WBCS Main level Arithmetic Solution set 1

First solution set for WBCS Main level Arithmetic

wbcs-mains-level-solution-set-1

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 first solution set on Arithmetic, easy and quick solution for 10 questions in the corresponding question set are explained. You should answer the question set first and then go through the following solutions.

1st WBCS main level Arithmetic solution set: time was 10 mins to answer

Problem 1

If A and B complete a job working together in 20 days, B and C together in 15 days and C and A in 12 days, all three working together will be able to complete the same job in,

  1. 9 days
  2. 10 days
  3. 10.5 days
  4. 6 days

Solution 1

For ease of solution we will use Work rate technique and define the work rates (or portion of work done in a day) by A, B and C as $A$, $B$ and $C$. This definition allows us to write the three equations as,

$20(A+B)= W$, where $W$ is assumed as the amount of work (whatever it may be, we won't have to determine this amount),

$15(B+C)=W$, and

$12(C+A)=W$.

As all three variables appear just twice in the three equations taken together, to get $(A+B+C)$ in terms of $W$, first we remove the 20, 15 and 12 by dividing $W$ on the RHS and then add the three equations together,

$2(A+B+C)=\left(\displaystyle\frac{1}{20}+\displaystyle\frac{1}{15}+\displaystyle\frac{1}{12}\right)W$.

Now our skill in adding fractions quickly is tested. We decide the LCM of 20, 15, 12 to be 60,

$A+B+C=\displaystyle\frac{1}{120}\left(3+4+5\right)W$, we have removed the 2 on LHS to the RHS by dividing it,

$=\displaystyle\frac{1}{10}W$,

Or, $10(A+B+C)=W$.

All three working together will complete the job in 10 days.

Answer: Option b: 10 days.

Solving in mind

While actually solving the problem, we won't write the steps as we wrote above. In MCQ test only the correct answer is required, not the steps.

So for convenience we may write straightway the fraction sum only,

$2(A+B+C)=\left(\displaystyle\frac{1}{20}+\displaystyle\frac{1}{15}+\displaystyle\frac{1}{12}\right)W$,

As the fraction sum is very easy, we arrive at the answer of 10 days without writing anything more.

With enough practice, these conceptual steps can be done in mind, without writing any steps at all.

Recommendation: After you absorb the right way to solve a problem, start practicing solving similar problems in mind. It will speed up the solution process greatly. As a precaution, with slightest discomfort with a step, write down the step.

Special Concept used: Work rate technique.

Problem 2

If $x$ is a prime number, the LCM of $x$ and $(x+1)$ is,

  1. $x^2$
  2. $(x+1)^2$
  3. $\displaystyle\frac{x(x+1)}{2}$
  4. $x(x+1)$

Solution 2

$x$ being a prime number, it has no factor except 1 and itself, and so it has no common factor with $(x+1)$. The LCM of two numbers being the smallest number that is fully divisible by each of the two numbers, by using the concept of definition of LCM, we can be sure that LCM of $x$ and $(x+1)$ is the product of the two.

Answer: Option d: $x(x+1)$.

Problem 3

$\left(2-\displaystyle\frac{1}{3}\right)\left(2-\displaystyle\frac{3}{5}\right)\left(2-\displaystyle\frac{5}{7}\right)...\left(2-\displaystyle\frac{999}{1001}\right)$ is equal to,

  1. $\displaystyle\frac{1003}{3}$
  2. $\displaystyle\frac{999}{1001}$
  3. $\displaystyle\frac{1001}{3}$
  4. None of these

Solution 3

Mentally carrying out the subtractions within brackets, we get the product terms as,

$\displaystyle\frac{5}{3}$, $\displaystyle\frac{7}{5}$, $\displaystyle\frac{9}{7}$, .... $\displaystyle\frac{1003}{1001}$.

It is easy to see, when the terms are multiplied together, every numerator is canceled out with next denominator. Only the first denominator 3 and last numerator 1003 are left out.

This is application of useful pattern identification.

Answer: Option a: $\displaystyle\frac{1003}{3}$.

Problem 4

The value of $\displaystyle\frac{\sqrt{32}+\sqrt{48}}{\sqrt{8}+\sqrt{12}}$ is,

  1. 2
  2. 4
  3. $\sqrt{2}$
  4. 8

Solution 4

Examining each of the four integers within square root, we detect if factor 4 is taken out of two square roots in numerator, the numerator equals the denominator, $\sqrt{8}+\sqrt{12}$.

While taking out 4 from the square roots we are left with 2 as answer.

Answer: Option a: 2.

Special Concept used: For simplification of surd fraction expression, we used Factoring out technique.

Problem 5

The difference between the squares of two consecutive even integers is always divisible by,

  1. 3
  2. 4
  3. 6
  4. 7

Solution 5

If $a$, and $a+2$ be two consecutive even integers, the difference of their squares is,

$(a+2)^2 - a^2 = 4a +4=4(a+1)$.

This is always divisible by 4 whatever the non-zero value of $a$ is.

Answer: Option b: 4.

This solution can again be arrived at in mind, if you are comfortable with number system concepts.

