# NCERT Solutions for Class 10 Maths Chapter 14 Exercise 14.1

Mathematics has always piqued the curiosity of students. Geometry is new to all students and may scare them even if Algebra, Trigonometry, and Statistics are all covered in Mathematics for Classes 9 and 10. While some kids struggle with it, others relish the challenges. Students may find it difficult to finish the NCERT Exercise because they may find it difficult to comprehend the themes on their first try. Despite the fact that learning something new is never simple, once students master it, they’ll feel tremendously satisfied.

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The information in this chapter has been broken down into a number of subtopics to make it more logically organised and simpler for students to understand. Mean of grouped data, mode of grouped data, median of grouped data and the graphical representation of cumulative frequency distribution are some of the subheadings for Mathematics Chapter 14 Class 10 as provided by the NCERT Solutions For Class 10 Maths Chapter 14 Exercise 14.1. These subjects all offer organised and condensed information regarding the chapter’s headers and subheadings. Students will find it simpler to compile the chapter’s notes as a result. Students will learn the fundamentals of statistics through the NCERT Solutions For Class 10 Maths Chapter 14 Exercise 14.1, which will aid them in understanding how  data analytics work

The NCERT Solutions For Class 10 Maths Chapter 14 Exercise 14.1 help students understand what the meaning of mean, median and modeand how the data is represented for cumulative frequency distribution. The NCERT Solutions For Class 10 Maths Chapter 14 Exercise 14.1 provide students with the knowledge of providing answers to various queries related to data analysis like mean. The mean of grouped data can be found using three formulas as mentioned in the chapter. The NCERT Solutions For Class 10 Maths Chapter 14 Exercise 14.1 also help students to figure out a way to find the median of grouped data. The formula to find mode of grouped data has also been given in the chapter and the NCERT Solutions For Class 10 Maths Chapter 14 Exercise 14.1. In addition to this, graphical representation for cumulative frequency distribution is also taught through the NCERT Solutions For Class 10 Maths Chapter 14 Exercise 14.1. Students can learn a lot through the Chapter Statistics of Class 10. It will help them get a deeper understanding of Statistics and help choose their stream of choice in Class 11.

## Access NCERT Solutions Maths Class 10 Chapter 14 – Statistics

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## NCERT Solutions for Class 10 Maths Chapter 14 Statistics Exercise 14.1

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Q.1 A survey was conducted by a group of students as a part of their environment awareness programme, in which they collected the following data regarding the number of plants in 20 houses in a locality. Find the mean number of plants per house.

 Number of plants 0-2 2-4 4-6 6-8 8-10 10-12 12-14 Number of houses 1 2 1 5 6 2 3

Which method did you use for finding the mean, and why?

Ans

$\begin{array}{l}\text{Here, we observe that class marks and frequencies are small}\\ \text{quantities.}\\ \text{So, we use direct method to compute the mean and proceed}\\ \text{as below.}\end{array}$

 Number of plants Number of houses (fi) xi fixi 0 – 2 1 1 1 2 – 4 2 3 6 4 – 6 1 5 5 6 – 8 5 7 35 8 – 10 6 9 54 10 – 12 2 11 22 12 – 14 3 13 39 Total 20 162

$\begin{array}{l}\text{Mean}=\overline{\mathrm{x}}=\frac{\sum {\mathrm{f}}_{\mathrm{i}}{\mathrm{x}}_{{}_{\mathrm{i}}}}{\sum {\mathrm{f}}_{\mathrm{i}}}=\frac{162}{20}=8.1\\ \text{Therefore, mean number of plants per house is 8.1.}\end{array}$

Q.2 Consider the following distribution of daily wages of 50 workers of a factory.

 Daily wages (in ₹) 500-120 520-140 540-160 560-180 580-200 Number of workers 12 14 8 6 10

Find the mean daily wages of the workers of the factory by using an appropriate method.

