Class 10

AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 3 Polynomials Ex 3.2 Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Maths Solutions 3rd Lesson Polynomials Exercise 3.2

10th Class Maths 3rd Lesson Polynomials Ex 3.2 Textbook Questions and Answers

Question 1.
The graphs of y = p(x) are given in the figure below, for some polynomials p(x). In each case, find the number of zeroes of p(x).

Answer:
i) There are no zeroes as the graph does not intersect the X – axis.
ii) The number of zeroes is one as the graph intersects the X – axis at one point only.
iii) The number of zeroes is three as the graph intersects the X – axis at three points.
iv) The number of zeroes is two as the graph intersects the X – axis at two points.
v) The number of zeroes is four as the graph intersects the X – axis at four points.
vi) The number of zeroes is three as the graph intersects the X – axis at three points.

Question 2.
Find the zeroes of the given polynomials,
(i) p(x) = 3x
(ii) p(x) = x² + 5x + 6
(iii) p(x) = (x + 2) (x + 3)
(iv) p(x) = x⁴ – 16
Answer:
i) Given p(x) = 3x
Let p(x) = 0
So, 3x = 0
x = \(\frac{0}{3}\) = 0,
Zeroes of p(x) = 3x is zero.
∴ No. of zeroes is one.

ii) Given p(x) = x² + 5x + 6 is a quadratic polynomial.
It has atmost two zeroes.
To find zeroes, let p(x) = 0
⇒ x² + 5x + 6 = 0
⇒ x² + 3x + 2x + 6 = 0
⇒ x(x + 3) + 2 (x + 3) = 0
⇒ (x + 3) (x + 2) = 0
⇒ x + 3 = 0 or x + 2 = 0
⇒ x = -3 or x = -2
Therefore the zeroes of the polynomial are -3 and -2.

iii) Given p(x) = (x + 2) (x + 3)
It is a quadratic polynomial.
It has atmost two zeroes.
Let p(x) = 0
⇒ (x + 2) (x + 3) = 0
⇒ (x + 2) = 0 or (x + 3) = 0
⇒ x = -2 or x = -3
Therefore the zeroes of the polynomial are -2 and – 3.

iv) Given p(x) = x⁴ – 16 is a biquadratic polynomial. It has atmost two zeroes.
Let p(x) = 0
⇒ x⁴ – 16 = 0
⇒ (x²)² – 4² = 0
⇒ (x² – 4) (x² + 4) = 0
⇒ (x + 2) (x – 2) (x² + 4) = 0
⇒ (x + 2) = 0 or (x – 2) = 0 or (x² + 4) = 0
⇒ x = -2 (or) x = 2 (or) x² = -4
Therefore the zeroes of the polynomial are 2, – 2, we do not consider √-4 since it is not real.

Question 3.
Draw the graphs of the given polynomial and find the zeroes. Justify the answers,
i) p(x) = x² – x – 12
ii) p(x) = x² – 6x + 9
iii) p(x) = x² – 4x + 5
iv) p(x) = x² + 3x – 4
v) p(x) = x² – 1
Answer:
i) Given polynomial p(x) = x² – x – 12.
List of values of p(x):

Now, let’s locate the points listed above on a graph paper and draw the graph.

Result: We observe that the graph cuts the X – axis at (-3, 0) and (4, 0).
So, the zeroes of the polynomial are -3 and 4.
Justification:
Given p(x) = x² – x – 12 = 0
⇒ x² – 4x + 3x – 12 = 0
⇒ x(x – 4) + 3(x – 4) = 0
⇒ (x – 4) (x + 3) = 0
⇒ x – 4 = 0 and x + 3 = 0
x = 4 and x = – 3

ii) Given polynomial p(x) = x² – 6x + 9
List of values of p(x):

Now, let’s locate the points listed above on a graph paper and draw the graph.

Result: We observe that the graph cuts the X – axis at (3, 0).
So, the zeroes of the given polynomial are same i.e., 3.
Justification:
Given p(x) = x² – 6x + 9
⇒ x² – 3x – 3x + 9 = 0
⇒ x(x – 3) – 3(x – 3) = 0
⇒ (x – 3) (x – 3) = 0
⇒ x – 3 = 0 and x – 3 = 0
x = 3 and x = 3

iii) Given polynomial p(x) = x² – 4x + 5
List of values of p(x):

Now, let’s locate the points listed above on a graph paper and draw the graph.

Result: We observe that the graph does not cut the X – axis at any point.
So, the quadratic polynomial p(x) has no zeroes.
Justification: For the given p(x) = x² – 4x + 5 not possible to split in factors.

iv) Given polynomial p(x) = x² + 3x – 4.
List of values of p(x):

Now, let’s locate the points listed above on a graph paper and draw the graph.

Result: We observe that the graph cuts the X – axis at (-4, 0) and (1, 0).
So, the zeroes of the polynomial are -4 and 1.
Justification:
Given p(x) = x² + 3x – 4 = 0
⇒ x² + 4x – x – 4 = 0
⇒ x(x + 4)- 1(x + 4) = 0
⇒ (x + 4) (x – 1) = 0
⇒ x + 4 = 0 and x – 1 = 0
x = – 4 and x = 1

v) Given polynomial p(x) = x² – 1
List of values of p(x):

Now, let’s locate the points listed above on a graph paper and draw the graph.

Result: We observe that the graph cuts the X – axis at (-1, 0) and (1,0).
So, the zeroes of the polynomial are – 1 and 1.
Justification:
Given p(x) = x² – 1 = 0
⇒ p(x) = (x + 1) (x – 1) = 0 [∵ a² – b² = (a + b) (a – b)]
⇒ x + 1 = 0 and x – 1 = 0
x = -1 and x = 1

Question 4.
Why are \(\frac{1}{4}\) and -1 zeroes of the polynomial p(x) = 4x² + 3x – 1 ?
Answer:
Given polynomial p(x) = 4x² + 3x – 1
Given zeroes are \(\frac{1}{4}\) and -1
Let x = \(\frac{1}{4}\)

Let x = -1
⇒ p(-1) = 4(-1)² + 3(-1)-1 = 4 – 3 – 1 = 4 – 4 = 0
∴ P(\(\frac{1}{4}\)) = 0 and p(-1) = 0
So these values are zeroes of the polynomial p(x).

AP State Board Syllabus AP SSC 10th Class English Textbook Solutions Chapter 3C What is My Name? Textbook Questions and Answers.

AP State Syllabus SSC 10th Class English Solutions Chapter 3C What is My Name?

10th Class English Chapter 3C What is My Name? Textbook Questions and Answers

Comprehension

Answer the following questions.

Question 1.
What made Mrs. Murthy so restless to know her name?
Answer:
In her scrubbing zeal, Mrs. Murthy had forgotten her name. At once she felt that she had lost her own identity. She had lost her self-respect. That made her so restless to know her name.

Question 2.
How did Mrs. Murthy’s husband look upon her desire to know her name?
Answer:
Mrs. Murthy’s husband laughed at her when she asked him about her name. He did not take it seriously. He wanted her to be called by his name ‘Mrs. Murthy’. He did not give any importance to her feelings.

Question 3.
Do you notice any change in Mrs.Murthy in the first picture and Sarada in the second picture?
Answer:
Yes. I have found a lot of difference in her being Mrs. Murthy and her being Sarada. After she got her name back she got her confidence back. After she recollected her name she felt like a real person because she got her identity and self-respect back. When she was a housewife she was different, and after she has remembered her name she has become flamboyant.

Question 4.
Do you find any similarities between Mrs.Murthy and the women in your family? If yes, list them.
Answer:
Yes. There are so many similarities between Sarada and the members of our family. Not only in our family but almost in all families we can find a woman like Sarada because she is a representative of women of any class of Indian society who are always confined to domestic work like cleaning floors, washing clothes, cooking, and looking after her children, her husband, and other family members.

Question 5.
Why do you think the writer decided to focus on the question of married women’s identity?
Answer:
The writer decided to focus on the question of married women’s identity because she wanted them to live with their own identity and self-respect. She did not want them to confine to their homes. She wanted them to be given equal rights, equal respect. She wanted that they should also assume some responsibility in nation building activity.

Question 6.
Do you really think a woman can forget her name? What do you think is the intention of the author here?
Answer:
No. I don’t think any woman can forget her own name. The intention of the author is that the women should be respected and should not be confined to the domestic work. She feels that women should not lose their own identity.

Question 7.
Which part of the story shows that Mrs. Murthy feels her identity restored?
Answer:
In the last part of the story, when she returned to her husband’s house Mrs. Murthy feels her identity restored. It is clear from her words “…. from now onwards don’t call me yemoi, geemoi. My name is Sarada – call me Sarada, understood ?”

Writing

I. Translation

Read the following news item in Telugu and compare it with its translation in English given after that.

The following is the translated version of the above Telugu news item.
Centre’s Nod to Kasturi Rangan Committee Recommendations on Western Ghats

NEW DELHI :
The Ministry of Environment has accepted the report made by the Kasturi Rangan Committee on the conservation of Western Ghats. The committee, in its recommendations, made it clear that no further development activities be undertaken in the Western Ghats spread across the 60 thousand square kilometers in six states. The committee was appointed by the Union Government and headed by Kasturi Rangan to suggest measures to conserve the rarest ecosystem of the Western Ghat forests. The committee submitted its Report on 15th of April. The Ministry of Environ¬ment , after taking opinions of the six state governments and the people of the states, accepted the recommendations. The Western Ghats extend in Gujarat, Maharashtra, Goa, Karnataka, Kerala and Tamil Nadu states.

Let’s think of the following :

Question 1.
Do you think that translation is just translation of language ? Or does it also include translation of ideas?
Answer:
No. I don’t think that translation is just a translation of language. It includes linguistic, pragmatic and cultural elements. A literary translation must reflect the imaginative intellectual and intuitive writing of the author. Literary translation must reflect all the literary features of the source text such as sound effects, selection of words, figures of speech, etc.

Question 2.
Which translation is better, true translation or free translation?
Answer:
True translation is a dynamic equivalent translation which focuses on creating an equivalent effect in the target languagae.

Free translation is a formal equivalent translation in which the form and content of the originaUmessage is to be preserved. Of these two types of translations, true translation is better. Since true translation yields in the equivalent effect and it conveys the message of the original to the receptor audience and are equivalent to the original text in a dynamic way, true translation is better.

Question 3.
Do you find any change in the order of the sentence? For e.g : We have Subject, Verb,Object in English but the order is Subject, Object, Verb in Telugu.
Answer:
Not only the structure of the sentence but the diction and style and the order of arrangement of phrases also changes from Telugu to English.
Languages have different pragmatic linguistic structures and norms transferring the norms of one language may well lead to pragmatic failure.

Question 4.
Do you think sometimes it creates problems in the choice of vocabulary while attempting to translate a text?
Answer:
Translation should implicate accurate meaning. It may be problematic for translators. Wrong choice of words may cause ambiguity. While choosing the apt words for translating a text the translator should consider the situationality, intentionality and acceptability.

Question 5.
Is it possible to translate a poem from one language to the other?
Answer:
No. Poetry is not possible to translate because no poem means just one thing. It is very difficult to translate a poem into another language because we may not be aware of many of the possible meanings of the poem.

Question 6.
Is it necessary to take cultural aspects into consideration?
Answer:
Yes. It is necessary to take cultural aspects into consideration while translating. Because translation is a kind of activity which inevitably involves at least two languages and two cultural traditions. There may be cultural difference between the source text and the target text.

The lesson, “What Is My Name?” is a translated version in English from Telugu.The following is a part of the Telugu version of the lesson. Read the Telugu version and observe how it was translated into English.

Activities :

Question 1.
Is this a good translation? Yes or no? Give reasons.
Answer:
Yes. This is a good translation because this translation is a formal equivalent to the original text in the form and the content. It is a good translation because it reflected the imaginative intellectual and intuitive writing to the author. But we have found some difficulties in the structure of the sentences, in diction and style.

Question 2.
Now translate the Telugu version on this page into English and list the difficulties you face.
Answer:
A young woman, before she became a housewife had been an educated, cultured, intelligent, capable, and quick-witted with a sense of humour and elegance.

A young man, who liked her beauty and intelligence and was attracted by the dowry offered by her father, tied the three sacred knots around her neck and made her a housewife. After making her his housewife he told her, “Look Ammadu. This house is yours”. On hearing his words the housewife at once pulled the edge of her sari and tucked it in at the waist and swabbed the entire house and decorated the floor with rangoli designs. On seeing this that young man praised her promptly by saying “Ammadu ! You are dexterous in swabbing floors and even more adept in drawing the muggulu. Sabash Keep it up.” He said it in English, giving her a pat on the shoulder in appreciation.

Overjoyed with this, the housewife continued to live with swabbing as the chief mission of her life. Always she used to scrub the floor spotlessly and decorate it with beautiful multi-coloured rangoli designs. Thus her life went on with a sumptuous and ceaseless supply of swabbing cloths and muggu baskets.

But one day while scrubbing the floor, the housewife suddenly asked herself, “What is my name?”. The query shook her up. Leaving the mopping cloth and the muggu basket there itself, she stood near the window scratching her head, lost in thoughts.

“What is my name ? What is my name ?”. The house across the street carried a name-board, Mrs. M. Suhasini, M.A., Ph.D., Principal, ‘X’ College. Yes, she too had a name as her neighbour did. “How could I forget like that? In my scrubbing zeal I have forgotten my name — what shall I do now?” The housewife was perturbed. Her mind became totally restless. Somehow she finished her daubing for the day.

Meanwhile, the maidservant came there. Hoping that at least she would remember her name, the housewife asked her, “Look ammayi, do you know my name ?”

Question 3.
Translate the following extract from the story into Telugu and compare it with the original story in Telugu. (Refer to teacher’s handbook for Telugu version.)
‘Sarada! My dear Sarada!’ she shouted and embraced her. The housewife felt like a person — totally parched and dried up, about to die of thirst — getting a drink of cool water from the new earthen kooja poured into her mouth with a spoon and given thus a new life.The friend did indeed give her a new life — ‘You are Sarada. You came first in our school in the tenth class. You came first in the music competition conducted by the college. You used to paint good pictures too. We were ten friends altogether — I meet all of them some time or other. We write letters to each other. Only you have gone out of our reach! Tell me why you are living incognito?’ her friend confronted her.
Answer:

Project Work

I. Influence of technical gadgets on human relations.
Visit five houses in your neighbourhood and collect the information in the given format related to human relations, i.e. spending quality time with the members of the family and friends, sharing and caring. Analyse the information and write a report by adding your opinion on how the modern gadgets are influencing human relations and present it before the class.

Family-1 (House-1) (Raja Rao’s Family)

Family-2 (House-2) (Rama Rao’s Family)

Family-3 (House-3) (Venkat’s Family)

Family-4 (House-4) (Nageswara Rao’s Family)

Family-5 (House-5) (Bhaskar’s Family)

We all know very well that electronic gadgets such as TVs, Mobile phones, Computers occupy a major place in our day-to-day lives. Though these gadgets have their own advantages, they also have a negative influence on the human relationships. They play a vital role in our lives today. Most of us are addicted to them. There are hundreds of channels which are viewed on TV by us. We start watching TV programmes from a very early time in the morning till midnight. Thus we don’t have any time to talk to our near and dear. In the same way, mobile phones too have advantages as well as disadvantages. Most of the children and youth are spending all their time in using their mobile phones. They talk, watch movies, listen to music, play games on their mobile phones. When they engage in using their mobile phones, how can they find time to spend with their family members? Today, computers have become indispensable to each one of us.

In most of the families, we find a complete different situation in managing human relationships before and after the accessibility of the electronic gadgets. Before the accessibility of these modern gadgets, people share most of their time to spend with their family members. They take care of all the family members. They try to share their feelings with their famiy members. They often visit their friends. Their relationships with their family members and friends are very cordial. They find a lot of time to do all the things leisurely. Thus family relations are hectic. They show love and affection for their family members. They find time to play with their friends. They help their family members in all the matters.

After the accessibility of the modern gadgets, they try to spend all their time in using them only. They spend with them hours together. They don’t find time to spend with their family members. They don’t visit their friends very often. They don’t care for their family members. They give importance to the gadgets only. They become mechanical. They don’t have any affection for their family members. They don’t understand the warmth of relationship. They don’t know how much they are missing the joyous family interaction. They should understand that no artificial media can substitute their family’s warmth and interaction. They should give importance to human relationships.

II. Nowadays, we can easily find children even as young as two years old playing with electronic devices and gadgets anywhere. It is not only the video games that make children stay, it also includes television, mobile phones, computers, tablet computers, PSP (Play Station Portable) games, etc. Parents may find it easier to make their children stay in one place by giving them a gadget to play with.

Work in groups and discuss the following:

Ways of managing children’s electronic devices consumption and preventing
Answer:
Group 1 :
The parents should make their children know the bad effects of spending more time with electronic devices.

Group 2:
The parents should monitor their children’s media consumption – television, mobile phones, computers, tablet computers, PSP games, etc.

Group 3 :
The parents should make their children aware about the advantages of playing games, doing exercises and yoga instead of their spending more time with electronic devices.

Group 4 :
The parents should stop themselves from using the electronic gadgets for longer times. Thus they can set an example for their children.

Group 5 :
The parents should make their children aware about the health problems that would arise with spending longer times with electronic gadgets.

Group 6 :
The parents should share thier feelings with their children. They should discuss with them the ill effects of the games they play. They should try to move closer to their children.

Sum up:
Today most of the children are addicted to the modern gadgets such as television, mobile phones, computers, tablet computers, PSP games, etc. These all become an integral part of children’s lives. Today children are heavily exposed to media. Parents may find it easier to make their children stay in one place by giving them a gadget to play with. The parents should watch carefully what their children are doing with the gadgets and how they are using it. They should prevent their children’s addiction to games. They should find ways to manage their children’s electronic device consumption.

