Category: CBSE Class 7

RS Aggarwal Class 7 Solutions Chapter 9 Unitary Method Ex 9B

These Solutions are part of RS Aggarwal Solutions Class 7. Here we have given RS Aggarwal Solutions Class 7 Chapter 9 Unitary Method Ex 9B.

Other Exercises

• RS Aggarwal Solutions Class 7 Chapter 9 Unitary Method Ex 9A
• RS Aggarwal Solutions Class 7 Chapter 9 Unitary Method Ex 9B
• RS Aggarwal Solutions Class 7 Chapter 9 Unitary Method Ex 9C
• RS Aggarwal Solutions Class 7 Chapter 9 Unitary Method CCE Test Paper

Question 1.
Solution:
48 men can dig a trench in = 14 days
1 man will dig the trench in = 14 x 48 days (less men more days)
28 men will dig the trench m = \(\frac { 14 x 48 }{ 28 }\) (more men less days)
= 24 days

Question 2.
Solution:
In 30 days, a field is reaped by = 16 men
In 1 day, it will be reaped by = 16 x 30 (less days, more men)
and in 24 days, it will be reaped by = \(\frac { 16 x 30 }{ 24 }\) men (more days, less men)
= \(\frac { 480 }{ 24 }\)
= 20 men

Question 3.
Solution:
In 13 days, a field is grazed by = 45 cows.
In 1 day, the field will be grazed by = 45 x 13 cows (less days, more cows)
and in 9 days the field will be grazed by = \(\frac { 45 x 13 }{ 9 }\) cows (more days, less cows)
= 5 x 13 = 65 cows

Question 4.
Solution:
16 horses can consume corn in = 25 days
1 horse will consume it in = 25 x 16 days (Less horse, more days)
and 40 horses will consume it in = \(\frac { 25 x 16 }{ 40 }\) days (more horses, less days)
= 10 days

Question 5.
Solution:
By reading 18 pages a day, a book is finished in = 25 days
By reading 1 page a day, it will be finished = 25 x 18 days (Less page, more days)
and by reading 15 pages a day, it will be finished in = \(\frac { 25 x 18 }{ 15 }\) days (more pages, less days)
= 5 x 6 = 30 days

Question 6.
Solution:
Reeta types a document by typing 40 words a minute in = 24 minutes
She will type it by typing 1 word a minute in = 24 x 40 minutes (Less speed, more time)
Her friend will type it by typing 48 words 24 x 40 a minute in = \(\frac { 24 x 40 }{ 48 }\) minutes
(more speed, less time)
= 20 minutes

Question 7.
Solution:
With a speed of 45 km/h, a bus covers a distance in = 3 hours 20 minutes
= 3\(\frac { 1 }{ 3 }\) = \(\frac { 10 }{ 3 }\) hours
With a speed of 1 km/h it will cover the distance m = \(\frac { 10 x 45 }{ 3 }\) h
(Less speed, more time)
and with a speed of 36 km/h, it will cover the distance in
= \(\frac { 10 x 45 }{ 3 x 36 }\) hr (more speed, less time)
= \(\frac { 25 }{ 6 }\) h
= 4\(\frac { 1 }{ 6 }\) h
= 4 hr 10 minutes

Question 8.
Solution:
To make 240 tonnes of steel, material is sufficient in = 1 month or 30 days
To make 1 tonne of steel, it will be sufficient in = 30 x 240 days (Less steel, more days)
To make 240 + 60 = 300 tonnes of steel it will be sufficient in = \(\frac { 30 x 240 }{ 300 }\) days
= 24 days (more steel, less days)

Question 9.
Solution:
In the beginning, number of men = 210
After 12 days, more men employed = 70
Total men = 210 + 70 = 280
Total period = 60 days.
After 12 days, remaining period = 60 – 12 = 48 days
Now 210 men can build the house in = 48 days
and 1 man can build the house in = 48 x 210 days (less men, more days) .
280 men can build the house in = \(\frac { 48 x 210 }{ 280 }\) days
(more men, less days)
= 36 days

Question 10.
Solution:
In 25 days, the food is sufficient for = 630 men
In 1 day, the food will be sufficient for = 630 x 25 men (less days, more men)
and in 30 days, the food will be sufficient for = \(\frac { 630 x 25 }{ 30 }\) hr
(more days less men)
= 525 men
Number of men to be transfered = 630 – 525 = 105 men

Question 11.
Solution:
Number of men in the beginning = 120
Number of men died = 30
Remaining = 120 – 30 = 90 men
Total period = 200 days
No. of days passed = 5
Remaining period = 200 – 5 = 195
Now, The food lasts for 120 men for = 195 days
The food will last for 1 man for = 195 x 120 days (Less men, more days)
The food will last for 90 men for = \(\frac { 195 x 120 }{ 90 }\)
(more men less days)
= 65 x 4 = 260 days

Question 12.
Solution:
Period in the beginning = 28 days
No. of days passed = 4 days.
Remaining period = 28 – 4 = 24 days
The food is sufficient for 24 days for = 1200 soldiers
The food will be sufficient for 1 day for = 1200 x 24 soldiers (Less days, more men)
and the food will be sufficient for 32 days = \(\frac { 1200 x 24 }{ 32 }\)
= 900 soldiers (more days, less men)
No. of soldiers who left the fort = 1200 – 900 = 300 soldiers

Hope given RS Aggarwal Solutions Class 7 Chapter 9 Unitary Method Ex 9B are helpful to complete your math homework.

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RS Aggarwal Class 7 Solutions Chapter 9 Unitary Method Ex 9A

These Solutions are part of RS Aggarwal Solutions Class 7. Here we have given RS Aggarwal Solutions Class 7 Chapter 9 Unitary Method Ex 9A.

