8, so we will get a GP with 5 terms 1,2,4,8,16. ILLUSTRATION 30) Insert 8 geometric means between 3 and 1536.Find the 6th GM ?

Common ratio = 1536

3

1

9= 2

GM1 = 3*2 = 6 GM2 = 3*22 =12 GM3 = 3*23 = 24…… GM6 = 3*26 = 192 Required answer = 192.



31) Find the number of GMs in a series where the ratio between first and last terms is 16:1 and common ratio of GP=2

Common ratio= 𝑏

𝑎

1

(𝑛+1)

2 = 161

𝑥

2x=16=>x=4 x=n+1=>n=4-1=3 Number of Geometric Means=3 C) Harmonic Mean (HM): Let a and b be two given quantities. It is required to insert n harmonic means h1, h2, h3,....hn between the quantities a and b

Harmonic mean of n numbers 𝑎1 , 𝑎2,……an is given by 𝑛

1

𝑎1+

1

𝑎2+

1

𝑎3+⋯.

1

𝑎𝑛

If a,b,c are in Harmonic Progression, then HM, b =2

1

𝑎+

1

𝑐

=2𝑎𝑐

𝑎 + 𝑐

Harmonic mean of 4 numbers a,b,c,d = 4

1

𝑎+

1

𝑏+

1

𝑐+

1

𝑑

i.e In an HP if you take three consecutive terms , middle term will

be the harmonic mean of the first term and the third term. Application of harmonic mean :- Harmonic mean helps us find out the average speed of a journey when the distances covered are equal

Average speed when the distances covered are equal = HM of the speeds

Let us say a person P, covers a distance A to B at a speed of “a” kmph and a distance B to A at a speed of “b” kmph.

As the distance is constant, we can find the average speed using the following formula Average speed=2𝑎𝑏

𝑎+𝑏 kmph

When the two quantities are in direct proportion, take the Arithmetic Mean When the two quantities are in inverse proportion, take the Harmonic Mean

ILLUSTRATION: 32) A person travels from A to B with a speed of 60 km/hr and returns from B to A at 40 km/hr. What is the average speed for the whole journey? Soln: Here since the distances covered are equal, average speed for the whole journey = Harmonic mean of the

speeds = HM of 60,40 = 2

1

60 +

1

40

=

2×60×40

60+40= 48 km/hr.



33) Amit covers first 𝟏

𝟑

𝒓𝒅 of the distance walking at 2 km/hr, second

𝟏

𝟑

𝒓𝒅 of the

Distance running at 3km/hr and the rest cycling at 6 km/hr. Find the average speed for the whole journey? a) 3 km/hr b) 3.66 km/hr c) 4 km/hr d) 5 km/hr. Sol: distances covered are equal for all 3 cases, each time 1/3rd of the total distance. So average speed = Harmonic mean of 2, 3 and 6

= 3

1

2 +

1

3 +

1

6

= 3 km/hr.

5.1.2) Median:- Median is the middle value of a group of numbers arranged in an ascending or descending order.

a) If the number of values (n) is odd, then median will be the 𝑛+1

2 th value

Take for example, 34) Find the median of the following numbers 51, 50, 49, 58, 48, 43 and 51 Arrange the group in the ascending order 43, 48, 49, 50, 51, 51, 58.

Median will be the (7+1)

2 th value = 4th value = 50

If the number of values (n) of a given set of data is even, then the median will be the mid value of the two middle numbers. Take for example, 35) Find the median of 54, 33, 23, 26, 25, 24 On arranging the values in ascending order, 23, 24, 25, 26, 33, 54. the two middle values are 25 and 26.

Median = 25+26

2 =

52

2 = 25.5

5.1.3) Mode:- Mode is the number that occurs most frequently in a given set of numbers. Take for example, 36) Find the mode of 1,1,1,1,3,4,5,6,2,2,3,3,1,1,6,6,7,7,7,8,8,8,8 The mode will be 1 as it occurs maximum number of times.

5.2) APPLICATIONS OF AVERAGES

5.2.1) Age related Averages

In questions where we need to assume the age of people involved, the present age is taken as a variable say x, and the present and past ages are based on the variable x, for example 5 years ago x-5 10 years from now x+10 15 years ago x-15 Using this, we can form equations easily and solve for the present age. Take for example the following questions



37) The average age of 5 sisters is 15 years. The youngest sister is 6 years old. When she was born, the average age of the remaining sisters was N years. What is the average age of the sisters excluding the youngest sister? Given that average age of sisters = 15 years. Sum of all their ages= 15x5= 75 years Sum of their ages excluding the youngest sister = 75-6 = 69 years

Average age of the remaining sisters = 69

4 = 17.25 years

5.2.2) Time, Speed, Distance

Average Speed = Total Distance Traveled

Total time taken

If the distance is constant, using the two speeds, say a and b, the Average speed can be calculated by finding out the harmonic mean of the speeds ‘a’ & ‘b’ (discussed in Harmonic Mean before)

Average Speed = Harmonic mean of a & b = 2ab

(𝑎+𝑏)

38) Athul travels from his home to office at 60 km/hr and returns back home from office at 40km/hr. If he

takes 3 hrs in all find the distance in km between his home and office? a) 48 km b) 72 km c) 60 km d) 92 km

Average speed = 2×60×40

100 = 48 kmph Speed =

Distance Traveled

Time taken

Hence, distance between home and office = 1.5 x 48= 72 kms

5.2.3) Application in Series of Numbers

Sum of 1st n consecutive natural numbers = 𝑛(𝑛+1)

2 Average of 1st n consecutive natural numbers =

(𝑛+1)

2

eg) Average of 1st 7 consecutive natural numbers = 8

2 =4

Sum of 1st n even consecutive natural numbers = n(n+1) Average of 1st n even consecutive natural numbers = (n+1) eg) Average of 1st 3 even natural numbers = 4 Sum of 1st n odd consecutive natural numbers = n2 Average of 1st n consecutive odd natural numbers = n eg) Average of 1st 3 odd natural numbers = 3 39) Consider a sequence of seven consecutive integers. The average of first five integers is n. The average of all seven integers is:

a) n b) n + 1 c) k x n, where k is a function of n. d) n + 𝟐

𝟕 .

