RS Aggarwal Class 6 Solutions Chapter 14 Constructions (Using Ruler and a Pairs of Compasses) Ex 14B

RS Aggarwal Class 6 Solutions Chapter 14 Constructions (Using Ruler and a Pairs of Compasses) Ex 14B

These Solutions are part of RS Aggarwal Solutions Class 6. Here we have given RS Aggarwal Solutions Class 6 Chapter 14 Constructions (Using Ruler and a Pairs of Compasses) Ex 14B.

Other Exercises

Question 1.
Solution:
(1) 60°
Steps of construction :
(i) Draw a ray OA.
RS Aggarwal Class 6 Solutions Chapter 14 Constructions (Using Ruler and a Pairs of Compasses) Ex 14B Q1.1
(ii) With centre O and with a suitable radius drawn an arc meeting OA at E.
(iii) With centre E and with same radius, draw another arc cutting the first arc at F.
(iv) Join OF and produce it to B Then ∠AOB = 60°
(2) 120°
Steps of construction :
(i) Draw a ray OA
(ii) With centre O and with a suitable radius draw an arc meeting OA at E
(iii) With centre E and with the same radius cut off the first arc firstly at F and then at G i.e. EF = FG.
(iv) Join OG and produce it to B.
Then, ∠AOB = 120°
RS Aggarwal Class 6 Solutions Chapter 14 Constructions (Using Ruler and a Pairs of Compasses) Ex 14B Q1.2
(3) 90°
Steps of construction :
(i) Draw a ray OA
RS Aggarwal Class 6 Solutions Chapter 14 Constructions (Using Ruler and a Pairs of Compasses) Ex 14B Q1.3
(ii) With centre O and a suitable radius draw an arc meeting OA at E.
(iii) With centre E and A with same radius cut off the arc first at F and then from F with same radius cut off arc at G.
(iv) With centres F and G with a suitable radius, draw two arcs intersecting each other at H.
(v) Join OH and produce it to B.
Then, ∠AOB = 90°.

Question 2.
Solution:
Steps of Construction :
(i) Draw a ray OA.
RS Aggarwal Class 6 Solutions Chapter 14 Constructions (Using Ruler and a Pairs of Compasses) Ex 14B Q2.1
(ii) With O as centre and any suitable radius draw an arc above OA, cutting it at a point B.
(iii) With B as centre and same radius as before draw another arc to cut the previous arc at C.
(iv) Join OC and produce it to D. Then ∠AOD = 60° is the required angle. To bisect the angle ∠AOD, with B as centre and radius more than half BC draw an arc. With C as centre and the same radius draw another are cutting the previous arc at E. Join OE and produce it. Then, OE is the required bisector of ∠AOD.

Question 3.
Solution:
Steps of constructions :
(i) Draw a ray OA.
(ii) With centre O and a suitable radius draw an arc meeting OA at E.
(iii) With centre E and with same radius, cut the first arc firstly at F and then from F with same radius cut act at G.
(iv) With centres F and G, with suitable radius, draw arcs intersecting each other at H.
RS Aggarwal Class 6 Solutions Chapter 14 Constructions (Using Ruler and a Pairs of Compasses) Ex 14B Q3.1
(v) Join OH intersecting the first arc at L and produce it to C.
(vi) With centre E and L and with suitable radius draw arcs intersecting each other at M.
(vii) Join OM and produce it to B.
Then ∠AOB = 45°

Question 4.
Solution:
(i) Steps of Construction :
1. Draw a ray OA.
2. With O as centre and any suitable radius draw an arc cutting OA at G.
3. With G as centre and same radius cut the arc at B and then B as centre and same radius cut the arc at C. Again, with C as centre and same radius cut the arc at D.
RS Aggarwal Class 6 Solutions Chapter 14 Constructions (Using Ruler and a Pairs of Compasses) Ex 14B Q4.1
4. With C as centre and radius more than half CD draw an arc.
5. With D as centre and same radius draw another arc to cut the previous arc at E.
6. Join OE and produce it to F.
Then ∠AOF = 150°
(ii) Steps of Construction :
1. Draw a ray OA.
2. With O as centre and any suitable radius draw an arc above OA, cutting it at B.
RS Aggarwal Class 6 Solutions Chapter 14 Constructions (Using Ruler and a Pairs of Compasses) Ex 14B Q4.2
3. With B as centre and same radius as before draw another arc to cut the previous arc at C. Join OC and prouce it to D.
4. Draw the bisector OE of ∠AQD. Then ∠AOE = 30°.
5. Draw the bisector OF of ∠AOE. Then ∠AOF = 15° is the required angle.
(iii) Steps of Construction :
1. Draw a ray OA.
RS Aggarwal Class 6 Solutions Chapter 14 Constructions (Using Ruler and a Pairs of Compasses) Ex 14B Q4.3
2. With O as centre and any suitable radius draw an arc above OA, cutting it at B.
3. With B as centre and same radius as before draw another arc to cut the previous arc at C. With C as centre and same radius draw the arc to cut it at D. Again with D as centre and same radius cut the arc at E.
4. Join OD and produce it to G. Then ∠AOG = 120°.
5. With D as centre and radius more than half DE draw an arc.
6. With E as centre and same radius draw another arc to cut the previous arc at F. Join OF.
7. Draw the bisector OH of ∠GOF. Then ∠AOH = 135° is the required angle.
(iv) Steps of Construction :
1. Draw a ray OA.
RS Aggarwal Class 6 Solutions Chapter 14 Constructions (Using Ruler and a Pairs of Compasses) Ex 14B Q4.4
2. With O as centre and any suitable radius draw an arc above OA, cutting it at B.
3. With B as centre and same radius cut the previous arc at C and then with C as centre and same radius cut the arc at D.
4. With C as centre and radius more than half CD draw an arc.
5. With D as centre and same radius draw another arc to cut the previous arc at E.
6. Join OE. Then ∠AOE = 90°.
7. Draw the bisector OF of ∠AOE.
8. Draw the bisector OG of ∠AOF.
Then ∠AOG = \(22 \frac { 1 }{ 2 } \) ° is the required angle.
(v) Steps of Construction :
1. Draw a ray OA.
RS Aggarwal Class 6 Solutions Chapter 14 Constructions (Using Ruler and a Pairs of Compasses) Ex 14B Q4.5
2. With O as centre and any suitable radius draw an arc cutting OA at B.
3. With B as centre and same radius cut the previous arc at C and then with C as centre and same radius cut the arc at D.
4. With C as centre and radius more than half CD draw an arc
5. With D as centre and same radius draw another arc to cut the previous arc at E.
6. Join OE. Also join OD and produce it to F.
7. Draw the bisector OG of ∠EOF Thus, ∠AOG = 105° is the required angle.
(vi) Steps of Construction :
1. Draw a ray OA.
RS Aggarwal Class 6 Solutions Chapter 14 Constructions (Using Ruler and a Pairs of Compasses) Ex 14B Q4.6
2. With O as centre and any suitable radius draw an arc cutting OA at B.
3. With B as centre and same radius* cut the previous arc at C and then with C as centre cut the arc at D.
4. With C as centre and radius more than half CD draw an arc.
5. With D as centre and same radius draw another arc to cut the previous arc at E.
6. Join OE. Also join OC and produce it to G.
7. Draw the bisector OF of ∠EOG. Then, ∠AOF = 75° is the required angle.
(vii) Steps of Construction :
1. Draw a ray OA.
RS Aggarwal Class 6 Solutions Chapter 14 Constructions (Using Ruler and a Pairs of Compasses) Ex 14B Q4.7
2. With O as centre and any suitable radius draw an arc above OA to cut it B.
With B as centre and same radius cut the previous arc at C and then with C as centre and same radius cut the arc at D.
4. With C as centre and radius more than half CD draw an arc.
5. With D as centre and same radius draw another arc to cut the previous arc at E.
6. Join OE. Then ∠AOE = 90°.
7. Draw the bisector OF of ∠AOE.
8. Draw the bisector OG of ∠EOF.
Then ∠AOG = \(67 \frac { 1 }{ 2 } \) ° is the required angle.
(viii) Steps of Construction :
1. Draw a ray OA.
RS Aggarwal Class 6 Solutions Chapter 14 Constructions (Using Ruler and a Pairs of Compasses) Ex 14B Q4.8
2. With O as centre and any su itable radius draw an arc above OA to cut it at B.
3. With B as centre and same radius cut the previous arc at C and then with C as centre and same radius cut the arc at D.
4. With C as centre and radius more than half CD, draw an arc.
5. With D as centre and same radius draw another arc to cut the previous arc at E.
6. Join OE. Then ∠AOE = 90°.
7. Draw the bisector OF of angle ∠AOE. Then, ∠AOF = 45° is the required angle.

Question 5.
Solution:
Steps of Construction :
1. Draw a line-segment AB = 5 cm with the help of a rular.
RS Aggarwal Class 6 Solutions Chapter 14 Constructions (Using Ruler and a Pairs of Compasses) Ex 14B Q5.1
2. With Aas centre and suitable radius draw an arc cutting AB at C.
3. With C as centre and same radius cut the previous arc at D and then with D as centre and same radius cut the arc at E
4. With D as centre and radius more than half DE draw an arc.
5. With E as centre and same radius draw another arc to cut the previous arc at F.
6. Join AF and produce it to G such that AG = 3.5 cm. Then ∠BAG = 90°.
7. With G as centre and radius equal to AB draw an arc. With B as centre and radius equal to AG draw another arc to cut the previous arc at H.
8. Join GH and BH. Then, AB HG is the required rectangle.

Question 6.
Solution:
Steps of Construction :
1. With the help of a ruler draw a line segment AB = 5 cm.
RS Aggarwal Class 6 Solutions Chapter 14 Constructions (Using Ruler and a Pairs of Compasses) Ex 14B Q6.1
2. With A as centre and any suitable radius draw an arc cutting AB at C.
3. With C as centre and same radius cut the previous arc at D and then with D as centre and same radius cut the arc at E.
4. With D as centre and radius more than half DE draw an arc.
5. With E as centre and same radius draw another arc to cut the previous arc at F.
6. Join AF and produce it to G such that AG = 5 cm.
7. With G as centre and radius equal to AB draw an arc. With B as centre and same radius draw another arc to cut the previous arc at H.
8. Join GH and BH. Then, AB HG is the required square.

