Maths M.C.Q. [1 Marks Each]

 Let there be $100$ students who took the examination.
The number of students who wrote answers in Hindi and in English are thus $40$ and $60$ respectively.
$\therefore$ average marks of all the candidates
$= (40 × 74 + 60 × 77) ÷ 100$
$= 7580 ÷ 100 = 75.8$

$ n = 25$
$u = 78.4 ($initial wrong mean$)$
$x = n × u = 25 × 78.4 = 1960$
Now $69$ was read as $96,$ so
$= 1960 - 96 + 69 = 1987$
$\therefore {\text{u'}} = \frac{1987}{25}$
thus $u = 79.48$

Average is the sum of the data values divided by the total values. Given that $13, 15, 11, 9, 8$ are the ages of $5$ children.
$\therefore$ average age $ = \frac{13 + 15 + 11 + 9 + 8}{5} = \frac{56}{5} = {11.2}$

Given observations: $994, 996, 998, 1000, 1002$
$\text{mean} = \frac{\text{sum}}{\text{number of observations}}$
$\text{mean} = \frac{{994+996+998+1000+1002}}{5}$
$\text{mean} = \frac{4990}{5}$
$\text{mean} = 998$

 Mean $ = \frac{2 + 3 + 4 + 5 +6}{5} = \frac{20}{5} = {4}$
Median is the middle most number $= 4$

We know that average of a bunch of numbers is the sum of the numbers divided by how many numbers in the bunch.
We can set up an equation for the average from the given question,
$\Rightarrow \frac{12+{\text{a}}+6+8+2+14}{6} = {6}$
$\Rightarrow\frac{{42}+{\text{a}}}{6} = {6}$
$\Rightarrow{42}+{\text{a}} = {36}$
$\text{a} = -{6}$

The first five natural numbers are: $1, 2, 3, 4, 5$
$\text{Mean}=\frac{\text{Sum of observations}}{\text{Number of observations}}$
$=\frac{1+2+3+4+5}{5}$
$=\frac{15}{5}$
$=3$
Thus, the mean of first five natural number is $3$
Hence, the correct option is $(c).$

After one value is removed:
Since all of the values are Integers, the sum here must be an integer. Sum $= ($number$) × ($average$).$
Since the average $ = {35}\frac{7}{17},$ and the sum must be an integer, the number of integers must be a multiple of $17.$
For any evenly spaced set,
average = median. After one of the consecutive integers is removed, most of the remaining set will still be evenly spaced.
As a result, the average of the remaining set $ ={37}\frac{7}{17}$ will still be close to the median. Implication: The number of integers $= 4 × 17 = 68$ with the result that $ = {35}\frac{7}{17},$
will be close to the median of the 68 mostly consecutive integers.
$\therefore{\text{sum}} = {68}\times{35}\frac{7}{17} = {2408}$
Original set:
Since $68$ integers remain after one of the integers is removed, the original set contains $69$ integers. Sum of the first n positive integers $ = \frac{(\text{n})({\text{n+1}})}{2}$
$\therefore{\text{sum}} = {69}\times\frac{70}{2} = {2415}.$
Removed integer = original sum - sum after one integer is removed $= 2415 - 2408 = 7.$

Mean $ = \frac{\text{Sum of observation}}{\text{Total}} = \frac{{5}\times{4}}{2(5)} = 2$ Sum of $1st\ 5$ whole no.
$ = \frac{5(4)}{2} = {10}$
$(0, 1, 2, 3, 4)$ First $5$ whole no.

Sum of $5$ numbers $= 27 \times 5 = 135$ when one of the numbers is excluded.
sum of remaining $4$ numbers $= 4 \times 25 = 100$ Excluded number $= 135 - 100 = 35.$

Let the $20$ numbers be $a_1​, a_2​, .... ,a_{20}$
Given that average of $20$ numbers is zero.
$\therefore\frac{ {\text{a}}_1+{\text{a}}_2+...+{\text{a}}_{20}}{20}$
$⇒ a_1​ + a_2​ + ... +a_{20}​=0$
$⇒ a_1​ + a_2​ + ... +a_{19}= -a_{20}$
$\therefore$ at the most $19$ numbers can be greater than zero.

