RS Aggarwal Class 6 Solutions Chapter 15 Polygons Ex 15

These Solutions are part of RS Aggarwal Solutions Class 6. Here we have given RS Aggarwal Solutions Class 6 Chapter 15 Polygons Ex 15

Question 1.
Solution:
(a), (b), (d) and (f) are simple closed figures.

Question 2.
Solution:
(a), (b) and (c) are polygons.

Question 3.
Solution:
(i) two
(ii) triangle
(iv) 3, 3
(v) 4, 4
(vi) closed figure.

Hope given RS Aggarwal Solutions Class 6 Chapter 15 Polygons Ex 15 are helpful to complete your math homework.

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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.

(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°

(3) 90°
Steps of construction :
(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 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.

(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.

(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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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RS Aggarwal Class 6 Solutions Chapter 11 Line Segment, Ray and Line Ex 11A

These Solutions are part of RS Aggarwal Solutions Class 6. Here we have given RS Aggarwal Solutions Class 6 Chapter 11 Line Segment, Ray and Line Ex 11A.

Other Exercises

Question 1.
Solution:
Line segment
(i) In figure (i) are $$\overline { XY }$$ and $$\overline { YZ }$$
(ii) In figure (ii) $$\overline { AD } ,\overline { AB } ,\overline { AC } ,\overline { AE } ,\overline { BD } ,\overline { BC } ,\overline { CE }$$
(iii) In figure (iii) $$\overline { PQ } ,\overline { PR } ,\overline { PS } ,\overline { QR } ,\overline { QS } ,\overline { RS }$$

Question 2.
Solution:
(i) In fig. (i), line segments is $$\overline { AB }$$, and
rays are $$\overrightarrow { AC }$$ and $$\overrightarrow { BD }$$.
(ii) In fig. (ii), line segments are
$$\overline { GE } ,\overline { GP } ,\overline { EP }$$ and rays are
$$\overrightarrow { EF } ,\overrightarrow { GH } ,\overrightarrow { PQ }$$
(iii) In fig. (iii), line segments are $$\overline { OL } ,\overline { OP }$$
and rays are $$\overrightarrow { LM } ,\overrightarrow { PQ }$$.

Question 3.
Solution:
(i) Four line segments are $$\overline { PR } ,\overline { QR } ,\overline { PQ } ,\overline { RS }$$.
(ii) Four ray can be $$\overrightarrow { PA },\overrightarrow { QC }, \overrightarrow { RB } ,\overrightarrow { SD }$$
(iii) $$\overline { PR } ,\overline { QS }$$ are two non-intersecting lines.

Question 4.
Solution:
(i) Three or more points in a plane are said to be collinear if they all lie on the same line.

(ii) In the figure given above, points A, B, C are collinear points.
We can draw exactly one line passing through three collinear points

Question 5.
Solution:
(i) Four pairs of intersecting lines are : (AB, PQ) ; (AB, RS) ; (CD, PQ) ; (CD, RS)
(ii) Four collinear points are : A, Q, S, B
(iii) Three non-collinear points are : A, Q, P
(iv) Three concurrent lines are : AB, PS and RS.
(v) Three lines whose point of intersection is P are : CD, PQ and PS.

Question 6.
Solution:
The lines drawn through given points A, B, C are as shown below. The names of these lines are AB, BC and AC.

Also it is clear that three different lines can be drawn.

Question 7.
Solution:
(i) In the the given figure, there are six line segments, namely AB, AC, AD, BD, BC, DC.

(ii) In the given figure, there are ten line segments, namely, AD, AB, AC, AO, OC, BC, BD, BO, OD, CD.

(iii) In the given figure, there are six line segments, namely AB, AF, BF, CD, DE, CE.

(iv) In the given figure, there are twelve line segments, namely, AB, BC, CD, AD, BF, CG, DH, AE, EF, FG, GH, EH.

