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Co-prime numbers - class-X

Description: co-prime numbers
Number of Questions: 24
Created by:
Tags: numbers maths integers, powers and roots revision of previous lessons be my multiple, i'll be your factor hightest common factor (hcf) and lowest common factor (lcm) hcf and lcm of three numbers reviewing number concepts factors and factorization of algebraic expressions factors and multiples relation among numbers hcf and lcm multiples and factors real numbers inside numbers
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Common factors of $9$ and $36$ are

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

  1. $1,3,9$

  2. $1,4,3,5,9$

  3. $1,4,5$

  4. none of these

Show answer
Correct Option: A
Explanation:

Factors of $ 9 = 1, 3, 9 $
Factors of $ 36 = 1, 3, 4, 6, 9, 12, 36 $

Common factors are $ 1, 3, 9 $

Two containers have $950$ litres and $570$ litres of petrol. Find the maximum capacity of a container which can measure the petrol of either contianer in exact number of times?

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

  1. $190$ litres

  2. $280$ litres

  3. $380$ litres

  4. $270$ litres

Show answer
Correct Option: A
Explanation:

Common factor of 950 and 570 is 190 , So A is the Correct answer

H.C.F. of 25, 75, 100 is

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

  1. 15

  2. 25

  3. 5

  4. 100

Show answer
Correct Option: B
Explanation:

Factors of 25 = 1, 5 and 25.
Factors of 75 = 1,3,5,15,25 and 75.
Factors of 100 = 1,2,4,5,10,20,25,50 and 100
Therefore, common factor of 25,75 and 100 = 1,5,25.
H.C.F of 25,75 and 100 = 25.
Option B is correct.

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

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

  1. $3^5$

  2. $3^9$

  3. $3^{14}$

  4. $3^{21}$

Show answer
Correct Option: A
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} $

Which of the following numbers has exactly 4 factors?

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

  1. $16$

  2. $14$

  3. $18$

  4. None of these

Show answer
Correct Option: B
Explanation:

(A) Factors of 16 are $1, 2, 4, 8, 16$
(B) Factors of 14 are $1, 2, 7, 14$
(C) Factors of  18 are $1, 2, 3, 6, 9, 18$

Find the common factors of the given terms:

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

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

  1. $6a^2b$

  2. $6ab^2$

  3. $6ab$

  4. $6$

Show answer
Correct Option: C
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 $2\times 3\times a\times b=6ab$

Find the common factors of the given terms:

$2x, 3x^2, 4$

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

  1. $3$

  2. $2$

  3. $6$

  4. $1$

Show answer
Correct Option: D
Explanation:

$2x$, $3x^2$, $4$
The factors of $2x=2\times x$
The factors of $3x^2=3\times x\times x$
The factors of $4=2\times 2$
Thus, the common factors is $1$

Find the common factors of the given terms:

$2y, 22xy$

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

  1. $2y$

  2. $y$

  3. $11y$

  4. $2$

Show answer
Correct Option: A
Explanation:

$2y$,$22xy$
The factors of $2y=2\times y$
The factors of $22xy=2\times 11\times x\times y$
Thus, the common factors are $2\times y=2y$

Find the common factors of the given terms:

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

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

  1. $x$

  2. $16$

  3. $4x$

  4. $32$

Show answer
Correct Option: C
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 $2\times 2 \times x=4x$

The expression $x^{2} - x - 30$ is positive for

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

  1. no value of $x$

  2. all values of $x$ between - 5 and 6

  3. all $x$

  4. $x > 6$ or $x < - 5$

Show answer
Correct Option: D

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

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

  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. $3^{\frac{1}{14}}\times 2^{\frac{1}{28}}$

Show answer
Correct Option: A
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.

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.

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

  1. $7:27:30$ hours

  2. $7:28:00$ hours

  3. $7:27:50$ hours

  4. $7:27:45$ hours

Show answer
Correct Option: D
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.

State true or false:

The common factors of $18$ and $24$ are $1,2,3,6$.

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

  1. True

  2. False

Show answer
Correct Option: A
Explanation:

Factors of $ 18 = 1, 2, 3, 6, 8, 9$ and $ 18 $

Factors of $ 24 = 1, 2, 3, 4, 6, 8, 12 $ and $ 24 $

Common factors are $ 1, 2, 3, 6 $

State the following statement is True or False
The common factors of $75$ and $50$ are $1,5,25$
maths be my multiple, i'll be your factor co-prime numbers lcm lowest common multiple (l.c.m.)

  1. True

  2. False

Show answer
Correct Option: A
Explanation:

Factors of $ 50 = 1, 2, 5, 10, 25 $ and $ 50 $

Factors of $ 75 = 1,3, 5, 15, 25 $ and $ 75 $

Common factors are $ 1, 5, 25 $

Common factors of $9$ and $36$ are 

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

  1. $1,3,9$

  2. $1,4,3,5,9$

  3. $1,4,5$

  4. None 

Show answer
Correct Option: A
Explanation:

Factors of $ 9 $  are $ 1, 3, 9 $
Factors of $ 36 $ are $ 1, 3, 4, 6, 9, 12, 36 $
$\therefore $ common factors are $ 1, 3, 9 $.

