WBCS Main level Arithmetic solution set 3 | SureSolv

WBCS Main level Arithmetic solution set 3

Third solution set for WBCS Main level Arithmetic

wbcs-mains-level-solution-set-3

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 third 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.

3rd WBCS Main level Arithmetic solution set: time to answer was 10 mins

Problem 1

The simplified value of $\displaystyle\frac{1}{56}+\displaystyle\frac{1}{72}+\displaystyle\frac{1}{90}+\displaystyle\frac{1}{110}+\displaystyle\frac{1}{132}$ would be,

  1. $\displaystyle\frac{5}{84}$
  2. $\displaystyle\frac{1}{12}$
  3. $\displaystyle\frac{1}{84}$
  4. $\displaystyle\frac{1}{28}$

Solution 1

For sum of fractions of this type, we must first jot down the factors of the denominators, because the LCM of the denominators is anyway needed directly or indirectly for evaluating such sum of fractions.

In school tests, you might adopt the conventional approach of summing up all the terms together in one step. But as we are in a MCQ test where steps are not needed to be written in the answer, we always look out for easier and faster way to solve.

Accordingly, using our basic level mental math skills in factorization, we immediately jot down the factors in pair of factors form (as it comes naturally, also as we can visualize a pattern even at this early stage) for each of the denominators. The factor pair list is,

(7,8), (8,9), (9,10), (10,11), and (11,12).

This shows LCM value will be large and single step summation will need lots of calculations.

So instead, we observe and use a pattern in the pairs of factors,

For such a pair, one of the four factors in two denominators appears common to both pairs of factors. This common factor property always makes simplification easy.

First and second pair of fractions follow this pattern.

Mentally we find that sum of the two pairs to be, 

$\displaystyle\frac{1}{56}+\displaystyle\frac{1}{72}$

$=\displaystyle\frac{16}{7\times{8}\times{9}}$

$=\displaystyle\frac{2}{7\times{9}}$, and

$\displaystyle\frac{1}{90}+\displaystyle\frac{1}{110}$

$=\displaystyle\frac{20}{9\times{10}\times{11}}$

$=\displaystyle\frac{2}{9\times{11}}$.

A this stage we jot down only the two results and find the pattern of common factor repeating again. The common factor is 9 at this stage. Adding up these two results,

$=\displaystyle\frac{2}{7\times{9}}+\displaystyle\frac{2}{9\times{11}}$

$=\displaystyle\frac{36}{7\times{9}\times{11}}$

$=\displaystyle\frac{4}{7\times{11}}$

We did this simplification mentally and jotted down only the final result. Only two terms remain, the result obtained till now and the fifth term.

We find the pattern of common factor in the two denominators persist even at this last stage. It is 11 this time. The final result is then,

$\displaystyle\frac{4}{7\times{11}}+\displaystyle\frac{1}{11\times{12}}$, factor 11 is common here

$=\displaystyle\frac{55}{7\times{11}\times{12}}$

$=\displaystyle\frac{5}{7\times{12}}$

$=\displaystyle\frac{5}{84}$.

All in mind and exploiting the pattern, step by step. The standard method of finding LCM and adding the five fraction expression would have been tedious and time-consuming by conventional approach of all terms summation at one go.

Answer: Option a: $\displaystyle\frac{5}{84}$.

Solving in mind

We have written down the pair of factor list for the denominators of the fraction terms in the first step,

(7,8), (8,9), (9,10), (10,11), and (11,12).

We added the first two pairs of fractions in mind and jotted down the result,

$=\displaystyle\frac{2}{7\times{9}}+\displaystyle\frac{2}{9\times{11}}$.

With same pattern of common factor in denominator persisting we do the sum in mind and write down the final sum of two fractions,

$=\displaystyle\frac{4}{7\times{11}}+\displaystyle\frac{1}{11\times{12}}$.

Using the same pattern we get the final result as,

$=\displaystyle\frac{5}{7\times{12}}$.

Only at this last stage we multiply the factors to get 84 as the final denominator. This is use of delayed evaluation technique, by which we delay a simplification (in this case multiplication) as late as possible.

Use of Pattern recognition technique or Key pattern discovery helped greatly.

We repeat the common factor pattern,

In a two fraction sum, out of four factors of two denominators, one factor is common to both the pairs of factors, and when summing up the two fractions, this common factor gets canceled out.

As we appreciate the usefulness of this two factor technique, we leave the two factors of each result as it is without multiplying except at the last stage. This is use of delayed evaluation technique.

