WBCS Main level Arithmetic solution set 8 | SureSolv

WBCS Main level Arithmetic solution set 8

Eighth solution set for WBCS Main level Arithmetic

wbcs main level arithmetic solution set 8

Compulsory paper VI of WBCS Main is on Arithmetic and Test of Reasoning. It has 200 questions to be answered in 180 minutes. In this eighth solution set on Arithmetic, easy and quick solutions for 10 questions in the corresponding question set are explained. You should answer the question set first and then go through the following solutions.

In each solution we will first describe how we solve the problem mentally in a few tens of seconds, and then will show the mathematical deductions, if any. When needed, we will also explain the supporting concepts.


8th WBCS Main level Arithmetic solution set: time to answer was 10 mins

Problem 1

Fraction equivalent of $\displaystyle\frac{2}{3}$% of $\displaystyle\frac{5}{8}$% is,

  1. $\displaystyle\frac{5}{12}$
  2. $\displaystyle\frac{1}{240}$
  3. $\displaystyle\frac{1}{24000}$
  4. $\displaystyle\frac{4}{37500}$

Solution 1: Solving in mind

The result will be product of the two percentage figures, $\displaystyle\frac{5}{12}$ divided by 100 twice. The final result is then,

$\displaystyle\frac{1}{24000}$.

Answer. Option c: $\displaystyle\frac{1}{24000}$.

Solution 1: Explanation 

In arithmetic, "of" represents multiplication. So the problem is transformed to the mathematical form,

$\displaystyle\frac{2}{3}\text{%} \times{\displaystyle\frac{5}{8}\text{%}}$

The result will be product of the two fractions, $\displaystyle\frac{5}{12}$ divided by 100 twice, one each for the each of the two percentages,

$\displaystyle\frac{5}{12}\times{\displaystyle\frac{1}{100\times{100}}}$

$=\displaystyle\frac{1}{24000}$

Answer: Option c: $\displaystyle\frac{1}{24000}$.

For understanding the concepts and use of percentage, one of the key elements of arithmetic problem solving, you should refer to our tutorial,

Basic and rich concepts on pecentage.

In any case following is the relevant percentage concepts.

Related percentage concepts

A percentage figure, say, 24 percent, expressed as 24% represents,

24 portions or 24 units for each 100 portion or 100 unit of the whole.

A single percentage figure is actually a ratio of two variables, the variable represented by the percentage, and the whole.

Percentage is used for compact comparison of two values.

For example, a frequently used statement is like, "I have got 80% in maths". The two variables here are the Maths marks, and the Total maths marks. When the total marks is 200, and the actual score is 160.

Percentage doesn't say anything about the actual values of the two variables, it just indicates a specific type of ratio, 80 for each 100.

As percentage is a ratio with denominator always as 100, to express a percentage as a fraction, we have to actually divide the percentage figure by 100 and reduce. In this case, 80% means, $\displaystyle\frac{80}{100}=\frac{4}{5}$.

Depending on the math problem, we automatically convert a percentage to its equivalent fraction, or many times, to its equivalent decimal.

Decimal equivalent of 80% will be, 0.8. Dividing by 100 makes percentage to decimal conversion instantaneous, just shift the decimal point two positions left.

Concepts: Percentage concepts -- Percentage to fraction conversion -- Percentage to decimal conversion -- Percentage of percentage.

Problem 2

$\displaystyle\frac{4}{5}$th of a number is 64. 70% of the number is,

  1. 44.8
  2. 70
  3. 42
  4. 56

Solution 2: Solving in mind

$\displaystyle\frac{4}{5}$th of the number is 64.

So, $\displaystyle\frac{8}{10}$th or 80% of the number is 64.

10% of the number is 8.

70% of the number is 56.

Answer: Option d: 56.

Solution 2: Explanation

To convert $\displaystyle\frac{4}{5}$ to percentage multiply by 100 and get 80%.

As percentage to fraction conversion is done by dividing by 100, for the reverse process of fraction to percentage conversion, we need to multiply the fraction by 100.