Problem 6

The least number which when divided by 15, 27, 35, and 42 leaves a remainder of 7 in each division is,

  1. 1883
  2. 2007
  3. 1897
  4. 1987

Solution 6

For the four given numbers to divide a number fully, the smallest such number will be the LCM of the four numbers.

First we will find the LCM and then will add 7 to it, so that by each division of this smallest such number by the series of four numbers, the remainder is 7.

The factors of the four numbers we identify quickly to be, 

(3, 5), (3, 3, 3), (5, 7), (2, 3, 7).

Taking non-redundant minimum number of common factors that cover all the factors of all these four numbers we have LCM as,

$3\times{5}\times{3}\times{3}\times{7}\times{2}=1890$. Just add 7 to this LCM to get the answer as 1897.

Answer: Option c: 1897.

Problem 7

$\left(\displaystyle\frac{243}{32}\right)^{-\displaystyle\frac{4}{5}}$ is equal to,

  1. $\displaystyle\frac{81}{16}$
  2. $\displaystyle\frac{16}{81}$
  3. $\displaystyle\frac{2}{9}$
  4. $\displaystyle\frac{9}{2}$

Solution 7

This is a problem on indices and we must be a bit careful in taking decisions.

We find a 5th root in the power of the bracketed fraction and ignore the minus sign for the moment. We will just reverse the result that we get without the minus sign. Main problem is the 5th root here.

Deductive reasoning

As none of the choices has any root, we are sure that the two integers 243 and 32 in the fraction within the bracket must each be 5th power of two integers. At this point we bring in our knowledge on numbers and remember that indeed $243=3^5$ and $32=2^5$.

This would simplify the given expression immediately to,

$\left(\displaystyle\frac{3}{2}\right)^{-4}$

=$\left(\displaystyle\frac{2}{3}\right)^{4}$, before applying the power we take care of the minus power,

=$\displaystyle\frac{16}{81}$.

Answer: Option: b: $\displaystyle\frac{16}{81}$.

In fact we do the whole process in mind.

Problem 8

$\sqrt{248 +\sqrt{52 + \sqrt{144}}}$ is equal to,

  1. 16.6
  2. 16
  3. 14
  4. 18.8

Solution 8

We reason that unless all the square roots are eliminated, the answer choices can't be without any square root. For the three consecutive roots to be eliminated, the third independent square root must have the term inside the square root, a square. Indeed, $144=12^2$.

So the second term under square root is transformed to,

$52+12=64=8^2$, again a favorable result in the right direction.

Now the third term within square root is transformed to,

$248+8=256=16^2$, as expected a square.

Answer: Option b: 16.

Notice that we have avoided writing step by step simplification of the given expression to highlight that the process would be carried out wholly in mind.

Problem 9

The ratio of cost price to sale price is 20 : 23. What is the profit percentage?

  1. 20%
  2. 15%
  3. 5%
  4. 6%

Solution 9

Basic profit and loss concept: profit or loss is a percentage on the Cost price.

Without going into any deductive steps we just multiply the 20 representing Cost price by 5 to make it 100 so that the figure 23 representing Sale price becomes 115.

It means, for each 100 portions of Cost price, Sale price exceeds the Cost price by 15 portions.

In other words, profit is 15 portions of every 100 portions of Cost price, that is, in short, 15%.

Whatever be the portion size in the ratios, above deduction holds. That is the basic property of Ratio and proporion.

Answer: Option b: 15%.

In this mathematical reasoning we have used, Basic profit and loss concept, the basic ratio and proportion concept and the basic percentage concept along with the fact that the requirement is to find a percentage, not any actual value.

In a large number of profit and loss problems you can get the result quickly by boldly transforming the Cost price to 100.

Problem 10

$\sqrt[3]{0.000216}$ is equal to,

  1. 0.6
  2. 0.006
  3. 0.06
  4. 0.0006

Solution 10

We take the first decision as to eliminate the decimal from the number under square root. $10^{-6}$ is thus factored out and, $10^{-2}=0.01$ is taken out of the cube root to leave 216 within the cube root.

Now we recognize, $216=6^3$. Taking 216 out of cube root results in 6.

Final result is then, 0.06.

Answer: Option: c: 0.06.

We have used Decimal elimination technique.

Takeaway

Notice that we have solved most of the problems in mind using basic concepts along with a few special techniques that we call rich concepts. This approach and skill in mathematical problem solving in mind greatly increases your speed and confidence in answering MCQ math problems.

Genrally the phrase "Mental maths" is used for indicating numerical calculations in mind. We have extended the scope of Mental maths to include other elements of fast math problem solving in mind by applying basic concepts, rich concepts and special techniques along with reasonably good mental ability on numbers.

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


Important tutorials on Arithmetic topics

Numbers, Number systems and basic arithmetic operations

Factorization

HCF and LCM

Fractions and decimals basic concepts part 1

Ratio and proportion

Arithmetic problems on mixing liquids and based on ages

How to solve Arithmetic problems on Work time, Work wages and Pipes and cisterns

Basic concepts on problems on speed time distance, Train running and Boats in rivers

Basic and rich concepts on Simple interest and Compound interest

Basic and rich percentage concepts


Question and Solution sets on WBCS Main Aritmetic

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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