Solution of Maths Question set 5 for WBCS

Submitted by Atanu Chaudhuri on Thu, 24/05/2018 - 23:09

$Solution of Maths Question set 5 for WBCS$

Solution of 5th WBCS math practice set

Solution of maths question set 5 for WBCS teaches how to solve 10 selected arithmetic aptitude questions in 10 minutes. Learn to solve arithmetic in mind.

Questions are on ratio, componendo dividendo, LCM, Simplification, simple geometry, decimals, time work, continued square root should be useful for prelims of other competitive exams also.

To take the test before going through the solutions, click here.

Solution of Maths Question set 5 for WBCS: time to answer was 10 mins

Problem 1

The product of two numbers is 1575 and their division is $\displaystyle\frac{7}{9}$. The two numbers then are,

45, 35
36, 35
35, 27
81, 63

Solution 1

HCF reintroduction technique in a ratio

In a ratio of two values, $a:b=7 : 9$, the result is expressed as a reduced fraction eliminating the HCF between the two values. Because it is division, we can multiply both the values in the ratio with cancelled out unknown HCF $x$, and express actual values as,

$a=7x$,

$b=9x$.

Product, $ab=63x^2=1575$,

Or, $x^2=25$,

Or, $x=5$.

The two numbers are, 45, 35.

All in mind using the rich concept of HCF reintroduction technique.

Answer: Option a: 45, 35.

How did we decide the solution path

But how did we decide that the division result is to be seen as a ratio and HCF reintroduction technique is to be applied to the ratio, thus getting the product in one unknown variable $x$?

Selection of this path is done through a combination of what we call End State Analysis Approach and Deductive reasoning.

End state analysis approach recommends focus on what is required, the end state, and what are given, the initial state. This is one of the most natural approach for any problem solving. And of course, find the shortest path from initial state to final state (generally by observing similarities between the two states).

The values of two variables are to be found out and two relations are given. Unfortunately both are non-linear. End state and initial state.

The main question is transformed to finding a way to combine the results of the two given equations. As the first equation is a product, how can we form a product from the second?

The answer was, first consider the division as a ratio and then apply the simple HCF reintroduction technique.

The problem was solved in 30 secs at most.

Problem 2

Two pipes A and B can fill a cistern in 6 minutes and 7 minutes respectively. The two pipes are opened alternately for 1 minute each starting with pipe A. In what time will they fill the cistern?

$1\displaystyle\frac{1}{4}$ minutes
$6\displaystyle\frac{3}{7}$ minutes
$5$ minutes
$5\displaystyle\frac{2}{3}$ minutes

Solution 2

This is a boundary condition problem with two events, opening of pipe A and opening of pipe B happening at regular intervals.

With such problems you can't get the solution by a formula, first you have to determine the boundary and then explore how the events affect the solution after crossing the boundary.

To know more about boundary condition problems, you may refer to our article,How to solve arithmetic boundary condition problems in a few simple steps.

These problems are quite interesting and a bit tricky.

Solution in 15 secs—Educated guess approach coupled with Free resource use principle

While we read through the problem, we read the choice values also.

In an instant we related the given values with the choice values and decided without any doubt that the answer must be within 6 minutes and 7 minutes. As Option c is only such choice, it is the answer.

Alternate minutes A and B opening will be nearly equivalent to A opened throughout (which will take 6 minutes) but because of a bit slower speed of B, we make a guess estimate of final time between 6 minutes and 7 minutes.

This is an educated guess based on firm ground of nathematical logic. Considering from pipe A's point of view, the final time has to be more than 6 minutes because pipe B is slowing down filling every alternate minute.

Alternately considering pipe B point of view, the final time must be less than 7 minutes as the faster pipe A speeds up the process in alternate minutes.

So it follows, the filling up time must be between 6 minutes and 7 minutes.

Just for the sake of curiosity, the final filling value is not guessed to be the average of 6 minutes and 7 minutes. For that matter we have not made a guess about the actual value, we estimated a range.

Just remember, we have not had gone this way of problem solving unless we had noticed that there is only one choice value larger than 6 minutes. This observation and guessing the more than 6 min final time were nearly simultaneous.

All in mind.

Answer: Option b: $6\displaystyle\frac{3}{7}$ minutes.

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

You need to be proficient in making educated guess based on given values and choice values.

ES: Educated guess estimate based on given pipe filling capacities

Considering from pipe A's point of view, the final time has to be more than 6 minutes because pipe B is slowing down filling every alternate minute.

Though it is common sense reasoning, one has to know which pipe is slower and which one faster (either by experience or by mathematical concepts).

Often you will reach the solution in MCQ questions by making educated guess estimate and comparing the estimate with given choice values.

