LCM and HCF 1/1

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Finding HCF and LCM HCF and LCM formulas HCF and LCM Exercise

1. Find the LCM of 16, 24, 32
a. 72
b. 64
c. 96
d. 120

Answer: C

Explanation:
$\qquad \begin{array}{|l}
\llap{2~~~~} 16, 24, 32 \\[4px] \hline
\llap{2~~~~} \phantom{0}8, 12, 16 \\[4px] \hline
\llap{2~~~~} \phantom{0}4, \phantom{0}6, \phantom{0}8 \\[4px] \hline
\llap{2~~~~} \phantom{0}2, \phantom{0}3, \phantom{0}4 \\[4px] \hline
\llap{~~~~} \phantom{0}1, \phantom{0}3, \phantom{0}2 \\[4px]
\end{array}$
LCM ＝ $2 \times 2 \times 2 \times 2 \times 3\times 2$ ＝ $ 2^5 \times 3$

2. Find the HCF of 120, 210
a. 6
b. 12
c. 30
d. 120

Answer: C

Explanation:
To find HCF the remainder in each divison becomes divisor and previous divisor becomes dividend. The last divisor which gives remainder $0$ is HCF of both the numbers.
$\quad\require{enclose}
\begin{array}{rll}
120 ){210}(1{\phantom{000000}} \\[2px]
\underline{120}{\phantom{00000000}} \\[1px]
90) 120 (1 {\phantom{000}}\\[1px]
\underline {90}{\phantom{00000}} \\[1px]
\boxed{30}\phantom{})90(3{\phantom{}} \\[1px]
\underline{90}\phantom{00} \\[1px]
0\phantom{00}
\end{array}$

3.LCM of $\displaystyle\frac{2}{7},\displaystyle\frac{3}{{14}}{\rm{ and }}\displaystyle\frac{5}{3}{\rm{ is}}$
a. 45
b. 35
c. 30
d. 25

Answer: C

Explanation:
$\dfrac{\text{LCM of numerators}}{\text{HCF of denominators}}$ ＝ $\dfrac{\text{LCM of 2, 3, 5}}{\text{HCF of 7, 14, 3}}$ ＝ $\dfrac{{30}}{1} = 30$

4. Find the HCF of $\dfrac{3}{{32}},\;\dfrac{6}{{10}},\;\dfrac{9}{{16}}$
a. 6
b. 12
c. 24
d. 120

Answer: C

Explanation:
HCF of fractions ＝ $\dfrac{{{\text{HCF of numerators}}}}{{{\text{LCM of numerators}}}}$ ＝ $\dfrac{{HCF(3,6,9)}}{{LCM(32,10,16)}}$ ＝ $\dfrac{3}{{160}}$

5. About the number of pairs which have 16 as their HCF and 136 as their LCM, the conclusion can be
a. only one such pair exists
b. only two such pairs exist
c. many such pairs exist
d. no such pair exists

Answer: D

Explanation:
HCF is always a factor of LCM. ie., HCF always divides LCM perfectly. So no such pair exists.

6. HCF of three numbers is 12. If they are in the ratio 1:2:3, then the numbers are
a. 12,24,36
b. 10,20,30
c. 5,10,15
d. 4,8,12

Answer: A

Explanation:
Let the numbers be $a, 2a , 3a$.
Then, HCF of above 3 numbers is = $a$
$ \Rightarrow $ $a=12$
The numbers are = $12,24,36$

7. The HCF of two numbers is 16 and their LCM is 160. If one of the numbers is 32, then the other number is
a. 48
b. 80
c. 96
d. 112

Answer: B

Explanation:
$HCF \times LCM = A \times B$
The number = $\displaystyle\frac{{{\rm{HCF \times LCM}}}}{{{\rm\text{Given number}}}}{\rm{ = }}\displaystyle\frac{{{\rm{16 \times 160}}}}{{{\rm{32}}}}{\rm{ = 80}}$

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Number System: BasicsNumber System: HCF and LCMNumber System: Factors and CoprimesNumber System: Divisbility Rules Number System: Power of a number in a Factorial Number System: Units digit of an expression Number System: Last two digits of an expressionNumber System: Base SystemNumber System: Last non zero digit of a factorial (LNZ)Number System: Last two non zero digits of a factorialNumber System: High level Exercise