They should make their children know the bad effects of spending more time with electronic devices. They should monitor their children’s media consumption. They should make them play games and do exercise and yoga. The parents should stop themselves from using the electronic gadgets and stand as an example to their children. They should share their feelings with their children. They should try to move closer to their children.

Thus, parents can manage their children’s electronic device consumption and prevent their addiction to games.

What is My Name? Summary in English

Sarada, before she got married, was a well-educated and cultured young woman. She was intelligent, capable, quick-witted and she had a sense of humour and elegance. She used to stand first in her class. She was good at music and dance. She used to paint good pictures.

Falling to her beauty, and intelligence and attracted by the dowry her father offered, a young man married her. Later he showed his house and told her that it was her house. Immediately she began to swab the floors and decorated the floor with rangoli designs. On seeing this, her husband praised that she was dexterous at swabbing the floor. Overjoyed by his applause, Sarada began living with swabbing as a mission of her life. Thus, her life went on scrubbing the house spotlessly and decorating the house with multi-coloured designs. In her scrubbing zeal she had forgotten her name. One day she tried to recollect what her name was. But she could not. She became restless. She asked her maid servant, her neighbours, her husband, and her children about her name. But they all told the name by which they used to call her by using their relation. Her husband laughed and did not take it seriously.

Finally the housewife decided to go to her parents’ house and look for her name in her certificates. But her certificates were kept on the attic. Meanwhile she met her classmate. She called the housewife by her name ‘Sarada’. At once Sarada felt like a person. Because our name gives us our personal identity and self-respect. Our name is our own- unique to us.

Here the author wants to tell that every woman has her own responsibility in nation-building. Women should be given equal rights with men. Women should not be confined to the four walls of the house. She should be let free. She can reach to the heights of sky. She can ascend to the pinnacles of any success and thus she can make any nation greater and stronger.

What is My Name? Glossary

quick-witted (adj) : intelligent; able to think quickly

elegance (n) : a satisfying or admirable neatness; ingenious simplicity or precision in something

dowry (n) : money and property paid by a bride’s family to the bridegroom at the time of marriage

swab (v) : clean

dexterous (adj) : skilful

appreciation (n) : recognizing and enjoying the good qualities of somebody or something

sumptuous (adj) : luxurious, splendid

cease (v) : stop

ceaseless (adj) : continuous

zeal (n) : great energy and enthusiasm

mopping (v) : cleaning/washing

perturb (v) : bother/disturb/trouble

daubing (n) : the act of spreading a substance such as mud thickly

take somebody aback : to shock or surprise somebody very much

immerse (v) : absorb oneself in something

urge (v) : to try hard to persuade

giggling (v) : laughing nervously

anguish (n) : severe pain, unhappiness

frantically (adv) : worriedly/anxiously

chore (n) : a task that you do regularly

attic (n) : a room or space just below the roof of a house often used for storing things ; loft

wail (v) : to make a long loud cry

maternity home (n) : hospital for deliveries

parch (v) : dehydrate

incognito (adv) : having a concealed identity

fish (v) : search

AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 4 Pair of Linear Equations in Two Variables Ex 4.3 Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Maths Solutions 4th Lesson Pair of Linear Equations in Two Variables Exercise 4.3

10th Class Maths 4th Lesson Pair of Linear Equations in Two Variables Ex 4.3 Textbook Questions and Answers

Question 1.
Solve each of the following pairs of equations by reducing them to a pair of linear equations.
i) \(\frac{5}{x-1}\) + \(\frac{1}{y-2}\) = 2
\(\frac{6}{x-1}\) + \(\frac{3}{y-2}\) = 1
Answer:
Given
\(\frac{5}{x-1}\) + \(\frac{1}{y-2}\) = 2
\(\frac{6}{x-1}\) + \(\frac{3}{y-2}\) = 1
Put \(\frac{1}{x-1}\) = a and \(\frac{1}{y-2}\) = b,
then the given equations reduce to
5a + b = 2 ……… (1)
6a – 3b = 1 ………. (2)

⇒ b = \(\frac{7}{21}\) = \(\frac{1}{3}\)
Substituting b = \(\frac{1}{3}\) in equation (1) we get

⇒ (x – 1) . 1 = 3 × 1
⇒ x – 1 = 3
⇒ x = 3 + 1 = 4
b = \(\frac{1}{y-2}\) ⇒ \(\frac{1}{3}\) = \(\frac{1}{y-2}\)
⇒ (y – 2) . 1 = 3 × 1
⇒ y – 2 = 3
⇒ y = 3 + 2 = 5
∴ Solution (x, y) = (4, 5)

ii) \(\frac{x+y}{xy}\) = 2;
\(\frac{x-y}{xy}\) = 6
Answer:
Given
\(\frac{x+y}{xy}\) = 2
⇒ \(\frac{x}{xy}\) + \(\frac{y}{xy}\) = 2
⇒ \(\frac{1}{y}\) + \(\frac{1}{x}\) = 2
\(\frac{x-y}{xy}\) = 6
⇒ \(\frac{x}{xy}\) – \(\frac{y}{xy}\) = 6
⇒ \(\frac{1}{y}\) – \(\frac{1}{x}\) = 6
Take \(\frac{1}{x}\) = a and \(\frac{1}{y}\) = b,
then the given equations reduces to

⇒ b = \(\frac{8}{2}\) = 4
Substituting b = 4 in equation (1) we get
a + 4 = 2 ⇒ a = 2 – 4 = -2
but a = \(\frac{1}{x}\) = -2 ⇒ x = \(\frac{-1}{2}\)
b = \(\frac{1}{y}\) = 4 ⇒ y = \(\frac{1}{4}\)
∴ Solution (x, y) = \(\left(\frac{-1}{2}, \frac{1}{4}\right)\)

iii) \(\frac{2}{\sqrt{x}}\) + \(\frac{3}{\sqrt{y}}\) = 2;
\(\frac{4}{\sqrt{x}}\) – \(\frac{9}{\sqrt{y}}\) = -1
Answer:
Given
\(\frac{2}{\sqrt{x}}\) + \(\frac{3}{\sqrt{y}}\) = 2 and \(\frac{4}{\sqrt{x}}\) – \(\frac{9}{\sqrt{y}}\) = -1
Take \(\frac{1}{\sqrt{x}}\) = a and \(\frac{1}{\sqrt{y}}\) = b,
then the given equations reduces to
2a + 3b = 2 …….. (1)
4a – 9b = – 1 …….. (2)

⇒ b = \(\frac{5}{15}\) = \(\frac{1}{3}\)
Substituting b = \(\frac{1}{3}\) in equation (1) we get
2a + 3\(\left(\frac{1}{3}\right)\) = 2
⇒ 2a + 1 = 2 ⇒ 2a = 2 – 1 ⇒ a = \(\frac{1}{2}\)

∴ Solution (x, y) = (4, 9)

iv) 6x + 3y = 6xy
2x + 4y = 5xy
Answer:
Given
6x + 3y = 6xy
⇒ \(\frac{6x+3y}{xy}\) = 6
⇒ \(\frac{6x}{xy}\) + \(\frac{3y}{xy}\) = 6
⇒ \(\frac{6}{y}\) + \(\frac{3}{x}\) = 6
2x + 4y = 5xy
⇒ \(\frac{2x+4y}{xy}\) = 5
⇒ \(\frac{2x}{xy}\) + \(\frac{4y}{xy}\) = 6
⇒ \(\frac{2}{y}\) + \(\frac{4}{x}\) = 6
Take \(\frac{1}{x}\) = a and \(\frac{1}{y}\) = b,
then the given equations reduces to
3a + 6b = 6 ……. (1)
4a + 2b = 5 ……. (2)

⇒ b = \(\frac{9}{18}\) = \(\frac{1}{2}\)
Substituting b = \(\frac{1}{2}\) in equation (1) we get
3a +6\(\left(\frac{1}{2}\right)\) = 6
⇒ 3a = 6 – 3
⇒ a = \(\frac{3}{3}\) = 1
but a = \(\frac{1}{x}\) = 1 ⇒ x = 1
b = \(\frac{1}{y}\) = \(\frac{1}{2}\) ⇒ y = 2
∴ Solution (x, y) = (1, 2)

v) \(\frac{5}{x+y}\) – \(\frac{2}{x-y}\) = -1
\(\frac{15}{x+y}\) + \(\frac{7}{x-y}\) = 10
where x ≠ 0, y ≠ 0
Answer:
Given
\(\frac{5}{x+y}\) – \(\frac{2}{x-y}\) = -1 and
\(\frac{15}{x+y}\) + \(\frac{7}{x-y}\) = 10
Take \(\frac{1}{x+y}\) = a and \(\frac{1}{x-y}\) = b, then
the given equations reduce to
5a – 2b = – 1 ……… (1)
15a + 7b = 10 ……… (2)

⇒ b = \(\frac{-13}{-13}\) = 1
Substituting b = 1 in equation (1) we get
5a – 2(1) = -1
⇒ 5a = -1 + 2
⇒ 5a = 1
⇒ a = \(\frac{1}{5}\)
but a = \(\frac{1}{x+y}\) = \(\frac{1}{5}\) ⇒ x + y = 5
b = \(\frac{1}{x-y}\) = 1 ⇒ x – y = 1
⇒ x = \(\frac{6}{2}\) = 3
Solving the above equations
Substituting x = 3 in x + y = 5 we get
3 + y = 5 ⇒ y = 5 – 3 = 2
∴ Solution (x, y) = (3, 2)

vi) \(\frac{2}{x}\) + \(\frac{3}{y}\) = 13
\(\frac{5}{x}\) – \(\frac{4}{y}\) = -2
where x ≠ 0, y ≠ 0
Answer:
Given
\(\frac{2}{x}\) + \(\frac{3}{y}\) = 13 and
\(\frac{5}{x}\) – \(\frac{4}{y}\) = -2
Take \(\frac{1}{x}\) = a and \(\frac{1}{y}\) = b, then
the given equations reduce to
2a + 3b = 13 ……… (1)
5a – 4b = -2 ……… (2)

⇒ b = \(\frac{69}{23}\) = 3
Substituting b = 3 in equation (1) we get
2a + 3 (3) = 13
⇒ 2a = 13 – 9
⇒ a = \(\frac{4}{2}\) = 2
but a = \(\frac{1}{x}\) = 2 ⇒ x = \(\frac{1}{2}\)
b = \(\frac{1}{y}\) = 3 ⇒ y = \(\frac{1}{3}\)
∴ Solution (x, y) = (\(\frac{1}{2}\), \(\frac{1}{3}\))

vii) \(\frac{10}{x+y}\) + \(\frac{2}{x-y}\) = 4
\(\frac{15}{x+y}\) – \(\frac{5}{x-y}\) = -2
Answer:
Given
\(\frac{10}{x+y}\) + \(\frac{2}{x-y}\) = 4 and
\(\frac{15}{x+y}\) – \(\frac{5}{x-y}\) = -2
Take \(\frac{1}{x+y}\) = a and \(\frac{1}{x-y}\) = b, then
the given equations reduce to
10a + 2b = 4 ……… (1)
15a – 5b = – 2 ……… (2)

⇒ b = \(\frac{16}{16}\) = 1
Substituting b = 1 in equation (1) we get
10a + 2(1) = 4
⇒ 10a = 4 – 2
⇒ a = \(\frac{2}{10}\) = \(\frac{1}{5}\)
but a = \(\frac{1}{x+y}\) = \(\frac{1}{5}\) ⇒ x + y = 5 ……. (3)
b = \(\frac{1}{x-y}\) = 1 ⇒ x – y = 1 …….. (4)
Adding (3) and (4)

⇒ x = \(\frac{6}{2}\) = 3
Substituting x = 3 in x + y = 5 we get
3 + y = 5 ⇒ y = 5 – 3 = 2
∴ Solution (x, y) = (3, 2)

viii) \(\frac{1}{3x+y}\) + \(\frac{1}{3x-y}\) = \(\frac{3}{4}\)
\(\frac{1}{2(3x+y)}\) – \(\frac{1}{2(3x-y)}\) = \(\frac{-1}{8}\)
Answer:
Given
\(\frac{1}{3x+y}\) + \(\frac{1}{3x-y}\) = \(\frac{3}{4}\) and
\(\frac{1}{2(3x+y)}\) – \(\frac{1}{2(3x-y)}\) = \(\frac{-1}{8}\)
Take \(\frac{1}{3x+y}\) = a and \(\frac{1}{3x-y}\) = b, then
the given equations reduce to

⇒ a = \(\frac{2}{8}\) = \(\frac{1}{4}\)
Substituting a = \(\frac{1}{4}\) in equation (1) we get

Solving (3) and (4)

⇒ x = \(\frac{6}{6}\) = 1
Substituting x = 1 in 3x + y = 4
⇒ 3(1) + y = 4
⇒ y = 4 – 3 = 1
∴ The solution (x, y) = (1, 1)

Question 2.
Formulate the following problems as a pair of equations and then find their solutions.
i) A boat goes 30 km upstream and 44 km downstream in 10 hours. In 13 hours it can go 40 km upstream and 55 km downstream. Determine the speed of the stream and that of the boat in still water.
Answer:
Let the speed of the boat in still water = x kmph
and the speed of the stream = y kmph
then speed in downstream = x + y
Speed in upstream = x – y
and time = \(\frac{\text { distance }}{\text { speed }}\)
By problem,
\(\frac{30}{x-y}\) + \(\frac{44}{x+y}\) = 10
\(\frac{40}{x-y}\) + \(\frac{55}{x+y}\) = 13
Take \(\frac{1}{x-y}\) = a and \(\frac{1}{x+y}\) = b, then
the given equations reduce to
30a + 44b = 10 ……… (1)
40a + 55b = 13 ……… (2)

⇒ b = \(\frac{1}{11}\)
Substituting b = \(\frac{1}{11}\) in equation (1) we get

⇒ x = 8
Substituting x = 8 in x – y = 5 we get
8 – y = 5
⇒ y = 8 – 5 = 3
∴ The solution (x, y) = (8, 3)
Speed of the boat in still water = 8 kmph
Speed of the stream = 3 kmph.

ii) Rahim travels 600 km to his home partly by train and partly by car. He takes 8 hours if he travels 120 km by train and rest by car. He takes 20 minutes more if he travels 200 km by train and rest by car. Find the speed of the train and the car.
Answer:
Let the speed of the train be x kmph
and the speed of the car = y kmph
By problem,

Take \(\frac{1}{x}\) = a and \(\frac{1}{y}\) = b, then
the given equations reduce to
15a + 60b = 1 ……… (1)
8a + 16b = \(\frac{1}{3}\) ⇒ 24a + 48b = 1 ……… (2)

⇒ a = \(\frac{-1}{-60}\) = \(\frac{1}{60}\)
Substituting a = \(\frac{1}{60}\) in equation (1) we get

but a = \(\frac{1}{x}\) = \(\frac{1}{60}\) ⇒ x = 60 kmph
b = \(\frac{1}{y}\) = \(\frac{1}{80}\) ⇒ y = 80 kmph
Speed of the train = 60 kmph and
speed of the car = 80 kmph

iii) 2 women and 5 men can together finish an embroidery work in 4 days while 3 women and 6 men can finish it in 3 days. Find the time taken by 1 woman alone and 1 man alone to finish the work.
Answer:
Let the time taken by 1 woman to complete the work = x days
and time taken by 1 man to complete the work = y days
∴ Work done by 1 woman in 1 day = \(\frac{1}{x}\)
Work done by 1 man in 1 day = \(\frac{1}{y}\)
By problem,

Take \(\frac{1}{x}\) = a and \(\frac{1}{y}\) = b,
then the above equations reduce to
2a + 5b = \(\frac{1}{4}\) and 3a + 6b = \(\frac{1}{3}\)
⇒ 8a + 20b = 1 …….. (1) and
9a + 18b = 1 ……… (2)

⇒ b = \(\frac{1}{36}\)
Substituting b = \(\frac{1}{36}\) in equation (1) we get

but a = \(\frac{1}{x}\) = \(\frac{1}{18}\) ⇒ x = 18 and
b = \(\frac{1}{y}\) = \(\frac{1}{36}\) ⇒ y = 36
∴ Time taken by 1 woman = 18 days
1 man = 36 days

AP State Board Syllabus AP SSC 10th Class English Textbook Solutions Chapter 3B Once Upon a Time Textbook Questions and Answers.

AP State Syllabus SSC 10th Class English Solutions Chapter 3B Once Upon a Time

10th Class English Chapter 3B Once Upon a Time Textbook Questions and Answers

Comprehension

I. Tick (✓) the option that will complete each of the following statements. In some cases more than one option may be possible.

1. In the first five stanzas the poet is talking about
a) the honest and innocent world of children.
b) the insincere world of adults.
c) the difference between the past and the present.
d) the old and the young.
Answer:
(b) ✓
(c) ✓

2. The last four lines of the poem suggest
a) hope.
b) regret.
c) a sense of loss.
d) eagerness to learn.
Answer:
(a) ✓
(b) ✓

3. The expression ice-cold-block eyes’ means
a) The eyes are wet with tears.
b) expressionless eyes,
c) a state of lack of feelings.
d) a dead man’s eyes.
Answer:
(b) ✓
(c) ✓

4. ‘They’ in line 4 of stanza 1 refers to
a) people in the past.
b) present day people.
c) all adults.
d) young children.
Answer:
(c) ✓

5. ‘Their …………. eyes search behind my shadow’ means
a) they avoid meeting his eyes.
b) they try to look at the darker side of the person.
c) they convey no emotions.
d) they try to see what is not there.
Answer:
(b) ✓

6. The poet has learnt
a) to shake hands.
b) the ways of the world.
c) to laugh.
d) to put on masks.
Answer:
(d) ✓

7. The poet wants to learn from his son because his son
a) is not corrupted by the ways of the world.
b) is more informed.
c) knows about good manners more than his father.
d) is more caring.
Answer:
(a) ✓

II. Answer the following questions in a sentence or two each.

Question 1.
When did people shake hands with their hearts?
Answer:
The people, in their childhood, when they didn’t know the falsehood and hypocrisies of the world, when they were not corrupted by the ways of the world, shook hands with their hearts.