Let us solve this by assumption. Let the 7 consecutive integers be 1,2,3,4,5,6,7 Average of first five integers = 3 Average of first seven integers = 4 Hence, answer is option (b)



5.3) WEIGHTED AVERAGE Observe that the average can be calculated directly only if the weights of all the factors are the same. If the values a,b,c and d are the individual averages of each of the groups and n1, n2, n3 and n4 are the number of observations or number of values , then the Weighted Average is calculated as

The term “weight” stands for the relative importance attached to the different values. 40) In Class I there are 12 students of average age 20 years and in Class II there are 16 students of average age 23 years. What is the average age of both the classes?

Solution = 12 ×20 + (16 ×23)

(12+16) =

240+368

28 =

608

28 = 20.7 years

NOTE:- Average always lies between the maximum and minimum value. It will be equal to the maximum or

minimum value if all the values are equal Average is the result of the net surplus + net deficit If the value of each quantity is increased or decreased by the same quantity “q”, then the average will also

increase or decrease by the quantity “q” respectively. If the value of each quantity is multiplied or divided by a quantity “q”, then the resultant average will also be

multiplied or divided by “q” respectively. When weights of all the quantities are the same, then the average can be calculated directly. If the weights

are different, we need to use the concept of weighted average

5.4) Shortcut Techniques in Averages Let us now see a useful shortcut technique to find the average; with the help of an example. 41) The average of a batsman in 16 innings is 36. In the next innings he is scoring 70 runs. What will be his new average? a) 44 b) 38 c) 40 d) 48 Conventionally solving

New average = old sum + new score

total number of innings =

16 ×36 +70

(16+1) = 38

Shortcut technique! Step 1) Take the difference between the new score and the old average = 70-36= 34

Step 2) This 34 extra runs is spread over 17 innings = 34

17= 2

Step 3) Hence, the average increases by 2 = >36+2 = 38 is the new average Let us try another question using the same technique 42) The average marks of 19 children in a particular school is 50. When a new student with marks 75 joins the

class, what will be the new average of the class? 1) Take the difference between the old average and the new marks = 75-50=25

2) This score of 25 is distributed over 20 students => 25

20 = 1.25

3) Hence, the average increases by 1.25=> 50+1.25 = 51.25 Let us try another question where the average dips



43) The average age of Mr Mark’s 3 children is 8 years. A new baby is born. Find the average age of all his children?

The new age will be 0 years. The difference between the old average and the new age = 0-8= -8

This age of 8 years is spread over 4 children => −8

4= −2

Hence, the average reduces to 8-2= 6 years Now let us look at a technique which will help us compute the new value when the average is given. Take this question for example 44) The average age of 29 students is 18. If the age of the teacher is also included the average age of the class

becomes 18.2. Find the age of the teacher? a) 28 b) 32 c) 22 d) 24 Conventionally solving, Let the average age of the teacher =x 29 × 18 + x × 1

30

Solving for x, we get x = 24 Using the shortcut, based on the same method we used previously Step 1: calculate the change in average = 18.2-18 = 0.2 This change in 0.2 is reflected over a sample size of 30

New age is an increased by 30 * 0.2 = 6 years above the average 18+6= 24

Think Quest- The average age of 26 students in an MBA school is 30. One student among these, quits the school in between. Can you find the age of that student if the new average is 29.8?

ILLUSTRATIONS 45) Average goals scored by 15 selected players in EPL is 16.Maximum goals scored by a player is 20 and

minimum is 12.Goals scored by players is between 12 and 20. What can be maximum number of players who scored at least 18 goals ?

a) 10 b) 5 c) 9 d) 6 e) none of these option (c) To maximize the number of players who scored 18 and above number of goals, we will assume that only one person has scored 20. To counter him, we will have one person who will score 12 goals. 15-2 =13 players left Now to maximize the 18 and above goals, for every two players who are scoring 18, we will have one player scoring 12. This is done, to arrive at the average of 16 We will have 8 players with a score of 18 and 4 players with a score of 12. The last player will have a score of 16 Thus, the maximum number of people with 18 and more goals = 9. 46) The average weight of a group of 8 girls is 50 kg. If 2 girls R and S replace P and Q, the new average weight

becomes 48 kg. The weight of P= Weight of Q and the weight of R= Weight of S. Another girl T, is included in the group and the new average weight becomes 48 kg. Weight of T= Weight of R. Find the weight of P? a) 48 kgs b) 52 kgs c) 46 kgs d) 56 kgs



option (d) 8 x 50 +R+S-P-Q= 48x8 R+S-P-Q=-16 P+Q-R-S= 16 R=S and P=Q P-R=8 One more person is included and the weight = 48 kg

Let the weight be a = 48 ×8+a 9

9 = 48

a=48 kg= weight of R weight of P= 48+8= 56 kg

5.5) MIXTURES & ALLIGATION

5.5.1) MIXTURES

When two or more components are mixed in a certain ratio, a mixture is created. Types of mixtures: Simple Mixtures:- When two or more different ingredients are mixed together, a simple mixture is formed Compound Mixtures:- When two or more simple mixtures are mixed together, a compound mixture is formed.

5.5.2) ALLIGATION

2.6.1) Concept of Alligation They say that examples are not the best way to teach a concept, they are the ONLY way. To understand the concept of alligations, let’s take an example and look at it from all angles possible. 47) 30 boys with an average weight of 60 kgs and 20 girls with an average weight of 40. Find the average weight of the whole class. To find the answer to the above question, we can use the formula for weighted average. Suppose the question was framed a bit differently The Average weight of boys is 60, average weight of girls is 40 and average weight of the whole class is 52. Find the ratio between the number of boys and girls The fastest way to solve this question would be by using Alligation So what exactly is Alligation? It is the reverse of weighted average; i.e. If the averages of two groups are separately given and the average of the whole group is given, then we can find out the ratio between the groups. For example, in the above question, the data given is Average weight of boys=60, average weight of girls=40 and average weight of the whole class is =52 We can represent the data in an Alligation chart as given below and use it to find the ratio between the number of boys and girls in the class.

(x=52-40=12) (y= 60-52=8) Take the difference across: x= 60-52=8 and y= 52-40=12. That is how we get the ratio between the boys and girls as 12:8 or 3:2.



Let us look at another question where you can use Alligation 48) In what ratio should a shopkeeper mix two types of rice, one costing 20 rupees/kg and another costing 10 rupees/kg to get a rice variety costing 14 rupees/kg Here also we can use Alligation as follows x=14-10=4 y=20-14=6 The ratio between the type 1 and type 2 rice is 4:6 or 2:3

5.5.3 FORMULAE IN MIXTURES & ALLIGATIONS

There is no necessity to memorize all the formula given here, as all of them can be easily deduced with the help of logic. Formula A,B and C are just algebraic representations of Alligation.