Special Concepts used: Pattern recognition technique and Delayed evaluation technique.

Problem 2

A man cultivates $\displaystyle\frac{1}{3}$rd of a land in 10 days while a second man cultivates $\displaystyle\frac{2}{5}$th of the same land in 6 days. In how many days would the two cultivate $\displaystyle\frac{4}{5}$th of this land if they work together?

  1. 4 days
  2. 10 days
  3. 5 days
  4. 8 days

Solution 2

Applying unitary method to the first two given statements we get the work rates of the first and the second man in terms of portion of land cultivated in a day as,

$\displaystyle\frac{1}{30}$, and

$\displaystyle\frac{1}{15}$.

Sum of these two, the total work done by the two men working together over a day, in terms of portion of land cultivated, is,

$\displaystyle\frac{1}{10}$.

To cultivate $\displaystyle\frac{4}{5}$th of land, the two together will then take 8 days.

All in mind, no jotting down on paper, if you are comfortable in doing simple sums of fractions mentally.

Answer: Option d: 8 days.

We have shown how we would calculate mentally, without writing.

Problem 3

The value of $(0.\overline{1})^2\left[1-9\times{(0.1\overline{6})^2}\right]$ is,

  1. $-\displaystyle\frac{1}{162}$
  2. $\displaystyle\frac{1}{109}$
  3. $\displaystyle\frac{1}{108}$
  4. $\displaystyle\frac{7696}{10^6}$

Solution 3

Though there are two methods to find out fractions equivalent to non-terminating repeating decimals, we recognize from our experience, $0.\overline{1}$ to be equal to $\displaystyle\frac{1}{9}$ and $0.1\overline{6}$ to be equal to $\displaystyle\frac{1}{6}$ because of the simplicity of the two terms.

So the second term under the brackets turns out to be $\displaystyle\frac{1}{4}$ with 9 cancelled out, and the value of the bracketed expression, $\displaystyle\frac{3}{4}$.

Final result is then the product of $\displaystyle\frac{1}{81}$ and $\displaystyle\frac{3}{4}$, that is,

$\displaystyle\frac{1}{108}$.

Answer: Option c: $\displaystyle\frac{1}{108}$.

We prefer to carry out as much of the simplification as possible in mind.


To know more about how to convert non-terminating repeating decimal or rational decimal to its equivalent fraction, refer to our Tutorial, Fractions and decimals basic concepts part 1 and solution set, SSC CGL level solution set 17 on fractions and decimals.

We reproduce the two methods here.

Conversion of repeating non-terminating rational decimal to a fraction, Conventional Method

First step: convert the number to a pure ovelined decimal with no leading digit before the repeating digits on their left.

Let us take up the term $0.3\overline{21}$ to show how the method works,

$0.3\overline{21} = \displaystyle\frac{3}{10} + \displaystyle\frac{0.\overline{21}}{10}$.

We have isolated the pure overlined decimal as $0.\overline{21}$. Now we will transform it to a fraction.

Second step: transformation of repeating non-terminating rational decimal to a fraction.

Let's assume,

$x = 0.\overline{21}$,

Multiplying both sides by 100,

$100x = 21.\overline{21} = 21 + x$,

Or $x = \displaystyle\frac{21}{99}$.

So,

$0.3\overline{21}=\displaystyle\frac{3}{10}+\displaystyle\frac{21}{990}=\displaystyle\frac{318}{990}$.

Just note, instead of $21$ if the repeating digits were $98$, the same method would have produced instead,

$x = \displaystyle\frac{98}{99}$.

Conversion of a non-terminating repeating rational decimal to fraction - faster method

We have to convert, $0.3\overline{21}$ to its equivalent fraction.

As numerator, consider all the digits to form initial numerator 321, then subtract 3 from 321 to get 318 as the final numerator.

As denominator, it will be number of 9s equal to number of digits under overline first. This will be suffixed by number of zeros equal to number of digits before the overline, up to the decimal point.

This will result in equivalent fraction as,

$0.3\overline{21}=\displaystyle\frac{318}{990}$.

How this math trix works, the concept

$.3\overline{21}=\displaystyle\frac{3}{10}+\displaystyle\frac{21}{990}$

$=\displaystyle\frac{3\times{99}+21}{990}$

$=\displaystyle\frac{3\times{100}+21 - 3}{990}$

$=\displaystyle\frac{318}{990}$.

This shortcut method is then a pefectly usable one as a math trix, because it follows from basic concepts, and is simple.