Question 2.
What is the poet crying over? What help does he want from his son?
Answer:
The poet regrets for losing the traits of his own character such as honesty, modesty, and sincerity. He laments over getting corrupted by the ways of the world. He regrets for his character being influenced by hypocrisy and fallacies of the world. The poet requests his son to help him regain his sincere and heartful, innocent and child-like smile.

Question 3.
“Most of all, I want to relearn
how to laugh, for my laugh in the mirror
shows only my teeth like a snake’s bare fangs !”
What does the poet mean by these lines?
Answer:
The poet feels his smile as fictitious, insincere, and hypocritic. He feels such a smile is dangerous. The comparison of his teeth to snake’s fangs makes false, mask-like smile seem dangerous.

Question 4.
What is the tone of the poem?
Answer:
The tone of the poem is roughly equivalent to the mood it creates in the reader. In Once Upon a Time’ the tone of the poem in the earlier stanzas is abashed, regretful but in the last stanza the poet ends the poem in an optimistic and hopeful tone. Thus the poet begins the poem in a negative tone i.e. somber but ends positively i.e. opti¬mism.

Question 5.
“Now they shake hands without hearts :
while their left hands search
my empty pockets.”
Why do the left hands search empty pockets now? What does this indicate?
Answer:
The poet expresses his concern for the influence of the western world on age-old African custom. He feels that the once enthusiastic and friendly society of Africa now treated its own people like strangers and looked at each other with suspicion and hostility. The white imperialists always exploited and plundered the wealth of their colonies. So their left hands search the empty pockets of their subjects in an endeavour to rob them further.

Question 6.
The poet uses certain words to express frustration and sorrow. Identify these words.
Answer:
The phrases “ice-block-cold eyes”, “shake hands without hearts”, “doors shut on me”, “learned to wear many faces”, “teeth like a snake’s bare fangs” are used to express the poet’s regret. The phrases or lines such as “…believe me, son. I want to be what 1 used to be”, “unlearn these muting things”, “want to relearn how to laugh” are the lines used to express his frustration.

Once Upon a Time Summary in English

Once upon a time, the people used to laugh with their hearts. There used to be sincerity in their laugh. Their laugh came from their hearts. There was genuinety in their actions and feelings. But people laugh superficially, in present. Their laugh is ficticious, feelingless. The eyes are dead like feelingless, and unsympathetic/apathetical. Even people shake hands mechanically and wish the people artificially but not heartfully.

In the third stanza the poet explains more about the changes the man possesses as he grows in age. He has noticed falsehood, superfluous feelings and deteriorating human relations in present day society. The poet also says that the people lie when they say the positive phrases like “Feel at home” and “Come again.” When the poet visits their house for the third time thinking that their words are genuine, the doors are shut on his face. In this material and artificial world the poet has learnt many things especially wearing many faces, like putting on many dresses. That means he changes his expressions and feelings to suit the situations and needs of the people with whom he is to deal with.

In behaving like that he loses his own character and traits of his self. As this is the way of the world the poet has also learnt to laugh with teeth but not with heart. He also has learnt to shake the hands of others but not with heart. He has learnt to say ‘Goodnight’ when he means Good riddance’. He has learnt to say Glad to meet you,’ when he is not glad and he says, ‘It’s nice talking to you’ when he is bored of talking.

But the writer is fed up with the forcible hypocrisy and pretension of falsehood. He wants to regain his real spirit and character. He wants to abandon all this falsehood. He wants to laugh sincerely as the children do. His laugh reveals all the fallacies of the world. When he looks at himself in the mirror his teeth are exposed and they appear like the fangs of a snake.

In the last stanza the poet appeals to his son to show him how to smile whole-heartedly. The poet’s desire to regain his original traits of his character, sincerity and to give up his falsehood and hypocrisy reveals his yearning for the innocence, faithfulness and sincerity.

Once Upon a Time Glossary

cock-tail (n) : a drink usually made from a mixture of one or more alcoholic drinks

conform (v) : to be and thinking the same way as most other people in a group or society; normally acceptable

portrait (n) : a painting, drawing or photograph of a person especially of the head and shoulders

good-riddance (n) : a feeling of relief when an unwanted person leaves

muting (adj) : changing all the time; expressionless/not expressed in speech

fangs (n) : long, sharp teeth of some animals like snakes and dogs

AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 4 Pair of Linear Equations in Two Variables Ex 4.2 Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Maths Solutions 4th Lesson Pair of Linear Equations in Two Variables Exercise 4.2

10th Class Maths 4th Lesson Pair of Linear Equations in Two Variables Ex 4.2 Textbook Questions and Answers

Form a pair of linear equations for each of the following problems and find their solution.
Question 1.
The ratio of incomes of two persons is 9 : 7 and the ratio of their expenditures is 4 : 3. If each of them manages to save Rs. 2000 per month, find their monthly income.
Answer:
Given ratio of incomes of two persons = 9 : 7
So let the incomes of each = Rs. 9x and Rs. 7x
and ratio of expenditures = 4 : 3
So let the expenditures of each = 4y and 3y
then earnings of each = (income – expenditure) of each
⇒ 9x – 4y = Rs. 2000 and 7x – 3y = 200
∴ 9x – 4y = 7x – 3y = 2000
⇒ 9x – 7x = 4y – 3y
⇒ y = 2x
now putting y = 2x in 9x – 4y = 2000 we get
9x – 4(2x) = 2000 ⇒ x = 2000
∴ Income of each = 9x = 9(2000) = 18000
and 7x = 7(2000) = 14,000

Question 2.
The sum of a two digit number and the number obtained by reversing the digits is 66. If the digits of the number differ by 2, find the number. How many such numbers are there?
Answer:
Let the digit in units place be x
and the digit in tens place be y
then the value of the number = 10y + x
Number obtained by reversing the digits = 10x + y
By problem,
(10y + x) + (10x + y) = 66
and x – y = 2
⇒ 11x – 11y = 66 and x – y = 2
⇒ x + y = 6 and x – y = 2
Solving these two equations
x + y = 6
x – y = 2
(+) 2x = 8
x = \(\frac{8}{2}\) = 4
Substituting x = 4 in x + y = 6
we get 4 + y = 6 ⇒ y = 2
Substituting x, y values in equations (10y + x) & (10x + y),
We get 10y + x
= 10(2) + 4 = 20 + 4 = 24
and 10x + y = 10(4) + 2
= 40 + 2 = 42
∴ The number is 42 or 24
Thus we have two such numbers.

Question 3.
The larger of two supplementary angles exceeds the smaller by 18°. Find the angles.
Answer:
Let the pair of supplementary angles be x and y [and x > y]
then we have x + y = 180° …… (1)
By problem, x = y + 18°
⇒ x – y = 18° …… (2)
Solving the equations (1) and (2) we get

and x = \(\frac{198}{2}\) = 99°
Substituting x = 99° in equation (2) we get
99° – y° = 18°
⇒ y° = 99° – 18 = 81°
∴ The angles are 99° and 81°.

Question 4.
The taxi charges in Hyderabad are fixed, along with the charge for the distance covered. For a distance of 10 km., the charge paid is Rs. 220. For a journey of 15 km. the charge paid is Rs. 310.
i) What are the fixed charges and charge per km?
ii) How much does a person have to pay for travelling a distance of 25km?
Answer:
Let the fixed charge be = Rs. x.
and the charge per one km = Rs. y.
By problem, x + 10y = 220 x + 15y = 310
Solving (1) and (2) we get

∴ y = \(\frac{-90}{-5}\) = 18
i.e., charge per one km = Rs. 18
Substituting y = 18 in equation (1) we get
x + 10 × 18 = 220
⇒ x = 220 – 180
⇒ x = Rs. 40
∴ Fixed charge = Rs. 40;
Charge per km = Rs. 18.

ii) Now, the charge for travelling a distance of 25 km = 25 × 18
= Rs. 450 + 40
= Rs. 490

Question 5.
A fraction becomes equal to \(\frac{4}{5}\) if 1 is added to both numerator and denominator. If, however, 5 is subtracted from both numerator and denominator, the fraction becomes equal to \(\frac{1}{2}\). What is the fraction?
Answer:
Let the numerator of the fraction = x
and the denominator of the fraction = y
By problem,
\(\frac{x+1}{y+1}\) = \(\frac{4}{5}\) and \(\frac{x-5}{y-5}\) = \(\frac{1}{2}\)
⇒ 5(x + 1) = 4(y + 1) and 2(x – 5) = 1(y – 5)
5x + 5 = 4y + 4 and 2x – 10 = y – 5
⇒ 5x – 4y = 4 – 5 and 2x – y = – 5 + 10
⇒ 5x – 4y = – 1 …… (1)
and 2x – y = 5 …… (2)

∴ y = \(\frac{-27}{-3}\) = 9
Substituting y = 9 in equation (2) we get
2x – 9 = 5
⇒ 2x = 5 + 9
⇒ 2x = 14 and
x = \(\frac{14}{2}\) = 7
Thus the fraction is \(\frac{x}{y}\) = \(\frac{7}{9}\)

Question 6.
Places A and B are 100 km apart on a highway One car starts from A and another from B at the same time at different speeds. If the cars travel in the same direction, they meet in 5 hours. If they travel towards each other, they meet in 1 hour. What are the speeds of the two cars?
Answer:

Let the speed of the car which started from the place A = x kmph
and B = y kmph
Distance travelled by first car in 5h = 5x and in 1h = x
The distance covered by second car in 5h = 5y and in 1h = y
By problem when travelled in same direction,
5x – 5y = 100 ⇒ x – y = 20 …… (1)
and when travelled towards each other
x + y = 100 ……. (2)
Solving (1) and (2),

∴ x = \(\frac{120}{2}\) = 60
Substituting x = 60 in equation (1) we get
60 – y = 20
⇒ y = 60 – 20 = 40 kmph
Thus the speed of the cars are 60 kmph and 40 kmph.

Question 7.
Two angles are complementary. The larger angle is 3° less than twice the measure of the smaller angle. Find the measure of each angle.
Answer:
Let the pair of complementary angles be x° and y° with x° > y°
then x° + y° = 90° and
By problem
x = 2y – 3° ⇒ x – 2y = – 3°
Solving these two equations we get,

∴ y = \(\frac{93}{3}\) = 31°
Substituting y = 31°in x + y = 90° we get
x + 31° = 90°
⇒ x = 90° – 31° = 59°
The angles are 59° and 31°.

Question 8.
An algebra textbook has a total of 1382 pages. It is broken up into two parts. The second part of the book has 64 pages more than the first part. How many pages are in each part of the book?
Answer:
Let the first part of the book contains x pages
and the second part of the book contains y pages By problem,
x + y = 1382 ….. (1)
y = x + 64 ⇒ x – y = -64 …… (2)
Solving equations (1) and (2) we get

∴ x = \(\frac{1318}{2}\) = 659
Substituting x = 659 in equation (1) we get
659 + y = 1382
⇒ y = 1382 – 659 = 723
∴ The number of pages in the first part = 659
Second part = 723

Question 9.
A chemist has two solutions of hydrochloric acid in stock. One is 50% solution and the other is 80% solution. How much of each should be used to obtain 100 ml of a 68% solution?
Answer:
Let the first solution contains 50% acid.
Second solution contains 80% acid.
Let x ml of 1st solution and y ml of second solution are added.
Then x + y = 100
Acid content in the ‘mix’ is 50% of x + 80% of y = 68%

∴ y = \(\frac{180}{3}\) = 60
Substituting y = 60 in equation (1) we get
x + 60 = 100
⇒ x = 100 – 60 = 40
∴ Quantity of first solution = 40 ml
Quantity of second solution = 60 ml

Question 10.
Suppose you have Rs. 12000 to invest. You have to invest some amount at 10% and the rest at 15%. How much should be invested at each rate to yield 12% on the total amount invested ?
Answer:
Let the amount to be invested @ 10% be Rs. x
and the amount to be invested @ 15% be Rs. y
By problem x + y = 12000 ……. (1)
Also 10% of x + 15% of y = 12% of 12000

⇒ y = \(\frac{-24000}{-5}\) = Rs. 4800
Substituting y = 4800 in equation (1) we get
x + 4800 = 12000
⇒ x = 12000 – 4800 = 7200
The invested @ 10% = Rs. 7200
@ 15% = Rs. 4800

AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 4 Pair of Linear Equations in Two Variables Ex 4.1 Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Maths Solutions 4th Lesson Pair of Linear Equations in Two Variables Exercise 4.1

10th Class Maths 4th Lesson Pair of Linear Equations in Two Variables Ex 4.1 Textbook Questions and Answers

Question 1.
By comparing the ratios \(\frac{a_{1}}{a_{2}}\), \(\frac{b_{1}}{b_{2}}\), \(\frac{c_{1}}{c_{2}}\) K find out whether the lines represented by the following pairs of linear equations intersect at a point, are parallel or are coincident.
a) 5x – 4y + 8 = 0
7x + 6y – 9 = 0
Answer:
Given: 5x – 4y + 8 = 0
7x + 6y – 9 = 0
\(\frac{a_{1}}{a_{2}}\) = \(\frac{5}{7}\); \(\frac{b_{1}}{b_{2}}\) = \(\frac{-4}{6}\); \(\frac{c_{1}}{c_{2}}\) = \(\frac{8}{-9}\)
∴ \(\frac{a_{1}}{a_{2}}\) ≠ \(\frac{b_{1}}{b_{2}}\)
Hence the given pair of linear equations represents a pair of intersecting lines.

b) 9x + 3y + 12 = 0
18x + 6y + 24 = 0
Answer:
Given : 9x + 3y + 12 = 0
18x + 6y + 24= 0
\(\frac{a_{1}}{a_{2}}\) = \(\frac{9}{18}\) = \(\frac{1}{2}\);
\(\frac{b_{1}}{b_{2}}\) = \(\frac{3}{6}\) = \(\frac{1}{2}\);
\(\frac{c_{1}}{c_{2}}\) = \(\frac{12}{24}\) = \(\frac{1}{2}\)
∴ \(\frac{a_{1}}{a_{2}}\) = \(\frac{b_{1}}{b_{2}}\) = \(\frac{c_{1}}{c_{2}}\)
The lines are coincident.

c) 6x – 3y + 10 = 0
2x – y + 9 = 0
Answer:
Given: 6x – 3y + 10 = 0
2x – y + 9 = 0
\(\frac{a_{1}}{a_{2}}\) = \(\frac{6}{2}\) = \(\frac{3}{1}\);
\(\frac{b_{1}}{b_{2}}\) = \(\frac{-3}{-1}\) = \(\frac{3}{1}\);
\(\frac{c_{1}}{c_{2}}\) = \(\frac{10}{9}\)
Here \(\frac{a_{1}}{a_{2}}\) = \(\frac{b_{1}}{b_{2}}\) ≠ \(\frac{c_{1}}{c_{2}}\)
∴ The lines are parallel.

Question 2.
Check whether the following equations are consistent or inconsistent. Solve them graphically. (AS2, AS5)
a) 3x + 2y = 8
2x – 3y = 1
Answer:
Given equaions are 3x + 2y = 8 and 2x – 3y = 1
\(\frac{a_{1}}{a_{2}}\) = \(\frac{3}{2}\);
\(\frac{b_{2}}{b_{-3}}\) = \(\frac{-4}{6}\);
\(\frac{a_{1}}{a_{2}}\) ≠ \(\frac{b_{1}}{b_{2}}\)
Hence the linear equations are consistent.


The lines intersect at (2, 1), so the solution is (2, 1).

b) 2x – 3y = 8
4x – 6y = 9
Answer:
Given: 2x – 3y = 8 and 4x – 6y = 9
\(\frac{a_{1}}{a_{2}}\) = \(\frac{2}{4}\) = \(\frac{1}{2}\);
\(\frac{b_{1}}{b_{2}}\) = \(\frac{-3}{-6}\) = \(\frac{1}{2}\);
\(\frac{c_{1}}{c_{2}}\) = \(\frac{8}{9}\)
∴ \(\frac{a_{1}}{a_{2}}\) = \(\frac{b_{1}}{b_{2}}\) ≠ \(\frac{c_{1}}{c_{2}}\)
Lines are inconsistent and have no solution.

Lines are parallel.

The lines are parallel and no solution exists.

c) \(\frac{3}{2}\)x + \(\frac{5}{3}\)y = 7
9x – 10y = 12
Answer:
Given pair of equations \(\frac{3}{2}\)x + \(\frac{5}{3}\)y = 7 and 9x – 10y = 12
Now take \(\frac{3}{2}\)x + \(\frac{5}{3}\)y = 7 ⇒ \(\frac{9x+10y}{6}\) = 7 ⇒ 9x + 10y = 42
and 9x – 10y =12
\(\frac{a_{1}}{a_{2}}\) = \(\frac{9}{9}\) = \(\frac{1}{1}\);
\(\frac{b_{1}}{b_{2}}\) = \(\frac{10}{-10}\) = \(\frac{1}{-1}\) and
\(\frac{c_{1}}{c_{2}}\) = \(\frac{-42}{-12}\) = \(\frac{7}{2}\)
Since \(\frac{a_{1}}{a_{2}}\) ≠ \(\frac{b_{1}}{b_{2}}\) they are intersecting lines and hence consistent pair of linear equations.


