## Seventh solution set for WBCS Main level Arithmetic

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 seventh 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 required we will also explain the supporting concepts.

### 7th WBCS Main level Arithmetic solution set: time to answer was 10 mins

#### Problem 1

A man gains 10% by selling an article at a certain price. If he sells it at double the price, the profit made is,

- 60%
- 20%
- 100%
- 120%

#### Solution 1: Solving in mind

With 10% profit, $SP = 1.1CP$.

$\text{Double of SP}=2.2CP$.

$\text{Profit}=120$%.

#### Solution 1: Detailed explanation

Basic profit concept,

Profit %$=\displaystyle\frac{\text{Sale price} - \text{Cost price}}{\text{Cost price}}\times{100}$%.

Profit is the excess of sale price over cost price. When we need **percent profit**, the percentage is always expressed **on the base of cost price**. Finally, for converting profit divided by cost price to a percent, we need to multiply it by 100. It means, when profit is 10%, for every 100 units of cost price, profit is 10 units.

In fraction form, this 10% profit means, actual profit is, $\displaystyle\frac{1}{10}=0.1$ of cost price.

In our problem as Profit is the extra amount of Sale price over Cost price, Sale price is,

$SP=CP + \text{Profit}=CP + 0.1CP=1.1CP$.

Generally we convert the SP immediately to this form for quick solution.

When sale price is doubled from the previous sale price, SP, the new sale price is,

$\text{New SP}=2SP=2.2CP$.

So the profit now is,

$2.2CP-CP=1.2CP=120$% of CP.

**Answer:** Option d: 120%.

**Concepts:** Profit and loss -- Percentage to decimal conversion -- Decimal to percentage conversion.

#### Problem 2

When 40% of a number is added to 42, the result is the number itself. The number is,

- 72
- 70
- 82
- 105

#### Solution 2: Solving in mind

40% added to 60% makes the whole 100%.

So 42 is 60% of the number; 10% of the number is 7, and the number is 70. Answer: Option b: 70.

#### Solution 2: Detailed explanation

If the number is $x$ (from school habits we are more comfortable in thinking in terms of variables and equations), by the problem statement,

$42 + 40\text{% of the number}=x$,

Or, $42=x(100\text{%}-40\text{%})=60\text{% of } x=0.6x$,

Or, $x=\displaystyle\frac{42}{0.6}=70$.

**Answer:** Option b: 70.

**Concepts:** Number system -- Percentage -- Percentage to decimal conversion.

#### Problem 3

If $\displaystyle\frac{4}{5}$th of a cistern is filled in 1 minute, how much more time will be required to fill the rest of it?

- 10 sec
- 20 sec
- 12 sec
- 15 sec

#### Solution 3: Solving in mind

As cistern filled amount or portion is proportional to filling time, rest to be filled $\displaystyle\frac{1}{5}$th of the cistern will be filled up in $\displaystyle\frac{1}{4}$th of a minute, that is, in 15 secs. Answer. Option d: 15 secs.

**Answer:** Option d: 15 secs.

**Concepts:** Pipes and cisterns -- Cistern portion filled to filling time direct proportionality with filling rate constant -- Unitary method.

#### 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: 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.

#### Solution 3: Detailed explanation

Applying unitary method (as filled portion is proportional to filling time),

$\displaystyle\frac{4}{5}$th of cistern is filled in 1 minute

Whole (1 unit) of cistern is filled up in $\displaystyle\frac{5}{4}$ minute

To be filled up $\displaystyle\frac{1}{5}$th portion of the cistern will be filled up in,

$\displaystyle\frac{5}{4}\times{\displaystyle\frac{1}{5}}=\frac{1}{4}\text{ minute}=15 \text{ secs}$.

**Answer:** Option d: 15 secs.

#### Problem 4

If $a:b=c:d=e:f=1:2$, then $(3a+5c+7e):(3b+5d+7f)=\text{?}$.

- $1:2$
- $2:1$
- $1:4$
- $8:7$

#### Solution 4: Solving in mind

As the coefficients of the three numerator variables of the first term (which are numerators of the given ratios) of the desired ratio are same as the coefficients of the denominator variables (which are corresponding denominators of the given ratios) of the second term of the desired ratio, with the given equality of the ratios, the given ratio value of $1:2$ will not be changed. Answer. Option a: $1:2$.

