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These NCERT Solutions for Class 6 Maths Chapter 11 Algebra Ex 11.5 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 6 Maths Chapter 11 Algebra Exercise 11.5

Question 1.
State which of the following are equations (with a variable). Give reason for your answer. Identify the variable from the equations with a variable.
(a) 17 = x + 7
(b) (t – 7) > 5
(c) \(\frac{4}{2}\) = 2
(d) (7 × 3) – 19 = 8
(e) 5 × 4 – 8 = 2x
(f) x – 2 = 0
(g) 2m < 30
(h) 2n + 1 = 11
(i) 7 = (11 × 5) – (12 × 4)
(j) 7 = (11 × 2) + p
(k) 20 = 5y
(l) \(\frac{3 \mathrm{q}}{2}\) < 5
(m) z + 12 > 24
(n) 20 – (10 – 5) = 3 × 5
(o) 7 – x = 5
Answer:
(a) It is an equation of variable as both the sides are equal. The variable is x.
(b) It is not an equation as L.H.S. is greater than R.H.S.
(c) It is an equation with no variable. But it is a false equation.
(d) It is an equation with no variable. But it is a false equation.
(e) It is an equation of variable as both the sides are equal. The variable is x.
(f) It is an equation of variable x.
(g) It is not an equation as L.H.S. is less than R.H.S.
(h) It is an equation of variable as both the sides are equal. The variable is n.
(i) It is an equation with no variable as its both sides are equal.
(j) It is an equation of variable p.
(k) It is an equation of variable y.
(l) It is not an equation as L.H.S. is less than R.H.S.
(m) It is not an equation as L.H.S. is greater than R.H.S.
(n) It is an equation with no variable.
(o) It is an equation of variable x

Question 2.
Complete the entries in the third column of the table.

Answer:

Question 3.
Pick out the solution from the values given in the bracket next to each equation. Show that the other values do not satisfy the equation.
(a) 5m = 60 (10,5,12,15)
(b) n + 12= 20(12,8,20,0)
(c) p – 5 = 5 (0, 10, 5,-5)
(d) \(\frac{\mathrm{q}}{2}\) = 7 (7, 2, 10, 14)
(e) r – 4 = 0 (4,-4, 8,0)
(f) x + 4 = 2 (-2, 0, 2, 4)
Answer:
(a) 5m = 60
Putting the given values in L.H.S.,
5 × 10 = 50
∵ L.H.S. ≠ R.H.S
∴ m = 10 is not the solution
5 × 5 = 25
∵ L.H.S. ≠ R.H.S
∴ m = 5 is not the solution
5 × 12 = 60
∵ L.H.S. = R.H.S
∴ m = 12 is a solution
5 × 15 = 75
∵ L.H.S. ≠ R.H.S
∴ m = 15 is not the solution

(b) n + 12 = 20
Putting the given values in L.H.S.,
12 + 12 = 24
∵ L.H.S. ≠ R.H.S
∴ n = 12 is not the solution
8 + 12 = 20
∵ L.H.S. = R.H.S
∴ n = 8 is a solution
20 + 12 = 32
∵ L.H.S. ≠ R.H.S
∴ n = 20 is not the solution
0 + 12 = 12
∵ L.H.S. ≠ R.H.S
∴ n = 0 is not the solution

(c) p – 5 = 5
Putting the given values in L.H.S.,
0 – 5 = -5
∵ L.H.S. ≠ R.H.S
∴ p = 0 is not the solution
10 – 5 = 5
∵ L.H.S. = R.H.S
∴ p = 10 is a solution
5 – 5 = 0
∵ L.H.S. ≠ R.H.S
∴ p = 5 is not the solution
-5 -5 = -10
∵ L.H.S. ≠ R.H.S
∴ p = -5 is not the solution

(d) \(\frac{\mathrm{q}}{2}\) = 7
Putting the given values in L.H.S.,
\(\frac{\mathrm{7}}{2}\)
∵ L.H.S. ≠ R.H.S
∴ q = 7 is not the solution
\(\frac{\mathrm{2}}{2}\) = 1
∵ L.H.S. ≠ R.H.S
∴ q = 2 is not the solution
\(\frac{\mathrm{10}}{2}\) = 5
∵ L.H.S. ≠ R.H.S
∴ q = 10 is not the solution
\(\frac{\mathrm{14}}{2}\) = 7
∵ L.H.S. = R.H.S
∴ q = 14 is a solution