Hope given RS Aggarwal Solutions Class 6 Chapter 14 Constructions Ex 14B are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RD Sharma Class 8 Solutions Chapter 7 Factorization Ex 7.2

RD Sharma Class 8 Solutions Chapter 7 Factorization Ex 7.2

These Solutions are part of RD Sharma Class 8 Solutions. Here we have given RD Sharma Class 8 Solutions Chapter 7 Factorizations Ex 7.2

Other Exercises

Factorize the following :

Question 1.
3x-9
Solution:
3x – 9 = 3 (x – 3)        (HCF of 3, 9 = 3)

Question 2.
5x – 15x2
Solution:
5x- 15x2 = 5x (1 – 3x)
{HCF of 5, 15 = 5 and of x, x2 = x}

Question 3.
20a12b2 – 15a8b4
Solution:
20a12b2 – 15a8b4
{HCF of 20, 15 = 5, a12, a8 = a8, b2, b4 = b2}
= 5a8 b2(4a4 – 3b2)

Question 4.
72xy – 96x7y6
Solution:
72xy – 96x7y6
HCF of 72, 96 = 24 of x6x7 = x6, y7,y6 = y6
∴ 72x7y6 – 96x7y6 = 24x6y6 (3y – 4x)

Question 5.
20X3 – 40x2 + 80x
Solution:
20x3 – 40x2 + 80x
HCF of 20, 40,80 = 20
HCF of x3, x2, x = x
∴ 20x3 – 40x2 + 80x = 20x (x2 – 2x + 4)

Question 6.
2x3y2 – 4x2y3 + 8xy4
Solution:
2x3y2 – 4x2y3 + 8xy4
HCF of 2, 4, 8 = 2
HCF of x3, x2, x = 1
and HCF of y2, y3, y4 = y2
∴ 2x3y2 – 4x2y3 + 8xy4
= 2xy2 (x2 – 2xy + 4y2)

Question 7.
10m3n2 + 15m4n – 20m2n3
Solution:
10m3n2 + 15m4n – 20m2n3
HCF of 10, 15, 20 = 5
HCF of m3, m4, m2 = m2
HCF of n2, n, n3 = n
10m3n2 + 15m4n – 20m2n3
5m2n(2mn + 3m2– 4n2)

Question 8.
2a4b4 – 3a3b5 + 4a2b5
Solution:
2a4b4 – 3a3b5 + 4a2b5
HCF of 2, 3, 4= 1
HCF of a4, a3, a2 = a2
HCF of b4, b5 b5 = b4
∴ 2a4b4 – 3a3b5 + 4a2b5 = a2b4
(2a2 – 3ab + 4b)

Question 9.
28a2 + 14a2b2 – 21a4
Solution:
28a2 + 14a2b2 – 21a4
HCF of 28, 14,21 =7
HCF of a2, a2, a4 = a2
HCF of 1, b2, 1 = 1
∴ 28a2 + 14a2b2-21a4 = 7a2
(4 + 2b2 – 3a2)

Question 10.
a4b – 3a2b2 – 6ab3
Solution:
a4b – 3a2b2 – 6ab3
HCF of 1,3,6 = 1
HCF of a4, a2, a = a
HCF of b, b2, b3 = b
∴ a4b – 3a2b2 – 6ab3 = ab (a3 – 3ab – 6b2)

Question 11.
2l2mn – 3lm2n + 4lmn2
Solution:
2l2mn – 3lm2n + 4lmn2
HCF 2, 3,4 = 1,
HCF of l2,l,l = l
HCF of m, m2, m = m
HCF of n, n, n2 = n
∴ 2l2 mn – 3lm2n + 4lmn2
= lmn (21 -3m + 4n)

Question 12.
x4y2 – x2y4 – x4y4
Solution:
x4y2 – x2y4 – x4y4
HCF of x4, x2, x4 = x2
HCF of y2, y4, y4 =y2
∴ x4y2 – x2y4 – x4y4 = x2y2 (x2 -y2 -x2y2)

Question 13.
9 x2y + 3 axy
Solution:
9 x2y + 3 axy
HCF of 9, 3 = 3
HCF of x2, x = x
HCF of y,y = y
HCF of 1,a = 1
∴ 9x2y + 3axy = 3xy (3x + a)

Question 14.
16m – 4m2
Solution:
16m – 4m2
HCF of 16, 4 = 4
HCF of m, m2 = m
∴ 16m – 4m2 = 4m (4 – m)

Question 15.
-4a2 + 4ab – 4ca
Solution:
-4a2 + 4ab – 4ca
HCF of 4, 4, 4 = 4
HCF of a2, a, a = a
∴ -4a2 + 4ab – 4ca = -4a (a – b + c)

Question 16.
x2yz + xy2z + xyz2
Solution:
x2yz + xy2z + xyz2
HCF of x2, x, x = x
HCF of y,y2,y=y
HCF of z, z,z2 = z
∴ x2yz + xy2z + xyz2 = xyz (x + y + z)

Question 17.
ax2y + bxy2 + cxyz
Solution:
ax2y + bxy2 + cxyz
HCF of x2, x, x = x,
HCF of y,y2,y = y
ax2y + bxy2 + cxyz = xy (ax + by + cz)

Hope given RD Sharma Class 8 Solutions Chapter 7 Factorizations Ex 7.2 are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RD Sharma Class 8 Solutions Chapter 7 Factorization Ex 7.1

RD Sharma Class 8 Solutions Chapter 7 Factorization Ex 7.1

These Solutions are part of RD Sharma Class 8 Solutions. Here we have given RD Sharma Class 8 Solutions Chapter 7 Factorizations Ex 7.1

Other Exercises

Find the greatest common factors (GCF / HCF) of the following polynomials : (1 – 14)

Question 1.
2x2 and 12x2
Solution:
2x2 and 12x2
HCF of 2 and 12 =2
HCF of x2,x2=x2
∴ HCF = 2x2

Question 2.
(6xy3 and 18x2y3
Solution:
6x3y and 18xy
HCF of 6, 18 = 6
HCF of x3 and x2 = x2
HCF of y and y3 -y
∴ HCF = 6x2y

Question 3.
7x, 21x2 and 14xy2
Solution:
7x, 21x2 and 14xy2
HCF of 7, 21 and 14 = 7
HCF of x, x2, x = x
∴ HCF = 7x

Question 4.
42x2yz and 63x3y2z3
Solution:
42x2yz and 63x3y2z3
HCF of 42 and 63 = 21
HCF of x2, x3 = x2
HCF of y,y2=y
HCF of z,z3 = z
∴ HCF = 21 x2yz

Question 5.
12ax2,6a2x3 and 2a3 x5
Solution:
12ax2, 6a2x3 and 2a3x5
HCF of 12, 6,2 = 2
HCF of a, a2, a3 = a
HCF of x2, x3, x5 = x2
∴ HCF = 2ax2

Question 6.
9x2, 15x2y3, 6xy2 and 21x2y2
Solution:
9x2, 15xV, 6xy2 and 21x2y2
HCF of 9, 15, 6,21 = 3
HCF of x2, x2, x, x2 = x
HCF of 1, y3, y2, y2 =2
∴ HCF = 3x

Question 7.
4a2b3 -12a3b, 18a4b3
Solution:
4a2b3, -12a3b, 18a4b3
HCF of 4, 12, 18 = 2
HCF of a2, a3, a4 = a2
HCF of b3,b, b3 = b
∴ HCF = 2a2b

Question 8.
6x2y2, 9xy3, 3x3y2
Solution:
6x2y2, 9xy3, 3x3y2
HCF of 6, 9, 3 = 3
HCF of x2, x, x3 = x
HCF of y2,y3,y2=y2
∴ HCF = 3xy2

Question 9.
a2b3, a3b2
Solution:
a2b3, a3b2
HCF of a2, a3 = a2
HCF of b3, b2 = b2
∴ HCF = a2b2

Question 10.
36a2b2c4, 54a5c2,90a4b2c2
Solution:
36a2b2c4, 54a5c2,90a4b2c2
HCF of 36, 54, 90 = 18
HCF of a2, a5, a4 = a2
HCF of b2, 1,b2= 1
HCF of c4,c2,c2 = c2
∴ HCF = 18a2 x 1 x c2 = 18a2c2

Question 11.
x3, – yx2
Solution:
x3, – yx2
HCF of x3, x2 = x2
HCF of 1, y= 1
∴ HCF = x2

Question 12.
15a3, -45a2, -150a
Solution:
15a3,-45a2,-150a
HCF of 15,45, 150 = 15
HCF of a3, a2, a = a
∴ HCF = 15a

Question 13.
2x3y2, 10x2y3, 14xy
Solution:
2x3y2, 10x2y3, 14xy
HCF of 2, 10, 14 = 2
HCF of x3, x2, x = x
HCF of y2,y3,y=y
∴ HCF = 2xy

Question 14.
14x3y5, 10x5y3, 2x2y2
Solution:
14x3y5, 10x5y3, 2x2y2
HCF of 14, 10, 2, = 2
HCF of x3, x5, x2 = x2
HCF of y5,y3,y2=y2
∴ HCF = 2xy

Find the greatest common factor of the terms in each of the following expressions:

Question 15.
5a4 + 10a3 – 15a2
Solution:
5a4 + 10a3– 15a2
HCF of 5, 10, 15 = 5
HCF of a4, a3, a2 = a2
∴ HCF = 5a2

Question 16.
2xyz + 3x2y + 4y2
Solution:
2xyz + 3x2y + 4y2
HCF of 2, 3,4 = 1
HCF of x, x2, 1 = 1
HCF of y,y,y2 =y
HCF of z, 1, 1 = 1
∴ HCF = y

