Class-8 Subject-Maths Chapter-2 Exercise2.1 Linears Equations in one Varibale

 EVENTS CONVENT HIGH SCHOOL
10/08/2021          CLASS-8         SESSION2021-22(SLOT-1)
Maths
Chapter-2
Linear Equations in one variable
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Class-7 Subject-Maths Chapter-1 Exercise1.3 INTEGERS

 
EVENTS CONVENT HIGH SCHOOL
10/08/2021          CLASS-7         SESSION2021-22(SLOT-1)
Maths
Chapter-1
INTEGERS EXERCISE 1.3
______________________________________

Question 1.Find each of the following products:

(a) 3 × (-1)

(b) (-1) × 225

(c) (-21) × (-30)

(d) (-316) × (-1)

(e) (-15) × 0 × (-18)

(f) (-12) × (-11) × (10)

(g) 9 × (-3) × (-6)

(h) (-18) × (-5) × (-4)

(i) (-1) ×(-2) × (-3) × 4

(j) (-3) × (-6) × (-2) × (-1)

Solution:

(a) 3 × (-1) = -3 × 1 = -3

(b) (-1) × 225 = -1 × 225 = -225

(c) (-21) × (-30) = (-) × (-) × 21 × 30 = 630

(d) (-316) × (-1) = (-) × (-) × 316 × 1 = 316

(e) (-15) × 0 × (-18) = 0 [∵ a × 0 = a]

(f) (-12) × (-11) × (10)

= (-) × (-) × 12 × 11 × 10 = 1320

(g) 9 × (-3) × (-6) = (-3) × (-6) × 9

= (—) × (-) × 3 × 6 × 9 = 162

(h) (-18) × (-5) × (-4)

= (-) × (-) × (-) × 18 × 5 × 4 = -360

(i) (-1) × (-2) × (-3) × 4

= (-) × (-) × (-) × 1 × 2 × 3 × 4 = -24

(j) (-3) × (-6) × (-2) × (-1)

= (-) × (-) × (-) × (-) × 3 × 6 × 2 × 1 = 36

 

Question 2.Verify the following:

(a) 18 × [7 + (-3)] = [18 × 7] + [18 × (-3)]

(b) (-21) × [(-4) + (-6)] = [(-21) × (-4)] + [(-21) × (-6)]

Solution:

(a) 18 × [7 + (-3)] = [18 × 7] + [18 × (-3)]

LHS = 18 × [7 + (-3)] = 18 × 4 = 72

RHS = [18 × 7] + [18 × (-3)] = 126 + (-54)

= 126 – 54 = 72

LHS = RHS

Hence, verified.

 

 (b) (-21) × [(-4) + (-6)] = [(-21) × (-4)] + [(-21) × (-6)]

LHS = (-21) × [(-4) + (-6)]

= (-21) × (-10)

= (-) × (-) × 21 × 10 = 210

RHS = [(-21) × (-4)] + [(-21) × (-6)]

= (84) + (126) = 84 + 126 = 210

LHS = RHS

Hence, verified.

 

Question 3.(i) For any integer a, what is (-1) × a equal to?

(ii) Determine the integer whose product with (-1) is 0.

(a) -22

(b) 37

(c) 0

Solution:

(i) (-1) × a = -a

(ii) (-1) × 0 = 0 [∵ a × 0 = 0]

Hence (c) 0 is the required integer.

 

Question 5.Find the product, using suitable properties:

(a) 26 × (-48) + (-48) × (-36)

(b) 8 × 53 × (-125)

(c) 15 × (-25) × (-4) × (-10)

(d) (-41) × 102

(e) 625 × (-35) + (-625) × 65

(f) 7 × (50 – 2)

(g) (-17) × (-29)

(h) (-57) × (-19) + 57

Solution:

(a) 26 × (-48) + (-48) × (-36)

= -48 × [26 + (-36)] = -48 × [26 – 36] = -48 × -10 = 480 [Distributive property of multiplication over

addition]

 

(b) 8 × 53 × (-125) = 53 × [8 × (-125)]

[Associative property of multiplication] = 53 × (-1000) = -53000

 

(c) 15 × (-25) × (-4) × (-10)

= [(-25) × (-4)] × [15 × (-10)]

[Regrouping the terms] = 100 × (-150) = -15000

 

(d) (-41) × 102 = (-41) × [100 + 2]

= (-41) × 100 + (-41) × 2

[Distributive property of multiplication over addition] = -4100 – 82 = -4182

 

(e) 625 × (-35) + (-625) × 65

= 625 × [(-35) + (-65)]

[Distributive property of multiplication over addition]

= 625 × (-100) = -62500

 

(f) 7 × (50 – 2) = 7 × 48 = 336 or

7 × (50 – 2) = 7 × 50 -7 × 2 = 350 – 14 = 336 [Distributive property of multiplication over addition]

 

(g) (-17) × (-29) = (-17) × [30 + (-1)]

= (-17) × 30 + (-17) × (-1)

= -510 + 17 = -493

[Distributive property of multiplication over addition]

 

(h) (-57) × (-19) + 57 = 57 × 19 + 57

= 57 × 19 + 57 × 1 [Y (-) × (-) = (+)] [Distributive property of multiplication over addition]

= 57 × (19 + 1) = 57 × 20 = 1140

 

Question 6.A certain freezing process requires that room temperature be lowered from 40°C at the rate of 5°C every hour. What will be the room temperature 10 hours after the process begins?