Conventional boundary condition approach of solution

We have to know the conventional solution path in case the choice values do not help. If all the choice values were more than 6 minutes we wouldn't have any other option but to go through the conventional solution.

Applying Work rate technique, every two minutes the two pipes are filling up,

$\displaystyle\frac{1}{6}+\displaystyle\frac{1}{7}=\displaystyle\frac{13}{42}$ portion of the cistern.

In other words, the two pipes will be filling up 13 portions out of total 42 portions every two minutes.

So in 6 minutes 39 portions out of 42 will be filled up and 3 portions will be left. On 7th minute pipe A is open, its rate is 7 portions out of 42 per minute and it will fill up these 3 portions in $\displaystyle\frac{3}{7}$ minutes in addition to 6 minutes.

ES: Essential mental math skill.

Problem 3

Value of $\sqrt{20+\sqrt{20+\sqrt{20+.........}}}$ is,

5
2
3
-4

Solution 3

This is an infinite series of square root of 20 one within another. With such infinite series, just like evaluation of fraction equivalent to non-terminating repeating decimals, we assume the given expression,

$\sqrt{20+\sqrt{20+\sqrt{20+.........}}}=x$,

Or $x^2 =20+x$.

Or, $x^2-x-20=(x-5)(x+4)=0$, the factorization is simple and done in mind

As $x$ cannot be negative, $x=5$.

Answer: Option a: 5.

ES: Finding roots of a quadratic equation by middle term split in mind

Finding roots of such simple quadratic equations is generally done by the method of middle term split.

In this method, we consider the numeric third term $-20$ as $-5\times{4}$. So absolute values of the roots will be 5 and 4.

Now we consider the middle term together with the possible root values,

$\text{Middle term }=-x=-5x+4x$, the only possibility. The factors $(x-5)$ and $(x+4)$ follow immediately.

All in mind.

ES: Essential mental math skill.

Problem 4

If $a:b=3:4$, then $(7a+3b):(7a-3b)$ is,

5 : 2
4 : 3
11 : 3
37 : 19

Solution 4

By noticing the target expression in the form of $(x+y):(x-y)$, we decide to convert the given ratio to $7a:3b$ and then apply Componendo and dividendo technique.

Whenever you find target expression in the form of $(x+y):(x-y)$, usually you can solve the problem quickly using Componendo dividendo technique on input relation $x:y$. This is application of End state analysis and selecting the best possible resource for quick solution. This skill in identifying the best possible path for quick solution comes through intelligent practice.

The given ratio,

$\displaystyle\frac{a}{b}=\frac{3}{4}$,

Or, $\displaystyle\frac{7a}{b}=\frac{21}{4}$,

Or, $\displaystyle\frac{7a}{3b}=\frac{7}{4}$.

By applying Componendo dividendo technique,

$\displaystyle\frac{(7a+3b)}{(7a-3b)}=\frac{7+4}{7-4}=\frac{11}{3}$.

All in mind.

Answer: Option c: $11:3$.

First thing to remember, Given a ratio, $ a:b=3:4$, you can always transform the ratio in many ways.

The second of course is to know and apply the Componendo dividendo technique which we consider as an Essential Mental Math Skill.

ES: Componendo dividendo technique in mind

We have arrived at the desired ratio,

$\displaystyle\frac{7a}{3b}=\frac{7}{4}$.

We will elaborate Componendo dividendo technique in three steps from this point.

First step: add 1 to both sides giving first result,

$\displaystyle\frac{7a+3b}{3b}=\frac{7+4}{4}$

Second step: subtract 1 from both sides giving second result,

$\displaystyle\frac{7a-3b}{3b}=\frac{7-4}{4}$.

Third step: divide first result by second result,

$\displaystyle\frac{(7a+3b)}{(7a-3b)}=\frac{7+4}{7-4}=\frac{11}{3}$.

It is pretty straightforward and easy to apply.

ES: Essential mental math skill.

Problem 5

The value of $(997)^2+(998)^2+(999)^2 -$

$\hspace{30mm} 997\times{998} - 998\times{999}-999\times{997}$ is,

Solution 5

Noticing that the given expression is of the form,

$a^2+b^2+c^2-ab-bc-ca$, we resort to simplification replacing, $a=997$, $b=998$ and $c=999$ and assuming the value of given expression as $x$. This is algebra, though simple, still algebra.

Multiplying the given expression by 2,

$2a^2+2b^2+2c^2-2ab-2bc-2ca=2x$,

Or, $(a-b)^2+(b-c)^2+(c-a)^2$

$= 1+1+4$

$=6$

$=2x$.

So, $x=3$.

Answer: Option c: 3.

To know more about algebra concepts and techniques in a short time, you may refer to Basic and rich algebra concpets.