Problem 4

The least number of square tiles required to pave the floor of a room 15m 17cm long and 9m 2cm broad is,

  1. 814
  2. 902
  3. 738
  4. 656

Solution 4

This is a straightforward Factorization and HCF problem.

Concept

To give you an idea about why it is so, let us consider the variation of the given problem—if the length and breadth were 24m and 16m, 8 would have been the HCF of the two so that putting 2 rows and 3 columns of 8m square tiles, that is $2\times{3}=6$ numbers of 8m square tiles, we would have perfectly covered the surface. That would be the minimum number of square tiles as we have used the largest square tile possible with largest side length of 8m which is the highest common factor of 24 and 16 resulting in smallest value of quotients 3 and 2. We could have used 4m square tiles with side length 4m, but then the number of tiles required would have been 24, four times the minimum number of 6.

covering-with-square-tiles.jpg

Finding HCF

We need to find then the HCF of 902 and 1517 (both converted to cm) and that would be the size of the largest square tile to cover the surface fully in rows and columns (covering the the breadth and the length) without using any tile fragments, which would have been equivalent to remainders.

Testing both the numbers quickly for factors 3, 7 and 11 we sense the HCF possibly to be a larger prime and decide to adopt the famous Euclid's Continued Division method taught in schools. We do it by hand,

$1517\div{902}$ ⇒ Remainder 615

$902\div{615}$ ⇒ Remainder 287, divisor in earlier division becomes dividend and remainder becomes divisor

$615\div{287}$ ⇒ Remainder 41

$287\div{41}$ ⇒ Remainder 0.

So 41 is the HCF of the two.

Now we divide mentally to get,

$1517=41\times{37}$, and

$902=41\times{22}$.

The required minimum number would then be, $22\times{37}=814$, multiplication done in mind.

This is a rare problem where we resorted to the school level HCF method to reach the answer in shortest possible time.

Answer: Option a: 814.

Methods used: Euclid's method to find HCF by continued division.


To know more about HCF and LCM, you may refer to our tutorial on HCF and LCM.


Problem 5

If $2A=3B=4C$, then $A : B : C$ is,

  1. 4 : 3 : 2
  2. 6 : 4 : 3
  3. 2 : 3 : 4
  4. 3 : 4 : 5

Solution 5

First we need to split the three term continued equation into two equations,

$2A=3B$, and

$3B=4C$.

And immediately the equivalent ratios,

$A:B=3:2$, and

$B:C=4:3$.

Here also we use a pattern or rule. In the equation, $2A=3B$, the ratio term for $A$ would be the coefficient of B, that is 3, and vice-versa. It follows simply as,

$2A=3B$,

Or, $\displaystyle\frac{A}{B}=\frac{3}{2}$.

This is a rich concept and considered as a useful pattern.

Using this technique, we arrive at the two ratios directly and without writing, remember them, so simple these are.

Now we know that to join the two ratios we have to transform one or both the ratios in such a way that the term values corresponding to the middle variable (which we can call ratio joining middle variable) same.

In this case the task is simple - just multiply the first ratio by 2 all through to get the joined ratio as,

$A:B:C=6:4:3$.

Ratio joining concepts

The technique of combining these two ratios into one, A : B : C is,

  • first to ensure the variable that appears as denominator in the first ratio appears as numerator in the second, and
  • the ratio term values of this middle or common variable are same in both the ratios.

In our problem, B is the middle or common variable that helps to join these two ratios.

We find that the first condition is already satisfied. It it were otherwise we would have had to invert the second ratio.

The ratio term values of B though differ. In the first it is 2, and in second it is 4.

To make the term value of B in the first ratio as 4, just multiply the ratio terms of the first ratio by 2 and get A : B = 6 : 4. This now satisfies the second condition also.

Now joining the two ratios we get the three variable ratio as, $A:B:C = 6:4:3$.

Answer: Option b: 6 : 4 : 3.

Explanation of the two ratio joining concept

After transformation, the first ratio means, "For every 6 of A, value of B is 4", and the second ratio means, "For every 4 of B value of C is 3." Joining these two English statements we can say, "For every 6 of A, B is 4, and so C is 3." This in fact is expressed as the three variable ratio, $A : B : C = 6 : 4 : 3$.

Problem 6

By selling an item for Rs.350 instead of Rs.400, the percentage loss increases by 5%. The cost price of the item is,

  1. Rs.1050
  2. Rs.1000
  3. Rs.433
  4. Rs.417.50

Solution 6

Basic profit and loss concept: Percentage Profit or loss is on Cost price.

