WBCS Main level Arithmetic solution set 6 | SureSolv

WBCS Main level Arithmetic solution set 6

Sixth solution set for WBCS Main level Arithmetic

wbcs-main-level-solution-set-6

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 sixth 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.

6th WBCS Main level Arithmetic solution set: time to answer was 10 mins

Problem 1

If the ratio of three numbers are $3:4:5$ and their LCM is 1200, then the smallest number is,

  1. 80
  2. 60
  3. 120
  4. 100

Solution 1

HCF reintroduction technique and concept of HCF and LCM

Introducing HCF of the three numbers as a common factor to the three ratio terms we get the actual values of the three numbers as , $3x$, $4x$ and $5x$.

In LCM, all unique factors of the numbers along with the single factor of HCF will be present. 

For example, if three numbers were 6, 8 and 10, their LCM is 120 and HCF 2. Unique uncommon factors in the three numbers are 3, 4 and 5 (with no common factors between any of these taken as pairs), multiplying each of which with HCF we get the numbers. That's why LCM becomes product of all unique uncommon factors of the numbers, in this case 3, 4 and 5 and the HCF 2, which equals 120.

The other way round, if we divide the LCM by the product of the unique uncommon factors, 60 we get the HCF as 2.

In our problem then we divide 1200 by 60 to get HCF $x=20$ and the desired smallest number $3x=60$.

To solve this problem you need clear concept of HCF and LCM.

Remember, when you are forming the product of three unique uncommon factors of the three numbers (without HCF as a factor) there might be a common factor between a pair of these factors. If it is so while forming the LCM dividing product, that common factor is to be taken once only.

For example, LCM of 4, 6 and 8 is 24 and HCF 2. Unique uncommon factors between the three (without HCF factor) seems to be 2 , 3, and 4. But, as first and third has 2 as a common factor, we have to take only one 2 making the dividing product not 24 but 12. Now if we divide the LCM by this product we will get the HCF as 2. This is needed as in LCM no factor of the three numbers can appear more number of times than it appears in any one number.

$4=2\times{2}$

$6=2\times{3}$

$8=2\times{2}\times{2}$

2 is the highest common factor, the least common multiple is,

$2\times{3}\times{2}\times{2}=24$.

As a second example for numbers, 6, 14 and 70, the factors are,

$6=2\times{3}$

$14=2\times{7}$

$70=2\times{5}\times{7}$.

HCF is 2 and LCM is,

$210=2\times{3}\times{7}\times{5}$, we drop one 7 common in the last two numbers while forming the LCM, though it is not common in all the three numbers.

If we had included two 7s in LCM it would have violated the basic rules in forming the LCM of a few numbers,

  1. No factor common in all numbers can appear more than once in the LCM, and
  2. No factor can appear more number of times than it appears in any single number. 7 does appear only once in 14 and 70, not twice.

This complication comes only in case of more than two numbers.

Answer: Option b: 60.

Problem 2

The least number which when diminished by 5 is divisible by each of 21, 28, 36, 45 is,

  1. 425
  2. 1260
  3. 1265
  4. 1259

Solution 2

Conventional Method: find the LCM of the four numbers and then add 5. It takes some time.

Alternatively, we subtract 5 from each of the choices and do factor test between visibly common factors of the four given numbers and the choice reduced by 5.

Fourth choice result 1254 not multiple of 5 (a factor in 45) and so invalid.

Second choice result 1255 not a multiple of 9 (a factor in 45) so invalid. Integer sum (sum of digits of 1255) 13 is not divisible by 9.

First choice result 420 is not also divisible by 9 and so invalid.

1265 must be the answer. Its result 1260 multiple of 9 and 5 as well as 7 and 4. These factor isolation and divisibiltiy test can be done quickly.

All in mind, and faster.

Answer: Option c: 1265.

You need to be proficient in mentally identifying commonly used factors of 2, 3, 4, 5, 7 and 9 in numbers. Efficient divisibility rules help.

ES: Method of applying divisibility rules mentally for identifying commonly used factors in numbers

Divisibility rule for 2 as a factor: number must be even.

Divisibility rule for 4 as a factor: last two digits of the number must be divisible by 4.

Divisibility rule for 8 as a factor: last three digits of the number must be divisible by 8.

Divisibility rule for 3 as a factor: Integer sum or sum of digits of the number must be divisible by 3.

Divisibility rule for 9 as a factor: Integer sum or sum of digits of the number must be divisible by 9.