(e) r – 4 = 0
Putting the given values in L.H.S.,
4 – 4 = 0
∵ L.H.S. = R.H.S
∴ r = 4 is a solution
– 4 -4 = -8
∵ L.H.S. ≠ R.H.S
∴ r = – 4 is not the solution
8 – 4 = 4
∵ L.H.S. ≠ R.H.S
∴ r = 8 is not the solution
0 – 4 = – 4
∵ L.H.S. ≠ R.H.S
∴ r = 0 is not the solution

(f) x + 4 = 2
Putting the given values in L.H.S.,
-2 + 4 = 2
∵ L.H.S. = R.H.S
∴ x = -2 is a solution
0 + 4 = 4
∵ L.H.S. ≠ R.H.S
∴ x = 0 is not the solution
2 + 4 = 6
∵ L.H.S. ≠ R.H.S
∴ x = 0 is not the solution
4 + 4 = 8
∵ L.H.S. ≠ R.H.S
∴ x = 4 is not the solution

Question 4.
(a) Complete the table and by inspection of the table, find the solution to the equation m + 10 = 16.

(b) Complete the table and by inspection of the table, find the solution to the equation 5t = 35.

(c) Complete the table and find the solution of the equation z/3 = 4 using the table.

(d) Complete the table and find the solution to the equation m – 7 = 3

Answer:
(a)

∵ At m = 6, m + 10 = 16 ∴ m = 6 is the solution.

(b)

∵ At t = 7, 5t = 35 ∴ t = 7 is the solution.

(c)

∵ At z = 12, z/3 = 4 ∴ z = 12 is the solution.

(d)

∵ At m = 10, m- 7= 3 ∴ m = 10 is the solution.

Question 5.
Solve the following riddles, you may yourself construct such riddles.
Who am I?
(i) Go round a square
Counting every corner
Thrice and no more!
Add the count to me
To get exactly thirty four!

(ii) For each day of the week
Make an upcount from me
If you make no mistake
You will get twenty three!

(iii) I am a special number
Take away from me a six!
A whole cricket team
You will still be able to fix!

(iv) Tell me who I am
I shall give a pretty clue!
You will get me back
If you take me out of twenty two!
Answer:
(i) According to the condition,
1 + 12 = 34 or × + 12 = 34
∴ By inspection, we have
22 + 12 = 34
So, I am 22.

(ii) Let I am ‘×’.
We know that there are 7 days in a week.
∴ up-counting from × for 7, the sum = 23
By inspections, we have
16 + 7 = 23
∴ × = 16
Thus I am 16.

(iii) Let the special number be x and there are 11 players in cricket team.
∴ Special Number -6 = 11
× – 6 = 11
By inspection, we get
17 – 6 = 11
∴ × = 17
Thus I am 17.

(iv) Suppose I am ‘×’
∴ 22 – 1 = 1
or 22 – × = ×
By inspection, we have
22 -11 = 11
∴ × = 11
Thus I am 11.

NCERT Solutions for Class 7 English

I Want Something in a Cage NCERT Solutions for Class 7 English An Alien Hand Chapter 6

I Want Something in a Cage NCERT Text Book Questions and Answers

I Want Something in a Cage Comprehension check – I

Question 1.
Write ‘True’or ‘False’ against each of following statements.

• Mr Purcell sold birds, cats, dogs and monkeys. True
• He was very concerned about the well-being of the birds and animals in his shop. False
• He was impressed by the customer who bought the two doves. False
• He was a successful shop owner, though insensitive and cold as a person. False

Question 2.
Why is Mr Purcell compared to an owl?
Answer:
Mr Purcell used to wear large glasses that magnified his eyes so as to give him the appearance of a wise and genial owl.

