Tag: Maths RS Aggarwal Solutions Class 7

RS Aggarwal Class 7 Solutions Chapter 2 Fractions Ex 2D

These Solutions are part of RS Aggarwal Solutions Class 7. Here we have given RS Aggarwal Solutions Class 7 Chapter 2 Fractions Ex 2D.

Other Exercises

• RS Aggarwal Solutions Class 7 Chapter 2 Fractions Ex 2A
• RS Aggarwal Solutions Class 7 Chapter 2 Fractions Ex 2B
• RS Aggarwal Solutions Class 7 Chapter 2 Fractions Ex 2C
• RS Aggarwal Solutions Class 7 Chapter 2 Fractions Ex 2D
• RS Aggarwal Solutions Class 7 Chapter 2 Fractions CCE Test Paper

OBJECTIVE QUESTIONS
Mark (√) against the correct answer in each of the following:
Question 1.
Solution:
(c)
Denominator in (a) and (b) is 10
These are decimal fractions
But denominator of (c) is 3
\(\frac { 10 }{ 3 }\) is a vulgar fraction

Question 2.
Solution:
(c)
\(\frac { 7 }{ 10 }\) and \(\frac { 7 }{ 9 }\) are proper fractions as each of these have numerator less than its denominator
\(\frac { 9 }{ 7 }\) is improper fraction

Question 3.
Solution:
(a)
\(\frac { 105 }{ 112 }\) is reducible fraction because HCF 112 of 105 and 112 is 7

Question 4.
Solution:
(c)

Question 5.
Solution:
(c)

Question 6.
Solution:
(d)

Question 7.
Solution:
(c)

Question 8.
Solution:
(d)

Question 9.
Solution:
(d)

Question 10.
Solution:
(b)

Question 11.
Solution:
(d)

Question 12.
Solution:
(c)

Question 13.
Solution:
(d)

Question 14.
Solution:
(d)

Question 15.
Solution:
(b)
The correct statement will be

Question 16.
Solution:
(c)
A car runs in 1 litre of petrol = 16 km

Question 17.
Solution:
(a)

Hope given RS Aggarwal Solutions Class 7 Chapter 2 Fractions Ex 2D are helpful to complete your math homework.

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RS Aggarwal Class 7 Solutions Chapter 3 Decimals CCE Test Paper

These Solutions are part of RS Aggarwal Solutions Class 7. Here we have given RS Aggarwal Solutions Class 7 Chapter 3 Decimals CCE Test Paper.

Other Exercises

• RS Aggarwal Solutions Class 7 Chapter 3 Decimals Ex 3A
• RS Aggarwal Solutions Class 7 Chapter 3 Decimals Ex 3B
• RS Aggarwal Solutions Class 7 Chapter 3 Decimals Ex 3C
• RS Aggarwal Solutions Class 7 Chapter 3 Decimals Ex 3D
• RS Aggarwal Solutions Class 7 Chapter 3 Decimals Ex 3E
• RS Aggarwal Solutions Class 7 Chapter 3 Decimals CCE Test Paper

Question 1.
Solution:
Cost of 1 pen = ₹ 32.50
Cost of 24 such pens = ₹ (32.50 x 24) = ₹ 80
Hence, the cost of 24 pens is ₹ 780.

Question 2.
Solution:
Distance covered by the bus in 1 hour = 64.5 km
Distance covered in 18h = (64.5 x 18) km = 1161 km
Hence, the bus can cover a distance of 1161 km in 18h.

Question 3.
Solution:
First, we will find the product 68 x 65 x 4
Now, 68 x 65 x 4 = 4420 x 4 = 17680
Sum of decimal places in the given decimals = (2 + 1 + 2) = 5
So, the product have five decimal places.
0.68 x 6.5 x 0.04 = 0.17680 = 0.1768

Question 4.
Solution:
Total weight of all the bags = 2231 kg
Weight of each bag = 48.5 kg

Question 5.
Solution:

Question 6.
Solution:
Product of the given decimals = 1.824
One decimal = 0.64
The other decimal = 1.824 ÷ 0.64

Hence, the other decimal is 2.85.

Question 7.
Solution:
Thickness of the pile of plywoods = 2.43 m = 2.43 x 100 cm = 243 cm
Thickness of one piece of plywood = 0.45 cm
Required no. of pieces of plywood

Hence, the required number of pieces of plywood is 540.

Question 8.
Solution:
Let the number of sides of the polygon be n.
Length of each side of the polygon = 3.8 cm
Perimeter of the polygon = (3.8 x n) cm
But it is given that its perimeter is 22.8 cm.

Hence, the given polygon has six sides.

