H.C.F and L.C.M — BSSC Quantitative Aptitude

• Step-by-step solution:
LCM of fractions = $\frac{\text{LCM of Numerators}}{\text{HCF of Denominators}}$.
Numerators are $2, 4$, and $5$. Their LCM is $20$.
Denominators are $3, 9$, and $6$. Their HCF is $3$.
Result = $\frac{20}{3}$.

• Exam Trick:
Since the options have different denominators, quickly check the LCM of the numerators ($2, 4, 5$), which is $20$. Only options with numerator $20$ are plausible. Then check the HCF of denominators ($3, 9, 6$), which is $3$. This directly gives $\frac{20}{3}$.

• Step-by-step solution:
The fundamental rule of HCF and LCM is: $\text{First Number} \times \text{Second Number} = \text{HCF} \times \text{LCM}$.
$84 \times \text{Second Number} = 12 \times 336$.
$\text{Second Number} = \frac{12 \times 336}{84}$.
Since $84 = 12 \times 7$, this simplifies to $\frac{336}{7} = 48$.

• Exam Trick:
Use the unit digit trick! The product of the unit digits of HCF and LCM ($2 \times 6 = 12$) ends in $2$. The product of the unit digits of the two numbers must also end in $2$. The first number ends in $4$. $4 \times \text{Option's unit digit}$ must end in $2$. Look at the options: $4 \times 8 = 32$ (ends in $2$). Option 1 ($48$) is the only one matching.

• Step-by-step solution:
The time after which the bells toll together again is the LCM of their individual intervals.
$\text{LCM of } 12, 15, 20, \text{ and } 30$.
$12 = 2^2 \times 3$
$15 = 3 \times 5$
$20 = 2^2 \times 5$
$30 = 2 \times 3 \times 5$
$\text{LCM} = 2^2 \times 3 \times 5 = 60 \text{ seconds}$.
$60 \text{ seconds} = 1 \text{ minute}$.
Time = $10:00\text{ AM} + 1 \text{ minute} = 10:01\text{ AM}$.

• Exam Trick:
The largest number is $30$. Multiples of $30$ are $30, 60, 90...$ Check which is divisible by all. $60$ is divisible by $12, 15, 20$, and $30$. So LCM is $60$ seconds ($1$ minute). Add it to the start time.

• Step-by-step solution:
The required number must perfectly divide $(411 - 3), (684 - 4), \text{ and } (821 - 5)$.
These numbers are $408, 680, \text{ and } 816$.
We need to find the HCF of $408, 680$, and $816$.
$408 = 136 \times 3$
$680 = 136 \times 5$
$816 = 136 \times 6$
The HCF is $136$.

• Exam Trick:
The numbers to be divided are $408, 680, \text{ and } 816$. Look at the difference between the first two numbers: $680 - 408 = 272$. The HCF must be a factor of the difference ($272$). Now check options: $272$ does not divide $680$ perfectly. The next largest factor of $272$ in the options is $136$. Check if $136$ divides all three: $408/136=3$, $680/136=5$, $816/136=6$. It works! So $136$ is the answer.

• Step-by-step solution:
First, find the LCM of the divisors $12, 15, 20$, and $35$.
$12 = 2^2 \times 3$
$15 = 3 \times 5$
$20 = 2^2 \times 5$
$35 = 5 \times 7$
$\text{LCM} = 2^2 \times 3 \times 5 \times 7 = 420$.
The general form of the required number is $420k + 5$.
For a $4$-digit number, $k$ must be at least $3$ (since $420 \times 2 = 840$, which is $3$-digit).
If $k = 3$, Number = $420 \times 3 + 5 = 1260 + 5 = 1265$.

• Exam Trick:
If the number leaves a remainder of $5$ when divided by $20$, the number minus $5$ must be perfectly divisible by $20$. Only option A ($1265 - 5 = 1260$) and option C ($1685 - 5 = 1680$) meet this. Since the question asks for the "smallest" 4-digit number, option A ($1265$) is the winner.

• Step-by-step solution:
Let the common ratio multiplier be $x$. The two numbers are $3x$ and $4x$.
The LCM of $3x$ and $4x$ is $12x$.
We are given that $\text{LCM} = 180$, so $12x = 180 \implies x = 15$.
The two numbers are $3 \times 15 = 45$ and $4 \times 15 = 60$.
Their sum is $45 + 60 = 105$.

