## Solution to 11th WBCS arithmetic practice set

*11th set of WBCS Arithmetic solution explains how the 10 questions in the arithmetic practice questions can be solved easy and quick in 10 minutes.*

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

### 11th WBCS arithmetic solution set: time to answer was 10 mins

#### Problem 1

What will be the percentage markup on cost price to achieve a profit of 4% after 20% discount?

- 10%
- 35%
- 25%
- 30%

#### Solution 1: Solving in mind: Profit or loss is on Cost price, Discount is on marked or list price

To achieve a profit of 4% on cost price $C$, sale price would be $1.04C$. After 20% discount on marked price $M$, sale price becomes, $0.8M$. This equals sale price,

$1.04C=0.8M$.

So percentage markup on cost price to reach the marked price is the percentage difference of marked price and cost price with reference to cost price,

From the above relation,

$1.04C=0.8M$,

Or, $M=1.3C$.

So, percentage markup would be 30% on cost price (**1.3C means C plus 30% of C**).

**Answer.** Option d: 30%.

**Concepts used:** Percentage to decimal conversion, dividing by 100 -- Profit concept, it is on cost price and additional to cost price -- Marked price, the listed price on which discount percentage reduces it to sale price -- Decimal to percentage conversion -- Solving in mind.

#### Problem 2

In a 200 litre solution of sugar fully dissolved in water, sugar is 20%. On heating for a while the solution gets more concentrated with sugar percentage increasing to 80% of the solution. What is the volume of water that evaporated?

- 140 litre
- 150 litre
- 130 litre
- 100 litre

#### Solution 2: Solving in mind: With drying up of solution, sugar amount in weight remains unchanged with only water amount getting reduced

Sugar is 20% of solution weight $W$ at the start with volume 200 litres. So sugar weight was 0.2W and it remains unchanged while water evaporated.

After evaporation, sugar of amount $0.2W$ becomes $0.8W_D$, $W_D$ being the new weight of the dried up solution.

So, $0.2W=0.8W_D$,

Or, $W_D=0.25W$, that is, present solution weight reduced to one-fourth of the original weight.

As weight of homogeneous solution is proportional to the original volume of 200 litres as well as the present volume, present dried up volume will be one-forth of 200 litres, that is 50 litres.

So evaporated water volume is,

$200-50=150$ litres.

**Answer:** Option b: 150 litres.

**Concepts used:** Drying up of homogeneous solution -- Percentage to decimal conversion -- Proportionality of volume to weight, sugar is in kgs but solution is in litres -- Solving in mind.

#### Problem 3

At what time between 1 O'clock and 2 O'clock both minute hand and hour hand will be together?

- $5\displaystyle\frac{6}{11}$ minutes past 1 O'clock
- $5\displaystyle\frac{4}{11}$ minutes past 1 O'clock
- $5\displaystyle\frac{5}{11}$ minutes past 1 O'clock
- $5\displaystyle\frac{3}{11}$ minutes past 1 O'clock

#### Solution 3: Solving in mind: Realtive angular speed of minute and hour hands

Minute hand moves round the clock face and traverses $360^0$ in $60$ minutes. Its speed is, $6^0$ per minute.

Hour hand traverses distance between two hour marks, that is, $30^0$ in 60 minutes. Its speed is $\displaystyle\frac{1}{2}^0$ per minute.

The *minute hand then approaches the hour hand* at a **relative speed** of,

$6^0-\displaystyle\frac{1}{2}^0=\displaystyle\frac{11}{2}^0$ per minute, with hour hand effectively standing still.

It is this speed at which the minute hand actually closes the gap between itself and the hour hand with both moving.

At 1 O'clock, minute hand stood at 00 or 12 hour mark and hour hand stood at 1 hour mark, fully $30^0$ ahead of the minute hand.

At a relative speed of $\displaystyle\frac{11}{2}^0$ per minute the minute hand will close this gap and will catch up with the hour hand in,

$\displaystyle\frac{\text{Distance}}{\text{Speed}}$

$=\displaystyle\frac{30}{\displaystyle\frac{11}{2}}$

$=\displaystyle\frac{60}{11}=5\displaystyle\frac{5}{11}$ minutes.