Mark (✓) against the correct answer in each of the following :
Question 9.
Solution:
(b) 2.04

Question 10.
Solution:
(b) 1\(\frac { 1 }{ 125 }\)

Question 11.
Solution:
(c) 2.005 kg
2 kg 5 g = (2 x 1000) g + 5 g = (2005)g
= \(\frac { 2005 }{ 1000 }\) kg = 2.005 kg

Question 12.
Solution:
(b) 0.08
We have :

Question 13.
Solution:
(c) 0.011
First, we will find the product 11 x 1 x 1
i.e. 11 x 1 x 1 = 11 x 1 = 11
Sum of decimal places in the given decimals = (1 + 1 + 2) = 4
1.1 x 0.1 x 0.01 = 0.0011 [4 places of decimal]

Question 14.
Solution:
(b) 2.03

Question 15.
Solution:
(c) is correct
Let the number added be x We have :
2.06 + x = 3.1
⇒ x = 3.1 – 2.06
Converting the given decimals into like decimals, we get:
2.06 and 3.10
Thus, required number = (3.10 – 2.06) = 1.04
Hence, 1.04 should be added to 2.06 to get 3.1.

Question 16.
Solution:
(b) 0.06 .
We have :
0.1 – x = 0.04
⇒ x = 0.1 – 0.04
Converting the given decimals into like decimals, we get:
0.10 and 0.04
Thus, required number = (0.10 – 0.04) = 0.06
Hence, 0.06 should be subtracted from 0.1 to get 0.04.

Question 17.
Solution:
(i) 1.001 ÷ 14 = 0.0715


47 x 53 = 2491
Sum of decimal places in the given decimals = (2 + 1) = 3
0.47 x 5.3 = 2.491
(iv) 0.023 x 0.03 = 0.69
Explanation first, we will multiply 23 by 3
23 x 3 = 69
Sum of decimal places in the given decimals = (3 + 2) = 5
0.023 x 0.03 =0.00069
(v) (0.7)² = 0.69
Explanation : (0.7)² = 0.7 x 0.7
First, we will find the product 0.7 x 0.7
Now, 7 x 7 = 49
Sum of decimal places in the given decimals = (1 + 1) = 2
So, the product must have two decimal places.
(0.7)² = 0.7 x 0.7 = 0.49
(vi) (0.05)³ = 0.000125
Explanation : First, we will find the
product 0.05 x 0.05 x 0.05
Now, 5 x 5 x 5 = 125
Sum of decimal places in the given decimals = (2 + 2 + 2) = 6
So, the product must have six decimal places.
(0.05)² = 0.05 x 0.05 x 0.05 = 0.000125

Question 18.
Solution:
(i) False
We have :
0.5 x 0.05 Now, 5 x 5 = 25
Sum of decimal places in the given decimals = (1 + 2) = 3
0.5 x 0.5 = 0.025
(ii) True
We have :
0.25 x 0.8
Now, 25 x 8 = 200
Sum of decimal places in the given decimals = (2 + 1) = 3
0.25 x 0.8 = 0.200 = 0.2
(iii) True
We have :
0.35 ÷ 0.7

(iv) False We have :
0.4 x 0.4 x 0.4
Now, 4 x 4 x 4 = 64
Sum of decimal places in the given decimals = (1 + 1 + 1) = 3
0.4 x 0.4 x 0.4 = 0.064
(v) True
6 cm = \(\frac { 6 }{ 100 }\) m = 0.06 m

Hope given RS Aggarwal Solutions Class 7 Chapter 3 Decimals CCE Test Paper are helpful to complete your math homework.

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RS Aggarwal Class 7 Solutions Chapter 3 Decimals Ex 3C

These Solutions are part of RS Aggarwal Solutions Class 7. Here we have given RS Aggarwal Solutions Class 7 Chapter 3 Decimals Ex 3C.

Other Exercises

• RS Aggarwal Solutions Class 7 Chapter 3 Decimals Ex 3A
• RS Aggarwal Solutions Class 7 Chapter 3 Decimals Ex 3B
• RS Aggarwal Solutions Class 7 Chapter 3 Decimals Ex 3C
• RS Aggarwal Solutions Class 7 Chapter 3 Decimals Ex 3D
• RS Aggarwal Solutions Class 7 Chapter 3 Decimals Ex 3E
• RS Aggarwal Solutions Class 7 Chapter 3 Decimals CCE Test Paper

Question 1.
Solution:
We know that by multiplying by 10, the decimal point is shifted one place to its right side.
(i) 73.92 x 10 = 739.2
(ii) 7.54 x 10 = 75.4
(iii) 84.003 x 10 = 840.03
(iv) 0.83 x 10 = 8.3
(v) 0.7 x 10 = 7.0
(vi) 0.032 x 10 = 0.32

Question 2.
Solution:
We know that by multiplying a decimal by 100, two decimal points are shifted to it right side
(i) 2.397 x 100 = 239.7
(ii) 6.83 x 100 = 683.0
(iii) 2.9 x 100 = 290
(iv) 0.08 x 100 = 8
(v) 0.6 x 100 = 60
(vi) 0.003 x 100 = 0.3