Solution: The unique solution of given pair of equations is (3.1, 1.4)

d) 5x – 3y = 11
-10x + 6y = -22
Answer:
Given pair of equations 5x – 3y = 11 and -10x + 6y = -22
\(\frac{a_{1}}{a_{2}}\) = \(\frac{5}{-10}\) = \(\frac{-1}{2}\);
\(\frac{b_{1}}{b_{2}}\) = \(\frac{-3}{6}\) = \(\frac{-1}{2}\) and
\(\frac{c_{1}}{c_{2}}\) = \(\frac{11}{-22}\) = \(\frac{-1}{2}\)
∴ \(\frac{a_{1}}{a_{2}}\) = \(\frac{b_{1}}{b_{2}}\) = \(\frac{c_{1}}{c_{2}}\)
∴ The lines are consistent.
∴ The given linear equations represent coincident lines.
Thus they have infinitely many solutions.

e) \(\frac{4}{3}\)x + 2y = 8
2x + 3y = 12
Answer:
Given pair of equations \(\frac{4}{3}\)x + 2y = 8 ⇒ \(\frac{4x+6y}{3}\) = 8 ⇒ 4x + 6y = 24 ⇒ 2x + 3y = 12
\(\frac{a_{1}}{a_{2}}\) = \(\frac{4}{2}\) = 2;
\(\frac{b_{1}}{b_{2}}\) = \(\frac{6}{3}\) = 2;
\(\frac{c_{1}}{c_{2}}\) = \(\frac{24}{12}\) = 2
∴ \(\frac{a_{1}}{a_{2}}\) = \(\frac{b_{1}}{b_{2}}\) = \(\frac{c_{1}}{c_{2}}\)
Thus the equations are consistent.
∴ The given equations have infinitely many solutions.

f) x + y = 5
2x + 2y = 10
Answer:
Given pair of equations x + y = 5 and 2x + 2y = 10
\(\frac{a_{1}}{a_{2}}\) = \(\frac{1}{2}\);
\(\frac{b_{1}}{b_{2}}\) = \(\frac{1}{2}\);
\(\frac{c_{1}}{c_{2}}\) = \(\frac{5}{10}\) = \(\frac{1}{2}\)
∴ \(\frac{a_{1}}{a_{2}}\) = \(\frac{b_{1}}{b_{2}}\) = \(\frac{c_{1}}{c_{2}}\)
Thus the equations are consistent and have infinitely many solutions.

g) x – y = 8
3x – 3y = 16
Answer:
Given pair of equations x – y = 8 and 3x – 3y = 16
\(\frac{a_{1}}{a_{2}}\) = \(\frac{1}{3}\);
\(\frac{b_{1}}{b_{2}}\) = \(\frac{-1}{-3}\) = \(\frac{1}{3}\) and
\(\frac{c_{1}}{c_{2}}\) = \(\frac{8}{16}\) = \(\frac{1}{2}\)
∴ \(\frac{a_{1}}{a_{2}}\) = \(\frac{b_{1}}{b_{2}}\) ≠ \(\frac{c_{1}}{c_{2}}\)
Thus the equations are inconsistent.
∴ They represent parallel lines and have no solution.

h) 2x + y – 6 = 0 and 4x – 2y – 4 = 0
Answer:
Given pair of equations 2x + y – 6 = 0 and 4x – 2y – 4 = 0
\(\frac{a_{1}}{a_{2}}\) = \(\frac{2}{4}\) = \(\frac{1}{2}\);
\(\frac{b_{1}}{b_{2}}\) = \(\frac{1}{-2}\) = \(\frac{-1}{2}\);
\(\frac{c_{1}}{c_{2}}\) = \(\frac{-6}{-4}\) = \(\frac{3}{2}\)
∴ \(\frac{a_{1}}{a_{2}}\) ≠ \(\frac{b_{1}}{b_{2}}\)
The equations are consistent.
∴ They intersect at one point giving only one solution.


The solution is x = 2 and y = 2

i) 2x – 2y – 2 = 0 and 4x – 4y – 5 = 0
Answer:
Given pair of equations 2x – 2y – 2 = 0 and 4x – 4y – 5 = 0
\(\frac{a_{1}}{a_{2}}\) = \(\frac{2}{4}\) = \(\frac{1}{2}\);
\(\frac{b_{1}}{b_{2}}\) = \(\frac{-2}{-4}\) = \(\frac{1}{2}\);
\(\frac{c_{1}}{c_{2}}\) = \(\frac{-2}{-5}\) = \(\frac{2}{5}\)
∴ \(\frac{a_{1}}{a_{2}}\) = \(\frac{b_{1}}{b_{2}}\) ≠ \(\frac{c_{1}}{c_{2}}\)
Thus the equations are inconsistent.
∴ They represent parallel lines and have no solution.

Question 3.
Neha went to a ‘sale’ to purchase some pants and skirts. When her friend asked her how many of each she had bought, she answered “The number of skirts are two less than twice the number of pants purchased. Also the number of skirts is four less than four times the number of pants purchased.”
Help her friend to find how many pants and skirts Neha bought.
Answer:
Let the number of pants = x and the number of skirts = y
By problem y = 2x – 2 ⇒ 2x – y = 2
y = 4x – 4 ⇒ 4x – y = 4


The two lines are intersecting at the point (1,0)
∴ x = 1; y = 0 is the required solution of the pair of linear equations.
i.e., pants =1
She did not buy any skirt.

Question 4.
10 students of Class-X took part in a mathematics quiz. If the number of girls is 4 more than the number of boys then, find the number of boys and the number of girls who took part in the quiz.
Answer:
Let the number of boys be x.
Then the number of girls = x + 4
By problem, x + x + 4 = 10
∴ 2x + 4 = 10
2x = 10-4
x = \(\frac{6}{2}\) = 3
∴ Boys = 3 Girls = 3 + 4 = 7 (or)
Boys = x, Girls = y
By problem x + y = 10 (total)
and y = x + 4 (girls)
⇒ x + y = 10 and x – y = – 4


∴ Number of boys = 3 and the number of girls = 7

Question 5.
5 pencils and 7 pens together cost Rs. 50 whereas 7 pencils and 5 pens together cost Rs. 46. Find the cost of one pencil and that of one pen.
Answer:
Let the cost of each pencil be Rs. x
and the cost of each pen be Rs. y.
By problem 5x + 7y = 50
7x + 5y = 46


The lines are intersecting at the point (3, 5).
x = 3 and y = 5 is the solution of given equations.
∴ Cost of one pencil = Rs. 3 and pen = Rs. 5

Question 6.
Half the perimeter of a rectangular garden, whose length is 4 m more than its width is 36 m. Find the dimensions of the garden.
Answer:
Let the width of the garden = x cm
then its length = x + 4 cm
Half the perimeter = \(\frac{1}{2}\) × 2(7+ b) = l + b
By problem, x + x + 4 = 36
2x + 4 = 36
2x = 36 – 4 = 32
∴ x = 16 and x + 4 = 16 + 4 = 20
i.e., length = 20 cm and breadth = 16 cm.
(or)
Let the breadth be x and length = y
then x + y = 36 ⇒ x + y = 36
y = x + 4 ⇒ x – y = -4


The two lines intersect at the point (16, 20)
i.e., length = 20 cm and the breadth = 16 cm.

Question 7.
We have a linear equation 2x + 3y – 8 = 0. Write another linear equation in two variables such that the geometrical representation of the pair so formed is intersect¬ing lines. Now, write two more linear equations so that one forms a pair of parallel lines and the second forms coincident line with the given equation.
Answer:
i) Given: 2x + 3y – 8 = 0
The lines are intersecting lines.
Let the other linear equation be ax + by + c = 0
∴ \(\frac{a_{1}}{a_{2}}\) ≠ \(\frac{b_{1}}{b_{2}}\); we have to choose appropriate values satisfying the condition above.
Thus the other equation may be 3x + 5y – 6 =0

ii) Parallel line \(\frac{a_{1}}{a_{2}}\) = \(\frac{b_{1}}{b_{2}}\) ≠ \(\frac{c_{1}}{c_{2}}\)
⇒ 2x + 3y – 8 = 0
4x + 6y – 10 = 0

iii) Coincident lines \(\frac{a_{1}}{a_{2}}\) = \(\frac{b_{1}}{b_{2}}\) = \(\frac{c_{1}}{c_{2}}\)
⇒ 2x + 3y – 8 = 0 ⇒ 8x + 12y – 32 = 0

Question 8.
The area of a rectangle gets reduced by 80 sq. units if its length is reduced by 5 units and breadth is increased by 2 units. If we increase the length by 10 units and decrease the breadth by 5 units, the area will increase by 50 sq. units. Find the length and breadth of the rectangle.
Answer:
Let the length of the rectangle = x units
breadth = y units Area = l . b = xy sq. units
By problem, (x – 5) (y + 2) = xy – 80 and          (x + 10) (y – 5) = xy + 50
⇒ xy + 2x – 5y – 10 = xy – 80 and                    xy – 5x + 10y – 50 = xy + 50
⇒ 2x – 5y = xy – 80 – xy + 10 and                   -5x + 10y = xy + 50 – xy + 50
⇒ 2x – 5y = – 70 and                                       -5x + 10y = 100


The two lines intersect at the point (40, 30)
∴ The solution is x = 40 and y = 30
i.e., length = 40 units; breadth = 30 units.

Question 9.
In X class, if three students sit on each bench, one student will be left. If four students sit on each bench, one bench will be left. Find the number of students and the number of benches in that class.
Answer:
Let the number of benches = x say and the number of students = y
By problem
y = 3x + 1 ⇒ 3x – y + 1 = 0
and y = 4(x – 1) ⇒ 4x – y – 4 = 0


The two lines intersect at (5, 16)
∴ The solution of the equation is x = 5 and y = 16
i.e., Number of benches = 5 and the number of students = 16

AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 5 Quadratic Equations Ex 5.4 Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Maths Solutions 5th Lesson Quadratic Equations Exercise 5.4

10th Class Maths 5th Lesson Quadratic Equations Ex 5.4 Textbook Questions and Answers

Question 1.
Find the nature of the roots of the following quadratic equations. If real roots exist, find them.
i) 2x² – 3x + 5 = 0
Answer:
Given: 2x² – 3x + 5 = 0
a = 2; b = -3; c = 5
Discriminant = b² – 4ac
b² – 4ac = (-3)² – 4(2)(5)
= 9 – 40
= -31 < 0
∴ Roots are imaginary.

ii) 3x² – 4√3x + 4 = 0
Answer:
Given: 3x² – 4√3x + 4 = 0
a = 3; b = -4√3; c = 4
b² – 4ac = (-4√3)² – 4(3)(4)
= 48 – 48 = 0
∴ Roots are real and equal and they
\(\frac{-b}{2a}\), \(\frac{-b}{2a}\)

iii) 2x² – 6x + 3 = 0
Answer:
Given: 2x² – 6x + 3 = 0
a = 2; b = -6; c = 3
b² – 4ac = (-6)² – 4(2)(3)
= 36 – 24
= 12 > 0
∴ The roots are real and distinct. They are

Question 2.
Find the values of k for each of the fol-lowing quadratic equations so that they have two equal roots.
i) 2x² + kx + 3 = 0
Answer:
Given : 2x² + kx + 3 = 0 has equal roots
∴ b² – 4ac = 0
Here a = 2; b = k; c = 3
b² – 4ac = (k)² – 4(2)(3) = 0
⇒ k² – 24 = 0
⇒ k² = 24
⇒ k = √24 = ± 2√6

ii) kx(x – 2) + 6 = 0
Answer:
Given: kx(x – 2) + 6 = 0
kx² – 2kx + 6 = 0
As this Q.E. has equal roots,
b² – 4ac = 0
Here
a = k; b = -2k; c = 6
∴ b² – 4ac = (-2k)² – 4(k)(6) = 0
⇒ 4k² – 24k = 0
⇒ 4k(k – 6) = 0
⇒ 4k = 0 (or) k – 6 = 0
⇒ k = 0 (or) 6
But k = 0 is trivial
∴ k = 6.

Question 3.
Is it possible to design a rectangular mango grove whose length is twice its breadth, and the area is 800 m²? If so, find its length and breadth.
Answer:
Let the breadth = x m
Then length = 2x m
Area = length x breadth = x.(2x)
= 2x² m²
By problem 2x² = 800 ⇒ x² = 400
and x = √400 = ± 20
∴ Breadth x = 20 m and
length 2x = 2 × 20 = 40 m.

Question 4.
The sum of the ages of two friends is 20 years. Four years ago, the product of their ages in years was 48. Is the above situation possible? If so, deter¬mine their present ages.
Answer:
Let the age one of the two friends be x years.
Then the age of the other = 20 – x
Then, 4 years ago their ages would be (x – 4) and (20 – x – 4) = 16 – x
∴ Product of their ages 4 years ago = (x – 4) (16 – x)
By problem (x – 4) (16 – x) = 48
⇒ x(16 – x) – 4(16 – x) = 48
⇒ 16x – x² – 64 + 4x = 48
⇒ x² – 20x + 112 = 0
Here a = 1; b = -20; c = 112
b² – 4ac = (-20)² – 4(1) (112)
= 400 – 448
= -48 < 0
Thus the roots are not real.
∴ The situation is not possible.

Question 5.
Is it possible to design a rectangular park of perimeter 80 m and area 400 m²? If so, find its length and breadth.
Answer:
Given: Perimeter of a rectangle 2(1 + b) = 80
⇒ 6 + b = \(\frac{80}{2}\) = 40
Area of the rectangle, l × b = 400
If possible, let us suppose that length of the rectangle = x m say
Then its breadth by equation (1) = 40 – x
By problem area = x . (40 – x) = 400
⇒ 40x – x² = 400
⇒ x² – 40x + 400 = 0
Here a = 1; b = -40; c = +400
b² – 4ac = (-40)² – 4(1)(+400)
= 1600 – 1600 = 0
∴ The roots are real and equal.
They are \(\frac{-b}{2a}\), \(\frac{-b}{2a}\)
i.e., \(\frac{-(-40)}{2 \times 1}\) = \(\frac{40}{2}\) = 20
∴ The dimensions are 20 m, 20 m.
(∴ The park is in square shape)

AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 5 Quadratic Equations Ex 5.3 Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Maths Solutions 5th Lesson Quadratic Equations Exercise 5.3

10th Class Maths 5th Lesson Quadratic Equations Ex 5.3 Textbook Questions and Answers

Question 1.
Find the roots of the following quadratic equations, if they exist, by the method of completing the square:
i) 2x² + x – 4 = 0
Answer:
Given: 2x² + x – 4 = 0
⇒ 2x² + x = 4
⇒ (√2x)² + x = 4
⇒ (√2x)² + 2.√2.x.\(\frac{1}{2 \sqrt{2}}\) = 4
Now LHS is in the form a² + 2ab
where b = \(\frac{1}{2 \sqrt{2}}\)
Adding b² = \(\left(\frac{1}{2 \sqrt{2}}\right)^{2}\) on both sides we get

ii) 4x² + 4√3x + 3 = 0
Answer:
Given: 4x² + 4√3x + 3 = 0
⇒ 4x² + 4√3x = -3
⇒ (2x)² + 2(2x)√3 = -3
LHS is of the form a² + 2ab where
where b = √3.
∴ Adding b² = (√3)² = 3 on both sides, we get
(2x)² + 2(2x)(√3) + (√3)² = -3 + (√3)²
⇒ (2x + √3)² = -3 + 3 = 0
∴ (2x + √3)² = 0
⇒ 2x + √3 = 0
⇒ 2x = -√3
⇒ x = \(\frac{-\sqrt{3}}{2}\)
∴ The roots are \(\frac{-\sqrt{3}}{2}\), \(\frac{-\sqrt{3}}{2}\).

iii) 5x² – 7x – 6 = 0
Given quardratic equation = 5x² – 7x – 6 = 0
∴ 5x² – 7x – 6
⇒ x² – \(\frac{7}{5}\)x = \(\frac{6}{5}\), it can be re-written as
x² – 2.\(\frac{7}{10}\)x = \(\frac{6}{5}\) now it is in the form
of a² – 2ab where a = x, and b = \(\frac{7}{10}\)
Now adding b² = \(\left(\frac{7}{10}\right)^{2}\) on both sides, we get

Note: If we take the Q.E. as 5x² – 7x + 6 = 0, then we get the T.B. answer.

iv) x² + 5 = -6x
Answer:
The given Q.E. is x² + 5 = -6x
⇒ x² + 6x = -5
⇒ (x)² + 2.(x).3 = -5
Now L.H.S. is of the form a² + 2ab where b = 3.
Adding b² = 3² on both sides we get
x² + 2(x)(3) + 3² = -5 + 3²
(x + 3)² = -5 + 9 = 4
∴ x + 3 = 74 = ± 2
⇒ x = +2 – 3 or – 2 – 3
= -1 or -5 are the roots of the given Q.E.