Question 17.
3a2b2 + 4b2c2 + 12a2b2c2
Solution:
3a2b2 + 4b2c2 + 12a2b2c2
HCF of 3, 4, 12 = 1
HCF of a2, 1, a2 = 1
HCF of b2, b2, b2 = b2
HCF of 1, c2, c2 = 1
∴ HCF = b2

 

Hope given RD Sharma Class 8 Solutions Chapter 7 Factorizations Ex 7.1 are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.4

RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.4

These Solutions are part of RD Sharma Class 8 Solutions. Here we have given RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.4

Other Exercises

Find the following products 

Question 1.
2a3 (3a + 5b)
Solution:
2a3 (3a + 5b) = 2a3 x 3a + 2a3 x 5b
= 6a3 +1 + 10a3b
= 6a4 + 10a3b

Question 2.
-11a (3a + 2b)
Solution:
-11a (3a + 2b) = -11a x 3a – 11a x 2b
= -33a2– 22ab

Question 3.
-5a (7a – 2b)
Solution:
-5a (7a – 2b) = -5a x 7a- 5a x (-2b)
= -35a2 + 10ab

Question 4.
-11y2 (3y + 7)
Solution:
-11y2 (3y + 7) = -11y2 x 3y – 11y2 x 7
= -33y2+1-77y2
= 33y3-77y2

Question 5.
\(\frac { 6x }{ 5 }\) (x3+y3)
Solution:
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.4 1

Question 6.
xy (x3-y3)
Solution:
xy (x3 – y3) =xy x x3 – xy x y3
= x1 + 3 x y – x x y1+3
= x4y – xy4

Question 7.
0.1y (0.1x5 + 0.1y)
Solution:
0.1y (0.1x5 + 0.1y) = 0.1y x 0.1x5 + 0.1y x 0.1y
= 0.01x5y + 0.01y2

Question 8.
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.4 2
Solution:
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.4 3

Question 9.
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.4 4
Solution:
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.4 5

Question 10.
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.4 6
Solution:
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.4 7

Question 11.
5x (10x2y – 100xy2)
Solution:
5x (10x2y – 100xy2)
= 1.5x x 10x2y – 1.5x x 100xy2
= 15x1 + 2y- 150x1+1 x y2
15 x3y- 150x2 y2

Question 12.
4.1xy (1.1x-y)
Solution:
4.1xy (1.1x-y) = 4.1xy x 1.1x – 4.1xy x y
= 4.51x2y-4.1xy2

Question 13.
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.4 8
Solution:
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.4 9

Question 14.
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.4 10
Solution:
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.4 11

Question 15.
\(\frac { 4 }{ 3 }\) a (a2 + 62 – 3c2)
Solution:
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.4 12
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.4 13

Question 16.
Find the product 24x2 (1 – 2x) and evaluate its value for x = 3.
Solution:
24x2 (1 – 2x) = 24x2 x 1 + 24x2 x (-2x)
= 24x2 + (-48x2+1)
= 24x2 – 48x3
If x = 3, then
= 24 (3)2 – 48 (3)3
= 24 x 9-48 x 27 = 216- 1296
= -1080

Question 17.
Find the product of -3y (xy +y2) and find its value for x = 4, and y = 5.
Solution:
-3y (xy + y2) = -3y x xy – 3y x y2
= -3xy2 -3y2 +1  = -3xy2 – 3y3
If x = 4, y = 5, then
= -3 x 4 (5)2 – 3 (5)3 = -12 x 25 – 3 x 125
= -300 – 375 = – 675

Question 18.
Multiply – \(\frac { 3 }{ 2 }\) x2y3 by (2x-y) and verify the answer for x = 1 and y = 2.
Solution:
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.4 14
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.4 15

Question 19.
Multiply the monomial by the binomial and find the value of each for x = -1, y = 25 and z =05 :
(i) 15y2 (2 – 3x)
(ii) -3x (y2 + z2)
(iii) z2 (x – y)
(iv) xz (x2 + y2)
Solution:
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.4 16
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.4 17

Question 20.
Simplify :
(i) 2x2 (at1 – x) – 3x (x4 + 2x) -2 (x4 – 3x2)
(ii) x3y (x2 – 2x) + 2xy (x3 – x4)
(iii) 3a2 + 2 (a + 2) – 3a (2a + 1)
(iv) x (x + 4) + 3x (2x2 – 1) + 4x2 + 4
(v) a (b-c) – b (c – a) – c (a – b)
(vi) a (b – c) + b (c – a) + c (a – b)
(vii) 4ab (a – b) – 6a2 (b – b2) -3b2 (2a2 – a) + 2ab (b-a)
(viii) x2 (x2 + 1) – x3 (x + 1) – x (x3 – x)
(ix) 2a2 + 3a (1 – 2a3) + a (a + 1)
(x) a2 (2a – 1) + 3a + a3 – 8
(xi) \(\frac { 3 }{ 2 }\)-x2 (x2 – 1) + \(\frac { 1 }{4 }\)-x2 (x2 + x) – \(\frac { 3 }{ 4 }\)x (x3 – 1)
(xii) a2b (a – b2) + ab2 (4ab – 2a2) – a3b (1 – 2b)
(xiii )a2b (a3 – a + 1) – ab (a4 – 2a2 + 2a) – b (a3– a2 -1)
Solution:
(i) 2x2 (x3 -x) – 3x (x4 + 2x) -2 (x4 – 3x2)
= 2x2 x x3-2x2x x-3x x x4-3x x 2x-2x4 + 6x2
= 2x2 + 3– 2x2 +1 – 3x,1+ 4-6x,1+1 -2x4 + 6x2
= 2x5 – 2x3 – 3x5 — 6x2 – 2x4 + 6x2
= 2x5 – 3x5 – 2a4 – 2x3 + 6x2 – 6x2
= -x5 – 2x4 – 2x3 + 0
= -x5-2x4-2x3
(ii) x3y (x2 – 2x) + 2xy (x3 – x4)
= x3y x x2 – x3y x 2x + 2ay x ac3 – 2xy x x4
= x3 + 2y-2x3 + 1 y + 2x1 + 3y – 2yx4+1
= x5y – 2x4y + 2x4y – 2yx5
= -x5y
(iii) 3a2 + 2 (a + 2) – 3a (2a + 1)
= 3a2 + 2a + 4 – 6a2 – 3a
= 3a2 – 6a2 + 2a – 3a + 4
= -3a2 – a + 4
(iv) x (x + 4) + 3x (2x2 – 1) + 4x2 + 4
= x2 + 4x + 3x x 2x2 – 3x x 1 + 4x2 + 4
= x2 + 4x + 6x2 +1 – 3x + 4x2 + 4
= x2 + 4x + 6x3 – 3x + 4x2 + 4
= 6a3 + 4x2 + x2 + 4x – 3x + 4
= 6x3 + 5x2 + x + 4
(v) a (b – c)-b (c – a) – c (a – b)
= ab – ac – be + ab – ac + bc
= 2ab – 2ac
(vi) a (b – c) + b (c – a) + c (a – b)
= ab – ac + bc – ab + ac – bc
= ab – ab + bc – be + ac – ac
= 0
(vii) 4ab (a – b) – 6a2 (b – b2) -3b2 (2a2 – a) + 2ab (b – a)
= 4a2b – 4ab2 – 6a2b + 6a2b2 – 6a2b2 + 3ab2 + 2ab2 – 2a2b
= 4a2b- 6a2b – 2 a2b – 4ab2 + 3 ab2 + 2ab2 + 6a2b2 – 6a2b2
= 4a2b – 8a2b – 4ab2 + 5 ab2 + 0
= – 4a2b + ab2
(viii) x2 (x2 + 1) – x3 (x + 1) – x (x3 – x)
= x2 + 2 + x2 – x3 + 1 – x3 – x1 + 3 + x1 + 1
= x4 + x2-x4-x3-x4 + x2
= x4-x4-x4-x3 + x2 + x2
= -x4 – x3 + 2x2
(ix) 2a2 + 3a (1 – 2a3) + a (a + 1)
= 2a2 + 3 a – 3 a x 2a3 + a2 + a
= 2a2 + 3a – 6a1 + 3 + a2 + a
= 2a2 + 3a – 6a4 + a2 + a
= -6a4 + 3a2 + 4a
(x) a2 (2a – 1) + 3a + a3 – 8
= 2 a2 x a – a2 x 1+3a + a3-8
= 2a3 – a2 + 3a + a3 – 8
= 2a3 + a3 – a2 + 3a – 8
= 3a3 – a2 + 3a – 8
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.4 18
(xii) a2b (a – b2) + ab2 (4ab – 2a2) – a3b (1 – 2b)
= a2b x a – a2b x b1 + ab2 x 4ab – ab1 x2a2 -a3b x 1 + a3b x 2b
= a2+1 b-a2b2 +1+ 4a1 +1 b2 +1 -2a2+1 b2-a3b + 2a3b1 +1
= a’b – a2b3 + 4a2b3 – 2a3b2 – a3b + 2a3b2
= a3b – a3b – a2b3 + 4a2b3 – 2a3b2 + 2a3b2
= 0 + 3a2b3 + 0 = 3 a2b3
(xiii) a2b (a3 – a + 1) – ab (a4 – 2a2 + 2a) – b (a3 -a2– 1)
= a2b x a3 – a2b x a + a2b – ab x a2 + ab x 2a2 – ab x 2a- ba3 + ba2 + b
= a2+ 3b – a2+1 b + a2b -a1 + 4b + 2a1 + 2b- 2a1+1 b- a3b + a2b + b
= a5b – a3b + a26 – a5b + 2a3b – 2a2b – a3b + a2b + b
= a5b – a3b + 2a3b – a36 – a3b + a2b – 2a2b + a2b + b
= a3b – a5b + 2a3b – 2a3b + 2a2b-2a2b + b
= 0 + 0 + 0 + b = b

Hope given RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.4 are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RS Aggarwal Class 6 Solutions Chapter 9 Linear Equations in One Variable Ex 9C

RS Aggarwal Class 6 Solutions Chapter 9 Linear Equations in One Variable Ex 9C

These Solutions are part of RS Aggarwal Solutions Class 6. Here we have given RS Aggarwal Solutions Class 6 Chapter 9 Linear Equations in One Variable Ex 9C.