Question 3.
Solution:
We know that by multiplying a decimal by 1000, three places of decimal are shifted to its right.
(i) 6.7314 x 1000 = 6731.4
(ii) 0.182 x 1000 = 182
(iii) 0.076 x 1000 = 76
(iv) 6.25 x 1000 = 6250
(v) 4.8 x 1000=4800
(vi) 0.06 x 1000 = 60

Question 4.
Solution:
(i) 5.4 x 16 = 86.4 (One place of decimal)
(ii) 3.65 x 19 = 69.35 (Two place of decimal)
(iii) 0.854 x 12 = 10.2468 (Three place of decimal)
(iv) 36.73 x 48 = 1763.04 (Two places of decimal)

(v) 4.125 x 86=354.750 (Three places of decimal)
= 354.75

Question 5.
Solution:
(i) 7.6 x 2.4= 18.24
{Sum of decimal places = 1 + 1 = 2}

Question 6.
Solution:
(i) 13 x 1.3 x 0.13 = 2.197
{Sum of decimal places = 1 + 2 = 3}

(ii) 2.4 x 1.5 x 2.5 = 9.000 = 9
{Sum of decimal places = 1 + 1 + 1 = 3}
(iii) 0.8 x 3.5 x 0.05 = 0.1400 = 0.14
{Sum of decimal places = 1 + 1 + 2 = 4}
(iv) 0.2 x 0.02 x 0.002 = 0.000008
{Sum of decimal places = 1 + 2 + 3 = 6}
(v) 11.1 x 1.1 x 0.11 = 1.3431
{Sum of decimal places = 1 + 1 + 2 = 4}

(vi) 2.1 x 0.21 x 0.021 = 0.00926
21 x 21 = 441
441 x 21 = 9261
{Sum of decimal places = 1 + 2 + 3 = 6}

Question 7.
Solution:
(i) (1.2)²= 1.2 x 1.2 = 1.44
{Sum of decimal places = 1 + 1 = 2}
(ii) (0.7)² = 0.7 x 0.7 = 0.49
{Sum of decimal places = 1 + 1 = 2}
(iii) (0.04)² = 0.04 x 0.04 = 0.0016
{Sum of decimal places = 2 + 2 = 4}
(iv) (0.11)² = 0.11 x 0.11 =0.0121
{Sum of decimal places = 2 + 2 = 4}

Question 8.
Solution:
(i) (0.3)³ = 0.3 x 0.3 x 0.3 = 0.027
{Sum of decimal places = 1 + 1 + 1 = 3}
(ii) (0.05)³= 0.05 x 0.05 x 0.05 = 0.000125
{Sum of decimal places = 2 + 2 + 2 = 6}
(iii) (1.5)³ = 1.5 x 1.5 x 1.5 = 3.375
{Sum of decimal places = 1 + 1 + 1 = 3}

Question 9.
Solution:
Distance covered in one hour = 62.5 km
Distance covered in 18 hours = 62.5 x 18 km = 1125.0 km

Question 10.
Solution:
Weight of one tin of oil = 16.8 kg
Weight of 45 tins = 16.8 x 45 kg = 756.0 kg = 756 kg

Question 11.
Solution:
Weight of wheat in one bag = 97.8 kg
weight of wheat in 500 bags = 97.8 x 500 kg = 48900.0 kg = 48900 kg

Question 12.
Solution:
Weight of one bag = 48.450 kg
Weight of 16 bags = 48.450 x 16 = 775.200 kg

Question 13.
Solution:
Quantity of sauce in one bottle = 0.845 kg
quantity of sauce in 72 bottles = 0.845 x 72 kg = 60.840 kg

Question 14.
Solution:
Quantity of jam in one bottle = 925 .
Quantity of jam in 25 bottles = 925 x 25 g = 23135 g = 23.125 kg

Question 15.
Solution:
Oil in one drum = 16.850 litres
Oil in 48 drums = 16.850 x 48 = 808.800 = 808.800 litres

Question 16.
Solution:
Cost of 1 kg rice = Rs 56.80
Cost of 16.25 kg of rice = Rs 56.80 x 16.25 = Rs 923.0000 = Rs 923

Question 17.
Solution:
Cost of one metre of cloth = Rs 108.5 0
Costof 18.5 metres of cloth = Rs 108.50 x 18.5 = Rs 2007.250 = Rs 2007.25

Question 18.
Solution:
Distance covered in one litre = 8.6 km
Distance covered in 36.5 litres = 8.6 x 36.5 km = 313.90 km = 313.9 km

Question 19.
Solution:
Charges for 1 km = Rs 9.80
Charges for 106.5 km = Rs 9.80 x 106.5 = Rs 1043.700 = Rs 1043.70

Hope given RS Aggarwal Solutions Class 7 Chapter 3 Decimals Ex 3C are helpful to complete your math homework.