A) Quantity of cheaper entity

Quantity of expensive entity =

Price of expensive entity – Mean price

Mean Price − Price of cheaper entity

B) Cost of Mixture(Cm) Quantity of Component A =xa Quantity of Component B = xb Cost of Component A= Ca Cost of Component B= Cb

Then, Cm= xa Ca +xb Cb

xa + xb

C) If you have 2 mixtures (M1 and M2), each having two components A and B 𝑀1 contains A & B in the ratio p:q 𝑀2 contains A & B in the ratio r:s

Quantity of A and B in the final mixture (xa & xb) D) If a vessel contains “a” liters of liquid A; and”b” liters are withdrawn and replaced by another liquid B of equal quantity and the operation is repeated n times, then:

1) Liquid A left after nth operation

Initial quantity of A in the vessel =

a−b n

(𝑎)𝑛 2)

Liquid A left after nth operation

Liquid B left after nth operation =

a−b n

(𝑎)𝑛

1− a−b n

(𝑎)𝑛

5.5.4) APPLICATIONS OF ALLIGATION

Consider the following common data to understand the application better. A= 20 B=10 Na =30 Nb= 45 The answer will be 14, in each of the following cases 1) Average of a class:- eg) Class A has 30 students scoring average marks of 20 and Class B has 45 students scoring average marks of 10. What is the average of both classes together (14) 2) Average price of goods eg) Sunita buys 30 kgs of sugar at Rs 20/kg and 45 kgs of sugar at 10/kg. What is the average price? (Rs 14/kg)



3) Percentage eg) A antique article salesman makes a profit of 20% by selling 30% of his articles and a profit of 10% by selling 45% of his goods. What is his net percentage profit? (14%) 4) Average Speed eg) A bus travels at 20kmph for 30 km and at 10 kmph for 45 km. Find the average speed of the bus for the entire journey (14kmph) 5) Mixtures A dishonest milkman dilutes two milk cans such that milk can A containing 30 liters is mixed with 20% water and milk can B containing 45 liters is mixed with 10% water. He combines both mixtures. Find the percentage of water in the final mixture (14%)

5.5.5) Application of Alligation- Illustrations

A) Application in profit & loss questions 49) How many kgs of rice costing Rs 8 per kg must be mixed with 36 kg of rice costing Rs 5.40 per kg so that

20% gain may be obtained by selling the mixture at Rs 7.20 per kg? a) 10 kg b) 12 kgc) 10.8 kg d) 8 kg Always remember that we need to take only the cost price and not the selling price when we consider the price of all entities in the question. In the question, the selling price of the mixture is given. We need to find out the cost price using the profit percentage given=20%. Cost price × 1.2=7.2 , implies that cost price of the mixture is 6. From the Alligation chart, X=6-5.40=0.6 Y=8-6=2 Thus the ratio between the quantity of type A rice and type B rice is 0.6:2 or 3:10

Therefore => x=10.8 kg

B) Application in mixtures of liquids 50) Vessel A contains milk and water in the ratio 4:5. Vessel B contains milk and water in the proportion 5:1.

In what proportion should quantities be taken from A & B to form a mixture in which milk and water are in the ratio 5:4?

a) 2:5 b) 3:2 c) 2:3 d) 5:2 For this question, we will consider the proportion of milk in each mixture. In Vessel A, the proportion of milk in

4

4+5 =

4

9. in vessel B, the proportion of milk is

5

5+1 =

5

6.

The amount of milk in the mixture= 5

(5+4) =

5

9

The ratio is



C) Application in percentage questions 51) A person has Rs.5000. He invests a part of it at 3% per annum and the remainder at 8% per annum simple

interest. His total income in 3 years is Rs.750. Find the sum invested at different rates of interest? a) 3500,1500 b) 1000,4000 c) 3000,2000 d) 1800,3200 He is investing part of it at 3% at annum and remaining at 8% per annum Since he is getting Rs 750 as profit for 3 years, rate percent of the whole amount is 5% . That can be calculated from 𝑃𝑁𝑅

100 = 750

(5000 ×3 ×𝑅)

100 = 750

R= 5% From the Alligation chart, The ratio of investment will be 3:2. Look in the answer options for that ratio. Only option (c) (3000, 2000) gives that ratio

TEST-Basic & Commercial Arithmetic No of questions-25 Time-50 min 1) In a public school there is an increase in number of admissions every year. The number of students in the year 2000 was 780 and it increased to 3000 in the year 2005. Find the annual increase? a) 25 % b) 31 % c) 39 % d) 20 % 2) Ram and Shyam invest Rs.15,000 and Rs.25,000, respectively, in a business and at the end of the year , make a profit of Rs.10,000. They agree to re-invest 12% of profit. Out of the remaining profit, each of them takes Rs.1000 and rest is divided in the ratio of their original investment. Find Ram’s total share in the profit . a) Rs. 4250 b) Rs.3000 c) Rs.2550 d) Rs.3550 3) At the end of year 1998, Ramesh bought 108 shares. Henceforth, every year he added p% of the shares at the beginning of the year and sold q% of the shares at the end of the year where p > 0 and q > 0. If Ramesh had 108 shares at the end of year 2002, after making the sales for that year, which of the following is true?

a) p = q b) p < q c) p > q d) 𝑝 =𝑞

2

4) A man purchased 40 kg of cotton at a rate of Rs 65 per kg. Then he extracted 20% waste (by weight) from it so that the quality of the cotton improved and he was able to sell it at Rs.80 per kg. Also the waste had 60% cottonseed by

weight which he was able to sell Rs.20 per kg. What percentage profit did he make? a) 8.65% b) 12.44% c) 14.70% d) 21.32% 5) Bangalore’s Metro urban planners expect the city’s population to increase by 10% per year over the next two years. If that projection were to come true, the population two years from now would be exactly double the population one year ago. Which of the following is closest to the percent population increase in Bangalore over the last year? a) 20 b) 40 c) 50 d) 65 6) 90 students represent x percent of the girls at Bangalore High School. If the girls at Bangalore High make up 40% of the total school population of x students, what is x? a) 125 b) 150 c) 225 d) 250 7) A fruit seller bought a few kgs of apples. First day he sold them at 100% profit and at the end of the day, he was able to recover his cost price. Next day he sold them at 50% profit and by the end of the day his net profit became 30%. On the third day he sold the remaining apples at 25% profit. His net profit on the sale of the apples is a)50% b)57.5% c)62.5% d)67.5% 8) A bought 2 apples and 3 oranges at Rs. X. He ate 1 orange and sold the remaining fruits to B at Rs. X. B ate one apple and sold the remaining fruits to C at Rs. X. If the profit % of A is 20%, then the profit % of B is a) 16.66% b) 20% c)24% d)25% 9) Let the rate of interest of money for the first year at CI be 10% which increases by 1% every successive year. Then the money will become double in a)third year b)fourth year c)fifth year d)sixth year.