Other Exercises

Question 1.
Solution:
Let the required number be x.
Then, x + 9 = 36
=> x = 36 – 9
(Transposing 9 to R.H.S.)
=> x = 27
The required number = 27.

Question 2.
Solution:
Let the required number be x. Then,
4x – 11 = 89
=> 4x = 89 + 11
RS Aggarwal Class 6 Solutions Chapter 9 Linear Equations in One Variable Ex 9C Q2.1

Question 3.
Solution:
Let the required number be x. Then,
x x 5 = x + 80
=> 5x = x + 80
=> 5x – x = 80
RS Aggarwal Class 6 Solutions Chapter 9 Linear Equations in One Variable Ex 9C Q3.1

Question 4.
Solution:
Let the three consecutive numbers be x,
x + 1 and x + 2. Then,
x + (x + 1) + (x + 2) = 114
=> x – x + 1 + x + 2 = 114
RS Aggarwal Class 6 Solutions Chapter 9 Linear Equations in One Variable Ex 9C Q4.1

Question 5.
Solution:
Let the required number be x. Then
x x 17 + 4 = 225
=> 17x + 4 = 225
=> 17x = 225 – 4
RS Aggarwal Class 6 Solutions Chapter 9 Linear Equations in One Variable Ex 9C Q5.1

Question 6.
Solution:
Let the required number be x. Then
3 x + 5 = 50
=> 3 x = 50 – 5
RS Aggarwal Class 6 Solutions Chapter 9 Linear Equations in One Variable Ex 9C Q6.1

Question 7.
Solution:
Let the smaller number be x.
Then the other number = x + 18
According to question,
x + (x + 18) = 92
=> 2 x + 18 = 92
=> 2 x = 92 – 18
(Transposing 18 to R.H.S.)
=> 2 x = 74
=> x = 37
(Dividing both sides by 2)
One number = 37
Another number = 37 + 18
= 55.

Question 8.
Solution:
Let the smaller number be x.
Then, the other number = 3 x
According to question, x + 3 x = 124
=> 4 x = 124
RS Aggarwal Class 6 Solutions Chapter 9 Linear Equations in One Variable Ex 9C Q8.1

Question 9.
Solution:
Let the smallest number be x.
Then, the other number = 5 x
According to question,
5 x – x = 132
=> 4 x = 132
RS Aggarwal Class 6 Solutions Chapter 9 Linear Equations in One Variable Ex 9C Q9.1

Question 10.
Solution:
Let the two consecutive even number be x and x + 2.
Then x + x + 2 = 74
=> 2 x + 2 = 74
=> 2 x = 74 – 2
(Transposing 2 to RHS.)
=> 2x = 72
=> x = 36
(Dividing both sides by 2)
The required numbers are 36 and (36 + 2) i.e. 36 and 38.

Question 11.
Solution:
Let the three consecutive odd numbers be x, (x + 2) and (x + 4).
According to the question,
x + (x + 2) + (x + 4) = 21
3x + 6 = 21
=> 3 x = 21 – 6
(Transposing 6 to R.H.S.)
=> 3 x = 15
=> x = 5
(Dividing both sides by 3)
The required numbers are 5, 5 + 2 and 5 + 4 i.e. 5, 7 and 9.

Question 12.
Solution:
Let the present age of Ajay be x years. Then, the present age of Reena = (x + 6) years
According to the question,
x + (x + 6) = 28
2x + 6 = 28
=> 2x = 28 – 6
(Transposing 6 to R.H.S.)
=> 2 x = 22
=> x = 11
(Dividing both sides by 2)
Present age of Ajay = 11 years
and present age of Reena = 11 + 6
= 17 years.

Question 13.
Solution:
Let the present age of Vikas be x years.
Then, the present age of Deepak = 2x years
According to the question,
2x – x = 11
=> x = 11
Present age of Vikas = 11 years
and present age of Deepak = 2 x 11
= 22 years.

Question 14.
Solution:
Let the present age of Rekha be x years
Then, the present age of Mrs. Goel = (x + 27) years
Rekha’s age after 8 years = (x + 8) years
Mrs. Goel’s age after 8 years = (x + 27 + 8) years
= (x + 35) years
According to the question,
x + 35 = 2 (x + 8)
=> x + 35 = 2x + 16
=> x – 2x = 16 – 35
(Transposing 2x to L.H.S. and 35 to R.H.S.)
=> – x = – 19
=> x = 19
Present age of Rekha =19 years and present age of Mr. Goel = 19 + 27 = 46 years.

Question 15.
Solution:
Let the present age of the son be A years.
Then, the present age of the man
= 4 x years
Son’s age after 16 years = (x + 16) years Man’s age after 16 years
= (4 x + 16) years According to the question,
4x + 16 = 2 (x + 16)
=> 4x + 16 = 2x + 32
=> 4x – 2x = 32 – 16
(Transposing 2 x to L.H.S. and 16 to R.H.S.)
=> 2x = 16
=> x = 8 (Dividing both sides by 2)
Present age of the son = 8 years and present age of the man = 8 x 4 = 32 years.

Question 16.
Solution:
Let the present age of the son be x years.
Then, the present age of the man = 3x years
five years ago, the age of the son = (x – 5) years
five years ago, the age of the man = (3x – 5) years
According to the question, 3x – 5 = 4(x – 5)
=>3x – 5 = 4x – 20
=>3x – 4x = – 20 + 5
(Transposing 4 x to L.H.S. and – 5 to R.H.S.)
=> – x = – 15 => x = 15
Present age of the son = 15 years and present age of the man = 3 x 15 = 45 years.

Question 17.
Solution:
Let the present age of Fatima be A years. According to the question,
x + 16 = 3 x
=> x – 3x = – 16
(Transposing 3 x to L.H.S. and 16 to R.H.S.)
=> – 2 x = – 16
=> x = 8
(Dividing both sides by – 2)
Present age of Fatima = 8 years.

Question 18.
Solution:
Let the present age of Rahim be x years
Rahim’s age after 32 years = (x + 32) years
Rahim’s age 8 years ago = (x – 8) years
According to the question, x + 32 = 5 (x – 8)
=> x + 32 = 5 x – 40
=> x – 5x = – 40 – 32
(Transposing 5 x to L.H.S. and 32 to R.H.S.)
– 4 x = – 72
x= 18
(Dividing both sides by – 4)
Present age of Rahim =18 years

Question 19.
Solution:
Let the number of 50 paisa coins be x.
Then, the number of 25 paisa coins = 4x
Total value of 50 paisa coins = 50 x paisa
and total value of 25 paisa coins
= 25 x 4 = 100 x paisa
But total value of both the coins
= Rs. 30 (Given)
= 30 x 100 paisa
= 3000 paisa
According to the question,
50 x + 100 x = 3000
=> 150 x = 3000
\(\\ \frac { 150x }{ 150 } \) = \(\\ \frac { 3000 }{ 150 } \)
(Dividing both sides by 150)
x = 20
Number of 50 paisa coins = 20
and number of 25 paisa coins = 4 x 20
= 80

Question 20.
Solution:
Let the price of the pen be x rupees. According to the question,
5 x = 3 x + 17
5 x – 3 x = 17
(Transposing 3 x to L.H.S.)
=> 2 x = 17
=> x = \(\\ \frac { 17 }{ 2 } \)
(Dividing both sides by 2)
Price of the pen = \(\\ \frac { 17 }{ 2 } \) rupees
= Rs. 8.50.

Question 21.
Solution:
Let the number of girls in the school be x.
Then, the number of boys in the school = (x + 334)
According to the question, x + (x + 334) = 572
=> 2 x + 334 = 572
=> 2 x = 572 – 334
(Transposing 334 to L.H.S.)
2 x = 238
=> x = \(\\ \frac { 238 }{ 2 } \)
(Dividing both sides by 2) => x = 119
Number of girls in the school = 119.

Question 22.
Solution:
Let the breadth of the park be x metres.
Then, the length of die park=3x metres.
According to the question,
Perimeter of the park = 168 metres
=> 2 (x + 3 x) = 168
=> 2 x 4x = 168
=> 8 x = 168
=> x = 21
(Dividing both sides by 8)
Breadth of the park = 21 metres and length of the park = 3 x 21 = 63 metres.

Question 23.
Solution:
Let the breadth of the hall be x metres.
Then, the length of the hall = (x + 5) metres .
According to question,
Perimeter of the hall = 74 metres
=> 2 (x + x + 5) = 74
=> 2 (2 x + 5) = 74
=> 4 x + 10 = 74
=> 4 x = 74 – 10
(Transposing 10 to R.H.S.)
4x = 64
=> \(\\ \frac { 4x }{ 4 } \) = \(\\ \frac { 64 }{ 4 } \)
(Dividing both sides by 4)
=> x = 16
Breadth of the hall = 16 metres
and length of the hall = 16 + 5 = 21 metres.

Question 24.
Solution:
Since a wire of length 86 cm is bent to form die rectangle, so the perimeter of the rectangle = 86 cm.
Let the breadth of the rectangle = x cm
Then, die length of the rectangle = (x + 7) cm
2 (x + x + 7) = 86
=> 2 (2 x + 7) = 86 .
=> 4 x + 14 = 86
=> 4x = 86 – 14
(Transposing 14 to R.H.S.)
=> 4 x = 72
\(\\ \frac { 4x }{ 4 } \) = \(\\ \frac { 72 }{ 4 } \) = 18
(Dividing both sides by 4)
Breadth of the rectangle = 18 cm
Length of the rectangle = (18 + 7) = 25 cm.

