Class 8

AP State Syllabus AP Board 8th Class Maths Solutions Chapter 11 Algebraic Expressions Ex 11.5 Textbook Questions and Answers.

AP State Syllabus 8th Class Maths Solutions 11th Lesson Algebraic Expressions Exercise 11.5

Question 1.
Verify the identity (a + b)² ≡ a² + 2ab + b² geometrically by taking
(i) a = 2 units, b = 4 units
(ii) a = 3 units, b = 1 unit
(iii) a = 5 units, b = 2 unit
Solution:
(i)

= 4 × 4 + 4 × 2 + 2 × 2 + 4 × 2
= 16+ 8 + 4 + 8 = 36 sq.units
[∵ (2 + 4)² = 6² = 36]

(ii)

Area of a square AEGI
= area of square ABCD + area of rectangle CDEF + area of square CFGH + area of rectangle BIHC.
= 3 × 3 + 3 × 1 + 1 × 1+3 × 1
= 9 + 3 + 1 + 3
= 16 sq. units
[∵ (3 + 1)2 = 4² = 16]

(iii)

= 5 × 5 + 2 × 5 + 2 × 2 + 5 × 2
= 25 + 10 + 4 + 10
= 49 sq.units
[∵ (5 + 2)² = 7² = 49]

Question 2.
Verify the identity (a – b)² ≡ a² – 2ab+ b² geometrically by taking
(i) a = 3 units, b= 1 unit
(ii) a = 5 units, b = 2 units
Solution:
(i)

Area of AIFE + Area of FGCH = (a – b)² = a² – 2ab + b² [area of AIFE – area of IBGF – area of EFHD + area of FGCH]
= 3 × 3 – 1 × 3 – 3 × 1 + 1 × 1
= 9 – 3 – 3 + 1 = 4
∴ (a – b)² = 4 sq. units
[∵ (3 – 1 )² = 2² = 4]

ii)

∴ (a – b)² = a² – 2ab + b²
Area of ABCD + Area of CYZS = a² – 2ab + b²
area of ABCD – area of BXYC – area of DCST + area of CYZS
=5 × 5 – 2 × 5 – 2 × 5 + 2 × 2
= 25 – 10 – 10 + 4
= 9 sq.units
[∵ (5 – 2)² = (3)² = 9]

Question 3.
Verify the identity(a + b)(a – b) ≡ a² – b² geometrically by taking
(i) a = 3 units, b = 2 units
(ii) a = 2 units, b = 1 unit
Solution:
i)

a² – b² = Area of Fig I. + Area of Fig II
= a(a – b) + b(a – b)
= (a – b) (a + b)
= 3 × 3 – 2 × 2
a² – b² = 9 – 4= 5sq . units
[ ∵ 3² – 2² = 9 – 4 = 5]

ii)

a² – b² = Area of Fig I. + Area of Fig II
= a(a – b) + b(a – b)
= (a + b) (a – b)
=(2 + 1)(2 – 1)
= 3 × 1 = 3
a² – b² = 3 sq. units
[∵ (2² – 1²) = 4 – 1 = 3]

AP State Syllabus AP Board 8th Class Maths Solutions Chapter 11 Algebraic Expressions Ex 11.4 Textbook Questions and Answers.

AP State Syllabus 8th Class Maths Solutions 11th Lesson Algebraic Expressions Exercise 11.4

Question 1.
Select a suitable identity and find the following products
(i) (3k + 4l)(3k + 4l)
(ii) (ax² + by²)(ax² + by²)
(iii) (7d – 9e)(7d – 9e)
(iv) (m² – n²)(m² + n²)
(v) (3t + 9s) (3t – 9s)
(vi) (kl – mn) (kl + mn)
(vii) (6x + 5)(6x + 6)
(viii) (2b – a)(2b +c)
Solution:
(3k + 4l) (3k 4l) = (3k + 4l)² is in the form of (a + b)².
=(3k)² + 2 × 3k × 4l+ (4l)² [ (a+ b)² = a² + 2ab + b²
= 3k × 3k + 24kl + 4l × 4l
= 9k² + 24kl + 16l²

ii) (ax² + by²) (ax² + by²) = (ax² + by²)² is in the form of (a + b)².
= (ax²)² + 2 × ax² × by² + (by²)² [ ∵ (a + b)² = a² + 2ab + b²]
= ax² × ax² + 2abx²y² + by² × by²
= a²x⁴ + 2ab x²y² + b²y⁴

iii) (7d – 9e) (7d – 9e)
= (7d – 9e)² is in the form of (a – b)².
= (7d)² – 2 × 7d × 9e + (9e)² [ ∵ (a – b)² = a² – 2ab + b²]
= 7d × 7d – 126de + 9e × 9e
= 49d² – 126de + 81e²

iv) (m² – n²) (m² + n²) is in the form of (a + b) (a – b).
∴ (a + b) (a – b) = a² – b²
∴ (m² + n²) (m² – n²) = (m²)² – (n²)² = m⁴ – n⁴

v) (3t + 9s) (3t – 9s) = (3t)² – (9s)² [ ∵ (a + b) (a – b) = a² – b² ]
= 3t × 3t – 9s × 9s
= 9t² – 81s²

vi) (kl – mn) (kl + mn) = (kl)² – (mn)² [ ∵(a + b) (a – b) = a² – b² ]
= kl × kl – mn × mn
= k²l² – m²n²

vii) (6x + 5) (6x + 6) is in the form of
(ax + b) (ax + c).
(ax + b) (ax + c) = a²x² + ax(b + c) + bc
(6x + 5) (6x + 6) = (6)²x² + 6x (5 + 6) + 5 × 6
= 36x² + 6x × 11 + 30
= 36x² + 66x + 30

viii) (2b – a) (2b + c) is in the form of (ax – b) (ax + c).
(ax – b) (ax + c) = a²x² + ax(c – b) – cb
(2b – a) (2b + c) = (2)²(b)² + 2b (c – a) – ca
= 4b² + 2bc – 2ab – ca

