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

When a dice is thrown, outcomes $1, 2, 3, 4, 5, 6$ are equally likely.

English → $200$
Science → $200$

Rolling an unbiased dice results in $6$ outcomes. All have equal probability of showing up.
Tossing a fair coin results in $2$ outcomes. Both have equal probabilities of showing up.
Drawing a card from from a well shuffled pack of $52$ cards can have $52$ outcomes in total with equal probabilities.
Hence, above all are example of random experiment.

Required number $= 8 + 10 = 18$

Number of red balls in the bag = favourable outcomes $= 5$
Total number of balls in the bag $= 5 - 4 = 9$
Probability (getting a red ball) $=\frac{\text{favourable outcomes}}{\text{Total outcomes}}$
P (red ball) $=\frac{5}{9}$

Number of students liking together yellow and green colours is $(14 + 20)\%$ i.e. $34\%$, which is approximately the same as those for red $(35\%).$

Number of possible outcomes is 36,
i.e. $(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6)$
$(3,1), (3, 2), (3, 3), … (3, 6)$
$(4, 1), (4, 2), … (4, 6)$
$(5, 1), (5, 2), … (5, 6)$
$(6,1), (6, 2), … (6, 6)$

A pie chart is in the shape of a circle.
Hence, the sum of all central angles is equal to the complete angle of a circle, which is $360^\circ .$
Thus, in any pie chart the sum of the central angles is $360^\circ .$

Required number $= 7 + 8 = 15$

Required number $= 4 + 2 = 6$

Central angle for blue $= 180^\circ $
Central angle for green $= 90^\circ $
$\therefore$ Difference $= 180^\circ - 90^\circ = 90^\circ $

If total number of votes $= 400$
Then, number of votes in favour of ‘Others’ $= 6\%$ of 400 $=\frac{6}{100}\times400$
$=\frac{3}{50}\times400$
$=24$

So, angle of the sector $\frac{27}{100}\times360^\circ = 97.2^\circ$

The probability of a sure event is $1.$

Expenditure done on conveyance
$= 6 \times 1000 = Rs. 6000$
Expenditure done on rent $= 4 \times 1000$
$= Rs. 4000$
$\therefore$ Difference $= Rs. 6000 – Rs. 4000$
$= Rs. 2000.$

The event that is sure to happen is called a certain event and probability of such an event is $1$ as this event is bound to happen.

Percentage of protein and other constituents in human body $= 16\% + 14\% = 30\%$
So, angle of the sector $= \frac{30}{100}\times 360^\circ = 108^\circ$

The last class has the upper limit of $500$, which is the upper limit for the whole table.

Blue → $180^\circ $

Distribution of protein in muscles $=\frac{1}{3}$
Distribution of protein in bones $=\frac{1}{6}$
Ratio of distribution of proteins in the muscles to that of proteins in the bones.
$=\frac{1}{3}:\frac{1}{5}$
$=\frac{1}{3}\times\frac{1}{5}$
$=2:1$

$\frac{5000}{1000} = 5\ cm$ is the hight of the bar corresponding to food.

Let the required number of students be $x.$
Then we have:
$\Rightarrow\frac{\text{x}}{1650}\times48$
$\Rightarrow\text{x}=\Big(48\times\frac{\text{1650}}{360}\Big)$
$\Rightarrow\text{x}=220$
Hence, the number of students who opted for arts stream is $220.$

$\frac{6000}{1000} = 6\ cm$ is the hight of the bar corresponding to conveyance.