The two hands will then meet at $\displaystyle\frac{5}{11}$ minutes past 1 O'clock.

**Answer:** Option c: $5\displaystyle\frac{5}{11}$ minutes past 1 O'clock.

**Concepts used:** Clock concepts -- Speed of hour hand and the minute hand -- Relative speed of minute hand and hour hand -- Race concept -- Solving mind.

The abovementioned tutorial on * How to solve clock problems* should make the concepts clear.

#### Problem 4

The simple interest over 2 years at a certain interest rate on a certain amount is Rs. 2400. If the difference between the compound interest compounded annually at same rate over same period on same amount, and the simple interest be Rs. 138, what is the percentage interest rate per annum?

- 13.5%
- 10.5%
- 12.5%
- 11.5%

#### Solution 4: Solving in mind: Concept of difference between simple interest and compound interest

Simple interest over 2 years assuming $x$ as the amount invested and $r$ as the interest rate is, $2xr$ and per year it is $xr=\text{Rs. }1200$.

Under compound interest the first year interest will be same as the simple interest over the first year. Compounding will happen over the second year and the extra compounded interest will be on the simple interest earned over the first year, that is, $xr$ at the rate of $r$. So the compound interest extra to simple interest is, $xr^2=\text{Rs. }138$.

Dividing the two we get,

$\displaystyle\frac{xr^2}{xr}$

$=r$

$=\displaystyle\frac{138}{1200}\times{100}$%

$=\displaystyle\frac{23}{2}=11.5$%.

**Answer:** Option d: 11.5%.

**Concepts:** Simple interest -- Compound interest -- Difference between simple interest and Compound interest -- Efficient simplification -- Solving in mind.

#### Problem 5

A basketful of oranges are counted in pairs, in 3s and in 5s, and every time one orange is left over. The least number of oranges in the basket is,

- 31
- 41
- 61
- 51

#### Solution 5: Solving in mind: Use of the concept underlying Euclid's division lemma

If a larger number $a$, the dividend, is divided by a smaller number $b$, the divisor, it holds that,

$a=bq+r$, where $q$ is the quotient and remainder $r$ may be 0 but must be less than $b$.

The three countings give rise to three divisions of the total number of oranges, say, $N$,

$N=2q_1+1$, Or, $N-1=2q_1$

$N=3q_2+1$, Or, $N-1=3q_2$, and

$N=5q_3+1$, Or, $N-1=5q_3$.

So if an integer $N$ reduced by 1 is a multiple of the LCM 30 of 2, 3 and 5, when divided by 2, 3 and 5, a remainder of 1 will be left in each case.

The smallest such number will then be,

$30+1=31$.

**Answer:** Option a: 31.

**Concepts used:** Remainder concept in Euclid's division lemma -- Key pattern identification -- LCM -- Solving in mind.

Naturally, for solving the problem mentally we didn't have to write the equations, and arriving directly on the **key pattern** that the number will be 1 more than the LCM of 2, 3 and 5.

#### Problem 6

HCF of $\displaystyle\frac{12}{13}$ and $\displaystyle\frac{3}{5}$ is,

- $3$
- $12$
- $\displaystyle\frac{12}{65}$
- $\displaystyle\frac{3}{65}$

#### Solution 6:

HCF of a pair of fractions is,

$\displaystyle\frac{\text{HCF of numerators}}{\text{LCM of denominators}}$

So in this problem, HCF of the two given fractions is,

$\displaystyle\frac{\text{HCF of 12 and 3}}{\text{LCM of 13 and 5}}$

$=\displaystyle\frac{3}{65}$.

