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

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 $= 6 + 9 + 8 = 23$