Other Exercises

• RS Aggarwal Solutions Class 7 Chapter 9 Unitary Method Ex 9A
• RS Aggarwal Solutions Class 7 Chapter 9 Unitary Method Ex 9B
• RS Aggarwal Solutions Class 7 Chapter 9 Unitary Method Ex 9C
• RS Aggarwal Solutions Class 7 Chapter 9 Unitary Method CCE Test Paper

Question 1.
Solution:
Cost of 15 oranges = Rs. 110
Cost of 1 orange = Rs. \(\frac { 110 }{ 15 }\)
and cost of 39 oranges = Rs. \(\frac { 110 }{ 15 }\) x 39
= Rs. 22 x 13 = Rs. 286

Question 2.
Solution:
In Rs. 260, the sugar is bought = 8 kg
and in Re. 1, the sugar is bought = \(\frac { 8 }{ 260 }\) kg
Then in Rs. 877.50, the sugar will be bought = \(\frac { 8 }{ 260 }\) x 877.50 kg
= \(\frac { 8 }{ 260 }\) x \(\frac { 87750 }{ 100 }\)
= 27 kg

Question 3.
Solution:
In Rs. 6290, silk is purchased = 37 m
and in Re. 1, silk is purchased = \(\frac { 37 }{ 6290 }\) m
and in Rs. 4420, silk will be purchased 37
= \(\frac { 37 }{ 6290 }\) x 4420 m = 26 m

Question 4.
Solution:
Rs. 1110 is wages for = 6 days.
Re. 1 will be wages for = \(\frac { 6 }{ 1110 }\) days
and Rs. 4625 will be wages for

Question 5.
Solution:
In 42 litres of petrol, a car covers = 357 km
and in 1 litre, car will cover = \(\frac { 357 }{ 42 }\) km
and in 12 litres, car will cover = \(\frac { 357 }{ 42 }\) x 12 = 102 km

Question 6.
Solution:
Cost of travelling 900 km is = Rs. 2520
and cost of 1 km will be = Rs. \(\frac { 2520 }{ 900 }\)
andcostof360kmwillbe = Rs. \(\frac { 2520 }{ 900 }\) x 360 = Rs. 1008

Question 7.
Solution:
To cover a distance of 51 km, time is taken = 45 minutes

Question 8.
Solution:
If weight is 85.5 kg, then length of iron rod = 22.5 m
If weight is 1 kg, then length of rod will be

Question 9.
Solution:
In 162 grams, sheets are = 6

Question 10.
Solution:
1152 bars of soap can be packed in 8 cartons
1 bar of soap coil be packed in

Question 11.
Solution:
In 44 mm of thickness, cardboards are = 16
In 1 mm of thickness, cardboards will be

Question 12.
Solution:
If length of shadow is 8.2 m, then
height of flag staff is = 7 m
If length of shadow is 1 m, then height will

Question 13.
Solution:
16.25 m long wall is build by = 15 men

Question 14.
Solution:
1350 litres of milk cm be consumed by = 60 patients
1 litres of milk can be consumed by = \(\frac { 60 }{ 1350 }\) patients
and 1710 litres of milk can be consumed

Question 15.
Solution:
2.8 cm extension is produced by = 150 g.
1 cm extension will be produced by = \(\frac { 150 }{ 2.8 }\) g
and 19.6 cm extension will be produced by

Hope given RS Aggarwal Solutions Class 7 Chapter 9 Unitary Method Ex 9A are helpful to complete your math homework.

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RS Aggarwal Class 7 Solutions Chapter 8 Ratio and Proportion CCE Test Paper

These Solutions are part of RS Aggarwal Solutions Class 7. Here we have given RS Aggarwal Solutions Class 7 Chapter 8 Ratio and Proportion CCE Test Paper.

Other Exercises

• RS Aggarwal Solutions Class 7 Chapter 8 Ratio and Proportion Ex 8A
• RS Aggarwal Solutions Class 7 Chapter 8 Ratio and Proportion Ex 8B
• RS Aggarwal Solutions Class 7 Chapter 8 Ratio and Proportion Ex 8C
• RS Aggarwal Solutions Class 7 Chapter 8 Ratio and Proportion CCE Test Paper

Question 1.
Solution:

Question 2.
Solution:
The sum of ratio terms is = 2 + 3 + 5 = 10
Then, we have :
A’s share = ₹ \(\frac { 2 }{ 10 }\) x 1100 = ₹ 220
B’s share = ₹ \(\frac { 3 }{ 10 }\) x 1100 = ₹ 330
C’s share = ₹ \(\frac { 5 }{ 10 }\) x 1100 = ₹ 550

Question 3.
Solution:
Product of the extremes = 25 x 6 = 150
Product of the means = 36 x 5 = 180
The product of the extremes is not equal to that of the means.
Hence, 25, 36, 5 and 6 are not in proportion.

Question 4.
Solution:
x : 18 :: 18 : 108
⇒ x x 108 = 18 x 18
(Product of extremes = Product of means)
⇒ 108x = 324
⇒ x = 3
Hence, the value of x is 3.

Question 5.
Solution:
Suppose that the numbers are 5x and 7x
Then, 5x + 7x = 84
12x = 84
x = 7
Hence, the numbers are (5 x 7) = 35 and (7 x 7) = 49.

Question 6.
Solution:
Suppose that the present ages of A and B are 4x yrs and 3x yrs, respectively
Eight years ago, age of A = (4x – 8) yrs
Eight years ago, age of B = (3x – 8) yrs
Then,
(4x – 8) : (3x – 8) = 10 : 7
⇒ \(\frac { 4x – 8 }{ 3x – 8 }\) = \(\frac { 10 }{ 7 }\)
⇒ 28x – 56 = 30x – 80
⇒ 2x = 24
⇒ x = 12

Question 7.
Solution:
Distance covered in 60 min = 54 km
Distance covered in 1 min = \(\frac { 54 }{ 60 }\) km
Distance covered in 40 min = \(\frac { 54 }{ 60 }\) x 40 = 36 km

Question 8.
Solution:
Suppose that the third proportional to 8 and 12 is x
Then, 8 : 12 :: 12 : x
⇒ 8x = 144 (Product of extremes = Product of means)
x = 18
Hence, the third proportional is 18.

Question 9.
Solution:
40 men can finish the work in 60 days
1 man can finish the work in 60 x 40 days [Less men, more days]
75 men will finish the work in = \(\frac { 60 x 40 }{ 75 }\) = 32 days
Hence, 75 men will finish the same work in 32 days.