Question 2.
Evaluate the following by using suitable identities:
(i) 304²
(ii) 509²
(iii) 992²
(iv) 799²
(v) 304 × 296
(vi) 83 × 77
(vii) 109 × 108
(viii) 204 × 206
Solution:
i) 304² = (300 + 4)² is in the form of (a + b)².
∵ (a+b)² = a² + 2ab + b²
a = 300, b = 4
(300 + 4)² = (300)² + 2 × 300 × 4 + (4)²
= 300 × 300+ 2400 + 4 × 4
= 90,000 + 2400 + 16
= 92,416

ii) 509² = (500 + 9)²
a  = 500, b = 9
= (500)² + 2 × 500 × 9 + (9)²
[ ∵ (a + b)² = a² + 2ab + b²]
= 500 × 500 + 9000 + 9 × 9
= 2,50,000 + 9000 + 81
= 2,59,081

iii) 992² = (1000 – 8)²
a = 1000, b = 8
= (1000)² – 2 × 1000 × 8 + (8)² [∵ (a-b)² = a² – 2ab + b²]
= 1000 × 1000 – 16,000 + 8 × 8
= 10,00,000 – 16000 + 64
= 10,00,064 – 1600
= 9,98,464

iv) 799² = (800 – 1)²
a = 800, b = 1
= (800)² – 2 × 800 × 1 + (1)²
= 800 × 800 – 1600 + 1
= 6,40,000 – 1600 + 1
= 6,40,001 – 1600
= 6,38,401

v) 304 × 296 = (300 + 4) (300 – 4) is in the form of (a + b) (a – b).
(a + b) (a – b) = a² – b²
∴ (300 + 4) (300 – 4) = (300)² – (4)²
= 300 × 300 – 4 × 4
= 90,000 – 16
= 89,984

vi) 83 × 77 = (80 + 3) (80 – 3)
= (80)² – (3)² [ ∵ (a + b) (a – b) = a² – b²]
= 80 × 80 – 3 × 3
= 6400 – 9
= 6391

vii) 109 × 108 = (100 + 9) (100 + 8)
= (100)² + (9 + 8)100 + 9 × 8
= 10,000 + 1700 + 72
= 11,772

viii) 204 × 206 = (205 – 1) (205 + 1)
= (205)² – (1)² [∵ (a + b)(a-b) = a² – b²]
= 205 × 205 – 1 × 1
= 42,025 -1
= 42,024

AP State Syllabus 8th Class Maths Solutions 3rd Lesson Construction of Quadrilaterals InText Questions

AP State Syllabus AP Board 8th Class Maths Solutions Chapter 3 Construction of Quadrilaterals InText Questions and Answers.

8th Class Maths 3rd Lesson Construction of Quadrilaterals InText Questions and Answers

Do this

Question 1.
Take a pair of sticks of equal length, say 8 cm. Take another pair of sticks of equal length, say 6 cm. Arrange them suitably to get a rectangle of length 8 cm and breadth 6 cm. This rectangle is created with the 4 available measurements. Now just push along the breadth of the rectangle. Does it still look alike? You will get a new shape of a rectangle Fig (ii), observe that the rectangle has now become a parallelogram. Have you altered the lengths of the sticks? No! The measurements of sides remain the same.

Give another push to the newly obtained shape in the opposite direction; what do you get? You again get a parallelogram again, which is altogether different Fig (iii). Yet the four measurements remain the same. This shows that 4 measurements of a quadrilateral cannot determine its uniqueness. So, how many measurements determine a unique quadrilateral? Let us go back to the activity!
You have constructed a rectangle with two sticks each of length 8 cm and other two sticks each of length 6 cm. Now introduce another stick of length equal to BD and put it along BD (Fig iv). If you push the breadth now, does the shape change? No!
It cannot, without making the figure open. The introduction of the fifth stick has fixed the rectangle uniquely, i.e., there is no other quadrilateral (with the given lengths of sides) possible now. Thus, we observe that five measurements can determine a quadrilateral uniquely. But will any five measurements (of sides and angles) be sufficient to draw a unique quadrilateral? (Page No. 60)
Answer:
Yes, any 5 individual measurements are needed to construct a quadrilateral.

Question 2.
Equipment (Page No. 61)
You need: a ruler, a set square, a protractor.
Remember: To check if the lines are parallel.
Slide set square from the first line to the second line as shown in adjacent figures.

Now let us investigate the following using proper instruments. For each quadrilateral,
a) Check to see if opposite sides are parallel.
b) Measure each angle.
c) Measure the length of each side.

Record your observations and complete the table below.
Answer:

Question 3.
Can you draw the angle of 60°?    (Page No. 63)

Answer:
Using a scale and compass,
we can construct 60°.

Question 4.
Construct the parallelogram above (Refer text book page no: 75) BELT by using other properties of parallelogram. (Page No. 75)
Answer:

We can construct a parallelogram using the measurements of a side, a diagonal and an angle.
BE = 5 cm ⇒ LT = 5 cm
∠B = 110° ⇒ ∠E = 180° – 110° = 70°
TE= 7.2 cm

Try these

Question 1.
Can you draw a parallelogram BATS where BA = 5 cm, AT = 6 cm and AS = 6.5 cm?    (Page No. 70)
Answer:

In a parallelogram BATS, opposite sides are equal.
BA = ST = 5 cms
AT = BS 6 cms
AS = 6.5 cms

∴ So, we can construct BATS parallelogram. It needs only three measurements.

Question 2.
A student attempted to draw a quadrilateral PLAY given that PL = 3 cm, LA = 4 cm, AY = 4.5 cm, PY = 2 cm and LY = 6 cm. But he was not able to draw it why ?
Try to draw the quadrilateral yourself and give reason. (Page No. 70)
Answer:
In a quadrilateral PLAY
PL = 3 cm LA = 4 cm AY = 4.5 cm
PY = 2 cm LY = 6 cm
Here YP + PL < YL [∵ 2 + 3 < 6 ⇒ 5 < 6]

But in a △YPL, the sum of two sides is less than the third side.
∴ We are unable to construct a quadrilateral PLAY [∵ YL > YP]
[∵ The arcs do not intersect which are drawn from L and P, also Y, P, L are collinear points]

Think, discuss and write

Question 1.
Is every rectangle a parallelogram? Is every parallelogram a rectangle?    (Page No. 63)
Answer:
Yes, every rectangle is a parallelogram. But every parallelogram is not a rectangle.

Question 2.
Uma has made a sweet chikki. She wanted it to be rectangular. In how many different ways can she verify that it is rectangular?    (Page No. 63)
Answer:
If the sweet chikki is to made into a rectangular shape, she has to verify the following shapes:

1. Quadrilateral
2. Trapezium
3. Parallelogram

Question 3.
Can you draw the quadrilateral ABCD with AB = 4.5 cm, BC = 5.2 cm, CD = 4.8 cm and diagonals AC = 5 cm, BD = 5.4 cm by constructing △ABD first and then fourth vertex ‘C’ ? Give reason.       (Page No. 72)
Answer:
We cannot construct △ABD. So, if we start first from △ABD, it is impossible to construct □ ABCD.
[∵ The length of \(\overline{\mathrm{AD}}\) is not given]

Question 4.
Construct a quadrilateral PQRS with PQ = 3 cm, RS = 3 cm, PS = 7.5 cm, PR = 8 cm and SQ = 4 cm. Justify your result.      (Page No. 72)
Answer:

PQ = 3 cm
RS = 3 cm
PS = 7.5 cm
PR = 8 cm
SQ = 4 cm
With the given measurements △PQS is not possible to construct.
∵ PQ + QS < PS
The arcs which drawn from P and Q are not intersecting.
∴ We can’t obtain vertex ‘S’.
∴ Without vertex ‘S’ we can’t get a quadrilateral PQRS.