On tuesday the number of workers in both the shift $= 70 + 45 = 115$
On thursday the number of workers in both the shift $= 60 + 30 = 90$
The total number of workers on both the days taken together $= 115 + 90 = 205$

Required difference $= 9 - 8 = 1$

The range of a data set is the difference between the highest and lowest value in a data set.
Data is $6, 14, 20, 16, 6, 5, 4, 18, 25, 15, 5$
The largest value of data set $= 25$
The smallest value of data set $= 4$
Range $= 25 - 4 = 21$

Number of cupcakes on monday $= 5 \times 8 = 40$
Number of cupcakes on tuesday $= 2\times8+\frac{1}{2}\times8 = 16 + 4 = 20$
The difference between the number of cupcakes sold on monday and tuesday $= 40 - 20 = 20$

Given, head appeared $120$ times, if a coin is tossed $200$ times. Then,
probability of getting a head in this experiment $=\frac{\text{Number of times head appeared}}{\text{Number of times coin is tossed}}$
$=\frac{120}{200}$
$=\frac{3}{5}$

Impossible event is an event that is non existing. There is no possibility that it can happen. Hence the probability of occurrence is $0.$

When two coins are tossed, then the possible outcomes will be:
$HH, HT, TH, TT$
Thus, probability of getting tails for both the coins $=\frac{1}{4}$

Two coins are tossed simultaneously, then possible outcomes $= 4$
Now probability of getting one head and one tail $=\frac{2}{4}=\frac{1}{2}$

Explanation: For any class, we can determine the class interval, by subtracting the lower limit from upper limit.
$500 - 450 = 50$
$450 - 400 = 50$

Highest frequency $= 7 \rightarrow 430-465$

$8$ cards are numbered $1, 2, 3, 4, 5, 6, 7, 8$
They are mixed and kept in a box one care is chosen at random, then probability of card having 9 number less than $4=\frac{3}{8}$

Total number of cards $= 5$
Number of cards having $2$ on it $= 2$
Probability of Rohan picking a card without seeing, that has the number $2$ on it $\frac{2}{5}$

Size $= 325 - 290 = 35$

Size $= 20 - 15 = 5$

Since the outcome of a throw of dice can never be $10$, the probability is $0.$

Total number of times coin tossed $= 12$
Total number of occurrence of head $= 5$
The chance of occurrence of head $=\frac{\text{Number of times headappeared}}{\text{Number of times a coinis tossed}}$
$=\frac{5}{12}$

Numbers on the slips are $1, 2, 3, 4$ and $5.$
Odd numbers $= 1, 3, 5$
Number of slips bears an odd number $= 3$
Probability that the slip bears an odd number $=\frac{\text{Number of slipsbearsan odd number}}{\text{Total number of slips}}$
$=\frac{3}{5}$

Required number
$=$ Frequency of class $(4 - 6) = 4$

Size of the class $150-175$
= Upper limit - Lower
Limit $= 175 - 150 = 25$

Red $\rightarrow 45\%$
Yellow $\rightarrow 45\%$

A die is thrown possible outcomes $= 6$
Now probability of getting an even number which are $2, 4, 6 =\frac{3}{6}=\frac{1}{2}$

Number of employees who joined company $A$ from january to april
$= 600 + 500 + 400 + 300 = 1800$
Number of employees who joined company $B$ from january to april
$= 400 + 200 + 100 + 100 = 800$
In company A there are $1000$ people more than company $B.$

Population of $USA$ in the year $1981$ and $1991$ together $= 45,000 + 60,000 = 105,000$
Population of $USA$ in the year $1996$ and $2001$ together $= 75,000 + 105,000 = 180,000$
The difference in the population of $USA$ in the year $1981$ and $1991$ together and in the year $1996$ and $2001$ together$= 180,000 - 105,000 = 75,000$

Lowest frequency $= 2 \rightarrow 325-360$

Central angle for sector $B$
$=\frac{40}{100}\times 360^\circ = 144^\circ$

Minimum number of students
$= 2 \rightarrow 8-10$

As, jar contains $6$ red, $5$ green, $4$ blue and $5$ yellow marbles of same size. Then,
Probability that the chosen marble is of red colour $=\frac{\text{Number of red marbles}}{\text{Total number of marbles}}$
$=\frac{6}{6+5+4+5}$
$=\frac{6}{20}$
$=\frac{3}{10}$