10) Arun buys oranges at Rs 16 for 9 and sells them at Rs 20 for 11. What is his gain or loss%? a) 4% b)2.7% c)4% d) 3.5% 11) It is given that an article is sold at 10% discount. The SP is Rs 81. What will be the SP when the discount is 20%? a) Rs. 80 b) Rs. 82 c) Rs. 72 d) Rs. 68 12) A confused student multiplies a number x by 6, instead of dividing it by 6. What is the percentage change in the result due to his mistake? a) 350% b) 35% c) 2550% d) none 13) In an election, the losing candidate secured 20% of the votes. The winner got 128000 votes totally. Assuming that no vote was invalid, what is the total number of votes cast? a) 160000 b) 240000 c) 330000 d)124000 14) A car which depreciates by Rs 230000 is worth Rs 392500 in the year 2000. By what percent has it depreciated? a) 37% b) 42% c) 39% d) 45% 15)If the price of coffee powder per kg is increased by 20%, then find by how much the consumption should be reduced so that the expenditure does not change? a) 25% b) 16.67% c) 19% d) 20% 16) Namita took a test in Physics, Chemistry, Math and Biology (P,C,M,B) to gain entrance into MSc. Science course. Each paper is of maximum 100 marks. She can get an admission into MSc. Biological Science if she has a minimum 60% marks in both Biology and Chemistry and a minimum of 55% overall. She can get an admission into MSc. Physical Science if she has a minimum 60% marks in Physics and Chemistry and 55% overall. She is also eligible to get both admissions. Given that she gets an admission into only MSc Biological Science and that she has scores 66 in Biology and 68 in Chemistry, what is the minimum marks she could have scored in Maths? a) 0 b) 26 c) 37 d) 27 17) From the data given below, find the Cost of Living Index for the year 2005, taking 2000 as the base year and base index as 100.

ITEM QUANTITY (kg)

Price/kg 2000

Price/ kg 2005

Wheat 25 12 15

Rice 10 8 12

Coffee 5 14 16

Sugar 2 40 36

Salt 1.5 60 90

a) 113.12 b) 156.24 c) 126.13 d) none of these

18) At a new index of 12 shares, the shares of HBL, Niposys and Brilliance have a weightage of 7%, 13% and 1% respectively. What is the increase in the price of the other shares, if the rise in these three is 9%, 10% and 4%, when the index rises by 6%? a) 4% b) 3.4% c) 6% d) none 19) Bindu put an amount x in the bank and obtained an interest of Rs 15 at the end of the year. She added Rs 85 to this amount and put the whole amount back in the bank for the second year at the end of the second year, Bindu obtained a total amount of Rs 420. What amount did she invest initially and what is the rate of interest being offered by the bank if the minimum amount that can be deposited is Rs 75? a)200, 10% b) 300, 5% c) 400, 6% d) 350, 5% 20) When a customer was bargaining with the milk man, the milk man argues that he is making a loss of 25%. The truth is that he adds water to the milk; he makes a 12.5% profit. When he sells 2 litres of milk, how much pure milk is there? a) 666.7 ml b) 1333.4 ml c) 987 ml d) 1654 ml 21) P sells 15 chocolates to Q at a profit of 10% who sells it to R at a loss of 20% who sells it to S at a profit of 20% who sells it to T at a profit of 10% who sells it to U at a 50% profit. If the cost at which U bought the chocolates is 1287 , then find the initial cost of each chocolate? a) 75 b)30 c) 40 d) 50 22) A publisher printed 3,000 copies of a book at a cost of Rs. 2,400. He gave 500 copies free to heads of institutions. He allowed a discount of 25% on published price and gave one extra copy when a customer buys 24 copies at a time. He sold all the copies in this scheme. If the published price is Rs. 3.25,find his gain or loss percentage. a) 143.75% gain b) 43.75% gain c) 43.75% loss d) None of these 23) In a parliamentary debate, a bill was being passed. 600 votes were cast initially but after some discussion, the opposition to the bill went up by 150%. The bill was then rejected by a majority 2 times as great as that by which it was passed previously. How many people rejected the bill initially? a) 500 b) 400 c) 300 d) 200 Questions 24-25 A dry ration retailer buys two grades of sugar from a wholesaler, one costing Rs 15/kg and another costing Rs 20/kg. He buys 10 kg of the first variety and some kg of the second variety and decides to make a profit of 20 % on the mixture. However, one day, he gets confused with the



numbers written on the sack and sells 10 kg of the mixture for an amount equal to the total quantity of the mixture. He realizes his mistake later in the day and sells the balance sugar at the predetermined selling price. He does not lose any money in this imbroglio. (Numbers on the sack indicate only quantity and price) 24) How many kg of the second variety of rice did he buy? a) 6 kg b) 4 kg c) 5 kg d) 10 kg 25) If he wants to get his profit of 20% after the mix-up, at what price should he sell the balance sugar? a) Rs 20/kg b) Rs 30/kg c) Rs 25/kg d) cannot be determined