Question 2.
Find the roots of the quadratic equations given in Q.1 above by applying the quadratic formula,
i) 2x² + x – 4 = 0
Answer:
Comparing this Q.E. with ax² + bx + c = 0
a = 2; b = 1; c = -4
x = \(\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}\)

ii) 4x² + 4√3x + 3 = 0
Answer:
Given: 4x² + 4√3x + 3 = 0
Here a = 4; b = 4√3 ; c = 3

iii) 5x² – 7x – 6 = 0
Answer:
Given: 5x² – 7x – 6 = 0
Here a = 5; b = -7 and c = -6

iv) x² + 5 = -6x
Answer:
Given: x² + 5 = -6x
⇒ x² + 6x + 5 = 0
Here a = 1; b = 6; c = 5

Question 3.
Find the roots of the following equations:
i) x – \(\frac{1}{x}\) = 3, x ≠ 0
Answer:
Given: x – \(\frac{1}{x}\) = 3
⇒ x² + 6x + 5 = 0
⇒ \(\frac{x^{2}-1}{x}\) = 3
⇒ x² – 1 = 3x
⇒ x² – 3x – 1 = 0
Here a = 1; b = -3; c = -1

ii) \(\frac{1}{x+4}\) – \(\frac{1}{x-7}\) = \(\frac{11}{30}\), x ≠ -4, 7
Answer:
Given: \(\frac{1}{x+4}\) – \(\frac{1}{x-7}\) = \(\frac{11}{30}\)

⇒ x² – 3x – 28 = -30
⇒ x² – 3x – 28 + 30 = 0
⇒ x² – 3x + 2 = 0
⇒ x² – 2x – x + 2 = 0
⇒ x(x – 2) – 1(x – 2) = 0
⇒ (x – 2) (x – 1) = 0
⇒ x – 2 = 0 (or) x – 1 = 0
⇒ x = 2 or x = 1
⇒ x = 2 or 1.

Question 4.
The sum of the reciprocals of Rehman’s ages, (in years) 3 years ago and 5 years from now is \(\frac{1}{3}\). Find his present age.
Answer:
Let the present age of Rehman be x years.
3 years ago Rehman’s age = x – 3 and its reciprocal is \(\frac{1}{x-3}\)
Rehman’s age 5 years from now = x + 5 and its reciprocal is \(\frac{1}{x+5}\)
The sum of the reciprocals


⇒ x² + 2x – 15 = 3(2x + 2)
⇒ x² + 2x – 15 = 6x + 6
⇒ x² + 2x – 15 – 6x – 6 = 0
⇒ x² – 4x – 21 =0
⇒ x² – 7x + 3x – 21 =0
⇒ x(x – 7) + 3(x – 7) 0
⇒ (x – 7) (x + 3) = 0
⇒ x – 7 = 0 or x + 3 = 0
⇒ x = 7 or x = -3
But x can’t be negative, x = 7
i.e., Present age of Rehman = 7 years.

Question 5.
In a class test, the sum of Moulika’s marks in Mathematics and English is 30. If she got 2 marks more in Mathematics and 3 marks less in English, the product of her marks would have been 210. Find her marks in the two subjects.
Answer:
Sum of the marks in Mathematics and English = 30
Let Moulika’s marks in Mathematics be x Then her marks in English = 30 – x
If she got 2 more marks in Mathematics, then her marks would be x + 2.
If she got 3 marks less in English then her marks would be 30 – x – 3 = 27 – x
By problem (x + 2) (27 – x) = 210
⇒ x(27 – x) + 2(27 – x) = 210
⇒ 27x – x² + 54 – 2x = 210
⇒ -x² + 25x + 54 = 210
⇒ x² – 25x – 54 + 210 = 0
⇒ x² – 25x + 156 = 0
⇒ x² – 12x – 13x + 156 = 0
⇒ x(x – 12) – 13(x 12) = 0
⇒ (x – 12) (x – 13) = 0
⇒ x – 12 = 0 or x – 13 = 0
⇒ x = 12 or x = 13
If x = 12, then marks in Mathematics = 12 English = 30 – 12 = 18
If x = 13, then marks in Mathematics = 13 English = 30 – 13 = 17

Question 6.
The diagonal of a rectangular field is 60 metres more than the shorter side. If the longer side is 30 metres more than the shorter side, find the sides of the field.
Answer:
Let the shorter side of the rectangular field = x m.
Then its longer side = x + 30 m.

The diagonal of a rectangle is also the hypotenuse of the lower triangle Here the diagonal = x + 60
∴ By Pythagoras Theorem
(side)² + (side)² = (hypotenuse)²
⇒ (x + 30)² + x² = (x + 60)²
⇒ x² + 60x + 900 + x² = x² + 120x + 3600
⇒ x² – 60x – 2700 = 0
⇒ x² – 90x + 30x – 2700 = 0
⇒ x(x – 90) + 30 (x – 90) = 0
⇒ (x – 90) (x + 30) = 0
⇒ x – 90 = 0 (or) x + 30 = 0
⇒ x = +90 (or) x = -30 But ‘x’ can’t be negative.
∴ x = 90 m
i.e., the shorter side x = 90 m Longer side x + 30 = 90 + 30 = 120 m.

Question 7.
The difference of squares of two numbers is 180. The square of the smaller number is 8 times the larger number. Find the two numbers.
Answer:
Let the large number be x.
8 times larger number = Square of the srnall number = 8x
Square of the larger number = x²
By problem, x² – 8x = 180
⇒ x² – 8x – 180 = 0
⇒ x²– 18x + 10x – 180 = 0
⇒ x(x – 18) + 10(x – 18) = 0
⇒ (x + 10)(x – 18) = 0
⇒ x + 10 = 0 (or) x – 18 = 0
⇒ x = -10 (or) x = 18
If x = 18, then larger number =18;
(small number)2 = 8 × (+18) = 144
∴ Small number = √144 = 12
The numbers are 18, 12
Note: Discard x = -10.

Question 8.
A train travels 360 km at a uniform speed. If the speed had been 5 km/h more, it would have taken 1 hour less for the same journey. Find the speed of the train.
Answer:
The distance travelled = 360 km.
Let the speed of the train = x kmph.
Time taken to complete a journey = \(\frac{\text { distance }}{\text { speed }}\)

⇒ x² + 5x = 1800
⇒ x² + 5x – 1800 = 0
⇒ x² + 45x – 40x – 1800 = 0
⇒ x(x + 45) – 40(x + 45) = 0
⇒ (x + 45) (x – 40) = 0
x + 45 = 0 or x -40 = 0
x = -45 or x = +40
But x can’t be negative.
∴ The speed of the train = 40 kmph.

Question 9.
Two water taps together can fill a tank in 9\(\frac{3}{8}\) hours. The tap of larger diameter takes 10 hours less than the smaller one to fill the tank separately. Find the time in which each tap can separately fill the tank.
Answer:
Let the time taken to fill the tank by smaller tap = x (hours)
So the part filled by smaller tap in
1 hour = \(\frac{1}{x}\) × \(\frac{75}{8}\) = \(\frac{75}{8x}\) ……. (1)
Again then the time taken to fill the tank by larger tap = (x – 10) hours
∴ the part of tank that can be filled by larger tap alone in one hour of time = \(\frac{1}{x-10}\)
∴ In \(\frac{75}{8}\) hours the part filled by larger tap = \(\frac{75}{8}\left(\frac{1}{x-10}\right)\)
∴ By both taps together

⇒ 150x – 750 = 8x² – 80x
⇒ 8x² – 80x – 150x + 750 = 0
⇒ 8x² – 230x + 750 = 0
⇒ 4x² – 115x + 375 = 0
⇒ 4x² – 100x – 15x + 375 = 0
⇒ 4x(x – 25) – 15(x – 25) = 0
∴ (4x – 15) (x – 25) = 0 15
⇒ 4x = 15, x = \(\frac{15}{4}\) or x = 25
x = 25 hours.
then time taken to fill by larger tap = x – 10 = 25 – 10 = 15 hours
(x cannot be \(\frac{15}{4}\) since we have considered ‘x’ as time taken by smaller tap, which is to be higher one)

Question 10.
An express train takes 1 hour less than a passenger train to travel 132 km between Mysore and Bangalore (without taking into consideration the time they stop at intermediate stations). If the average speed of the express train is 11 km/hr more than that of the passenger train, find the average speed of the two trains.
Answer:
Let the speed of the passenger train = x kmph.
Then speed of the express train = x + 11 kmph.
Distance travelled = 132 km
We know that time = \(\frac{\text { distance }}{\text { speed }}\)

⇒ x² + 11x = 13 × 11
⇒ x² + 11x – 1452 = 0
⇒ x² + 44x – 33x – 1452 = 0
⇒ x(x + 44) – 33 (x + 44) = 0
⇒ (x + 44) (x – 33) = 0
⇒ x + 44 = 0 (or) x – 33 = 0
⇒ x = -44 (or) x = 33
But x can’t be negative.
∴ Speed of the passenger train = x = 33 kmph.
Speed of the express train = x + 11 = 44 kmph.

Question 11.
Sum of the areas of two squares is 468 m². If the difference of their perimeters is 24m, find the sides of the two squares.
(OR)
If the sum of the areas of two squares is 468 m² and the difference of their perimeters is 24m, then find the measurements of their sides.
Answer:
Let the side of first square = x m say Then perimeter of the first square = 4x [∵ P = 4 . side]
By problem, perimeter of the second square = 4x + 24 (or) 4x – 24
∴ Side of the second square =

Now sum of the areas of the two squares is given as 468 m²
x² + (x + 6)² = 468
⇒ x² + x² + 12x + 36 = 468
⇒ 2x² + 12x + 36 – 468 = 0
⇒ 2x² + 12x – 432 = 0
⇒ x² + 6x – 216 = 0
⇒ x² + 18x – 12x – 216 = 0
⇒ x(x + 18)- 12(x + 18) = 0
⇒ (x + 18) (x – 12) = 0
⇒ x + 18 = 0 (or) x – 12 = 0
⇒ x = -18 (or) 12
But x can’t be negative.
∴ x = 12
i.e., side of the first square = 12
∴ Perimeter = 4 × 12 = 48
∴ Perimeter of the second square = 48 + 24 = 72
∴ Side of the second square = \(\frac{72}{4}\) = 18 m.
(or)
x² + (x – 6)² = 468
⇒ x² + x² – 12x + 36 = 468
⇒ 2x² – 12x – 432 – 0
⇒ x² – 6x – 216 = 0
⇒ x² – 18x + 12x – 216 = 0
⇒ x(x-18) + 12(x-18) = 0
⇒ (x – 18) (x + 12) = 0
⇒ x – 18 = 0 (or) x + 12 = 0
⇒ x = 18 (or) – 12
But x can’t be negative.
∴ x = 18
i.e., side of the first square = 18 m
∴ Perimeter = 4 × 18 = 72
Perimeter of the second square = 72 – 24 = 48
∴ Side of the second square = \(\frac{48}{4}\) = 12 m.
i.e., In any way, the sides of the squares are 12m, 18m.

Question 12.
If a polygon of ‘n’ sides has \(\frac{1}{2}\)n(n – 3) diagonals. How many sides will a polygon having 65 diagonals? Is there a polygon with 50 diagonals?
Answer:
Given: Number of diagonals of a polygon with n-sides = \(\frac{n(n-3)}{2}\)
No. of diagonals of a given polygon = 65
i.e., \(\frac{n(n-3)}{2}\) = 65
where n is number of sides of the polygon
⇒ n² – 3n = 2 × 65
⇒ n² – 3n – 130 = 0
⇒ n² – 13n + 10n – 130 = 0
⇒ n(n – 13) + 10(n – 13) = 0
⇒ (n – 13) (n + 10) = O
⇒ n – 13 = 0 (or) n + 10 = 0
⇒ n = 13 (or) n = -10
But n can’t be negative.
∴ n = 13 (i.e.) number of sides = 13.
Also to check 50 as the number of diagonals of a polygon
∴ \(\frac{n(n-3)}{2}\) = 50
⇒ n² – 3n = 100
⇒ n² – 3n – 100 = 0
There is no real value of n for which the above equation is satisfied.
∴ There can’t be a polygon with 50 diagonals.

SCERT AP 10th Class Social Study Material Pdf 11th Lesson ఆహార భద్రత Textbook Questions and Answers.

AP State Syllabus 10th Class Social Solutions 11th Lesson ఆహార భద్రత

10th Class Social Studies 11th Lesson ఆహార భద్రత Textbook Questions and Answers

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ప్రశ్న 1.
తప్పు వాక్యాలను సరి చేయండి. (AS1)
* ఆహార భద్రత సాధించటానికి ఆహార ఉత్పత్తిని మాత్రమే పెంచితే సరిపోతుంది.
* ఆహార భద్రత సాధించటానికి ఒకే పంటసాగును ప్రోత్సహించాలి.
* తక్కువ ఆదాయం ఉన్న ప్రజలలో తక్కువ కాలరీల వినియోగం ఎక్కువగా ఉంటుంది.
* ఆహార భద్రతను సాధించటంలో చట్ట సభల ప్రాధాన్యత ఎక్కువ.
* పిల్లల్లో పోషకాహార లోపాన్ని సరిచెయ్యటానికి ప్రజా పంపిణీ వ్యవస్థను ఉపయోగించవచ్చు.
జవాబు:

• ఆహార భద్రత సాధించటానికి ఆహార ఉత్పత్తి ఒక్కటే పెంచితే సరిపోదు; ఉత్పత్తితో పాటు ఆహార ధాన్యాల లభ్యతా, ఆహార అందుబాటు కూడా ముఖ్యం.
• ఆహార భద్రత సాధించటానికి ఒకే పంటసాగును కాక ఇతర పంటల దిగుబడి కూడా పెంచేలా ప్రోత్సహించాలి. ఉదాహరణకు వరి, గోధుమలతో పాటు జొన్న, నూనెగింజల దిగుబడులు కూడా పెరుగుతున్నాయి.
• తక్కువ ఆదాయం ఉన్న ప్రజలలో ‘తక్కువ కాలరీల వినియోగం’ ఉంటుంది. తక్కువ ఆదాయం ఉన్న (పేద) ప్రజలకు కొనుగోలు శక్తి తక్కువ ఉంటుంది. ఆహార ధాన్యాల కొనుగోలు, వినియోగం తక్కువ ఉంటుంది. ఆ ఆహారం వల్ల వారు పొందే కాలరీలు కూడా సహజంగా తక్కువగానే ఉంటాయి.
• ఆహార భద్రతను సాధించటంలో చట్ట సభలతో పాటు న్యాయస్థానాల ప్రాధాన్యత కూడా ఎక్కువగానే ఉంది. ఉదాహరణకు మధ్యాహ్న భోజన పథకం అమలుపై న్యాయ వ్యవస్థ ఆదేశాలు.
• పిల్లల్లో పోషకాహార లోపాన్ని సరిచెయ్యటానికి ప్రజాపంపిణీ వ్యవస్థ కంటే ఎక్కువగా అంగన్‌వాడీలు (ICDS), మధ్యాహ్న భోజన పథకమును ఉపయోగిస్తున్నారు.

ప్రశ్న 2.
గ్రామీణ ప్రాంతాలలో కాలరీల వినియోగం గత కొద్ది కాలంగా ……. 2004-05 లో తలసరి సగటు కాలరీల వినియోగం అవసరమైన దానికంటే ….. ఉంది. పట్టణ ప్రాంతంలో ఉంటున్న వ్యక్తికి రోజుకు కనీసం 2100 కాలరీలు అవసరం. పటణ ప్రాంతంలో 2004-05 లో కాలరీల అవసరం, వినియోగం మధ్య అంతరం ……… (AS1)
జవాబు:
తగ్గుతోంది. తక్కువగా, పెరిగింది.

ప్రశ్న 3.
ప్రకృతి వైపరీత్యం వల్ల ఒక సంవత్సరం ఆహార ధాన్యాల ఉత్పత్తి తగ్గిందని అనుకుందాం. ఆ సంవత్సరంలో ఆహార ధాన్యాల లభ్యత పెరగటానికి ప్రభుత్వం ఏ చర్యలు తీసుకోవాలి? (AS4)
(లేదా)
ఆహార ధాన్యాల ఉత్పత్తి తగ్గినపుడు, ఆహార ధాన్యాల లభ్యత పెరగడానికి ప్రభుత్వం ఏ చర్యలు తీసుకోవాలని నీవు అనుకుంటున్నావు?
జవాబు:
ఆహార ధాన్యాల లభ్యత పెరగటానికి ప్రభుత్వం తీసుకోవలసిన చర్యలు :

1. ఆహార ధాన్యాల లభ్యత పెంచటానికి ‘దిగుమతులు’ ఒక ముఖ్య మార్గం.
2. ఆహార లభ్యతను పెంచటానికి ప్రభుత్వ నిల్వల (బఫర్ నిల్వలు)’ ను ఉపయోగించుకోవటం మరో ముఖ్యమైన మార్గం. (FCI ధాన్యాగారాలలోని నిల్వలను ఉపయోగించుకోవాలి.)
3. ఇతర ప్రత్యామ్నాయ ఆహార పదార్థాల లభ్యతను అందుబాటులోకి తేవాలి.
4. నల్ల బజారు (Black Market), అక్రమ నిల్వలను అరికట్టాలి.
5. ఎగుమతులను నిషేధించుట మరియు అవసరమైన ఆంక్షలు విధించుట. (ధరలను అదుపులో ఉంచాలి.)
6. తర్వాతి సంవత్సరంలో మంచి దిగుబడులు సాధించటానికి అవసరమైన చర్యలు చేపట్టడం చేయాలి.
   ఉదా : హరిత విప్లవం

ప్రశ్న 4.
బరువు తక్కువగా ఉండటానికి, ఆహార అందుబాటుకు మధ్య గల సంబంధాన్ని తెలియజేయటానికి మీ ప్రాంతం నుంచి ఒక ఊహాజనిత ఉదాహరణను ఇవ్వండి. జ. సరిపడా ఆహారం ఉంటే ఎవరూ ఉండవలసిన దానికంటే తక్కువ బరువు కానీ, తక్కువ ఎత్తుగానీ ఉండరు. దీనికి ఉదాహరణ, మా గ్రామం ప్రాంతంలోని సంఘటన. (AS4)
జవాబు:

1. మా ప్రాంతంలోని ప్రజా వైద్యశాలకు తక్కువ బరువు ఉన్న రోగులు ప్రతిరోజు పెద్ద సంఖ్యలో వస్తారు.
2. అక్కడి డాక్టరు ఈ పరిస్థితిని వివరించారు.
3. కుటుంబానికి నెలకు రేషను దుకాణం ద్వారా లభించే ఆహార ధాన్యాలు అయిదుగురు ఉన్న కుటుంబంలో 11 రోజులకు సరిపోతాయి.
4. మిగిలిన రోజులకు వాళ్లు తాము పండించిన ఆహారంపైన లేదా మార్కెట్లో కొన్న దానిపైన ఆధారపడాలి.
5. వ్యవసాయ కూలీల, ఆదాయంలో ఎక్కువ భాగం ఇంటి అద్దె, కరెంటు ఇతరత్రా అవసరాలకు ఖర్చు అయిపోతుంది. కాబట్టి వీరు రిటైల్ మార్కెట్లో ఆహారధాన్యాల కొనుగోలు చేయలేకపోతున్నారు.
6. ఈ విధంగా సరిపడినంత ఆహారం తీసుకోలేకపోతున్నారు, కనుక వీరు తీవ్ర శక్తి లోపం (BMI-18.5) కలిగి ఉన్నారు. (తక్కువ బరువు సమస్య తీవ్రంగా ఉంది.)
7. అధిక శాతం ప్రజలు వారికి కావల్సిన దానికంటే తక్కువ కాలరీలు తీసుకుంటున్నారు, కారణం పేదరికం వల్ల ఆహారం అందుబాటులో లేకపోవడమేనని నేను గుర్తించాను.

