WBCS Maths Solved Arithmetic Questions Set 9

Submitted by Atanu Chaudhuri on Sat, 15/12/2018 - 21:27

$WBCS Maths Solved Arithmetic Questions Set 9$

Quick solution to 9th WBCS math question set

Learn how to solve arithmetic questions quickly from WBCS maths solved arithmetic questions. Included are boats and streams concepts, fractions and more.

All solutions focus on solving the question mentally.

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WBCS Maths Solved Arithmetic Questions Set 9: time to answer was 10 mins

Problem 1

What will be the value of $8^{0.28}\times{4^{0.08}}$?

$1$
$4^{0.36}$
$2$
$8^{0.36}$

Solution 1: Solving in mind

Identifying the pattern that 3 times 0.28 plus 2 times 0.08 is 1, we get the answer straightway as 2.

Answer. Option c: $2$.

Solution 1: Mathematical deduction

We have to equalize the base so that the powers of two product terms can be added. We go for the base of 2 directly,

$8^{0.28}\times{4^{0.08}}$

$=(2^3)^{0.28}\times{(2^2)^{0.08}}$

$=2^{3\times{0.28}}\times{2^{2\times{0.08}}}$

$=2^{0.84}\times{2^{0.16}}$

$=2$.

Answer: Option c: $2$.

Concepts used: Indices concepts -- Key pattern identification -- Base equalization technique --- Deductive reasoning -- Solving in mind.

Explanation of concepts and process of solving

First pattern identification: The problem is on indices. Two term product is to be simplified.

First conclusion by deductive reasoning: To combine the two product terms, the bases 8 and 4 must be transformed to the same value so that the powers of the two can be added as in,

$a^x\times{a^y}=a^{x+y}$.

This is a specific application of the powerful general Base equalization technique. To know more refer to our article,

Base equalization technique in solving math problems on indices elegantly.

Second conclusion and solution

$8=2^3$, and $4=2^2$. So common value base can only be 2. Decision made then was to convert 8 and 4 to powers of 2 and reach a simple solution because of the hidden pattern in the powers of the two product terms that we identified in the beginning itself.

Problem 2

The selling price of an article is $\displaystyle\frac{4}{3}$ times its cost price. The gain percentage is,

$33\displaystyle\frac{1}{3}$%
$25\displaystyle\frac{1}{4}$%
$20\displaystyle\frac{1}{2}$%
$20\displaystyle\frac{1}{3}$%

Solution 2: Solving in mind

Gain is difference of selling price and cost price, that is, $\displaystyle\frac{1}{3}$ of cost price, and so gain percentage is, $33\displaystyle\frac{1}{3}$%

Answer: Option a: $33\displaystyle\frac{1}{3}$%.

Solution 2: Explanation

Profit or actual gain is,

$\text{Selling price}-\text{Cost price}$

$=\displaystyle\frac{4}{3}\text{ of Cost price}-\text{Cost price}$

$=\displaystyle\frac{1}{3}\text{ of Cost price}$.

So percentage gain is,

$\displaystyle\frac{\text{Actual gain}}{\text{Cost price}}\times{100}$

$=\displaystyle\frac{100}{3}$

$=33\displaystyle\frac{1}{3}$%.

Answer: Option a: $33\displaystyle\frac{1}{3}$%.

Concepts used: Profit and loss -- Fraction to percentage conversion -- Percentage gain is actual gain as percentage of cost price -- Solving in mind.

Problem 3

In a mixture of 60 litres, the ratio of milk and water is 2 : 1. If the ratio of milk and water is to be 1: 2, then the amount of water (in litres) to be added is,

Solution 3: Solving in mind

In 60 litres, that is in 3 portions, 2 portions or 40 litres is milk and 20 litres is water. In new mixture milk amount remains unchanged, but the water amount need to be double of 40 litres, that is 80 litres. In the present 20 litres of water then 60 litres of water is to be added.

Answer: Option d: 60.

Concepts used: Ratio proportion -- Mixture or alligation -- Portion concept -- Solving in mind.

Problem 4

If $\text{20% of a} = \text{30% of b} = \displaystyle\frac{1}{6} \text{of c}$, then $a:b:c$ is,

$2:3:16$
$3:2:16$
$10:15:18$
$15:10:18$

Solution 4: Solving in mind

Taking the first pair, $a:b=15:10$, and the second pair, $b:c=10:18$, and joining the two by equal link term value 10 of $b$, $a:b:c=15:10:18$.