TEST-MISCELLANEOUS ARITHMETIC No. of questions- 25 Time-50 minutes 1)A bottle contains 100% phenyl. 1/3 of it is replaced with water. The operation is repeated 4 times. What is the ratio of phenyl: water after the last operation a) 16:81 b) 1:5 c) 14:82 d)none of these 2) A mixture of 125 litres of milk and water contains 20% water. What amount of water needs to be added to this milk-water mixture in order to increase the percentage of water to 25% of the new mixture? a) 7 b) 5.66 c) 8.33 d) none of these 3)There is a bottle of Coke . 25% of it is replaced by Pepsi. Again 25% is replaced by Pepsi and finally another 25% is replaced by Pepsi. What is the approximate final percentage of Coke in the bottle? a) 56 b) 52 c) 66 d) 42 4)Two casks contain mixtures of petrol (P) and kerosene (K). In the first cask P:K=7:3 and in the second cask P:K= 3:1. In what ratio should the mixtures from the two casks be taken to give a mixture in which P:K is 11:4? a) 2:3 b) 2:2 c) 1:2 d) 1:3 5)A rice trader sells one type of rice at Rs 2.70/ kg and loses 10%. He sells another type of rice at 4.5/kg and gains 12 ½ %. His objective is to mix these two types and sell the mixture at 3.95/kg and get a profit of 25%. What should be the ratio of quantities of rice mixed? a) 23:6 b) 84:16 c) 6:1 d) none of these 6) A student finds the average of 10 positive integers. Each integer contains two digits. By mistake, the boy interchanges the digits of one number say ba for ab. Due to this, the average becomes 1.8 less than the previous one. What was the difference of the two digits a and b?

a) 2 b)4 c)6 d)none of these 7) C and D are the children of A&B, who are married to each other. D is 5 years younger than C. Five years ago, A was twice as old as C. The current age of B is the average of A and C’s ages. If the average age of all of them is 34 years, find the current age of D? a) 18 b) 21 c) 22 d) 27 8) Arun invests in bonds worth Rs. 40000 and he receives a 15% interest per annum. How much money should he invest in bonds that give 10% per annum so that he gains overall 12% per annum? a) 50000 b) 65000 c) 60000 d) 62500 9) The average weight of a group of 8 girls is 50 kg. If 2 girls R and S replace P and Q, the new average weight becomes 48 kg. The weight of P= Weight of Q and the weight of R= Weight of S. Another girl T, is included in the group and the new average weight becomes 48 kg. Weight of T= Weight of R. Find the weight of P? a) 48 b) 52 c) 46 d) none 10) Disha wants to paint her room a unique brown. For that she uses a certain violet paint which contains 30 % blue pigment and 70 % red pigment by weight. She mixes that violet paint with a certain green paint which contains 50 % blue pigment and 50 % yellow pigment. When these paints are mixed to produce the brown paint which she uses, it contains 40 % blue pigment. If the brown paint weighs 10 grams, then the red pigment contributes how many grams of that weight? a) 2.8 gms b) 3.5 gms c) 4gms (d) 5.3 gms 11) A set S of real values consists of the following 5 elements, {4, 8, 12, 16, x} in any order. For how many values of x does the mean of set S equal the median of set S? a) 0 b) 1 c) 2 d) 3 12) The average age of ‘n’ persons is 55. Two persons aged 41 and 47 respectively leave the group and three new persons aged between 60 and 65 join the group. If the average age of the group goes up by one year and if the initial number of people in the group is a multiple of 7, what is the number of persons now in the group? a) 42 b) 43 c) 44 d) 45 13) Manju, who works at Ganesh Fruit Juice Centre, prepares 51 glasses of milkshake in 4 min, 18 sec in his first shift. After having a sandwich, he prepares 73 glasses in 7 min 13 sec. After another sandwich, he prepares 112 glasses in 12 min, 24 sec. The container in which the milkshake is prepared contains a maximum of 9 glasses of a drink at a time. Find his approximate average time of



preparation of one container of the drink, if he starts with a fresh container each time. a) 48s b) 50s c) 40 s d) 51 s 14) There is a bag with a mixture of sugar, rice and wheat grains. The quantity of rice is 4/5

th that of wheat and the

quantity of sugar is 3/4th

that of rice. In 12 kg of the mixture how much wheat and sugar will be there? a) 7 kg b) 10 kg c) 8 kg d) none of these 15) At a gold shop’s annual promotion, gold coins are distributed with each purchase. The higher the purchase, the higher the value of the gold coin. The value of the gold coin is directly proportional to the thickness & the square of the diameter. Two gold coins are given to a customer. The value of coin 1: coin 2 =4:1. The diameter of the coins are in a ratio 4:3. Find the ratio of the thickness of the coins? a) 9:6 b) 11:7 c) 9:4 d) 12:5 16) There are 5 friends in a class who get marks in the ratio 3:4:5:6:7. the maximum marks are the same in each case. The total marks obtained by the 5 are 3/5

th the maximum

obtainable marks. How many friends got more than 70% of the total marks? a) 1 b) 2 c) 3 d) 4 17) The charges for a graphics designer are partially fixed and partially variable with the number of hours. The charge is Rs 550 for 4.5 hours and Rs 300 for 2 hours. Find the charges of 1 hour? a) 100 b) 200 c) 300 d) 400 Questions 18-19 In Little flowers school, for a school drill, the students are divided into 2 groups of lilies and daisies. The ratio of lilies: daisies = 8:3. the ratio of boys: girls is 7:4. 60% of the Daisy group is boys. 18) What is the ratio of the number of girls in the lily group and the number of boys in the daisy group? a) 20:13 b) 11:20 c) 14:9 d) cannot be determined 19) What is the difference in the number of boys who are in the lily group and the number of girls in the daisy group, given that there are 48 girls in the daisy group? a) 200 b) 240 c) 160 d) none of these 20) Till the age of 21, Arun grows such that his height varies as the square root of his age. When Arun is 9 years old, his height is 4 feet. What is his height at 16 years? a) 5 feet b) 5 feet 4 inches c) 6 feet d) 6 feet 2 inches

21) The height of an apple tree is given by the sum of two terms. The first term is directly proportional to the weight of the tree and the second term is directly proportional to the square of the weight of the tree. The height of the tree was 6 m and 8 m, when its weight was 40 kg and 50 kg respectively. Find the approximate weight of the tree when its height was 5 m. a) 39 kg b) 25 kg c) 32 kg d) 35 kg 22) A picnic invites two kinds of charges: bus fare, which is independent of the number of people attending the picnic and buffet lunch, which increases directly with an increase in the number of people. The charges are calculated to be Rs. 165 per head when there are 200 invitees and Rs. 170 per head when there are 150 invitees. What would be the charges per head when there are 100 invitees? a) Rs.175 b) Rs.180 c) Rs.185 d) Rs.190 23) Praveen buys two cans of milk from a manufacturer. The two cans are diluted to the same extent. To can 1 Praveen replaces 10 litres of the solution with pure milk and the concentration of milk becomes twice of what it was. To can 2 he replaces 20 litres with pure milk, what is the ratio of concentration of can 1 initially to that in can 2 finally. a) 1:1.5 b) 1:2 c) 1:3 d) cannot be determined 24) The receipts on railway travel vary as the excess of speed of the train over 30 kmph. The expenses vary as the square of that excess. What is the speed at which the profits will be greatest if at 60 kmph is the expenses are just covered? a) 40 b) 35 c) 45 d) None of these 25) There are 3 types of Coffee powder available at La Café. They are graded based on the chicory content. The three grades of coffee contain 1% chicory, 2% chicory and 3% chicory in the coffee blend. If p kgs of the 1% grade and q kgs of the 2% grade and r kgs of the 3% grade are mixed to give p+q+r kgs of 1.5% grade, what is p in terms of q and r?