Total amount spent on shirts is $5 \times 450 = 2250$
amount on trousers is $4 \times 750 = 3000$ and on shoes $12 \times 750 = 9000$
hence total amount spent is $2250 + 3000 + 9000 = 14250$
so average per article is $\frac{14250}{21} = 678.57$

Average is the sum of the data values divided by the total values.The first $4$ prime numbers are $2, 3, 5, 7$
$\therefore$ average of first $4$ prime numbers $ = \frac{2 + 3 + 5 + 7}{4} = \frac{17}{4} = {4.25}$

 Required average
$ = \frac{\big(50.25\times{16} + 45.15 \times{8}\big)}{16 + 8}$
$ =\frac{ \big(804 + 361.20\big)}{24}$
$ = \frac{1165.20}{24}$
$ = {48.55}$

Given, Rahul and $34$ mean score in $5$ test was $84$. And his means score in the first $4$ test of these test was $87.$ Then total score in $5$ tests $= 84 × 5 = 420$ And total score in $4$ tests $= 87 × 4 = 348$ So, the score in fifth test $= 420 - 348 = 72.$

$\text{mean}\frac{\text{sum of the terms}}{\text{total number of terms}}$
$\Rightarrow \frac{{2+4+6+8+}{\text{x+y}}}{6} = {5}$
$\frac{{20+}{\text{x+y}}}{6} = {5}$
$20 + x + y = 5 (6)$
$x + y = 30 - 20 = 10$

$ = \frac{33.5+30.4+25.6+31.5+29}{5} = \frac{150}{5} = {30}$

Let the grade he got on final test be $x$ The average of all four grades is $90.$
So, we have $ = \frac{{92+89+86+}{\text{x}}}{4} = {90}$
$\Rightarrow x = 360 - 267 = 93$

Since the weight of each student increases by $3\ kgs,$ their mean will also increase by $3\ kgs,$ hence new mean becomes $235 + 3 = 238\ kg$

 The sum of two numbers is twice their average.
$x + y = 100 $
$y + z = 160$
$x = 100 - y $
$z = 160 - y$
substitute these expressions for $z$ and $x : z - x = (160 - y) - (100- y) = 160 - y - 100 + y= 160 - 100 = 60$
alternatively, pick smart numbers for $x$ and $y.$
Let $x = 50$ and $y = 50 ($this is an easy way to make their average equal to $50).$
since the average of $y$ and $z$ must be $80,$ you have $z = 110$
$\therefore z - x = 110 - 50 = 60$

Given is that mean age of all males is $15$ years and age of all females is $19$ years in a colony, when we find the mean age mm of the combined group of males and females in colony, we can conclude that mean age will be more than $15$ but less than. so $15 < m < 19$

We know that Mean $ = \frac{\text{sum of observations}}{\text{total numer of observations}}$
$\therefore$ Mean of $10$ natural numbers $ = \frac{\text{sum of 10 natural number}}{10}$
We know that the sum of $′n′$ natural numbers $ = \frac{\text{n}{(\text{n+1})}}{2}$
$\therefore$ Sum of $10$ natural numbers $ = \frac{{10}\times{11}}{2} = {55}$
Mean of natural numbers $ = \frac{55}{10} = \frac{11}{2}$

The first $50$ natural numbers are $1, 2, 3, ..... ,50$
We know that, the sum of first nn natural numbers is $ = \frac{\text{n}({\text{n+1}})}{2}$
$\therefore$ sum of first $50$ natural numbers is $\frac{{50}{(51)}}{2} = {25}(51) = {1275}$
the average of first $50$ natural numbers is $\frac{1275}{50} = {25.5}$

Let the numbers be $x - 2, x, x + 2, x + 4$ according to the question,
$ = \frac{\text{x-2+x+x+2+x+4}}{4} = {15}$
$4x + 4 = 60$
$4x = 56$
$⇒ x = 14$
hence, the second highest number $i.e. x + 2 = 14 + 2 = 16$

$\frac{\text{(x + 2)}\times{60}+\text{x}\times{120}+\text{(x - 2)}\times{180}}{{\text{x + 2}}+{\text{x + x}}-{2}} = {100}$
$\frac{{60}\text{x}+{120}+{120}{\text{x}}+{180}{\text{x}} - {360}}{\text{3x}} = {100}$
$360x - 240 = 300x$
$60x = 240$
$x = 4$

Let the average age of the whole team by $x$ years.
$\therefore 11x - (26 + 29) = 9(x - 1)$
$⇒ 11x - 9x = 46$
$⇒ 2x = 46$
$⇒ x = 23.$
So, average age of the team is $23$ years.