Answer: Option d: $15:10:18$.

Concepts: Percentage to fraction -- ratio concepts -- Ratio joining.

Explanation

Given are two equations joined together by two equality symbols. We will break them into two parts. Taking the first part,

$\text{20% of a} = \text{30% of b}$,

Or, $\displaystyle\frac{a}{5}=\frac{3b}{10}$, converting the percentages to fractions dividing by 100 in each case,

Or, $\displaystyle\frac{a}{b}=\displaystyle\frac{15}{10}$, we have not minimized the fraction because we noticed the term value of 10 in the second part.

Similarly from the second part we get,

$\text{30% of b} = \displaystyle\frac{1}{6} \text{ of c}$,

Or, $\displaystyle\frac{3b}{10}=\frac{c}{6}$,

Or, $\displaystyle\frac{b}{c}=\frac{10}{18}$.

Middle value of 10 for common link term $b$ being equal we join the two ratios directly without any transformation and get,

$a : b: c=15:10:18$.

Problem 5

Which fraction comes next in the sequence,

$\displaystyle\frac{1}{2}$, $\displaystyle\frac{3}{4}$, $\displaystyle\frac{5}{8}$, $\displaystyle\frac{7}{16}$,..?

$\displaystyle\frac{12}{35}$
$\displaystyle\frac{9}{32}$
$\displaystyle\frac{11}{34}$
$\displaystyle\frac{10}{17}$

Solution 5: Solving in mind by key pattern identification

Identifying the key patterns—numerator values are in sequence 1, 3, 5, 7..., and denominator values in another sequence, 2, 4, 8, 16..., we deduce easily the next term in the given sequence as, $\displaystyle\frac{9}{32}$.

Answer: Option b: $\displaystyle\frac{9}{32}$.

Concepts used: Number series -- Fraction series -- Key pattern identification -- Consistent relation between numerator values -- Consistent relation between denominator values -- Solving in mind.

Solution 5: Explanation of solving the series problem

In the given problem on series of fractions, either there would be a consistent relation between two consecutive terms, or in this case of series of fractions, a consistent relation between the consecutive numerator values as well as another consistent relation between the two consecutive denominator values. It takes practically no time with this approach to recognize that consecutive two values of numerator values differ by 2, and consecutive two values of denominator increase by a factor of 2.

Problem 6

The simple interest on a sum for 5 years is two-fifth of the sum. The percentage interest rate per annum is then,

$10$%
$6$%
$8$%
$12\displaystyle\frac{1}{2}$%

Solution 6: Solving in mind

Under simple interest scheme interest remains constant every year. So per year interest in this case is, $\displaystyle\frac{2}{25}$th of sum, or $8$%.

Answer: Option c: $8$%.

Concepts used: Simple interest remains constant every year so that in 1 year it will be one-fifth of total interest in 5 years -- Simple interest per annum is the increase in percentage of initial sum -- Fraction to percentage conversion by multiplying with 100 -- Solving in mind.

Problem 7

A clock is started at noon. By 10 minutes past 5, the hour hand has turned through an angle of

$155^0$
$145^0$
$150^0$
$160^0$

Solution 7: Solving in mind: Using Clock hour hand movement speed

The hour hand moves from one hour mark to the next hour mark in 1 hour, that is, in 60 minutes. The difference or distance between two consecutive hour marks is, $\displaystyle\frac{360^0}{12}=30^0$.

So in 5 hours the hour hand moves by $5\times{30^0}=150^0$ and in 10 minutes more it moves by another $30^0\times{\displaystyle\frac{10}{60}}=5^0$, a total of $155^0$.

Answer: Option a: $155^0$.

Concepts: Clock concepts -- Clock hour hand speed in degrees -- Twelve hours of a clock make $360^0$ -- Solving in mind.

Problem 8

The speed of a boat in still waters is 15 km/hr. It goes 30 km upstream and returns to the starting point downstream in 4 hrs 30 mins. The speed of current of the stream is,

8 km/hr
5 km/hr
10 km/hr
15 km/hr

Solution 8: Upstream speed and downstream speed concepts

Speed of boat moving upstream reduces from its still water speed by the stream speed, whereas its speed downstream is increased by the stream speed,

$\text{Upstream speed}=V-S$, where $V$ and $S$ are the still water speed of the boat and stream speed,

$\text{Downstream speed}=V+S$.

In our problem $V=15$ km/hr.