Question 8.
Solution:
$$\overleftrightarrow { PQ }$$ is a line

(i) False, as M does not lie on $$\overrightarrow { NQ }$$
(ii) True
(iii) True
(iv) True
(v) True

Question 9.
Solution:
(i) False
Point has no dimensions.
(ii) False
A line segment has a length.
(iii) False
A ray has no finite length.
(iv) False
If $$\overrightarrow { AB }$$ and ray $$\overrightarrow { BA }$$ have opposite directions.
(v) True
Length of $$\overline { AB }$$ and $$\overline { BA }$$ is same.
(vi) True
Line $$\overleftrightarrow { AB }$$ and $$\overleftrightarrow { BA }$$ are same.
(vii) True
Distance between A and B or B and A is same, so they determine a unique line segment.
(viii) True
Two lines intersect each other at one point.
(ix) False
Two intersecting planes intersect at one line not at one point.
(x) False
If A, B, C are collinear and points D, E are collinear then it is not necessary that there points A, B, C, D and E are collinear.
(xi) False
Infinite number of rays can be drawn with a given end point.
(xii) True
We can draw one and only one line passing through two given points.-
(xiii) True
We can draw infinite number of lines pass through a given point.

Question 10.
Solution:
(i) definite
(ii) one
(iii) no
(iv) definite
(v) cannot. Ans.

Hope given RS Aggarwal Solutions Class 6 Chapter 11 Line Segment, Ray and Line Ex 11A are helpful to complete your math homework.

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RS Aggarwal Class 6 Solutions Chapter 10 Ratio, Proportion and Unitary Method Ex 10D

These Solutions are part of RS Aggarwal Solutions Class 6. Here we have given RS Aggarwal Solutions Class 6 Chapter 10 Ratio, Proportion and Unitary Method Ex 10D.

Other Exercises

OBJECTIVE QUESTIONS
Mark against the correct answer in each of the following :

Question 1
Solution:
(d) $$\\ \frac { 92 }{ 115 }$$

Question 2.
Solution:
(a) $$\\ \frac { 57 }{ x }$$
= $$\\ \frac { 51 }{ 85 }$$

Question 3.
Solution:
(a) $$\\ \frac { 25 }{ 35 }$$
= $$\\ \frac { 45 }{ x }$$

Question 4.
Solution:
(c) $$\\ \frac { 4 }{ 5 }$$
= $$\\ \frac { x }{ 35 }$$

Question 5.
Solution:
(b) $$\\ \frac { a }{ b }$$
= $$\\ \frac { c }{ d }$$

Question 6.
Solution:
(b) a : b :: b : c

Question 7.
Solution:
(b) $$\\ \frac { 5 }{ 8 }$$ < $$\\ \frac { 3 }{ 4 }$$ => 4 x 5 < 3 x 8 => 20 < 24

Question 8.
Solution:
Total amount = Rs 760
Ratio A : B = 8 : 11

Question 9.
Solution:
(d) largest = $$\\ \frac { 252\times 7 }{ 5+7 }$$
= $$\\ \frac { 252\times 7 }{ 12 }$$
= 21 x 7
= 147

Question 10.
Solution:
(b) largest = $$\\ \frac { 90\times 5 }{ 1+3+5 }$$
= $$\\ \frac { 90\times 5 }{ 9 }$$
= 50 cm

Question 11.
Solution:
(c) total strength of school
largest = $$\\ \frac { 840 }{ 5 }$$ x (12 + 5)
= $$\\ \frac { 840\times 17 }{ 5 }$$
= 168 x 17
= 2856

Question 12.
Solution:
(b) Cost of 12 pens = Rs 138
Cost of 1 pen = Rs $$\\ \frac { 138\times 14 }{ 12 }$$
and cost of 14 pens = Rs 161

Question 13.
Solution:
(b) $$\\ \frac { 24\times 15 }{ 8 }$$
= 45 days

Question 14.
Solution:
(a) $$\\ \frac { 26\times 40 }{ 20 }$$
= 52 men

Question 15.
Solution:
(b) In 6 L of petrol, a car covers = 111 km
In 1 L, it will cover = $$\\ \frac { 111 }{ 6 }$$ km
and in 10 L it will cover = $$\\ \frac { 111\times 10 }{ 6 }$$ km
= 185 km