Problem 6

If the sides of a triangle are in the ratio $3:4:5$, the largest angle of the triangle is,

$75^0$
$90^0$
$120^0$
$50^0$

Solution 6

We know from our experience that it is not a simple task to relate side lengths with angles in a triangle in general. So we focus our attention on the ratio of the sides itself and discover immediately the pattern of $5^2=3^2+4^2$.

This is a frequently encountered pattern of side lengths of a right triangle where hypotenuse is 5, and the relation is the foundation of Trigonometry, the Pythagoras theorem.

We also know that the total of three angles in a triangle being $180^0$, in a right triangle, the right angle is the largest angle.

All in mind.

Answer: Option b: $90^0$.

Concepts used: Domain mapping from pattern in three numbers to nature of the triangle using Pythagoras theorem and HCF reintroduction technique implicitly. After all we needed to have the actual values of side lengths that could be considered as $3x$, $4x$ and $5x$ where $x$ has been the cancelled out HCF.

Problem 7

The largest possible length that can be used for measuring exactly the three lengths, 7m, 3m 85cm and 12m 95cm is,

25cm
15cm
42cm
35cm

Solution 7

This is a straightforward HCF determination problem for the three numbers 700, 385 and 1295 (all three lengths converted to common unit cm).

Taking out factor 5 at the first step the numbers are transformed to,

140, 77, 259.

It seems from the first two that 7 can be another common factor, and on dividing 259 we find it indeed so.

Taking out factor 7 at the second step the numbers are transformed to,

20, 11, 37.

No further common factors.

So HCF is $5\times{7}=35$, and 35cm is the largest required length.

Answer: Option: d: 35cm.

We needed to write down the three sets of three numbers,

700, 385, 1295

140, 77, 259 and

20, 11, 37.

Rest done in mind.

Concept used: HCF by factorization.

Problem 8

Area of a square field is 6050 sq m. What is the length of the diagonal?

110m
120m
135m
112m

Solution 8

If $a$ is the side length of a square, its diagonal is,

$d^2=2a^2$, as the diagonal forms the hypotenuse of a right triangle with two equal sides.

So,

$d^2=2\times{6050}=12100$,

Or, $d=110$m

Answer: Option a: 110m.

Key concepts used: Relation between area of a square and its diagonal length.

Problem 9

Value of $0.\overline{3}+0.\overline{4} +0.\overline{7}+0.\overline{8}$ is,

$2.4$
$2.\overline{4}$
$2.44$
$2.444$

Solution 9

As the first two terms match in repeating digit length, decimal distance and sum of the digits is less than or equal to 9, we can add up the two to get,

$0.\overline{3}+0.\overline{4} +0.\overline{7}+0.\overline{8}$

$=0.\overline{7} +0.\overline{7}+0.\overline{8}$.

For further progress we need to convert the repeating decimals to equivalent fractions and express the sum as,

$\displaystyle\frac{7}{9}+\displaystyle\frac{7}{9}+\displaystyle\frac{8}{9}$

$=\displaystyle\frac{22}{9}$

$=2\displaystyle\frac{4}{9}$

$=2.\overline{4}$.

Answer: Option b: $2.\overline{4}$.

Concepts used: Non-terminating repeating decimal to fraction conversion.

To know more about how this conversion works you may refer to WBCS main level Arithmetic solution set 3 and Fractions and decimals basic concepts 1.

Problem 10

40 men take 8 days to earn Rs.2000. How many men will earn Rs.200 in 2 days?

Solution 10

This is a Time work wage problem where the wage or earning per day for a worker is fixed and is directly proportional to the work rate or portion of work done in a day.

We will use mandays technique to solve the problem in a few steps easily.

By basic time work wage concept, the given statement means, 320 mandays worth of work resulted in an earning of Rs.2000. So Rs.200 will be one-tenth or 32 mandays worth of work.

Now we will apply the time and work concept. as the earning of Rs.200 is to be made in 2 days, for 32 mandays equivalent work, 16 men have to work.

Answer: Option: a: 16.

Special concept used: Time work wage concept, Mandays concept.

Takeaway

Along with the Number system, Ratio and Proportion, Basic algebra concepts, Pythagoras theorem, HCF and LCM, Factorization, Basic time and work concepts, Basic work and wages concepts we have used the important additional concepts and techniques of Deductive reasoning, Boundary condition problem, Educated guess and test, Principle of free resource use, Mental maths, Essential mental math skills, Roots of quadratic equation, Pattern identification technique, Componendo dividendo technique, Repeating decimal to fraction conversion technique,Work rate technique, Mandays technique and HCF reintroduction technique. These techniques 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 the process we have highlighted this time three important Essential Mental math skills and indicated how the skills work.