ప్రశ్న 5.
వారం రోజుల మీ కుటుంబ ఆహార అలవాట్లను విశ్లేషించండి. దాంట్లోని పోషకాలను వివరించటానికి ఒక పట్టిక తయారుచెయ్యండి.
జవాబు:
విద్యార్థులు స్వయంగా చేయగలరు.
ఆధారం:

ప్రశ్న 6.
ఆహార ఉత్పత్తి పెరగటానికి, ఆహార భద్రతకు మధ్యగల సంబంధాన్ని వివరించండి. (AS1)
(లేదా)
ఆహార ఉత్పత్తి, ఆహార భద్రతల మధ్య సంబంధాన్ని వివరించుము.
జవాబు:
ఆహార ఉత్పత్తి పెరగటానికి, ఆహార భద్రతకు మధ్య అవినాభావ సంబంధముందని చెప్పవచ్చు.

1. రోజువారీ కనీస ఆహార అవసరాలు తీర్చటానికి సరిపడేటంత ఆహార పదార్థాల ఉత్పత్తి కచ్చితంగా ఉండటం ఆహార భద్రతకు ముఖ్యమైన అవసరం.
2. ఆహార ఉత్పత్తి పెరిగినట్లయితే దేశంలో తలసరి సగటు ఆహార ధాన్యాల లభ్యత పెరుగుతుంది.
3. ఆహార లభ్యత పెరిగినట్లయితే ప్రజలకు ఆహారం అందుబాటులో ఉంటుంది.
4. ఆహారం అందుబాటులో ఉంటే ఆహార భద్రత సాధించినట్లే.
5. ఆహార ఉత్పత్తి పెరిగితే బఫర్ నిల్వలు పెరుగుతాయి; ప్రజాపంపిణీ వ్యవస్థ సమర్థంగా పని చేస్తుంది, కొనుగోలు శక్తి తద్వారా వినియోగించే స్థితి పెరుగుతుంది. పోషకాహార స్థాయి పెరుగుతుంది. ఈ విధంగా ఆహార భద్రత సమర్ధవంతంగా కల్పించవచ్చు.

ప్రశ్న 7.
“ప్రజాపంపిణీ వ్యవస్థ ప్రజలకు ఆహార భద్రత ఉండేలా చూడగలదు.” ఈ వ్యాఖ్యానానికి మద్దతుగా వాదనలు పేర్కొనండి. (AS1)
(లేదా)
“ప్రజా పంపిణీ వ్యవస్థ ప్రజలకు ఆహార భద్రత ఉండేలా చూడగలదనే” వాక్యాన్ని ఎలా సమర్థించగలవు?
జవాబు:

1. ప్రజాపంపిణీ వ్యవస్థలో చౌక ధరల దుకాణాలు ప్రధానమయినవి. నిత్యావసర వస్తువులను తక్కువ ధరకు అందించేవి చౌకధరల దుకాణాలు.
2. భారతదేశంలో ఆహారధాన్యాలు అందుబాటులో ఉండటానికి, ప్రజలకు చౌకధరల దుకాణాలు ఎంతో ముఖ్యమైనవి.
3. చౌకధరల దుకాణాల నుంచి కొనుగోలు చేసే ఆహారధాన్యాలు, వాళ్ళ మొత్తం ఆహార ధాన్యాల వినియోగంలో ఎక్కువ శాతమే ఉంది.
4. అన్ని వర్గాల ప్రజలకు, అన్ని రకాల నిత్యావసర వస్తువులను కొనుగోలు చేసే శక్తి పెంచుటకుగాను (ధరల యంత్రాంగం ద్వారా) ధరలను అదుపులో ఉంచుతుంది. తద్వారా ఆహార పదార్థాల అందుబాటు గుణాత్మకంగాను, పరిమాణాత్మకం గాను పెరుగుతుంది.
5. పేదలకు, అత్యంత పేదలకు సబ్సిడీ ధరకు ఆహార ధాన్యాలను సరఫరా చేస్తూ (PDS ద్వారా) వారికి ఆహార భద్రత కల్పిస్తుంది.
6. గ్రామీణ ప్రాంతాల్లో అత్యధికంగా 75 శాతానికి, పట్టణ జనాభాలో 50 శాతానికి ప్రజాపంపిణీ వ్యవస్థ నుంచి ఆహార ధాన్యాలను కొనుగోలు చేసే హక్కు ఉంది.

ప్రశ్న 8.

ఆహార భద్రత గురించి పై పోస్టరు ఏమి తెలియజేస్తున్నది? (AS1)
జవాబు:
ఆహార భద్రత గురించి పై పోస్టరు మనలో ప్రతి ఏడుగురిలో ఒకరు ఆకలితో నిద్రపోతున్నారని, ఇంక ఇలా ఉండాల్సిన అవసరం లేదని తెలియచేస్తోంది. ఆ ఒక్కరూ కూడా ఆహారాన్ని తీసుకొని హాయిగా నిద్రిస్తారని తెలియచేస్తోంది.

ప్రశ్న 9.
ఆహార భద్రత గురించి ఇటువంటిదే ఒక పోస్టరు తయారుచెయ్యండి. (AS6)
జవాబు:
ఆధారం:

10th Class Social Studies 11th Lesson ఆహార భద్రత InText Questions and Answers

10th Class Social Textbook Page No.147

ప్రశ్న 1.
కింది వాక్యాలను చదివి హెక్టారుకు వరి, గోధుమల దిగుబడులు ఎలా ఉన్నాయో వివరించండి.
…………….., ……….. పంటల దిగుబడులు వరి, గోధుమలతో పోలిస్తే ఎప్పుడూ తక్కువగానే ఉన్నాయి. అయితే ఇటీవల కాలంలో ఈ పంటల దిగుబడులు మెల్లగా పెరుగుతున్నాయి.
జవాబు:
జొన్న, నూనెగింజలు.

10th Class Social Textbook Page No.147

ప్రశ్న 2.
దీర్ఘకాలం పాటు వరి, గోధుమ దిగుబడులు గణనీయంగా పెరగటానికి ఏ అంశాలు దోహదం చేశాయి?
జవాబు:
దీర్ఘకాలం పాటు వరి, గోధుమ దిగుబడులు గణనీయంగా పెరగటానికి దోహదం చేసిన అంశాలు

• మేలు జాతి, సంకర జాతి విత్తనాలు వాడటం (ఉదా : ‘SRI’ వరి)
• నాణ్యమైన పురుగు మందులు వాడుట ద్వారా సస్యరక్షణ చేపట్టడం.
• విస్తృతంగా రసాయన ఎరువులను వాడటం. (ఉదా : పొటాషియం, నైట్రోజన్ ఎరువులు)
• సాగునీటి వసతులను విస్తరించటం. (ఉదా : కాలువలు, గొట్టపు బావుల తవ్వకం)
• విత్తుటకు, దున్నుటకు, పంట కోత మొ||న వాటికి యంత్రాల వాడకం. (ఉదా : ట్రాక్టర్, కంబైన్డ్ హార్వెస్టర్)

10th Class Social Textbook Page No.152

ప్రశ్న 3.
దేశంలో అధిక శాతం ప్రజలకు ఆహార ధాన్యాలు అందుబాటులో లేని నేపథ్యంలో స్వల్ప ఆదాయం కోసం ఆహార ధాన్యాలను ఎగుమతి చెయ్యటం సరైనదేనా?
జవాబు:
సరియైన విధానం కాదు, దేశంలో అధిక ప్రజలకు ఆహార ధాన్యాలు అందుబాటులో లేని నేపథ్యంలో స్వల్ప ఆదాయం కోసం ఆహార ధాన్యాలను ఎగుమతి చెయ్యడం సరికాదు.

• ఎగుమతులు పెరిగినట్లయితే ఆహార ధాన్యాల ధరలు పెరిగి, పేదలకు ఆహార అందుబాటు ఇంకా దూరమవుతుంది.
• ఆహార ధాన్యాల నిల్వలు పెంచకుండా, ఎగుమతులు చేసినట్లయితే PDS ద్వారా పంపిణీకి ధాన్యాల కొరత ఏర్పడుతుంది.
• కరవుకాటకాలు, ఇతర ప్రకృతి విపత్తులు సంభవించినట్లయితే ఆహార ధాన్యాల దిగుబడి తగ్గుతుంది, లభ్యత, అందుబాటు కూడా తక్కువగా ఉంటుంది. ఇలాంటి సమయాల్లో అవసరమైతే ఆహార ధాన్యాలను దిగుమతి చేసుకోవాలి.

10th Class Social Textbook Page No.152

ప్రశ్న 4.
క్రింది రేఖాచిత్రపటాన్ని పరిశీలించండి.
రేఖాచిత్రపటం : 2009-10 లో బియ్యం, గోధుమల కొనుగోళ్ళలో రేషను దుకాణాల నుంచి కొన్న శాతం

ఖాళీలను పూరించండి :
అఖిల భారతానికి ప్రజల మొత్తం వినియోగంలో …….. (1)……… శాతం బియ్యం , ……… (2)…….. శాతం గోధుమ చౌక ధరల దుకాణాల నుంచి కొనుగోలు చేస్తారు. దీని అర్థం ప్రజలు తమ ఆహార ధాన్యాల అవసరంలో అధిక భాగం …. (3)…… నుంచి కొనుక్కోవాలి. అయితే …… (4)…………….. (5)…….. రాష్ట్రాలలో పరిస్థితి బాగుంది. …… (6)…….. ……(7)……… …. (8)……. రాష్ట్రాలలో ప్రజాపంపిణీ వ్యవస్థ ప్రజల ఆహార ధాన్యాల అవసరాలను నామమాత్రంగా తీరుస్తోంది.
జవాబు:

1. 39,
2. 28,
3. రిటైల్ మార్కెట్,
4. తమిళనాడు,
5. ఉమ్మడి ఆంధ్రప్రదేశ్
6. బీహార్,
7. రాజస్థాన్,
8. పంజాబ్.

10th Class Social Textbook Page No.155

ప్రశ్న 5.
బడిలో చేరే వయస్సు రాని పిల్లలకు పోషకాహార శాస్త్రజ్ఞులు మూడు చార్జులను ఉపయోగిస్తారు. ఈ మూడు వేరు వేరు సూచికలు మనకు పిల్లల పోషకాహార స్థాయికి సంబంధించి పూర్తి వివరాలను అందిస్తాయి. వాటిని కింద ఇచ్చాం.

జవాబు:

10th Class Social Textbook Page No.155

ప్రశ్న 6.
ఈ గణాంకాల నుంచి ఎటువంటి నిర్ధారణలకు వస్తారు ? ఒక పేరా రాయండి.
జవాబు:
ఈ గణాంకాల నుంచి నిర్ధారణలకు వచ్చిన అంశాలు :

• 45% మంది పిల్లలు వయస్సుకు తగ్గ బరువు ఉండటం లేదు.
• 41% మంది పిల్లలు వయస్సుకు తగ్గ ఎత్తు లేరు.
• 21% మంది పిల్లలు ఎత్తుకు తగ్గ బరువు లేరు. అంటే సరైన BMI కలిగి లేరు.
• ఎక్కువ మంది పిల్లలు పోషకాహార లోపం కలిగి ఉన్నారు.
• చాలా మంది పిల్లల్లో వ్యాధినిరోధక శక్తి తక్కువగా ఉంది.
• అంగన్‌వాడీ (ICDS) లాంటి పథకాలు సమర్థంగా అమలు అయ్యేలా చూడాలి.

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ప్రశ్న 7.
పాఠం 9 (రాంపురం : గ్రామ ఆర్థిక వ్యవస్థ) లోని “భూమి, ఇతర సహజ వనరులు” అనే భాగాన్ని మళ్ళీ చదవండి. భూమి నుంచి పంట ఉత్పత్తి పెంచటానికి ఏ ఏ విధానాలు ఉన్నాయి?
జవాబు:
గత కొద్ది దశాబ్దాలుగా సాగు కింద ఉన్న భూమి ఇంచుమించు స్థిరంగా ఉందని మనకు తెలుసు, కాబట్టి భూమి నుంచి పంట ఉత్పత్తి పెంచటం ముఖ్యం.

• హెక్టారుకు లభించే పంట దిగుబడిని పెంచటానికి అవసరమైన ఉత్పాదకాలను (ఉదా : HYV విత్తనాలు) సక్రమంగా వినియోగించుకోవాలి.
• సాగునీటి వసతులను (సక్రమ జల నిర్వహణ ద్వారా) పెంచటం. అయితే ఈ కీలక వనరు. అందరికీ అందేలా పంచుకునే పద్ధతిలో వినియోగించాలి. (ఉదా : గొట్టపు బావులు త్రవ్వడం).
• వర్షాభావ పంట రకాలను స్థానిక పరిస్థితులకు అనుగుణంగా విత్తటం, వర్షపు నీటిని నిల్వ చేయటం, పంట మార్పిడి వంటి పద్ధతుల ద్వారా దిగుబడులు పెంచాలి.
• సరియైన ‘సస్య రక్షణ’ చర్యలు చేపట్టాలి. ఉదా : నాణ్యమైన పురుగు మందులు వాడుట.
• అవసరమైన మేర ‘ఎరువుల వాడకం చేపట్టడం. ఉదా : రసాయన, సేంద్రియ ఎరువులను వాడుట.
• నేల సారాన్ని పెంచి, దిగుబడులను పెంచే బహుళ పంటల నమూనాను అనుసరించాలి. అంటే ఒకే పంట పొలంలో – గోధుమ, సజ్జ, బంగాళదుంప మొ||న పంటలను ఒకేసారి పండించటం.
• కంబైన్డ్ హార్వెస్టర్ లాంటి ఆధునిక యంత్రాలను వాడుట ద్వారా పంటకాలము ఆదా అవుతుంది. నూర్పిడి సమయంలో జరిగే ధాన్యాల వృథాను తగ్గిస్తుంది.

10th Class Social Textbook Page No.146

ప్రశ్న 8.
ఇవ్వబడిన రేఖాచిత్ర పటమును పరిశీలించి ఖాళీలను పూరించండి (ప్రతి బిందువు దగ్గర కచ్చితమైన విలువను తెలుసుకోటానికి ‘వై’ అక్షం మీది స్కేలుని ఉపయోగించండి).

ఆహారధాన్యాల ఉత్పత్తి 1970-71 నుండి ……(1)…… కు పెరిగింది. వరి ఉత్పత్తి 1970-71 నాటి 40 మిలియన్ టన్నుల నుండి 2010-11 నాటికి ….. (2)…… మిలియన్ టన్నులకు పెరిగింది. ఈ 40 ఏళ్ల కాలంలో ఉత్పత్తి వేగంగా పెరిగిన మరొక ముఖ్యమైన ఆహారపంట …. (3)…… . వరి, గోధుమలతో పోలిస్తే 1970-2011 కాలంలో …..(4)……. ఉత్పత్తి పెరగలేదు. దీనికి కారణం …. (5)……. అయి ఉండవచ్చు.
జవాబు:
1) 2010-11 వరకు
2) 95 3 ) గోధుమ
4) జొన్న, నూనె గింజలు
5) i) ప్రాధాన్యతనివ్వకపోవడం,
ii) వర్షాధార పంటలు కావడం,

10th Class Social Textbook Page No.147

ప్రశ్న 9.
జొన్న దిగుబడులను పెంచటంపై దృష్టి ఎందుకు పెట్టాలి? చర్చించండి.
జవాబు:
జొన్న దిగుబడులను పెంచటంపై దృష్టి ఎందుకు పెట్టాలంటే –

• జొన్నను మంచి పోషక ధాన్యంగా వ్యవహరిస్తున్నారు.
• జొన్న పంటను వర్షాధార ప్రాంతాలలో కూడా సాగుచేయవచ్చు.
• నేల, ఇతర సహజ వనరులు అంతరించిపోకుండా, క్షీణతకు గురికాకుండా చూడటానికి.
• ఆహార భద్రత, ఆహారధాన్యాల అందుబాటు పెంచుటకు.
• పురుగు మందులు, రసాయన ఎరువులు ఎక్కువగా వాడనవసరం లేదు.
• సాగునీటి లభ్యత తక్కువగా ఉన్న ప్రాంతాలలో కూడా జొన్న పంటను పండించవచ్చు.
• జొన్నలకు మార్కెట్ కూడా బాగా ఉంది. లక్షలాది ప్రజల ప్రధాన ఆహారం జొన్న.