Divisibility rule for 5 as a factor: last digit of the number must be 5 or 0.

Factor identification of 7: Recommendation: directly divide by 7. There is a rule for this, but it is an extra mental load as it is not straightforward like the rest.

ES: Essential mental math skill.

Problem 3

A containter contains $x$ litres of milk. $y$ litres of milk was taken out and replaced by water. How much milk will be there in the container after carrying out this process $(n-1)$ times?

  1. $x\left(1-\displaystyle\frac{y}{x}\right)^n$ litres
  2. $x\left(1-\displaystyle\frac{y}{x}\right)^2$ litres
  3. $x\left(1-\displaystyle\frac{y}{x}\right)^{n-1}$ litres
  4. $x\left(1-\displaystyle\frac{y}{x}\right)^3$ litres

Solution 3

Basic concept of milk replacement with water

Before 1st replacement, milk was $x$ litres, after replacement it became, $x-y=x\left(1-\displaystyle\frac{y}{x}\right)=m_1$ litres.

Total volume of mixture still remains $x$ litre.

So in $x$ litre of mixture, milk is $m_1$ litre.

Thus, in every litre of mixture, milk is now, $\displaystyle\frac{m_1}{x}$ litre. Assumption here is, the mixture is a homogeneous liquid with milk and water uniformly distributed throughout the mixture.

2nd replacement: now out of this mixture again $y$ litre of mixture is replaced with water. The milk amount reduces by the milk contained in this $y$ litre of mixture.

At this stage, in 1 litre of mixture milk was $\displaystyle\frac{m_1}{x}$ litre, so in $y$ litre milk was $\displaystyle\frac{ym_1}{x}$ litre.

So reduced amount of milk after second replacement will be,

$m_1-\displaystyle\frac{ym_1}{x}=m_1\left(1-\displaystyle\frac{y}{x}\right)=x\left(1-\displaystyle\frac{y}{x}\right)^2$ litre.

Likewise, after this process is carried out $(n-1)$ times, the milk in the container will be,

$x\left(1-\displaystyle\frac{y}{x}\right)^{n-1}$ litre.

Answer: Option c: $x\left(1-\displaystyle\frac{y}{x}\right)^{n-1}$ litre. 

Problem 4

What number should come next in the sequence, 6, 18, 72, 360, 2160...?

  1. 15120
  2. 12120
  3. 13120
  4. 14120

Solution 4

The pattern identified is, 3 times 6 is 18; 4 times 18 is 72; 5 times 72 is 360 and 6 times 360 is 2160.

So the next term in the sequence should be, $7\times{2160} =15120$.

Answer: Option a: 15120.

Problem 5

A pump can fill a tank with water in 2 hrs. Because of a leak developed in the tank, it takes now $2\displaystyle\frac{1}{2}$ hrs to fill the tank. The leak will empty the full tank with pump not running, in,

  1. $8$ hrs
  2. $10$ hrs
  3. $14$ hrs
  4. $2\displaystyle\frac{1}{3}$ hrs

Solution 5

Rate of filling the tank without leak, $\displaystyle\frac{1}{2}$ of tank in 1 hr.

Rate of filling the tank with leak, $\displaystyle\frac{2}{5}$ of tank in 1 hr.

Expressing the relation of leak and pump working together in 1 hr,

Rate of filling by pump - Rate of emptying by leak = Effective rate of filling,

Thus,

Rate of emptying the tank by the leak is then, 

$\displaystyle\frac{1}{2} - \displaystyle\frac{2}{5}=\displaystyle\frac{1}{10}$ of tank in 1 hr.

Reversing, time taken for the leak to empty the full tank working independently is then 10 hrs.

Answer: Option b: 10 hrs.

Problem 6

Gold is 19 times as heavy as water and copper is 9 times as heavy as water. The ratio in which these two metals are to be mixed so that the mixed alloy becomes 15 times as heavy as water is,

  1. 2 : 3
  2. 1 : 2
  3. 3 : 2
  4. 19 : 135

Solution 6

Assuming $w$ as the weight of water, actual weights of gold and copper are, $19w$ and $9w$.

If $a$ portions of gold is mixed with $b$ portions of copper to get the alloy 15 times as heavy as water, the weight of $a+b$ portions of alloy is $(a+b)15w$.

So,

$19aw+9bw=(a+b)15w$,

Or, $19a+9b=15a + 15b$,

Or, $4a=6b$,

Or, $a:b=3:2$.