Question 16.
Solution:
(a) $$\\ \frac { 28\times 550 }{ 700 }$$
= 22 days

Question 17.
Solution:
Ratio in the angles of triangle
= 3 : 1 : 2

Question 18.
Solution:
(b) Ratio in length and breadth of a rectangle = 5 : 4
Width = 36 m
Length = $$\\ \frac { 5 }{ 4 }$$ x 36 m
= 45m

Question 19.
Solution:
Bus covers in 3 hrs = 195 km
It will cover in 1 hr = $$\\ \frac { 195 }{ 3 }$$ = 65 km

Question 20.
Solution:
1 dozen = 12 bars
Cost of 5 bars of soap = Rs.82.50

Question 21.
Solution:
Total pencils in 30 packets of 8 pencils
= 30 x 8= 240
and total pencils of 25 packets of 12
pencils = 25 x 12 = 300
Now cost of 240 pencils = Rs. 600
Then cost of 1 percent = Rs. $$\\ \frac { 600 }{ 24 }$$
and cost of 300 pencils = Rs. $$\\ \frac { 600\times 300 }{ 240 }$$
= Rs 750 (b)

Question 22.
Solution:
Journey of 75 km costs = Rs 215
Cost of 1 km = Rs $$\\ \frac { 215 }{ 75 }$$

Question 23.
Solution:
1st term = 12
2nd term = 21
fourth term = 14

Question 24.
Solution:
10 boys dig a patch in = 12 hrs
1 boy will dig it in = 12 x 10 hours

Hope given RS Aggarwal Solutions Class 6 Chapter 10 Ratio, Proportion and Unitary Method Ex 10D are helpful to complete your math homework.

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RS Aggarwal Class 6 Solutions Chapter 10 Ratio, Proportion and Unitary Method Ex 10C

These Solutions are part of RS Aggarwal Solutions Class 6. Here we have given RS Aggarwal Solutions Class 6 Chapter 10 Ratio, Proportion and Unitary Method Ex 10C.

Other Exercises

Question 1.
Solution:
Cost of 14 m of cloth = Rs. 1890
Cost of 1 m = Rs. $$\\ \frac { 1890 }{ 14 }$$
and cost of 6 m = Rs. $$\\ \frac { 1890\times 6 }{ 14 }$$
= Rs. 135 x 6
= Rs. 810

Question 2.
Solution:
Cost of 1 dozen or 12 soaps = Rs. 285.60
Cost of 1 soap = Rs. $$\\ \frac { 285.60 }{ 12 }$$
Cost of 15 soaps = Rs. $$\\ \frac { 285.60\times 15 }{ 12 }$$
= Rs. 357.00

Question 3.
Solution:
Cost of 9 kg of rice = Rs. 327.60
Cost of 1 kg = Rs. $$\\ \frac { 327.60 }{ 9 }$$
and cost of 50 kg = Rs. $$\\ \frac { 327.60\times 50 }{ 9 }$$
= Rs. 36.40 x 50
= Rs. 1820

Question 4.
Solution:
Weight of 22.5 metres of the iron rod: = 85.5 kg
Weight of 1 metre of the iron rod

Question 5.
Solution:
Quantity of oil in 15 tins = 234 kg
Quantity of oil in 1 tin = $$\\ \frac { 234 }{ 15 }$$ kg
Quantity of oil in 10 tins

Question 6.
Solution:
Distance covered by the car in 12 litres of diesel = 222 kms
Distance covered by the car in 1 litre of diesel = $$\\ \frac { 222 }{ 12 }$$ km

Question 7.
Solution:
Charges of 25 tonnes of weight = Rs. 540
charges of 1 ton = Rs.$$\\ \frac { 540 }{ 25 }$$

Question 8.
Solution:
Weight of copper in 4.5 g of alloy = 3.5g
Weight of copper in 1 g of alloy

Question 9.
Solution:
In Rs. 87.50, the inland letter are purchased = 35
In Re. 1, letters can be purchased
= $$\\ \frac { 35 }{ 87.50 }$$
and in Rs. 315, letters can be purchased

Question 10.
Solution:
4 dozen = 4 x 12 = 48 bananas
In Rs. 104, banana are purchased = 48