**Answer:** Option d: $\displaystyle\frac{3}{65}$.

**Concepts used:** HCF of fractions -- HCF of integers -- LCM of integers -- Solving in mind.

#### Problem 7

Two rational numbers lying between $\displaystyle\frac{4}{5}$ and $\displaystyle\frac{6}{7}$ are,

- $\displaystyle\frac{29}{35}$, $\displaystyle\frac{62}{70}$
- $\displaystyle\frac{28}{34}$, $\displaystyle\frac{35}{39}$
- $\displaystyle\frac{29}{35}$, $\displaystyle\frac{5}{6}$
- $\displaystyle\frac{65}{84}$, $\displaystyle\frac{5}{6}$

#### Solution 7: Solving in mind: Check and ensure the range of two fractions to be in ascending order: Comparison of fractions

To reduce number of comparisons we check and ensure that the range values are indeed in ascending order.

We apply **specific rules** here **for comparing** and deciding, which fraction between the two is the larger one.

We will apply the **First rule or technique of fraction comparison **for comparing $\displaystyle\frac{4}{5}$ and $\displaystyle\frac{6}{7}$:

If the difference between the numerator and denominator of two fractions are same, the fraction with larger numerator will be the larger one.

So without actually doing a subtraction, we can conclude,

$\displaystyle\frac{4}{5} \lt \displaystyle\frac{6}{7}$.

#### Solution 7: Strategy of checking choice values with range limit values

With two range limit fractions in ascending order, we will follow the **principle of checking valid or invalid choice** as,

If any choice value is less than the lower limit or greater than the higher limit, then the choice values can't be placed within the range and the choice is invalid.

Keeping this basic principle in mind, we adopt a **systematic approach**,

We will compare one value of a

suitable choicewith the lower limit of the range. If invalid, we will select second suitable choice, otherwise we will check the second value of the choice with the higher limit.

As $\displaystyle\frac{29}{35}$ appears in two choice values we will check it first. As suitable choice, we will select the third choice with second fraction $\displaystyle\frac{5}{6}$ rightaway being identified to be less than the higher limit of $\displaystyle\frac{6}{7}$ (by rule 1).

**Solution 7: Comparing $\displaystyle\frac{4}{5}$ with $\displaystyle\frac{29}{35}$**

To compare these two fractions we will use the generally known rule of base equalization.

**Second rule and technique of fraction comparison: Denominator equalization**

If denominators of two fractions are same, the fraction with larger numerator will be the larger.

By multiplying the numerator and denominator of $\displaystyle\frac{4}{5}$ by 7, the fraction is converted to $\displaystyle\frac{28}{35}$, thus equalizing the two denominators. Now we can easily identify,

$\displaystyle\frac{29}{35} \gt \displaystyle\frac{4}{5}$.

With both values of the third choice within the given range, this must be the correct answer (as, in the MCQ problem, answer can only be one).

These decisions can be quickly taken if you know the techniques and follow a systematic approach.

**Answer:** Option c: $\displaystyle\frac{29}{35}$, $\displaystyle\frac{5}{6}$.

**Concepts:** Fraction range placement problem -- Strategic problem solving approach -- Fraction comparison -- Base equalization technique -- Solving in mind.

**Note:** Rule 3, the **third powerful rule for fraction comparison** might also have been used if required,

If denominator numerator difference of two fractions is smaller for the fraction with larger numerator, it is the larger of the two.

For example, $\displaystyle\frac{7}{9} \gt \displaystyle\frac{4}{7}$.

This is true because of **percentage difference between denominator and numerator is smaller for the larger fraction**.

For example, to compare $\displaystyle\frac{4}{5}$ with $\displaystyle\frac{65}{84}$ we will take the denominator to 85, as close to 84 as possible, by multiplying both numerator and denominator by 17. The converted fraction $\displaystyle\frac{68}{85}$ is larger than $\displaystyle\frac{64}{84}$ by rule 3.

**Caution:** Unless you are clear on these techniques and are experienced in fraction comparison, attempting a fraction range placement problem may take a lot of your valuable time, and so should be skipped.

#### Problem 8

What is the unit's digit of the product of all the prime numbers between 10 and 30?

- 7
- 9
- 3
- 1

#### Solution 8: Solving in mind: Multiplying the unit's digit of the prime numbers in the range

The concept used in this problem is,

Unit's digit of the product of two integers will be the unit's digit of the product of the unit's digit of the two integers.

