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Multiple choice maths construction of angles identifying angles acute and obtuse angles types of angle

$\displaystyle 179^{o}$ is an example of:

  1. obtuse angle

  2. acute angle

  3. right angle

  4. None of the above
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
An angle with measure $=90^{o}$ is a right angle
An angle with measure $>90^{o}$ is an obtuse angle
An angle with measure $<90^{o}$ is an acute angle

Here, $179^o$ is greater than $90^o$ and less than $180^{o}$. 

Hence, it is an obtuse angle.

Multiple choice maths construction of angles identifying angles acute and obtuse angles types of angle

An angle which measures $\displaystyle 180^{o}$ is called:

  1. zero angle

  2. right angle

  3. straight angle

  4. acute angle
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

An angle which measure $0^0$ is called zero angle.

An angle which measure more than $0^o$ and less than $90^o$ is an acute angle.
An angle which measures $90^o$ is called a right angle.

An angle which measure more than $90^o$ and less than $180^o$ is an obtuse angle.
An straight line makes the angle of $180^o$.
Hence, the answer is straight angle.

Multiple choice maths measure of central tendency assumed mean method assumed mean method of finding mean mean and median

If the mean of x and $\displaystyle \frac{1}{x}$ is M, then the mean of x$^3$ and $\displaystyle \frac{1}{x^3}$ is

  1. $\displaystyle \frac{M (M^2 - 3)}{2}$
  2. $M (4M^2 - 3)$
  3. $M^3$
  4. $M^3 + 3$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Given,

$M=\dfrac{x+\dfrac{1}{x}}{2}$
$2M=x+\dfrac{1}{x}$                  $ ....... (1)$

On cubing both sides, we get
$(2M)^3=(x+\dfrac{1}{x})^3$
$8M^3=x^3+\dfrac{1}{x^3}+3(x+\dfrac{1}{x})$
$8M^3=x^3+\dfrac{1}{x^3}+3(2M)$
$8M^3=x^3+\dfrac{1}{x^3}+6M$
$x^3+\dfrac{1}{x^3}=8M^3-6M$
$\dfrac{x^3+\dfrac{1}{x^3}}{2}=M(4M^2-3)$

Hence, this is the answer.

Multiple choice maths measure of central tendency assumed mean method assumed mean method of finding mean mean and median

Sum of squares of deviation of $10$ observations measured from $5$ is $17$ and sum of squares of observations is $170$ then mean of observation is

  1. $40.3$
  2. $4.5$
  3. $4$
  4. $4.03$
  5. $4.3$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$x _1,x _2,x _3,x _4...x _10\x _1^2+x _2^2+x _3^2+x _4^2+....x _10^2=17 (Given) \rightarrow (i)$

$(x _1-5)^2+(x _2-5)^2+(x _3-5)^2+....+(x _10-5)^2=17(Given)\rightarrow (i)$
$Mean (M)=\cfrac{x _1+x-2+x _3+....x _10}{10}\ \Rightarrow x _1+x _2+x _3+....+x _10=10M\rightarrow(iii)$
From equation $(ii)$ we get
$x _1^2+25-10x _1+x _2^2+25-10x _2+x _3^2+25-10x _3+...+x _10^2+25-10x _10=17\ \Rightarrow (x _1^2+x _2^2+....x _10^2)-10(x _1+x _2+x _3+....+x _10)+25\times10=17\ \Rightarrow 170-10(10M)+250=17\ \Rightarrow420-100M=17\ \Rightarrow100M=403\ \Rightarrow M=\cfrac{403}{100}=4.03$

Multiple choice maths measure of central tendency assumed mean method assumed mean method of finding mean mean and median

The sum of the deviations of a set of values $x 1, x _2$, ...... $x _n$ measured from $50$ is $-10$ and the sum of deviations of the values from $46$ is $70$. The mean is __________.

  1. $49$
  2. $49.5$
  3. $49.75$
  4. $50$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Sum of deviations from $50$ is $-10$


$\Rightarrow \sum(xi - 50) = -10$
     $\sum x _i - 50\sum1 = -10$
     $\sum x _i - 50n = -10$

$\therefore y-50n=-10.....(1)$

Sum of deviations from $46$ is $70$

$\Rightarrow \sum(x _i - 46) = 70$
     $\sum x _i - 46\sum1 = 70$
     $\sum x _i - 46n = 70$

$\therefore y-46n=70.....(2)$


Solving $(1)$ and $(2)$, we get
$4n = 80$ i.e. $n=20$

Putting value of $n$ in $(1)$, we get
$y=990$

Mean $= \dfrac{\sum x _i}{n} = \dfrac{y}{n} = \dfrac{990}{20} = 49.5$

Multiple choice maths measures of central tendency assumed mean method assumed mean method of finding mean mean and median

The exam scores of all 500 students were recorded and it was determined that these scores were normally distributed. If Jane's score is 0.8 standard deviation above the mean, then how many, to the nearest unit, students scored above Jane?