Question 3.
From the third paragraph pick out
i. words associated with cries of birds,
ii. words associated with noise,
iii. words suggestive of confusion and fear.
Answer:
i. whispered twitters; squeals, cheeps, and sudden squeaks
ii. constant stir of movement; whispered, sly rustling
iii. frantic, frightened, bewildered, blindly seeking

I Want Something in a Cage Comprehension check – II

Question 1.
Do you think the atmosphere of Mr Purcell’s shop was cheerful or depressing? Give reasons for your answer.
Answer:
I think the atmosphere of Mr Purcell’s shop was depressing because the various animals that were inside the cages kept squealing, twittering and squeaking constantly. These animals were kept on display as if they were things of show, and perhaps they were making so much noise just to be let out of their cages.

The stranger who came to the pet shop was wearing shiny shoes that squeaked. His suit was cheap, ill-fitting but looked new. He had a shuttling glance and close-cropped hair. The man wanted to buy something in a cage.

Question 2.
i. The man insisted on buying the doves because he was fond of birds. Do you agree?
Answer:
No, the man was not fond of the birds. It did not matter to the man whether he was buying doves or any other bird. All he knew was that he wanted to buy birds that were in a cage. It was only when he looked at the doves in the cage that he got the idea to buy them.

ii. How had he earned the five dollars he had?
Answer:
He had earned the five dollars he had by spending ten years in the prison. He had to put in really hard labour to earn these five dollars – that is, fifty cents a year.

Question 3.
Was the customer interested in the care and feeding of the doves he had bought? If not, why?
Answer:
No, the customer was not interested in the care and feeding of the doves he had bought, because he didn’t need to. He had something even more valuable to offer them. The man wanted to free the birds right after he had bought them.

I Want Something in a Cage Exercises Question and Answer 

Discuss the following topics in groups.

Question 1.
Why, in your opinion, did the man set the doves free?
(Encourage the students to express their opinions clearly in groups.)
Answer:
I believe that the man set the doves free because he had begun to understand the real value of freedom after spending ten years in prison. He knew what it felt to be imprisoned and not have the liberty to spread your wings far and wide. He wanted the birds to take their new flight to freedom. He gave the birds what he himself valued the most.

Question 2.
Why did it make Mr Purcell feel “vaguely insulted’’?
Answer:
Mr Purcell felt “vaguely insulted” because he was making money by selling the caged birds and other animals. And here was a man who was willing to give up all his hard-earned money of ten years to buy the birds their freedom. Mr Purcell must have felt very small in comparison to the man. His actions made him think and realize the inherent cruelty of his profession.

In other words, while Mr Purcell was taking away the animals’ right to live freely, here was a man who was ready to give away everything he owned just so he could give the birds back their right.

These NCERT Solutions for Class 6 Maths Chapter 6 Integers Ex 6.3 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 6 Maths Chapter 6 Integers Exercise 6.3

Question 1.
Find:
(a) 35 – (20) = ……………..
(b) 72-(90) = ………………
(c) (-15)-(-18) = ………………..
(d) (-20)-(13) = ………………….
(e) 23-(-12) = ……………………
(f) (-32) – (-40) = ……………………..
Answer:
(a) 35 – (20) = 15
(b) 72-(90) = -18
(c) (-15)-(-18) = -15+ 18 = 3
(d) (-20)-(13) = -20- 13 = -33
(e) 23-(-12) = 23 + 12 = 35
(f) (-32) – (-40) = -32 + 40 = 8

Question 2.
Fill in the blanks with >, < or = sign:
Answer:
(a) (-3) + (-6) ………….. (-3) – (-6)
(b) (-21) – ( – 10) ………….. (-31) + (-11)
(c) 45 – (-11) ………….. 57 +(-4)
(d) (-25) – (-42) ………….. (-42) – (-25)
Answer:
(a) (-3) + (-6) < (-3) – (-6)
(b) (-21) – ( – 10) > (-31) + (-11)
(c) 45 – (-11) > 57 +(-4)
(d) (-25) – (-42)_>_ (-42) – (-25)

Question 3.
Fill in the blanks:
(a) (-8) + ……….. = 0
(b) 13 + ……….. =0
(c) 12 + ……….. = 0
(d) (-4) + ……….. =-12
(e) ……….. – 15 = -10
Answer:
(a) (-8) + 8 = 0
(b) 13+ (-13) =0
(c) 12+ (-12) =0
(d) (-4) + (-8) =-12
(e) 5 – 15 = -10