Question 5.
Can you construct the quadrilateral PQRS, if we have an angle of 100° at P instead of 75°? Give reason. (Page No. 74)
Answer:

PQ = 4 cm,
QR = 4.8 cm,
ZP = 100°,
ZQ = 100°,
ZR = 120°
∴ We can construct a quadrilateral with the given measurements.
Since the sum of 4 angles is equal to 360°.

Question 6.
Can you construct the quadrilateral PLAN if PL = 6 cm, ∠A = 9.5 cm, ∠P = 75°, ∠L = 15° and ∠A = 140°?
(Draw a rough sketch in each case and analyse the figure). State the reasons for your conclusion. (Page No. 74)
Answer:
PL = 6 cm, ∠A = 9.5 cm, ∠P = 75°, ∠L = 15°, ∠A = 140°

∴ With the given measurements it is not possible to construct a quadrilateral.

Question 7.
Do you construct the given quadrilateral ABCD with AB = 5 cm, BC = 4.5 cm, CD = 6 cm, ∠B = 100°, ∠C = 75° by taking BC as base instead of AB? If so, draw a rough sketch and explain the various steps involved in the construction. (Page No. 77)
Answer:
AB = 5 cm, BC = 4.5 cm, CD = 6 cm, ∠B = 100°, ∠C = 75°

Construction Steps:

1. Construct a line segment with radius 4.5 cms as \(\overline{\mathrm{BC}}\)
2. With the centres B and C draw two rays with 100°, 75° respectively.
3. With the centres B and C, two arcs are drawn with radius 5 cm and 6 cm respectively. The arcs and the rays are intersected.
4. Let the intersecting points be keep as A, D.
5. Join A, D.
6. ∴ ABCD quadrilateral is formed.
   

Question 8.
Can you construct the given AC = 4.5 cm and BD = 6 cm quadrilateral (rhombus) taking BD as a base instead of AC? If not give reason. (Page No. 79)
Answer:

We can construct a rhombus taking BD as base instead of base AC.

Question 9.
Suppose the two diagonals of this rhombus are equal in length, what figure do you obtain? Draw a rough sketch for it. State reasons. (Page No. 79)
Answer:
In a rhombus if the two diagonals are equal then it becomes a square.
∴ ABCD is a square.
[∵ AB = BC = CD = DA Also AC = BD]

AP State Syllabus AP Board 8th Class Maths Solutions Chapter 11 Algebraic Expressions Ex 11.3 Textbook Questions and Answers.

AP State Syllabus 8th Class Maths Solutions 11th Lesson Algebraic Expressions Exercise 11.3

Question 1.
Multiply the binomials:
(i) 2a – 9 and 3a + 4
(ii) x – 2y and 2x – y
(iii) kl + lm and k – l
(iv) m² – n² and m + n
Solution:
i) 2a – 9 and 3a + 4
(2a – 9) (3a + 4) = 2a (3a + 4) – 9(3a + 4)
= 6a² + 8a – 27a – 36
= 6a² – 19a – 36

ii) x – 2y and 2x – y
(x – 2y) (2x – y) = x(2x – y) – 2y(2x – y)
= 2x² – xy – 4xy + 2y²
= 2x² – 5xy + 2y²

iii) kl + lm and k – l
(kl + lm) (k – l) = kl(k – l) + lm(k – l)
= k²l – l²k + klm – l²m

iv) m² – n² and m + n
(m² – n²) (m + n) = m²(m + n) – n²(m + n)
= m³ + m²n – n²m – n³

Question 2.
Find the product:
(i) (x + y)(2x – 5y + 3xy)
(ii) (mn – kl + km) (kl – lm)
(iii) (a – 2b + 3c)(ab² – a²b)
(iv) (p³ + q³)(p – 5q+6r)
Solution:
i) (x + y) (2x – 5y + 3xy)
= x(2x – 5y + 3xy) + y(2x – 5y + 3xy)
= 2x² – 5xy + 3x²y + 2xy – 5y² + 3xy²
= 2x² – 5y² – 3xy + 3x²y + 3xy²

ii) (mn – kl + km) (kl – lm)
= kl(mn – kl + km) – lm(mn – kl + km)
= klmn – k²l² + k²lm – lm²n + kl²m – klm²

iii) (a – 2b + 3c) (ab² – a²b) = a(ab² – a²b) – 2b(ab² – a²b) + 3c(ab²– a²b)
= a²b² – a³b – 2ab³ + 2a²b² + 3ab²c – 3a²bc
= 3a²b² – a³b – 2ab³ + 3ab²c – 3a²bc

iv) (p³ + q³) (p – 5q + 6r) = p³(p – 5q + 6r) + q³(p – 5q + 6r)
= p⁴ – 5p³q + 6p³r + pq³ – 5q⁴ + 6rq³
= p⁴ – 5q⁴ – 5p³q + 6p³r + pq³ + 6rq³

Question 3.
Simplify the following:
(i) (x-2y) (y – 3x) + (x+y) (x-3y) – (y – 3x) (4x – 5y)
(ii) (m + n) (m² – mn + n²)
(iii) (a – 2b + 5c) (a – b) – (a – b – c) (2a + 3c) + (6a + b) (2c – 3a – 5b)
(iv) (pq-qr-i-pr) (pq-i-qr) – (pr-i-pq) (p-i-q – r)
Solution:
i) (x – 2y) (y – 3x) + (x + y) (x – 3y) – (y – 3x) (4x – 5y)
= (y – 3x) [x – 2y – (4x – 5y)] + (x + y)(x – 3y)
= (y – 3x) [x – 2y – 4x + 5y] + (x + y) (x – 3y)
= (y – 3x) (3y – 3x) + (x + y) (x – 3y)
= y(3y – 3x) – 3x(3y – 3x) + x(x – 3y) + y(x – 3y)
= 3y² – 3xy – 9xy + 9x² + x² – 3xy + xy – 3y²
= 10x² – 14xy

ii) (m + n) (m²– mn + n²)
= m(m² – mn + n²) + n(m² – mn + n²)
= m³ – m²n + n²m + nm² – mn² + n³
= m³ + n³

iii) (a – 2b + 5c) (a – b) – (a – b – c) (2a + 3c) + (6a + b) (2c – 3a – 5b)
= a(a – 2b + 5c) – b(a – 2b + 5c) – 2a(a – b – c) – 3c(a – b – c) + 6a(2c – 3a – 5b) + b(2c – 3a – 5b)
= a² – 2ab + 5ac – ab + 2b² – 5bc – 2a² + 2ab + 2ac – 3ac + 3bc + 3c² + 12ac – 18a² – 30ab + 2bc – 3ab – 5b²
= – 19a² – 3b² – 34ab + 16ac + 3c²

iv) (pq – qr + pr) (pq + qr) – (pr + pq) (p + q – r)
= pq(pq – qr + pr) + qr(pq – qr + pr) – pr(p + q – r) – pq(p + q – r)
= p²q² – pq²r + p²qr + pq²r – q²r² + pqr² – p²r – pqr + pr² – p²q – pq² + pqr
= p²q² – q²r² + p²qr + pqr² – p²r + pr² – p²q – pq ²

