WBCS Main level Arithmetic solution set 2

Second 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 second 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.

2nd WBCS Main level Arithmetic solution set: time to answer was 10 mins

Problem 1

If sum of two numbers is 10 and the sum of their reciprocals is $\displaystyle\frac{5}{12}$, the numbers would be,

  1. 6 and 4
  2. 8 and 2
  3. 9 and 1
  4. 7 and 3

Solution 1

If we assume the two numbers as $a$ and $b$, $a+b=10$, and $\displaystyle\frac{a+b}{ab}=\frac{5}{12}$, which gives $ab=24$.

Knowing the nature of the problem, we know this approach won't give us the answer as we don't have two linear equations necessary for getting values of two unknowns.

Instead we take a problem solving approach and examine the choice values and test which pair can produce 24 as product.

Answer comes easily as 6 and 4.

Answer: Option a: 6 and 4.

This is an example application of Principle of free resource use, where we have used the free resource of the choice values and tested an intermediate result with the choice values to get the answer.

In school maths this is not acceptable, but in competitive tests, rather than your mathematical skill, your problem solving skill is tested.

Special Concept used: Principle of free resource use.

Problem 2

A man bought 4 trousers at Rs.750 each, 4 shirts at Rs.450 each and 12 pairs of shoes at Rs.750 each. What is the average expenditure per item?

  1. Rs.900
  2. Rs.1000
  3. Rs.690
  4. Rs.800

Solution 2

As per item price of trousers and a pair of shoes is same, we can say for 16 items average price is Rs.750 which gives a total of $\text{Rs.}16000-\text{Rs.}4000=\text{Rs.}12000$.

Total expenditure for 4 shirts is, Rs.1800.

So for a total of 20 items cost is, Rs.13800.

Average expenditure per item is,



Answer: Option c: Rs.690.

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

Problem 3

A dishonest dealer declares that he sells his goods at cost price, but actually he uses a false weight of 900gms for 1kg weight. His profit percentage is,

  1. $13%$
  2. $11.25%$
  3. $12\displaystyle\frac{1}{9}%$
  4. $11\displaystyle\frac{1}{9}%$

Solution 3

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

Assuming the dealer sells at the rate of 100 money units per kg as declared, he actually bought the goods at the Cost price of 90 money units, because the goods he sold as 1kg was actually is 90% of 1 kg.

Effectively then the dealer gains 10 money units on his Cost price of 90 money units.

So the profit percent is, 


Answer: Option d: $11\displaystyle\frac{1}{9}%$.

We have used the basic profit and loss concept and Cost price base transformation technique in transforming Cost price to 100 money units whatever the unit may be. It is possible because all values are in percentages and ratios. Cost price is the base and assuming it as 100 produces immediate solution in many profit and loss problems.

Problem 4

Which of $\sqrt[3]{3}$ and $\sqrt{2}$ is larger?

  1. $\sqrt[3]{3}$
  2. $\sqrt{2}$
  3. The two are equal
  4. None of the above

Solution 4

The first is a cube root of power $\frac{1}{3}$ whereas the second is a square root of power $\frac{1}{2}$. To compare the two, the bases must be equalized by raising both terms to power of 6. Here base is the power that is transformed to 1 for both terms.

Thus, after raising to the power of 6, we get the first term as $3^2=9$ and the second as $2^3=8$.

So, $\sqrt[3]{3}$ is larger.

Answer: Option a: $\sqrt[3]{3}$.

Special Concept used: For comparison of terms with different powers, we used Base equalization technique.

Problem 5

If P is 50% of Q and Q is 50% of R then, P : Q : R is,

  1. 4 : 2 : 1
  2. 1 : 2 : 4
  3. 2 : 1 : 4
  4. 1 : 4 : 2

Solution 5

By first condition, P : Q = 1 : 2. 50% means one-half, and P is half of Q.

By second condition, Q : R = 1 : 2. Q is half of R.

The technique of combining these two ratios into one, P : Q : R 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, Q 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 Q though differ. In the first it is 2, and in second it is 1.

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

Now joining the two ratios we get the three variable ratio as, P : Q : R = 1 : 2 : 4.

Answer: Option b: 1 : 2 : 4.

Explanation of the two ratio joining concept

After transformation, the first ratio means, "For every 1 of P, value of Q is 2", and the second ratio means, "For every 2 of Q value of R is 4." Joining these two English statements we can say, "For every 1 of P, Q is 2, and so R is 4." This in fact is expressed as the three variable ratio, P : Q : R = 1 : 2 : 4.

Problem 6

The number of digits of the square root of 0.00059049 is,

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

Solution 6

$\sqrt{0.00059049}= 10^{-4}(\sqrt{59049})$.