Question 11.
Solution:
In Rs. 22770, chairs are purchased =18
In Re. 1, chairs will be purchased
= $$\\ \frac { 18 }{ 22770 }$$
and in Rs. 10120, chairs will be
purchased = $$\\ \frac { 18\times 10120 }{ 22770 }$$
= 8

Question 12.
Solution:
(i) A car travels 195 km distance in = 3 hours
It will travel 1 km distance in = $$\\ \frac { 3 }{ 195 }$$ hr.
and it will travel 520 km distance in
= $$\\ \frac { 3\times 520 }{ 195 }$$
= 8 hr
(ii) A car travels in 3hr = 195 km

Question 13.
Solution:
(i) A laborer earn in 12 days = Rs. 1980
He will earn in 1 day = Rs. $$\\ \frac { 1980 }{ 12 }$$
and he will earn in 7 days

Question 14.
Solution:
(i) Weight of 65 books = 13 kg
Then weight of 1 book = $$\\ \frac { 13 }{ 65 }$$ kg

Question 15.
Solution:
Number of boxes needed for 6000 pens = 48
Number of boxes needed for 1 pen 48 = $$\\ \frac { 48 }{ 6000 }$$

Question 16.
Solution:
Clearly, less workers will build the wall in more days.
And, more workers will build the wall in less days.
24 workers can build the wall in 15 days
1 worker can build the wall in (15 x 24) days
(less worker, more days)
9 workers will build the wall in

Question 17.
Solution:
Men needed to finish a piece of work in 26 days = 40
Men needed to finish a piece of work in 1 day = 40 x 26 (less days, more men)

Question 18.
Solution:
Clearly, less men will take more days to consume the food.
And, more men will take less days to consume the food.
550 men have provisions for 28 days
1 men has provisions for (28 x 550) days [less men, more days]
700 men will have provisions for

Question 19.
Solution:
Clearly, less persons will consume the rice in more days.
And more persons will consume the rice in less days.
60 persons consume the bag of rice in 3 days.
1 person will consume the bag of rice in
(3 x 60) days (less persons, more days)
18 persons will consume the bag of rice

Hope given RS Aggarwal Solutions Class 6 Chapter 10 Ratio, Proportion and Unitary Method Ex 10C are helpful to complete your math homework.

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RS Aggarwal Class 6 Solutions Chapter 10 Ratio, Proportion and Unitary Method Ex 10B

These Solutions are part of RS Aggarwal Solutions Class 6. Here we have given RS Aggarwal Solutions Class 6 Chapter 10 Ratio, Proportion and Unitary Method Ex 10B.

Other Exercises

Question 1.
Solution:
(i) 4, 6, 8, 12
If it is in proportion, then
If ad = bc if 4 x 12 = 6 x 8
If 48 = 48
Which is true
∴ 4, 6, 8, 12 are in proportion
(ii) 7, 42, 13, 78
7 : 42 :: 13 : 78
If ad = bc if 7 x 78 = 42 x 13
If 546 = 546
Which is true
∴ 7, 42, 13, 78 are in proportion
(iii) 33, 121, 9, 96 or 33 : 121 :: 9 : 96
are in proportion
If 33 x 96 = 121 x 9
If 3168 = 1089
Which is not true
∴ 33, 121,9, 96 are not in proportion
(iv) 22, 33, 42, 63 or 22 : 33 :: 42 : 63
are in proportion
If 22 x 63 = 33 x 42
If 1386 = 1386
Which is true
∴ 22, 33, 42, 63 are in proportion
(v) 32, 48, 70, 210 or 32 : 48 :: 70 : 210
are in proportion
If 32 x 210 = 48 x 70
If 6720 = 3360
Which is not true
∴ 32, 48, 70, 210 are not in proportion
(vi) 150, 200, 250, 300 or
150 : 200 :: 250 : 300 are in proportion
If ad = bc if 150 x 300 = 200 x 250
If 45000 = 50000
Which is not true
∴ 150, 200,250, 300 are not in proportion