#### Solution 4: Detailed explanation

The given ratio is a running ratio consisting of three ratio equations. The first can be isolated as,

$a:b=1:2$,

Or, $b=2a$.

Similarly,

$d=2c$, and,

$f=2e$.

Substituting these into the desired ratio, its numerator and denominator variables turn to same expression, $(3a+5c+7e)$ leaving $1:2$ as the resultant ratio value.

**Answer:** Option a: $1:2$.

**Concepts:** Ratio -- Running ratios -- Pattern identification technique.

#### Problem 5

The speed of a train 150m long is 50km/h. How much time will it take to pass a platform 600m long?

- 64 sec
- 60 sec
- 50 sec
- 54 sec

#### Solution 5: Solving in mind

To pass the platform, the train has to cover its own length plus the platform length (by basic train passing concept; it follows common sense). This total length is 750m or $\displaystyle\frac{3}{4}$th of a km. We convert to km as speed is in km/hr.

The train speed being 50km/h, it covers 1km in $\displaystyle\frac{60}{50}=\frac{6}{5}$th of a minute, that is, 72 secs. We convert to minute and then secs as desired choice values are in secs.

So to cover $\displaystyle\frac{3}{4}$th of a km, it will take,

$\displaystyle\frac{3}{4}\times{72}=54$ secs.

**Simple unitary method** is applied.

**Answer: **Option d: 54 secs.

**Concepts:** Time distance -- Train passing -- Unit conversion -- Distance time direct proportionality with speed constant -- Unitary method.

#### Problem 6

A can do a work in 20 days and B can do it in 12 days. B worked at it for 9 days. In how many days can A alone finish the remaining work?

- 3
- 5
- 7
- 11

#### Solution 6: Solving in mind

In 9 days B finishes $\displaystyle\frac{3}{4}$th of the work, leaving $\displaystyle\frac{1}{4}$th for A to finish in 5 more days.

**Answer:** Option b: 5 days.

#### Solution 6: Detailed explanation

Work portion completed by a worker is proportional to days he worked when he works at a constant rate.

As B finishes the work in 12 days, if he works for 9 days he will finish, $\displaystyle\frac{9}{12}=\frac{3}{4}$ portion of the work.

The work left will then be, $\displaystyle\frac{1}{4}$ of the work.

Again we apply the concept of proportionality of work portion to work days on A's work now.

A completes the whole work in 20 days,

So he will complete $\displaystyle\frac{1}{4}$th of the work in,

$\displaystyle\frac{20}{4}=5$ days.

**Concepts:** Work time -- Work done to days of work proportionality -- unitary method.

With direct proportionality, unitary method in three steps is the fastest and safest approach.

#### Problem 7

If $(1^2 + 2^2 + 3^2+.....+10^2)=385$, then the value of $(2^2+4^2+6^2+.....+20^2)$ is,

- $770$
- $1155$
- $385\times{385}$
- $1540$

#### Solution 7: Solving in mind

We compare the first term of the target expression with the first term of the given expression, and then the second term of the target expression with the second term of the given expression, and quickly * discover the pattern* that each term of the given expression is multiplied by $2^2=4$ to get the corresponding term of the target expression.

So the target expression value will be 4 times the value of the given expression, $4\times{385}=1540$.

**Answer:** Option d: 1540.

**Concepts:** Number system -- Pattern identification technique.

Problem 8

If $1.5x=0.04y$, then the value of $\left(\displaystyle\frac{y-x}{y+x}\right)$ is,

- $\displaystyle\frac{73}{77}$
- $\displaystyle\frac{703}{77}$
- $\displaystyle\frac{73}{770}$
- $\displaystyle\frac{730}{77}$

#### Solution 8: Solving in mind

First we find $\displaystyle\frac{y}{x}=\frac{1.5}{0.04}=\frac{75}{2}$.

In the second step we apply Componendo dividendo to get the final result directly,

$\displaystyle\frac{y-x}{y+x}=\frac{75-2}{75+2}=\frac{73}{77}$.

**Answer:** Option a: $\displaystyle\frac{73}{77}$.

**Concepts:** Pattern identification -- Input transformation for applying componendo dividendo (keeping $y$ in numerator and so on) -- Componendo dividendo.