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RS Aggarwal Class 7 Solutions Chapter 2 Fractions Ex 2C

These Solutions are part of RS Aggarwal Solutions Class 7. Here we have given RS Aggarwal Solutions Class 7 Chapter 2 Fractions Ex 2C.

Other Exercises

• RS Aggarwal Solutions Class 7 Chapter 2 Fractions Ex 2A
• RS Aggarwal Solutions Class 7 Chapter 2 Fractions Ex 2B
• RS Aggarwal Solutions Class 7 Chapter 2 Fractions Ex 2C
• RS Aggarwal Solutions Class 7 Chapter 2 Fractions Ex 2D
• RS Aggarwal Solutions Class 7 Chapter 2 Fractions CCE Test Paper

Question 1.
Solution:

Question 2.
Solution:

Question 3.
Solution:

Question 4.
Solution:

Question 5.
Solution:
Total weight of 18 boxes

Question 6.
Solution:
Total amount of oranges=Rs 378

Question 7.
Solution:

Question 8.
Solution:

Question 9.
Solution:

Question 10.
Solution:

Question 11.
Solution:

Question 12.
Solution:

Question 13.
Solution:
Total quantity of milk = 24 litres
and quantity of milk got by one student = \(\frac { 2 }{ 5 }\) litres

Question 14.
Solution:
Quantity of water in a bucket

Question 15.
Solution:

Question 16.
Solution:
Product of two numbers = 42

Question 17.
Solution:

Hope given RS Aggarwal Solutions Class 7 Chapter 2 Fractions Ex 2C are helpful to complete your math homework.

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RS Aggarwal Class 7 Solutions Chapter 2 Fractions Ex 2B

These Solutions are part of RS Aggarwal Solutions Class 7. Here we have given RS Aggarwal Solutions Class 7 Chapter 2 Fractions Ex 2B.

Other Exercises

• RS Aggarwal Solutions Class 7 Chapter 2 Fractions Ex 2A
• RS Aggarwal Solutions Class 7 Chapter 2 Fractions Ex 2B
• RS Aggarwal Solutions Class 7 Chapter 2 Fractions Ex 2C
• RS Aggarwal Solutions Class 7 Chapter 2 Fractions Ex 2D
• RS Aggarwal Solutions Class 7 Chapter 2 Fractions CCE Test Paper

Question 1.
Solution:

Question 2.
Solution:

Question 3.
Solution:

Question 4.
Solution:

Question 5.
Solution:
Cost of 1 metre pf cloth

Question 6.
Solution:
Distance covered in 1 hour

Question 7.
Solution:

Question 8.
Solution:

Question 9.
Solution:

Question 10.
Solution:

Question 11.
Solution:
Weight of Amit = 35 kg.

Question 12.
Solution:
Total number of students in a class = 42
Number of boys = \(\frac { 5 }{ 7 }\) of 42 = 5 x 6 = 30
Number of girls = 42 – 30 = 12

Question 13.
Solution:
Sapna total income for one month = Rs 24000
Amount spent = \(\frac { 7 }{ 8 }\) of her income
= \(\frac { 7 }{ 8 }\) x 24000
= Rs (7 x 3000) = Rs 21000
Amount deposited in the bank per month = Rs 24000 – 21000 = Rs 3000

Question 14.
Solution:
Length of each side of a square

Question 15.
Solution:
Length of rectangular field (l)

Hope given RS Aggarwal Solutions Class 7 Chapter 2 Fractions Ex 2B are helpful to complete your math homework.

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RS Aggarwal Class 7 Solutions Chapter 1 Integers CCE Test Paper

These Solutions are part of RS Aggarwal Solutions Class 7. Here we have given RS Aggarwal Solutions Class 7 Chapter 1 Integers CCE Test Paper.

Other Exercises

• RS Aggarwal Solutions Class 7 Chapter 1 Integers Ex 1A
• RS Aggarwal Solutions Class 7 Chapter 1 Integers Ex 1B
• RS Aggarwal Solutions Class 7 Chapter 1 Integers Ex 1C
• RS Aggarwal Solutions Class 7 Chapter 1 Integers Ex 1D
• RS Aggarwal Solutions Class 7 Chapter 1 Integers CCE Test Paper

Question 1.
Solution:
Let the other integer be a. Then, we have;
a + (-12) = 43
⇒ a = 43 – (-12) = 55
Hence, the other integer is 55.

Question 2.
Solution:
p – (-8) = 3
⇒ p = 3 + (-8)
⇒ p = -5
Hence, the value of p is -5.

