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Maths 5 Marks Questions

Stastics · CBSE STD 10 (CBSE - English Medium). 16 questions.

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Question 15 Marks
250 apples of a box were weighed and the distribution of masses of the apples is given in the following table:

Mass (in grams)80-100100-120120-140140-160160-180
Number of apples206070x60

(i)Find the value of X and the mean mass of the apples.
(ii) Find the model mass of the apples.

Answer
(i) Here, the value of x= 250-210=40

Mass (in gm)FiXiFiXi
80-10020901800
100-12060106600
120-140701309100
140-160401506000
160-180601701020

$\Sigma F_i=250 \Sigma F_2 X_1=33700$
We know that, 
- Mean $=\frac{\Sigma F _{ i } X _{ i }}{\Sigma F _{ i }}$
$So\Rightarrow Mean =\frac{33700}{250}$
→Mean = 134.8.g

(ii)We know that,
Model class is the group with highest frequency 
So,model class=120-140
Hence, modal mass = 70 g

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Question 25 Marks
The distribution below gives the weight of $30$ students of a class. Find the median weight of the students:
Weight in kg $40-45$ $45-50$ $50-55$ $55-60$ $60-65$ $65-70$ $70-75$
Number of Students $2$ $3$ $8$ $6$ $6$ $3$ $2$
Answer
Weight $($in $kg)$ Frequency Cumulative frequency
$40-45$ $2$ $2$
$45-150$ $3$ $5$
$50-355$ $8$ $13$
$55-60$ $6$ $19$
$60-65$ $6$ $25$
$65-70$ $3$ $28$
$70-75$ $2$ $30$
  $N = 30$  
Wehave $N =30$ and $\frac{ N }{2}=15$.
Median class $=55-60$,
So $, l =55 . f =6, cf =13$ and $h =60-55=5$
Median $=1+\frac{\frac{N}{2}-fc}{f} \times h$
$\Rightarrow $ Median
$\Rightarrow 55+\frac{15-13}{6} \times 5$
Median $=55+1.67$
$=56.67 \ kg$
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Question 35 Marks
The houses in a row are numbered consecutively from 1 to 49. Show that there exists a value of X such that sum of houses preceding the house numbered X is equal to sum of the numbers of houses following X.
Answer
Assume a value of $x$ such that the sum of houses preceding the house numbered X is equal to sum of the numbers of houses following X .
That is, $1+2+3 \ldots .+(x-1)=(x+1)+(x+2)$ $+\ldots .+49$
So,
$\begin{array}{l}1+2+3+\ldots .+(x-1)=\{1+2+\ldots .+x+(x+1) \\ +\ldots .+49\}-(1+2+3+\ldots .+x)\end{array}$
$\begin{array}{l}\frac{(x-1)}{2}(1+x-1)=\frac{49}{2}(1+49)-\frac{x}{2}(1+x) \\x(x-1)=49 \times 50-x(1+x) \\x(x-1)+1(1+x)=49 \times 50 \\x^2-x+x+x^2=49 \times 50 \\x^2=49 \times 25\end{array}$
Taking square root,
$x=7 \times 5=35$
Since $x$ is not a fraction, the value of $x$ satisfying the given condition exists and has a value of 35.
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Question 45 Marks
Answer
Calculation of Mean
MarksFrequencyMid Values xi$\begin{array}{l}d_i=x_i-A \\ =x_i-112.5\end{array}$$u_i=\frac{x_i-A}{h}$fiui
0-2510 12.5-100-4-40
25-501537.5-75-3-45
50-752262.5-50-2-44
75-1003087.5-25-1-30
100-12528112.5000
125-15027137.525127
150-17512162.550224
175-2006187.575318
$A=112.5$ is the assumed Mean, $h=25$
Total of the frequency $N=150, \sum f_i u_i=-90$
Using the formula, Mean $=\bar{X}=A+h\left(\frac{1}{N} \Sigma f_i u_i\right)$
$\bar{X}=112.5+25\left(\frac{1}{150} \times-90\right)=112.5+25\left(\frac{-90}{150}\right)=112.5+25\left(\frac{-3}{5}\right)=112.5-15$
Mean $=97.5$
Calculation of Mode
Marks0-2525-5050-7575-100100-125125-150150-175175-200
Number of students101522302827126
Here the maximum frequency is 30 and corresponding class is $75-100$
$l=75$ Lower limit of modal class
$f=30$ Frequency of modal class
$h=25$ Width of modal class
$f_1=22$ Frequency of class preceding the modal class
$f_2=28$ Frequency of class following the modal class
Putting value is the formula
$\begin{array}{l}\text { Mode }=l+\frac{f-f_1}{2 f-f_1-f_2} \times h \\=75+\frac{30-22}{2 \times 30-22-28} \times 25 \\=75+\frac{8}{60-22-28} \times 25 \\=75+\frac{8}{10} \times 25 \\=75+20 \\\text { Mode }=95\end{array}$
Hence, the mean is $97.5$ and mode is $95.$
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Question 55 Marks
Weights of new born babies in a hospital are as follows:
Weight (in kg)1.3-1.51.5-1.71.7-1.91.9-2.12.1-2.32.3-2.52.5-2.72.7-2.9
Number of new born babies14 6910x83