Question 4.
Find:
(a) (-7)-8-(-25)
(b) (-13)+ 32-8-1
(c) (-7) + (-8) + (-90)
(d) 50 – (-40) – (-2)
Answer:
(a) (-7) – 8 – (-25) = -7 – 8 + 25
= -15 + 25= 10
(b) (-13)+ 32-8-1
= -13 + 32 – 8 – 1 = 32 -22 = 10
(c) (-7) + (-8) + (-90) = -7 – 8 – 90 = -105
(d) 50 – (-40) – (-2) = 50 + 40 + 2 = 92

These NCERT Solutions for Class 6 Maths Chapter 6 Integers Ex 6.2 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 6 Maths Chapter 6 Integers Exercise 6.2

Question 1.
Using the number line write the integer which is:
(a) 3 more than 5
(b) 5 more than -5
(c) 6 less than 2
(d) 3 less than -2
Answer:
(a) 8

Question 2.
Use number line and add the following integers:
(a) 9 +(-6)
(b) 5 + (-11)
(c) (-1) + (-7)
(d) (-5) + 10
(e) (-1) + (-2) + (-3)
(f) (-2)+ 8 +(-4)
Answer:
(a) 9 + (-6) = 3

(b) 5 + (-11) = -6

(c) ( – 1) + ( – 7) = -8

(d) (-5) + 10 = 5

(e) ( – 1) + ( – 2) + ( – 3) = -6

(f) ( – 2) + 8 + ( – 4) = 2

Question 3.
Add without using number line:
(a) 11 +(-7)
(b) (-13)+ (+18)
(c) (-10)+ (+19)
(d) (-250) + (+ 150)
(e) (-380) + (-270)
(f) (-217) + (-100)
Answer:
(a) 11 +(-7) = 11-7 = 4
(b) (-13)+ 18 = 5
(c) (-10)+ (+19) = -10+19 = 9
(d) (-250) + (+150) = -250 + 150 = -100
(e) (-380) + (-270) = -380 – 270 = -650
(f) (-217)+ (-100) = -217-100 = -317

Question 4.
Find the sum of:
(a) 137 and -354
(b) -52 and 52
(c) -312,39 and 192
(d) -50,-200 and 300
Answer:
(a) 137 + (-354) = 137 – 354 = -217
(b) (-52)+ 52 = 0
(c) (-312)+ 39 +192
= -312 + 231 = -81
(d) (-50) + (-200) + 300
= -50 – 200 + 300 = -250 + 300 = 50

Question 5.
Find the sum:
(a) (-7) + (-9) + 4 + 16
(b) (37) + (-2) + (-65) + (-8)
Answer:
(a) (-7) + (-9) + 4 + 16
= -7-9 + 4+ 16 =-16 + 20 = 4
(b) 37 +(-2) + (-65) + (-8)
= 37 – 2 – 65 – 8 = 37 – 75 = -38

NCERT Solutions for Class 7 English

Golu Grows a Nose NCERT Solutions for Class 7 English An Alien Hand Chapter 5

Golu Grows a Nose NCERT Text Book Questions and Answers

Golu Grows a Nose Exercises Question and Answer

Answer the following questions.

Question 1.
Whom does Golu ask, “Why don’t you ever fly like other birds? ”
Answer:
Golu posed this question to the ostrich.

Question 2.
Which uncle of Golu had red eyes?
Answer:
Golu’s uncle, the hippopotamus, had red eyes.

Question 3.
Golu relatives did not answer his questions because
Answer:
i. they were shy.
ii. the questions were too difficult.
iii. Golu was a naughty baby.
Answer:
ii. the questions were too difficult.

Question 4.
Who advised Golu to go to the Limpopo river?
Answer:
The mynah advised Golu to go to the Limpopo river.

Question 5.
Why did Golu go to the river?
Answer:
Golu was a curious child. He went to the river to meet a crocodile and ask him what crocodiles have for dinner.