Next is to apply three step Unitary method as below,

80% of the whole number is 64,

So, 10% of the whole number is $\displaystyle\frac{64}{8}=8$,

Then, 70% of the whole number is $7\times{8}=56$.

Note: We don't have to find the number, we find the desired 70% of the number directly.

Answer: Option d: 56.

Concepts: Fraction to percentage conversion -- Unitary method.

Applying Unitary method is a basic skill in arithmetic problem solving. In the following section we explain the method in some detail. If you are fully conversant with method, just skip the section.


Unitary method: Essential math skill: ES

This is one of the most basic tools in your arithmetic problem solving toolset and you should be thoroughly clear about the concept and its applications.

Problem example: If Ramesh purchased at Rs.30 twenty number of pens, how much he would have to pay for purchasing 50 pens?

Situation of use: In Unitary method, there are two variables (of different units) that are proportional to each other. Let us assume number of pens and its total cost of purchase are the two variables. The price per pen is constant and the number of pens and its total cost of purchase are directly proportional to each other. If you purchase now double the number of pens you had purchased in an earlier occasion, you have to pay double the amount you had paid in the first occasion. If follows common sense as well as mathematical requirements. There as innumerable pairs of such variables in real life situations.

Method of use: Unitary method is a simple three-step method that can be processed mentally with ease.

First step: Describe the initial first case situation: Ramesh purchased 20 pens at 30 rupees.

Note: The two variables are, number of pens purchased and total cost of purchase that are proportional to each other.

An important rule: In the first case description, the variable for which we want to find the value must be placed second in sequence, with independent variable first in sequence. Here we have mentioned independent variable number of pens first and dependent variable, total cost of pens second, as we want to find the total cost of purchase on a later second occasion for a given value of the number of pens.

Second step: Find the value of dependent second variable for 1 unit of independent first variable; this is why the name of Unitary method: Ramesh purchased 1 pen at $\displaystyle\frac{30}{20}=1.5$ rupees; divide the right position variable value by left position variable value.

Third step: Find the desired value of total cost by multiplying cost of unit number of pens by number of pens purchased this time: Total cost of 50 pens is, $50\times{1.5}=75$ rupees.

Usually if you follow the rules, and know where unitary method needs to be applied, you can carry out the three steps in mind. A large number of Arithmetic problems need application of unitary method.


Problem 3

If $\displaystyle\frac{1}{5}:\displaystyle\frac{1}{x}::\displaystyle\frac{1}{x}:\displaystyle\frac{1}{1.25}$, then $x$ is,

  1. 2.5
  2. 1.5
  3. 2
  4. 3.5

Solution 3: Solving in mind

While solving in mind, for avoiding the inconvenience of dealing with inverses, we substitute, $\displaystyle\frac{1}{x}=a$ and convert the proportion to the equation form,

$\displaystyle\frac{1}{5}:a=a:\displaystyle\frac{1}{1.25}$.

Cross-multiplying, we obtain, $a^2=\displaystyle\frac{1}{6.25}$. 

As $2.5^2=6.25$, $x$ gets the value 2.5.

Answer: Option a: 2.5.

Solution 3: Deduction

Given,

$\displaystyle\frac{1}{5}:\displaystyle\frac{1}{x}::\displaystyle\frac{1}{x}:\displaystyle\frac{1}{1.25}$

Or, $\left(\displaystyle\frac{1}{x}\right)^2=\displaystyle\frac{1}{5}\times{\displaystyle\frac{1}{1.25}}=\displaystyle\frac{1}{6.25}$

Or, $\left(\displaystyle\frac{1}{x}\right)^2=\left(\displaystyle\frac{1}{2.5}\right)^2$

Or, $x=2.5$.

Concepts: Ratio proportion -- Fraction arithmetic.

Problem 4

What number should come next in the sequence, 4, 5, 7, 10, 11, 13, 16, —?

  1. 21
  2. 19
  3. 18
  4. 17

Solution 4: Solving in mind

To discover the inherent pattern in a given series of numbers, we calculate the difference between the adjacent pair of numbers as well as note the sign of the result.