Question 3.
Solution:
Product of (-16) and (-9) = (-16) x (-9) = 144
Now, (-132) ÷ 6 gives the quotient -22.
144 + (-22) = 122

Question 4.
Solution:
Suppose that a divides -240 to obtain 16. Then, we have:
(-240) ÷ a =16
⇒ a = (-240) ÷ 16 = -15
Hence, -15 should divide -240 to obtain 16.

Question 5.
Solution:
Let a be divided by (-7) to obtain 12. Then, we have:
a ÷ (-7) = 12
a = \(\frac { -7 }{ 12 }\)
Hence, \(\frac { -7 }{ 12 }\) should be divided by -7 to obtain 12.

Question 6.
Solution:
(i) -450
(ii) 360
(iii) -1080
(iv) -600
(v) (-5)5 = -3125
(vi) (-1)25 = -1

Question 7.
Solution:
(i) (-16) x 12 + (-16) x 8 = (-16) x (12 + 8) [Associative property]
= (-16) x 20
= -320
(ii) 25 x (-33) + 25 x (-17)
= 25 x [(-33) + (-17)] [Associative property]
= 25 x (-50)
= -1250
(iii) (-19) x (-25) + (-19) x (-15)
= (-19) x [(-25) + (-15)] [Associative property]
= (-19) x (-40)
= 760
(iv) (-47) x 68 – (-47) x 38
= (-47) x (68 – 38) [Asssociative property]
= (-47) x 30
= -1410
(v) (-105) ÷ 21
(-105) ÷ 21 = -5
(vi) (-168) ÷ (-14) = 12
(vii) 0 ÷ (-34)
= 0 (zero).
Dividing 0 by any integer gives 0.
(viii) 37 ÷ 0
Not defined.
Dividing any integer by zero is not defined.

Mark (√) against the correct answer in each of the following:
Question 8.
Solution:
(d) -8
Let the other integer be a. Then, we have:
2 + a = -6
⇒ a = -6 – 2 = -8
⇒ The other integer is -8.

Question 9.
Solution:
(b) 8
Suppose that a is subtracted from (-7). Then, we have
-7 – a = -15
a = -7 + 15 = 8
8 must be subtracted from -7 to obtain-15.

Question 10.
Solution:
(b) 108
(108) + (-18) = -6

Question 11.
Solution:
(a) 370
= (-37) x (-7) + (-37) x (-3)
= (-37) x [(-7) + (-3)] [Associative property]
= (-37) x (-10)
= 370

Question 12.
Solution:
(c) -250
=(-25) x 8 + (-25) x 2
= (-25) x (8 + 2) [Associative property]
= (-25) x 10 = -250

Question 13.
Solution:
(b) -3
(-9) – (-6)
= (-9) + 6
= -3

Question 14.
Solution:
(b) -6
-8 – (-6) = 2
Hence, -8 is -6 less than -2.

Question 15.
Solution:
(i) (-35) x -1 = 35
(ii) (-53) x (1) = -53
(iii) (-14) x (-1) = (-16) x (-14) [Commutative property]
(iv) (-21) x (0) = 0 [Property of zero]
(v) (-119) ÷ 17 = (-7)
(vi) (-247) ÷ (-19) = 13
(vii) (0) ÷ 31 = 0
(viii) (152) ÷ (-19) = -8

Question 16.
Solution:
(i) True (T)
(ii) (-8) ÷ 0 = 0
False (F). Dividing any integer by zero is not defined.
(iii) False (F).
(-1) ÷ (-1) = 1
(iv) True (T)
(v) True(T)
(vi) False (T).
68 ÷ (-17) = -4

Hope given RS Aggarwal Solutions Class 7 Chapter 1 Integers CCE Test Paper are helpful to complete your math homework.

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RS Aggarwal Class 7 Solutions Chapter 3 Decimals Ex 3E

These Solutions are part of RS Aggarwal Solutions Class 7. Here we have given RS Aggarwal Solutions Class 7 Chapter 3 Decimals Ex 3E.

Other Exercises

• RS Aggarwal Solutions Class 7 Chapter 3 Decimals Ex 3A
• RS Aggarwal Solutions Class 7 Chapter 3 Decimals Ex 3B
• RS Aggarwal Solutions Class 7 Chapter 3 Decimals Ex 3C
• RS Aggarwal Solutions Class 7 Chapter 3 Decimals Ex 3D
• RS Aggarwal Solutions Class 7 Chapter 3 Decimals Ex 3E
• RS Aggarwal Solutions Class 7 Chapter 3 Decimals CCE Test Paper

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

Question 2.
Solution:
(c)

Question 3.
Solution:
(b)

= \(\frac { 208 }{ 100 }\) = 2.08

Question 4.
Solution:

Question 5.
Solution:
(b) 70g = \(\frac { 70 }{ 1000 }\) = 0.07 kg

Question 6.
Solution:
(c) 5 kg 6 g = 5\(\frac { 6 }{ 1000 }\) kg = 5.006 kg

Question 7.
Solution:
(c) 2 km 5 m = 2\(\frac { 5 }{ 1000 }\) km = 2.005 km

Question 8.
Solution:
(c)
1.007 – 0.7 = 1.007 – 0.700 = 0.307

Question 9.
Solution:
(b)
0.1 – 0.03 = 0.10 – 0.03 = 0.07

Question 10.
Solution:
(c)
3.5 – 3.07 = 3.50 – 3.07 = 0.43

Question 11.
Solution:
(c)
0.23 x 0.3 = 0.069

Question 12.
Solution:
(b)
0.02 x 30 = .60 = .6

Question 13.
Solution:
(b)
0.25 x 0.8 = 0.200 = 0.2

Question 14.
Solution:
(c)
0.4 x 0.4 x 0.4 = 0.064

Question 15.
Solution:
(b)
1.1 x .1 x .01 = .0011

Question 16.
Solution:
(a)

Question 17.
Solution:
(b)
1.02 ÷ 6 = \(\frac { 1.02 }{ 6 }\) = 0.17

Question 18.
Solution:

Question 19.
Solution:
(b)

Question 20.
Solution:
(a)

Question 21.
Solution:
(c)

Hope given RS Aggarwal Solutions Class 7 Chapter 3 Decimals Ex 3E are helpful to complete your math homework.

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RS Aggarwal Class 7 Solutions Chapter 4 Rational Numbers Ex 4D

These Solutions are part of RS Aggarwal Solutions Class 7. Here we have given RS Aggarwal Solutions Class 7 Chapter 4 Rational Numbers Ex 4D.

Other Exercises

• RS Aggarwal Solutions Class 7 Chapter 4 Rational Numbers Ex 4A
• RS Aggarwal Solutions Class 7 Chapter 4 Rational Numbers Ex 4B
• RS Aggarwal Solutions Class 7 Chapter 4 Rational Numbers Ex 4C
• RS Aggarwal Solutions Class 7 Chapter 4 Rational Numbers Ex 4D
• RS Aggarwal Solutions Class 7 Chapter 4 Rational Numbers Ex 4E
• RS Aggarwal Solutions Class 7 Chapter 4 Rational Numbers Ex 4F
• RS Aggarwal Solutions Class 7 Chapter 4 Rational Numbers Ex 4G
• RS Aggarwal Solutions Class 7 Chapter 4 Rational Numbers CCE Test Paper

Question 1.
Solution:
(i) Additive inverse of 5 = -5
(ii) Additive inverse of -9 = – (-9) = 9

Question 2.
Solution:

Question 3.
Solution:

Question 4.
Solution:

Question 5.
Solution:

Question 6.
Solution:

Question 7.
Solution:

Question 8.
Solution:

Question 9.
Solution:

Question 10.
Solution:

Question 11.
Solution:

Question 12.
Solution:
The required number = \(\frac { -1 }{ 1 }\) – \(\frac { 2 }{ 9 }\)

Question 13.
Solution:

Question 14.
Solution:

Question 15.
Solution:

Question 16.
Solution:

Hope given RS Aggarwal Solutions Class 7 Chapter 4 Rational Numbers Ex 4D are helpful to complete your math homework.

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RS Aggarwal Class 7 Solutions Chapter 2 Fractions CCE Test Paper

These Solutions are part of RS Aggarwal Solutions Class 7. Here we have given RS Aggarwal Solutions Class 7 Chapter 2 Fractions CCE Test Paper.

Other Exercises

• RS Aggarwal Solutions Class 7 Chapter 2 Fractions Ex 2A
• RS Aggarwal Solutions Class 7 Chapter 2 Fractions Ex 2B
• RS Aggarwal Solutions Class 7 Chapter 2 Fractions Ex 2C
• RS Aggarwal Solutions Class 7 Chapter 2 Fractions Ex 2D
• RS Aggarwal Solutions Class 7 Chapter 2 Fractions CCE Test Paper

Question 1.
Solution:
(i) A number of the form \(\frac { a }{ b }\), where a and b are rational numbers, is called a natural number.
Here, a is the numerator and b is the denominator.
(a) \(\frac { 2 }{ 3 }\) is a fraction with 2 as the numerator and 3 as the denominator.
(b) \(\frac { 12 }{ 5 }\) is a fraction with 12 as the numerator and 5 as the denominator.
(ii) A fraction whose denominator is a whole number other than 10, 100, 1000, etc., is called a vulgar faction.
Examples : \(\frac { 2 }{ 5 }\) and \(\frac { 4 }{ 15 }\).
(iii) A fraction whose numerator is greater than or equal to its denominator is called an improper fraction.
Example : \(\frac { 11 }{ 3 }\) and \(\frac { 41 }{ 35 }\)

Question 2.
Solution:
Required number to be added

Hence, the required number is 8\(\frac { 2 }{ 5 }\)

Question 3.
Solution:

Question 4.
Solution:

Question 5.
Solution:

Question 6.
Solution:

Question 7.
Solution:

Question 8.
Solution:

Question 9.
Solution:

Mark (√) against the correct answer in each of the following:

Question 10.

