WBCS Main Level Arithmetic Solution Set 10

Submitted by Atanu Chaudhuri on Mon, 17/12/2018 - 01:00

$WBCS Arithmetic Solution Set 10$

Solution to 10th WBCS arithmetic practice set

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

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

Problem 1

a, b, c and d are four consecutive odd integers and their average is 42. What is the product of a and c?

1845
1890
1677
1860

Solution 1: Solving in mind: Average and odd number property

If $x$ be the first odd integer in the series of 4, the average is, $x+3=42$. So $x=39$ and the third number is 43. Product of the two is, $39\times{43}=1677$.

The product expression we needed to write down for accuracy.

Answer. Option c: 1677.

Solution 1: Mathematical deduction

The four odd integers are, $a=x$, $b=x+2$, $c=x+4$ and $d=x+6$.

Sum of the four is, $4x+12$, and average,

$\displaystyle\frac{4x+12}{4}=x+3=42$.

So, $x=39$, and,

$ac=39\times{(39+4)}=39\times{43}=1677$.

Concepts used: Average -- Odd integer property -- Solving in mind.

Problem 2

The HCF and LCM of two numbers $x$ and $y$ are respectively 3 and 105. If value of $x+y=36$, then $\displaystyle\frac{1}{x}+\displaystyle\frac{1}{y}$ is,

$\displaystyle\frac{3}{5}$
$\displaystyle\frac{4}{35}$
$\displaystyle\frac{1}{35}$
$35$

Solution 2: Solving in mind: Product relation between HCF and LCM and basic algebra

By the nature of HCF and LCM of two numbers $x$ and $y$, the product of the numbers is equal to the product of the HCF and LCM. So we have here,

$xy=\text{HCF}\times{\text{LCM}}=3\times{105}=315$.

Dividing $x+y=36$ by $xy=315$, we get,

$\displaystyle\frac{x+y}{xy}=\displaystyle\frac{1}{x}+\displaystyle\frac{1}{y}=\displaystyle\frac{36}{315}=\frac{4}{35}$.

Answer: Option b: $\displaystyle\frac{4}{35}$.

Concepts used: HCF and LCM product property -- Basic algebra -- Solving in mind.

Problem 3

A gardener plants 17956 trees in such a way that there are as many rows as there are trees in a row. The number of trees in a row are,

Solution 3: Solution using method of square root of a number

As the desired number of trees must be the square root of 17956, we choose the fastest way to get it by applying method of square root on 17956 as below,

square root of 17956

Answer: Option d: 134.

Concepts used: Square root of an integer.

To know more about the square root method refer to our tutorial,

How to find perfect and approximate square root of integers or decimals.

Problem 4

$\displaystyle\frac{(0.73)^3+(0.27)^3}{(0.73)^2+(0.27)^2-(0.73)\times{(0.27)}}$=?.

$0.73$
$1$
$0.4087$
$0.46$

Solution 4: Solving in mind: Key pattern identification and solution

Knowing that $a^3+b^3=(a+b)(a^2-ab+b^2)$ and identifying that the second factor in the denominator will be eliminated by expanding the numerator leaving $(a+b)=0.73+0.27=1$, solution is reached in a few seconds. This key pattern identification by assuming $a=0.73$, and $b=0.27$ and arriving at the solution have been nearly simultaneous.

Answer: Option b: $1$.

Solution 4: Mathematical deduction

As $a^3+b^3=(a+b)(a^2-ab+b^2)$, assuming $a=0.73$ and $b=0.27$, the given expression is transformed to,

$\displaystyle\frac{(0.73)^3+(0.27)^3}{(0.73)^2+(0.27)^2-(0.73)\times{(0.27)}}$

$=\displaystyle\frac{a^3+b^3}{a^2-ab+b^2}$

$=\displaystyle\frac{(a+b)(a^2-ab+b^2)}{a^2-ab+b^2}$

$=a+b$

$=0.73+0.27$

$=1$.

Concepts: Sum of cubes expansion -- Key pattern identification -- Basic algebraic concepts -- Substitution -- Solving in mind.

Problem 5

The simplified value of $\displaystyle\frac{1}{1000}\left(\displaystyle\frac{1}{5}+999\displaystyle\frac{494}{495}\times{99}\right)$ is,

99
99900
990
9900

Solution 5: Solving in mind by breaking up of mixed fraction and key pattern identification

The key pattern of $495=5\times{99}$ was identified first and then the mixed fraction $999\displaystyle\frac{494}{495}$ involving large number multiplication broken up as,

$999\displaystyle\frac{494}{495}=999+\displaystyle\frac{494}{495}$.

The result in the brackets simplifies to, $999\times{99}+99=99\times{1000}$ in the numerator when taken out of the brackets. Denominator being 1000, it cancels out leaving 99.

Answer: Option a: 99.

Concepts used: Mixed fraction breakup -- Key pattern identification -- Efficient simplification -- Solving in mind.

Mathematical deduction

$\displaystyle\frac{1}{1000}\left(\displaystyle\frac{1}{5}+999\displaystyle\frac{494}{495}\times{99}\right)$

$=\displaystyle\frac{1}{1000}\left(\displaystyle\frac{1}{5}+999\times{99}+\displaystyle\frac{494}{5}\right)$

$=\displaystyle\frac{1}{1000}\left(999\times{99}+\displaystyle\frac{495}{5}\right)$

$=\displaystyle\frac{1}{1000}\left(999\times{99}+99\right)$

$=\displaystyle\frac{1}{1000}(99\times{1000})$

$=99$.