If the mode of the data is 2.2 kg, find the unknown frequency x.
Answer
Here the unknown frequency is $x$ $\text {So}, l=2$,
$h=0.2, f=10, f_1=9, f_2=x$
As we know
$\begin{array}{l}\text { Mode }=l+\frac{f-f_1}{2 f-f_1-f_2} \times h \\2.2=2.1+\frac{10-9}{20-9-x} \times 0.2 \\2.2-2.1=\left(\frac{1}{11-x}\right) \times 0.2 \\0.1=\left(\frac{1}{11-x}\right) \times 0.2 \\\Rightarrow 11-x=2 \\\Rightarrow x=9\end{array}$
Hence, the missing frequency is $9.$
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Question 65 Marks
The mean of the following data is 42. Find the missing frequencies x and y if the total frequency is 100.
Classes0-1010-2020-3030-4040-5050-6060-7070-80
Frequency710x13y10149
Answer
From question, total frequency is 100 .
$\begin{array}{l} 7+10+x+13+y+10+14+9=100 \\\Rightarrow x+y=37\quad\ldots \text {(i)}\end{array}$
ClassesxiFrequency(fi)
0-1057
10-201510
20-30 25X
30-403513
40-5045Y
50-605510
60-706514
70-80759


The Mean will be given by the following formula:
Image
$\begin{aligned}& (5 \times 7)+(15 \times 10)+(25 \times x)+(35 \times 13) \\= & \frac{+(45 \times y)+(55 \times 10)+(65 \times 14)+(75 \times 9)}{(7+10+x+13+y+10+14+9)}\end{aligned}$
Simplifying we get:
$\begin{array}{l}\Rightarrow \frac{25 x+45 y+2775}{x+y+63}=42 \\\Rightarrow 25 x+45 y+2775=42(x+y+63) \\\Rightarrow 25 x+45 y+2775=42 x+42 y+2646 \\\Rightarrow 17 x-3 y=129 \quad\ldots \text {(ii)}\end{array}$
Multiply (i) by 3,
$\Rightarrow 3 x+3 y=111\quad \ldots \text {(iii)}$
Add (ii) and (iii),
$20 x=240$
Divide the above equation by 20,
$\Rightarrow x=12$
Putting the above value in $x+y=37$,
We get $y=25$
Therefore, $x=12$ and $y=25$.
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Question 75 Marks
Find mode of the following data:
Classes0-2020-4040-6060-8080-100100-120120-140
Frequency681012653
Answer
Find out the modal group (the group with highest frequency), which is $60-80$.
The formula to estimate the Mode is:
$\text { Mode }=L+\frac{\left(f_m-f_{m-1}\right)}{\left(f_m-f_{m-1}\right)+\left(f_m-f_{m+1}\right)} \times w$
Where,
$L=$ the lower boundary of the modal group $=60$
$f_{m-1}=$ the frequency of the group before the modal group = 10
$f_m=$ the frequency of the modal group $=12$
$f_{m+1}=$ the frequency of the group after the modal group $=6$
$w=$ the group width $=20$
Mode $=60+\frac{(12-10)}{(12-10)+(12-6)} \times 20$
Mode $=60+\frac{2}{2+6} \times 20$
$\Rightarrow \text { Mode }=60+5=65$
Hence, mode is $65.$
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Question 85 Marks
Answer
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Question 95 Marks
Find the missing frequency f1 and f2 in the following distribution table, if N = 100 and median is 32.
Class:0-1010-2020-3030-4040-5050-60Total
Frequency:10F12530F210100
Answer
We know that the frequency of the class $10-20$ is $f_1$ and that of the class $40-50$ is $f_2$.
Also, the total frequency is 100 .
Therefore, $10+f_1+25+30+f_2+10=100$
$\begin{array}{l}f_1+f_2=100-75 \\
f_1+f_2=25\quad \quad \ldots \ldots \text {(i)}\end{array}$
ClassFrequencyCumulative Frequency
0-101010
10-20f110 + f1
20-302525 + 10 + f1 = 35 + f1
30-403030 + 35 + f1 = 65 + f1
40-50f2f2 + 65 + f1 = 65 + f1 + f2
50-601010 + f2 + 65 + f1 = 75 + f1 + f2