Question 6.
The crocodile lay on the bank of the Limpopo river. Golu thought it was
Answer:

• a living crocodile.
• a dead crocodile.
• a log of wood.
• a log of wood.

Question 7.
What did the crocodile do to show that it was a real crocodile?
Answer:
The crocodile shed crocodile tears in front of Golu to show that it was a real crocodile.

Question 8.
“Come here, little one, and I’ll whisper the answer to you.” The crocodile said this because
i. he couldn’t stand up.
ii. he wanted to eat Golu.
iii. Golu was deaf.
Answer:
ii. he wanted to eat Golu.

Question 9.
Who helped Golu on the bank of the river?
Answer:
It was the python who Golu had met on his way to the Limpopo river who helped Golu on the river bank. He advised Golu to pull his nose as much as he can and even helped him in doing so.

Question 10.
Name two things the elephant can do with his trunk, and two he cannot.
Answer:
The elephant can lift his nose to keep the flies away and also pluck grass or bananas with his trunk to eat them. Two things that he cannot do with his trunk include eating and drinking.

These NCERT Solutions for Class 9 Maths Chapter 9 Areas of Parallelograms and Triangles Ex 9.2 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 9 Maths Chapter 9 Areas of Parallelograms and Triangles Exercise 9.2

Question 1.
In fig. 9.15, ABCD is a parallelogram and AE ⊥ DC and CF ⊥ AD. If AB = 16 cm, AE = 8 cm and CF = 10 cm. Find AD.

Solution:
In this fig. we have given AB = 16 cm, AE = 8 cm, CF = 10 cm and ABCD is a parallelogram.
∴ AB = CD (opp side of parallelogram)
∴ CD = 16 cm
Again we know that,
Area of parallelogram = Base × Corresponding altitude
∴ Area of parallelogram ABCD = CD × AE
= 16 × 8
= 128 cm² …..(i)
Again Area of parallelogram = AD × CF
128 = AD × 10 (from equ. (i))
AD = \(\frac{128}{10}\) = 12.8 cm
AD = 12.8 cm

Question 2.
If E, F, G and H are respectively the mid points of the sides of a parallelogram ABCD. show that ar(EFGH) = \(\frac {1}{2}\) ar(ABCD).
Solution:
Given: ABCD is a parallelogram, in which E, F, G, and H are the midpoints of sides AB, BC, CD, and AD respectively.

To prove that
ar(||gm EFGH) = \(\frac {1}{2}\) ar (||gm ABCD)
Construction: Join HF
Proof: ABCD is a ||gm
∴ AD || BC and AD = BC
∴ AH = BF and AH || BF
Therefore, ABFH is a ||gm.
Since, ∆HEF and || gm HABF are on the same base HF and between the same parallel lines, AB and HF
∴ ar(∆HEF) = \(\frac {1}{2}\) ar(||gm HABF) …….(i)
Similarly, ∆HGF and ||gm HDCF are on same base HF and between the same parallel lines, DC and HF.
∴ ar (∆HGF) = \(\frac {1}{2}\) ar (|| gm HDCF) …….(ii)
Adding (i) and (ii) we get
ar(∆HEF) + ar(∆HGF)
= \(\frac {1}{2}\) ar(|| gm HABF) + \(\frac {1}{2}\) ar(||gm HDCF) + ar (|| gm FFGH)
= \(\frac {1}{2}\) [ar(||gm HABF) + ar (||gm HDCF)]
or, ar (||gm EFGH) = \(\frac {1}{2}\) ar (|| gm ABCD)

Question 3.
P and Q are any two points lying on the sides DC and AD respectively of a parallelogram ABCD shows that ar(∆APB) = ar(∆BQC).
Solution:
Given: ABCD is a || gm in which P and Q are the points on side DC and AD respectively.
To prove that: ar(∆APB) = ar(∆BQC)
Proof: ABQC and || gm ABCD lie on the same base BC and between the same parallel lines.