For the first three pairs, the differences we find to be, 1, 2, and 3. Next three pairs also produce the same difference 1, 2 and 3 thus confirming the inherent pattern of forming each term of the sequence,

Every consecutive three pairs of terms in the number series will have positive difference of 1, 2 and 3.

Next number in the sequence, the seventh term, being the first term of difference triplet 1, 2 and 3,  will then have a positive difference of 1 with its previous term 16.

So the next term in the sequence must be 17.

Answer: Option d: 17.

Concepts: Identification of inherent pattern in a series of numbers -- number system -- Pattern identification technique.

Problem 5

The perfect square number of least value divisible by 3, 4, 5, 6, and 8 is,

  1. 900
  2. 3600
  3. 1600
  4. 1200

Solution 5: Solving in mind: First method by examining choice values

By examining the given three square value choices, 900 does not have 8 as a factor and 1600 does not have 3 as a factor. So the remaining square value choice 3600 must be the answer. On test with given numbers indeed it is found to be divisible by all five given numbers.

Solution 5: Solving in mind: Second method of counting square factors in the product

Among the given numbers, the factors 2, 3, and 4, occur twice, but 5 occurs only once, whereas $2^2=4$ is already present in 4. So multiplying the product of the numbers by 5 and dividing by $2^2=4$ will be the desired least square number that is divisible by all the five given numbers.

The desired number is,

$3\times{5}\times{6}\times{8}\times{5}$

$=3^2\times{4^2}\times{5^2}$

$=3600$.

Answer: Option b: 3600.

Concepts: Factorization -- Squares of numbers -- Choice test -- Counting of square factors and eliminating common square factors in the product of the numbers. We call this as Factor analysis.

Problem 6

A tree increases annually by $\displaystyle\frac{1}{8}$th of its height. What will be its height after 2 years if it stands today 64 ft high?

  1. 72 ft
  2. 74 ft
  3. 81 ft
  4. 80 ft

Solution 6: Solving in mind

In first year the growth will be 8 ft and total height will become 72 ft. In second year growth will be another 9 ft and total height will be 81 ft.

Answer: Option c: 81 ft.

Solution 6: Explanation

In a year as the tree grows $\displaystyle\frac{1}{8}$ of its height at the start of the year, the growth is Compound growth same as Compound interest. 

First year growth is easy to determine. It will be 8 ft.

But in the second year growth will not only be $\displaystyle\frac{1}{8}$ of the first year beginning height of 64 ft, that is 8 ft. There will be an additional growth of $\displaystyle\frac{1}{8}$th of the first year growth of 8 ft, that is, 1 ft. In total then in two years the tree will grow, $8+8+1=17$ft from its original height of 64 ft.

If you use the compound interest or compound growth formula,

$\text{Final height}=\left(1+\displaystyle\frac{1}{8}\right)^2\text{Original height}$

$=\left(\displaystyle\frac{9}{8}\right)^2\times{64}$ft

$=81$ft.

Concepts: Compound growth same as compound interest -- Compound growth as growth on growth -- Growth always as compound growth.

Problem 7

In a school, the ratio of boys and girls is $4:5$. When 100 girls leave the school, the ratio becomes, $6:7$. How many boys are there in the school?

  1. 1200
  2. 1300
  3. 1600
  4. 1500

Solution 7: Solving in mind: Using factor multiple concept on the ratio values

For quickly solving such problems mentally we often use the factor multiple concept on the ratio values.

Let us explain how.

The number of boys does not change. So it must be a multiple of both the ratio numerators, 4 and 6. Only 1200 and 1500 of the four choice values satisfy the criterion. Testing with 1200, in the first case number of girls must be, $1200\times{\displaystyle\frac{5}{4}}=1500$. When it is decreased by 100, it becomes 1400 and the ratio becomes, $\displaystyle\frac{1200}{1400}=\frac{6}{7}$.

We are sure of the answer as 1200, but still for curiosity, also check 1500 as number of boys.

In this case, the number of girls must be, $1500\times{\displaystyle\frac{5}{4}}=1875$. Reducing it by 100 result is 1775 which is not divisible by 7. So 1500 cannot be the number of boys.