Mark (✓) against the correct answer in each of the following :
Question 10.
Solution:
(d) 6 : 4 : 3
A = \(\frac { 3 }{ 2 }\) B
C = \(\frac { 3 }{ 4 }\) B
A : B : C = \(\frac { 3 }{ 2 }\) B : B : \(\frac { 3 }{ 4 }\) B
= 6 : 4 : 3

Question 11.
Solution:
(a) 2 : 3 : 4

Question 12.
Solution:
(c) 11 : 3

Question 13.
Solution:
(a) 3
Let us assume that the number to be subtracted is x
Then, (15 – x) : (19 – x) = 15 : 3
⇒ \(\frac { 15 – x }{ 19 – x }\) = \(\frac { 3 }{ 4 }\)
⇒ 60 – 4x = 57 – 3x
⇒ x = 3

Question 14.
Solution:
(b) ₹ 360
Sum of the ratio terms = 4 + 3 = 7
B’s share = ₹ 840 x \(\frac { 3 }{ 7 }\) = ₹ 360

Question 15.
Solution:
(c) 40 years
Suppose that the present ages of A and B are 5x yrs and 2x yrs, respectively
After 5 years, the ages of A and B will be (5x + 5) yrs and (2x + 5) yrs, respectively
Then, (5x+ 5) : (2x + 5) = 15 : 7
⇒ \(\frac { 5x+ 5 }{ 2x + 5 }\) = \(\frac { 15 }{ 7 }\)
Cross multiplying; we get:
⇒ 35x + 35 = 30x + 75
⇒ 5x = 40
⇒ x = 8
Then, the present age of A is 5 x 8 = 40 yrs.

Question 16.
Solution:
(a) 896
Suppose that the number of boys in the school is x
Then, x : 320 = 9 : 5
⇒ 5x = 2880
⇒ x = 576
Hence, total strength of the school = 576 + 320 = 896

Question 17.
Solution:

Question 18.
Solution:
(i) True
Suppose that the men proportional is x
Then, 0.4 : x :: x : 0.9
⇒ 0.9 x 0.4 = x x x (Product of extremes = Product of means)
⇒ x² = 0.36
⇒ x = 0.6
(ii) False
Suppose that the third proportional is x.
Then, 9: 12 :: 12 : x
⇒ 9x = 144 (Product of extremes = Product of means)
⇒ x = 16
(iii) True
8 : x :: 48 : 18
⇒ 144 = 48x (Product of extremes = Product of means)
⇒ x = 3
(iv) True

Hope given RS Aggarwal Solutions Class 7 Chapter 8 Ratio and Proportion CCE Test Paper 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 7 Solutions Chapter 8 Ratio and Proportion Ex 8C

These Solutions are part of RS Aggarwal Solutions Class 7. Here we have given RS Aggarwal Solutions Class 7 Chapter 8 Ratio and Proportion Ex 8C.

Other Exercises

• RS Aggarwal Solutions Class 7 Chapter 8 Ratio and Proportion Ex 8A
• RS Aggarwal Solutions Class 7 Chapter 8 Ratio and Proportion Ex 8B
• RS Aggarwal Solutions Class 7 Chapter 8 Ratio and Proportion Ex 8C
• RS Aggarwal Solutions Class 7 Chapter 8 Ratio and Proportion CCE Test Paper

Objective questions :
Mark (✓) against the correct answers in each of the following :
Question 1.
Solution:
(d) a : b = 3 : 4, b : c = 8 : 9

Question 2
Solution:
(a) A : B = 2 : 3, B : C = 4 : 5

Question 3.
Solution:
(d)

Question 4.
Solution:
(b) 15% of A = 20% of B

Question 5.
Solution:
(a)

Question 6.
Solution:
(b) A : B = 5 : 7, B : C = 6 : 11
LCM of 7, 6 = 42

Question 7.
Solution:
(c) 2A = 3B = 4C = x

Question 8.
Solution:
(a)
\(\frac { A }{ 3 }\) = \(\frac { B }{ 4 }\) = \(\frac { C }{ 5 }\) = 1(suppose)
A = 3, B = 4, C = 5
A : B : C = 3 : 4 : 5

Question 9.
Solution:
(b)

Question 10.
Solution:
(c)

Question 11.
Solution:
(c) (3a + 5b) : (3a – 5b) = 5 : 1

Question 12.
Solution:
(c) 7 : x :: 35 : 45

x = 9

Question 13.
Solution:
(b) Let x to be added to each term of 3 : 5
Then \(\frac { 3 + x }{ 5 + x }\) = \(\frac { 5 }{ 6 }\)
By cross multiplication
18 + 6x = 25 + 5x
6x – 5x = 25 – 18
x = 7
7 is to be added

Question 14.
Solution:
(d) Ratio in two numbers = 3 : 5
Let first number = 3x
Then second number = 5x
According to the condition,
\(\frac { 3x + 10 }{ 5x + 10 }\) = \(\frac { 5 }{ 7 }\)
(By cross multiplication)
25x + 50 = 21x + 70
25x – 21x = 70 – 50
4x = 20
x = 5
First number = 3 x 5 = 15
and second number = 5 x 5 = 25
Sum of numbers = 15 + 25 = 40

Question 15.
Solution:
(a)
Let x be subtracted from each of the term
\(\frac { 15 – x }{ 19 – x }\) = \(\frac { 3 }{ 4 }\)
⇒ 4 (15 – x) = 3 (19 – x)
⇒ 60 – 4x = 57 – 3x
⇒ -4x + 3x = 57 – 60
⇒ -x = -3
x = 3
Required number = 3

Question 16.
Solution:
(a)
Amount = Rs. 420
and ratio = 3 : 4
Sum of ratios = 3 + 4 = 7
A’s share = \(\frac { 420 x 3 }{ 7 }\) = Rs. 60 x 3 = Rs. 180

Question 17.
Solution:
(d)
Let number of boys = x, then
x : 160 : : 8 : 5
⇒ x x 5 = 160 x 8
x = \(\frac { 160 x 8 }{ 5 }\) = 32 x 8 = 256
Number of total students of the school = 256 + 160 = 416

Question 18.
Solution:
(a)

Question 19.
Solution:
(c)
Let x be the third proportional to 9 and 12 then
9 : 12 :: x : 12
⇒ 9 x x = 12 x 12
⇒ x = \(\frac { 12 x 12 }{ 9 }\) = \(\frac { 144 }{ 9 }\) = 16
Third proportional = 16

Question 20.
Solution:
Answer = (b)
Mean proportional of 9 and 16 = √(9 x 16) = √144 = 12

Question 21.
Solution:
(a)
Let age of A = 3x
and age of B = 8x
6 years hence, their ages will be 3x + 6 and 8x + 6
\(\frac { 3x + 6 }{ 8x + 6 }\) = \(\frac { 4 }{ 9 }\)
⇒ 9 (3x + 6) = 4 (8x + 6)
⇒ 27x + 54 = 32x + 24
⇒ 32x – 27x = 54 – 24
⇒ 5x = 30
⇒ x = 6
A’s age = 3x = 3 x 6 = 18 years

Hope given RS Aggarwal Solutions Class 7 Chapter 8 Ratio and Proportion 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.