a) q+3r b) 𝑞+𝑟

4 c) 2q + 3r d) 4.5q+3r



BASIC &COMMERCIAL ARITHMETIC-

SOLUTIONS 1) We need to use Compounded Annual Growth Rate (CAGR) for this question Conventionally, CAGR is computed using the formula

[ 𝐹𝑖𝑛𝑎𝑙

𝐼𝑛𝑖𝑡𝑖𝑎𝑙

1

𝑛 − 1]

Using the reverse gear approach, we can obtain the answer much faster 𝐹𝑖𝑛𝑎𝑙

𝐼𝑛𝑖𝑡𝑖𝑎𝑙=

3000

780≈ 3.8

We need to look for something close to 3.8 in the answer options (Here n=5) Choose a middle answer option, assuming the answer is 30% (1.3)

5 = (1.3)

2 ×(1.3)

3 = 17×17×1.3 = 3.75. Hence, the answer

will be slightly higher than this. = 31%. Option (b) 2) 12% of profit = 1200 =>Remaining = 8800 After taking 1000 each, amount remaining = 6800. Ram’s

share = 3

8× 6800 = 2550

Ram’s profit = 2550+1000= 3550. Answer is option (d) 3) If the number of shares remain constant, then p has to be greater than q. answer is option (c) You can verify this by taking some simple numbers 4) option (d) Cost price= 40x65= 2600 and Cotton at 80 per kg= 32 x 80 = 2560 Cotton seed at 20 per kg = 4.8 x 20 =96. Therefore, %

change = 56

2600× 100 = 21.32%

5) If we assume that today’s population is 100, next year it would be 1.1 × 100 = 110, and the following year it would be 1.1 × 110 = 121. If this is double the population of one year ago, the population at that time must have been 0.5 × 121 = 60.5. Because the problem seeks the “closest” answer choice, we can round 60.5 to 60. In this scenario, the population has increased from 60 to 100 over the last year, a net increase of 40 residents. To determine the percentage increase over the last year,

divide the net increase by the initial population: 40/60 = 4

6=

2

3, or roughly 67%. Option (d)

6) Let x be the number of students, let n be the number of girls. Then 40/100*x =n& 90 =x/100 *n. Thus, x= 150 7) Conventional method: Let x be the quantity of apples bought and p be their cost price per kg.Let the amount he sold on the first day be y,

hence 2py=px, 𝑦 =𝑥

2= 0.5 x

Let him sell z apples on the second day where 2py+1.5pz=1.3 px=>1.5 z=0.3 x=>z=0.2x

So his net profit ={ 1.3 𝑝𝑥 + 0.3 𝑥 1.25 𝑝

𝑝𝑥 -1} X100 = (1.3 + 0.375

-1)X100 = 67.5% . Shortcut:- Assumption: Let him buy 3 kg of apples at Rs 10 per kg. Hence his total cost price is Rs. 30.On the first day he should sell at 100% profit, i.e. at Rs 20 per kg, and he is supposed to break even, hence he can sell 1.5 kg of apples. Now on day 2 he sells at 50% profit, i.e. at Rs 15 per kg, and makes a profit of 30% overall. i.e. at the end of day 2 he should have Rs. 39. From day 1 he has Rs 30, so on day 2 he gets Rs. 9.Therefore he sells 6 kg at Rs. 15 to get 9 Rs. He is left with 0.9 kg which he has to sell at 12.5 Rs so he gets 11.25 Rs. In total he gets 30 + 9 + 11.25 = 50.25. Hence his profit is

20.25

30 × 100 = 67.5 %.

8) Let the cost of an apple be a and that of an orange be b.

Then𝑏

2𝑎+3𝑏 =

𝑝𝑟𝑜𝑓𝑖𝑡 𝑃𝑒𝑟𝑐𝑒𝑛𝑡𝑎𝑔𝑒𝑜𝑓𝐴

100 =

20

100=

1

5

a = b.Let x and y be the cost price of an apple and an orange respectively for B. As a = b , x = y ( as CP of both fruits would have risen by the same ratio)

Now profit % of B = 𝑥

2𝑥+2𝑦 × 100 =

100

4= 25%

9) The amount after n years will be (1.1)(1.2)(1.3)….. (1+ 0.n) times the initial amount after n years. By observation we can see that (1.1)(1.2)(1.3)(1.4)>2 and(1.1)(1.2)(1.3) <2. Hence, the amount becomes double in the fourth year. 10) To make calculation simpler, take a number common to 11 and 9 99.

CP of 99 oranges= 99𝑥16

9 = Rs 176 SP of 99

oranges= 99 ×20

11= 𝑅𝑠 180

Profit on 99 ORANGES= Rs 4 and % profit = 4

176× 100 =

2.272% 11) Given that 10% discount = 0.9 MP= Rs 81 =>MP= 90. SP, when discount is 20% = 0.8x 90 =Rs 72



12) Here we can assume a value for x. Say x= 36. Dividing 36

6= 6 Multiplying 36x6 = 216

% change= 216−6

6× 100 =

210

6× 100= 3500%

13) The winner will get 100-20= 80% of the votes 80% of the votes represent 128000