Answer: Option a: 1200.

Concepts: Number system -- Ratio proportion -- Factor multiple test on ratio term values against choice values.

Solution 7: Mathematical deduction which is more familiar

Assuming $x$ as the cancelled out HCF we will reintroduce it by multiplying both the first ratio term values to get actual number of boys as $4x$ and number of girls as $5x$.

When 100 girls leave we have in the changed situation,

$\displaystyle\frac{4x}{5x-100}=\frac{6}{7}$.

Cross multiplying,

$28x=30x-600$,

Or, $2x=600$,

Or $x=300$.

So number of boys is $4x=1200$.

You can follow this method also to arrive at solution mentally in a few tens of seconds.

This is a familiar and so safer method.

Problem 8

A commission agent allows a rebate of 2%  to an investor while the company pays an interest of 15% on the investment. What rate of interest does the investor actually earn on his investment?

  1. $17$%
  2. $16\displaystyle\frac{3}{8}$%
  3. $15$%
  4. $15\displaystyle\frac{15}{49}$%

Solution 8: Roles of investor, commission agent and company

Being an agent, the commission agent sits between the two main parties, the investor and the company. Investor intends to invest the money on the company (every company needs outside money to run its business and grow). In return for the investment, the company assures a payment of Rs.15 for every Rs.100 invested.

Investor $\rightarrow$ Commission agent $\rightarrow$ Company

As the company and the investor don't know each other, the commission agent finds the interested investor for the company and helps to complete the transaction. For his efforts, he takes 2% of the investment amount, that is Rs.2 for every Rs.100 invested.

For every 100 rupees the investor invests with the company, the commission agent rebates or reduces that by 2%, or 2 rupees, and transfers Rs. 98 to the company. Effectively then, the company pays the commission of 2% to the commission agent and interest of 15% to the investor. It is important to note,

The percentages are always on the investment amount. For every Rs.100 invested, the commission agent gets Rs.2 as commission and the investor gets Rs.15 as interest.

Solution 8: Solving in mind

As the company pays an interest at the rate of 15%, it will pay the interest on Rs.100 and the interest will be Rs.15. Effectively by investing Rs.98 the investor will get a return of Rs.15. So the effective interest rate will be,

$\displaystyle\frac{1500}{98}=\frac{750}{49}=15\displaystyle\frac{15}{49}$%.

Answer: Option d: $15\displaystyle\frac{15}{49}$%.

Concepts: Investment -- Investment rebate -- Commission -- actual return on investment.

Note: From the company, the investor gets 15% on every Rs.100 he invests, but from the business of the company he gets an enhanced interest of $15\displaystyle\frac{15}{49}$% because he knows, out of every Rs.100 an amount of Rs.98 is actually invested in the business of the company.

Problem 9

In what proportion must water be added to spirit to gain 20% by selling it at cost price?

  1. $1:5$
  2. $2:3$
  3. $4:5$
  4. $3:5$

Solution 9: Solving in mind: Mechanism of mixing water and gaining profit

Finally 100 units of spirit mixed with water is sold at the cost price of 100 units of spirit to get a gain of 20%, which is $\displaystyle\frac{1}{5}$. So 100 units of spirit mixed with water must be, $\left(1+\displaystyle\frac{1}{5}\right)=\displaystyle\frac{6}{5}$ of the actual spirit volume in the mixture.

It follows then, for every 5 portions of spirit 1 portion of water is mixed with the spirit, that is, the ratio of mixing is, $1:5$.

Answer: Option a: $1:5$.

This is the shortest way to the solution based on clear understanding of mixing and concept of profit, which is,

Profit is always on the cost price.

Selling price is for 6 units of spirit, whereas its cost price is equal to 5 units of spirit. Gain is 1 price unit for evety 5 cost price unit, that is, 20%. We do not need to calculate any actual values, or even the sale price.

Concepts: Profit -- Ratio proportion -- Visualization of actually what happens -- considering all actions based on portions or suitable units, whatever that may be. This is possible because all important elements, including percentages are in terms of ratios.