RS Aggarwal Class 7 Solutions Chapter 8 Ratio and Proportion Ex 8B

These Solutions are part of RS Aggarwal Solutions Class 7. Here we have given RS Aggarwal Solutions Class 7 Chapter 8 Ratio and Proportion Ex 8B.

Other Exercises

• RS Aggarwal Solutions Class 7 Chapter 8 Ratio and Proportion Ex 8A
• RS Aggarwal Solutions Class 7 Chapter 8 Ratio and Proportion Ex 8B
• RS Aggarwal Solutions Class 7 Chapter 8 Ratio and Proportion Ex 8C
• RS Aggarwal Solutions Class 7 Chapter 8 Ratio and Proportion CCE Test Paper

Question 1.
Solution:
We know that a, b, c, d are in proportion if ad = bc
Now 30, 40, 45, 60 are in proportion
if 30 x 60 = 40 x 45
if 1800= 1800
which is true
30, 40, 45, 60 are in proportion.

Question 2.
Solution:
We know that if a, b, c, d are in proportion if ad = bc
Now 36, 49, 6, 7 are in proportion
if 36 x 7 = 49 x 6
if 252 = 294
But 252 ≠ 294
36, 49, 6, 7 are not in proportion

Question 3.
Solution:
2 : 9 : : x : 27
9 x x = 2 x 27
x = \(\frac { 2 x 27 }{ 9 }\) = 2 x 3 = 6

Question 4.
Solution:
8 : x : : 16 : 35
x x 16 = 8 x 35
x = \(\frac { 8 x 35 }{ 6 }\) = \(\frac { 35 }{ 2 }\) = 17.5

Question 5.
Solution:
x : 35 : : 48 : 60
x x 60 = 35 x 48
x = 7 x 4 = 28

Question 6.
Solution:
Let x be the fourth proportional, then

x = \(\frac { 35 }{ 2 }\) = 17.5
Fourth proportional = 17.5

Question 7.
Solution:
36, 54, x are in continued proportion
36 : 54 : : 54 : x
⇒ 36 x x = 54 x 54

Question 8.
Solution:
27, 36, x are in continued proportion
27 : 36 :: 36 : x
27 x x = 36 x 36

Question 9.
Solution:
Let x be the third proportional, then
(i) 8 : 12 : : 12 : x

Question 10.
Solution:
Third proportional = 28 then
7 : x :: x : 28
⇒ 7 x 28 = x x x
⇒ x² = 28 x 7 = 196
⇒ x = √196 = 14
x = 14

Question 11.
Solution:
Let x be the mean proportional, then
(i) 6 : x :: x : 24
⇒ x² = 6 x 24 = 144
x = √144 = 12
Mean proportional = 12
(ii) 3 : x : : x : 27
⇒ x² = 3 x 27 = 81
x = √81 = 9
Mean proportional = 9
(iii) 0.4 : x :: x : 0.9
⇒ x² = 0.4 x 0.9
x = √o.36 = 0.6
Mean proportional = 0.6

Question 12.
Solution:
Let x be added to each of the given numbers then
5 + x, 9 + x, 7 + x, 12 + x are in proportion
\(\frac { 5 + x }{ 9 + x }\) = \(\frac { 7 + x }{ 12 + x }\)
By cross multiplication :
(5 + x) (12 + x) = (7 + x) (9 + x)
⇒ 60 + 5x + 12x + x² = 63 + 7x + 9x + x²
⇒ 60 + 17x + x² = 63 + 16x + x²
⇒ 17x + x² – 16x – x² = 63 – 60
⇒ x = 3
Required number = 3

Question 13.
Solution:
Let x be subtracted from each of the given number, then
10 – x, 12 – x, 19 – x and 24 – x are in proportion
\(\frac { 10 – x }{ 12 – x }\) = \(\frac { 19 – x }{ 24 – x }\)
By cross multiplication :
(10 – x) (24 – x) = (19 – x) (12 – x)
⇒ 240 – 10x – 24x + x² = 228 – 19x – 12x + x²
⇒ 240 – 34x + x² = 228 – 31x + x²
⇒ -34x + x² + 31x – x² = 228 – 240
⇒ -3x = -2
⇒ 3x = 12
⇒ x = 4
Required number = 4

Question 14.
Solution:
Scale of map = 1 : 5000000
Distance between two town on the map = 4 cm

Question 15.
Solution:
Height of a tree = 6 cm
and its shadow at same time = 8 m
Shadow of a pole = 20 m
Let height of pole = x m
6 : 8 = x : 20
⇒ x= \(\frac { 6 x 20 }{ 8 }\) = 15 m
Height of pole = 15 m

Hope given RS Aggarwal Solutions Class 7 Chapter 8 Ratio and Proportion Ex 8B 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 7 Solutions Chapter 8 Ratio and Proportion Ex 8A

These Solutions are part of RS Aggarwal Solutions Class 7. Here we have given RS Aggarwal Solutions Class 7 Chapter 8 Ratio and Proportion Ex 8A.

Other Exercises

• RS Aggarwal Solutions Class 7 Chapter 8 Ratio and Proportion Ex 8A
• RS Aggarwal Solutions Class 7 Chapter 8 Ratio and Proportion Ex 8B
• RS Aggarwal Solutions Class 7 Chapter 8 Ratio and Proportion Ex 8C
• RS Aggarwal Solutions Class 7 Chapter 8 Ratio and Proportion CCE Test Paper

Question 1.
Solution:
(i) 24 : 40
HCF of 24, 40 = 8
24 : 40 = 24 ÷ 8 : 40 ÷ 8 = 3 : 5 (Dividing by 8)
(ii) 13.5 : 15 or 135 : 150
HCF of 135 and 150 = 15
135 ÷ 15 : 150 ÷ 15 (Dividing by 15)
= 9 : 10