Answer: Option c: 3 : 2.

Concepts used: HCF reintroduction technique -- mixture and alligation -- ratio and proportion -- portion concept.

Problem 7

A student was asked to divide a number by 3, but instead of dividing he multiplied it by 3 and got 29.7 as the result. The correct answer would be,

  1. 9.9
  2. 9.3
  3. 9.8
  4. 3.3

Solution 7

Assuming the number as $x$,

$3x=29.7$,

So, $x=9.9$.

Dividing 3 would give the result, 3.3.

Answer: Option d: 3.3.

Alternatively without full division of 29.7 by 3 we get the value of $x$ as 9 plus. Only the fourth choice then satisfies division by 3. This is combined use of number estimation and choice value test. In case of a longer division, this approach would have given the answer faster.

Problem 8

Which of the following fractions is the largest?

  1. $\displaystyle\frac{7}{8}$
  2. $\displaystyle\frac{63}{80}$
  3. $\displaystyle\frac{31}{40}$
  4. $\displaystyle\frac{13}{16}$

Solution 8

In comparing first with second we equalize denominator by multiplying the numerator and denominator of the first by 10. As numerator of the transformed first numerator 70 is larger than 63, the numerator of the second, the first fraction is larger.

Same way we find the first larger than the third, and fourth.

We identified the pattern that denominator of the first is a factor of all the other three and used denominator equalization in all comparisons.

Answer: Option a: $\displaystyle\frac{7}{8}$

Key concepts used: Pattern identification -- Fraction comparison by denominator equalization.

Fraction comparison

Fraction comparison can also be done by numerator equalization if it is more convenient. In this case, the fraction with smaller denominator will be the larger fraction.

Occasionally when both methods cannot be applied but the numerator and denominator difference is same, the fraction with larger numerator is the larger fraction.

All these three methods are shortcut quick methods for fraction comparison. If none of these methods can be applied, the only path left is to form the LCM of denominators and cross multiply with suitable factors following conventional method of fraction subtraction or addition.

Problem 9

A fisherman can row 2 km against the stream in 20 minutes and return in 15 minutes. The speed of the boat in still water is,

  1. $2$ km per hour
  2. $3$ km per hour
  3. $1\displaystyle\frac{1}{2}$ km per hour
  4. $7$ km per hour

Solution 9

Quick conceptual solution by getting against stream speed as 6 km per hour which is still water speed less the stream speed. So the still water speed must be more than 6 km per hour. Only option d satisfies this condition. Pattern identification also used along with choice value test, usual norm for answering MCQ tests questions quickly.

Answer: Option d: $7$ km per hour.

Conventional solution

If sill water speed of boat is $B$ km per hour and stream speed is $S$ km per hour,

Against stream speed or upstream speed is, $B-S$ km per hour and,

With the stream or downstream speed is, $B+S$ km per hour.

From first statement,

$B-S=\displaystyle\frac{2}{\displaystyle\frac{1}{3}}=6$ km per hour

From second statement,

$B+S=\displaystyle\frac{2}{\displaystyle\frac{1}{4}}=8$ km per hour.

So still water boat speed is,

$B=\displaystyle\frac{6+8}{2}=7$ km per hour.

Problem 10

Two trains running in opposite direction cross a man standing on the platform in 27 secs and 17 secs respectively and they cross each other in 23 secs. The ratio of their speeds is,

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

Solution 10

Train running concept

When a running train crosses a man standing on the platform, efffectively it traverses its length, and when two trains running in opposite directions cross each other, they cross sum of their lengths in speed that is a sum of their speeds.

Problem solving 10

If $A$ and $B$ are speeds and $L$ and $M$ be the lengths of two trains,

$27A = L$,

$17B=M$, normalizing length unit according to time unit in seconds

Adding,

$27A+17B=L+M$.

By second statement,

$23(A+B)=L+M=27A+17B$,

Or, $4A=6B$,

Or, $A:B=3:2$.

Answer: Option b: 3 : 2.

Concepts used: Train running -- Unit normalization -- speed time distance.

Takeaway

Along with the Boats in rivers concepts, Fractions, Factorization, HCF and LCM, Basic profit and loss concepts, Ratio and Proportion, Mixture or alligation, and Pipes and cisterns, we have used the important additional concepts and techniques of Fraction comparison techniques, LCM rules, Divisibility rules, Unit normalization, Number estimation skill 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 one of the important Essential Mental math skills of factor identification using a short set of divisibility rules.


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