2 Marks Questions - Page 2 - Maths STD 10 Questions - Vidyadip

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2 Marks Questions

Question 512 Marks

Write down the decimal expansions of the following rational numbers by writing their denominators in the form $2^m× 5^n$, where, m, n are non-negative integers.
$\frac{13}{125}$

Answer

The given number is $\frac{13}{125}.$
$\frac{13}{125} = \frac{13}{5^{3}} = \frac{13}{5^{3} \times 2^0}$
$= \frac{13\times 2^{3}}{5^{3}\times 2^{3}} = \frac{13\times 8}{(10)^3} = \frac{104}{1000} = 0.104$
Thus, the decimal expansion of $\frac{13}{125}$ is 0.014.

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Question 522 Marks

Without actually performing the long division, state whether state whether the following rational numbers will have a terminating decimal expansion or a non terminating repeating decimal expansion.
$\frac{129}{2^2\times5^7\times7^{17}}$

Answer

Since the denominator is not of the form $2^m,5^n$, and it also has $7$ as its factor, the decimal expansion of $\frac{129}{2^2\times5^77^{17}}$ is non-terminating repeating.

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Question 532 Marks

Write 98 as product of its prime factors.

Answer

Using factors tree for prime factorisation of $98$.

prime factoris of $98 = 2 \times 7 \times 7$

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Question 542 Marks

Define $HCF$ of two positive integers and find the $HCF$ of the following pairs of numbers:
$240$ and $6552$

Answer

By applying Euclid’s division lemma
$6552 = 240 \times 27 + 72$
Since remainder ≠ 0, apply division lemma on divisor of $240$ and remainder $72.$
$240 = 72 \times 3 + 24$
Since remainder ≠ 0, apply division lemma on divisor of $72$ and remainder $24.$
$72 = 24 \times 3 + 0$
$\therefore$ $H.C.F$ of $6552$ and $240$ is $= 24.$

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Question 552 Marks

Find the $HCF$ of the following pairs of integers and express it as a linear combination of them.
$1288$ and $575$

Answer

By applying Euclid’s division lemma
$1288 = 575 × 2 + 138 …..(i)$
Since remainder $≠ 0,$ apply division lemma on division $575$ and remainder $138.$
$575 = 138 × 4 + 23 …..(ii)$
Since remainder $≠ 0,$ apply division lemma on division $138$ and remainder $23 …..(iii)$
$HCF = 23$
Now, $23 = 575 - 138 × 4 [$from $(ii)]$
$= 575 - [1288 - 575 × 2] × 4 [$from $(i)]$
$= 575 - 1288 × 4 + 575 × 8$
$= 575 × 9 - 1288 × 4$

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Question 562 Marks

Use Euclid's division algorithm to find the $HCF$ of:
$136, 170$ and $255$

Answer

Given integers are $136, 170$ and $255.$
Let us first find the $HCF$ of $136, 170$ by Euclid lemma.
Clearly, $170 > 136.$ So, we will apply Euclid’s division lemma to $136$ and $170.$
$170 = 136 × 1 + 34$
Remainder is $34$ which is a non-zero number.
Now, apply Euclid’s division lemma to $136$ and $34.$
$136 = 34 × 4 + 0$
The remainder at this stage is zero.
Therefore, $34$ is the $HCF$ of $136$ and $170.$
Now, again use Euclid’s division lemma to find the $HCF$ of $34$ and $255.$
$255 = 34 × 7 + 17$
Remainder is $17$ which is a non-zero number.
Now, apply Euclid’s division lemma to $34$ and $17.$
$34 = 17 × 2 + 0$
The remainder at this stage is zero.
Therefore, $17$ is the $HCF$ of $136, 170$ and $255.$

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Question 572 Marks

What can you say about the prime factorisations of the $43$ of the following rationals:
$0.120120012000120000...$

Answer

We have,
$0.120120012000120000...$
We can see that it is a non-terminating repeating decimal expersion.
So, its denominator has factors others them $2$ or $5$.