HCF of 25, 65, 80 = 5
Dividing by 5,
5 : 13 : 16

Question 2.
Solution:
(i) 75 paise : 3 rupees = 75 paise : 300 paise
(converting into same kind)
HCF of 75, 300 = 75
75 : 300 = 75 ÷ 75 : 300 ÷ 75 (Dividing by 75) = 1 : 4
(ii) 1 m 5 cm : 63 cm = 105 cm : 63 cm
(converting into same kind)
HCF of 105 and 63 = 21
105 ÷ 21 : 63 ÷ 21 (Dividing by 21)
= 5 : 3
(iii) 1 hour 5 minutes : 45 minutes = 65 minutes : 45 minutes
(converting into minutes)
13 : 9 (dividing by 5)
= 13 : 9
(iv) 8 months : 1 year = 8 months : 12 months
(converting into the same kind)
HCF of 8 and 12 = 4
Dividing by 4
8 ÷ 4 : 12 ÷ 4
= 2 : 3
(v) 2 kg 250 g : 3 kg = 2250 g : 3000 g (converting into the same kind)
HCF of 2250 and 3000 = 750
Dividing by 750,
2250 ÷ 750 : 3000 ÷ 750 = 3 : 4
(vi) 1 km : 750 m = 1000 m : 750 m
(converting into metre)
= 4 : 3 (dividing by 250)
= 4 : 3

Question 3.
Solution:

Question 4.
Solution:
A : B = 5 : 8, B : C = 16 : 25

Question 5.
Solution:
A : B = 3 : 5, B : C = 10 : 13

A : B : C = 6 : 10 : 13

Question 6.
Solution:
A : B = 5 : 6 and B : C = 4 : 7

Question 7.
Solution:
Total amount = Rs. 360
Sum of ratios = 7 + 8 = 15

Question 8.
Solution:
Total amount = Rs. 880

Question 9.
Solution:
Total amount = Rs. 5600
Ratio in A : B : C = 1 : 3 : 4

Question 10.
Solution:
Let x be added to each term Then
9 + x : 16 + x = 2 : 3

Question 11.
Solution:
Let x be subtracted from each term Then
(17 – x) : (33 – x) = 7 : 15

Question 12.
Solution:
Ratio in two numbers = 7 : 11
Let first number = 7x
Then second number = 11x
Then adding 7 to each number, the ratio is 2 : 3
\(\frac { 7x + 7 }{ 11x + 7 }\) = \(\frac { 2 }{ 3 }\)
By cross multiplying:
3 (7x + 7) = 2 (11x + 7)
⇒ 21x + 21 = 22x + 14
⇒ 21 – 14 = 22x – 21x
⇒ x = 7
First number = 7x = 7 x 7 = 49
and second number = 11x = 11 x 7 = 77
Hence numbers are 49, 77

Question 13.
Solution:
The ratio in two numbers = 5 : 9
Let the first number = 5x
Then second number = 9x
By subtracting 3 from each number the ratio is 1 : 2
\(\frac { 5x – 3 }{ 9x – 3 }\) = \(\frac { 1 }{ 2 }\)
By cross multiplication,
2 (5x – 3) = 1 (9x – 3)
⇒ 10x – 6 = 9x – 3
⇒ 10x – 9x = -3 + 6
⇒ x = 3
First number = 5x = 5 x 3 = 15
and second .number = 9x = 9 x 3 = 27
Hence numbers are 15, 27

Question 14.
Solution:
Ratio in two numbers = 3 : 4
LCM = 180
Let first number = 3x
Then second number = 4x
Now LCM = 3 x 4 x x = 12x
12x = 180
⇒ x = 15
Numbers will be 3 x 15 = 45 and 4 x 15 = 60

Question 15.
Solution:
Ratio in present ages of A and B = 8 : 3
Let A’s age = 8x
Then B’s age = 3x
6 years hence,
A’s age will be = 8x + 6
and B’s will be = 3x + 6
\(\frac { 8x + 6 }{ 3x + 6 }\) = \(\frac { 9 }{ 4 }\)
(By cross multiplication)
4 (8x + 6) = 9 (3x + 6)
⇒ 32x + 24 = 27x + 54
⇒ 32x – 27x = 54 – 24
⇒ 5x = 30
⇒ x = 6
A’s present age = 8x = 8 x 6 = 48 years
and B’s age = 3x = 3 x 6 = 18 years

Question 16.
Solution:
Ratio in copper and zinc = 9 : 5
Let alloy = x gm

Question 17.
Solution:
Ratio in boys and girls = 8 : 3
and total number of girls = 375
Let number of boys = 8x
Then number of girls = 3x
3x = 375
⇒ x = 125
Number of boys = 8x = 8 x 125 = 1000

Question 18.
Solution:
Ratio in income and savings = 11 : 2
Let income = 11x
Then savings = 2x
But savings = Rs. 2500
2x = 2500
⇒ x = 1250
Then income = 1250 x 11 = Rs. 13750
and expenditure = Total income – savings = 13750 – 2500 = Rs. 11250

Question 19.
Solution:
Total amount = Rs. 750
Ratio in rupee, 50 P and 25 P coins =5 : 8 : 4
Let number of rupees = 5x
Number of 50 P coins = 8x
and number of 25 coins = 4x
According to the condition,

Number of 1 Re coins = 5x = 5 x 75 = 375
Number of 50 P coins = 8x = 8 x 75 = 600
and number of 25 P coins = 4x = 4 x 75 = 300

Question 20.
Solution:
(4x + 5) : (3x + 11) = 13 : 17
\(\frac { 4x + 5 }{ 3x + 11 }\) = \(\frac { 13 }{ 17 }\)
By cross multiplication,
68x + 85 = 39x + 143
⇒ 68x – 39x = 143 – 85
⇒ 29x = 58
x = 2
Hence x = 2

Question 21.
Solution:
x : y = 3 : 4

Question 22.
Solution:
x : y = 6 : 11
\(\frac { x }{ y }\) = \(\frac { 6 }{ 11 }\)
Now (8x – 3y) : (3x + 2y)

Question 23.
Solution:
Sum of two numbers = 720
Ratio of two numbers = 5 : 7
Let first number = 5x
Then second number = 7x
5x + 7x = 720
⇒ 12x = 720
⇒ x = 60
First number = 5x = 5 x 60 = 300
and second number = 7x = 7 x 60 = 420

Question 24.
Solution:
(i) (5 : 6) or (7 : 9)

Question 25.
Solution:
(i) (5 : 6), (8 : 9), (11 : 18)

Hope given RS Aggarwal Solutions Class 7 Chapter 8 Ratio and Proportion Ex 8A are helpful to complete your math homework.