Hope given RS Aggarwal Solutions Class 6 Chapter 9 Linear Equations in One Variable Ex 9C are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

CA Foundation Business Economics Study Material – Internal and External Economies

CA Foundation Business Economics Study Material Chapter 3 Theory of Production and Cost – Internal and External Economies

Internal Economies and Diseconomies

  • Internal economies are those benefits which accrue to a firm when it expands the scale of production.
  • Internal economies are the result of the firm’s own efforts independent of the actions of other firms.
  • These economies are particular to the individual firms and are different for different firms depending upon the size of the firm.

The main types of internal economies are as follows

1. Technical Economies:

– The large scale production is associated with technical economies.
– As the firm increases its scale of production, it becomes possible to use better plant, machinery, equipment and techniques of production.
– Following are the main forms (causes/reasons) of technical economies

  • Economies of superior techniques
    – A large sized firm can use sophisticated and costly machines and equipments.
    – Use of superior techniques reduces the cost of production per unit and increases aggregate output.
  • Economies of increased dimensions
    – A large firm can get the mechanical advantage in using large machines and other mechanical units to produce more output.
    – E.g. A Large boiler, large furnace, etc. can be operated by same team as required by smaller boiler, furnace, etc.
  • Economies of linked processes
    – A large sized firm can develop its own sources of raw material, means of transportation, distribution system, etc.
  • Economies of the use of By-products
    – A large sized firm can avoid all kinds of wastage of materials. The firm can use its by- products and waste material to produce another material.
    – E.g.- Sugar industry can make alcohol out of the molasses.
  • Economies of specialization
    – A large sized firm can introduce greater degree of division of labour and specialisation.

2. Managerial Economies:

  • Large sized firms can introduce division of labour in managerial tasks.
  • They can employ business executive of high skill and qualification to look after the functioning of various departments like production, finance, sales, advertising, personnel, etc.
  • This helps to increase the efficiency and productivity of managers resulting in reduction in managerial costs.

3. Commercial Economies:

  • A large sized firm is able to reap economies of bulk purchases.
  • It can get discounts from suppliers, railways, transport companies, etc.
  • It enjoys prompt and regular supply of raw materials.
  • A large sized firm can also afford to spend large amount of money on advertising, publicity, etc.
  • It can also give various concessions to wholesale and retail dealers and customers and thus capture markets for its product.

4. Financial Economies:

  • A big firm enjoys goodwill among lenders or investors.
  • For raising finance it can either borrow from bank as it can offer better security or it can raise finance by issuing shares, debentures and by inviting public deposits. Such opportunities are not available to small firms.

5. Risk Bearing Economies:

  • A large firm is better placed to face the uncertainties and risks of business.
  • A big firm producing many variety of goods is in a better position to withstand economic ups and downs. Therefore, it enjoys economies of risk bearing.

Internal diseconomies means all those factors which raise the cost of production per unit of a particular firm when the scale of production is expanded beyond the point of optimal capacity.

Such diseconomies of scale are as follows

1. Production Diseconomies:

  • Production diseconomies sets in when expansion of firm’s production beyond optimum size leads to rise in the cost per unit of output.
  • E.g. Use of inferior or less efficient factors due to non-availability of efficient factors raises the per unit cost of output.

2. Managerial Diseconomies:

  • As the scale of production increases burden on management also increases.
  • Co-ordination of work among different departments becomes difficult. Supervision and control over the activities of subordinates becomes difficult, decision taking is delayed, etc.
  • As a result, wastage increase and the efficiency and productivity decrease.
  • Per unit cost starts rising.

3. Technical Diseconomies:

  • Every equipment has an optimum point at which it works more efficiently and economically.
  • Beyond optimum point they are overworked and may result in breakdowns, heavy cost of maintenance, etc.

4. Financial Diseconomies:

  • Expansion of production beyond the optimum scale results in increase in the cost of capital.
  • It may be due to increased dependence on external finances.

5. Marketing Diseconomies:

  • Selling diseconomies set in if the scale of production is expanded beyond optimum level.
  • The advertisement expenditure and marketing overheads increase more proportionately with the scale.

External Economies and Diseconomies

  • External economies are those benefits which accrue to all the firms operating in a given industry from the growth and expansion of that industry.
  • External economies are not related to an individual firm’s own cost reduction efforts.
  • These are common to all the firms in an industry and shared by many firms or industries.

The main types of external economies are as follows

1. Technological Economies:

  • When the whole industry expands, it may result in the discovery of new technical knowledge, firms pool manpower and finance for research and development resulting in new and improved methods of production and new inventions.
  • Use of improved and better machinery improves production function and cost of production per unit falls.

2. Economies of Localization:

  • When in an area, many firms producing the same commodity are set up, it is called localization of an industry.
  • Due to localization there is expansion of railways, post & telegraph, banking services, insurance, setting up of booking offices by transport, companies, setting § up of powerful transformer by electricity department, etc.
  • All the firms get these facilities at low prices.

3. Economies of Information:

  • As pointed earlier, firms pool their resources for research and development.
  • All firms get the benefit of the research in terms of market information, technical information, information about governments economic policies, information about availability of new source of raw material, etc.
  • Also, specialized journals give information about latest developments.

4. Cheaper Inputs:

  • When an industry expands its needs for raw materials, machines, etc. also expand.
  • This may result in exploration of new and cheaper sources of raw materials, machinery, etc.
  • Also, the industries producing such inputs also expand in scale.
  • Therefore, they can supply these inputs at lower prices.
  • As a result the cost of production per unit of the firm using these inputs falls.

5. Growth of Ancillary Industries:

  • With the growth of an industry, many firms specialized in the production of inputs like raw material, tools, machinery, etc. come up.
  • Such firms are called ancillary units which provides inputs at lower cost to the main industry.
  • Likewise, some firms may get developed by processing the waste products of the industry.
  • Thus, wastes are converted into by-products. This reduces the cost of production in general.

6. Development of Skilled Labour:

  • When an industry expands specialized institutions like colleges, training centers, management institutes, etc. develop.
  • This results in continuous availability of skilled labour like technicians, engineers, management experts, etc.

7. Better transportation & Marketing Facilities:-

  • When an industry expands many specialized transporters also develop.
  • The firm in need of specialized transport service can get them easily at cheaper rates.
  • Also many new marketing outlets and specialized marketing institutions develop. The firm need not spend on developing its own marketing outlets.
  • This reduces the cost.

The growth and expansion of an industry in a particular area beyond optimum level results in many disadvantages for firms in the industry. Such disadvantages increases the costs of production of each firm. Therefore, they are called external diseconomies. Some of the external diseconomies are as follows:

1. Diseconomies of Scarcity of Inputs:

  • When an industry expands its need for raw materials, machines, tools and equipments, etc. also expands.
  • Some inputs are such which cannot be totally substituted.
  • The firms supplying these inputs come under pressure and may supply inputs at a higher price.
  • This raises the cost of production per unit of the firm who uses these inputs.

2. Diseconomies of Strains on Infrastructure:

  • Due to concentration of firms in an area infrastructural facilities become inadequate over a time.
  • E.g. Excessive pressure on transport system results in delayed transportation of raw materials and finished goods.
  • Other facilities like electric power supply, communication system, water supply, etc. are also over taxed.
  • This puts strain on infrastructural facilities resulting in increased cost of production. ’

3. Diseconomies of High Factor Prices:

  • With the concentration of an industry in a particular area, the demand for factors of production rises.
  • Thus, the prices of the factors of production go up resulting in increased cost of production.

4. Diseconomies of Expenditure on Advertising:

  • Expansion of an industry also means increase in the number of firms.
  • This means increase in competition among the firms.
  • This forces a firm to spend more and more on advertising.
  • This raises per unit cost.

Internal and External Economies

S.No INTERNAL ECONOMIES EXTERNAL ECONOMIES
1.
  • Internal economies are the benefits which accrue to a firm when it expands the scale of production.
  • External economies are those benefits which accrue to all the firms operating in a given industry from the growth and expansion of that industry.
2.
  • Internal economies are called ‘internal’ because these arise due to the internal efforts of the firm.
  • These economies are specific to the individual firm and are different for different firms depending upon the size of the firm.
  • External economies are called ‘external’ because they accrue to a firm as a result of factors that are entirely outside the firm i.e. from the expansion of the industry.
3.
  • Internal economies are the result of the firm’s OWN EFFORTS INDEPENDENT OF THE ACTIONS OF OTHER FIRMS.
  • These economies are peculiar to each fir m.
  • It reflects the working pattern of the firm.
  • External economies are independent of firm’s own efforts and output.
  • They are dependent on the general development of the industry.
  • They are not restricted to a single firm but are shared by a number of firms.
4.
  • Internal economies cause the long-run average cost to fall in the initial stage and internal diseconomies cause the long-run average cost to rise at the later stage.
  • Thus, the shape of LAC curve is determined by internal economies and diseconomies as scale expands.
  • External economies and diseconomies cause the LAC curve to shift down or up as the case may be.
  • When external economies increase, the cost per unit of output falls.
  • So, LAC curve shift downwards.
  • When external diseconomies are more, the cost per unit of output rises.
  • So, LAC curve shift upwards.
5. CA Foundation Business Economics Study Material Internal and External Economies 1 CA Foundation Business Economics Study Material Internal and External Economies 2
6.
  • If every thing is effectively managed, internal economies can be of long term in nature.
  • External economies depend upon the conditions of the entire industry and economy.
  • Thus, it can be of short term in nature.
7.
  • Internal economies are in the form of technical economies like superior techniques, use of by- products, etc.; managerial economies; commercial economies; financial economies and risk-bearing economies.
  • External economies are in the form of cheaper inputs; discovery of new technical knowledge; development of skilled labour; economies of information; growth of ancillary units; better transport and marketing facilities.

RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.7

RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.7

These Solutions are part of RD Sharma Class 8 Solutions. Here we have given RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.7

Other Exercises

Question 1.
Find the following products :
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.7 1
Solution:
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.7 2
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.7 3
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.7 4

Question 2.
Evaluate the following :
(i) 102 x 106
(ii) 109 x 107
(iii) 35 x 37
(iv) 53 x 55
(v) 103 x 96
(vi) 34 x 36
(vii) 994 x 1006
Solution:
(i) 102 x 106 = (100 + 2) (100 + 6)
= (100)2 + (2 + 6) x 100 + 2 x 6
= 10000 + 800 + 12 = 10812

(ii) 109 x 107 = (100 + 9) (100 + 7)
= (100)2 + (9 + 7) x 100 + 9 x 7
=10000 + 1600 + 63 = 11663

(iii) 35 x 37 = (30 + 5) (30 + 7)
= (30)2 + (5 + 7) x 30 + 5 x 7
= 900 + 12 x 30 + 35
= 900 + 360 + 35 = 1295

(iv) 53 x 55 = (50 + 3) (50 + 5)
= (50)2 + (3 + 5) x 50 + 3 x 5
= 2500 + 8 x 50 + 15
= 2500 + 400+ 15 = 2915

(v)103 x 96 = (100 + 3) (100-4)
= (100)2 + (3 – 4) x 100 + 3 x (-4)
= 10000+ (-1) x 100-12
= 10000 – 100 – 12 = 10000 – 112 = 9888

(vi) 34 x 36 = (30 + 4) (30 + 6)
= (30)2 + (4 + 6) x 30 + 4 x 6
= 900 + 10 x 30 + 24
= 900 + 300 + 24 = 1224

(vii) 994 x 1006 = (1000 – 6) (1000 + 6)
= (1000)2 + (-6 + 6) x 1000 + (-6) x 6
= 1000000 + 0-36
= 1000000-36 = 999964

Hope given RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.7 are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RS Aggarwal Class 6 Solutions Chapter 9 Linear Equations in One Variable Ex 9B

RS Aggarwal Class 6 Solutions Chapter 9 Linear Equations in One Variable Ex 9B

These Solutions are part of RS Aggarwal Solutions Class 6. Here we have given RS Aggarwal Solutions Class 6 Chapter 9 Linear Equations in One Variable Ex 9B.

Other Exercises

Solve each of the following equations and verify the answer in each case :

Question 1.
Solution:
x + 5 = 12
=> x + 5 – 5 = 12 – 5
(Subtracting 5 from both sides)
=> x = 7
.’. x = 7 is the solution of the given equation.
Check : Substituting x = 7 in the given equation, we get
L.H.S. = 7 + 5 = 12 and R.H.S. = 12
∴When x = 7, we have L.H.S. = R.H.S.

Question 2.
Solution:
x + 3 = – 2
=>x + 3 – 3 = – 2 – 3
(Subtracting 3 from both sides)
=> x = – 5
∴ x = – 5 is the solution of the given equation.
Check : Substituting x = – 5 in the given equation, we get:
L.H.S. = – 5 + 3 = – 2 and R.H.S. = – 2
When x = – 5,
∴we have L.H.S. = R.H.S.

Question 3.
Solution:
x – 7 = 6
=>x – 7 + 7 = 6 + 7
(Adding 7 to both sides)
=> x – 13
So, x = 13 is the solution of the given equation.
Check : Substituting x – 13 in the given equation, we get
L.H.S.= 13 – 7 = 6 and R.H.S. = 6
∴When x = 13, we have L.H.S. = R.H.S.

Question 4.
Solution:
x – 2 = – 5
=> x – 2 + 2 = – 5 + 2
(Adding 2 on both sides)
=> x = – 3
So, x = – 3 is the solution of the given equation.
Check : Substituting x = – 3 in the given equation, we get
L.H.S. = – 3 – 2 = – 5 and R.H.S. = – 5 When x = – 3,
we have
L.H.S. = R.H.S.

Question 5.
Solution:
3x – 5 = 13
=>3x – 5 + 5 = 13 + 5
(Adding 5 on both sides)
=> 3x = 18
=>\(\\ \frac { 3x }{ 3 } \) = \(\\ \frac { 18 }{ 3 } \)
(Dividing both sides by 3)
=> x = 6
x = 6 is the solution of the given equation.
Check : Substituting x = 6 in the given equation, we get
L.H.S. = 3 x 6 – 5 = 18 – 5 = 13 and R.H.S. = 13
∴ When x = 6, we have L.H.S. = R.H.S

Question 6.
Solution:
4x + 7 = 15
=> 4x + 7 – 7 = 15 – 7
(Subtracting 7 from both sides)
=> 4x = 8
=> \(\\ \frac { 4x }{ 4 } \) = \(\\ \frac { 8 }{ 4 } \)
(Dividing both sides by 4)
=> x = 2
x = 2 is the solution of the given equation.
Check : Substituting x = 2 in the given equation, we get
L.H.S. = 4 x 2 + 7 = 8 + 7 = 15 and R.H.S. = 15
∴When x = 2, we have L.H.S. = R.H.S.

Question 7.
Solution:
\(\\ \frac { x }{ 5 } \) = 12
=> \(\\ \frac { x }{ 5 } \) x 5 = 12 x 5
(Multiplying both sides by 5)
=> x = 60
x = 60 is the solution of the given equation.
Check : Substituting x = 60 in the given equation, we get
L.H.S. = \(\\ \frac { 60 }{ 5 } \) = 12 and R.H.S. = 12
When x = 60, we have
∴L.H.S. = R.H.S.

Question 8.
Solution:
\(\\ \frac { 3x }{ 5 } \) = 15
=> \(\\ \frac { 3x }{ 5 } \) x \(\\ \frac { 5 }{ 3 } \) = 15 x \(\\ \frac { 5 }{ 3 } \)
RS Aggarwal Class 6 Solutions Chapter 9 Linear Equations in One Variable Ex 9B Q8.1

Question 9.
Solution:
5x – 3 = x + 17
=> 5x – x = 17 + 3
RS Aggarwal Class 6 Solutions Chapter 9 Linear Equations in One Variable Ex 9B Q9.1
So. x = 5 is a solution of the given
equation.
Check : Substituting x = 5 in the given
equation, we get
L.H.S. = 5 x 5 – 3 = 25 – 3 = 22
R.H.S. 5 + 17 = 22
∴When x = 5, we have L.H.S. = R.H.S.

Question 10.
Solution:
2x – \(\\ \frac { 1 }{ 2 } \) = 3
=> 2x = 3 + \(\\ \frac { 1 }{ 2 } \)
RS Aggarwal Class 6 Solutions Chapter 9 Linear Equations in One Variable Ex 9B Q10.1
RS Aggarwal Class 6 Solutions Chapter 9 Linear Equations in One Variable Ex 9B Q10.2

Question 11.
Solution:
3(x + 6) = 24
=> \(\\ \frac { 3(x+6) }{ 3 } \) = \(\\ \frac { 24 }{ 3 } \)
(Dividing both sides by 3)
x + 6 = 8
=> x = 8 – 6
(Transposing 6 to R.H.S.)
=> x = 2
x = 2 is a solution of the given equation.
Check : Substituting the value of x = 2
in the given equation, we get
L.H.S. = 3(2 + 6 ) = 3 x 8 = 24
and RH.S. = 24
∴When x = 2, we have L.H.S. = R.H.S.

Question 12.
Solution:
6x + 5 = 2x + 17
=> 6x – 2x = 17 – 5
(Transposing 2 x to L.H.S. and 5 to R.H.S.)
=> 4x = 12
=> \(\\ \frac { 4x }{ 4 } \) = \(\\ \frac { 12 }{ 4 } \)
(Dividing both sides by 4)
=> x = 3
x = 3 is a solution of the given
equation.
Check : Substituting x = 3 in the given
equation, we get
L.H.S.= 6 x 3 + 5 = 18 + 5 = 23
R.H.S.= 2 x 3 + 17 = 6 + 17 = 23
∴When x = 3, we have L.H.S. = R.H.S.

Question 13.
Solution:
\(\\ \frac { x }{ 4 } \) – 8 = 1
=> \(\\ \frac { x }{ 4 } \) = 1 + 8
RS Aggarwal Class 6 Solutions Chapter 9 Linear Equations in One Variable Ex 9B Q13.1
R.H.S = 1
∴When x = 36,we have L.H.S. = R.H.S.

Question 14.
Solution:
\(\\ \frac { x }{ 2 } \) = \(\\ \frac { x }{ 2 } \) + 1
=> \(\\ \frac { x }{ 2 } \) – \(\\ \frac { x }{ 3 } \) = 1
RS Aggarwal Class 6 Solutions Chapter 9 Linear Equations in One Variable Ex 9B Q14.1

Question 15.
Solution:
3(x + 2) – 2(x – 1) = 7
=> 3x + 6 – 2x + 2 = 7
(Removing brackets)
3x – 2x + 6 + 2 = 7
x + 8 = 7
x = 7 – 8
(Transposing 8 to R.H.S.)
x = – 1 is a solution of the given
equation.
Check : Substituting x = – 1 in the given
equation, we get
L.H.S. = 3 ( – 1 + 2) – 2( – 1 – 1)
= 3 x 1 + ( – 2 x – 2)
= 3 + 4 = 7 and
R.H.S. = 7
When x = – 1, we have
L.H.S. = R.H.S.

