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Multiple choice maths be my multiple, i'll be your factor co-prime numbers lcm lowest common multiple (l.c.m.)

The HCF of $3^5, 3^9$, and $3^{14}$ is

  1. $3^5$
  2. $3^9$
  3. $3^{14}$
  4. $3^{21}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

To  find  the  Highest  Common  Factor  (HCF)  of  two  or  more  numbers, 
find  prime  factors  of  the  numbers , and  then  identify  the  common  prime  factors.
Then  the  HCF  is  the  product  of  the  common  prime  factors.
$3^{5}= 1 \times 3^{5}$
$3^{9} = 1 \times 3^{5} \times 3^{4}$
$3^{14} = 1 \times 3^{5} \times 3^{9}$
Hence the HCF is $ 3^{5}.$
Another  method  to  find  the  answer  is  to find  the  largest  divisor  of all  three  numbers,
 which  is $ 3^{5} $

Multiple choice maths be my multiple, i'll be your factor co-prime numbers lcm lowest common multiple (l.c.m.)

Find the common factors of the given terms:

$6 abc, 24ab^2, 12 a^2b$

  1. $6a^2b$
  2. $6ab^2$
  3. $6ab$
  4. $6$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$6 abc$, $24ab^2$, $12 a^2b$
The factors of $6abc=2\times 3\times a\times b\times c$
The factors of $24ab^2=2\times 2\times 2\times 3\times a\times b\times b$
The factors of $12 a^2b=2\times 2\times 3\times a\times a\times b$
The common factors are $2\times 3\times a\times b=6ab$

Multiple choice maths be my multiple, i'll be your factor co-prime numbers lcm lowest common multiple (l.c.m.)

Find the common factors of the given terms:

$16 x^3, 4x^2, 32x$

  1. $x$
  2. $16$
  3. $4x$
  4. $32$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$16 x^3$,  $4x^2$, $32x$
The factors of $16 x^3=2\times 2\times 2\times 2\times x\times x\times x$
The factors of $4x^2=2\times 2 \times x\times x$
The factors of $32x=2\times 2\times 2\times 2\times 2\times x$
The common factors are $2\times 2 \times x=4x$

Multiple choice maths be my multiple, i'll be your factor co-prime numbers lcm lowest common multiple (l.c.m.)

Solve the given exponent:
$\sqrt[4]{12} \times \sqrt[7]{6}$  

  1. $2^{\frac{9}{14}}\times 3^{\frac{11}{28}}$
  2. $3^{\frac{9}{14}}\times 2^{\frac{11}{28}}$
  3. $2^{\frac{1}{14}}\times 3^{\frac{1}{28}}$
  4. $3^{\frac{1}{14}}\times 2^{\frac{1}{28}}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$\sqrt[4]{12} \ \times \ \sqrt[7]{6}$


$=(12)^{\frac{1}{4}} \ \times \ (6)^{\frac{1}{7}}$

$=(2\times2\times3)^{\frac{1}{4}} \ \times \ (2\times3)^{\frac{1}{7}}$

$=(2^2\times3)^{\frac{1}{4}} \ \times \ (2\times3)^{\frac{1}{7}}$

$=2^{2\times(\frac{1}{4})}\times 3^{\frac{1}{4}} \ \times2^\frac{1}{7}\times3^\frac{1}{7}$

$=2^{\frac{1}{2}}\times 3^{\frac{1}{4}} \ \times2^\frac{1}{7}\times3^\frac{1}{7}$

$=2^{(\frac{1}{2}+\frac{1}{7})}\times 3^{(\frac{1}{4}+\frac{1}{7})}$-----If base is same, then their powers can be added, by product law.

$=2^{\frac{9}{14}}\times 3^{\frac{11}{28}}$

Option A.

Multiple choice maths be my multiple, i'll be your factor co-prime numbers lcm lowest common multiple (l.c.m.)

Choose the most appropriate option.
The traffic lights at three different signal points change after every $45$ seconds, $75$ seconds and $90$ seconds respectively. If all change simultaneously at $7:20:15$ hours, then they will change again simultaneoulsy at.

  1. $7:27:30$ hours
  2. $7:28:00$ hours
  3. $7:27:50$ hours
  4. $7:27:45$ hours
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
The $3$ signals (at $3$ points) change every $45 s,\, 75 s,\, 90 s$

So, they will change simultaneously for a common time, which is the common multiple or $L.C.M$ of the three

$\Rightarrow$  $45 = 5\times 9 = 3^2 \times 5$ 

$\Rightarrow$  $75 = 3\times 25 = 3\times 5^2$

$\Rightarrow$  $90 = 9\times 10 = 2\times 3^2\times 5$

$L.C.M= 2 \times 3^2 \times 5^2 = 2\times 9\times 25 = 450s$

So, they will change simultaneously every $450s$ or $7\,mins\,30 \,sec$

$\Rightarrow$  So, next they will change together at $7:27:45$ hours.