Question 4.
If a, b, care positive real numbers such that \(\frac{a+b-c}{c}=\frac{a-b+c}{b}=\frac{-a+b+c}{a}\) ,find the value of \(\frac{(a+b)(b+c)(c+a)}{a b c}\)
Solution:
\(\frac{a+b-c}{c}=\frac{a-b+c}{b}=\frac{-a+b+c}{a}\) = k then
\(\frac{a+b-c}{c}\) = k ⇒ a + b – c = kc
⇒ a + b = (ck + c) = c(k + 1) …………… (1)
Similarly b + c = a(k + 1) ……………(2)
c + a = b(k + 1) ………………..(3)

AP State Syllabus AP Board 8th Class Maths Solutions Chapter 11 Algebraic Expressions Ex 11.2 Textbook Questions and Answers.

AP State Syllabus 8th Class Maths Solutions 11th Lesson Algebraic Expressions Exercise 11.2

Question 1.
Complete the table:

Solution:

Question 2.
Simplify: 4y(3y + 4)
Solution:
4y(3y + 4) = 4y × 3y + 4y × 4
= 12y² + 6y

Question 3.
Simplify x(2x² – 7x + 3) and find the values of it for (i) x = 1 and (ii) x = 0
Solution:
x(2x² – 7x + 3)
= x × 2x² – x × 7x + x × 3
= 2x³ – 7x² + 3x
= 3 × 3 – 1 × 3 – 3 × 1 + 1 × 1
=9 – 3 – 3 + 1 = 4
∴(a – b)² = 4sq.units
[∵ (3 – 1)² = 2² = 4]

(ii)

∴ (a – b)² = a² – 2ab + b²
Area of ABCD + Area of CYZS
= a² – 2ab + b²
area of ABCD – area of BXYC – area of DCST + area of CYZS
= 5 × 5 – 2 × 5 – 2 × 5 + 2 × 2
= 25 – 10 – 10 + 4
= 9 sq.units
[∵ (5 – 2)² = (3)² = 9]

Question 4.
Add the product: a(a – b), b(b – c), c(c – a)
Solution:
a(a – b) + b(b – c) + c(c – a)
=a × a – a × b + b × b – b × c + c × c – c × a
=a² – ab + b² – bc + c² – ca
=a² + b² + c² – ab – bc – ca

Question 5.
Add the product: x(x + y – r), y(x – y+r), z(x – y – z)
Solution:
x(x + y – r) +y(x – y + r) + z(x – y – z)
= x² + xy – xr + xy – y² + yr + zx – yz – z²
= x² – y² – z² + 2xy – xr + yr + zx – yz

Question 6.
Subtract the product of 2x(5x – y) from product of 3x(x+2y)
Solution:
3x(x + 2y) – 2x(5x – y)
=(3x × x + 3x × 2y)-(2x × 5x – 2x × y)
= 3x² + 6xy – (10x² – 2xy)
= 3x² + 6xy- 10x² + 2xy
= 8xy – 7x²

Question 7.
Subtract 3k(5k – l + 3rn) from 6k(2k + 3l – 2rn)
Solution:
6k(2k + 3l – 2m) – 3k(5k – l + 3m)
= 12k²+ 18kl – 12km – 15k² + 3kl – 9km
= -3k² + 21kl – 21km

Question 8.
Simplify: a²(a – b + c) + b²(a + b – c) – c²(a – b – c)
Solution:
a²(a – b + c) + b²(a + b – c) – c²(a – b – c)
= a³ – a²b + a²c + ab² + b³ – b²c – ac² + bc² + c³
= a³ + b³ + c³ – a²b + a²c + ab² – b²c – ac² – bc²

AP State Syllabus AP Board 8th Class Maths Solutions Chapter 11 Algebraic Expressions Ex 11.1 Textbook Questions and Answers.

AP State Syllabus 8th Class Maths Solutions 11th Lesson Algebraic Expressions Exercise 11.1

Question 1.
Find the product of the following pairs:
(i) 6, 7k
(ii) – 31, – 2m
(iii) -5t² – 3t²
(iv) 6n, 3m
(v) – 5p², – 2p
Solution:
The product of 6, 7k = 6 × 7k = 42k
ii) The product of – 3l, – 2m = (- 3l) × (- 2m) = 6/m
iii) The product of – 5t², – 3t² = (- 5t²) × (- 3t²) = 15t⁴
iv) The product of 6n, 3m = 6n × 3m = 18mn
v) The product of – 5p², – 2p = (- 5p²) × (- 2p) = 10p³

Question 2.
Complete the table of the products.

Solution:

Question 3.
Find the volumes of rectangular boxes with given length, breadth and height in the following table.

Solution:

Question 4.
Find the product of the following monomials
(i) xy, x²y , xy, x
(ii) a, b, ab, a³ b, ab³
(iii) kl, lm, km, klm
(iv) pq ,pqr, r
(v) – 3a, 4ab, – 6c, d
Solution:
i) The product of xy, x²y, xy, x = xy × x²y × xy × x
= x⁵ × y³= x⁵y³

ii) The product of a, b, ab, a³b, ab³ = a × b × ab × a³b × ab³
= a⁶ × b⁶ = a⁶ b⁶

iii) The product of kl, lm, km, klm = kl × lm × km × klm
k³ × l³ × m³ =k³l³m³

iv) The product of pq, pqr, r = pq × pqr × r
= p² × q² × r² – p²q²r²

v) The product of – 3a, 4ab, – 6c, d = (- 3a) × 4ab × (- 6c) x d
= + 72a² × b × c × d
= 72a²bcd

Question 5.
If A = xy,B = yz and C = zx, then find ABC=
Solution:
ABC = xy × yz × zx = x²y²z²

Question 6.
If P = 4x², T = 5x and R = 5y, then \(\frac{\mathrm{PTR}}{100}\) =
Solution:
\(\frac{P^{\prime} \Gamma R}{100}=\frac{4 x^{2} \times 5 x \times 5 y}{100}=\frac{100 x^{3} y}{100}\) = x³ y

Question 7.
Write some monomials of your own and find their products.
Solution:
The product of,some monomials is given below :
i) abc × a²bc = a³b²c²
ii) xy × x²z × yz² = x³y²z³
iii) p × q × r = p³q³r³

AP State Syllabus 8th Class Maths Solutions 2nd Lesson Linear Equations in One Variable InText Questions

AP State Syllabus AP Board 8th Class Maths Solutions Chapter 2 Linear Equations in One Variable InText Questions and Answers.