Ans

$\begin{array}{l}\text{We find class mark for each interval by using the following}\\ \text{relation.}\\ {\text{x}}_{\mathrm{i}}=\frac{\text{Upper class limit}+\text{Lower class limit}}{2}\\ \text{Class size(h) of the given data}=20\\ \text{Here, we take assumed mean(a)}=550\text{and proceed as below.}\end{array}$

 Daily wages (in ₹) Number of workers (fi) xi di = xi – 550 ${\mathrm{u}}_{\mathrm{i}}=\frac{{\mathrm{x}}_{\mathrm{i}}-550}{\mathrm{h}}$ fiui 500 – 520 12 510 -40 -2 -24 520 – 540 14 530 -20 -1 -14 540 – 560 8 550 0 0 0 560 – 580 6 570 20 1 6 580 – 600 10 590 40 2 20 Total 50 -12

$\begin{array}{l}\text{Mean}=\overline{\mathrm{x}}=\mathrm{a}+\left(\frac{\sum {\mathrm{f}}_{\mathrm{i}}{\mathrm{u}}_{{}_{\mathrm{i}}}}{\sum {\mathrm{f}}_{\mathrm{i}}}\right)\mathrm{h}=550+\left(\frac{-12}{50}\right)×20=545.2\end{array}$

$\begin{array}{l}\text{Therefore, mean daily wages of the workers is}₹545.20\end{array}$

Q.3 The following distribution shows the daily pocket allowance of children of a locality. The mean pocket allowance is ₹ 18. Find the missing frequency f.

 Daily pocket allowance (in ₹) 11-13 13-15 15-17 17-19 18-21 21-23 23-25 Number of children 7 6 9 13 f 5 4

Ans

$\begin{array}{l}\text{We find class mark for each interval by using the following}\\ \text{relation.}\\ {\text{x}}_{\mathrm{i}}=\frac{\text{Upper class limit}+\text{Lower class limit}}{2}\\ \text{Class size(h) of the given data}=2\\ \text{Given mean(a)}=18\\ \text{We proceed as below to find}{\mathrm{d}}_{\mathrm{i}},\text{}{\mathrm{f}}_{\mathrm{i}}{\mathrm{d}}_{\mathrm{i}}\text{.}\end{array}$

 Daily pocket allowance (in ₹) Number of children (fi) xi di = xi – 18 fidi 11 – 13 7 12 -6 -42 13 – 15 6 14 -4 -24 15 – 17 9 16 -2 -18 17 – 19 13 18 0 0 19 – 21 f 20 2 2f 21 – 23 5 22 4 20 23 – 25 4 24 6 24 Total $\sum {\mathrm{f}}_{\mathrm{i}}=44+\mathrm{f}$ 2f – 40

$\begin{array}{l}\text{Mean}=\overline{\mathrm{x}}=\mathrm{a}+\frac{\sum {\mathrm{f}}_{\mathrm{i}}{\mathrm{d}}_{{}_{\mathrm{i}}}}{\sum {\mathrm{f}}_{\mathrm{i}}}\\ \text{or}18=18+\frac{2\mathrm{f}-40}{44+\mathrm{f}}\\ \text{or \hspace{0.17em}}\mathrm{f}=20\\ \text{Therefore, missing frequency}\mathrm{f}\text{is}20.\end{array}$

Q.4 Thirty women were examined in a hospital by a doctor and the number of heartbeats per minute were recorded and summarised as follows. Find the mean heartbeats per minute for these women, choosing a suitable method.

 Number of heartbeats per minute 65-68 68-71 71-74 74-77 77-80 80-83 83-86 Number of women 2 4 3 8 7 4 2

Ans

$\begin{array}{l}\text{We find class mark for each interval by using the following}\\ \text{relation.}\\ {\text{x}}_{\mathrm{i}}=\frac{\text{Upper class limit}+\text{Lower class limit}}{2}\\ \text{Class size(h) of the given data}=3\\ \text{Here, we take assumed mean(a)}=75.5\text{and proceed as below}\\ \text{to find}{\mathrm{d}}_{\mathrm{i}},\text{}{\mathrm{u}}_{\mathrm{i}}\text{and}{\mathrm{f}}_{\mathrm{i}}{\mathrm{u}}_{\mathrm{i}}\text{.}\end{array}$