Dividing 59049 by 9 we get, 6561. Dividing 6561 by 9 again we get, 729 which we know as square of 27.

So, $\sqrt{59049}=\sqrt{9^2\times{27^2}}=243$, and so


Answer: Option d: 4.

We have used the decimal elimination technique first by taking $10^{-4}$ out of the square root and then with the final objective of finding square root of 59049, started to factorize it.

In factorization, we always start with divisibilty test of 3, and then 9 if the number is odd. Divisibility test of 9 is, if sum of digits is divisible by 9 the number is divisible by 9.

Problem 7

What is the ratio of areas of two squares if the ratio of their diagonals is 5 : 2?

  1. 125 : 8
  2. 5 : 2
  3. 4 : 25
  4. 25 : 4

Solution 7

Diagonal of a square is, $d^2=2a^2$, where $d$ and $a$ are diagonal and side lengths of a square. This is because in a square, all sides are equal and a diagonal forms the hypotenuse of a right triangle.

As area of a square is square of its side, ratio of areas of the two given squares would be proportional to ratio of squares of its diagonals, that is, $\displaystyle\frac{25}{4}$.

Answer: Option: d: $\displaystyle\frac{25}{4}$.

Problem 8

Two trains, each 100m long and moving in opposite directions, cross each other in 8 secs. If one is moving twice as fast as the other, the speed of the faster train is,

  1. 60km/hr
  2. 70km/hr
  3. 50km/hr
  4. 40km/hr

Solution 8

Realative speed of the two trains moving in opposite directions is the sum of their speeds. If $S$ is the speed of the slower train in km/hr, $2S$ is the speed of the faster tain, and the relative speed is, $3S$.

A total length of 200m is crossed in 8 secs at speed $3S$.

$3S=\displaystyle\frac{\text{distance}}{\text{time}}=\frac{200}{8}\text{ m/sec}$


$=25\times{3.6}\text{km/hr}$, $1\text{m/sec}=\displaystyle\frac{\displaystyle\frac{1}{1000}\text{km}}{\displaystyle\frac{1}{3600}\text{hr}}=3.6\text{km/hr}$

Or, $2S= 60\text{km/hr}$.

We have transformed $3S$ to $2S$ which is the target speed of the faster train.

Answer: Option a: 60 km/hr.

Problem 9

In what ratio must water be mixed with milk to make a profit of $16\displaystyle\frac{2}{3}$% on selling the diluted milk at cost price?

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

Solution 9

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

Cost price transformation to 100 units: Assume cost price of 100 volume units (any) of pure milk is 100 money units.

On this cost price profit is, $16\displaystyle\frac{2}{3}$ money units.

This is equivalent to $16\displaystyle\frac{2}{3}$ volume units of pure milk, which he didn't have to pay for because of replacing this much milk by mixing water.

So he mixed $16\displaystyle\frac{2}{3}$ volume units of water for every 100 volume units of pure milk.

Thus ratio of water to milk is,

$\displaystyle\frac{16\displaystyle\frac{2}{3}}{100}=1 : 6$

Answer: Option d: 1 : 6.

In this conceptual 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 ratio, not any actual value.

Most important though was the use of the Base tranformation technique in assuming 100 money units as the base of Cost price as well as 100 volume units of milk purchased at this base price. We could make this assumption only because all components are in percentages or ratios.

Base transformation technique: 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

A policeman, while running to catch a thief, is 114m behind the thief. The policeman runs 21m and the thief runs 15m every minute. In what time will the policeman catch the thief?

  1. 18 minutes
  2. 16 minutes
  3. 19 minutes
  4. 17 minutes

Solution 10

This is a problem on Race where the faster person is behind the slower person.

By applying Race concept we deduce that the policeman is approaching the thief at a relative speed of $\text{(21m/minute - 15m/minute) = 6m/minute}$.

At this relative speed, the policeman will cover the 114m starting gap in,

$\displaystyle\frac{114}{6}\text{ minutes}=19\text{ minutes}$.

Answer: Option: c: 19 minutes.

When the policeman approaches the thief at the relative speed, the speed of the thief effectively becomes zero and he stands still.


Along with the basic number system concepts, profit and loss concepts, ratio and proportion concepts, speed time distance concepts and mixing or alligation concepts, we have used the important additional concepts of Base transformation of cost price, Base equalization, Principle of free resource use, Ratio joining and Race. These we call as the rich concepts that help to solve relevant problems easily and quickly.

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

Important tutorials on Arithmetic topics

Numbers, Number systems and basic arithmetic operations



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

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