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## Solving 10 WBCS arithmetic practice problems in mind

4th set of WBCS Arithmetic Practice Problems solved using basic concepts and advanced techniques for solving arithmetic questions in minimum time.

*Topics of the questions cover boats rivers, number system, profit loss, fraction range, ratio, factorization, simplification by basic algebra* should be useful for *prelims of other competitive exams* also.

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

### 4th set of WBCS Arithmetic Practice problems solved in mind: time to answer was 10 mins

#### Problem 1

In a river a man takes 3 hrs in rowing 3kms upstream, but rowing downstream he covers 15kms in same 3 hrs. The speed of the river current is,

- 2km/hr
- 4km/hr
- 9km/hr
- 6km/hr

#### Solution 1

#### Basic boats in rivers concept

If $V$ is the boat speed in still waters and $R$ be the river current speed,

- Effective downstream speed of the boat is, $V+R$, the boat will move faster as the river current speed will aid the boat movement, and
- Effective upstream speed of the boat is, $V-R$, the boat will move slower as the river current speed will oppose the boat movement.

In this problem, downstream speed, $V+R=\displaystyle\frac{15}{3}=5\text{km/hr}$, and

Upstream speed, $V-R=\displaystyle\frac{3}{3}=1\text{km/hr}$.

Subtracting second from the first and dividing by 2 we get the river current speed as,

$R=\displaystyle\frac{4}{2}=2\text{km/hr}$

All in mind and using the simple concept of Boats in Rivers.

**Note. This class of problems fall in a more general class, Moving platform problems.**

**Answer:** Option a: 2km/hr.

#### Problem 2

The product of two successive positive integers is 1980. The smaller of the number is,

- 34
- 44
- 45
- 35

#### Solution 2

We will use estimate of squares near the choice numbers and compare with number 1980, as this number is $(\text{Choice number})^2 +(\text{Choice number})$.

We can see that $40^2=1600$ is much less than 1980. So the choice number must be either 44 or 45 (because both 34 and 35 are less than 40).

Taking 45 we form its square as 2025, larger than 1980. So 44 must be the answer.

If we first take 44 as the test choice, we form the square as 1936 and add 44 to get 1980.

All in mind.

**Answer:** Option b: 44.

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

You need to be proficient in mental multiplication of two numbers of two digit numbers.

### ES: Method of mental multiplication of two numbers of 2 digit integers

We would calculate, $79\times{48}$ in mind.

**Step 1: ** *Multiply unit's digits* 8 and 9 $\Rightarrow$ result: $8\times{9}=72$ $\Rightarrow$ unit's digit 2 carry 7

**Step 2: ** *Cross multiply and add* 8 with 7 and 4 with 9 $\Rightarrow$ $8\times{7}+4\times{9}=92$ add carry 7 $\Rightarrow$ 99 $\Rightarrow$ ten's digit 9 carry 9

**Step 3: ** *Multiply ten's digits and add carry* $\Rightarrow$ $4\times{7}+9=37$ $\Rightarrow$ result 3792.

orDaily practicecombined withdrill of two-digit number multiplicationimproves speed of mental calculation dramatically. This speed is essential for success in any competitive test math.number table drill

#### ES: Essential mental math skill.

#### Problem 3

A shop owner gains the selling price of 20 chairs by selling 100 chairs. What is his gain percent?

- 15%
- 12%
- 25%
- 20%

#### Solution 3

**Basic profit and loss concept:** Profit or loss is on Cost price ($CP$), and so, $\text{Sale price } SP = \text{Cost price } CP+\text{Profit}$.

In this problem if the gain is $20SP$ and total sale price is $100SP$, $CP=80SP$, and gain percent is, 25%.

All in mind, using basic concepts only, *profit and loss concept* and *percentage concep*t.

**Answer:** Option c: 25%.

#### Problem 4

The rational numbers lying between $\displaystyle\frac{1}{4}$ and $\displaystyle\frac{3}{4}$ are,

- $\displaystyle\frac{262}{1000}$, $\displaystyle\frac{752}{1000}$
- $\displaystyle\frac{9}{40}$, $\displaystyle\frac{31}{41}$
- $\displaystyle\frac{13}{50}$, $\displaystyle\frac{264}{350}$
- $\displaystyle\frac{63}{250}$, $\displaystyle\frac{187}{250}$

#### Solution 4

#### Range test and ES: Fraction comparison

We will compare the *first fraction of a choice pair* with $\displaystyle\frac{1}{4}$ to see whether the first fraction is less than range start $\displaystyle\frac{1}{4}$.