100% will represent = 128000 𝑥100

80= 160000

14) Original value = 392500 + 230000 = 622500. Final value = 3925000

Percentage depreciation = 𝑑𝑒𝑝𝑟𝑒𝑐𝑖𝑎𝑡𝑖𝑜𝑛

𝑖𝑛𝑖𝑡𝑖𝑎𝑙 𝑣𝑎𝑙𝑢𝑒x 100 =

230000

622500𝑥 100 = 36.95%

15) Use the formula 𝑟

100+𝑟× 100 % =

20

120 𝑥100 =

100

6 =

16.67%. SHORTCUT **You can also use the constant product rule. Increase of

20% 1

5 will result in a decrease of

1

6= 16.66% so that

the expenditure remains constant. 16) option (d) She has scored 66+ 68 = 134 marks already. She needs 220 marks to gain admission into MSc Biological Science = 220-134 = 86 marks more. Given that she gains admission only into Biological Science, hence she can’t get more than 59 in Physics. To minimize Math, we need to maximize Physics = 59. Marks left = 86-59 = 27 marks for Math as a minimum 17) option (c)

ITEM QUANTITY (kg)

Price/kg 2000

Price/ kg 2005

Wheat 25 12 15

Rice 10 8 12

Coffee 5 14 16

Sugar 2 40 36

Salt 1.5 60 90

Cost of living in 2000 = 25(12) +10(8) + 5(14) + 2(40) + 1.5 (60) = 620 Cost of living in 2005 = 25(15) + 10(12) + 5(16) + 2(36) + 1.5 (90) = 782 Taking 2000 as the base year, cost of living index for

2005 =782

620× 100 = 126.3

18) option (d). Let’s say the value of the index = 100. Then HBL = 7, Niposys = 13 & Brilliance= 1 Others = 100-21 = 79 Increased index value = 106 Increased individual values HBL = 1.09x7=7.63, Niposys = 14.3 & Brilliance = 1.04 Total = 22.97 Others = 106- 22.97=83.03

Increase = 83.1−79

79 𝑥 100 = 5.2%

19) option b Go from answer options If x= 300, and interest = Rs 15

Using SI= 𝑃𝑥𝑅𝑥𝑇

100, we get R=5%

Look at the answer options. By plugging in values, only 300 & 5% will match. All other values will give a different rate of interest value. 20) option (b) Let the CP of 1 litre = Rs 100, SP = Rs 75 SP of milk man includes = Milk man’s CP + 12.5% of milk man’s CP

75 = 1 +1

8𝐶𝑃

9

8𝐶𝑃 = 75 =>CP= 66.67.

For 100 rupees, he actually sells 666.7 ml. every litre has 666.7 ml of pure milk. For 2 litres = 666.7 x 2 = 1333.4 ml 21) option a Go from answer options Let the Price of each chocolate= 50 Therefore, 15 chocolates= 750. The order will be as follows 750(P) 825 (Q) 650 (R) 780 (S) 858 (T) 1287 (U) 22) option (a) Out of 3000 books, 500 are given free and from the balance 2500, for every 24 books one book is given free. Hence, 2400 books are sold at 75% of 3.25. Thus, SP = 2400 x 3.25 x 0.75, CP = 2400. Gain percentage = 143.75. 23) option (d) There are totally 600 voters, so after the number of people who initially rejected increases by 150%, it must still be below 600. That is true only for option (d). Verifying the option: Initially, Against = 200 & For = 400. Hence, majority = 400-200 = 200 After discussions, against = 200 + 1.5x200 = 500, For = 600-500 = 100. Hence, majority = 500-100 = 400, which is twice the initial majority. Thus, this is the correct answer. 24) option (c) The best way to solve this question is by substituting answer options. Let’s assume that he buys 5 kgs of 2

nd variety of sugar.

Total CP = 10x15 + 5x20 = 250 Profit of 20% implies a SP of 1.2 x 250 = 300 The mixture of 15 kg would be labeled as Rs 300 or Rs 20/kg He sells 10 kgs at Rs 15/kg. he gets Rs 150. His balance of 5 kgs, he sells at Rs 20, from which he gets Rs 100 the total cost (150 +100=250) Hence, option (c)



25) Option (b) To get a profit of 20%, he needs to get 300- 150 = Rs. 150

more from 5 kgs . 150

5 =Rs 30 / kg. Answer is option (b)

MISCELLANEOUS ARITHMETIC-

SOLUTIONS 1) option (d) 𝐿𝑖𝑞𝑢𝑖𝑑 𝐴 𝑙𝑒𝑓𝑡 𝑎𝑓𝑡𝑒𝑟 𝑛𝑡𝑕 𝑜𝑝𝑒𝑟𝑎𝑡𝑖𝑜𝑛

𝐼𝑛𝑖𝑡𝑖𝑎𝑙 𝑞𝑢𝑎𝑛𝑡𝑖𝑡𝑦 𝑜𝑓 𝐴 𝑖𝑛 𝑡𝑕𝑒 𝑣𝑒𝑠𝑠𝑒𝑙 =

𝑎−𝑏 𝑛

𝑎𝑛

Amount of phenyl remaining = 1 −1

3

4

=16

81

Therefore, ratio or phenyl:water = 16:65. (81-16=65) 2) option (c) We need to find out how much of a solution of 100% water needs to be added to a solution containing 20% water to attain a dilution of 25%. This can be found out as follows

Ratio = 75:5 = 15:1 That is for 15 parts of a 20% water solution, one part of 100% water solution needs to be added. Therefore, for a

solution of 125 litres, 125

15= 8.33 litres needs to be added

3) option (d) Answer is based on (100-25)

3= 75

3 . This translates to

42.18% 4) option (b) Using alligation, we can directly solve this question Take either petrol or kerosene.

Petrol in cask 1 = 7

10 Petrol in cask 2 =

3

4

Petrol in mixture = 11

15

Ratio to take the mixture =

1

60:

1

30= 1: 2

5) option b This question should be solved using alligation. For alligation; always consider the cost price only.