Question 2.
Solution:
(i) We have 60 : 105 :: 84 : 147
Product of means = 105 x 84 = 8820
Product of extremes = 60 x 147 = 8820
∴ Product of means = Product of extremes
Hence 60 : 105 :: 84 : 147 is verified.
(ii) We have 91 : 104 :: 119 : 136
Product of means = 104 x 119 = 12376
Product of extremes = 91 x 136 = 12376
Product of means = Product of extremes
Hence 91 : 104 :: 119 : 136 is verified.
(iii) We have 108 : 72 :: 129 : 86
Product of means = 72 x 129 = 9288
Product of extremes = 108 x 86 = 9288
Product of means = Product of extremes
Hence 108 : 72 :: 129 : 86 is verified.
(iv) We have 39 : 65 :: 141 : 235
Product of means = 65 x 141 = 9165
Product of extremes = 39 x 235 = 9165
∴ Product of means = Product of extremes
Hence 39 : 65 :: 141 : 235 is verified.

Question 3.
Solution:
(i) We have 55 : 11 :: x : 6
Product of means = 11 × x = 11x
Product of extremes = 55 x 6 = 330

Question 4.
Solution:
(i) We have, 51 : 68 = $$\\ \frac { 51 }{ 68 }$$
= $$\\ \frac { 3 }{ 4 }$$

Question 5.
Solution:
(i) 25 cm : 1 m and Rs. 40 : Rs. 160
= $$\\ \frac { 25cm }{ 1000cm }$$ = $$\\ \frac { 1 }{ 4 }$$,
$$\\ \frac { Rs.40 }{ Rs.160 }$$ = $$\\ \frac { 1 }{ 4 }$$

Question 6.
Solution:
Let the third term be x.
Then 51 : 68 :: x : 108
Now, product of means = x × 68
And, product of extremes = 51 × 108
x × 68 = 51 × 108
=> x = $$\\ \frac { 51\times 108 }{ 68 }$$
= 3 × 27 = 81
x = 81
Hence the third term of the given proportion is 81

Question 7.
Solution:
1st term =12, third term = 8 and fourth term = 14
Let 2nd term = x, then

Question 8.
Solution:
(i) The given numbers 48, 60, 75 are in
continued proportion if 48 : 60 :: 60 : 75.
Now, product of means = 60 x 60 = 3600
And, product of extremes = 48 x 75 = 3600
∴ Product of means = Product of extremes
So, 48 : 60 :: 60 : 75
Hence, the numbers 48, 60, 75 are in continued proportion.
(ii) The given numbers 36, 90, 225 are in
continued proportion of 36 : 90 :: 90 : 225
Now, product of means = 90 x 90 = 8100
And, product of extremes = 36 x 225 = 8100
∴ Product of means = Product of extremes
So, 36 : 90 :: 90 : 225
Hence, the numbers 36, 90, 225 are in continued proportion.
(iii)The given numbers 16, 84, 441 are in
continued proportion if 16 : 84 :: 84 : 441.
Now, product of means = 84 x 84 = 7056
And, product of extremes = 16 x 441 = 7056
Product of means = Product of extremes.
So, 16 : 845 :: 84 : 441
Hence 16, 84, 441 are in continued proportion.
(iv) The given numbers 27, 36, 48 are
in continued proportion if 27 : 36 :: 36 : 48
Now, product of means = 36 x 36 = 1296
And, product of extremes = 27 x 48 = 1296
∴ Product of means = Product of extremes.
So, 27 : 36 :: 36 : 48
Hence, the numbers 27, 36, 48 are in continued proportional.

Question 9.
Solution:
It is given that 9, x, x, 49 are in proportion, that is, 9 : x :: x : 49
∴ Product of means = Product of extremes

Question 10.
Solution:
Let the height of the pole be x metres.

Question 11.
Solution:
5 : 3 :: x : 6

Hope given RS Aggarwal Solutions Class 6 Chapter 10 Ratio, Proportion and Unitary Method Ex 10B are helpful to complete your math homework.