• Exam Trick:
The sum of the two numbers is proportional to $3 + 4 = 7$. This means the final sum must be a multiple of $7$. Looking at the options ($105, 120, 90, 75$), $105$ is the ONLY number perfectly divisible by $7$ ($105 / 7 = 15$). Solved without calculating the LCM!

• Step-by-step solution:
Let the two numbers be $33a$ and $33b$, where $a$ and $b$ are co-prime integers.
According to the question, $33a + 33b = 528$.
$33(a + b) = 528 \implies a + b = \frac{528}{33} = 16$.
Now, we find pairs of co-prime numbers $(a, b)$ that sum up to $16$:
The pairs are $(1, 15), (3, 13), (5, 11), \text{ and } (7, 9)$.
(Pairs like $(2, 14), (4, 12), (6, 10), (8, 8)$ are not co-prime as they share a common factor $2$).
Thus, there are $4$ such pairs.

• Exam Trick:
No shortcut, standard method is already fastest. Just divide the sum by the HCF ($528/33 = 16$) and count the co-prime pairs. Remember that evens summing to an even number will share a factor of $2$, so only odd pairs work for $16$.

• Step-by-step solution:
Make the number of decimal places equal: $1.75$, $5.60$, and $7.00$.
Now, temporarily remove the decimals and find the HCF of the integers $175, 560, \text{ and } 700$.
$175 = 25 \times 7 = 5^2 \times 7$
$560 = 35 \times 16 = 2^4 \times 5 \times 7$
$700 = 100 \times 7 = 2^2 \times 5^2 \times 7$
The common factors are $5$ and $7$. So, $\text{HCF} = 5 \times 7 = 35$.
Since we adjusted for two decimal places, place the decimal back: $\text{HCF} = 0.35$.

• Exam Trick:
The HCF must perfectly divide the smallest number, $1.75$. Eliminate $3.5$ as it is larger than $1.75$. Test $0.7$: $0.7 \times 2 = 1.4$ and $0.7 \times 3 = 2.1$, so it does not divide $1.75$. This leaves $0.35$ and $0.07$. Since $0.35 \times 5 = 1.75, 0.35 \times 16 = 5.6, \text{ and } 0.35 \times 20 = 7$, $0.35$ works and is the highest.

• Step-by-step solution:
Convert dimensions to cm: Length = $1517\text{ cm}$, Breadth = $902\text{ cm}$.
To find the minimum number of tiles, the size of each square tile must be maximum. This is the HCF of $1517$ and $902$.
Difference method for HCF: $1517 - 902 = 615$. Factors of $615$ are $5 \times 3 \times 41$.
Since neither $1517$ nor $902$ ends in $0$ or $5$, $5$ is not a factor. Sum of digits of $902$ is $11$ (not div by $3$). Thus, HCF is $41$.
Side of largest square tile = $41\text{ cm}$.
Number of tiles = $\frac{\text{Area of floor}}{\text{Area of 1 tile}} = \frac{1517 \times 902}{41 \times 41} = 37 \times 22 = 814$.

• Exam Trick:
Unit digit shortcut: The number of tiles along the length is $\frac{1517}{41}$, which must end in $7$ (since $1 \times 7 = 7$). The number of tiles along the breadth is $\frac{902}{41}$, which must end in $2$ (since $1 \times 2 = 2$). Total tiles = $(\dots 7) \times (\dots 2)$, so the final answer must end in $4$. Only $814$ ends in $4$!

• Step-by-step solution:
To find the LCM of algebraic expressions, multiply the LCM of their numerical coefficients by the highest power of each variable present.
Numerical coefficients: $12$ and $18$. Their LCM is $36$.
Variables:
Highest power of $x$ is $x^3$.
Highest power of $y$ is $y^3$.
Highest power of $z$ is $z^1$ (or just $z$).
LCM = $36 \times x^3 \times y^3 \times z = 36x^3y^3z$.

• Exam Trick:
Visual inspection is enough here. Look for the largest number that $12$ and $18$ both divide into (which is $36$) and simply pick the highest exponents you see for every letter ($x^3, y^3, z$). Distractor $6x^2y^2$ is actually the HCF, not LCM.