Question 16.
Solution:
5 (x- 1) + 2 (x + 3) + 6 = 0
= 5 (x – 1) + 2 (x + 3) = – 6
(Transposing 6 to R.H.S.)
= 5x – 5 + 2x + 6 = – 6
RS Aggarwal Class 6 Solutions Chapter 9 Linear Equations in One Variable Ex 9B Q16.1
RS Aggarwal Class 6 Solutions Chapter 9 Linear Equations in One Variable Ex 9B Q16.2

Question 17.
Solution:
6(1 – 4 x) + 7 (2 + 5 x) – 53
=> 6 – 24x + 14 + 35 x = 53
(Removing brackets)
=> – 24 x + 35 x + 14 + 6 = 53
=> 11 x + 20 = 53
=> 11 x = 53 – 20
=> 11 x = 33
(Transposing 20 to R.H.S.)
RS Aggarwal Class 6 Solutions Chapter 9 Linear Equations in One Variable Ex 9B Q17.1

Question 18.
Solution:
16 (3x – 5) – 10 (4x – 8) = 40
=> 48x – 80 – 40x + 80 = 40
(Removing brackets)
=> 48x – 40 x – 80 + 80 = 40
=> 8x = 40
RS Aggarwal Class 6 Solutions Chapter 9 Linear Equations in One Variable Ex 9B Q18.1

Question 19.
Solution:
3 (x + 6) + 2 (x + 3) = 64
=> 3x + 18 + 2x + 6 = 64
(Removing brackets)
=> 3x + 2x + 18 + 6 = 64
=> 5x + 24 = 64
=> 5x = 64 – 24
RS Aggarwal Class 6 Solutions Chapter 9 Linear Equations in One Variable Ex 9B Q19.1

Question 20.
Solution:
3(2 – 5x) – 2 (1 – 6x) = 1
=> 6 – 15x – 2 + 12x = 1
(Removing brackets)
=> 6 – 2 – 15x + 12x = 1
=> 4 – 3x = 1
– 3x = 1 – 4
RS Aggarwal Class 6 Solutions Chapter 9 Linear Equations in One Variable Ex 9B Q20.1

Question 21.
Solution:
\(\\ \frac { n }{ 4 } -5\) = \(\\ \frac { n }{ 6 } \) + \(\\ \frac { 1 }{ 2 } \)
Multiplying each term by 12, the L.C.M. of 4, 6, 2, we get
RS Aggarwal Class 6 Solutions Chapter 9 Linear Equations in One Variable Ex 9B Q21.1

Question 22.
Solution:
\(\\ \frac { 2m }{ 3 } +8\) = \(\\ \frac { m }{ 2 } -1\)
Multiplying each term by 6, the L.C.M. of 2 and 3, we get
\(\\ \frac { 2m }{ 3 } \) x 6 + 8 x 6 = \(\\ \frac { m }{ 2 } \) x 6 – 1 x 6
RS Aggarwal Class 6 Solutions Chapter 9 Linear Equations in One Variable Ex 9B Q22.1

Question 23.
Solution:
\(\\ \frac { 2x }{ 5 } \) – \(\\ \frac { 3 }{ 2 } \) = \(\\ \frac { x }{ 2 } +1\)
Multiplying each term by 10, the L.C.M. of 5 and 2, we get
RS Aggarwal Class 6 Solutions Chapter 9 Linear Equations in One Variable Ex 9B Q23.1
RS Aggarwal Class 6 Solutions Chapter 9 Linear Equations in One Variable Ex 9B Q23.2

Question 24.
Solution:
\(\\ \frac { x-3 }{ 5 } \) – 2 = \(\\ \frac { 2x }{ 5 } \)
multiplying each term by 5, we get
RS Aggarwal Class 6 Solutions Chapter 9 Linear Equations in One Variable Ex 9B Q24.1
RS Aggarwal Class 6 Solutions Chapter 9 Linear Equations in One Variable Ex 9B Q24.2

Question 25.
Solution:
\(\\ \frac { 3x }{ 10 } \) – 4 = 14
=> \(\\ \frac { 3x }{ 10 } \) = 14 + 4
RS Aggarwal Class 6 Solutions Chapter 9 Linear Equations in One Variable Ex 9B Q25.1
RS Aggarwal Class 6 Solutions Chapter 9 Linear Equations in One Variable Ex 9B Q25.2

Question 26.
Solution:
\(\\ \frac { 3 }{ 4 } (x-1)\) = (x – 3)
RS Aggarwal Class 6 Solutions Chapter 9 Linear Equations in One Variable Ex 9B Q26.1

Hope given RS Aggarwal Solutions Class 6 Chapter 9 Linear Equations in One Variable Ex 9B are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

CA Foundation Business Economics Study Material – Production Optimisation

CA Foundation Business Economics Study Material Chapter 3 Theory of Production and Cost – Production Optimisation

Production Optimisation
Isoquants:

An iso-product curve or isoquant is a curve, which represents the various combinations of two variable inputs that give the same level of output. As all combinations on the iso-product curve give the same level of output, the producer becomes indifferent to these combinations. That is why iso-product curve are also called ‘production indifference curve’ or ‘equal product curve’. To understand consider the following production isoquant schedule.

CA Foundation Business Economics Study Material Production Optimisation 1

In the schedule I above, the producer is indifferent whether he gets combination A, B, C, D or E. This is because all the combinations of capital and labour give the same level of output i.e. 100 units.

By plotting the above combinations on a graph, we can derive an iso-product curve as shown in the following figure:

CA Foundation Business Economics Study Material Production Optimisation 2

In the diagram, quantity of capital is measured on X-axis and quantity of labour on Y-axis.

The various combinations A, B, C, D, E of capital and labour are plotted and on joining them we derive an iso-product curve. All combinations lying on the iso-product curve yield the same level of output i.e. 100 units and hence technically equally efficient.

If the production schedule II is also plotted on the graph, we will get another iso-product curve IQ200. This will lie above the IQ100 as the combinations contain greater quantities of capital and labour. A set of iso-product curves is called iso-product curve map.

CA Foundation Business Economics Study Material Production Optimisation 3

In the diagram, it can be observed that each iso-product curve is labelled in terms of output. All combinations lying of IQ100 give the output of 100 units and all the combinations lying on IQ200 give the output of 200 units. Higher iso-product curve represent higher level of output. Also it indicates how much more output can be achieved.

Marginal Rate of Technical Substitution
The rate at which one factor of production is substituted in place of the other factor without any change in the level of output is called as the marginal rate of technical substitution. Consider the following schedule.

CA Foundation Business Economics Study Material Production Optimisation 4

Each of the factor combinations in the table above yields same level of output. Moving from combination A to B, one unit of capital replaces 4 units of labour. Similarly, moving from B to C, one unit of capital now replaces only 3 units of labour and so on. It implies that labour and capital are imperfect substitutes. That is why MRTSKL is continuously diminishing. We can measure MRTSKL on an iso-product curve.

‘Iso-Cost Line’ OR ‘Equal Cost Lines’
Iso-cost line (also known Equal Cost Line; Price Line; Outlay Line; Factor Price Line) shows the various combinations of two factor inputs which the firm can purchase with a given outlay (i.e. budget) and at given prices of two inputs.

Example. A firm has with itself Rs. 1,000 which it would like to spend on factor ‘X’ and factor ‘Y’.
Price of factor ‘X’ is Rs. 20 per unit.
Price of factor ‘Y’ is Rs. 10 per unit.
Therefore, if the firm spends the whole amount on factor X, it can buy 50 units of X and if the whole amount is spent on factor Y, it can buy 100 units of Y. However, in between these two extreme limits, it can have many combinations of X and Y for the outlay of Rs. 1,000. Graphically it can be shown as follows –

CA Foundation Business Economics Study Material Production Optimisation 5

In the diagram OP shows 100 units of Y and OM shows 50 units of X. When we join the two points P and M, we get the iso-cost line. All the combinations of factor X and factor Y lying on iso-cost line can be purchased by the firm with an outlay of Rs. 1,000. If the firm increases the outlay to Rs. 2,000, the iso-cost line shifts to the right, if prices of two factors remains unchanged. The slope of the iso-cost line is equal to the ratio of the prices of two factors. Thus,
CA Foundation Business Economics Study Material Production Optimisation 6

Producer’s Equilibrium OR Production Optimization
A firm always try to produce a given level of output at minimum cost. For this it has to use that combination of inputs which minimizes the cost of production. This ensures maximization of profits and produce a given level of output with least cost combination of inputs. The least-cost combination of inputs or factors is called producer’s equilibrium or production optimization. This is determined with the help of (a) isoquants, & (b) iso-cost line.

An isoquant or iso-product curve is a curve which shows the various combinations of two inputs that produce same level of output. The isoquants are negatively sloped and convex to origin. The slope of isoquants shows the marginal rate of technical substitution which diminishes. Thus, MRTSxy
CA Foundation Business Economics Study Material Production Optimisation 7
Iso-cost line shows the various combination of two factor inputs which the firm can purchase with a given outlay and at given prices of inputs. There can be different outlays and hence different iso-cost lines. Slope of iso-cost line shows the ratio of the price of two inputs i.e. Px/Py

CA Foundation Business Economics Study Material Production Optimisation 8

Which will be the least cost combination can be understood with the help of following figure. Suppose firm wants to produce 300 units of a commodity. It will first see the isoquant that represents 300 units.

In the adjoining diagram we find that all combinations a, b, c, d and e can produce 300 units of output. In order to produce 300 units firm with try to find out least cost combination. For this it will super impose the various iso-cost lines on isoquant as shown in the diagram. The diagram shows that combination ‘C’ is,the least cost combination as here isoquant is tangent to iso-cost line HI. All other combinations a, b, d and e lying on isoquant cost more as these points lie on higher iso-cost lines. Hence, the point of tangency of isoquant and iso-cost line shows least cost combination. At the point of tangency.

Slope of iso-quant = Slope of iso-cost line

CA Foundation Business Economics Study Material Production Optimisation 9
Thus, the firm will choose OM units of factor X and ON units of factor Y and be at equilibrium as the marginal physical products of two factors are proportional to the factor prices.

CA Foundation Business Economics Study Material – Law of Returns to Scale

CA Foundation Business Economics Study Material Chapter 3 Theory of Production and Cost – Law of Returns to Scale

Law of Returns to Scale

  • The Law of Returns to Scale examines the production function i.e. the input – output relation in long run where increase in output can be achieved by varying the units of ALL FACTORS IN THE SAME PROPORTION.
  • Thus, in long run all factors become variable.
  • It means that in long run the scale of production and the size of the firm can be increased.

