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Initial position = 10m final position = -350 m
Total distance elevator has to travel = initial position – final position
= 10 – (-350)
= 10 + 350 = 360 m
Now,
Elevator descends at rate 6m/min
∴ Elevator travels 6m in = 1 minute
Elevator travels 1 m in: = \(\frac{1}{6}\)
Elevator travels 360m in = \(\frac{1}{6}\) × 360 minutes
= 60 minutes
= 1 hour
∴ It will take the elevator 1 hour.

80 – 56 ÷ 8 × 9 (Division) 

= 80 – 7 × 9 (Multiplication) 

= 80 – 63 (Subtraction) 

= 17

Decrease in water level per every week = 5 inches = (- 5) 

Decrease in water level per 6 weeks = (-5) × 6 . 

= – (5 × 6) 

= – 30 inches 

Change in the level of the water in the pond per 6 weeks is 30 inches decreased.

Profit on each notebook = ₹5 

Profit on 1500 notebooks = 1500 × 5 

Total profit on 1500 note books = ₹ 7500/- 

Loss on each pen = ₹3 

Which we denoted by – 3. 

Loss on 1500 pens = 1500 × -3 

= – ₹ 4500 

profit > loss

So, he will get profit. 

Total profit = 7500 – 4500 

= ₹ 3000/-

a) -11, -10, -5, -1,

b) -7, -15, -18, -20.

1 + 2 + 3 + 6 + (– 2) + (– 3) (One possible answer).

i) 7 and 12 

Integers lying between 7 and 12 are 8, 9, 10, ll. 

ii) -5 and -1 

Integers lying between -5 and -1 are -2, -3 and -4. 

iii) -3 and 3 

Integers lying between -3 and 3 are -2, -1, 0, 1, 2. 

iv) -6 and 0 

Integers lying between -6 and 0 are -5, -4, -3, -2, -1.

Profit on Monday = + ₹ 250 

Loss on Tuesday = – ₹ 120 

Loss on Wednesday = – ₹ 180 

Total loss two days = – 120 + (- 180) 

= – 120 – 180 = – 300 

Loss is greater than profit. So, he will get loss. 

∴ Total loss on three days = + 250 + (-300) 

= + 250 – 300

= – 50 

= – ₹ 50

Given profit per kg of tomato = ₹ 7 

Loss per kg of brinjal = ₹ 4 

Weight of tomatoes sold = 68 kgs 

Let number of kgs of tomato be x kg. 

Number of kgs of brinjals be y kg. 

7x – 4y = 0 

7 × 68 – 4y = 0

– 4y = – 7 × 68

y = -7 x 68/-4

∴ Weight of brinjals sold = + 7 × 17 

= + 119 kgs.

(i) Total marks scored by Sona = 20

Total correct answers = 10

∴ Marks for correct answers = 10 × 3 = 30 but she got 20 marks.

∴ Marks for incorrect answers = 20 – 30 = – 10 

-1 mark is given for every incorrect answer.

∴  Total incorrect answers  = -10/-1 = 10

(ii) Total correct answers = 10

Total incorrect answers = 10 (From (i) part)

∴ Total questions given in the test = 10 + 10 = 20

We can take any two negative integers, i.c., -2 and -4.

∴ Sum = -2 + (-4) = -2 – 4 = -6,

which is less than both -2 and -4.

Taking profit as a positive integer and loss as negative integer, we have grocer’s net profit or loss in 3 days

= ₹(47-12-8) = ₹27

Hence, the grocer has a profit of ₹ 27.

(i) 18 x [7 + (-3)] = 18 x 4 = 72

and [18 x 7] + [18 x (-3)] = 126 – 54 = 72

so 18 x [7 + (-3)] = [18 x 7] + [18 x (-3)]

(ii) (-21) x [(- 4) +(6)] = (-21) x (- 4 – 6) |

= (-21) x (-10) = 210

and [(21) x(-4)] + [(-21) x (-6)] = 84 + 126 = 210

so (-21) x [(- 4) + (-6)] = [(-21) x (- 4)] + [(-21) x (-6)]

Let x be the required integer.

According to question,

x = 4 + (-x), where (-x) is the additive inverse of x.

⇒ x = 4 – x 

⇒ x + x = 4 

=> 2x = 4 

=> x = 2

∴ The required integer is 2.

0 is the integer which is its own additive inverse.

Let the required integer be x.

According to question,

x = (-x) – 2, where -x is the additive inverse of x.

⇒ x = -x – 2 

⇒ x + x = -2

⇒ 2x = -2 

⇒ x = -1

(i) We have, a * b = a × b +(a × a + b × b)

Now, put a = (-3) and b = (-5)

(-3)* (-5) = (- 3) × (- 5)+ [(- 3) × (- 3)+ (- 5) × (- 5)]

= 15 + (9 + 25)= 15 + 34 = 49

(ii) Now, put a = – 6 and b = 2

(-6)*2 = (-6) × 2 +[(-6) × (-6) + 2 × 2

= -6 × 2 + (36 + 4)= -12 + 40= 28

Initial temperature = 40°C
Room temperature decreases at 5°C per hour
Temperature change in 1 hour = -5°C
Temperature change in 10 hour = -5 × 10°C = -50°C
Room temperature after 10 hours = Initial temperature + Temperature change in 10 hour
= 40 + (-50)
= 40 – 50
∴ Room temperature after 10 hours = -10°C

On the number line, the value of a number increases as we move to right and decreases as we move to the left. 

