Tag: maths

Questions Related to maths

Calculate the probability that a spinner, having the numbers one through five evenly spaced, will land on an odd number exactly once if the spinner is used three times.

maths introduction of probability theory types of events some more terms in probability events and its algebra

  1. $\dfrac {12}{125}$

  2. $\dfrac {18}{125}$

  3. $\dfrac {27}{125}$

  4. $\dfrac {36}{125}$

  5. $\dfrac {54}{125}$

Show answer
Correct Option: D
Explanation:

Total possible combinations when spinner used three times is $5 \times 5 \times 5 = 125$.
Out of three times, inexactly one number is odd implies the other two numbers are even.
The possible combinations such that exactly one number is odd is $3 \times 2 \times 2 + 2 \times 3 \times 2 + 2 \times 2 \times 3 = 36$.
The probability is $\dfrac {36}{125}$.

A bag contains $10$ balls, each labelled with a different integer from $1$ to $10$, inclusive. If $2$ balls are drawn simultaneously from the bag at random, calculate the probability that the sum of the integers on the balls drawn will be greater than $6$.

maths introduction of probability theory types of events some more terms in probability events and its algebra

  1. $0.41$

  2. $0.43$

  3. $0.60$

  4. $0.76$

  5. $0.87$

Show answer
Correct Option: E
Explanation:

Out of $10$ balls , $2$ balls can be selected in ${ ^{ 10 }{ C } } _{ 2 } = 45$
Number of ways of selecting $2$ balls such that sum is less than or equal to $6$ is $6$
Probability that the sum of integers on the balls drawn will be greater than $6$ is $1-\dfrac {6}{45} = \dfrac {39}{45} = 0.87$

What is the H.C.F . of $348$ and $319$

maths hcf-lcm common factors and hcf hcf highest common factor (h.c.f.)

  1. $348$

  2. $319$

  3. $56$

  4. $29$

Show answer
Correct Option: D
Explanation:

$348=$$2\times$$2\times$$3\times$$29$

$319=11$$\times29$
thus $HCF=29$

HCF of $144$ and $198$ is

maths hcf-lcm common factors and hcf hcf highest common factor (h.c.f.)

  1. $9$

  2. $18$

  3. $6$

  4. $12$

Show answer
Correct Option: A
Explanation:

HCF of 144 and 198 is:

$ 198 = 2 \times 3^2 \times 11$
$ 144 = 2^4 \times 3^2$

Highest Common factor is 9.
Option A is correct

HCF of $24 $ and $36$ is ..............

maths hcf-lcm common factors and hcf hcf highest common factor (h.c.f.)

  1. $6$

  2. $4$

  3. $9$

  4. $12$

Show answer
Correct Option: D
Explanation:
$2\underline {|24} {\;} 2\underline {|36}$
$2\underline {|12} {\;} 2\underline {|18}$
$2\underline {|6} {\;} 3\underline {|9}$
$3\underline {|3} {\;} 3\underline {|3}$
    1    1
$24 = 2 \times 2 \times 2 \times 3$
$36 = 2 \times 2 \times 3 \times 3$
$\therefore H.C.F=2\times 2 \times 3=12$

HCF of the numbers $36$ and $144$ is 

maths hcf-lcm common factors and hcf hcf highest common factor (h.c.f.)

  1. $36$

  2. $144$

  3. $4$

  4. $2$

Show answer
Correct Option: A
Explanation:

Factors of the given numbers are,

$36= 2\times 2 \times 3 \times 3$
$144 = 2 \times 2 \times 2 \times 2 \times 3 \times 3 $
$\therefore $ HCF of $36$ and $144$ $ = 2 \times 2 \times 3 \times 3 = 36$


The number of ordered pairs $(a, b)$ of positive integers, such that $a + b = 90$ and their greatest common division is $6$, equals

maths hcf-lcm common factors and hcf hcf highest common factor (h.c.f.)

  1. $5$

  2. $4$

  3. $8$

  4. $10$

Show answer
Correct Option: C
Explanation:

Let's look at products of $6$
$6, 12, 18, 24, 30, 36, 42, 48, 54, 60, 66, 72, 78, 84, 90$
The pairs whose sums equal to $90$ are:
$6 + 84=90$
$12 + 78=90$
$18 + 72=90$
$66 + 24=90$
$60 + 30=90$
$54 + 36=90$
$48 + 42=90$
Total number of pairs are $7$.
But there can be  $7$ more pairs when the numbers are reversed.
$\therefore$ total number of ordered pair are $14$

If $x =2^3 \times 3 \times 5^2,Y=2^2 \times 3^3$, then HCF(x,y)is:

maths hcf-lcm common factors and hcf hcf highest common factor (h.c.f.)

  1. 12

  2. 108

  3. 6

  4. 36

Show answer
Correct Option: A
Explanation:

$x= 2^3 \times 3 \times 5^2$
$y = 2^2 \times 3^3$
HCF (x,) $=2^2 \times 3$= 12

The product of two numbers is $2240$ and their HCF is $14$. Which of the following is not the possible pair.

maths hcf-lcm common factors and hcf hcf highest common factor (h.c.f.)

  1. $(14,160)$

  2. $(28,80)$

  3. $(42, 80)$

  4. $(56,40)$

Show answer
Correct Option: C
Explanation:

Pair in Option C is not possible.
because in option C pair is $(42,80)$
Since, Product of numbers $=42\times 80 =3360$
Option C is correct.

G.C.D. of $(a + b -c)^6$ and $(a + b -c)^4$ is

maths hcf-lcm common factors and hcf hcf highest common factor (h.c.f.)

  1. $(a+b-c)^6$

  2. $(a+b-c)^{10}$

  3. $(a+b-c)^2$

  4. $(a+b-c)^4$

Show answer
Correct Option: D
Explanation:

Since, $(a + b -c)^6 = (a + b -c)^4 \times (a + b -c)^2 $
$\therefore$  G.C.D. of $(a + b -c)^6$ and $(a + b -c)^4$ = $(a + b -c)^4$
$\because (a + b -c)^4$ is greatest common in $(a + b -c)^4$ and $(a + b -c)^6$.
Option D is correct.