Find the H.C.F. of 2.1, 1.05 and 0.63.
Answer: C
Clear the decimals by scaling all three by 100, then rescale the answer.
\[ \text{H.C.F.}(210, 105, 63) = 21 \]
\[ \frac{21}{100} = 0.21 \]
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If the sum of two numbers is 55 and the H.C.F. and L.C.M. of these numbers are 5 and 120 respectively, then the sum of the reciprocals of the numbers is equal to:
Answer: C
Let the numbers be \(a\) and \(b\).
Then,
\(a + b = 55\) and
\(ab = 5 \times 120 = 600\)
The required sum
\(=\dfrac{1}{a} + \dfrac{1}{b}\\~\\= \dfrac{a+b}{ab}\\~\\=\dfrac{55}{600}\\~\\= \dfrac{11}{120}\)
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Find the greatest number that will divide 400, 435 and 541 leaving 9, 10 and 14 as remainders respectively
Answer: B
Subtract each remainder, then take the H.C.F. of the results.
\[ 400 - 9 = 391, \quad 435 - 10 = 425, \quad 541 - 14 = 527 \]
\[ 391 = 17 \times 23, \quad 425 = 17 \times 25, \quad 527 = 17 \times 31 \]
\[ \text{H.C.F.} = 17 \]
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The Greatest Common Divisor for two numbers is 8 while their Least Common Multiple is 144. Find the other number if one number is 16.
Answer: C
For any two numbers,
\[ \text{H.C.F.} \times \text{L.C.M.} = \text{product of the numbers} \]
\[ 8 \times 144 = 16 \times n \quad\Longrightarrow\quad n = \frac{1152}{16} = 72 \]
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The least number, which when divided by 12, 15, 20 and 54 leaves in each case a remainder of 8 is:
Answer: D
The number is 8 more than a common multiple of all four.
\[ 12 = 2^{2}\cdot 3, \ 15 = 3\cdot 5, \ 20 = 2^{2}\cdot 5, \ 54 = 2\cdot 3^{3} \]
\[ \text{L.C.M.} = 2^{2} \cdot 3^{3} \cdot 5 = 540 \]
\[ 540 + 8 = 548 \]
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Find the largest number which divides 62,132,237 to leave the same reminder.
Answer: C
When the remainder is the same but unknown, take the H.C.F. of the differences - the remainder cancels out of each.
\[ 132 - 62 = 70, \quad 237 - 132 = 105, \quad 237 - 62 = 175 \]
\[ \text{H.C.F.}(70, 105, 175) = 35 \]
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Find the H.C.F. of \(\dfrac{2}{3}, \dfrac{4}{6}, \dfrac{8}{27}\)
Answer: A
Formula,
\(H.C.F. = \dfrac{H.C.F. of Numerators}{L.C.M. of Denominators}\)
So,
\(\dfrac{H.C.F. of (2,4,8) }{L.C.M. of (3,6,27)} \\~\\=\dfrac{2}{27}\)
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The least multiple of 7, which leaves a remainder of 4, when divided by 6, 9, 15 and 18 is:
Answer: D
L.C.M. of 6, 9, 15 and 18 is 90.
Let required number be \(90k + 4\), which is multiple of \(7\).
Least value of k for which \((90k + 4)\) is divisible by \(7\) is \(k = 4.\)
Required number,
\(= (90 \times 4) + 4\)
\(= 364\)
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The sum and HCF of two numbers are 156 and 13. The numbers of such number pairs is
Answer: A
If the HCF is 13, both numbers are 13 times something, and those somethings share no factor. Write the pair as \( 13a \) and \( 13b \) with \( \gcd(a,b)=1 \):
\[ 13a + 13b = 156 \quad\Longrightarrow\quad a + b = 12 \]
So the question becomes: how many coprime pairs sum to 12?
Two pairs qualify. The coprimality condition is the whole point: without it the HCF would be larger than 13 and the given data would be contradicted.
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If the LCM of two numbers is 138. But their GCD is 23. The numbers are in a ratio 1:6. Which is the largest number amongst the two?
Answer: B
With H.C.F. 23 the numbers are \( 23m \) and \( 23n \) with \( m \) and \( n \) coprime, and the ratio \( 1 : 6 \) makes them 23 and 138.
\[ \text{L.C.M.} = 23 \times 1 \times 6 = 138 \quad\checkmark \]
So the larger number is 138.
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