M.C.Q. [1 Marks Each] - MATHS STD 7 Questions - Vidyadip

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M.C.Q. [1 Marks Each]

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16 questions · auto-graded multiple-choice test.

MCQ 11 Mark

In a school, only $2$ out of $5$ students can participate in a quiz. What is the chance that a student picked at random makes it to the competition?

A
$20\%$
✓
$40\%$
C
$50\%$
D
$30\%$

Answer

Correct option: B.

$40\%$

Total number of outcomes = Total number of students $= 5$
Number of possible outcomes = Students participating in a quiz $= 2$
$\therefore\text{Probability}=\frac{\text{Number of possible outcomes}}{\text{Total number of outcomes}}=\frac{2}{5}$
to To find percentage, we havemultiply it by hundred $=\frac{2}{5}\times100=40\%$

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MCQ 21 Mark

There are $2$ aces in each of the given set of cards placed face down. From which set are you certain to pick the two aces in the first go?

A
B
✓
D

Answer

Correct option: C.

From third set, we are certain to pick the two aces in the first go because it has only 2 cards and it is given that every set has $2$ aces.

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MCQ 31 Mark

Which of the following has the same mean, median and mode?

A
$6, 2, 5, 4, 3, 4, 1$
B
$4, 2, 2, 1, 3, 2, 3$
C
$2, 3, 7, 3, 8, 3, 2$
✓
$4, 3, 4, 3, 4, 6, 4$

Answer

Correct option: D.

$4, 3, 4, 3, 4, 6, 4$

$a.$ Data $($in ascending order$) \rightarrow 1, 2, 3, 4, 4, 5, 6$
Here, $n = 7 ($odd$)$
Median $=$ Value of $\Big(\frac{\text{n+1}}{2}\Big)^\text{th}$ observation $=$ Value of $\Big(\frac{8}{2}\Big)^\text{th}$ observation $= 4$
$\text{Mean}=\frac{\text{Sum of observation}}{\text{n}}$
$=\frac{1+2+3+4+4+5+6}{7}$
$=\frac{25}{7}$
$=3.57$
Mode $=$ Most frequent observation $= 4$
Hence,
Mean $\neq $ Median $=$ Mode
$b.$ Data $($in ascending order$) \rightarrow 1, 2, 2, 2, 3, 3, 4$
Here, $n = 7 ($odd$)$
Median $=$ Value of $\Big(\frac{\text{n+1}}{2}\Big)^\text{th}$ observation $=$ Value of $\Big(\frac{7+1}{2}\Big)^\text{th}$ observation $= 2$
$\text{Mean}=\frac{\text{Sum of observation}}{\text{n}}$
$=\frac{1+2+2+2+3+3+4}{7}$
$=\frac{17}{7}$
$=2.428$
Mode $=$ Most frequent observation $= 2$
Hence,
Mean $\neq$ Median $=$ Mode
$c.$ Data $($in ascending order$) \rightarrow 2, 2, 3, 3, 3, 7, 8$
Here, $n = 7 ($odd$)$
Median $=$ Value of $\Big(\frac{\text{n+1}}{2}\Big)^\text{th}$ observation $=$ Value of $\Big(\frac{7+1}{2}\Big)^\text{th}$ observation $= 3$
$\text{Mean}=\frac{\text{Sum of observation}}{\text{n}}$
$=\frac{2+2+3+3+3+7+8}{7}$
$=\frac{28}{7}$
$=4$
Hence,
Mean $\neq$ Median $=$ Mode
$d.$ Data $($in ascending order$) \rightarrow 3, 3, 4, 4, 4, 4, 6$
Here, $n = 7 ($odd$)$
Median $=$ Value of $\Big(\frac{\text{n+1}}{2}\Big)^\text{th}$ observation $=$ Value of $\Big(\frac{7+1}{2}\Big)^\text{th}$ observation $= 4$
$\text{Mean}=\frac{\text{Sum of observation}}{\text{n}}$
$=\frac{3+3+4+4+4+4+6}{7}$
$=\frac{28}{7}$
$=4$
Hence,
Mean $=$ Mode $=$ Median

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MCQ 41 Mark

The mean of three numbers is $40$. All the three numbers are different natural numbers. If lowest is $19$, what could be highest possible number of remaining two numbers?

✓
$81$
B
$40$
C
$100$
D
$71$

Answer

Correct option: A.

$81$

Mean of three numbers $= 40$ and lowest number $= 19...[$given$]$
Let the three observations be $19, x$ and $y,$ respectively.
$\text{Mean}=\frac{\text{Sum of all observations}}{\text{Total number of observations}}$
$\Rightarrow40=\frac{19+\text{x}+\text{y}}{3}$ [$\because$ mean = 40, given]
$\Rightarrow3\times40=19+\text{x}+\text{y}$
$\Rightarrow120=19+\text{x}+\text{y}$
$\Rightarrow\text{x}+\text{y}=120-19$
$\Rightarrow\text{x}+\text{y}=101\ ...(\text{i})$
Since, $19$ is the lowest observation.
Hence, for highest possible value of remaining two numbers, one must be 20.
Let $x = 20$
From Eq.$(i)$, we get
$20 + y = 101$
$\Rightarrow y = 101 - 20$
$\Rightarrow y = 81$

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MCQ 51 Mark

The range of the data : $21, 6, 17, 18, 12, 8, 4, 13$ is:

✓
$17$
B
$12$
C
$8$
D
$15$

Answer

Correct option: A.