We know that median is 32 , so it lies in the class 30 - 40 .
Hence $30-40$ is the median class.
We have, $l=$ Lower limit of median class $=30$, $h =10, f=$ Frequency of median class $=30$,
$c f=$ Cumulative frequency of class preceding median class $=35+f_1, N=100$
We know,
$\begin{array}{l}\text { Median }=l+\frac{\frac{N}{2}-c f}{f} \times h \\
32=30+\frac{\frac{100}{2}-\left(35+f_1\right)}{30} \times 10 \\\Rightarrow 2=\frac{50-\left(35+f_1\right)}{3} \\
\Rightarrow 6=15-f_1 \Rightarrow f_1=9\end{array}$
We know that, $f_1+f_2=25\quad\quad\quad\quad \text {(Using (i))}$
$\therefore f_2=16$
The missing frequency $f_1=9$ and $f_2=16$.
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Question 105 Marks
If the median of the following data is 525. Find the values of x and y if the sum of the frequencies is 100.
C.I.0-100100-200200-300300-400400-500
Frequency25X1217

C.I.500-600600-700700-800800-900900-1000
Frequency20Y974
Answer
Construct the following table.
Class IntervalFrequenc yCumulative frequency
0-10022
100-20052+5=7
200-300X7 + x
300-4001212 + 7 + x = 19 + x
400-5001719 + x + 17 = 36 + x
500-6002020 + 36 + x = 56 + x
600-700Y56 + x + y
700-80099 + 56 + x + y = 65 + x + y
800-90077 + 65 + x + y = 72 + x + y
900-100044 + 72 + x + y = 76 + x + y

As median is 525 , the median class is $500-600$.
$l=$ Lower limit of median class $=500$
$h=$ Class Size $=100$
$n=$ Sum of frequencies $=100$
$c f=$ Cumulative frequency of class before median class $=36+x$
$f=$ Frequency of median class $=20$
Median $=l+\left(\frac{\frac{n}{2}-c f}{f}\right) \times h$
$525=500+\frac{\frac{100}{2}-(36+x)}{20} \times 100$
$\begin{array}{l}\Rightarrow 525-500=(50-(36+x)) \times 5 \\
\Rightarrow 25=(14-x) \times 5 \Rightarrow 5=(14-x) \\\Rightarrow x=14-5=9\end{array}$
The sum of frequencies is 100 .
$\Rightarrow 100=76+x+y$
Substituting $x=9$, we get
$\begin{array}{l}100=76+9+y \\\Rightarrow 100=85+y \Rightarrow y=100-85=15\end{array}$
Hence, the value of $x$ is 9 and value of $y$ is 15.
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Question 115 Marks
The following table gives production yield per hectare of wheat of 100 farms of a village :
Production yield40-4545-5050-5555-6060-6565-70
No. of farms4616203024
Change the distribution to a 'more than' type distribution and draw its ogive.
Answer
Production yieldcf
more than 6524
more than 6054
more than 5574
more than 5090
more than 4596
more than 40100
For 'more than' ogive, On X-axis - Lower class unit, On Y-axis-cf
$\therefore$ Coordinates are- $A(40, 100), B(45, 96), C(50,90), D(55, 74), E(60, 54), F(65, 24)$
Image
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Question 125 Marks
The mean of the following frequency distribution is 18. The frequency f in the class interval 19-21 is missing. Determine f.
Class interval11-1313-1515-1717-1919-2121-2323-25
Frequency36913f54
Answer
Classfixifixi
11-1331236
13-1561484
15-17916144
17-191318234
19-21f2020f
21-23522110
23-2542496
 $\Sigma f i=40+f$ $\Sigma f_i x_i=704+20 f$