Problem 6

If $\displaystyle\frac{1}{25}:\displaystyle\frac{1}{x}::\displaystyle\frac{1}{x^2}:\displaystyle\frac{1}{78.125}$, then $x$ is,

1.5
12.5
2
3.5

Solution 6: Solution by factorization

Each of the ratios are divisions of two terms and inverting, the two are converted to, $\displaystyle\frac{x}{25}$, and $\displaystyle\frac{x^2}{78.125}$. The proportionality equates the two fractions transposing which we get,

$x^3=25\times{78.125}$.

Let us show this mathematically,

$\displaystyle\frac{1}{25}:\displaystyle\frac{1}{x}::\displaystyle\frac{1}{x^2}:\displaystyle\frac{1}{78.125}$

Or, $\displaystyle\frac{\displaystyle\frac{1}{25}}{\displaystyle\frac{1}{x}}=\displaystyle\frac{\displaystyle\frac{1}{x^2}}{\displaystyle\frac{1}{78.125}}$

Or, $x^3=25\times{78.125}$

We needed to write this result and its next factorization of 78.125 by 25 because 78.125 is divisible by 25,

$x^3=25\times{25}\times{3.125}$

$=25\times{25}\times{0.5}\times{6.25}$

$=5^3\times{2.5^3}$.

So $x=5\times{2.5}=12.5$.

Answer: Option b: 12.5.

Concepts used: Ratios are fractions -- Divisibility -- Factorization -- Proportionality.

Problem 7

The product of two co-primes is 117. Their LCM should be,

117
1
equal to their HCF
cannot be determined

Solution 7: Solving in mind: Using factorization and multiplying the factors

Knowing $117=9\times{13}$, and 117 being product of two co-primes having no factors common between the two, the factors must be 9 and 13 and not 3 and 39. So LCM must be the product of 9 and 13, that is 117.

Solution 7: Solving in mind: by definition of co-primes: Quicker method

As co-primes do not have any factor common to them, their LCM must be their product which is 117. By this faster method, you won't need to know the factors.

Answer: Option a: 117.

Concepts: Co-prime concept -- LCM -- Solving in mind.

Problem 8

Which pair of numbers should come next in the sequence, 9 12 11 14 13 16 15...?

18 21
18 17
17 18
14 13

Solution 8: Solving in mind by breaking up of the series into two and key pattern identification

As two numbers are to be found out, we break up the given series in alternate numbers resulting into two numbers of series,

9 11 13 15..., and

12 14 16...

The next two numbers should then be, 18 17, the fourth of the second series first and the fifth of the first series next.

You need to be careful at this point.

Answer: Option b: 18 17.

Concepts: Number series -- Key pattern identification -- Strategic problem solving -- Problem breakdown technique -- Solving in mind.

Problem 9

The compound interest on Rs. 2000 for two years at 8% per annum is,

Rs. 228.80
Rs. 232.80
Rs. 532.80
Rs. 332.80

Solution 9: Solving in mind: Summing up simple interest and interest on interest of 1st year

At simple interest of 8% on Rs. 2000 for 2 years, the interest per annum would be Rs. 160 and for two years, Rs. 320. When under compound interest scheme, the first year interest will accrue an additional interest at 8% on the interest of first year amounting to, $\text{Rs. 160}\times{0.08}=\text{Rs. 12.80}$.

Total interest would then be, Rs. 332.80.

Solution 9: By Compound interest formula

By compound interest formula, after 2 years at compound interest of 8%, the principal $P$ at the start will become,

$P_2=P(1+r)^2$.

So interest after two years will be,

$P_2-P=2Pr+Pr^2=320+160\times{0.08}=320+12.80=332.80$.

Answer: Option d: Rs. 332.80.

Concepts: Compound interest concept -- Growth concept -- Solving in mind.

Problem 10

Reflex angle between the hands of a clock at 10.25 is,

$180^0$
$195^0$
$192\displaystyle\frac{1}{2}^0$
$197\displaystyle\frac{1}{2}^0$

Solution 10: Solving in mind: Speed of hour hand

At 10.25, the minute hand will be at 5 hour mark, where $360^0$ of the circular face of the clock is divided into 12 parts of $30^0$ each.

So if at 10.25 the hour hand were at 10 hour mark, the reflex angle, the larger angle more than $180^0$, would have been $7\times{30^0}=210^0$.

But hour hand was at 10 mark at 10.00 and moved by $\displaystyle\frac{30^0}{12}=2.5^0$ every five minutes (as it moves by $30^0$ in 60 minutes or 1 hour, that is, $0.5^0$ in 1 minute). In 25 minutes then it had moved from 10 hour mark by $5\times{2.5^0}=12.5^0$.

The reflex angle would then be reduced by this movement and would be,

$210^0-12.5^0=197\displaystyle\frac{1}{2}^0$.

The following figure should make the situation clearer.

Answer: Option d: $197\displaystyle\frac{1}{2}^0$.

Concept: Clock problems -- Speed of hour hand movement in degrees -- Reflex angle -- Solving in mind.

Takeaway

Concepts and techniques used and explained

Average, Number system, Number series, HCF and LCM, Basic algebra, Key pattern identification, Solving in mind, Sum of cubes, Mixed fraction breakup technique, Ratio and proportion, Divisibility, Co-prime numbers, Compound interest, Clock concepts, Clock hour hand speed in degrees, Reflex angle, Efficient simplification, 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 square root in problem 3 and ratio and proportion problem 6, each of the other eight problems could be solved in a few tens of seconds mentally and we explained how.

Concept tutorials and articles on Arithmetic

Question and Solution sets on WBCS Arithmetic

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Solution to 10th WBCS Math Question set for Arithmetic Practice