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RS Aggarwal Class 7 Solutions Chapter 10 Percentage Ex 10B

These Solutions are part of RS Aggarwal Solutions Class 7. Here we have given RS Aggarwal Solutions Class 7 Chapter 10 Percentage Ex 10B.

Other Exercises

• RS Aggarwal Solutions Class 7 Chapter 10 Percentage Ex 10A
• RS Aggarwal Solutions Class 7 Chapter 10 Percentage Ex 10B
• RS Aggarwal Solutions Class 7 Chapter 10 Percentage Ex 10C
• RS Aggarwal Solutions Class 7 Chapter 10 Percentage CCE Test Paper

Question 1.
Solution:
Rupesh seemed 495 marks out of 750
Percentage of marks = \(\frac { 495 }{ 750 }\) x 100 = 66%

Question 2.
Solution:
Monthly salary = Rs. 15625
Increase = 12%
Amount of increase = \(\frac { 15625 x 12 }{ 100 }\) = Rs. 1875
New salary = Rs. 15625 + Rs. 1875 = Rs. 17500

Question 3.
Solution:
Excise duty in the beginning = Rs. 950
Reduced duty = Rs. 760
Reduction = Rs. 950 – Rs. 760 = Rs. 190
Reduction percent = \(\frac { 190 x 100 }{ 950 }\) = 20%

Question 4.
Solution:
Let x be the total cost of the T.V
96% of x = 10464

Total cost of T.V = Rs. 10900

Question 5.
Solution:
Let number of students = x
In a school boys = 70%
girls = 100 – 70 = 30%
Now 30% of x = 504

Number of boys = 1680 – 504 = 1176
and number of total students = 1680

Question 6.
Solution:
Copper required = 69 kg
copper in ore = 12%
Let quantity of ore = x kg
12% of x = 69

Quantity of ore = 575kg

Question 7.
Solution:
Pass marks = 36%
A students gets marks = 123
But failed by 39 marks
Pass marks = 123 + 39 = 162
Now, 36% of maximum marks = 162
Maximum marks = \(\frac { 162 x 100 }{ 36 }\) = 450 marks

Question 8.
Solution:
Let number of apples = x
Number of apples sold = 40% of x

Hence number of apples = 700

Question 9.
Solution:
Let total number of examinees = x
the numbers of examinees who passed = 72% of x
= \(\frac { x x 72 }{ 100 }\)
= \(\frac { 18x }{ 25 }\)

Question 10.
Solution:
Let the gross value of moped = x
Amount of commission = 5% of x

Question 11.
Solution:
Total gunpowder = 8 kg
Amount of nitre = 75%
amount of sulphur = 10%
Rest of powder which is charcoal = 100 – (75 + 10) = 100 – 85 = 15 = 15%
Amount of charcoal = 8 x \(\frac { 15 }{ 100 }\) = \(\frac { 120 }{ 100 }\)
= \(\frac { 6 }{ 5 }\) kg = 1 kg 200 grams = 1.2 kg

Question 12.
Solution:
Quantity of chalk = 1 kg or 1000 g

Question 13.
Solution:
Let total number of days on which the
school open = x
and Sonal’s attendance = 75%
x x 75% = 219

No. of days on which was school open = 292 days

Question 14.
Solution:
Rate of commission = 3%
Amount of commission = Rs. 42660
Let value of property = x
then 3% of x = 42660

Question 15.
Solution:
Total votes of the constituency = 60000
Votes polled = 80% of total votes
= \(\frac { 80 }{ 100 }\) x 60000 = 48000
Votes polled in favour of A = 60% of polled votes
= \(\frac { 60 }{ 100 }\) x 48000 = 28800
Votes polled in favour of B = 48000 – 28800 = 19200

Question 16.
Solution:
Let original price of shirt = Rs. x
Discount = 12%

Original price of shirt = Rs. 1350

Question 17.
Solution:
Let original price of sweater = x
Rate of increase = 8%
Increased price = x + 8% of x

Hence, original price of sweater = Rs. 1450

Question 18.
Solution:
Let total income = x

Question 19.
Solution:
Let the given number = 100
Then increase % = 20%

Decrease = 100 – 96 = 4
Decrease per cent = 4%

Question 20.
Solution:
Let original salary of the officer = Rs. 100
Increase = 20%

Question 21.
Solution:
Rate of commission = 2% on first Rs. 200000
1 % on next Rs. 200000 and 0.5% on remaining price
Sale price of property = Rs.200000 + 200000 +140000 = Rs. 540000
Now commission earned by the
= Rs. 200000 x 2% + Rs. 200000 x 1% + 140000 x 0.5%

Question 22.
Solution:
Let Akhil’s income = Rs. 100
Then income of Nikhil’s will be = Rs. 100 – 20 = Rs. 80
Amount which is more than that of Akhil’s = 100 – 80 = Rs. 20
% age = \(\frac { 20 x 100 }{ 80 }\) = 25%

Question 23.
Solution:
Let income of Mr Thomas = Rs. 100
then income of John = Rs. 100 + 20 = Rs. 120
Income of Mr Thomas is less than John = Rs. 120 – 100 = Rs. 20
% age = \(\frac { 20 x 100 }{ 120 }\)
= \(\frac { 50 }{ 3 }\) = 16\(\frac { 2 }{ 3 }\) %

Question 24.
Solution:
Present value of machine = Rs. 387000
Rate of depreciation = 10%
Let 1 year ago the value of machine was = x

1 year ago, value of machine = Rs. 430000

Question 25.
Solution:
Present value of car = Rs. 450000
Rate of decreasing of value = 20%
Value after 2 years

Question 26.
Solution:
Present population = 60000
Rate of increase = 10%
Increased population after 2 years

Question 27.
Solution:
Let the price of sugar = Rs. 100
and consumption = 100 kg.
Increase price of 100 kg = Rs. 100 + 25 = Rs. 125
Now increased amount on 100 kg = Rs. 125

Hope given RS Aggarwal Solutions Class 7 Chapter 10 Percentage Ex 10B 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 7 Solutions Chapter 7 Linear Equations in One Variable CCE Test Paper

These Solutions are part of RS Aggarwal Solutions Class 7. Here we have given RS Aggarwal Solutions Class 7 Chapter 7 Linear Equations in One Variable CCE Test Paper.