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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

Question 3.
Solution:
Let the required number be x. Then,
x x 5 = x + 80
=> 5x = x + 80
=> 5x – x = 80

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

Question 5.
Solution:
Let the required number be x. Then
x x 17 + 4 = 225
=> 17x + 4 = 225
=> 17x = 225 – 4

Question 6.
Solution:
Let the required number be x. Then
3 x + 5 = 50
=> 3 x = 50 – 5

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

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

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.

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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
=> 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
=> 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
=> 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 $$\\ \frac { 5 }{ 3 }$$

Question 9.
Solution:
5x – 3 = x + 17
=> 5x – x = 17 + 3

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 + $$\\ \frac { 1 }{ 2 }$$

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

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

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

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.)

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

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

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

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

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

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

Question 24.
Solution:
$$\\ \frac { x-3 }{ 5 }$$ – 2 = $$\\ \frac { 2x }{ 5 }$$
multiplying each term by 5, we get

Question 25.
Solution:
$$\\ \frac { 3x }{ 10 }$$ – 4 = 14
=> $$\\ \frac { 3x }{ 10 }$$ = 14 + 4

Question 26.
Solution:
$$\\ \frac { 3 }{ 4 } (x-1)$$ = (x – 3)

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

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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.

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.

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.

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.

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.

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.

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

.’. 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.

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.

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

When z = 2, we have L.H.S. = R.H.S. So, z = 2 is the solution of the given equation.

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

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RS Aggarwal Class 6 Solutions Chapter 8 Algebraic Expressions Ex 8D

These Solutions are part of RS Aggarwal Solutions Class 6. Here we have given RS Aggarwal Solutions Class 6 Chapter 8 Algebraic Expressions Ex 8D.

Other Exercises

Simplify :

Question 1.
Solution:
We have : a – (b – 2a)
= a – b + 2a
= a + 2a – b
= (1 + 2) a – b
= 3a – b.

Question 2.
Solution:
We have : 4x – (3y – x + 2z)
= 4x – 3y + x – 2z
= 4x + x – 2y – 2z
= 5x – 3y – 2z

Question 3.
Solution:
We have :
(a2 + b2 + 2ab) – (a2 + b2 – 2ab)
= a2 + b2 + 2ab – a2 – b2 + 2ab
= a2 – a2 + b2 – b2 + 2ab + 2ab
= 0 + 0 + (2 + 2) ab
= 4 ab

Question 4.
Solution:
We have :
– 3 (a + b) + 4 (2a – 3b) – (2a – b)
= – 3a – 3b + 8a – 12b – 2a + b
= – 3a + 8a – 2a – 3b – 12b + b
= ( – 3 + 8 – 2) a + ( – 3 – 12 + 1) b
= 3a – 14 b.

Question 5.
Solution:
We have :
– 4x2 + {(2x2 – 3) – (4 – 3x2)}
= – 4x2 + {2x2 – 3 – 4 + 3x2}
[removing grouping symbol]
= – 4x2 + {5x2 – 7)
= – 4x2 + 5x2 – 7
(removing grouping symbol {})
= x2 – 7

Question 6.
Solution:
We have :
– 2 (x2 – y+ xy) – 3 (x2 + y2 – xy)
= – 2x2 + 2y2 – 2xy – 3x2 – 3y2 + 3xy
= – 2x2 – 3x2 + 2y2 – 3y2 – 2xy + 3xy
= ( – 2 – 3)x2 + (2 – 3) y2 + ( – 2 + 3)xy
= – 5x2 – y2 + xy

Question 7.
Solution:
a – [2b – {3a – (2b – 3c)}]
= a – [2b – {3a – 2b + 3c}]
[removing grouping symbol( )]
= a – [2b – 3a + 2b – 3c]
(removing grouping symbol {})
= a – [4b – 3a – 3c]
= a – 4b + 3a + 3c
(removing grouping symbol [ ])
= 4a – 4b + 3c

Question 8.
Solution:
Removing the innermost grouping symbol ( ) first, then { } and then [ ], we have :
– x + [5y – {x – (5y – 2x)}]
= – x + [5y – {x – 5y + 2x}]
= – x + [5y – {3x – 5y}]
= – x + [5y – 3x + 5y]
= – x + [ 10y – 3x]
= – x + 10y – 3x
= – x – 3x + 10y
= – 4x + 10y