 Number of heartbeats per minute Number of women (fi) xi di = xi – 75.5 ${\mathrm{u}}_{\mathrm{i}}=\frac{{\mathrm{x}}_{\mathrm{i}}-75.5}{\mathrm{h}}$ fiui 65-68 2 66.5 -9 -3 -6 68-71 4 69.5 -6 -2 -8 71-74 3 72.5 -3 -1 -3 74-77 8 75.5 0 0 0 77-80 7 78.5 3 1 7 80-83 4 81.5 6 2 8 83-86 2 84.5 9 3 6 Total 30 4

$\begin{array}{l}\text{Mean}=\overline{\mathrm{x}}=\mathrm{a}+\left(\frac{\sum {\mathrm{f}}_{\mathrm{i}}{\mathrm{u}}_{{}_{\mathrm{i}}}}{\sum {\mathrm{f}}_{\mathrm{i}}}\right)\mathrm{h}=75.5+\left(\frac{4}{30}\right)×3=75.9\\ \text{Therefore, mean heartbeats per minute is}75.9.\end{array}$

Q.5 In a retail market, fruit vendors were selling mangoes kept in packing boxes. These boxes contained varying number of mangoes. The following was the distribution of mangoes according to the number of boxes.

 Number of mangoes 50-52 53-55 56-58 59-61 62-64 Number of boxes 15 110 135 115 25

Find the mean number of mangoes kept in a packing box. Which method of finding the mean did you choose?

Ans

$\begin{array}{l}\text{The given class intervals are not continuous. So, we add 0.5 to upper class limit and subtract 0.5 to lower class limit}\\ \text{of each interval. We find class mark for each interval by using the following}\\ \text{relation.}\\ {\text{x}}_{\mathrm{i}}=\frac{\text{Upper class limit}+\text{Lower class limit}}{2}\\ \text{Class size(h) of the given data}=3\\ \text{Here, we take assumed mean(a)}=57\text{and proceed as below}\\ \text{to find}{\mathrm{d}}_{\mathrm{i}},\text{}{\mathrm{u}}_{\mathrm{i}}\text{and}{\mathrm{f}}_{\mathrm{i}}{\mathrm{u}}_{\mathrm{i}}\text{.}\end{array}$

 Class interval fi xi di = xi – 57 ${\mathrm{u}}_{\mathrm{i}}=\frac{{\mathrm{x}}_{\mathrm{i}}-57}{\mathrm{h}}$ ${f}_{i}{u}_{i}$ 49.5 – 52.5 15 51 -6 -2 -30 52.5 – 55.5 110 54 -3 -1 -110 55.5 – 58.5 135 57 0 0 0 58.5 – 61.5 115 60 3 1 115 61.5 – 64.5 25 63 6 2 50 Total 400 25

$\begin{array}{l}\text{Mean}=\overline{\mathrm{x}}=\mathrm{a}+\left(\frac{\sum {\mathrm{f}}_{\mathrm{i}}{\mathrm{u}}_{{}_{\mathrm{i}}}}{\sum {\mathrm{f}}_{\mathrm{i}}}\right)\mathrm{h}=57+\left(\frac{25}{400}\right)×3=57.1875\approx 57.19\\ \text{Therefore, mean number of mangoes is}57.19.\\ \text{We have chosen step deviation method as values of}{\mathrm{f}}_{\mathrm{i}}\text{,}{\mathrm{d}}_{\mathrm{i}}\text{are big}\\ \text{and there is a common multiple between all}{\mathrm{d}}_{\mathrm{i}}.\end{array}$

Q.6 The table below shows the daily expenditure on food of 25 households in a locality.

 Daily Expenditure (in ₹) 100-150 150-200 200-250 250-300 300-350 Number of households 4 5 12 2 2

Find the mean daily expenditure on food by a suitable method.