Solution:
(d) \(\frac { 5 }{ 8 }\)
\(\frac { 5 }{ 8 }\) is a vulgar fraction, because its denominator is other than 10,100, 1000, etc.

Question 11.
Solution:
(c) \(\frac { 46 }{ 63 }\)
A fraction \(\frac { a }{ b }\) is said to be irreducible or in its lowest terms if the HCF of a and b is 1
46 = 2 x 23 x 1
63 = 3 x 3 x 21 x 1
Clearly, the HCF of 46 and 63 is 1.
Hence, \(\frac { 46 }{ 63 }\) is an irreducible fraction.

Question 12.
Solution:
(d) None of these
Reciprocal of 1\(\frac { 3 }{ 5 }\) = Reciprocal of \(\frac { 8 }{ 5 }\) = \(\frac { 5 }{ 8 }\)

Question 13.
Solution:

Question 14.
Solution:

Question 15.
Solution:

Question 16.
Solution:
(b) 33 km
Distance covered by the car on

Question 17.
Solution:

Question 18.
Solution:
(i) False.
By cross multiplication, we have:
9 x 24 = 216 and 13 x 16 = 208
However, 216 > 208
\(\frac { 9 }{ 16 }\) > \(\frac { 13 }{ 24 }\)

Hope given RS Aggarwal Solutions Class 7 Chapter 2 Fractions 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 1 Integers Ex 1D

These Solutions are part of RS Aggarwal Solutions Class 7. Here we have given RS Aggarwal Solutions Class 7 Chapter 1 Integers Ex 1D.

Other Exercises

• RS Aggarwal Solutions Class 7 Chapter 1 Integers Ex 1A
• RS Aggarwal Solutions Class 7 Chapter 1 Integers Ex 1B
• RS Aggarwal Solutions Class 7 Chapter 1 Integers Ex 1C
• RS Aggarwal Solutions Class 7 Chapter 1 Integers Ex 1D
• RS Aggarwal Solutions Class 7 Chapter 1 Integers CCE Test Paper

OBJECTIVE QUESTIONS
Mark (√) against the correct answer in each of the following.
Question 1.
Solution:
(c)
6 – (-8) = 6 + 8 = 14

Question 2.
Solution:
(b)
-9 – (-6) = -9 + 6 = -3

Question 3.
Solution:
(d)
-3 + 5 = 2

Question 4.
Solution:
(a)
-1 – (+5) = -1 – 5 = -6

Question 5.
Solution:
(a)
-2 – (4) = -2 – 4 = -6

Question 6.
Solution:
(b)
-4 – (+4) = -4 – 4 = -8

Question 7.
Solution:
(b)
-3 – (-5) = -3 + 5 = 2

Question 8.
Solution:
(c)
-3 – (-9) = -3 + 9 = 6

Question 9.
Solution:
(c)
-5 – (6) = -5 – 6 = -11

Question 10.
Solution:
(c)
-8 – (-13) = -8 + 13 = 5

Question 11.
Solution:
(a)
(-36) ÷ (-9) = 4

Question 12.
Solution:
(b)
0 ÷ (-5) = 0
(Zero divided by any integer other than zero, is zero)

Question 13.
Solution:
(c)
Division by zero is not defined

Question 14.
Solution:
(b)

Question 15.
Solution:
(b)
-3 + 9 = 6

Question 16.
Solution:
(a)
-4 – (-10) = -4 + 10 = 6

Question 17.
Solution:
(a)
Sum = 14
One integer = -8
Second = 14 – (-8) = 14 + 8 = 22

Question 18.
Solution:
(c)

Question 19.
Solution:
(b)
(-15) x 8 + (-15) x 2
= (-15) {8 + 2}
= -15 x 10 = -150

Question 20.
Solution:
(b)
(-12) x 6 – (-12) x 4 = (-12) (6 – 4) = -12 x 2 = -24

Question 21.
Solution:
(b)
(-27) x (-16)+ (-27) x (-14)
= (-27) {-16 – 14}
= (-27) x (-30)
= 810

Question 22.
Solution:
(a)
30 x (-23) + 30 x 14
= 30 x (-23 + 14)
= 30 x (-9)
= -270

Question 23.
Solution:
(c)
Sum of two integers = 93
One integer = -59
Second = 93 – (-59) = 93 + 59 = 152

Question 24.
Solution:
(b)
(?) ÷ (-18) = -5
Let x ÷ (-18) = -5
⇒ \(\frac { x }{ -18 }\) = -5
⇒ x = (-5) x (-18) = 90

Hope given RS Aggarwal Solutions Class 7 Chapter 1 Integers Ex 1D 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 3 Decimals Ex 3B

These Solutions are part of RS Aggarwal Solutions Class 7. Here we have given RS Aggarwal Solutions Class 7 Chapter 3 Decimals Ex 3B.