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Question 582 Marks

Define $HCF$ of two positive integers and find the $HCF$ of the following pairs of numbers:
$75$ and $243$

Answer

Prime factors of $75$ and $243$ are
$75 = 3 × 5 × 5 = 3 × 5^2$
$243 = 3 × 3 × 3 × 3 × 3 = 3^5$
$HCF (75, 243) =$ Product of the smallest power of each common prime factor in the number
$= 3$

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Question 592 Marks

The following table gives production yield per hectare of wheat of 100 farms of a village:

Production yield	50-55	55-60	60-65	65-70	70-75	75-80 in kg per hecctare
Number of farms	2	8	12	24	38	16

Draw ‘less than’ ogive and ‘more than’ ogive.

Answer

Prepare a table for less than type

Production yield	No. of farms	Production yild (less than)	Cumulative frequency	Suitable points
50-55	2	55	2	(55, 2)
55-60	8	60	10	(60, 10)
60-65	12	65	22	(65, 22)
65-70	24	70	46	(70, 46)
70-75	38	75	84	(75, 84)
75-80	16	80	100	(80, 100)

Now, plot the less than ogive using suitable points.

Again, prepare a table for more than type.

Production yield	No. of farms	Production yield (more than)	Cumulative frequency	Suitable points
50-55	2	50	100	(50, 100)
55-60	8	55	98	(55, 98)
60-65	12	60	90	(60, 90)
65-70	24	65	78	(65, 78)
70-75	38	70	54	(70, 54)
75-80	13	75	16	(75, 16)

Now, plot the more than ogive with suitable points.

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Question 602 Marks

The shirt sizes worn by a group of 200 persons, who bought the shirt from a store, are as follows:

Shirt size	37	38	39	40	41	42	43	44
Number of persons	15	25	39	41	36	14	15	12

Find the model shirt size worn by the group.

Answer

Shirt size	37	38	39	40	41	42	43	44
Number of persons	15	25	39	41	36	17	15	12

We see that frequency of 40 is maximum which is 41
$\therefore\ \text{Mode}=40$

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Question 612 Marks

Write the median class for the following frequency distribution:

Class-interval	0-10	10-20	20-30	30-40	40-50	50-60	60-70	70-80
Frequency	5	8	7	12	28	20	10	10

Answer

We are given the following table.

Class interval	Frequency	Cumulative Frequency
0-10	5	5
10-20	8	13
20-30	7	20
30-40	12	32
40-50	28	60
50-60	20	80
60-70	10	90
70-80	10	100
	N = 100

Here, N = 100 $\therefore\frac{\text{N}}{2}=50$ The cumulative frequency just greater than 50 is 60.

So, the median class is 40−50.

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Question 622 Marks

Write the median class of the following distribution:

Classes	0-10	10-20	20-30	30-40	40-50	50-60	60-70
Frequency	4	4	8	10	12	8	4

Answer

Consider the following distribution table.

Class	Frequency	C.F.
0-10	4	4
10-20	4	8
20-30	8	16
30-40	10	26
40-50	12	38
50-60	8	46
60-70	4	50
	N = 50

Here, N = 50 $\frac{\text{N}}{2}=25$ The cumulative frequency just greater than 25 is 26. So, the meian class is 30-40

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Question 632 Marks

A student draws a cumulative frequency curve for the marks obtained by 40 students of a class as shown below. Find the median marks obtained by the students of the class.

Answer

Median marks Here N = 40, then $\frac{\text{N}}{2}=\frac{40}{2}=20$ From 20 on y-axis, draw a line parallel to the x-axis meeting the curve at P and from P, draw a perpendicular on x-axis meeting it at M. Then M is the median which is 50.

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Question 642 Marks

Calculate the mean for the following distribution:

x	5	6	7	8	9
f	4	8	14	11	3

Answer

x	f	fx
5	4	20
6	8	48
7	14	98
8	11	88
9	3	27
	N = 40	$\sum\text{fx}=281$

$\text{Mean}\ \bar{\text{X}}=\frac{\sum\text{fx}}{\text{N}}$
$\bar{\text{X}}=\frac{281}{40}=7.025$

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Question 652 Marks

Draw an ogive to represent the following frequency distribution:

Class-interval	0-4	5-9	10-14	15-19	20-24
No. of students	2	6	10	5	3

Answer

Firstly, prepare the cumulative frequency table.

Class interval	No. of students	Less than	Cumulative frequency	Suitable points
0-4	2	4	2	(4,2)
5-9	6	9	8	(9,8)
10-14	10	14	18	(14,18)
15-19	5	19	23	(19,23)
20-24	3	24	26	(24,26)

Now, plot the less than ogive using the suitable points.

