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

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.

  1. $\dfrac {12}{125}$
  2. $\dfrac {18}{125}$
  3. $\dfrac {27}{125}$
  4. $\dfrac {36}{125}$
  5. $\dfrac {54}{125}$
Reveal answer Fill a bubble to check yourself
D Correct answer
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 $\dfrac {36}{125}$.

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

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 $6$.

  1. $0.41$
  2. $0.43$
  3. $0.60$
  4. $0.76$
  5. $0.87$
Reveal answer Fill a bubble to check yourself
E Correct answer
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 $1-\dfrac {6}{45} = \dfrac {39}{45} = 0.87$

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

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

  1. $6$
  2. $4$
  3. $9$
  4. $12$
Reveal answer Fill a bubble to check yourself
D Correct answer
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
$24 = 2 \times 2 \times 2 \times 3$
$36 = 2 \times 2 \times 3 \times 3$
$\therefore H.C.F=2\times 2 \times 3=12$
Multiple choice maths hcf-lcm common factors and hcf hcf highest common factor (h.c.f.)

HCF of the numbers $36$ and $144$ is 

  1. $36$
  2. $144$
  3. $4$
  4. $2$
Reveal answer Fill a bubble to check yourself
A Correct answer
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$


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

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

  1. $5$
  2. $4$
  3. $8$
  4. $10$
Reveal answer Fill a bubble to check yourself
C Correct answer
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 $14$

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

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

  1. $(14,160)$
  2. $(28,80)$
  3. $(42, 80)$
  4. $(56,40)$
Reveal answer Fill a bubble to check yourself
C Correct answer
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.

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

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

  1. $(a+b-c)^6$
  2. $(a+b-c)^{10}$
  3. $(a+b-c)^2$
  4. $(a+b-c)^4$
Reveal answer Fill a bubble to check yourself
D Correct answer
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.