Other Exercises

• RS Aggarwal Solutions Class 7 Chapter 3 Decimals Ex 3A
• RS Aggarwal Solutions Class 7 Chapter 3 Decimals Ex 3B
• RS Aggarwal Solutions Class 7 Chapter 3 Decimals Ex 3C
• RS Aggarwal Solutions Class 7 Chapter 3 Decimals Ex 3D
• RS Aggarwal Solutions Class 7 Chapter 3 Decimals Ex 3E
• RS Aggarwal Solutions Class 7 Chapter 3 Decimals CCE Test Paper

Add:
Question 1.
Solution:
Converting them into like decimals 16.00, 8.70, 0.94, 6.80 and 7.77
Now, adding them,
16.0 + 8.70 + 0.94 + 6.80 + 7.77 = 40.21

Question 2.
Solution:
Converting them into like decimals 18.600, 206.370, 8.008, 26.400, 6.900
Adding we get
18.600 + 206.370 + 8.008 + 26.400 + 6.900 = 266.278

Question 3.
Solution:
Converting them into like decimals, 63.50, 9.70, 0.80, 26.66, 12.17
Adding we get:
63.50 + 9.70 + 0.80 + 26.66 + 12.17 = 112.83

Question 4.
Solution:
Converting them into like decimals 17.400, 86.390, 9.435, 8.800, 0.060
Adding we get:
17.400 + 86.390 + 9.435 + 8.800 + 0.060 = 122.085

Question 5.
Solution:
Converting them into like decimals 26.900, 19.740, 231.769, 0.048
Now adding we get:
26.900 + 19.740 + 231.769 + 0.048 = 278.457

Question 6.
Solution:
Converting them into like decimals 23.800, 8.940, 0.078 and 214.600
Now adding we get:
23.800 + 8.940 + 0.078 + 214.600 = 247.418

Question 7.
Solution:
Converting them into like decimals.
6.606, 66.600, 666.000,0.066, 0.660
Now adding we get:
6.606 + 66.600 + 666.000 + 0.066 + 0,660 = 739.932

Question 8.
Solution:
9.090, 0.909, 99.900, 9.990, 0.099
Now adding we get:
9.090 + 0.909 + 99.900 + 9.990 + 0.099 = 119.988

Subtract:
Question 9.
Solution:
14.79 from 72.43
72.43 – 14.79 = 57.64

Question 10.
Solution:
Converting them into like decimals, We get
36.74 and 52.60
Now 52.60 – 36.74 = 15.86

Question 11.
Solution:
Converting them into like decimals, We get
13.876 and 22.000
22.000 – 13.876 = 8.124

Question 12.
Solution:
Converting them into like decimals, We get
15.079 and 24.160
24.160 – 15.079 = 9.081

Question 13.
Solution:
Converting them into like decimals We get
0.680 and 1.007
1.007 – 0.680 = 0.327

Question 14.
Solution:
Converting them into like decimals,
We get 0.4678 and 5.0500
5.0500 – 0.4678 = 4.5822

Question 15.
Solution:
Converting them into like decimals,
We get 2.5307 and 8.0000
8.0 – 2.5307 = 5.4693

Question 16.
Solution:
There are like decimals
9.1 – 6.732 = 2.269

Question 17.
Solution:
Converting them into like decimals,
We get 5.746 and 9.100
9.100 – 5.746 = 3.354

Question 18.
Solution:
Converting into like decimals, we get,
63.59 and 92.00
Required number = 92.00 – 63.58 = 28.42

Question 19.
Solution:
Converting into like decimals, we get:
8.100 and 0.813
Required number = 8.100 – 0.813 = 7.287

Question 20.
Solution:
Converting them into like decimals, we get: 32.67 and 60.10
Required number = 60.10 – 32.67 = 27.43

Question 21.
Solution:
Converting into like decimals, we get 74.3 and 26.87
Required number = 74.30 – 26.87 = 47.43

Question 22.
Solution:
Cost of notebook = Rs. 23.75
Cost ofpencil = Rs. 2.85
Costofpen =Rs. 15.90
Total cost = Rs. 42.50
Amount gave to the shop keeper = 50 rupees
Balance amount got = Rs 50.00 – Rs 42.50 = 7.50

Hope given RS Aggarwal Solutions Class 7 Chapter 3 Decimals Ex 3B 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.