In a school, $351$ boys and $273$ girls have been divided into the largest possible equal classes with each class having equal boys and girls. Find the number of classes

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

  1. $11$

  2. $13$

  3. $10$

  4. $9$

Show answer
Correct Option: B

What is the least number by which 2352 is to be multiplied to make it a perfect square?

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

  1. $6$

  2. $4$

  3. $3$

  4. $8$

Show answer
Correct Option: C
Explanation:
$2$ $2352$
$2$ $1176$
$2$ $588$
$2$ $294$
$3$ $147$
$7$ $49$
$7$

L.C.M of $2352=2^2\times 2^2\times 7^2\times 3$
To make $2352$ a perfect square it must be multiplied by $3$

Find the 1st common multiple of $6$ and $8$.

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

  1. $24$

  2. $16$

  3. $12$

  4. $2$

Show answer
Correct Option: A
Explanation:

$1$st common multiple of $6,8$ is same as LCM of these numbers.


$6 = 2\times3$
$8 = 2^{3}$

$\therefore$ LCM $= 2^{3}\times3 = 24$

the first four common multiple of numbers $6,8,10$ are

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

  1. $10,20,30,40$

  2. $120,240,360,480$

  3. $8,40,80,120$

  4. $6,60,120,240$

Show answer
Correct Option: B
Explanation:
$6 = 2\times3$
$8 = 2^{3}$
$10 = 2\times5$

$\Rightarrow$ LCM of $6,8,10 = 2^{3}\times3\times5 = 120$

$\therefore 120$ is the least common multiple of $6,8,10$. Thus, all multiples of $120$ are common multiples of $6,8$ and $10$.

$\therefore$ First four common multiples $= 120,240,360,480$

Find two common multiples of $12,15$

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

  1. $48,96$

  2. $60,120$

  3. $10,20$

  4. $24,30$

Show answer
Correct Option: B
Explanation:

$12= 2^{2}\times 3$

$15 = 3\times5$

$\Rightarrow$ LCM$= 2^{2}\times3\times5 = 60$

$\therefore 60$ is the least common multiple of $12,15$. Thus, all multiples of $60$ are common multiples of $12$ and $15$.

Answer $= 60,120$

The 1st three common multiple of numbers $12,8,16 $ are

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

  1. $12,24,36$

  2. $8,16,24$

  3. $16,32,48$

  4. $48,96,144$

Show answer
Correct Option: D
Explanation:
$12 = 2^{2}\times3$
$8 = 2^{3}$
$16 = 2^{4}$

$\Rightarrow$ LCM of $12,8,16 = 2^{4}\times3 = 48$

$\therefore 48$ is the least common multiple of $12,8,16$. Thus, all multiples of $48$ are common multiples of $12,8$ and $16$.

$\therefore$ First three common multiples $= 48,96,144$

Find first five common multiples of $1,2$ and $3$.

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

  1. $2,4,8,10,20$

  2. $3,6,12,30,60$

  3. $6,12,18,24,30$

  4. $1,2,3,4,5$

Show answer
Correct Option: C
Explanation:
$\Rightarrow$ LCM of $1,2,3 = 1\times2\times3 = 6$

$\therefore 6$ is the least common multiple of $1,2,3$. Thus, all multiples of $6$ are common multiples of $1,2$ and $3$.

$\therefore$ First five common multiples $= 6,12,18,24,30$

Select the correct option.
The HCF and the LCM of $12, 21, 15$ respectively are

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

  1. $3, 140$

  2. $12, 420$

  3. $3, 420$

  4. $420, 3$

Show answer
Correct Option: A
Explanation:

Numbers $= 12, 15, 21$


$12 =  2 \times 2 \times 3$


$15 = 3 \times 5$

$21 = 3 \times 7$

HCF = Product of smallest power of each common prime factor $= 3' = 3$
LCM = Product of greatest power of each prime factor 

$2^2 \times 3 \times 5 \times 7 = 4 \times 3 \times 5 \times 7 = 420$

$(C) \,\, 3, 420$

The greatest number which when subtracted from 5834, gives a number exactly divisible by each of 20, 28, 32 and 35 is

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

  1. 1120

  2. 4714

  3. 5200

  4. 5600

Show answer
Correct Option: B
Explanation:

 Number which is exactly divisible by 20, 28, 32 and 35 should be the common multiple of all these.
$20 = 2^2 \times 5$
$28 = 2^2 \times 7 \Rightarrow 32 = 2^5$
35.=5 $\times $ 7
LCM = $2^5 \times 5 \times 7 = 1120$
Hence the greatest number that should be subtracted
=5834-1120 = 4714

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