Question 9.
Solution:
Removing the innermost grouping symbol ( ) first, then { } and then [ ], we have :
86 – [15x – 7 (6x – 9) – 2 {10x – 5(2 – 3x)}]
= 86 – [15x – 42x + 63 – 2 {10x – 10 + 15x}
= 86 – [ 15x – 42x + 63 – 2 {25x – 10}]
= 86 – [15x – 42x + 63 – 50x + 20]
= 86 – 15x + 42x – 63 + 50x – 20
= (86 – 63 – 20) – 15x + 42x + 50x
= (86 – 83) + (- 15 + 42 + 50) x
= 3 + 77x

Question 10.
Solution:
Removing the innermost grouping ‘ symbol () first, then { } and then [ ], we have :
12x – [3x3 + 5x2 – {7x2 – (4 – 3x – x3) + 6x3} – 3x]
= 12x – [3x3 – 5x2 – {7x2 – 4 + 3x + x3 + 6x3} – 3x]
= 12x – [3x3 + 5x2 – {7x2 – 4 + 3x + 7x3} – 3x]
= 12x – [3x3 + 5x2 – 7x2 + 4 – 3x – 7x3 – 3x]
= 12x – [3x3 – 7x3 + 5x2 – 7x2 + 4 – 3x – 3x]
= 12x – [ – 4x3 + 2x2 + 4 – 6x]
= 12x + 4x3 + 2x2 – 4 + 6x
= 12x + 6x + 4x3 + 2x2 – 4
= 18x + 4x3 + 2x2 – 4
= 4x3 + 2x2 + 18x – 4

Question 11.
Solution:
Removing the innermost grouping symbol ( ) first, then { } and then [ ], we have
5a – [a2 – {2a (1 – a + 4a2) – 3a (a2 – 5a – 3)}] – 8a
= 5a – [a2 – {2a – 2a2 + 8a3 – 3a3 + 15a2 + 9a}] – 8a
= 5a – [a2 – {2a + 9a – 2a2 + 15a2 + 8a3 – 3a3}] – 8a
= 5a – [a2 – {11a + 13a2 + 5a3}] – 8a
= 5a – [a2 – 11a – 13a2 – 5a3] – 8a
= 5a – a2 + 11a + 13a2 + 5a3 – 8a
= 5a + 11a – 8a – a2 + 13a2 + 5a3
= 8a + 12a2 + 5a3
= 5a3 + 12a2 + 8a.

Question 12.
Solution:
Removing the innermost grouping symbol ( ) first, then { } and then [ ], we have :
3 – [x – {2y – (5x + y – 3) + 2x2} – (x2 – 3y)]
= 3 – [x – {2y – 5x – y + 3 + 2x2} – x2 + 3y]
= 3 – [x – {y – 5x + 3 + 2x2} – x2 + 3y]
= 3 – [x – y + 5x – 3 – 2x2 – x2 + 3y]
= 3 – [6x + 2y – 3 – 3x2]
= 3 – 6x – 2y + 3 + 3x2
= 6 – 6x – 2y + 3x2

Question 13.
Solution:
Removing the innermost grouping symbol ( ) first, then { } and then [ ], we have :
xy – [yz – zx – {yx – (3y – xz} – (xy – zy)}]
= xy – [yz – zx – {yx – 3y + xz – xy + zy}]
= xv – [yz – zx – {- 3y + xz + zy}]
= xy – [yz – zx + 3y – xz – zy]
= xy – [ – 2xz + 3y]
= xy + 2xz – 3y

Question 14.
Solution:
Removing the innermost grouping symbol ( ) first, then { } and then [ ], we have
2a – 3b – [3a – 2b – {a – c – (a – 2b)}]
= 2a – 3b – [3a – 2b – {a – c – a + 2b}]
= 2a – 3b – [3a – 2b – { – c + 2b}]
= 2a – 3b – [3a – 2b + c – 2b]
= 2a – 3b – 3a + 2b – c + 2b
= 2a – 3a – 3b + 2b + 2b – c
= – a + b – c