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Question 662 Marks

find the class marks of classes 10-25 and 35-55.

Answer

We know that Class mark $=\frac{\text{Sum of its limits}}{2}$ $\therefore$ Class mark of 10-25 $=\frac{10+25}{2}=\frac{35}{2}=17.5$ and class mark of 35-55 $=\frac{35+55}{2}=\frac{90}{2}=45$

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Question 672 Marks

Find the mode of the following distribution.

Class-interval	10-15	15-20	02-25	25-30	30-35	35-40
Frequency	30	45	75	35	25	15

Answer

Class interval	10-15	15-20	20-25	25-30	30-35	35-40
No. of persons	30	45	75	35	25	15

Here the maximum frequency is 75 then the corresponding class 20-25 is the moidel class
$\text{L}=25,\text{h}=25-20=5,\text{f}=75,\text{f}_1=45,\text{f}_2=35$
$\text{Model}=\text{L}+\frac{\text{f}-\text{f}_1}{2\text{f}-\text{f}_1-\text{f}_2}\times\text{h}$
$=20+\frac{75-45}{2\times75-45-35}\times5$
$=20+\frac{30\times5}{70}$
$=20+2.14$
$=22.14$

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Question 682 Marks

n the graphical representation of a frequency distribution, if the distance between mode and mean isk times the distance between median and mean, then write the value of k.

Answer

We know that Mode = 3 median – 2 mean ….(i) Now mode – mean = k (median – mean), ….(ii) But mode – mean = 3 median – 2 mean [from (i)] ⇒ Mode – mean = 3 (median – mean) ….(iii) Comparing (ii) and (iii) k = 3

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Question 692 Marks

Find the mode of the following distribution:

Class-interval	0-10	10-20	20-30	30-40	40-50	50-60	60-70	70-80
Frequency	5	8	7	12	28	20	10	10

Answer

Class interval	0-10	10-20	20-30	30-40	40-50	50-60	60-70	70-80
No.of persons	5	8	7	12	28	20	10	10

Here the maximum frequency is 28 then the corresponding class 40-50 is the model class
$\text{L}=40,\text{h}=50-40=10,\text{f}=28,\text{f}_1=12,\text{f}_2=20$
$\text{Model}=\text{L}+\frac{\text{f}-\text{f}_1}{2\text{f}-\text{f}_1-\text{f}_2}\times\text{h}$
$=40+\frac{28-12}{2\times28-12-20}\times10$
$=40+\frac{16\times10}{24}$
$=40+16.67=46.67$

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Question 702 Marks

Find the mode of the following distribution.

Class-interval	25-30	30-35	35-40	40-45	45-50	50-60
Frequency	25	34	50	42	38	14

Answer

Class interval	25-30	30-35	35-40	40-45	45-50	50-60
No. of persons	25	34	50	42	38	14

Here the maximum frequency is 28 then the corresponding class 40-45 is the model class
$\text{L}=35,\text{h}=40-35=5,\text{f}=50,\text{f}_1=34,\text{f}_2=42$
$\text{Model}=\text{L}+\frac{\text{f}-\text{h}}{2\text{f}-\text{f}_1-\text{f}_2}\times\text{h}$
$=35+\frac{50-34}{2(50)-34-42}\times5$
$=35+\frac{16\times5}{24}$
$=35+3.33$
$=35.33$

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Question 712 Marks

The monthly profits (in Rs.) of 100 shops are distributed as follows:

Profits per shop	0-50	50-100	100-150	150-200	200-250	250-300
No. of shops	12	18	27	20	17	6

Draw the frequency polygor it.

Answer

Profits per shop	No. of shops (f)	c.f.
0-50	12	12
50-100	18	30
100-150	27	57
150-200	20	77
200-250	17	94
250-300	6	100

Represent profits per shop along x-axis and no. of shop (c.f.) along y-axis.
Plot the points (50, 12), (100, 30), (150, 57), (200, 77), (250, 94) and (300, 100) on the graph and join them with ruler. This is the cumulative polygon as shown.

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2 Marks Questions - Page 2 - Maths STD 10 Questions - Vidyadip