Ans

$\begin{array}{l}\text{We find class mark for each interval by using the following relation.}\\ {\text{x}}_{\mathrm{i}}=\frac{\text{Upper class limit}+\text{Lower class limit}}{2}\\ \text{Class size(h) of the given data}=50\\ \text{Here, we take assumed mean(a)}=225\text{and proceed as below}\\ \text{to find}{\mathrm{d}}_{\mathrm{i}},\text{}{\mathrm{u}}_{\mathrm{i}}\text{and}{\mathrm{f}}_{\mathrm{i}}{\mathrm{u}}_{\mathrm{i}}\text{.}\end{array}$

 Class interval fi xi di = xi – 225 ${\mathrm{u}}_{\mathrm{i}}=\frac{{\mathrm{x}}_{\mathrm{i}}-57}{\mathrm{h}}$ fiui 100 – 150 4 125 -100 -2 -8 150 – 200 5 175 -50 -1 -5 200 – 250 12 225 0 0 0 250 – 300 2 275 50 1 2 300 – 350 2 325 100 2 4 Total -7

$\begin{array}{l}\text{Mean}=\overline{\mathrm{x}}=\mathrm{a}+\left(\frac{\sum {\mathrm{f}}_{\mathrm{i}}{\mathrm{u}}_{{}_{\mathrm{i}}}}{\sum {\mathrm{f}}_{\mathrm{i}}}\right)\mathrm{h}=225+\left(\frac{-7}{25}\right)×50=211\\ \text{Therefore, mean daily expenditure is}₹211.\end{array}$

Q.7 To find out the concentration of SO­2 in the air (in parts per million, i.e., ppm), the data was collected for 30 localities in a certain city and is presented below.

 Concentration of SO­2 (in ppm) Frequency 0.00 – 0.04 0.04– 0.08 0.08 – 0.12 0.12–0.16 0.16–0.20 0.20–0.24 4 9 9 2 4 2

Find the mean concentration of SO2 in the air.

Ans

$\begin{array}{l}\text{We find class mark for each interval by using the following}\\ \text{relation.}\\ {\text{x}}_{\mathrm{i}}=\frac{\text{Upper class limit}+\text{Lower class limit}}{2}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\\ \text{Class size(h) of the given data}=0.04\\ \text{Here, we take assumed mean(a)}=0.14\text{and proceed as below}\\ \text{to find}{\mathrm{d}}_{\mathrm{i}},\text{}{\mathrm{u}}_{\mathrm{i}}\text{and}{\mathrm{f}}_{\mathrm{i}}{\mathrm{u}}_{\mathrm{i}}\text{.}\end{array}$

 Concentration of SO2 (in ppm) Frequency (fi) xi di = xi – 0.14 ${\mathrm{u}}_{\mathrm{i}}=\frac{{\mathrm{x}}_{\mathrm{i}}-0.14}{\mathrm{h}}$ b fiui 0.00 – 0.04 4 0.02 -0.12 -3 -12 0.04 – 0.08 9 0.06 -0.08 -2 -18 0.08 – 0.12 9 0.10 -0.04 -1 -9 0.12 – 0.16 2 0.14 0 0 0 0.16 – 0.20 4 0.18 0.04 1 4 0.20 – 0.24 2 0.22 0.08 2 4 Total 30 -31

$\begin{array}{l}\text{Mean}=\overline{\mathrm{x}}=\mathrm{a}+\left(\frac{\sum {\mathrm{f}}_{\mathrm{i}}{\mathrm{u}}_{{}_{\mathrm{i}}}}{\sum {\mathrm{f}}_{\mathrm{i}}}\right)\mathrm{h}=0.14+\left(\frac{-31}{30}\right)×0.04\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}\approx 0.099\text{ppm}\\ {\text{Therefore, mean concentration of SO}}_{\text{2}}\text{in the air is}0.099\text{ppm}.\end{array}$

Q.8 A class teacher has the following absentee record of 40 students of a class for the whole term. Find the mean number of days a student was absent.

 Number of days 0-6 6-10 10-14 14-20 20-28 28-38 38-40 Number of students 11 10 7 4 4 3 1

Ans

$\begin{array}{l}\text{We find class mark for each interval by using the following}\\ \text{relation.}\\ {\text{x}}_{\mathrm{i}}=\frac{\text{Upper class limit}+\text{Lower class limit}}{2}\\ \text{Assumed mean(a)}=17\\ \text{We proceed as below to find}{\mathrm{d}}_{\mathrm{i}},\text{}{\mathrm{f}}_{\mathrm{i}}{\mathrm{d}}_{\mathrm{i}}\text{.}\end{array}$