  1. $109$
  2. $106$
  3. $150$
  4. $160$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let m be the mean and s be the standard deviation and find the z score.
$z = (x - m) /s = (0.8 s + m - m) / s = 0.8$
The percentage of student who scored above Jane is (from table of normal distribution).
1 - 0.7881 = 0.2119 = 21.19%
The number of student who scored above Jane is (from table of normal distribution).
21.19% 0f 500 = 106

Multiple choice maths measures of central tendency assumed mean method assumed mean method of finding mean mean and median

Following table gives frequency distribution of milk (in litres) given per week by 50 cows.
Find average (mean) amount of milk given by a cow by 'shift of origin method'.

Milk (in litres) 24 - 30 30 - 36 36 - 42 42 - 48 48 - 54 54 - 60 60 - 66 66 - 72  72 - 78 78 - 84 84 - 90
No. of cows 1 3 8 5 5 5 8 4 6 2 3
  1. $51.12$ litres
  2. $54.12$ litres
  3. $57.12$ litres
  4. $60.12$ litres
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Consider the following table, to calculate mean by "shift of origin method":

$x _i$=mid value of class interval
Assumed mean $a=57$

 $ci$  $f _i$  $x _i$ $d _i=x _i-a$  $f _id _i$
24-30  1 27  27-57= -30  -30
30-36  3 33  33-57= -24  -72
36-42  8 39  39-57= -18  -144
42-48   5 45   45-57= -12  -60
48-54   5 51  51-57= -6  -30
54-60   5 57  57-57=0  0
60-66   8 63   63-57=6   48
66-72   4 69  69-57=12  48
72-78   6 75   75-57=18  108
78-84  2 81   81-57=24  48
84-90 3  87   87-57=30  90
 $N=\Sigma f _i=50$          
 $\Sigma f _id _i=6$

Mean $\overline x=a +\dfrac {\Sigma f _id _i}{N}$
$\therefore \overline x=57 + \dfrac{6}{50}=57.12$

Average amount of milk given by cow is $ 57.12$ litres
Hence, option $C$ is correct.

Multiple choice maths repeated multiplication exponentiation index notation and products of prime factors powers

The value of $\displaystyle\frac{2^{m+3}\times3^{2m-n}\times5^{m+n+3}6^{n+1}}{6^{m+1}\times10^{n+3}\times15^m}$ is equal to

  1. 0

  2. 1
  3. $2^m$

  4. none of these
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Numerator $=2^{m+3}\cdot3^{2m-n}\cdot5^{m+n+3}\cdot2^{n+1}\cdot3^{n+1}$
$=2^{m+n+4}\cdot3^{2m+1}\cdot5^{m+n+3}$ (i)
Denominator $=2^{m+1}\cdot3^{m+1}\cdot2^{n+3}\cdot5^{n+3}\cdot3^m\cdot5^m$
$=2^{m+n+4}\cdot3^{2m+1}\cdot5^{m+n+3}$ (ii)
Given expression $=1$

Multiple choice maths repeated multiplication exponentiation index notation and products of prime factors powers

What is the unit digit in ${({6374}^{1793}\times {625}^{317}\times{341}^{491})}$?

  1. $0$
  2. $2$
  3. $3$
  4. $5$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Unit digit in ${6374}^{1993}=$ Unit digit in ${(4)}^{1793}$
=Unit digit in $[{({4}^{2})}^{896}\times 4]$
=Unit digit in $(6\times 4)=4$
Unit digit in ${(625)}^{317}=$ Unit digit in ${(5)}^{317}=5$
Unit digit in ${(341)}^{491}=$ Unit digit in ${(1)}^{491}=1$
Required digit$=$ Unit digit in $(4\times 5\times1)=0$

Multiple choice maths repeated multiplication exponentiation index notation and products of prime factors powers

The number of values of $x\ \epsilon \ [0,5]$ at which $f(x)=|x-\dfrac{1}{4}|+|x-2|+\tan{x}$ is not differentiable are

  1. $0$
  2. $2$
  3. $3$
  4. $4$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The function contains absolute value terms and a trigonometric term. The absolute value expressions are not differentiable at their critical points where the inside equals zero, and tan(x) has discontinuities and points of non-differentiability within the interval [0, 5], culminating in 4 points total.