8th Class Maths 2nd Lesson Linear Equations in One Variable InText Questions and Answers

Do this

Question 1.
Which of the following are linear equations:        [Page No. 35]
i) 4x + 6 = 8
ii) 4x – 5y = 9
iii) 5x² + 6xy – 4y² = 16
iv) xy + yz + zx = 11
v) 3x + 2y – 6 = 0
vi) 3 = 2x + y
vii) 7p + 6q + 13s = 11
Answer:
(i), (ii), (v), (vi), (vii) are the linear equations.

Question 2.
Which of the following are simple equations?        [Page No. 36]
i) 3x + 5 = 14
ii) 3x – 6 = x + 2
iii) 3 = 2x + y
iv) \(\frac{x}{3}\) + 5 = 0
v) x² + 5x + 3 = 0
vi) 5m – 6n = 0
vii) 7p + 6q + 13s = 11
viii) 13t – 26 = 39
Answer:
(i), (ii), (iv), (viii) are the simple equations.
Since these are all in the form of ax + b = 0.

AP State Syllabus 8th Class Maths Solutions 1st Lesson Rational Numbers InText Questions

AP State Syllabus AP Board 8th Class Maths Solutions Chapter 1 Rational Numbers InText Questions and Answers.

8th Class Maths 1st Lesson Rational Numbers InText Questions and Answers

Do this

Question 1.
Consider the following collection of numbers 1, \(\frac{1}{2}\), -2, 0.5, 4\(\frac{1}{2}\), \(\frac{-33}{7}\), 0, \(\frac{4}{7}\), \(0 . \overline{3}\), 22, -5, \(\frac{2}{19}\), 0.125. Write these numbers under the appropriate category. [A number can be written in more than one group]  (Page No. 2)
Answer:
i) Natural numbers 1, 22
ii) Whole numbers 0, 1, 22
iii) Integers 0, 1, 22, -5, -2
iv) Rational numbers 1, \(\frac{1}{2}\), -2, 0.5, 4\(\frac{1}{2}\), \(\frac{-33}{7}\), 0, \(\frac{4}{7}\), \(0 . \overline{3}\), 22, -5, \(\frac{2}{19}\), 0.125 etc.
Would you leave out any of the given numbers from rational numbers? No
Is every natural number, whole number and integer is a rational number? Yes

Question 2.
Fill the blanks in the table.     (Page No. 6)
Answer:

Question 3.
Complete the following table.     (Page No. 9)
Answer:

Question 4.
Complete the following table.      (Page No. 13)
Answer:

Question 5.
Complete the following table.      (Page No. 16)
Answer:

Question 6.
Complete the following table.      (Page No. 17)
Answer:

Question 7.
Represent – \(\frac{13}{5}\) on the number line.     (Page No. 22)
Answer:
Representing – \(\frac{13}{5}\) on the number line.

Try These

Question 1.
Hamid says \(\frac{5}{3}\) is a rational number and 5 is only a natural number. Shikha says both are rational numbers. With whom do you agree?       (Page No. 3)
Answer:
I would not agree with Hamid’s argument. Since \(\frac{5}{3}\) is a rational number. But ‘5’ is not only
a natural number, it is also a rational number.
Since every natural number is a rational number,
According to Shikha’s opinion \(\frac{5}{3}\), 5 are rational numbers.
∴ I agree with Shikha’s opinion.

Question 2.
Give an example to satisfy the following statements.        (Page No.3)
i) All natural numbers are whole numbers but all whole numbers need not be natural numbers.
ii) All whole numbers are integers but all integers are not whole numbers.
iii) All integers are rational numbers but all rational numbers need not be integers.
Answer:
i) ‘0’ is not a natural number.
∴ Every whole number is not a natural number. (∵ N ⊂ W)
ii) -2, -3, -4 are not whole numbers.
∴ All integers are not whole numbers. (∵ W ⊂ Z)
iii) \(\frac{2}{3}\), \(\frac{7}{4}\) are not integers.
∴ Every rational number is not an integer. (∵ Z ⊂ Q)

Question 3.
If we exclude zero from the set of integers is it closed under division? Check the same for natural numbers.    (Page No. 6)
Answer:
If ‘0’ is subtracted from the set of integers then it becomes Z – {0}.
Closure property under division on integers.
Ex: -4 ÷ 2 = -2 is an integer.
3 ÷ 5 = \(\frac{3}{5}\) is not an integer.
∴ Set of integers doesn’t satisfy closure property under division.
Closure property under division on natural numbers.
Ex: 2 ÷ 4 = \(\frac{1}{2}\) is not a natural number.
∴ Set of natural numbers doesn’t satisfy closure property under division.

Question 4.
Find using distributivity.     (Page No. 16)
A) \(\left\{\frac{7}{5} \times\left(\frac{-3}{10}\right)\right\}+\left\{\frac{7}{5} \times\left(\frac{9}{10}\right)\right\}\)
B) \(\left\{\frac{9}{16} \times 3\right\}+\left\{\frac{9}{16} \times-19\right\}\)
Answer:
Distributive law: a × (b + c) = ab + ac
A)

B)

Question 5.
Write the rational number for the points labelled with letters, on the number line.       (Page No. 22)
i)

ii)

Answer:
i) A = \(\frac{1}{5}\), B = \(\frac{4}{5}\), C = \(\frac{5}{5}\) = 1, D = \(\frac{7}{5}\), E = \(\frac{8}{5}\), F = \(\frac{10}{5}\) = 2.
ii) S = \(\frac{-6}{4}\), R = \(\frac{-6}{4}\), Q = \(\frac{-3}{4}\), P = \(\frac{-1}{4}\)

Think, discuss and write

Question 1.
If a property holds good with respect to addition for rational numbers, whether it holds good for integers? And for whole numbers? Which one holds good and which doesn’t hold good?     (Page No. 15)
Answer:
Under addition the properties which are followed by set of rational numbers are also followed by integers.

Question 2.
Write the numbers whose multiplicative inverses are the numbers themselves.      (Page No. 15)
Answer:
The number T is multiplicative inverse of itself.
∵ 1 × \(\frac{1}{1}\) = 1 ⇒ 1 × 1 = 1
∴ The multiplicative inverse of 1 is 1.