 Number of days Number of students fi xi di = xi – 17 fidi 0 – 6 11 3 -14 -154 6 – 10 10 8 -9 -90 10 – 14 7 12 -5 -35 14 – 20 4 17 0 0 20 – 28 4 24 7 28 28 – 38 3 33 16 48 38 – 40 1 39 22 22 Total 40 -181

$\begin{array}{l}\text{Mean}=\overline{\mathrm{x}}=\mathrm{a}+\frac{\sum {\mathrm{f}}_{\mathrm{i}}{\mathrm{d}}_{{}_{\mathrm{i}}}}{\sum {\mathrm{f}}_{\mathrm{i}}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=17+\frac{-181}{40}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=12.475\approx 12.48\\ \text{Therefore, mean number of days for which a student was absent is}12.48.\end{array}$

Q.9 The following table gives the literacy rate (in percentage) of 35 cities. Find the mean literacy rate.

 Literacy rate (in %) 45-55 55-65 65-75 75-85 85-95 Number of cities 3 10 11 8 3

Ans

$\begin{array}{l}\text{We find class mark for each interval by using the following relation.}\\ {\text{x}}_{\mathrm{i}}=\frac{\text{Upper class limit}+\text{Lower class limit}}{2}\\ \text{Class size(h) of the given data}=10\\ \text{Here, we take assumed mean(a)}=70\text{and proceed as below}\\ \text{to find}{\mathrm{d}}_{\mathrm{i}},\text{}{\mathrm{u}}_{\mathrm{i}}\text{and}{\mathrm{f}}_{\mathrm{i}}{\mathrm{u}}_{\mathrm{i}}\text{.}\end{array}$

 Literacy rate (in %) Number of cities fi xi di = xi – 70 ${\mathrm{u}}_{\mathrm{i}}=\frac{{\mathrm{x}}_{\mathrm{i}}-70}{\mathrm{h}}$ fiui 45 – 55 3 50 -20 -2 -6 55 – 65 10 60 -10 -1 -10 65 – 75 11 70 0 0 0 75 – 85 8 80 10 1 8 85 – 95 3 90 20 2 6 Total 35 -2

$\begin{array}{l}\text{Mean}=\overline{\mathrm{x}}=\mathrm{a}+\left(\frac{\sum {\mathrm{f}}_{\mathrm{i}}{\mathrm{u}}_{{}_{\mathrm{i}}}}{\sum {\mathrm{f}}_{\mathrm{i}}}\right)\mathrm{h}=70+\left(\frac{-2}{35}\right)×10=69.43\\ \text{Therefore, mean literacy rate is}69.43.\end{array}$

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### 1. Where can students find NCERT Solutions For Class 10 Maths Chapter 14 Exercise 14.1?

For students who study Mathematics in Class 10, the NCERT Solutions For Class 10 Maths Chapter 14 Exercise 14.1 are available on the Extramarks website and learning app. For the benefit of students in schools, especially CBSE board affiliated schools, they have been made for studying and comprehending solutions simpler and more enjoyable. One of the many NCERT solutions for Class 10 accessible on Extramarks are the NCERT Solutions For Class 10 Maths Chapter 14 Exercise 14.1. For a variety of tests, including the TOEFL, SATs, Olympiad, Board Exam, JEE, NEET, CUET, etc., Extramarks offers study materials. Students can search for study materials for any exam they are studying for on Extramarks. The NCERT Solutions For Class 10 Maths Chapter 14 Exercise 14.1 are also accessible on Extramarks in addition to solutions to questions of exercises of other chapters and subjects. The NCERT Solutions For Class 10 Maths Chapter 14 Exercise 14.1 were written by academic experts to assist students with any questions they may be having.

### 2. Are NCERT Solutions For Class 10 Maths Chapter 14 Exercise 14.1 helpful for students?

The NCERT Solutions For Class 10 Maths Chapter 14 Exercise 14.1 are considered to be helpful for students getting ready for board exams (like CBSE board exams or ICSE board exam). The NCERT Solutions For Class 10 Maths Chapter 14 Exercise 14.1 are really accurate and will help students with their doubts with that accuracy and efficiency. The NCERT solutions offered by Extramarks should be utilised by students to gain better scores in their exams. The students’ grades will improve because of the NCERT Solutions For Class 10 Maths Chapter 14 Exercise 14.1. They will be able to better understand the chapter as a result. It will aid students in arousing interest in the field and subject and help them make wiser choices about their future. Students may learn a lot about the NCERT Solutions For Class 10 Maths Chapter 14 Exercise 14.1.