The law of returns to scale analyse the effects of scale on the level of output as-

  1. Increasing Returns to Scale:
    • When the output increases by a greater proportion than the proportion increases in all the factor inputs, it is increasing returns to scale.
    • E.g. When all inputs are increased by 10% and output rises by 30%.
    • The reasons of increasing returns to scale are – internal and external economies of scale; indivisibility of fixed factors; improved organisation; division of labour and specialisation; better supervision and control; adequate supply of productive factors, etc.
  2. Constant Returns to Scale:
    • When the output increases exactly in the same proportion as that of increase in all factor inputs, it is constant returns to scale.
    • E.g. – When all inputs are increased by 10% and output also rises by 10%.
    • The reason of constant returns to scale is that beyond a certain point, internal and external economies are NEUTRALISED by growing internal and external diseconomies.
  3. Diminishing Returns to Scale:
    • When the output increases by a lesser proportion than the proportion increase in all the factor inputs, it is diminishing returns to scale.
    • E.g. When all inputs are increased by 20% but output rises by 10%.
    • The reason of diminishing returns to scale is increased internal and external diseconomies of production.
    • Internal diseconomies like difficulties in management, lack of supervision and control, delay in decision-making etc.
    • External diseconomies like insufficient transport system, high freights, high prices of raw materials, power cuts, etc.

The law of returns to scale can also be illustrated with the help of the following schedule and diagram.
CA Foundation Business Economics Study Material Law of Returns to Scale 1
CA Foundation Business Economics Study Material Law of Returns to Scale 2

Returns to Factor and Returns to Scale

Returns to Factor Returns to Scale
1. Meaning
  • Returns to factor refers to the various production sizes where one factor is variable and other factor of production are fixed.
  • In other words, it examines production function when the output is increased by varying the quantity of one input.
  • It examines the effect of CHANGE IN THE PROPORTIONS between inputs on output.
  • Returns to scale refers to the various production sizes where increase in output can be achieved by varying the units of ALL FACTORS in the SAME PROPORTIONS.
  • It show the effects on output when all factor inputs are varied in the same proportion simultaneously.
2. Nature of Inputs
  • Quantities of some inputs are fixed while the quantities of other inputs vary.
  • In other words, there are FIXED and VARIABLE factors of production.
  • Quantities of all inputs can be varied.
  • In other words, all factors of production are VARIABLE.
3. Time Element
  • Returns to factor is called a SHORT RUN production function.
  • Returns to scale is called a LONG RUN production function.
4. Application
  • It does not apply where the factors must be used in fixed proportion to produce a commodity.
  • It does apply where the factors must be used in fixed proportions to produce a commodity.
5. Stages of Law
  • The law has three stages namely –
    (a)    Increasing Returns to factor,
    (b)   Diminishing Returns to Factor, &
    (c)   Negative Returns to factor ‘
  • Of the three stages, diminishing returns pre-dominate.
  • The law has three stages namely –
    (a)    Increasing Returns to Scale,
    (b)   Constant Returns to Scale,
    (c)   Diminishing Returns to Scale.
  • All the three stages of return appear.
6. Causes of Operation
  • Increasing returns to factor is due to indivisibility of fixed factors and division of labour and specialisation.
  • Diminishing returns is due to non- optimal factor proportion and imperfect substitutability of factors.
  • Negative returns fall in the efficiency of fixed and variable factors.
  • Increasing returns to scale is due to increased internal and external economies.
  • Constant returns to scale is due to the fact that internal and external economies are neutralised by growing internal and external diseconomies.
  • Diminishing returns is due to internal and external diseconomies of scale.
7. Scale of Production
  • The scale of output is unchanged and the production plant or the size and efficiency of the firm remain constant.
  • This is because, only one factor is variable and all other factors are fixed.
  • The scale of output can be increased and so the size of the firm too can be expanded.
  • This is because all factors are variable and hence can be increased in the same proportion simultaneously.

RS Aggarwal Class 6 Solutions Chapter 9 Linear Equations in One Variable Ex 9A

RS Aggarwal Class 6 Solutions Chapter 9 Linear Equations in One Variable Ex 9A

These Solutions are part of RS Aggarwal Solutions Class 6. Here we have given RS Aggarwal Solutions Class 6 Chapter 9 Linear Equations in One Variable Ex 9A.

Other Exercises

Question 1.
Solution:
Let x be the given number, then
(i) 5x = 40
(ii) x + 8 = 15
(iii) 25 – x = 7
(iv) x – 5 = 3
(v) 3x – 5 = 16
(vi) x – 12 = 24
(vii) 19 – 2x = 11
(viii) \(\\ \frac { x }{ 8 } \) = 7
(ix) 4x – 3 = 17
(x) 6x = x + 5

Question 2.
Solution:
(i) 7 less than from the number x is 14.
(ii) Twice the number y is 18.
(iii) 11 increased by thrice the number x is 17.
(iv) 3 less than twice the number x is 13.
(v) 30 less than 12 times the number is 6.
(vi) Quotient of twice the number z and 3 is

Question 3.
Solution:
(i) The given equation is 3x – 5 = 7
Substituting x = 4, we get
L.H.S. = 3 x – 5
= 3 x 4 – 5
= 12 – 5
= 7 = R.H.S.
It is verified that x = 4 is the root of the given equation.
(ii) The given equation is 3 + 2 x = 9
Substituting x = 3, we get L.H.S. = 3 + 2x
= 3 + 2 x 3
= 3 + 6 = 9
= R.H.S.
It is verified that x = 3 is the root of the given equation.
(iii) The given equation is 5x – 8 = 2x – 2
Substituting x = 2, we get
L.H.S. = 5x – 8
=5 x 2 – 8
= 10 – 8
= 2
R.H.S. = 2x – 2
= 2 x 2 – 2
= 4 – 2
= 2
L.H.S. = R.H.S.
Hence, it is verified that x = 2, is the root of the given equation.
(iv) The given equation is 8 – 7y = 1 Substituting y = 1, we get L.H.S. = 8 – 7y
= 8 – 7 x 1
= 8 – 7
= 1
= R.H.S.
Hence, it verified that y = 1 is the root of the given equation.
(v) The given equation is \(\\ \frac { z }{ 7 } \) = 8
Substituting the value of z = 56, we get
L.H.S.= \(\\ \frac { 56 }{ 7 } \)
= 8
= R.H.S.
Hence, it is verified that z = 56 is the root of the given equation.

Question 4.
Solution:
(i) The given equation is y + 9 = 13
We try several values of y and find L.H.S. and the R.H.S. and stop when L.H.S. = R.H.S.
RS Aggarwal Class 6 Solutions Chapter 9 Linear Equations in One Variable Ex 9A Q4.1
When y = 4, we have L.H.S. = R.H.S.
So y = 4 is the solution of the given equation.
(ii) The given equation is x – 1 = 10
We guess and try several values of x to find L.H.S. and R.H.S. and stop when L.H.S. = R.H.S.
RS Aggarwal Class 6 Solutions Chapter 9 Linear Equations in One Variable Ex 9A Q4.2
When x = 17, we hive L.H.S. = R.H.S
So x = 17 is the solution of the given equation.
(iii) The given equation is 4x = 28
We guess and try several values of x to find L.H.S. and R.H.S. and stop when L.H.S. = R.H.S.
RS Aggarwal Class 6 Solutions Chapter 9 Linear Equations in One Variable Ex 9A Q4.3
When x = 7, we have L.H.S. = R.H.S.
So x = 7 is the solution of the given equation.
(iv) The given equation is 3y = 36
We guess and try several values of y to find L.H.S. and R.H.S. and stop when L.H.S. = R.H.S.
RS Aggarwal Class 6 Solutions Chapter 9 Linear Equations in One Variable Ex 9A Q4.4
When y = 12, we have L.H.S. = R.H.S.
So y = 12 is the solution of the given equation.
(v) The given equation is 11 + x = 19
We guess and try several values of x to find L.H.S. and R.H.S. and stop when L.H.S. = R.H.S.
RS Aggarwal Class 6 Solutions Chapter 9 Linear Equations in One Variable Ex 9A Q4.5
When x = 8, we have L.H.S. = R.H.S.
So, x = 8 is the solution of the given equation.
(vi) The given equation is \(\\ \frac { x }{ 3 } \) = 4
We guess and try several values of x to find L.H.S. and R.H.S. and stop when L.H.S. = R.H.S.
RS Aggarwal Class 6 Solutions Chapter 9 Linear Equations in One Variable Ex 9A Q4.6
When x = 12, we have L.H.S. = R.H.S.
So, x = 12 is the solution of the given equation.
(vii) The given equation is 2 x – 3 = 9
We guess and try several values of x to find L.H.S. and R.H.S. and stop when
RS Aggarwal Class 6 Solutions Chapter 9 Linear Equations in One Variable Ex 9A Q4.7
.’. When x = 6, we have L.H.S. = R.H.S.
So, x = 6 is the solution of the given equation.
L.H.S. = R.H.S.
(viii) The given equation is \(\\ \frac { 1 }{ 2 } \) x + 7 = 11
We guess and try several values of x to find L.H.S. and R.H.S. and stop when L.H.S. = R.H.S.
RS Aggarwal Class 6 Solutions Chapter 9 Linear Equations in One Variable Ex 9A Q4.8
When x = 8, we have L.H.S. = R.H.S.
So, x = 8 is the solution of the given equation.
(ix) The given equation is 2y + 4 = 3y (x)
We guess and try several values of z to find L.H.S. and R.H.S. and stop when L.H.S. = R.H.S.
RS Aggarwal Class 6 Solutions Chapter 9 Linear Equations in One Variable Ex 9A Q4.9
When y = 4, we have L.H.S. = R.H.S. So, y = 4 is the solution of the given equation
(x) The given equation is z – 3 = 2z – 5
We guess and try several values of z to find L.H.S. and R.H.S. and stop when L.H.S. = R.H.S
RS Aggarwal Class 6 Solutions Chapter 9 Linear Equations in One Variable Ex 9A Q4.10
When z = 2, we have L.H.S. = R.H.S. So, z = 2 is the solution of the given equation.

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