Solution:

Temperature of the room in the beginning = 40°C

Temperature after 1 hour

= 40°C – 1 × 5°C = 40°C – 5°C – 35°C

Similarly, temperature of the room after 10 hours

= 40°C – 10 × 5°C = 40°C – 50°C = -10°C

 

Class-6 Subject-Maths Chapter-1 Exercise1.3 Knowing our Numbers

 

EVENTS CONVENT HIGH SCHOOL
10/08/2021          CLASS-6         SESSION2021-22(SLOT-1)
Maths
Chapter-1
Knowing our Numbers
______________________________________

QUESTION-1 EXPLANATION

Question 1.Estimate each of the following using general rule:

(a) 730 + 998

(b) 796 – 314

(c) 12,904 + 2,888

(d) 28,292 – 21,496

Make ten more such examples of addition, subtraction and estimation of their outcome.

Solution:

(a) 730 + 998

Rounding off 730 nearest to hundreds = 700

Rounding off 998 nearest to hundreds = 1,000

∴ 730 + 998 = 700 + 1000 = 1700

 

(b) 796 – 314

Rounding off 796 nearest to hundreds = 800

Rounding off 314 nearest to hundreds = 300

∴ 796 – 314 = 800 – 300 = 500

 

(c) 12,904 + 2,888

Rounding off 12,904 nearest to thousands = 13000

Rounding off 2888 nearest to thousands = 3000

∴ 12,904 + 2,888 = 13000 + 3000 = 16000

 

(d) 28,292 – 21,496

Rounding off 28,292 nearest to thousands = 28,000

Rounding off 21,496 nearest to thousands = 21,000

∴ 28,292 – 21,496 = 28,000 – 21,000 = 7,000

 

Example 1: 1210 + 2365 = 1200 + 2400 = 3600

Example 2: 3853 + 6524 = 4000 + 7000 = 11,000

Example 3: 8752 – 3654 = 9,000 – 4,000 = 5,000

Example 4: 4538 – 2965 = 5,000 – 3,000 = 2,000

Example 5: 1927 + 3185 = 2000 + 3,000 = 5,000

Example 6: 3258 – 1698 = 3000 – 2000 = 1,000

Example 7: 8735 + 6232 = 9000 + 6000 = 15,000

Example 8: 1038 – 1028 = 1000 – 1000 = 0

Example 9: 6352 + 5830 = 6,000 + 6,000 = 12,000

Example 10: 9854 – 6385 = 10,000 – 6000 = 4,000

QUESTION-2 EXPLANATION

Question 2.Give a rough estimate (by rounding off to nearest hundreds) and also a closer estimate (by rounding off to nearest tens):

(a) 439 + 334 + 4,317

(b) 1,08,734-47,599

(c) 8,325-491

(d) 4,89,348-48,365

Make four such examples:

Solution:

(a)439 + 334 + 4,317

(i) Rough estimate (Rounding off to nearest hundreds)

439 + 334 + 4,317 = 400 + 300 + 4300 = 5,000

(ii) Closer estimate (Rounding off to nearest tens)

439 + 334 + 4317 = 440 + 330 + 4320 = 5090.

 

(b) 1,08,734 – 47,599

(i) Rough estimate (Rounding off to nearest hundreds)

1,08,734 – 47,599 = 1,08,700 – 47,600 = 61,100

(ii) Closer estimate (Rounding off to nearest tens)

1,08,734 – 47,599 = 1,08,730 – 47,600 = 61,130.

 

(c) 8325 – 491

(i) Rough estimate (Rounding off to nearest hundreds)

8325 – 491 = 8300 – 500 = 7800

(ii) Closer estimate (Rounding off to nearest tens)

8325 – 491 = 8330 – 490 = 7840.

 

(d) 4,89,348 – 48,365

(i) Rough estimate (Rounding off to nearest hundreds)

4,89,348 – 48,365 = 4,89,300 – 48,400 = 4,40,900

(ii) Closer estimate (Rounding off to nearest tens)

4,89,348 – 48,365 = 4,89,350 – 48,370 = 4,40,980

 

Example 1:384 + 562

Solution:(i) Rough estimate (Rounding off to nearest hundreds)

384 + 562 = 400 + 600

= 1,000

(ii) Closer estimate (Rounding off to nearest tens)

384 + 562 = 380 + 560

= 940

 

Example 2:8765 – 3820

Solution:(i) Rough estimate (Rounding off to nearest hundreds)

8765 – 3820 = 8800 – 3900

= 4900

(ii) Closer estimate (Rounding off to nearest tens)

8765 – 3820 = 8770 – 3820

= 4950

 

Example 3:6653 – 8265

Solution:(i) Rough estimate (Rounding off to nearest hundreds)

6653 + 8265 = 6700 + 8300

= 15,000

(ii) Closer estimate (Rounding off to nearest tens)

6653 + 8265 = 6650 + 8270

= 14920

 

Example 4:3826 – 1262

Solution:(i) Rough estimate (Rounding off to nearest hundreds)

3826 – 1262 = 3800 – 1300

= 2500

(ii) Closer estimate (Rounding off to nearest tens)

3826 – 1262 = 3830 – 1260

= 2570

QUESTION-3 EXPLANATION

Question 3.Estimate the following products using general rule:

(a) 578 x 161

(b)5281 x 3491

(c) 1291 x 592

(d) 9250 x 29

Solution:

(a) 578 x 161 = 600 x 200 = 1,20,000

(b) 5281 x 3491 = 5000 x 3000 = 1,50,00,000

(c) 1291 x 592 = 1300 x 600 = 7,80,000

(d) 9250 x 29 = 9000 x 30 = 2,70,000