If you want to refresh concepts on this highly effective method, read our article, **Componendo dividendo explained.**

#### Problem 9

The population of a town increases annually by 5% . If it is 15435 now, the population two years ago was,

- 13700
- 14000
- 15000
- 14800

#### Solution 9: Solving in mind

We know population growth is a compound growth exactly same as compound interest. Then,

$\text{Final population}=(1.05)^2\times{\text{Initial population}}$.

We divided final population of 15435 by 1.05 twice, getting initial population as, 14000.

**Note:** Instead of calculating $(1.05)^2$ and dividing 15435 by the result it is much easier to divide by 1.05 twice.

**Answer: **Option b: 14000.

#### Solution 9: Detailed explanation

**Truth:** In general, unless word "simple" is mentioned, all growths are compound growths exactly same as Compound interest.

**Compounding concept**: in the first year, growth will be 5% of initial value, total value will then be, $1.05\text{(Initial value)}$. At the end of second year, growth will be on this increased initial value as the year-starting value, that is,

$\text{5% of 1.05(Initial value)}$

$=0.05\times{1.05\text{(Initial value)}}$.

The final value at the end of two years will then be,

$\text{Final value}$

$=1.05\text{(Initial value)}+0.05\times{1.05\text{(Initial value)}}$

$=(1.05)^2\text{(Initial value)}$

So,

$\text{Initial value}=\displaystyle\frac{\text{Final value}}{1.05\times{1.05}}$

$=\displaystyle\frac{15435}{1.05\times{1.05}}$

$=14000$.

**Concepts:** Growth concept -- percentage to decimal conversion -- easy division by avoiding multiplication of denominator factors.

#### Problem 10

In a 100m race, A beats B by 10m and C by 13m. In a race of 180m between the three, B will beat C by,

- 6m
- 5m
- 5.4m
- 4.5m

#### Solution 10: Solving in mind

In the time A covers 100m, B covers 90m and C covers 87m.

Assuming A, B and C run at the same speed as the first time, when A covers 180m taking **1.8 times the time of the first race**, B covers $1.8\times{90}=162$m, and C will cover, $1.8\times{87}=156.6$m, behind by 5.4m. This happens because of Time distance proportionality when speed is constant.

A quicker way to the solution is—in a 100m race between the three, C falls behind B by 3m, so in a 180m race, C will fall behind by $3\times{\displaystyle\frac{180}{100}}=5.4$m. Unitary method is applied because of distance time proportionality.

#### Detailed explanation of mechanism of proportionality in this case

In the first race between the three, let's assume A covers 100m in time of $T$ units, B covers 90m in time $T$ units and in same time C covers 87m.

In the second instance the distance to be covered is 1.8 times 100m and A covers it in $1.8T$ units of time.

So with same speeds as before, in time 1.8T, B will cover $1.8\times{90}=162$m and C will cover $1.8\times{87}=156.6$m, because, each with their constant speeds will cover a distance that is proportional to the time. The time is determined in the races by A.

**Answer:** Option c: 5.4m.

**Concept:** Speed time distance -- Time distance proportionality when speed is constant -- Visualization of exactly what happens in each case.

**Note:** If the second race were between only B and C, B would have beaten C by 6m, and not by 5.4m, because time taken by B to cover 180m would have been 2T units. This time would have been determined by B this time, not by A.

### Takeaway

We have used and explained the concepts and methods,

*Profit and loss concepts, Percentage, Number system, Pipes and cisterns, unitary method, Ratio and Proportion, Running ratio, Pattern identification, Time distance, Train passing, Time distance proportionality when speed is constant, Work time, Work done to time worked proportionality, Componendo dividendo, Growth, How to save time in calculation, 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 this time one of the important * Essential Mental math skills* of Unitary method.

### Concept tutorials and articles on Arithmetic

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**Numbers, Number system and basic arithmetic operations**

**Factorization or finding out factors**

**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**

**How to find perfect square root of integers or decimals New**

#### Ratio and proportion and mixing liquids

**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**

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**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**

#### Clocks and Calendars

**How to solve Clock problems New**

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### Question and Solution sets on WBCS Main Arithmetic

**WBCS main level Arithmetic Solution set 11**

**WBCS main level Arithmetic Question set 11**

**WBCS main level Arithmetic Solution set 10**

**WBCS main level Arithmetic Question set 10**

**WBCS main level Arithmetic Solution set 9**

**WBCS main level Arithmetic Question set 9**

**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**