If the fraction tested is less than $\displaystyle\frac{1}{4}$, the choice pair of will be out of range and so invalid. We reject the choice pair.

If the first fraction tested is more than $\displaystyle\frac{1}{4}$, we compare the second fraction with $\displaystyle\frac{3}{4}$ to see whether this second fraction is less than $\displaystyle\frac{1}{3}$ which is the range end.

If this second fraction of a choice fraction pair is less than $\displaystyle\frac{3}{4}$, the choice pir of fractions is the answer, otherwise the choice pair is out of range and we take up a new choice.

This is **Range test based upon fraction comparison.**

For first pair, numerator 262 is more than 250, the one-fourth of 1000. So this choice might be the answer and we must test the second fraction. But 752 is more than 750 which is three-fourth of 1000. So this choice is invalid.

For second choice pair, the first fraction fails the test as 9 is less than 10 the one-fourth of 40.

For third choice pair, 13 is larger than 12.5 the one-fourth of 50. To test the second fraction we notice the numerator to be a multiple of 3, the other factor being 88. 4 times 88, that is, 352 is larger than 350. So $\displaystyle\frac{264}{350}$ is larger than $\displaystyle\frac{264}{352}=\frac{3}{4}$, (equal numerator, smaller denominator means larger fraction).

The fourth choice must then be the answer. If you have time you might test it for validity.

**Answer:** Option d: $\displaystyle\frac{63}{250}$, $\displaystyle\frac{187}{250}$.

Though we did all the calculations and estimates in mind, this problem might take nearly a minute.

To know more about fraction comparison you may refer to our tutorial on **Fractions and decimals basic concepts part 1.**

#### ES: Essential mental math skill.

#### Problem 5

The ratio of two numbers is 5 : 8 and their difference is 69. The numbers are,

- 69, 128
- 115, 184
- 128, 197
- 43, 112

#### Solution 5

**Basic ratio concepts:** In ratio $A:B=5:8$, the fraction is minimized with HCF between the actual values of $A$ and $B$ eliminated.

**HCF reintroduction technique:** Based on fundamental ratio concept we reintroduce the HCF as $x$ multiplying it with both numerator and denominator of the fraction, giving,

$\displaystyle\frac{A}{B}=\frac{5x}{8x}$, where actual values of $A$ and $B$ are $5x$ and $8x$ respectively.

The difference between the numbers is then,

$3x=69$,

Or, $x=23$.

The numbers are,

115, 184.

**Answer: **Option b: 115, 184.

#### Problem 6

By what least number must 21600 be multiplied to make it a perfect cube?

- 10
- 30
- 6
- 60

Solution 6

We recognize 216 to be $6^3$. So we multiply the number 21600 by 10 to make the number as 216 multiplied by $10^6$, a cube of 100.

The cube root of transformed number 216000 would be 600.

**Answer:** Option a: 10.

**Concepts used:** Cube root, Indices.

#### Problem 7

The value of $\left[\left\{\left(-\displaystyle\frac{1}{2}\right)^2\right\}^{-2}\right]^{-1}$ is,

- $\displaystyle\frac{1}{4}$
- $\displaystyle\frac{1}{16}$
- $4$
- $16$

#### Solution 7

We evaluate the innermost term first. It becomes $\displaystyle\frac{1}{4}$. The negative power inverts it to 4, square of which is 16. The last power of $-1$ just inverts the result again to finally, $\displaystyle\frac{1}{16}$.

**Answer:** Option: b: $\displaystyle\frac{1}{16}$.

**Concept used:** Indices.

#### Problem 8

Each side of a rhombus is 5cm. Its area is,

- 23 cm sq
- 24 cm sq
- 25 cm sq
- data insufficient

#### Solution 8

#### Area of a Rhombus

Let us see how a rhombus behaves and what are its basic characteristics. In the figure below ABCD is a rhombus all four sides of which are of equal length $a$,

In a rhombus, its area = base side $\times{}$ height = a $\times{}$ h. If you push the rhombus from right to left holding its corner A and keeping its base CD fixed, it adjusts its shape to the rectangle shape of CDGF without losing or adding any area. The area of the triangles $\triangle AGD$ and $\triangle BFC$ are same.

As area of the rectangle = a $\times{}$ h, we get the area of the rhombus as = base $\times{}$ height.

In our problem we have side length $a=5$cm, but we have not been given any clue to find the height $h$. As you can see from the $\triangle ADG$, DG is height $h$ and is different from side length of the rhombus $AD=a$.