For example unit's digit of $1467\times{459}$ will be unit's digit of $7\times{9}$ which will be 3.

Using this concept we take the first two prime numbers 11 and 13 and remember the unit's digit of the product as 3.

Then we continue to evaluate the unit's digit of $3\times{17}$ giving 1, then $1\times{19}$ giving 9, then unit's digit of $9\times{23}$ giving 7 and finishing with, unit's digit of $7\times{29}$ giving 3.

**Answer:** Option c: 3.

**Concepts:** Unit's digit of the product of two integers as the unit's digit of the product of their unit's digits -- Prime numbers -- Solving in mind.

#### Problem 9

Present ages (in years) of Romi and Runu are in the ratio of 5 : 6. Three years ago their ages were in the ratio 9 : 11. What will the ratio of their ages after 6 years from now?

- 6 : 7
- 1 : 2
- 7 : 8
- 3 : 4

#### Solution 9: Solving in mind: Ratio concept and HCF reintroduction technique

Introducing the cancelled out HCF as a product of both terms of the ratio we have the ratio $5:6=5x:6x$.

Thre years ago,

$\displaystyle\frac{5x-3}{6x-3}=\frac{9}{11}$.

Cross-multiplying,

$55x-33=54x-27$,

Or, $x=6$.

So 6 years from now the ratio of their ages will be,

$\displaystyle\frac{30+6}{36+6}=\frac{6}{7}$, the ages of both increasing by 6 years.

**Answer:** Option a: 6 : 7.

**Concepts:** Ratio concept -- HCF reintroduction technique -- Age concept: $x$ years from now age of all increases by $x$ years -- Solving in mind.

#### Problem 10

Inlet pipes A and B can fill a tank independently in 15 mins and 12 mins respectively. If both are opened simultaneously and B is closed after 3 mins how much more time will A take to fill the tank?

- 6 mins
- 8 mins 15 secs
- 10 mins 30 secs
- 9 mins 15 secs

#### Solution 10: Solving in mind: Fill rate in terms of portion of tank filled per unit time

Pipe A fills the tank at a fill rate of $\displaystyle\frac{1}{15}$ portion of tank per minute and pipe B fills at a fill rate of $\displaystyle\frac{1}{12}$ portion of tank per minute. So together in 3 minutes the two pipes will fill,

$\displaystyle\frac{1}{5}+\displaystyle\frac{1}{4}=\displaystyle\frac{9}{20}$ portion of tank.

Left will be then $\displaystyle\frac{11}{20}$ portion of tank.

Applying unitary method,

Pipe A fills up 1 tank or full tank in 15 minutes,

So, pipe A will fill up $\displaystyle\frac{11}{20}$ portion of tank in,

$15\times{\displaystyle\frac{11}{20}}=\displaystyle\frac{33}{4}=8.25$ mins, or, 8 mins 15 secs.

**Answer:** Option b: 8 mins 15 secs.

**Concept:** Pipes and cistern problems -- Fill rate as portion of tank filled in unit time -- Working together concept -- Unitary method -- Solving in mind.

### Takeaway

We have used and explained the **concepts and methods:**

*Percentage, Percentage to decimal conversion, Decimal to percentage conversion, Profit and loss concepts, Marked price, Discount, Drying up of homogeneous solution, Concentration of dissolved matter,**Clock concepts, Speed of hour hand and the minute hand, Relative speed of minute hand and hour hand, Race concept, Simple interest, Compound interest, Difference between simple interest and Compound interest, Remainder concept in Euclid's division lemma, Key pattern identification, LCM of integers, HCF of fractions, HCF of integers, Fraction range placement problem, Strategic problem solving approach, Fraction comparison, Base equalization technique, Unit's digit of product of two integers, Prime numbers, Ratio concept, HCF reintroduction technique, Age problems, Pipes and cistern problems, Fill rate as portion of tank filled in unit time, Working together concept, Unitary method,** Solving mind.*

All problems could be solved in mind in a few tens of seconds, but with use of requisite concepts, identification of patterns and application of powerful methods.

We include this approach of solving a problem in mind in the **expanded scope of mental maths.**

### 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.**