$17$

Here,
Highest observation $= 21$
Lowest observation $= 4$
Range = Highest observation - Lowest observation
$= 21 - 4 = 17$

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MCQ 61 Mark

Khilona earned scores of $97, 73$ and $88$ respectively in her first three examinations. If she scored $80$ in the fourth examination, then her average score will be:

A
Increased by $1$
B
Increased by $1.5$
C
Decreased by $1$
✓
Decreased by $1.5$

Answer

Correct option: D.

Decreased by $1.5$

$\text{Average score}=\frac{\text{Sum of scores in all exams}}{\text{Total number of exams}}$
$\therefore$ Average score in first three examination $=\frac{97+73+88}{3}=\frac{258}{3}=86$
Also, average score in four examination $=\frac{97+73+88+80}{4}=\frac{338}{4}=84.5$
Hence, average score is decreased by $(86 - 84.5) = 1.5$

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MCQ 71 Mark

The number of trees in different parks of a city are $33, 38, 48, 33, 34, 34, 33$ and $24$. The mode of this data is:

A
$24$
B
$34$
✓
$33$
D
$48$

Answer

Correct option: C.

$33$

We have, $33, 38, 48, 33, 34, 34, 33$ and $24.$
On arranging the data in ascending order, we get $24, 33, 33, 33, 34, 34, 38$ and $48.$
Here, $33$ occurs more frequently, i.e. $3$ times.
Mode of data $= 33$
Note: Mode is the observation that occurs most frequently in the data.

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MCQ 81 Mark

Let x, y, z be three observations. The mean of these observations is:

A
$\frac{\text{x}\times\text{y}\times\text{z}}{3}$
✓
$\frac{\text{x}+\text{y}+\text{z}}{3}$
C
$\frac{\text{x}-\text{y}-\text{z}}{3}$
D
$\frac{\text{x}\times\text{y}+\text{z}}{3}$

Answer

Correct option: B.

$\frac{\text{x}+\text{y}+\text{z}}{3}$

Here $x, y$ and $z$ be three observations.
We know that, $\text{Mean}=\frac{\text{Sum of observations}}{\text{Number of observations}}$
$=\frac{\text{x+y+z}}{3}$

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MCQ 91 Mark

Some integers are marked on a board. What is the range of these integers?

A
$31$
✓
$37$
C
$20$
D
$3$

Answer

Correct option: B.

$37$

Here, highest observation $= +20$ and lowest observation $= -17$
As we know,
Range = Highest observation – Lowest observation $= +20 - (-17) = 20 + 17 = 37$

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MCQ 101 Mark

The difference between the highest and the lowest observations in a data is its:

A
Frequency.
B
Width.
✓
Range.
D
Mode.

Answer

Correct option: C.

Range.

The difference between the highest and the lowest observations in a data is its range.

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MCQ 111 Mark

On tossing a coin, the outcome is:

A
Only head.
B
Only tail.
C
Neither head nor tail.
✓
Either head or tail.

Answer

Correct option: D.

Either head or tail.

When we toss a coin, two outcomes are possible, i.e. head or tail.

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MCQ 121 Mark

Which measures of central tendency get affected if the extreme observations on both the ends of a data arranged in descending order are removed?

✓
Mean and mode.
B
Mean and Median.
C
Mode and Median.
D
Mean, Median and Mode.

Answer

Correct option: A.

Mean and mode.

Mean is defined as follows:
$\text{Mean}=\frac{\text{Sum of observation}}{\text{Number of observations}}$
So, if we remove the extrema values that both sum and total number of observations will change. Hence, mean wiii aiso change.
Mode is that observation which occurs the most. So, if extreme value of those values which occurs mostly than mode can affect it they are removed.
Median is the mid value. So, if extreme values are removed than the mid value remains same. Hence, median will not change.

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MCQ 131 Mark

Out of $5$ brands of chocolates in a shop, a boy has to purchase the brand which is most liked by children. What measure of central tendency would be most appropriate if the data is provided to him?

A
Mean.
✓
Mode.
C
Median.
D
Any of the three.

Answer

Correct option: B.

Mode.

Mode is the most appropriate central tendency because it is the observation that occurs most frequently.
Here, by the measurement of mode, we can find out the chocolates which is most liked by children.

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MCQ 141 Mark

The median of the data $3, 4, 5, 6, 7, 3, 4$ is:

✓
$5$
B
$3$
C
$4$
D
$6$

Answer

Correct option: A.

$5$

We know that, median is the middle most observation.
For finding the median of the data firstly, we arrange the data in ascending order, i.e. Ascending order
i.e $3, 3, 4, 4, 5, 6, 7.$
$n = 7$ (odd)
$\therefore$ Median = Value of $\Big(\frac{\text{n+1}}{2}\Big)^{\text{th}}$ observation = Value of $\Big(\frac{7+1}{2}\Big)^{\text{th}}$ observation
$= 4th$ observation $= 4$

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MCQ 151 Mark

In the previous question, what is the probability of picking up an ace from set $(d)$?

A
$\frac{1}{6}$
✓
$\frac{2}{6}$
C
$\frac{3}{6}$
D
$\frac{4}{6}$

Answer

Correct option: B.

$\frac{2}{6}$

$\text{Probability}=\frac{\text{Number of possible outcomes}}{\text{Total number of outcomes}}$
Total number of cards in set $(d) = 6$
Number of possible outcomes $= 2$ [$\because$ $2$ aces in every set, given]
So, probability $=\frac{2}{6}$

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MCQ 161 Mark

Which measure of central tendency best represents the data of the most popular politician after a debate?

A
Mean.
B
Median.
✓
Mode.
D
Any of the above.

Answer

Correct option: C.

Mode.

Mode is the most frequent observation in a data. So, the measure of central tendency best represents the data of most popular politician after a debate.

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