Question 3.
Can you find the reciprocal of ‘0’ (zero)? Is there any rational number such that when it is multiplied by ‘0’ gives ‘1’?
          (Page No. 15)
Answer:
The reciprocal of ‘0’ is \(\frac{1}{0}\).
But the value of \(\frac{1}{0}\) is not defined.
∴ There is no number is found when it is multiplied ‘0’ gives 1.
∵ 0 × (Any number) = 0

∴ No, there is no number is found in place of ‘A’.

Question 4.
Express the following in decimal form.     (Page No. 28)
i) \(\frac{7}{5}\), \(\frac{3}{4}\), \(\frac{23}{10}\), \(\frac{5}{3}\),\(\frac{17}{6}\),\(\frac{22}{7}\)
ii) Which of the above are terminating and which are non-terminating decimals?
iii) Write the denominators of above rational numbers as the product of primes.
iv) If the denominators of the above simplest rational numbers has no prime divisors other than 2 and 5 what do you observe?
Answer:
i) \(\frac{7}{5}\) = 0.4,
\(\frac{3}{4}\) = 0.75,
\(\frac{23}{10}\) = 2.3,
\(\frac{5}{3}\) = 1.66… = \(1 . \overline{6}\),
\(\frac{17}{6}\) = 2.833… = \(2.8 \overline{3}\),
\(\frac{22}{7}\) = 3.142
ii) From the above decimals \(\frac{7}{5}\), \(\frac{3}{4}\), \(\frac{23}{10}\) are terminating decimals.
While \(\frac{5}{3}\),\(\frac{17}{6}\),\(\frac{22}{7}\) are non-terminating decimals
iii) By writing the denominators of above decimals as a product of primes is

iv) If the denominators of integers doesn’t have factors other than 2 or 5 and both are called terminating decimals.

Question 5.
Convert the decimals \(0 . \overline{9}\), \(14 . \overline{5}\) and \(1.2 \overline{4}\) to rational form. Can you find any easy method other than formal method?     (Page No. 31)
Answer:
Let x = \(0 . \overline{9}\)
⇒ x = 0.999 ……. (1)
The periodicity of the above equation is ‘1’. So it is to be multiplied by 10 on both sides.
⇒ 10 × x = 10 × 0.999
10x = 9.999 …….. (2)
From (1) & (2)

∴ x = 1 or \(0 . \overline{9}\) = 1
Second Method:
\(0 . \overline{9}\) = 0 + \(0 . \overline{9}\)
= 0 + \(\frac{9}{9}\)
= 0 + 1
= 1

Let x = \(14 . \overline{5}\)
⇒ x = 14.55 …….. (1)
The periodicity of the equation (1) is 1.
So it should be multiplied by 10 on both sides.
⇒ 10 × x = 10 × 14.55
10x = 145.55 …….. (2)

Second Method:

Let x = \(1.2 \overline{4}\)
⇒ x= 1.244 …….. (1)
Here periodicity of equation (1) is 1. So it should be multiplied by 10 on both sides.
⇒ 10 × x = 10 × 1.244
10 x = 12.44 …….. (2)

Second Method:

AP State Syllabus AP Board 8th Class Maths Solutions Chapter 10 Direct and Inverse Proportions Ex 10.4 Textbook Questions and Answers.

AP State Syllabus 8th Class Maths Solutions 10th Lesson Direct and Inverse Proportions Exercise 10.4

Question 1.
Rice costing ₹480 is needed for 8 members for 20 days. What is the cost of rice required for 12 members for 15 days?
Solution:
Method – 1: Number of men and rice required to them are in inverse proportion.
Number of men ∝ \(\frac{1}{\text { No. of days }}\)

⇒ Compound ratio of 8:12 and 20: 15
= \(\frac{8}{12}=\frac{20}{15}\) = \(\frac{8}{9}\) …………….. (2)
From (1), (2)
480 : x = 8 : 9
⇒ \(\frac{480}{x}=\frac{8}{9}\)
⇒ x = \(\frac{480 \times 9}{8}\) = ₹540
∴ The cost of required rice is ₹ 540

Method – II :
\(\frac{M_{1} D_{1}}{W_{1}}=\frac{M_{2} D_{2}}{W_{2}}\)
M₁ = No. of men
D₁ = No .of days
W₁ = Cost of rice
∴ M₁ = 8
D₁ = 20
W₁ = ₹ 480
M₂ = 12
D₂ = 15
W₂ = ? (x)

⇒ x = 45 x 12 = ₹ 540
The cost of required rice = ₹ 540/-

Question 2.
10 men can lay a road 75 km. long in 5 days. In how many days can 15 men lay a road 45 km. long?
Solution:
\(\frac{M_{1} D_{1}}{W_{1}}=\frac{M_{2} D_{2}}{W_{2}}\)
∴ M₁ = 10
D₁ = 5
W₁ = 75
M₂ = 15
D₂ = ?
W₂ = 45

∴ x = 2
∴ No. of days are required = 2

Question 3.
24 men working at 8 hours per day can do a piece of work in 15 days. In how many days can 20 men working at 9 hours per day do the same work?
Solution:
M₁D₁H₁ = M₂D₂H₂
∴ M₁ = 24
D₁ = 15 days
H₁ = 8 hrs
M₂ = 20
D₂ = ?
H₂ = 9 hrs
⇒ 24 × 15 × 8 = 20 × x × 9

∴ No. of days are required = 16
[ ∵ No. of men and working hours are in inverse]

Question 4.
175 men can dig a canal 3150 m long in 36 days. How many men are required to dig a canal 3900 m. long in 24 days?
Solution:
\(\frac{M_{1} D_{1}}{W_{1}}=\frac{M_{2} D_{2}}{W_{2}}\)
M₁ = 175
D₁ = 36
W₁ = 3150
M₂ = ?
D₂ = 24
W₂ = 3900

∴ No. of workers are required = 325

Question 5.
If 14 typists typing 6 hours a day can take 12 days to complete the manuscript of a book, then how many days will 4 typists, working 7 hours a day, can take to do the same job?
Solution:
M₁D₁H₁ = M₂D₂H₂
M₁ = 14
D₁ = 12 days
H₁ = 6
M₂ = 4
D₂ = ?
H₂ = 7
⇒ 14 × 12 × 6 = 4 × x × 7

⇒ x = 36
∴ No. of days are required = 36
[ ∵ No of men and working hours are in inverse proportion]

AP State Syllabus AP Board 8th Class Maths Solutions Chapter 10 Direct and Inverse Proportions Ex 10.3 Textbook Questions and Answers.