10th Class Social Textbook Page No.148

ప్రశ్న 10.
1971కి చూపించిన విధంగా, 1991, 2011 సంవత్సరాలకు తలసరి సగటు ఆహార ధాన్యాల అందుబాటును లెక్కగట్టండి.

గమనిక : 1 టన్ను = 1000 కిలోలు; 1 కిలో = 1000 గ్రాములు; # = మిలియన్ టన్నులు

* మీరు లెక్కించిన దాని ఆధారంగా ఖాళీలను పూరించండి : 1971 నుంచి 1991 నాటికి తలసరి ఆహారధాన్యాల లభ్యత ………….. (పెరిగింది/తగ్గింది), కానీ 2011లో ఇది ……………. (ఎక్కువ | తక్కువగా) ఉంది. ఇటీవల దశాబ్దాలలో జనాభావృద్ధిలో తగ్గుదల ఉన్నప్పటికీ ఇలా జరిగింది. భవిష్యత్తులో ఆహారధాన్యాల లభ్యత పెరగటానికి ప్రభుత్వం. ………………. చర్యలు చేపట్టాలి.
జవాబు:

గమనిక : 1 టన్ను = 1000 కిలోలు; 1 కిలో = 1000 గ్రాములు; # = మిలియన్ టన్నులు.

మీరు లెక్కించిన దాని ఆధారంగా ఖాళీలను పూరించండి : 1971 నుంచి 1991 నాటికి తలసరి ఆహారధాన్యాల లభ్యత పెరిగింది (పెరిగింది/తగ్గింది), కానీ 2011లో ఇది తక్కువగా (ఎక్కువ తక్కువగా) ఉంది. ఇటీవల దశాబ్దాలలో జనాభావృద్ధిలో తగ్గుదల ఉన్నప్పటికీ ఇలా జరిగింది. భవిష్యత్తులో ఆహారధాన్యాల లభ్యత పెరగటానికి ప్రభుత్వం ఉత్పత్తి పెంచటం, , దిగుమతులు పెంచటం లాంటి చర్యలు చేపట్టాలి.

10th Class Social Textbook Page No.150

ప్రశ్న 11.
వ్యవసాయ వైవిధ్యీకరణకు సంబంధించిన పదాలు, వాక్యాల కింద గీత గీసి భారతీయ రైతులకు ఇది ఎందుకు అవసరమో వివరించండి.
జవాబు:
భారతీయ రైతులకు వ్యవసాయ వైవిధ్యీకరణ అవసరం; ఎందుకంటే,

• భారతీయ రైతులు ఎక్కువ శాతం మంది చిన్న, సన్నకారు రైతులే.
• ఎక్కువ మంది పేద రైతులే, వారి ఆదాయం పెరగాలంటే ఇది అవసరం.
• అధిక దిగుబడులు పొందడానికి,
• భారతదేశంలో వ్యవసాయం ఋతుపవనాలపై ఆధారపడి ఉంది కనుక వర్షాభావ కాలంలో, వర్షాభావ ప్రాంతంలో ఈ విధమైన వ్యవసాయం ఎంతో అవసరం.
• అల్ప ఉపాధి, ప్రచ్ఛన్న నిరుద్యోగిత నివారణకు కూడా ఇది ఎంతో అవసరం.
• అల్ప ఆదాయ సన్న, చిన్నకారు రైతులకు ఆహార భద్రత కల్పించుటకు.

10th Class Social Textbook Page No.150

ప్రశ్న 12.
మీ గ్రామంలో లేదా మీకు తెలిసిన గ్రామంలో వ్యవసాయ వైవిధ్యీకరణను వివరించండి.
జవాబు:
నాకు తెలిసిన ‘బేతపూడి’ గ్రామంలోని వ్యవసాయ వైవిధ్యీకరణ గురించి వివరిస్తాను.

• ఈ గ్రామంలోని అన్ని భూములకు సాగునీటి వసతులు (కాలువలు లేదా బోరుబావులు) ఉన్నాయి.
• ఈ గ్రామంలో వ్యవసాయ పెట్టుబడులకుగాను ఋణసౌకర్యం అందించుటకు బ్యాంక్ సౌకర్యం కలదు.
• ఆధునిక వ్యవసాయ పద్ధతులను ఉపయోగిస్తున్నారు. సాంకేతిక విజ్ఞానం చాలా బాగుంది.
• వరి, జొన్న, మొక్కజొన్న, మినుము, పెసర వంటి ఆహార ధాన్యాలతోపాటు ప్రత్తి, మిరప వంటి వాణిజ్య పంటలు సాగుచేస్తున్నారు.
• పంటల మధ్య కాలంలో కూరగాయలు పండిస్తున్నారు. బెండ, వంగ, దోస, టమోట పండిస్తున్నారు.
• అంతర్ పంటలుగా ‘కందులు’ (కందిపప్పు) ను పండిస్తున్నారు.
• దాదాపుగా అందరికి 2-3 గేదెలు ఉన్నాయి. పాడి పరిశ్రమ బాగా అభివృద్ధి చెందింది.
• గ్రామంలో 4 కంబైన్డ్ హార్వెస్టర్లు, దాదాపు 16 ట్రాక్టర్లు ఉన్నాయి. వ్యవసాయ యాంత్రీకరణ జరిగింది.
• వివిధ మార్కెట్ల సమాచారం అందుబాటులో ఉంటుంది. మెరుగైన రవాణా సౌకర్యాలు కలిగి ఉన్నది.
  సూచన : విద్యార్థులు తాము చూసిన గ్రామం గురించి స్వయంగా రాయగలరు.

10th Class Social Textbook Page No.150

ప్రశ్న 13.
ఎనిమిదవ తరగతిలోని ప్రజా పంపిణీ వ్యవస్థపై చర్చను గుర్తుకు తెచ్చుకోండి. ప్రజాపంపిణీ వ్యవస్థకు, ప్రజల ఆహార భద్రతకు గల సంబంధం ఏమిటి?
(లేదా)
భారతదేశ దక్షిణాది రాష్ట్రాలలో ప్రజా పంపిణీ వ్యవస్థ బాగుందని అధ్యయనాలు వెల్లడి చేస్తున్నాయి. ఇది ప్రజా పంపిణి వ్యవస్థను అందరికి వర్తింపచేసిన రాష్ట్రాలు కావటము గమనించవలసిన విషయం. ఇవి అందరికి తక్కువ ధరలలో ఆహార ధాన్యాలను అందించాయి. ఇందుకు విరుద్ధంగా ఇతర రాష్ట్రాలు పేదవాళ్ళను గుర్తించి ఆహార ధాన్యాలను పేదలకు, పేదలు కాని వాళ్ళకు వేరు వేరు ధరలకు అమ్మాయి. పేదలలో కూడ అత్యంత పేదలకు కూడ వేరే హక్కులు ఉన్నాయి. వాళ్ళకు అందించే మోతాదు వేరు.
ప్రజాపంపిణీ వ్యవస్థకు, ప్రజల ఆహార భద్రతకు మధ్య గల సంబంధాన్ని వివరించండి.
జవాబు:
ప్రజాపంపిణీ వ్యవస్థ (PDS)కు, ప్రజల ఆహార భద్రతకు ఎంతో దగ్గరి సంబంధముంది.

• ప్రజాపంపిణీ వ్యవస్థలో చౌకధరల దుకాణాలు ఎంతో ముఖ్యమైనవి.
• పేదలకు, అత్యంత పేదలకు సబ్సిడి ధరకు నిత్యావసర వస్తువులను సరఫరా చేస్తూ ఆహార పదార్థాల అందుబాటుకు తద్వారా ఆహారభద్రతకు PDS ఎంతో తోడ్పడుతుంది.
• వివిధ పథకాలు (మధ్యాహ్న భోజన పథకం, MNREP, FFW, AAY, ICDS మొ||నవి) ద్వారా పేద ప్రజలకు ఆహార ధాన్యాలను పంపిణీ చేస్తూ ఆహారం అందుబాటులోకి తెస్తుంది.
• ధరలను అదుపులో ఉంచుట ద్వారా అల్ప ఆదాయ వర్గాల కొనుగోలు శక్తి పెరుగుతుంది. తద్వారా ఆహార ధాన్యాల అందుబాటు పరిమాణాత్మకంగా, గుణాత్మకంగా పెరుగుతుంది.

10th Class Social Textbook Page No.153

ప్రశ్న 14.
సమర్థంగా పనిచేసే అంగన్‌వాడీ కేంద్రం ఈ పరిస్థితిని ఎలా సరిదిద్దగలదు ? చర్చించండి.
జవాబు:
మన దేశంలో మొత్తం మీద 16% పిల్లల్లో బరువు చాలా తక్కువగా ఉన్న తీవ్ర పరిస్థితి ఉంది. మొత్తంగా 45% పిల్లలు తక్కువ బరువు ఉన్నారు. 1-3 సం||రాల పిల్లలతో పోలిస్తే, 3-5 సం||రాల పిల్లల్లో ఇది చాలా ఎక్కువగా ఉంది. సమరంగా పనిచేసే అంగన్‌వాడీ కేంద్రం ఈ పరిస్తితిని చాలా వరకు సరిదిద్దగలదు.

• అంగన్‌వాడీల్లో 1-5 సం||రాల పిల్లలు ఎక్కువగా ఉంటారు, వీరు శిక్షణ పొందిన ఆయాల సంరక్షణలో ఉంటారు.
• ప్రభుత్వం పౌష్టికాహారం (పోషకాహారం)ను అంగన్‌వాడీల ద్వారానే సరఫరా చేస్తుంది.
• పిల్లల యొక్క ఎత్తు, బరువులను ఎప్పటికప్పుడు పరీక్షించి, తగుచర్యలు తీసుకుంటారు.
• పిల్లలకు అవసరమైన వైద్య, ఆరోగ్య సూచనలు అందించబడతాయి, వ్యాక్సినేషన్ ఉంటుంది.
• ఆహ్లాదకరమైన, పరిశుభ్రమైన వాతావరణం ఉండటం వలన పిల్లల ఎదుగుదల చక్కగా ఉంటుంది.
• అంగన్‌వాడీ కేంద్రంలో సరఫరా చేయు గ్రుడ్లు, ప్రోటీన్స్ (సోయాబీన్స్ పొడి), సమతౌల్య ఆహారం పొడి మొ||నవి పిల్లల ఎదుగుదలకు, అభివృద్ధికి ఎంతో తోడ్పడతాయి.

AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 5 Quadratic Equations Ex 5.2 Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Maths Solutions 5th Lesson Quadratic Equations Exercise 5.2

10th Class Maths 5th Lesson Quadratic Equations Ex 5.2 Textbook Questions and Answers

Question 1.
Find the roots of the following quadratic equations by factorisation,
i) x² – 3x – 10 = 0
Answer:
Given: x² – 3x – 10 = 0
x² – 5x + 2x- 10 = 0
⇒ x(x – 5) + 2 (x – 5) = 0
⇒ (x – 5) (x + 2) = 0
⇒ x – 5 = 0 or x + 2 = 0
⇒ x = 5 or x = -2
⇒ x = 5 or -2
are the roots of the given Q.E.

ii) 2x² + x – 6 = 0
Answer:
Given: 2x² + x – 6 = 0
⇒ 2x² + 4x – 3x – 6 = 0
⇒ 2x(x + 2) – 3(x + 2) = 0
⇒ (x + 2) (2x – 3) = 0
⇒ (x + 2) or 2x – 3 = 0
⇒ x = -2 or 2x = 3
⇒ x = -2 or \(\frac{3}{2}\)
are the roots of the given Q.E.

iii) √2x² + 7x + 5√2 =0
Answer:
Given: √2x² + 7x + 5√2 =0
⇒ √2x² + 5x + 2x + 5√2 = 0
⇒ x(√2x + 5)+ √2(√2x + 5) = 0
⇒ (√2x + 5) (x + √2) = 0
⇒ √2x + 5 = 0 or x + √2 = 0
⇒ √2x = -5 or x = -√2
⇒ x = \(\frac{-5}{\sqrt{2}}\) = or -√2
are the roots of √2 the given Q.E.

iv) 2x² – x + \(\frac{1}{8}\) = 0
Answer:
Given: 2x² – x + \(\frac{1}{8}\) = 0
⇒ \(\frac{16 x^{2}-8 x+1}{8}\) = 0
⇒ 16x² – 8x + 1 =0
⇒ 16x² – 4x – 4x + 1 = 0
⇒ 4x(4x – 1) – l(4x – 1) = 0
⇒ (4x – 1) (4x – 1) – 0
⇒ 4x – 1 = 0
⇒ 4x = l
⇒ x = \(\frac{1}{4}\), \(\frac{1}{4}\)
are the roots of given Q.E.

v) 100x² – 20x + 1 = 0
Answer:
Given : 100x² – 20x + 1 =0
⇒ 100x² – 10x – 10x + 1 = 0
⇒ 10x(10x – 1) – l(10x – 1) = 0
⇒ (10x – 1) (10x – l) = 0
⇒ 10x – 1 = 0
⇒ 10x = 1
⇒ x = \(\frac{1}{10}\), \(\frac{1}{10}\)
are the roots of the given Q.E.

vi) x(x + 4) = 12
Answer:
Given: x(x + 4) = 12
⇒ x² + 4x = 12
⇒ x² + 4x – 12 = 0
⇒ x² + 6x – 2x – 12 = 0
⇒ x(x + 6) – 2(x + 6) = 0
⇒ (x + 6) (x – 2) = 0
⇒ x + 6 = 0 or x – 2 = 0
⇒ x = -6 or x = 2
⇒ x = -6 or 2
are the roots of the given Q.E.

vii) 3x² – 5x + 2 = 0
Answer:
Given: 3x² – 5x + 2 = 0
⇒ 3x² – 3x – 2x + 2 = 0
⇒ 3x(x – 1) – 2(x – 1) = 0
⇒ (x – 1) (3x – 2) = 0
⇒ x – 1 = 0 or 3x – 2 = 0
⇒ x = 1 or \(\frac{2}{3}\),
⇒ x = 1 or \(\frac{2}{3}\) are the roots of the given Q.E.

viii) x – \(\frac{3}{x}\) = 2
Answer:
Given: x – \(\frac{3}{x}\) = 2
⇒ \(\frac{x^{2}-3}{x}\) = 2
⇒ x² – 3 = 2x
⇒ x² – 2x – 3 = 0
⇒ x² – 3x + x – 3 = 0
⇒ x(x – 3) + l(x – 3) = 0
⇒ (x – 3) (x + 1) = 0
⇒ (x – 3) = 0 or (x + 1) = 0
⇒ x = 3 or x = -1
⇒ x = 3 or -1 are the roots of the given Q.E.

ix) 3(x – 4)² – 5(x – 4) = 12
Answer:
Take (x – 4) = a, then the given Q.E. reduces to 3a² – 5a = 12
⇒ 3a² – 5a – 12 = 0
⇒ 3a² – 9a + 4a – 12 = 0
⇒ 3a(a – 3) + 4(a – 3) = 0
⇒ (a – 3) (3a + 4) = 0
⇒ a – 3 = 0 or 3a + 4 = 0
⇒ a = 3 or a = \(\frac{-4}{3}\)
but a = x – 4
x – 4 = 3 (or) x – 4 = \(\frac{-4}{3}\)
⇒ x = 7 or x = 4 – \(\frac{-4}{3}\) = \(\frac{8}{3}\)
∴ x = 7 or \(\frac{8}{3}\)
are the roots of the given Q.E.

Question 2.
Find two numbers whose sum is 27 and product is 182.
Answer:
Let a number be x.
Then the other number = 27 – x
Product of the numbers = x(27 – x) = 27x – x²
By problem 27x – x² = 182
⇒ x² – 27x + 182 = 0
⇒ x² – 14x – 13x + 182 = 0
⇒ x(x- 14) – 13(x – 14) = 0
⇒ (x – 13) (x – 14) = 0
⇒ x – 13 = 0 or x – 14 = 0
⇒ x = 13 or 14.
∴ The numbers are 13; 27 – 13 = 14 or 14 and 27 – 14 = 13.

Question 3.
Find two consecutive positive integers, sum of whose squares is 613.
Answer:
Let a positive integer be x.
Then the second integer = x + 1
Sum of the squares of the above integers = x² + (x + 1)²
= x² + x² + 2x + 1
= 2x² + 2x + 1
By problem 2x² + 2x + 1 = 613
⇒ 2x² + 2x – 612 = 0
⇒ x² + x – 306 = 0
⇒ x² + 18x – 17x – 306 = 0
⇒ x(x + 18) – 17(x + 18) = 0
⇒ (x – 17) (x + 18) = 0
⇒ x – 17 = 0 (or) x + 18 = 0
⇒ x = 17 (or) -18,
we do not consider -18
Then the numbers are (17, 17 + 1)
i.e., 17, 18 are the required two consecutive positive integers.