Maximum number of students$= 9 \rightarrow 6 – 8$

The number of student having height more than equal to $130\ cm$ but less than $145\ cm= 8 + 14 + 18 = 40$

Arranging the data in the tally marks we get:

Required number $= 9 + 2 = 11$

A coin has two sides, Head and tails.
Probability = Number of possible outcomes/Total number of outcomes
We know that the tail is present on one side of a coin and in total there could be two outcomes $H \& T.$
Hence, the probability of getting a tail $=\frac{1}{2}$

Required number
$= 2 + 16 + 14 + 4 + 6 + 14 + 8 + 10 = 74$

Total number of students $= 10 + 14 + 12 + 13 + 11 = 60$

$200 - 200 = 0$

The grades better than grade $'C'$ is grade $'A'$ and $'B'.$
Percentage of students getting grade $'A' = 12\%$
Percentage of students getting grade $'B' = 29\%$
Total percentage of students getting grade better than $'C' = 12\% + 29\% = 41\%$

Required number $= 5 + 2 + 4 + 6 + 7 + 5$
$= 29$

Number of possible outcomes is $8,$
i.e. $HHH, HHT, HTH, THH, TTH, THT, HTT, TTT.$

Intervals are $60 - 65, 65 - 70, 70 - 75, 75 - 80, 80 - 85, 85 - 90$
Class width = Lower class limit - Upper class limit
So the class width $= 65 - 60 = 5$

Number of students from $300 - 350 = 12$
Number of students from $350 - 400 = 14$
Thus, the total number of students from $300 - 400 = 12 + 14 = 26$

The number of buses produced in april $= 1200$
The number of buses produced in may $= 1400$
The number of buses produced in june $= 1600$
The total number of buses produced over the $3$ years $= 1200 + 1400 + 1600 = 4200$

Number of possible outcomes is $4,$
i.e. $HH, HT, TH, TT.$

Total number of buttons $= 4 + 7 + 3 + 6 = 20$
Probability that button on the floor is blue $=\frac{\text{Number of blue button}}{\text{Total number of button}}$
$=\frac{4}{20}$
$=\frac{1}{5}$

Total expenditure$= (1 + 6 + 4 + 5 + 2 + 3) \times 1000 = Rs. 21000$

Frequency of class $395 – 430$ is $6$

The sample space obtained on throwing a dice is: $S = \{1, 2, 3, 4, 5, 6\}$
The composite numbers from the sample space are: Composite $No. = \{4, 6\}$

$144^\circ - 108^\circ = 36^\circ$

Blue $(25\%)$, i.e. $\frac{25}{100}=\frac{1}{4}$ of votes received by blue.
Red $(35\%)$, i.e. $\frac{35}{100}=\frac{7}{20}$ of votes received by red.
Green $(20\%)$, i.e. $\frac{20}{100}=\frac{1}{5}$ of votes received by green.
Yellow $(14\%)$, i.e $\frac{14}{100}=\frac{7}{50}$ of votes received by yellow.
Others $(6\%)$, i.e. $\frac{6}{100}=\frac{3}{50}$ of votes received by others.
Hence, green is the required colour.

Required number $= 6 + 7 + 5 = 18$

one card is drawn from a well shuffled deck of 52 cards, possible out comes $= 52$
The probability of card which is a queen $\frac{4}{52}=\frac{1}{13}$

Probability $\frac{2}{6} = \frac{1}{3}$

Given, a bag contains $3$ red, $5$ black and $4$ blue balls.
Then, total number of outcomes = Total number of balls $= 3 + 5 + 4 = 12$

Total number of balls in bag $= 4 + 4 = 8$
Total number of red balls $= 4$
Probability of getting a red ball = $\frac{4}{8} = \frac{1}{2}$

For a die, the possible outcomes are, $1, 2, 3, 4, 5, 6$
From the above outcomes, only $6$ is greater than $5$
So, the possible outcome is $6.$

Number of possible outcomes of each dies $= 6$
For two dies, number of possible outcomes $= 6 \times 6 = 36$

Upper limit of class interval $75-85$ is $85.$
Note: The upper value of class interval is called its upper class limit and lower value of a class interval is called lower class limit.