AP State Syllabus 8th Class Maths Solutions 10th Lesson Direct and Inverse Proportions Exercise 10.3

Question 1.
Siri has enough money to buy 5 kg of potatoes at the price of ₹ 8 per kg. How much can she buy for the same amount if the price is increased to ₹ 10 per kg?
Solution:
Number of kgs of potatoes to their price are in inverse proportion.
∴ x₁y₁ = x₂ y₂
⇒ 8 × 5 = 10 × x
⇒ x = \(\frac{8 \times 5}{10}\) = 4 kgs
∴ 4 kgs of potatoes will be purchased at the rate of ₹ 10 per kg.

Question 2.
A camp has food stock for 500 people for 70 days. ¡f200 more people join the camp, how long will the stock last?
Solution:
Number of persons and their food stock are in inverse proportion.
⇒ x₁y₁ = x₂ y₂ (Let y₂ = x say)
⇒ 500 × 70 = (500 + 200) × x
⇒  x = \(\frac{500 \times 70}{700}\) = 5 × 10
∴ x = 50
∴ The food will be stock for (200 + 500) 700 men = 50 days

Question 3.
36 men can do a piece of work in 12 days. ¡n how many days 9 men can do the same work?
Solution:
Number of workers and number of days are in inverse proportion
∴ x₁y₁ = x₂ y₂ let y₂ = x (say)
= 36 × 12 = 9 × x
x = \(\frac{36 \times 12}{9}\) = 48
∴ x = 48 days

Question 4.
A cyclist covers a distance of28 km in 2 hours. Find the time taken by him to cover a distance of 56 km with the same speed.
Solution:
Time and distance are in direct proportion.
∴ \(\frac{x_{1}}{y_{1}}=\frac{x_{2}}{y_{2}}\) , Let y₂ = x (say)
⇒ \(\frac{28}{2}\) = \(\frac{56}{x}\)
⇒ x = \(\frac{56}{14}\)
∴ x = 4 hours

Question 5.
A ship can cover a certain distance in 10 hours at a speed of 16 nautical miles per hour. By how much should its speed be increased so that it takes only 8 hours to cover the same distance? (A nautical mile in a unit of measurement used at sea distance or sea water i.e. 1852 metres).
Solution:
Speed and distance are in inverse proportion.
⇒ x₁y₁ = x₂ y_{2 ,} Let x₂ = x (say)
⇒ 16 × 10 = x × 8
⇒ x = \(\frac{16 \times 10}{8}\)= 20
∴ x = 20
∴ The speed to be increased
= 20 – 16 = 4 nautical miles

Question 6.
5 pumps are required to fill a tank in 1\(\frac { 1 }{ 2 }\) hours. How many pumps of the same type are used to fill the tank in half an hour.
Solution:
Number of pumps and time to fill the tanks are in inverse proportion.
⇒ x₁y₁ = x₂ y₂
⇒ 5 × 1\(\frac { 1 }{ 2 }\) = x x 1\(\frac { 1 }{ 2 }\)
⇒ 5 × \(\frac { 3 }{ 2 }\) = x x \(\frac { 1 }{ 2 }\)
⇒ x = 5 × 3 = 15
∴ Number of pumps required = 15

Question 7.
If 15 workers can build a wall in 48 hours, how many workers will be required to do the same work in 30 hours?
Solution:
Number of workers and time are in inverse proportion.
⇒ x₁y₁ = x₂ y₂
⇒ 15 × 48 = x × 30
⇒ x = \(\frac{15 \times 48}{30}\) = 24
∴ Number of workers required = 24

Question 8.
A School has 8 periods a day each of45 minutes duration. How long would each period become ,if the school has 6 periods a day? ( assuming the number of school hours to be the same)
Solution:
Time and number of periods are in inverse proportion.
⇒ x₁y₁ = x₂ y₂
⇒ 45 × 8 = x × 6
⇒ \(\frac{45 \times 8}{6}\)
⇒ 60 minutes

Question 9.
If z varies directly as xand inversely as y. Find the percentage increase in z due to an increase of 12% in x and a decrease of 20% in y.
Solution:
Given that
z varies directly as x and inversely as y So, z ∝ x (1); z ∝ 1/y ……………… (2)
From (1) & (2), z ∝ \(\frac{\mathrm{x}}{\mathrm{y}}\)

Let x₁ = 100x, x₂ = 112x
(∵ It increases 12%)
y₁ = 100y, y₂ = 80y
(∵ It decreases 20%)
From (3),

∴ z is increased in 40%

Question 10.
If x + 1 men will do the work in x + 1 days, find the number of days that (x + 2) men can finish the same work.
Solution:
Number of workers and number of days are in inverse proportion.
⇒ x₁y₁ = x₂ y₂
⇒ (x + 1) (x + 1) = (x + 2) x k
⇒ k = \(\frac{(x+1)(x+1)}{(x+2)}\)
∴ k = \(\frac{(x+1)^{2}}{(x+2)}\)

Question 11.
Given a rectangle with a fixed perimeter of 24 meters, if we increase the length by 1 m the width and area will vary accordingly. Use the following table of values to look at how the width and area vary as the length varies.
What do you observe? Write your observations in your note books

Solution:

AP State Syllabus AP Board 8th Class Maths Solutions Chapter 10 Direct and Inverse Proportions Ex 10.2 Textbook Questions and Answers.

AP State Syllabus 8th Class Maths Solutions 10th Lesson Direct and Inverse Proportions Exercise 10.2

Question 1.
Observe the following tables and fmd which pair of variables (x and y) are in inverse proportion

Solution:
i) From the given table if the value of x is decreases then the value of ‘y’ is increases.
∴ x, y are in inverse proportion.
ii) From the given table if the value of x is increases then the value of y is decreases.
∴ x, y are in inverse proportion.
iii) From the given table the value of x is decreases then the value of y is increases.
∴ x, y are in inverse proportion.

Question 2.
A school wants to spend ₹6000 to purchase books. Using this data, fill the following table.

Solution:

Question 3.
Take a squared paper and arrange 48 squares in different number of rows as shown below.

What do you observe? As R increases, C decreases
(i) Is R₁:R₂ = C₂:C₁?
(ii) Is R₃:R₄ = C₄:C₃?
(iii) Is R and C inversely proportional to each other?
(iv) Do this activity with 36 squares.
Solution:
(i) Is R₁:R₂ = C₂:C₁
⇒ 2 : 3 = 16 : 24

(ii) Is R₃:R₄ = C₄:C₃

R₁:R₂ = C₂:C₁

(iii) R₃:R₄ = C₄:C₃
⇒ 4 : 6 = 8 : 12
\(\frac{4}{6}=\frac{8}{12}=\frac{4 \times 2}{6 \times 2}=\frac{4}{6} \Rightarrow \frac{4}{6}=\frac{4}{6}\) =
∴ R₃:R₄ = C₄:C₃

(iv) Do this activity with 36 squares.