The basic speed time distance concept tells us,

$\text{time}=\displaystyle\frac{\text{distance}}{\text{speed}}$.

Adding up the upstream and downstream travel times we get,

$\displaystyle\frac{30}{15-S}+\displaystyle\frac{30}{15+S}=4.5$,

Or, $\displaystyle\frac{900}{225-S^2}=4.5$,

Or, $\displaystyle\frac{1}{225-S^2}=\frac{1}{200}$.

So by observation we identify, $S=5$ km/hr.

Answer: Option b: 5 km/hr.

Concepts: Boats in rivers -- Upstream speed -- Downstream speed -- Speed time distance relation -- Efficient simplification -- Delayed evaluation technique.

Efficient simplification

First we combined the two LHS terms using algebraic relation, $(a+b)(a-b)=a^2-b^2$ for denominator, and $30\times{2a}$ for numerator.

Next we made the LHS numeraror value as 1 and simplified the RHS by removing the decimal by multiplying by 2 and common factor 9 between the numerator and denominator.

The possible value of $S=5$ was immediately apparent from the simplified equation. Thus we avoided solving a quadratic equation or early evaluation of products.

Problem 9

Which of the following fractions lies between $\displaystyle\frac{3}{5}$ and $\displaystyle\frac{2}{3}$?

$\displaystyle\frac{1}{15}$
$\displaystyle\frac{1}{3}$
$\displaystyle\frac{2}{5}$
$\displaystyle\frac{31}{50}$

Solution 9: Solving in mind: Placement of a fraction within a range

First step: Verify the given range to be in ascending order. Indeed,

$\displaystyle\frac{2}{3}-\displaystyle\frac{3}{5}=\displaystyle\frac{1}{15} \gt 0$.

Second step: Examining and comparing the choice values with one of the two given fractions and then if required, with the second factor. If a fraction is greater than the higher given value or smaller than the lower given value, it is out of range. Only when a fraction is greater than the smaller given fraction and lesser than the larger given fraction, the choice falls within the range.

Solution 9: Solving in mind: Choosing option values of fraction for comparison suitably

By inspection we find, $\displaystyle\frac{1}{15}$, $\displaystyle\frac{2}{5}$ and $\displaystyle\frac{1}{3}$ to be smaller than the lower range boundary. So, the last choice, $\displaystyle\frac{31}{50}$ must be the answer. We verify for certainty,

$\displaystyle\frac{31}{50}-\displaystyle\frac{3}{5}=\displaystyle\frac{1}{50} \gt 0$, and,

$\displaystyle\frac{2}{3}-\displaystyle\frac{31}{50}=\displaystyle\frac{7}{150} \gt 0$.

Answer: Option d: $\displaystyle\frac{31}{50}$.

Concepts: Fraction range placement -- Fraction comparison -- Solving in mind -- Problem solving strategies -- Choice value test.

With right problem solving strategy we could solve the problem in mind easily in quick time. You should also be able to do it.

Problem 10

How many digits are required for numbering the pages of a 300 page book ?

Solution 10: Solving in mind

By digit length we divide the numbers from 1 to 300 in three groups,

1 to 9: 9 numbers, each 1 digit long: number of total digits 9;

10 to 99: 90 numbers, each 2 digit long: total digits 180;

100 to 300: 201 numbers, each 3 digit long: total digits 603.

So grand total is,

$9+180+603=792$.

Answer: Option b: 792.

Concept: Total number of digits in a range of numbers -- Problem breakdown technique in breaking up the range into three ranges each of which with numbers of equal length.

Takeaway

We have used and explained the concepts and methods,

Percentage, Percentage to fraction, Fraction to percentage, Number system, Number series, Choice value test, Indices concepts, Key pattern identification, Base equalization technique, Deductive reasoning, Solving in mind, Profit and loss, Percentage gain, Ratio proportion, Mixture or alligation, Portion concept, Ratio joining, Simple interest, Clock concepts, Clock hour hand speed in degrees, Boats in rivers, Upstream speed, Downstream speed, Speed time distance relation, Efficient simplification, Fraction range, Fraction comparison, Problem solving strategies, Problem breakdown technique.

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.

Except the boats in rivers problem 8, each of the other nine problems could be solved in a few tens of seconds and we explained how.

Concept tutorials and articles on Arithmetic

Question and Solution sets on WBCS Arithmetic

For all WBCS main arithmetic question sets click here.

For all WBCS main arithmetic solution sets click here.

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WBCS Main Level Arithmetic Solution Set 9