Question 4.
The altitude of a right triangle is 7 cm less than its base. If the hypotenuse is 13 cm, find the other two sides.
Answer:
Let the base of the right triangle = x cm
Then its altitude = x – 7 cm
By Pythagoras Theorem

(base)² + (height)² = (hypotenuse)²
⇒ x² + (x – 7)² = 13²
⇒ x² + x² – 14x + 49 = 169 .
⇒ 2x² – 14x + 49 – 169 = 0
⇒ 2x² – 14x – 120 = 0
⇒ x² – 7x – 60 = 0
⇒ x² – 12x + 5x – 60 = 0
⇒ x(x – 12) + 5(x – 12) = 0
⇒ (x – 12) (x + 5) = 0
⇒ x – 12 = 0 (or) x + 5 = 0
⇒ x = 12 (or) x = -5 But x can’t be negative.
∴ x = 12
x – 7 = 12 – 7 = 5
The two sides are 12 cm and 5 cm.

Question 5.
A cottage industry produces a certain number of pottery articles in a day. It was observed on a particular day that the cost of production of each article (in rupees) was 3 more than twice the number of articles produced on that day. If the total cost of production on that day was Rs. 90, find the number of articles produced and the cost of each article.
Answer:
Let the number of articles produced be x.
Then the cost of each article = 2x + 3
Total cost of the articles produced = x [2x + 3] = 2x² + 3x
By problem 2x² + 3x = 90
⇒ 2x² + 3x – 90 = 0
⇒ 2x² + 15x – 12x – 90 = 0
⇒ x (2x + 15) – 6 (2x + 15) = 0
⇒ (2x + 15) (x – 6) = 0
⇒ 2x + 15 = 0 (or) x – 6 = 0
⇒ x = \(\frac{-15}{2}\) or x = 6
But x can’t be negative.
∴ x = 6
2x + 3 = 2 × 6 + 3 = 15
∴ Number of articles produced = 6 Cost of each article = Rs. 15.

Question 6.
Find the dimensions of a rectangle whose perimeter is 28 meters and whose area is 40 square meters.
Answer:
Let the length of the rectangle = x
Given perimeter = 2(1 + b) = 28
⇒ (1 + b) = \(\frac{28}{2}\) = 14
Breadth of the rectangle = 14 – x
Area = length . breadth = x (14 – x)
= 14x – x²
By problem, 14x – x² = 40.
⇒ x² – 14x + 40 = 0
⇒ x² – 10x – 4x + 40 = 0
⇒ x(x – 10) – 4(x – 10) = 0
⇒ (x – 10) (x – 4) = 0
⇒ x – 10 = 0 (or) x – 4 = 0
⇒ x = 10 (or) 4
∴ Length = 10 m or 4 m
Then breadth = 14 – 10 = 4 m (or) 14 – 4 = 10 m

Question 7.
The base of a triangle is 4 cm longer than its altitude. If the area of the triangle is 48 sq.cm, then find its base and altitude.
Answer:
Let the altitude of the triangle h = x cm
Then its base ‘b’ = x + 4.
Area = \(\frac{1}{2}\) × base × height
= \(\frac{1}{2}\)(x + 4)(x)
= \(\frac{x^{2}+4 x}{2}\)
By problem \(\frac{x^{2}+4 x}{2}\) = 48
⇒ x² + 4x = 2 × 48
⇒ x² + 4x – 96 = 0
⇒ x² + 12x – 8x – 96 = 0
⇒ x(x + 12) – 8(x + 12) = 0
⇒ (x + 12)(x – 8) = 0
⇒ x + 12 = 0 (or) x – 8 = 0
⇒ x = -12 (or) x = 8
But x can’t be negative.
∴ x = 8 and x + 4 = 8 + 4 = 12
Hence altitude = 8 cm and base = 12 cm.

Question 8.
Two trains leave a railway station at the same time. The first train travels towards west and the second train towards north. The first train travels 5 km/hr faster than the second train. If after two hours they are 50 km. apart, find the average speed of each train.
Answer:
Let the speed of the slower train = x kmph
Then speed of the faster train = x + 5 kmph.

Distance = Speed × Time
Distance travelled by the first train = 2(x + 5) = 2x + 10
Distance travelled by the second train = 2.x = 2x
By Pythagoras Theorem
(hypotenuse)² = (side)² + (side)²
⇒ (2x)² + (2x + 10)2² = 50²
⇒ 4x² + (4x² + 40x + 100) = 2500
⇒ 4x² + 4x² + 40x + 100 = 2500
⇒ 8x² + 40x – 2400 = 0
⇒ x² + 5x – 300 = 0
⇒ x² + 20x – 15x – 300 = 0
⇒ x (x + 20) – 15 (x + 20) = 0
⇒ (x + 20) (x – 15) = 0
∴ x – 15 = 0 (or) x + 20 = 0
⇒ x = 15 (or) – 20
But x can’t be negative.
∴ Speed of the slower train x = 15 kmph.
Speed of the faster train x + 5 = 15 + 5 = 20 kmph.

Question 9.
In a class of 60 students, each boy contributed rupees equal to the number of girls and each girl contributed rupees equal to the number of boys. If the total money then collected was Rs. 1600, how many boys are there in the class?
Answer:
Let the number of boys in the class = x
Then number of girls in the class = 60 – x [∵ total students = 60]
Money contributed by the boys = x(60 – x) = 60x – x² [∵ given]
Money contributed by the girls = (60 – x)x = 60x – x²
∴ Money contributed by the class = 120x – 2x²
By problem 120x -2x² = 1600
⇒ 2x²– 120x + 1600 = 0
⇒ x² – 60x + 800 = 0
⇒ x² – 40x – 20x + 800 = 0
⇒ x(x – 40) – 20 (x – 40) = 0
⇒ (x – 40) (x – 20) = 0
⇒ x = 40 (or) 20
∴ Boys = 40 or 20 Girls = 20 or 40.

Question 10.
A motor boat heads upstream a distance of 24 km on a river whose current is running at 3 km per hour. The trip up and back takes 6 hours. Assuming that the motor boat maintained a constant speed, what was its speed ?
Answer:
Let the speed of the boat in still water be x kmph.
Speed of the current = 3 kmph
Then speed of the boat in upstream = (x – 3) kmph
Speed of the boat in downstream = (x + 3) kmph
By problem total time taken = 6h.

⇒ 24(2x) = 6(x² – 9)
⇒ 8x = x² – 9
⇒ x² – 8x – 9 = 0
⇒ x² – 9x + x-9 = 0
⇒ x (x – 9) + 1 (x – 9) = 6
⇒ (x – 9) (x + 1) = 0
⇒ x – 9 = 0 or x + 1 = 0
x can’t be negative,
∴ x = 9
i.e., speed of the boat in still water = 9 kmph.

AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 5 Quadratic Equations Ex 5.1 Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Maths Solutions 5th Lesson Quadratic Equations Exercise 5.1

10th Class Maths 5th Lesson Quadratic Equations Ex 5.1 Textbook Questions and Answers

Question 1.
Check whether the following are quadratic equations.
i) (x + l)² = 2(x-3)
Answer:
Given: (x + l)² = 2(x – 3)
⇒ x² + 2x + 1 = 2(x – 3) = 2x – 6
⇒ x² + 2x + l – 2x + 6 = 0
⇒ x² + 7 = 0 is a Q.E.

ii) x² – 2x = (-2) (3 – x)
Answer:
Given: x² – 2x = -2(3 – x)
⇒ x² – 2x = -6 + 2x
⇒ x² – 4x + 6 = 0 is a Q.E.

iii) (x-2) (x + 1) = (x- 1) (x + 3)
Answer:
Given: (x – 2) (x + 1) = (x – 1) (x + 3)
⇒ x (x + 1) – 2 (x +1)
= x (x + 3) – 1 (x + 3)
Note : Compare the coefficients of x2 on both sides. If they are equal it is not a Q.E.
⇒ x² + x – 2x – 2 = x² + 3x – x -3
⇒ x² – x – 2 = x² + 2x – 3
⇒ 3x – 1 = 0 is not a Q.E.

iv) (x – 3) (2x + 1) = x(x + 5)
Answer:
Given: (x – 3) (2x + 1) = x(x + 5)
⇒ x (2x + 1) – 3 (2x + 1) = x . x + 5 . x
⇒ 2x² + x – 6x – 3 = x² + 5x
⇒ 2x² – 5x – 3 – x² – 5x = 0
⇒ x² – 10x – 3 = 0 is a Q.E.
(or)
Comparing the coefficients of x2 on both sides.
x . 2x and x . x
⇒ 2x² and x²
2x² ≠ x²
Hence it’s a Q.E.

v) (2x – 1) (x – 3) = (x + 5) (x – 1)
Answer:
Given: (2x – 1) (x – 3) = (x + 5) (x – 1)
⇒ 2x (x – 3) -1 (x – 3) = x (x – 1) + 5(x – 1)
⇒ 2x² – 6x – x + 3 = x² – x + 5x – 5
⇒ 2x² -7x + 3 – x² – 4x + 5 = 0
⇒ x² – 11x + 8 = 0
Hence it’s a Q.E.
(or)
Co.eff. of x² on L.H.S. = 2 × 1 = 2
Co.eff. of x² on R.H.S = 1 × 1 = 1
LHS ≠ RHS Hence it is a Q.E.

vi) x² + 3x + 1 = (x – 2)²
Answer:
Given: x² + 3x + 1 = (x – 2)²
⇒ x² + 3x + 1 = x² – 4x + 4
⇒ 7x – 3 = 0 is not a Q.E.

vii) (x + 2)³ = 2x (x² – 1)
Answer:
Given: (x + 2)³ = 2x(x² – 1)
⇒ x³ + 6x² + 12x + 8 = 2x³ – 2x [∵ (a + b)³ = a³ + 3a²b + 3ab² + b³]
⇒ -x³ + 6x² + 14x + 8 = 0
is not a Q.E. [∵ degree = 3]

viii) x³ – 4x² – x + 1 = (x – 2)³
Answer:
Given : x³ – 4x² – x + 1 = (x – 2)³
⇒ x³ – 4x² – x + 1 = x³ – 6x² + 12x – 8
⇒ 6x² – 12x + 8 – 4x² – x + 1 = 0
⇒ 2x² – 13x + 9 = 0 is a Q.E.

Question 2.
Represent the following situations in the form of quadratic equations:
i) The area of a rectangular plot is 528 m². The length of the plot (in meters) is one more than twice its breadth. We need to find the length and breadth of the plot.
Answer:
Let the breadth of the rectangular plot be x m.
Then its length (by problem) = 2x + 1.
Area = l . b = (2x + 1) . x = 2x² + x
But area = 528 m² (∵ given)
∴ 2x² + x = 528
⇒ 2x² + x – 528 = 0 where x is the breadth of the rectangle.

ii) The product of two consecutive positive integers is 306. We need to find the integers.
Answer:
Let the consecutive integers be x and x + 1.
Their product = x(x + 1) = x² + x
By problem x² + x = 306
⇒ x² + x – 306 = 0
where x is the smaller integer.

iii) Rohan’s mother is 26 years older than him. The product of their ages after 3 years will be 360 years. We need to find Rohan’s present age.
Answer:
Let the present age of Rohan be x years.
Then age of Rohan’s mother = x + 26
After 3 years:
Age of Rohan would be = x + 3
Rohan’s mother’s age would be = (x + 26) + 3 = x + 29
By problem (x + 3) (x + 29) = 360
⇒ x(x + 29) + 3(x + 29) = 360
⇒ x² + 29x + 3x + 87 = 360
⇒ x² + 32x + 87 – 360 = 0
⇒ x² + 32x – 273 = 0
⇒ x² + 39x – 7x – 273 = 0
⇒ x (x + 39) – 7 (x + 39) = 0
⇒ (x – 7) (x + 39) = 0
⇒ x = 7 or x = -39 ‘x’ being age cannot be negative.
∴ x = Present age of Rohan = 7 years.

iv) A train travels a distance of 480 km at a uniform speed. If the speed had been 8km/h less, then it would have taken 3 hours more to cover the same distance. We need to find the speed of the train.
Answer:
Let the speed of the train be x km/h.
Then time taken to travel a distance of distance of 480 km = \(\frac{\text { distance }}{\text { speed }}\) = \(\frac{480}{x}\)
If the speed is 8km/h less, then time needed to cover the same distance would be \(\frac{480}{x-8}\)

⇒ x² – 8x = 1280
⇒ x² – 8x – 1280 = 0
where x is the speed of the train.

AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 6 Progressions Ex 6.5 Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Maths Solutions 6th Lesson Progressions Exercise 6.5

10th Class Maths 6th Lesson Progressions Ex 6.5 Textbook Questions and Answers

Question 1.
For each geometric progression find the common ratio ‘r’, and then find a_n.
i) 3, \(\frac{3}{2}\), \(\frac{3}{4}\), \(\frac{3}{8}\), …….
Answer:
Given G.P.: 3, \(\frac{3}{2}\), \(\frac{3}{4}\), \(\frac{3}{8}\), …….

ii) 2, -6, 18, -54, …….
Answer:
Given G.P. = 2, -6, 18, -54, …….
a = 2, r = \(\frac{a_{2}}{a_{1}}\) = \(\frac{-6}{2}\) = -3
a_n = a . r^n-1 = 2 × (-3)^n-1
∴ r = -3; a_n = 2(-3)^n-1

iii) -1, -3, -9, -27, ……
Given G.P. = -1, -3, -9, -27, ……
a = -1, r = \(\frac{a_{2}}{a_{1}}\) = \(\frac{-3}{-1}\) = 3
a_n = a . r^n-1 = (-1) × 3^n-1
∴ r = 3; a_n = (-1) × 3^n-1

iv) 5, 2, \(\frac{4}{5}\), \(\frac{8}{25}\), …….
Given G.P. = 5, 2, \(\frac{4}{5}\), \(\frac{8}{25}\), …….
a = 5, r = \(\frac{a_{2}}{a_{1}}\) = \(\frac{2}{5}\)
a_n = a . r^n-1 = 5 × \(\left(\frac{2}{5}\right)^{n-1}\)
∴ r = \(\frac{2}{5}\); a_n = 5\(\left(\frac{2}{5}\right)^{n-1}\)

Question 2.
Find the 10^th and nth term of G.P.: 5, 25, 125,…..
Answer:
Given G.P.: 5, 25, 125,…..
a = 5, r = \(\frac{a_{2}}{a_{1}}\) = \(\frac{25}{5}\) = 5
a_n = a . r^n-1 = 5 × (5)^n-1 = 5^1+n-1 = 5ⁿ
a₁₀ = a . r⁹ = 5 × 5⁹ = 5¹⁰
∴ a₁₀ = 5¹⁰; a_n = 5ⁿ

Question 3.
Find the indicated term of each geometric progression.
i) a₁ = 9; r = \(\frac{1}{3}\); find a₇.
Answer:
a_n = a . r^n-1

ii) a₁ = -12; r = \(\frac{1}{3}\); find a₆.
Answer:
a_n = a . r^n-1

Question 4.
Which term of the G.P.
i) 2, 8, 32,….. is 512?
Answer:
Given G.P.: 2, 8, 32,….. is 512
a = 2, r = \(\frac{a_{2}}{a_{1}}\) = \(\frac{8}{2}\) = 4
Let the n^th term of G.P. be 512

512 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2
= 2⁹
∴ 2n – 1 = 9
[∵ bases are equal, exponents are also equal]
∴ 2n = 9 + 1 = 10
n = \(\frac{10}{2}\) = 5
∴ 512 is the 5^th term of the given G.P.

ii) √3, 3, 3√3, …….. is 729?
Answer:
Given G.P.: √3, 3, 3√3, …….. is 729
a = √3, r = \(\frac{a_{2}}{a_{1}}\) = \(\frac{3}{\sqrt{3}}\) = √3
now a_n = a . r^n-1 = 729
⇒ (√3)(√3)^n-1 = 729
⇒ (√3)ⁿ = 3⁶ = (√3)¹²
⇒ n = 12
So 12th term of GP √3, 3, 3√3, …….. is 729.

iii) \(\frac{1}{3}\), \(\frac{1}{9}\), \(\frac{1}{27}\), ……. is \(\frac{1}{2187}\)?
Answer:
Given G.P.: \(\frac{1}{3}\), \(\frac{1}{9}\), \(\frac{1}{27}\), ……. is \(\frac{1}{2187}\)

Let \(\frac{1}{2187}\) be the n^th term of the G.P., then
a_n = a . r^n-1 = \(\frac{1}{2187}\)

[∵ bases are equal, exponents are also equal]
7^th term of G.P is \(\frac{1}{2187}\).

Question 5.
Find the 12^th term of a G.P. whose 8 term is 192 and the common ratio is 2.
Answer:
Given a G.P. such that a₈ = 192 and r = 2
a_n = a . r^n-1
a₈ = a . (2)^8-1 = 192

= 3 × 2¹⁰ = 3 × 1024 = 3072.

Question 6.
The 4^th term of a geometric progression is \(\frac{2}{3}\) and the seventh term is \(\frac{16}{81}\). Find the geometric series.
Answer:
Given: In a G.P.

Now substituting r = \(\frac{2}{3}\) in equation (1)
we get,

Question 7.
If the geometric progressions 162, 54, 18, ….. and \(\frac{2}{81}\), \(\frac{2}{27}\), \(\frac{2}{9}\),….. have their n^th term equal, find the value of n.
Answer:
Given G.P.: 162, 54, 18, ….. and \(\frac{2}{81}\), \(\frac{2}{27}\), \(\frac{2}{9}\),……

Given that n^th terms are equal
a_n = a . r^n-1

⇒ 3^n-1+n-1 = 81 × 81
⇒ 3^2n-2 = 3⁴ × 3⁴
⇒ 3^2n-2 = 3⁸ [∵ a^m . aⁿ = a^m+n]
⇒ 2n – 2 = 8
[∵ bases are equal, exponents are also equal]
2n = 8 + 2
⇒ n = \(\frac{10}{2}\) = 5
The 5^th terms of the two G.P.s are equal.