From the above table we can conclude that if number of rows are increases then number of columns are decreases.

AP State Syllabus AP Board 8th Class Maths Solutions Chapter 10 Direct and Inverse Proportions Ex 10.1 Textbook Questions and Answers.

AP State Syllabus 8th Class Maths Solutions 10th Lesson Direct and Inverse Proportions Exercise 10.1

Question 1.
The cost of 5 meters of a particular quality of cloth is ₹ 210. Find the cost of(i) 2 (ii) 4
(iii) 10 (iv) 13 meters of cloth of the same quality.
Solution:
The cost of 5 m of a cloth = ₹ 210
The length of a cloth and its price are in direct proportion.
i) \(\frac{\mathrm{x}_{1}}{\mathrm{y}_{1}}=\frac{\mathrm{x}_{2}}{\mathrm{y}_{2}}\)
Here x₁ = 5, y₁ = 210
x₂ = 2, y₂ = ?

Question 2.
Fill the table.

Solution:

Question 3.
48 bags of paddy costs ₹ 16, 800 then find the cost of 36 bags of paddy.
Solution:
Number of bags of paddy and their cost are in direct proportion.
⇒ \(\frac{x_{1}}{y_{1}}=\frac{x_{2}}{y_{2}}\) , x₁ = 16,800
x₂ = 36 y₂ = ?

= 3 × 4200
y₂ = ₹ 12600
∴ The cost of 36 bags of paddy = ₹ 12600

Question 4.
The monthly average expenditure of a family with 4 members is 2,800. Find the
monthly average expenditure ofa family with only 3 members.
Solution:
Number of family members and their expenditure are in direct proportion.
⇒ \(\frac{x_{1}}{y_{1}}=\frac{x_{2}}{y_{2}}\) , x₁ = 4
y₁ = 2,800
x₂ = 3 y₂ = ?

= 3 × 700 = 2100
y₂ = ₹ 2100
The expenditure for 3 members = ₹ 2100

Question 5.
In a ship of length 28 m, height of its mast is 12 m. If the height of the mast in its model is
9 cm what is the length of the model ship?
Solution:
The length of ship and the height of its mast are in direct proportion.
⇒ \(\frac{x_{1}}{y_{1}}=\frac{x_{2}}{y_{2}}\) , x₁ = 28
y₁ = 12
x₂ = ? y₂ = 9

x₂ = 7 × 3 = 21
∴ The length of model ship = 21 m

Question 6.
A vertical pole of 5.6 m height casts a shadow 3.2 m long. At the same time find (j) the
length of the shadow cast by another pole 10.5 m high (ii) the height of a pole which casts
a shadow 5m long.
Solution:
length of a vertical pole and length of its shadow are in direct proportion.
⇒ \(\frac{x_{1}}{y_{1}}=\frac{x_{2}}{y_{2}}\)

i) x₁ = 5.6
y₁ = 3.2
x₂ = 10.5 y₂ = ?

∴The length of the shadow = 6 cm

ii) x₁ = 5.6 m x₂ = ?
y₁ = 3.2 m y₂ = 5
⇒ \(\frac{x_{1}}{y_{1}}=\frac{x_{2}}{y_{2}}\)

∴ x₂ = 8.75

Question 7.
A loaded truck travels 14 km in 25 minutes. If the speed remains the same, how far can it travel in 5 hours?
Solution:
Time and distance are in direct proportion
⇒ \(\frac{x_{1}}{y_{1}}=\frac{x_{2}}{y_{2}}\)
x₁ = 14 km , x₂ = ?
y₁ = 25min = \(\frac{25}{60} \mathrm{hr}=\frac{5}{12} \mathrm{hr}\) = y₂ = 5hrs
⇒ \(x_{2}=\frac{x_{1} \times y_{2}}{y_{1}}=\frac{14 \times 5}{5}=\frac{14 \times \not 5 \times 12}{\not 5}\)
= 168 km
∴ Lorry travelled in 5 hrs = 168km

Question 8.
If the weight of 12 sheets of thick paper is 40 grams, how many sheets of the same paper would weigh 16 \(\frac { 2 }{ 3 }\) kilograms?
Solution:
Number of pages and their weight are in direct proportion.
⇒ \(\frac{x_{1}}{y_{1}}=\frac{x_{2}}{y_{2}}\)
x₁ = 12 km , x₂ = ?
y₁ = 40 gm
y₂ = 16 \(\frac { 2 }{ 3 }\) gm = \(\frac { 50 }{ 3 }\) x 1000 gm
= \(\frac { 50000 }{ 3 }\) gm

From (1)

∴ Number of pages = 5000

Question 9.
A train moves at a constant speed of 75 km/hr.
(i) How far will it travel in 20 minutes?
(ii) Find the time required to cover a distance of 250 km.
Solution:
Speed of the train = 75 km/hr
i) The distance travelled in 20 min.
d = s x t = 75 x 20 min
= 75 x = 25 km
= \(75 \times \frac{20}{60}=\frac{75}{3}\) = 25 km

ii) Time taken to travel 250 km
t = \(\frac{d}{s}=\frac{250}{75}\)
t = \(\frac{10}{3}\) hrs

Question 10.
The design of a microchip has the scale 40:1. The length of the design is 18cm, find the actual length of the micro chip?
Solution:
The scale of the design of a microchip
= 40 : 1
The length of the design = 18 cm
The actual length of microchip = ?
The length of the design and actual length of the microchip are in direct proportion.
⇒ \(\frac{x_{1}}{y_{1}}=\frac{x_{2}}{y_{2}}\)
x₁ = 40 km , x₂ = 18
y₁ = 1
y₂ = ?
⇒ \(\frac{40}{1}=\frac{18}{y_{2}}\)
⇒ \(\frac{18}{40}=\frac{9}{20}\) cm
∴ The original (actual) length of the microchip = [latexs]\frac{9}{20}[/latex]cm

Question 11.
The average age of consisting doctors and lawyers is 40. If the doctors average age is 35 and the lawyers average age is 50, fmd the ratio of the number of doctors to the number of lawyers.
Solution:
Let the number of doctors = x
Number of lawyers = y
The average age of doctors = 35
The total age of doctors = 35 × x
= 35 x years
The average age of lawyers = 50
∴ The total age of lawyers = 50 x y
= 50y
According to the sum
\(\frac{35 x+50 y}{x+y}\) = 40
⇒ 35x + 50y = 40x + 40y
⇒ 40x – 35x = 50y – 40y
⇒ 5x = lOy
⇒ \(\frac{x}{y}=\frac{10}{5}\) (or)
x : y = 2 : 1
∴ The ratio of number of doctors to lawyers = 2:1