Maximum frequency $= 16 \rightarrow 30-40$

Central angle of the sector representing travel expenses $=\Big(\frac{\text{value of expenses on travel}}{\text{monthly income}}\times360\Big)^\circ$
$=\Big(\frac{250}{2400}\times360\Big)^{\circ}$
$=37\frac{1}{2}^{\circ}$

Given, $O$ represents $5$ eggs eggs
$\therefore OOOO$ represents $5 \times 4 = 20$ eggs.

A sure event is the one that is defined to happen for sure.
Getting a head or tail while tossing a coin once is a sure event as the outcome will definitely be either head or tail.
Thus, the probability of a sure event is $1.$
Also, the probability of an impossible event is $0.$

Given classes are $0-10$ and $10-20.$
As, the class of given classes is $10$, so the next classes will be $20-30$ and $30-40.$
As, the fourth class is $30-40.$
Hence, the lower limit of fourth class is $30.$

The number of students getting less than $8$ marks $= 6 + 9 + 8 = 23$

Social Science $\rightarrow 100.$

Arranging the data in the tally marks we get:

The most liked animal is cat which is liked by $6$ students.

A symbol is used to represent $10$ flowers
Number of symbols to be drawn to represent $60$ flowers $=\frac{60}{10}=6$

One card is drawn from a well-shuffled deck of $52$ card, possible out comes $= 52$ probability of a card being a black $6$
(Which are two) $=\frac{2}{52}=\frac{1}{26}$

Width of class interval = upper class limit-lower class limit$= 40 - 30$
$= 10$

Minimum frequency $= 2 \rightarrow 20-30$

Frequency of class $20-25 =$ Frequency of class $30-35 = 20$

Central angle of the sector representing the sikh community $=\Big(\frac{\text{value(in%)of each community}}{\text{100}}\times360\Big)^\circ$
$=\Big(\frac{35}{100}\times360\Big)^{\circ}$
$=126^\circ$

Even prime number$ = 2$
$\therefore$ Probability = $\frac{1}{6}$

The total number of people $= 2400$
So, number of people going to office by walking = $\frac{75}{360}\times2400 = 500$

When a die is thrown,
Number of total possible outcomes $= 6$
Number of even prime numbers $= 1 (2$ is the only even prime number)
Thus, the probability of getting a even prime number,
$\text{P} = \frac{1}{6}$

A die has six sides each consisting numbers from $1$ to $6$, i.e. $\{1, 2, 3, 4, 5, 6\}.$
Total possible outcomes $= 6$
Number of even numbers on a die $= 3$
i.e. $\{2, 4, 6\}$
Probability $=\frac{3}{6}=\frac{1}{2}$

Expenditure done on food $= 5 \times 1000$
$= Rs. 5000$

We know, histogram is a graphical representation of data using bars with no gap in between, i.e. it shows continuous data.
Also, the height of the bar shows the frequency of the class-interval.
Hence, option $B$ is correct.

$300 \rightarrow $ Maths.

The upper value of specific class interval is called the Upper limit of that class. And, the lower value of specific class interval is called the Lower limit of that class. Thus, the Upper Limit of Class Interval $35 – 45$ is $45.$

Class mark $= \frac{\text{Upper limit+ Lower limit}}{2}$
$\therefore $ Class mark of the class 20-30 $= \frac{30+ 20}{2}= \frac{50}{2}=25$

In a bag, there are $3$ white and $2$ red balls possible outcomes $= 3 + 2 = 5$
Now probability of a red ball drawn$=\frac{2}{5}$

Greatest percentage $= 40\% \rightarrow B$

Range of data = Maximum value - Minimum value $= 61 - 20 = 41$

Required number $= 5 + 2 = 7$