As -3 lies right to -4 on the number line. So, -3 < -4 is not true.

Taking winning by time be a positive integer and losing by time be a negative integer.

∴ Sana’s total time (winning/losing)

= (10 – 60 + 20 – 25 – 37 + 12) seconds

= (42 – 122) seconds = -80 seconds

Hence, Sana lost the race by 80 seconds or 1 min. 20 seconds.

i.e., Fatima won the race finally.

We know, |- x| = x 

|- 700| = 700

(a) 18 × [7 – 3] = [126] + [18 × -3] 

18 × [4] = 126 + [-54] 

72 = 72 (verified) 

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

-21 × -10 = [84] + [126] + 210 = 210 

LHS = RHS (verified)

Additive inverse of + 2 = – (+2) = – 2 

Additive inverse of – 3 = – (-3) = + 3.

-1:

a x (-1) = -a = additive inverse of (a)

a Δ b = a × b – 2 × a × b + b × b (-a) × b + b × b

(i) 4 Δ (-3) 

= 4 × (-3) – 2 × 4 × (-3) + (-3) × (-3)(-4) × (-3) + (-3) × (-3) 

= -12 + 24 + 108 + 9 

= -12 + 141 

= 129

(ii) (-7) Δ (-1) 

= (-7) × (-1) – 2 × (-7) × (-1) + (-1) × (-1) (7) × (-1) + (-1) × (-1) 

= 7 -14 - 7 + 1 

= 8 - 21

= -13

Now, (-3) Δ 4 

= (-3) × 4 – 2 × (-3) ×(4) + 4 × 4(3) × 4 + 4 × 4

= -12 + 24 + 192 + 16 

= -12 + 232 

= 220

But 4 Δ (-3) = 129

∴ 4 Δ (-3) ≠ (-3) A 4

And, (-1) Δ (-7) 

= (-1) × (-7) – 2 × (-1) × (-7) + (-7) × (-7)(1) × (-7) + (-7) × (-7)

= 7 -14 – 343 + 49 

= 56 – 357 

= -301

But (-7) Δ (-1) = -13

∴ (-7) Δ (-1) ≠ (-1) Δ (-7)

i) + ₹ 500

ii) – 5° C

(a) As u × v = u and u = -4 ∴ -4 × v = -4

⇒ v = l (Multiplicative identity)

(b) As x × w = w. Given that x ≠ 1

∴ x × w = w is only possible when w = 0

(c) As u + x = w, Put u = – 4 and w = 0

∴ -4 + x = 0

⇒ x = 4 (Transposing -4 to R.H.S.)

1

1 is the multiplicative identity for integers.

i.e. 1 x a = a

6 × 1/6 = 1 (OR) 6 ÷ 6

But 1/6 is not a integer.

Multiplicative identity of 6 is 1/6

∴ 1/6 does not exist in integers.

-3, 8, -8, 8:

[(–8) + (-3) ] + 8 

= (-8) + [(–3) + 8] 

= –3

0:

(– 43) + 0 = – 43

1 is the multiplicative identity for integers.

y, x, z:

By associative property of integers, we have

(x + y) + z = x + (y + z)

True

As 1 is multiplicative identity for integers.

True

a × (b – c) = (a × b) – (a × c)

(Distributive property of multiplication over subtraction)

83:

83 ÷ ( – 1 ) = – 83

12, 5:

[12 × ( – 7)] × 5 = 12 × [(– 7) × 5] 

(Associative property of integers)

True

99 = 100 – 1 and 101 = 100 + 1

So, 99 × 101 = (100 – 1) × (100 + 1)

(a) The additive inverse of 3 is –3. 

So, – 4 – 3 = – 4 + (–3) = – (4 + 3) = –7 

(b) The additive inverse of –3 is + 3. 

So, – 4 – (–3) = – 4 + (+3) = –1

(i) 7 + 4 = 11 

(ii) 8 +( – 3) = 5 

(iii) 11 + 3 = 14 

(iv) 14 + ( – 6) = 8

(v) 9 + ( – 7) = 2 

(vi) 14 +( – 10) = 4 

(vii) 13 + ( – 15) = -2 

(viii) 4 + ( – 4) = 0 

(ix) 10 +( – 2) = 8 (x) 100 +( – 80) = 20 

(xi) 225 +( – 145) = 80

The sum of a negative integer and a positive integer is always a positive integer.

False (F),

Because we find the difference between the number and assign the bigger number sign.

23, 1, -100, 1:

23 × ( – 99) = 23 × ( – 100 + 1 ) = 23 × (-100) + 23 × 1 (Distributive property of integers)