∴ ar (∆BQC) = \(\frac {1}{2}\) ar(||gm ABCD) ……(i)
Similarly, ∆APB and ||gm ABCD lie on the same base AB and between same parallel lines.
∴ ar(∆APB) = \(\frac {1}{2}\) ar (||gm ABCD) ……(ii)
From equation (i) and (ii), we can say
ar(∆BQC) = ar(∆APB)

Question 4.
In fig. 9.16, P is a point in the interior of a parallelogram ABCD. Show that
(i) ar(∆APB) + ar(∆PCD) = \(\frac {1}{2}\) ar (||gm ABCD)
(ii) ar(∆APD) + ar(∆PBC) = ar(∆APB) + ar(∆PCD)

Solution:
Given: ABCD is a || gm in which P is any point interior to it.
To prove that:
(i) ar(∆APB) + ar(∆PCD) = \(\frac {1}{2}\) ar (||gm ABCD)
(ii) ar (∆APD) + ar(∆PBC) = ar(∆APB) + ar(∆PCD)

Construction: Through P draw a line parallel to AB which interacts with AD at M and BC at N.
Proof: In parallelogram ABCD,
AB || CD (opp sides of a ||gm)
But, AB || MN (by construction)
∴ MM || CD
In ABNM
AB || MN and AM || BN (opp sides of || gm)
∴ ABNM is a || gm
Again, ∆APB and || ABNM lie on the same base AB and between the same parallel lines AB and MN.
∴ ar (∆APB) = \(\frac {1}{2}\) ar (|| gm ABNM) ……(i)
Similarly, MNCD is a parallelogram.
Since ∆PCD and || gm MNCD lie on same base CD and between same parallel lines CD and MN.
Therefore, ar (∆PCD) = \(\frac {1}{2}\) ar (|| gm MNCD) ……(ii)
Adding (i) and (ii) we get,
ar(∆APB) + ar(∆PCD) = \(\frac {1}{2}\) [ar (|| gm ABNM) + ar (|| gm MNCD)]
∴ ar(∆APB) + ar(∆PCD) = \(\frac {1}{2}\) [ar (|| gm ABCD) ……(iii)

(ii) Similarly, itive draw a line through P and parallel to D.
Then, ar(∆APD) + ar(∆PBC) = \(\frac {1}{2}\) ar (|| gm ABCD) ……(iv)
From equation (iii) and (iv) we can say
ar(∆APP) + ar(∆PBC) = ar(∆APB) + ar(∆PCD)

Question 5.
In fig. 9.17, PQRS and ABRS are parallelograms and X is any point on side BR. Show that
(i) ar (|| gm PQRS) = ar (|| gm ABRS)
(ii) ar (∆AXS) = \(\frac {1}{2}\) ar (|| gm PQRS)

Solution:
Parallelograms PQRS and ABRS are lying on the same base SR and in between the same parallel lines SR and PB.

We know that parallelograms lie on the same base and between the same parallel lines are equal in area,
∴ ar(|| gm PQRS) = ar(|| gm ABRS) ……(i)

(ii) Again, ∆AXS and || gm ABRS lie on the same base AS and between the same parallel lines AS and BR.
∴ ar (∆AXS) = \(\frac {1}{2}\) ar (|| gm ABRS)
or, ar (∆AXS) = \(\frac {1}{2}\) ar (|| gm PQRS) (from equ. (i))

Question 6.
A farmer was having a field in the form of a parallelogram PQRS. She took any point A on RS and joined it to point P and Q. In how many parts the field is divided? What is the shape of these parts? The fanner wants to saw wheat and pulses in equal portions of the field separately. How should she do it?
Solution:
According to the question, fields is divided into three-part. The shape of the three parts is triangular as ∆PSA, ∆PAQ, and ∆QRA.

Again, according to the question, the farmer wants to sow wheat and pulses in an equal portion of the field separately.
We know that if a parallelogram and triangle are on the same base and between the same parallels, then the area of the triangle is half the area of the parallelogram.
∴ ar(∆PAQ) = \(\frac {1}{2}\) ar (|| gm PQRS)
or, ar(∆PAQ) = ar(∆PSA) + ar(∆QRA)
So, farmers must sow wheat in triangular parts PAQ and pluses in other two triangular parts PSA and QRA, or, pulses in triangular parts PAQ and wheat in other two triangular parts PSA and QRA.