 Let the man spends $Rs. x$ on purchasing each kind of pens at their respective rates. Total cost $= x + x + x = 3x$
Now, number of pens bought of $Rs. 5$ will be $\frac{\text{x}}{5}$
Number of pens bought of $Rs. 10$ will be $\frac{\text{x}}{10}$
Number of pens bought of $Rs. 20$ will be $\frac{\text{x}}{20}$
So, total number of pens bought $ = \frac{\text{x}}{5}+\frac{\text{x}}{10}+\frac{\text{x}}{20}$
Average cost $ =\frac{ \text{Total cost}}{\text{No. of total pens}}= \frac{ \text{x+x+x}}{\frac{\text{x}}{5}+\frac{\text{x}}{10}+\frac{\text{x}}{20}}$
$ = \frac{{3}}{\frac{1}{5}+\frac{1}{10}+\frac{1}{20}} = \frac{60}{7}$

 $\text{Mean} = \frac{\text{x + x + 2 + x + 4 + x + 6 + x + 8}}{8} $
$= 205x = 80x = 16$

 First $726$ natural numbers $= 1, 2, 3, 4 .... 726$
sum of $726$ numbers $= 1 + 2 + 3 + 4 .... 72$
this forms an $AP,$ with first term $= 1,$ last term $= 726$ and number of terms $= 1$
$\text{sum of AP} = \frac{\text{n}}{{2}} (\text{a + 1})$
$\text{sum of AP} = \frac{726}{2} (1+726)$
$\text{sum of AP} =263901$
$\text{mean} = \frac{\text{sum}}{\text{number of turns}} = \frac{263901}{726} = 363.5$

 The factor of $24$ are $1, 2, 3, 4, 6, 8, 12, 24$
$\therefore{\text{A.M}} = \frac{\sum{\text{x}}}{\text{n}}$
$\Rightarrow\frac{1+2+3+4+6+8+12+24+}{8}\Rightarrow\frac{60}{8} = \frac{15}{2}$

 Given $3$ Quantities: $12 - n, 12, 12 + n$ Average of the above $3$ Quantities
$ = \frac{\text{sum of the above 3 quantities}}{\text {number of quantities}} $
$ = \frac{{12 }-{\text{n}} +12+12+{\text{n}}}{3}$
$ = \frac{{12}\times{3}}{3} = {12}$
$\therefore$ average of the Quantities listed above is $12.$

 $\frac{2+4+6+8+10+12+}{6} = \frac{42}{6} = 7$

Mean of n observations $= 12$
Sum of observations $= 132$
$\text{Mean}=\frac{\text{Sum of observations}}{\text{Number of observations}}$
$\Rightarrow12=\frac{132}{\text{n}}$
$\Rightarrow\text{n}=\frac{132}{12}=11$
Thus, the value of $n$ is $11$
Hence, the correct option is $(c).$

Let $a,b$ and $c$ are three numbers $ = \frac{\text{a + b + c}}{3} = {60}$
$a + b + c = 180$
and $a = \frac{1}{4} \text{(b+c)} = \frac{1}{4} (180) - {\text{a}}$
$4a = 180 - a$
$\text{a} = \frac{180}{5} = {36}$

The first four natural numbers are $1, 2, 3, 4$
$\text{A.M} = \frac{\sum{\text{x}}}{\text{n}} = \frac{{1}^{3} +{2}^{3} + {3}^{3}+{4}^{3}}{4}$
$\frac{1+4+27+64}{4} = \frac{100}{4} = {25}$

 There are $52$ cards in a standard deck. There are four different suits Diamonds (red), Clubs (black), Hearts (red), and Spades (black) each containing $13$ cards.
$\therefore $ Number of red cards $($favourable outcomes$) = 13 + 13 = 26$
Therefore
Probability of getting a red card $=\frac{26}{52}=\frac{1}{2}$
Hence, the correct option is $(b).$

 We know that sum of $1$ to $n$ numbers $ = \frac{\text{n}(\text{n+1})}{1}$
Then sum of $1$ to $100 = \frac{{100}(100+1)}{2} = \frac{{100}\times{101}}{2} = \frac{10100}{2} = 5050$
Then average of $1$ to $100$ positive numbers $ = \frac{5050}{100} = 50.5$

 Given observations $25, 37, 84$ No .of observations $3$ The mean of
observation $\frac{25+37+84}{3}=\frac{146}{3}=48.6$

 Mean of $5$ number $= 27.$ If $1$ of the numbers is excluded,
$\therefore M′ = M - 2$
$= 27 - 2$
$= 25 × 4$
$= 100M = 27 × 5$
$= 135$ Excluded number $= 135 - 100 = 35.$