Selling Price (SP) Cost Price(CP) SP of rice 1 = 2.70/kg CP = 2.7/0.9 = 3/kg SP of rice 2= 4.5/kg CP= 4.5/1.125 = 4/kg SP of mixture = 3.95/kg CP= 3.95/1.25 = 3.16/kg

Required Ratio = 84:16 6) option a We can infer that the difference between ab and ba is 18. Now all numbers of the type 13, 24,35. I.e. the difference is 2 will give a difference of 18 when reversed. Therefore difference of a and b will always be 2. 7) Let the present age of each of them be the letters themselves. Present age of D= C-5 Present age of A will be 2(C-5)+5 =2C-5 and the present age of B will be (C+2C-5)/2 = (3C-5)/2 Average age of members =34

𝐶 + 𝐶 − 5 + 2𝐶 − 5 +3𝐶−5

2= 34𝑥4

4𝐶 +3𝐶−5

2= 146 11C= 297

C=27 D= 27-5=22 8) Using alligation

He should invest in the ratio 2:3. He should invest 60000 9) option (d) 8 x 50 +R+S-P-Q= 48x8 R+S-P-Q=-16 P+Q-R-S= 16 R=S and P=Q P-R=8 One more person is included and the weight = 48 kg Let the weight be a 48𝑥8 +𝑎

9= 48

a=48 kg= weight of R weight of P= 48+8= 56 kg



10) Option (b) Use Alligation Considering only the blue pigment in all the paints

Thus, 5 gms of each paint is used. Weight of red paint = 7

10𝑥 5 = 3.5 gms. Option (b)

11) option (d) To solve this problem quickly, you can come up with likely values for x that would make the mean equal to the median. Three values that would make the set symmetrical are 0, 10, and 20: {0, 4, 8, 12, 16} {4, 8, 10, 12, 16} {4, 8, 12, 16, 20} To verify if there are any other possibilities for x, we need to evaluate each range of “x” (1) If x is less than or equal to 8, then the median is equal to 8. We now set the mean equal to the median: 40 + 𝑥

5= 8 40 + x = 40 x = 0

(2) If x is between 8 and 12, then the median is equal to x. 40 + 𝑥

5= 𝑥 40 + x = 5x 40 = 4x x = 10

(3) If x is greater than 12, then the median is equal to 12. 40 + 𝑥

5= 12 40 + x = 60 x = 20

12) Option (b) Let 𝑥, 𝑦, 𝑧 be the ages of 3 persons which lies between 60 and 65

The equation governing the problem 55𝑛−41−47+ 𝑥+𝑦+𝑧

𝑛+1=

56 Also given n is a multiple of 7 55n – 88 +k = 56n + 56 where k =𝑥+𝑦+𝑧, n+ 144=k Now k ranges between 3x61 and 3x64 183 ≤n +44 ≤ 192 39 ≤n ≤4 Thus n =42 The latest number n + 1 =42+1=43 Shortcut! Go from answer options. The initial number of people in the group is a multiple of 7. The answer options give the number of initial people + 1. Thus the correct answer option is option (b) 13) option (d) In the first shift, he uses the container 6 times. 45 glasses in 5 rounds and the remaining 6 glasses in the 6

th

round. Similarly, he prepared 73 glasses in 9 rounds, and 112 glasses in 13 rounds. Hence, the required average

= total time/ total number of rounds = 4 min 18𝑠 + 7 min 13𝑠 + 12 min 24𝑠

6+9+13=

23 𝑚𝑖𝑛 +55𝑠

28 = 51 seconds

14) answer=option c If we take the amount of sugar as 3x, then rice = 4x and Wheat = 5x (ratio = 3:4:5; as rice= 4/5 wheat and sugar = 3/4 rice) 12x= 12 x=1 Wheat = 5 kg and Sugar = 3 kg. Total = 8 kg 15) option c V Value d diameter t thickness given that V α d

2t ratio of diameters = 4:3 and

thickness = 𝑡1: 𝑡2 𝑉1

𝑉2=

16 𝑥 𝑡1

9 𝑥 𝑡2 4 =

16𝑡1

9𝑡2 𝑡1: 𝑡2 = 9: 4

16) Lets assume that the maximum marks are 100. Total marks available =500

Aggregate = 3

5𝑥 500 = 300

Which is divided in the ratio 3:4:5:6:7 Marks are 36, 48,60, 72 and 84. Only 2 students get more than 70% 17) The total charge = Fixed component +Variable component Z= F+kx Where F and x are fixed 550 =F +k(4.5) 300=F + k(2) Solving simultaneously, F= 100 and k= 100 Charges for 1 hour = 100 +1(100) =200 Questions 18-19 Since 8+3=7+4=11, the safest assumption is that there are 110 students in total. Then, the table will look like this

BOYS GIRLS

LILIES 52 28 80

DAISIES 18 12 30

70 40 110

Question 18 28:18 = 14:9 Question 19 1.2x=48 => x= 40 So the number of boys in lilies will be 5.2 × 40 = 208 So the difference will be 208-48 = 160 20) option (b)

Relationship given:- 𝐻 𝛼 𝐴

𝐻 = 𝐾 𝐴

4 = 𝐾 9 𝐾 =4

3

Height of Arun at 16 years H=4

3× 4 =

16

3= 5 feet 4

inches



21) option (d) Let the weight and height of the tree be ‘w’ and ‘h’ respectively. Thus, 𝑕 = 𝑎𝑤 + 𝑏𝑤2 where a and b are constants. Here, 6 = 40𝑎 + 402 × 𝑏 8 = 50𝑎 + 502 × 𝑏

Hence, 𝑎 =11

100𝑎𝑛𝑑 𝑏 =

1

1000

If h = 5,

5 = 11

100× 𝑤 +

1

1000× 𝑤2

Solving, w = 35 approximately. 22) option (b) Let the bus fare = Rs. x and buffet lunch per head = Rs. y x + 200y = 165 × 200 …(i) x + 150y = 170 × 150 …(ii) Then, x + 100y = 2 × (ii) – (i) = 51000 – 33000 = 18000 Therefore, Cost per head = Rs. 180 23) If the addition of 10 litres of pure milk makes the concentration twice of what it was before, then the amount added will be the amount of milk present initially. Replacement of 20 litres, will be the replacement with two 10 litres, making the concentration thrice of what it was initially. Therefore, the ratio = 1:3 24) Let the excess of speed over 30 kmph=S Receipts= R and Expenses=E Then R=K1 S And E= K2S

2

Also, at 60kmph, R=E Thus, K1=30 K2 We need to maximize R-E= K1S- K2S

2=SK2(30-S)

S+30-S, the sum is constant, the product will be maximum when they are equal Thus S=30-S => S=15. Speed = 30+15= 45 25) option a Put r=0, then if 1 kg of 1% grade is mixed with 1 kg of 2% grade, we get 2 kg of 1.5%grade Hence p=1, q=1 and r=0. Substitute in answer options to get answer as option (a)

Basic & Commercial Arithmetic - Byju'sbyjus.com/.../ias_files/basic_and_commercial_arithmetic.pdfThen (c + d) = 1.2 (a + b) Also b = 30, d = 40, c = 80 80 + 40 = 1.2 (a + 30) => 120

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