**Answer:** Option d : data insufficient.

**Key concepts used:** Basic concepts on area of rhombus, and data insufficiency concept.

#### Problem 9

If $A:B=\displaystyle\frac{1}{6} : \displaystyle\frac{1}{5}$, $B:C=\displaystyle\frac{1}{4} : \displaystyle\frac{1}{3}$, and $C:D=\displaystyle\frac{1}{3} : \displaystyle\frac{1}{5}$, then $A:D$ is,

- $\displaystyle\frac{1}{24} : \displaystyle\frac{1}{25}$
- $\displaystyle\frac{1}{25} : \displaystyle\frac{1}{29}$
- $\displaystyle\frac{1}{24} : \displaystyle\frac{1}{29}$
- $\displaystyle\frac{1}{27} : \displaystyle\frac{1}{25}$

#### Solution 9

**Conversion of fraction ratio terms:** To conveniently join or relate two ratios in a series of ratios, we will convert all the fraction ratio terms to integer terms by multiplying by the LCM of the corresponding fraction denominators,

$A:B=5:6$, $B:C=3:4=6:8$, $C:D=5:3=40:24$, first we equalized B values of first and second to 6, then proactively for equalization of C values of third and second we converted C value of the third ratio to 40,

Or, $A:B=25:30$, $B:C=30:40$, $C:D=40:24$, at second stage all values of joining variables are equalized; 40 of C is transmitted back to the first getting 25 for A.

S, $A:B:C:D=25:30:40:24$,

Or, $A:D=25:24$,

Or, $A:D=\displaystyle\frac{1}{24} : \displaystyle\frac{1}{25}$, dividing by LCM $25\times{24}$, to convert the $A$ and $D$ ratio term values to fractions.

**Answer: **Option a: $\displaystyle\frac{1}{24} : \displaystyle\frac{1}{25}$

**Concepts used: ** *Advanced Ratio joining*.

#### Problem 10

3 men or 5 women can do a work in 12 days. In how many days will 6 men and 5 women will take to complete the work?

- 4 days
- 15 days
- 20 days
- 10 days

#### Solution 10

If we take 3 men as the first team and 5 women as the second team of workers, both teams do work at same rate to complete the work independently in 12 days, and so these two are *equivalent teams as far as work rate is concerned*.

In 6 men and 5 women thus we have * three such equivalent teams* who will finish the work while working together in one-third of 12 days, that is, 4 days.

**Answer:** Option: a: 4 days.

**Special concept used:** *Work team concept*.

### Takeaway

Along with the *Boats in rivers concepts, Profit and loss concepts, basic number system concepts, Fractions, Factorization, HCF and LCM, Basic profit and loss concepts, Ratio and Proportion, Time and work problems, **Indices, Rhombus area, and Data insufficiency,* we have used the important additional techniques of *Fraction comparison techniques, Fraction ratio terms, Work rate technique, **Work teams, **Ratio joining, Advanced ratio joining,* 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 two of the important * Essential Mental math skills* and indicated how the skills can be improved.

### Concept tutorials and articles on Arithmetic

#### Number system, fractions and surds

**Numbers, Number system and basic arithmetic operations**

**Factorization or finding out factors**

**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**

**How to find perfect square root of integers or decimals**

#### Ratio and proportion and mixing liquids

**Arithmetic problems on mixing liquids and based ages**

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**How to solve arithmetic mixture problems in a few steps 2**

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**Basic and rich concepts on simple interest and compound interest**

#### Work time, work wages and pipes and cisterns

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#### Speed time distance, train problems and boats in rivers

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**How to solve time distance problems in a few simple steps 1**

**How to solve time distance problems in a few simple steps 2**

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**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**

#### Clocks and Calendars

**How to solve Calendar problems**

### Question and Solution sets on WBCS Arithmetic

**10 WBCS Math Questions - 12th Solved Arithmetic Practice Set**

**Solution to 11th WBCS Math Question set for Arithmetic Practice**

**11th WBCS Math Question set for Arithmetic Practice**

**Solution to 10th WBCS Math Question set for Arithmetic Practice**

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**4th WBCS Math Question set for Arithmetic Practice**

**Solution to 3rd WBCS Math Question set for Arithmetic Practice**

**3rd WBCS Math Question set for Arithmetic Practice**

**Solution to 2nd WBCS Math Question set for Arithmetic Practice**

**2nd WBCS Math Question set for Arithmetic Practice**

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**WBCS Main level Arithmetic Question set 1**