Other Exercises

• RS Aggarwal Solutions Class 7 Chapter 7 Linear Equations in One Variable Ex 7A
• RS Aggarwal Solutions Class 7 Chapter 7 Linear Equations in One Variable Ex 7B
• RS Aggarwal Solutions Class 7 Chapter 7 Linear Equations in One Variable Ex 7C
• RS Aggarwal Solutions Class 7 Chapter 7 Linear Equations in One Variable CCE Test Paper

Question 1.
Solution:
We have:
x³ + y³ + z³ – 3xyz
= (-2)³ + (-1)³ + (3)³ – 3 x (-2) x (-1) x 3
= – 8 – 1 + 27 – 18 = -27 + 27 = 0

Question 2.
Solution:
Co-efficient of x in the given numbers are
(i) -5y
(ii) 2y² z
(iii) \(\frac { -3 }{ 2 }\) ab

Question 3.
Solution:
We have:
(4xy – 5x² – y² + 6) – (x² – 2xy + 5y² – 4)
= 4xy – 5x² – y² + 6 – x² + 2xy – 5y² + 4
= -6x² – 6y² + 6xy +10
= -2 (3x² + 3y – 3xy – 5)

Question 4.
Solution:
We have:
(2x² – 3y² + xy) – (x² – 2xy + 3y²)
= 2x² – 3y² +xy – x² + 2xy – 3y²
= 2x² – x² – 3y² – 3y² + xy + 2xy
= x² – 6y² + 3xy
x² – 2xy + 3y² is less than 2x² – 3y² + xy by x² – 6y + 3xy.

Question 5.
Solution:
We have:

Question 6.
Solution:
We have:
(3a + 4) (2a – 3) + (5a – 4) (a + 2)
= {3a (2a – 3) + 4 (2a – 3)} + {5a (a + 2) – 4 (a + 2)}
= (6a² – 9a + 8a – 12) + (5a² + 10a – 4a – 8)
= (6a² – a – 12) + (5a² + 6a – 8)
= (11a² + 5a – 20)

Question 7.
Solution:

Question 8.
Solution:
We have:

Question 9.
Solution:
Let the consecutive odd number be
x and (x + 2)
x + (x + 2) = 68
2x + 2 = 68
2x = 68 – 2 = 66
x = 33
The required numbers are 33 and (33 + 2), i.e., 35.

Question 10.
Solution:
Let Reenu’s present age be x
Then, her father’s present age will be 3x
Reenu’s age after 12 years = (x + 12)
3x + 12 = 2x + 24
x = 12
Reenu’s present age = x = 12 yrs
And her father’s age = 3x = (3 x 12) = 36 yrs

Mark (✓) against the correct answer in each of the following :
Question 11.
Solution:

Question 12.
Solution:

Question 13.
Solution:

Question 14.
Solution:
(c) 18
Let the number be x.
According to the equation, we have:
4x = x + 54
⇒ 3x = 54
⇒ x = 18

Question 15.
Solution:
(b) 52°
Let the two complementary angles be x° and (90 – x)°.
According to the equation, we have:
x – (90 – x) = 14
⇒ 2x = 104
⇒ x = 52
(90° – x)° = 90° – 52° = 38°
The larger angle is 52°.

Question 16.
Solution:
(c) 32 m
Let the length and breadth of the rectangle be l m and b m, respectively.
According to the questions, we have:
l = 2b ……(i)
2 (l + b) = 96 …..(ii)
Now, 2 (2b+ b) = 96
⇒ 6b = 96
⇒ b = 16
Length = 16 x 2 m = 32 m

Question 17.
Solution:
(b) 12 years
Let the ages of A and B be x and y years respectively,

Question 18.
Solution:
(i) -2a² b is a monomial.
(ii) (a² – 2b²) is a binomial.
(iii) (a + 2b – 3c) is a trinomial.
(iv) In -5ab, the coefficient of a is -5.
(v) In x² + 2x – 5, the constant term is -5.

Question 19.
Solution:
(i) False.
The coefficient of x is -1.
(ii) False.
The coefficient of x in – 3.
(iii) False.
LHS = (5x – 7) – (3x – 5) = 5x – 7 – 3x + 5 = 2x – 2.
(iv) True.
LHS = (3x + 5y) (3x – 5y)
= 3x (3x – 5y) + 5y (3x – 5y)
= 9x² – 15xy + 15xy – 25y²
= 9x² – 25y²
(v) True
(a² + b²)

Hope given RS Aggarwal Solutions Class 7 Chapter 7 Linear Equations in One Variable CCE Test Paper are helpful to complete your math homework.

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RS Aggarwal Class 7 Solutions Chapter 5 Exponents CCE Test Paper

These Solutions are part of RS Aggarwal Solutions Class 7. Here we have given RS Aggarwal Solutions Class 7 Chapter 5 Exponents CCE Test Paper.

Other Exercises

• RS Aggarwal Solutions Class 7 Chapter 5 Exponents Ex 5A
• RS Aggarwal Solutions Class 7 Chapter 5 Exponents Ex 5B
• RS Aggarwal Solutions Class 7 Chapter 5 Exponents Ex 5C
• RS Aggarwal Solutions Class 7 Chapter 5 Exponents CCE Test Paper

Question 1.
Solution:

Question 2.
Solution:
Let the required number be x

Question 3.
Solution:
Let the required number be x.
(-20)^-1 ÷ x = (-10)^-1

Question 4.
Solution:
(i) 2000000 = 2.000000 x 10⁶ [Since the decimal point is moved 6 to the left]
= 2 x 10⁶
(ii) 6.4 x 10⁵ = 6.4 x 100000 = 640000

Question 5.
Solution:
We have :

Question 6.
Solution:
We have :

Mark (✓) against the correct answer in each of the following :
Question 7.
Solution:

Question 8.
Solution:

Question 9.
Solution:

Question 10.
Solution:

Question 11.
Solution:

Question 12.
Solution:
(c) 3.263 x 10⁵
A given number is said to be in standard form if it can be expressed as
k x 10n, where k is a real number such that 1 < k < 10 and n is a positive integer.
For example : 3.263 x 10⁵

Question 13.
Solution:

Question 14.
Solution:
(i) True
645 = 6.45 x 10²
[Since the decimal point is moved 2 places to the left]
(ii) False
27000 = 2.7 x 10⁴
[Since the decimal point is moved 4 places to the left]
(iii) False
(3° + 4° + 5°) = 1
(iv) False
Reciprocal of 56 = Reciprocal of

Hope given RS Aggarwal Solutions Class 7 Chapter 5 Exponents CCE Test Paper are helpful to complete your math homework.