Problem 10

A water tank can be filled up by two pumps in 20 minutes and 30 minutes respectively. They were started together but due to some technical fault, the first pump stopped after some time and the second pump worked alone for the last 10 minutes. How long did they work together?

  1. 4 minutes
  2. 8 minutes
  3. 10 minutes
  4. 6 minutes

Solution 10: Solving in mind

As the second pump fills the tank in 30 minutes, it would have filled $\displaystyle\frac{1}{3}$rd of the tank in the last 10 minutes. So, the rest $\displaystyle\frac{2}{3}$rd of the tank must have been filled by the two pumps working together till the first pump was working.

Together in every minute the two pumps fill,

$\displaystyle\frac{1}{20}+\displaystyle\frac{1}{30}=\displaystyle\frac{1}{12}$ th of the tank.

So to fill up $\displaystyle\frac{2}{3}=\frac{8}{12}$th of the tak the two pumps must have worked together for 8 minutes.

Answer: Option b: 8 minutes.

Concept: Work time -- Pipes cisterns -- Working together -- Work leftout.

Takeaway

We have used and explained the concepts and methods,

Percentage, Percentage to fraction, Fraction to percentage, Percentage to decimal, Percentage of percentage, Ratio proportion, Unitary method, Number system, Pattern identification technique, Number series, Factor analysis for least square number determination, Factorization, Choice value test, Growth as compound growth, Compound growth same as compound interest, Factor multiple technique for ratios, HCF reintroduction technique for ratios, Investment and commission, Profit by liquid dilution, Pipes and cisterns, Working together, Leftout work portion, and Visualization of events.

We have also shown how we process the steps mostly in mind thus speeding up the problem solving. This we consider to be within expanded scope of Mental maths.

In the process we have highlighted one of the important Essential Mental math skills of Unitary method.


Concept tutorials and articles on Arithmetic

Number system, fractions and surds

Numbers, Number system and basic arithmetic operations

Factorization or finding out factors

HCF and LCM

Fractions and decimals, basic concepts part 1

How to solve surds part 1 rationalization

How to solve surds part 2 double square root surds and surd term factoring

Ratio and proportion and mixing liquids

Ratio and proportion

Arithmetic problems on mixing liquids and based ages

How to solve arithmetic mixture problems in a few steps 1

How to solve arithmetic mixture problems in a few steps 2

Percentage

Basic and rich percentage concepts

Componendo dividendo

Componendo dividendo explained

Simple interest and compound interest

Basic and rich concepts on simple interest and compound interest

Work time, work wages and pipes and cisterns

How to solve arithmetic problems on work time, work wages and pipes and cisterns

How to solve time work problems in simpler steps type 1

How to solve time work problems in simpler steps type 2

Speed time distance, train problems and boats in rivers

Basic concepts on speed time distance, train running and boats and rivers

How to solve time distance problems in a few simple steps 1

How to solve time distance problems in a few simple steps 2

Profit and loss

How to solve in a few steps profit and loss problems 1

How to solve similar problems in a few seconds profit and loss problems 2

How to solve difficult profit and loss problems in a few steps 3

How to solve difficult profit and loss problems in a few steps 4


Question and Solution sets on WBCS Main Aritmetic

WBCS main level Arithmetic Solution set 8

WBCS main level Arithmetic Question set 8

WBCS main level Arithmetic Solution set 7

WBCS Main level Arithmetic Question set 7

WBCS Main level Arithmetic Solution set 6

WBCS Main level Arithmetic Question set 6

WBCS Main level Arithmetic Solution set 5

WBCS Main level Arithmetic Question set 5

WBCS Main level Arithmetic Solution set 4

WBCS Main level Arithmetic Question set 4

WBCS Main level Arithmetic Solution set 3

WBCS Main level Arithmetic Question set 3

WBCS Main level Arithmetic Solution set 2

WBCS Main level Arithmetic Question set 2

WBCS Main level Arithmetic Solution set 1

WBCS Main level Arithmetic Question set 1