$\begin{aligned} & \bar{x}=18, \bar{x}=\frac{\Sigma f i x i}{\Sigma f i} \Rightarrow \frac{704+20 f}{40+f}=18 \\ \Rightarrow & 704+20 f=760+18 f \Rightarrow 2 f=56 \Rightarrow f=28\end{aligned}$
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Question 135 Marks
Electric buses are becoming popular nowadays. These buses have the electricity stored in a battery. Electric buses have a range of approximately 280 km with just one charge. These buses are superior to diesel buses as they reduce brake wear and also reduce pollution. Transport department of a city wants to buy some electric buses for the city. So, the department wants to know to buy some electric buses for the city. So, the department wants to know the distance travelled by existing public transport buses in a day.
The following data shows the distance travelled by 50 existing public transport buses in a day.
Image
Daily Distance Travelled (in km)100-120120-140140-160160-180180-200
Number of Buses12148610

(a) Find the 'median' distance travelled by a bus.
(b) Find the 'mean (average)' distance travelled by a bus.
Answer
Distance (in km)No. of Buses (fi)Mid- point (xi)Cffixi
100-12012110 121320
120-14014130261820
140-1608150341200
160-1806170401020
180-20010190501900
$\begin{array}{l}N=\Sigma f_i =50\end{array}$ $\begin{array}{l}\Sigma f_i X i  =7260\end{array}$

Here, $N =50 \Rightarrow \frac{N}{2}=25$
So, median class is $120-140$
Here, $l =120, F=12, f =14$ and $h =20$
$\begin{aligned}\text { Median } & =\lambda+\left(\frac{\frac{N}{2}-F}{f}\right) \times h \\& =120+\left(\frac{25-12}{14}\right) \times 20 \\& =120+18.57 \\& =138.57\end{aligned}$
Hence, median distance travelled by the bus is 138.57 km .
$\begin{aligned}\text {(b)}\text { Mean } & =\frac{\sum f_{i} x_{i}}{\sum f_{i}} \\& =\frac{7260}{50}=145.2\end{aligned}$
Hence, the mean distance travelled by the bus is 145.2 km.
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Question 145 Marks
Draw a 'less than' ogive for the following frequency distribution:
Classes: 0-1010-2020-3030-4040-5050-6060-7070-80
Frequency:71413122011158
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Question 155 Marks
The following table gives production yield in kg per hectare of wheat of 100 farms of a village:
Production yield (kg/ hectare):40-4545-5050-5555-6060-6565-70
Number of farms:4616203024

Change the distribution to a 'more than type' distribution, and draw its ogive.

Answer
Production yield (in kg/ha)Number of farms (f)
40-454
45-506
50-5516
55-6020
60-6530
65-7024
 $\Sigma f _{ i }=100$

 
Production yield (X-axis)Number of farms (y-axis)
More than 40$100$
More than 45$100 – 4 = 96$
More than 50$90 - 6 = 90$
More than 55$90-16 = 74$
More than 60$74-20 = 54$
More than 65$54-30 = 24$

Image
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Question 165 Marks
Calculate the mean of the following frequency distribution:
Class:10-3030-5050-7070-9090-110110-130
Frequency:5 8122032
Answer
ClassFrequency(fi)Xidi = xi - A$\mu_{ i }=\frac{ x _{ i }- A }{ h }$fiui
10-30520-60-3-15
30-50840-40-2-16
50-701260-20-1-12
70-9020A(80)000
90-11031002013
110-13021204024
 $\Sigma f _{ i }=50$   $\Sigma f _{ i } \mu_{ i }=-36$

$\begin{array}{l}\text { Mean }= A +\frac{\sum f _{ i } \mu_{ i }}{\sum f _{ i }} \times h \\ \text { Here, } A =80, \Sigma f _{ i } \mu_{ i }=-36, \Sigma f _{ i }=50 \\ \text { and } h =30-10=20 \\ \begin{aligned} \therefore \text { Mean } & =80+\frac{(-36)}{50} \times 5 \\ & =80+\frac{(-36)}{10} \\ & =80-3.6=76.4\end{aligned}\end{array}$
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5 Marks Questions - Maths STD 10 Questions - Vidyadip