As it is a changing situation, we would use Change analysis technique to say,

Reduction of Rs.50 resulted in the increase of 5% loss, and so these are equal. Thus,

$\text{Rs.}50=5\text{% loss }=5\text{% of Cost price}$.

So Cost price = Rs.1000.

$5$% is equivalent to $\displaystyle\frac{1}{20}$th. So Cost price is 20 times 50 which is 1000. This is the way we calculate in mind.

Answer: Option B: Rs.1000.

If there are two related variables and one changes, using this change only we may arrive at the solution very quickly. This is change analysis technique.

Problem 7

The average of the first 100 natural numbers is,

  1. 50.5
  2. 51
  3. 50
  4. 51.5

Solution 7

Conceptually we know, average of 1, 2, 3, 4, 5 is the middle number 3 (for all odd numbered such series of numbers this is true).

Similarly, the average of first 99 natural numbers would be the middle number 50.

So the total of first 100 natural numbers would be, $50\times{99}+100=5050$, and the average,

$\displaystyle\frac{5050}{100}=50.5$.

There are formulas for average of first $n$ natural numbers. But we do like to go conceptually in mind rather than use formulae.

Answer: Option: a: 50.5.


To know more about addition of first $n$ natural numbers, you may refer to our tutorial on Numbers, number system and basic arithmetic operations.


Problem 8

The bell of a wall-clock requires 3 seconds to ring 5 times. The time required for the bell to ring 6 times is,

  1. $3\displaystyle\frac{1}{2}$ seconds
  2. $4$ seconds
  3. $5$ seconds
  4. $3\displaystyle\frac{3}{4}$ seconds

Solution 8

In 5 rings there were 4 ring intervales that consumed 3 secs. So each ring interval took, $\displaystyle\frac{3}{4}$ secs.

In 6 rings there will be 5 ring intervals which would then take,

$5\times{\displaystyle\frac{3}{4}}=3\displaystyle\frac{3}{4}$ seconds.

Here we have used Unitary method assuming each ring interval takes same time.

Answer: Option d: $3\displaystyle\frac{3}{4}$ seconds.

Problem 9

Two numbers are in the ratio of 3 : 5. If each number is increased by 10, the ratio becomes 5 : 7. The numbers are,

  1. 3, 5
  2. 7, 9
  3. 15, 25
  4. 13, 22

Solution 9

Basic ratio concepts: If a ratio is, $a:b=2:3$, the ratio is expressed in minimzed fraction form with no common factors. The common factor that has been eliminated to reduce the fraction is the HCF of the two values of $a$ and $b$.

HCF reintroduction technique: In many ratio problems we assume $x$ as the cancelled out HCF and express such a ratio as, say, $a:b=2x:3x$. Advantage in this approach is, instead of two variables we need to deal with one variable $x$.

Using HCF reintroduction technique, the ratio to start with is,

$a:b=3x:5x$,

Or, $\displaystyle\frac{a}{b}=\frac{3x}{5x}$.

And after increasing each term by 10,

$\displaystyle\frac{3x+10}{5x+10}=\frac{5}{7}$

With awareness of the concepts this step can be written down directly.

In conventional approach, we transpose the two sides to form,

$21x+70=25x+50$,

Or, $4x=20$,

Or, $x=5$.

So the numbers are, $3x=15$ and $5x=25$.

Answer: Option c: 15, 25.

This is the simplest approach in this case.

Problem 10

The area of a square is equal to the area of a circle. The ratio between a side of the square and the radius of the circle is,

  1. $1 : \sqrt{\pi}$
  2. $\sqrt{\pi}:1$
  3. $\pi : 1$
  4. $1:\pi$

Solution 10

Area of a square is,

$A=a^2$, where $a$ is the side length of the square

And area of a circle with same area is,

$A=\pi{r^2}$, where $r$ is the radius

So ratio of the side length of the square and the radius of the circle is,

$\displaystyle\frac{a}{r}=\frac{\sqrt{\pi}}{1}$.

Answer: Option: b: $\sqrt{\pi}:1$.

Takeaway

Along with the basic number system concepts, fractions and decimals, factorization, HCF, LCM, profit and loss concepts and ratio and proportion concepts, we have used the important additional concepts of Common factor pattern, Pattern recognition technique, Delayed evaluation technique, Unitary method, Work rate technique, Decimal conversion fast method or math trix, Euclid's HCF method of repeated division, Ratio joining, Change analysis technique and HCF reintroduction technique. These we call as the rich concepts that help to solve relevant problems easily and quickly.

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 most exams, use of concepts and problem solving ability is judged.


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