Question 15.
Solution:
Removing the innermost grouping symbol () first, then { } and ten [ ], we have:
– a – [a + {a + b – 2a – (a – 2b)} – b]
= – a – [a + {a + b – 2a – a + 2b} – b]
= – a – [a + { – 2a + 3b} – b]
= – a – [a – 2a + 3b – b]
= – a – a + 2a – 3b + b
= – 2a + 2a – 2b
= – 2 b

Question 16.
Solution:
Removing the innermost grouping symbol ‘—’ first, then ( ), then { } and then [ ], we have
2a – [4b – {4a – (3b – $$\overline { 2a+2b }$$)}]
= 2a – [4b – {4a – (3b – 2a – 2b)}]
= 2a – [4b – {4a – (b – 2a)}]
= 2a – [4b – {4a – b + 2a}]
= 2a – [4b – {6a – b}]
= 2a – [4b – 6a + b]
= 2a – [5b – 6a]
= 2a – 5b + 6a
= 8a – 5b.

Question 17.
Solution:
Removing the innermost grouping < symbol ( ) first, then { } and then [ ], we have :
5x – [4y – {7x – (3z – 2y) + 4z – 3(x + 3y – 2z)}]
= 5x – [4y – {7x – 3z + 2y + 4z – 3x – 9y + 6z}]
= 5x – [4y – {4x + 7z – 7y}]
= 5x – [4y – 4x – 7z + 7y]
= 5x – [11y – 4x – 7z]
= 5x – 11y + 4x + 7z
= 9x – 11y + 7z

Hope given RS Aggarwal Solutions Class 6 Chapter 8 Algebraic Expressions Ex 8D 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 8 Algebraic Expressions Ex 8C

These Solutions are part of RS Aggarwal Solutions Class 6. Here we have given RS Aggarwal Solutions Class 6 Chapter 8 Algebraic Expressions Ex 8C.

Other Exercises

Question 1.
Solution:
(i) The required sum
= 3x + 7x
= (3 + 7) x
= 10x

Question 2.
Solution:
we get

Question 3.
Solution:
(i) Arranging the like terms columnwise and adding, we get :

Question 4.
Solution:
(i) We have :
2x – 5x = (2 – 5)x = – 3x
(ii) We have :
6x – y – (- xy) = 6xy + xy = 7xy
(iii) We have : 5b – 3a
(iv) We have : 9y – ( – 7x) = 9y + 7x
(v) We have : – 7x2 – 10x2 = ( – 7 – 10)x2
= – 17x2
(vi) We have : b2 – a2 – (a2 – b2)
= b2 – a2 – a2 + b2
= b2 + b2 – a2 – a2
= (1 + 1) b2 + ( – 1 – 1) a2
= 2b2 – 2a2

Question 5.
Solution:
(i) Arranging the like terms columnwise, we get :

Question 6.
Solution:
(i) Rearranging and collecting the like terms, we get :

Question 7.
Solution:
We have:

Question 8.
Solution:
We have :
A = 7x2 + 5xy – 9y2
B = – 4x2 + xy + 5y2
C = 4y2 – 3x2 – 6xy

= 0+0+0 = 0
Hence the result

Question 9.
Solution:
Required expression

Question 10.
Solution:
Substituting the values of P, Q, R and S, we have :
P + Q + R + S = (a2 – b2 + 2ab)

Question 11.
Solution:
Required expression

Question 12.
Solution:
Required expression

Question 13.
Solution:
Required expression

Question 14.
Solution:
Required expression

Question 15.
Solution:
Sum of 5x – 4y + 6z and – 8x + y – 2z
= 5x – 4y + 6z – 8x + y – 2z
= 5x – 8x – 4y + y + 6z – 2z
= – 3x – 3y + 4z
Sum of 12x – y + 3z and – 3x + 5y – 8z
= 12x – y + 3z – 3x + 5y – 8z
= 12x – 3x – y + 5y + 3z – 8z
= 9x + 4y – 5z

Question 16.
Solution:
Required expression

Question 17.
Solution:
Required expression

Hope given RS Aggarwal Solutions Class 6 Chapter 8 Algebraic Expressions Ex 8C 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.