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RS Aggarwal Class 7 Solutions Chapter 7 Linear Equations in One Variable Ex 7C

These Solutions are part of RS Aggarwal Solutions Class 7. Here we have given RS Aggarwal Solutions Class 7 Chapter 7 Linear Equations in One Variable Ex 7C.

Other Exercises

• RS Aggarwal Solutions Class 7 Chapter 7 Linear Equations in One Variable Ex 7A
• RS Aggarwal Solutions Class 7 Chapter 7 Linear Equations in One Variable Ex 7B
• RS Aggarwal Solutions Class 7 Chapter 7 Linear Equations in One Variable Ex 7C
• RS Aggarwal Solutions Class 7 Chapter 7 Linear Equations in One Variable CCE Test Paper

Objective Questions :
Mark (✓) against the correct answer in each of the following :
Question 1.
Solution:
(d)

Question 2.
Solution:
(d)

Question 3.
Solution:
(a)
2n + 5 = 3 (3n – 10)
⇒ 2n + 5 = 9n – 30
⇒ 9n – 2n = 5 + 30
⇒ 7n = 35
⇒ n = 5

Question 4.
Solution:
(c)

Question 5.
Solution:
(c)
8 (2x – 5) – 6 (3x – 7) = 1
⇒ 16x – 40 – 18x + 42 = 1
⇒ -2x + 2 = 1
⇒ -2x = 1 – 2 = -1
x = \(\frac { 1 }{ 2 }\)

Question 6.
Solution:
(d)

Question 7.
Solution:
(a)

Question 8.
Solution:
(b)
Let first whole number=x
Then second number = x + 1
and sum = 53
x + x + 1 = 53
⇒ 2x = 53 – 1
⇒ 2x = 52
⇒ x = 26
Smaller number = 26

Question 9.
Solution:
Let first even number = 2x
Then second number = 2x + 2
and sum = 86
2x + 2x + 2 – 86
⇒ 4x = 86 – 2 = 84
⇒ x = 21
Larger even number = 2x + 2 = 2 x 21 + 2 = 42 + 2 = 44

Question 10.
Solution:
(b)
Let first odd number = 2x + 1
Second number = 2x + 3
2x + 1 + 2x + 3 = 36
⇒ 4x + 4 = 36
⇒ 4x = 36 – 4 = 32
⇒ x = 8
Smaller number = 2x + 1 = 2 x 8 + 1 = 16 + 1 = 17

Question 11.
Solution:
(d)
Let number = x
2x + 9 = 31
⇒ 2x = 31 – 9 = 22
⇒ x = 11

Question 12.
Solution:
(a)
Let number = x then
3x + 6 = 24
⇒ 3x = 24 – 6 = 18
⇒ x = 6
Number = 6

Question 13.
Solution:
(a)

Question 14.
Solution:
(b)
Let first angle = x
Then second = 90° – x
x – (90° – x) = 10
⇒ x – 90° + x = 10°
⇒ 2x = 10° + 90° = 100°
x = 50°
Second angle = 90° – 50° = 40°
Larger angle = 50°

Question 15.
Solution:
(b)
Let first angle = x
Then second = 180° – x
x – (180° – x) = 20°
⇒ x – 180° + x = 20°
⇒ 2x = 20° + 180° = 200°
x = 100°
Second angle = 180° – 100° = 80°
Smaller angle = 80°

Question 16.
Solution:
(c)
Let age of A = 5x
Then age of B = 3x
After 6 years,
A’s age = 5x + 6
and B’s age = 3x + 6
\(\frac { 5x + 6 }{ 3x + 6 }\) = \(\frac { 7 }{ 5 }\)
⇒ 25x + 30 = 21x + 42
⇒ 25x – 21x = 42 – 30
⇒ 4x = 12
⇒ x = 3
A’s age = 5x = 5 x 3 = 15 years

Question 17.
Solution:
(b)
Let the number = x
According to the condition,
5x = 80 + x
⇒ 5x – x = 80
⇒ 4x = 80
⇒ x = 20
Number = 20

Question 18.
Solution:
(c)
Let width of rectangle = x m
Then length = 3x m
Perimeter = 96 m
2 (x + 3x) = 96
⇒ x + 3x = \(\frac { 96 }{ 2 }\) = 48
⇒ 4x = 48
⇒ x = 12
Length = 3x = 12 x 3 = 36 m

Hope given RS Aggarwal Solutions Class 7 Chapter 7 Linear Equations in One Variable Ex 7C are helpful to complete your math homework.

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RS Aggarwal Class 7 Solutions Chapter 5 Exponents Ex 5C

These Solutions are part of RS Aggarwal Solutions Class 7. Here we have given RS Aggarwal Solutions Class 7 Chapter 5 Exponents Ex 5C.

Other Exercises

• RS Aggarwal Solutions Class 7 Chapter 5 Exponents Ex 5A
• RS Aggarwal Solutions Class 7 Chapter 5 Exponents Ex 5B
• RS Aggarwal Solutions Class 7 Chapter 5 Exponents Ex 5C
• RS Aggarwal Solutions Class 7 Chapter 5 Exponents CCE Test Paper

OBJECTIVE QUESTIONS
Mark (✓) tick against the correct answer in each of the following :
Question 1.
Solution:
(d)

Question 2.
Solution:
(c)

Question 3.
Solution:
(c)

Question 4.
Solution:
(b)

Question 5.
Solution:
(c)

Question 6.
Solution:
(b)

Question 7.
Solution:
(b)

Question 8.
Solution:
(a)

Question 9.
Solution:
(c)

Question 10.
Solution:
(b)

Question 11.
Solution:
(b)

Question 12.
Solution:
(b)

Question 13.
Solution:
(d)

Question 14.
Solution:
(a)

Question 15.
Solution:
(c)

Question 16.
Solution:
(a)

Question 17.
Solution:
(c)

Question 18.
Solution:
(b)

Question 19.
Solution:
(c)

Question 20.
Solution:
(c)
Required number = 10^-1 ÷ (-8)^-1

Question 21.
Solution:
(c)
The number which is in standard form is 2.156 x 10⁶

Hope given RS Aggarwal Solutions Class 7 Chapter 5 Exponents Ex 5C are helpful to complete your math homework.

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