### 3. Are there additional resources available for Class 10 students like the NCERT Solutions For Class 10 Maths Chapter 14 Exercise 14.1?

To make it simpler for students receiving their education for Class 10, Extramarks offers resources like NCERT Solutions For Class 10 Maths Chapter 14 Exercise 14.1. Students can easily go through the NCERT Solutions For Class 10 Maths Chapter 14 Exercise 14.1, making it a fantastic resource for review prior to exams. Extramarks provides NCERT Solutions for various subjects of Class 10 like English, Hindi, Mathematics, Science, Social Science, etc. All of this material like NCERT Solutions, study notes and important concepts and questions can be accessed by students through Extramarks portal.

### 4. How might the NCERT Solutions For Class 10 Maths Chapter 14 Exercise 14.1 help Students prepare for their board exam?

Students can learn how to write and respond to questions during board examination by using the NCERT Solutions For Class 10 Maths Chapter 14 Exercise 14.1. When students are having trouble with a question in NCERT, it is helpful to practise with help of the NCERT Solutions For Class 10 Maths Chapter 14 Exercise 14.1 as they will solve doubts faster. Students can move over their errors in answering a question and move on to the next one with the aid of the NCERT Solutions For Class 10 Maths Chapter 14 Exercise 14.1. They can use it to make changes to their own responses or perhaps to figure out a simpler method to respond to the question which will save them time in their board examinations. Students can raise their performance either way. They must use all the resources available to them for help and faster results.

### 5. Why do students struggle with questions when working through Chapter 10 Statistics of Mathematics Class 10?

It is common to run into problems when trying to solve these questions in Mathematics. The formulas are challenging to fully comprehend and use correctly on the first try. To advance in their studies, students can seek assistance from their peers, teachers, and the NCERT Solutions For Class 10 Maths Chapter 14 Exercise 14.1. They can enlist Extramarks’ assistance for any doubt that may arise. It offers preparation for board exams and admission exams for various colleges and universities. Extramarks offers sessions for clearing up confusion. Students can evaluate the effectiveness of the NCERT Solutions For Class 10 Maths Chapter 14 Exercise 14.1. Students can think about the NCERT Solutions For Class 10 Maths Chapter 14 Exercise 14.1 as a result of their hard work for studying and refer to it before the exam. It can be used by students to review right before an exam and to check their own work for errors.

### 6. How can students study effectively for the board exams to increase their performance?

Board exams have always been emphasised to students. However, this does not necessarily imply that they should be afraid of it, feel concerned about it, or try to make up for any preparation errors they might make. Their performance could suffer if they get stressed before tests, so they should make every effort to relax. Here are some pointers for studying for board exams:

• First, instead of studying for long periods of time, strive to study effectively. Ten hours a day of studying could be detrimental to their physical and mental wellbeing. If they can do the same amount of work in less time, students are not expected to study for long periods of time. This gives them more time for review after the day’s work or for their extracurricular activities or interests. This will effectively increase their concentration and focus on studies, while also keeping them motivated and healthy.
• Create a planner and write down your daily objectives. To build a schedule for themselves, they can write down daily, weekly, and monthly objectives. Set academic goals that are in line with their course schedule to allow enough time for review before tests. They don’t have to be academic objectives. They can structure their days around personal objectives they want to accomplish each day, but it must include time for studies. Setting and achieving goals will keep kids motivated during their study period. Additionally, it will assist them maintain their schedule and finish their coursework on time. This will enhance their performance during the board exam by building confidence in their ability to answer questions and managing anxiety.
• Additionally, they have the option of studying in groups. They may benefit from group study if they have trouble focusing or are easily distracted by their surroundings. Members can hold one another accountable and help one another in tracking their academics and syllabus. They can also review each other’s performance in mock tests and sample papers and help each other grow. They can support one another in overcoming doubts.