Page 164 – 165

Q. Find the unknown weights in the following cases:

Solution:
(i) Fig 7.1:
Total weight = 16, Weight on either side = 8.

(ii) Fig 7.2:
Total weight = 24, Weight on either side = 12.

(iii) Fig 7.3:
Total weight = 8, Weight on either side = 4.

(iv) Fig 7.4:
Total weight = 18, Weight on either side = 9.

(v) Fig 7.5:
Total weight = 40, Weight on either side = 20.

(vi) Fig 7.6:

(vii) Fig 7.7:

(viii) Fig 7.8:

Page 165 – 166

Q. Find the unknown weight of the sack in the following cases. In Fig. 7.10, all the sacks have the same weight.

[Hint: If we remove equal weights from both the plates, will the weighing scale still be balanced? Remove one sack from each plate for Fig. 7.10.]

[Hint: Can you remove objects so that the sacks are only on one plate?]

Solution:
(i) Fig 7.9:
After removing 2 kg of weight from both plates, we get
1 sack = 10 kg

(ii) Fig 7.10:
After removing 1 sack from both plates, we get
1 sack = 14 kg

(iii) Fig 7.11:
After removing 2 sacks from both plates, we get
3 sacks = 21 kg
∴ 1 sack = $\frac{21}{3}$ = 7 kg.

(iv) Fig 7.12:
After removing 60 sacks and 50 kg of weight from both plates, we get
30 sacks = 450 kg
1 sack = $\frac{450}{30}$ = 15 kg

Figure it Out (Page 172)

1. Solve these equations and check the solutions.
(a)  3x– 10 = 35
(b)  5s = 3s
(c)  3u – 7 = 2u + 3
(d)  4 (m + 6) – 8 = 2m – 4
(e) $\frac{u}{15}$ = 6
Solution:
(a)  3x– 10 = 35
Adding 10 on both sides, we get
3x – 10 + 10 = 35 + 10
3x = 45
Dividing both sides by 3, we get
$\frac{3x}{3}$ = $\frac{45}{3}$
∴ x = 15.
Check: Substituting u = 15, we get
3(15) – 10 = 35
45 – 10 = 35
35 = 35.
∴ L.H.S = R.H.S

(b) 5s = 3s
Subtracting 3s from both sides, we get
5 s – 3s = 3 s – 3s
2s = 0
Dividing both by 2, we get
$\frac{2s}{2}$ = $\frac{0}{2}$
∴ s = 0.
Check: Substituting s = 0, we get
5(0) = 3(0)
0 = 0
∴ L.H.S = R.H.S

(c) 3u – 7 = 2u + 3
Adding 7 on both sides, we get
3u – 7 + 7 = 2u + 3 + 7
3u = 2u + 10
Subtracting 2u from both sides, we get
3u – 2u = 2u + 10 – 2u
∴ u = 10.
Check: Substituting u = 10, we get
3u – 7 = 2u + 3
3(10) – 7 = 2(10) + 3
30 – 7 = 20 + 3
23 = 23
∴ L.H.S = R.H.S

(d) 4 (m + 6) – 8 = 2m – 4
4m + 24 – 8 = 2m – 4
4m + 16 = 2m – 4
Subtracting 16 from both sides, we get
4m + 16 – 16 = 2m – 4 – 16
4m = 2m – 20
Subtracting 2m from both sides, we get
4m – 2m = 2m – 20 – 2m
2m = -20
Dividing both sides by 2, we get
$\frac{2m}{2}$ = $\frac{-20}{2}$
m = -10.
Check: Substituting m = -10, we get
4 {(-10) + 6} – 8 = 2(-10) – 4
4 {(-4)} – 8 = -20 – 4
-16 – 8 = – 24
-24 = -24
∴ L.H.S = R.H.S

(e) $\frac{u}{15}$ = 6
Multiplying both sides by 15, we get
$\frac{u}{15}$ × 15 = 6 × 15
u = 90.
Check: Substituting u = 90, we get
$\frac{90}{15}$ = 6
6 = 6
∴ L.H.S = R.H.S

2. Frame an equation that has no solution.
[Hint: 4 more than a number, and 5 more than a number can never be equal!]
Solution:
Let the number be x.
4 more than a number = $x+4$
5 more than a number = $x+5$
Framing an equation by making them equal, we get
x + 4 = x + 5
Subtracting x from both sides, we get
x + 4 – x = x + 5 – x
4 = 5
This is not true, so the equation is impossible. Therefore, the equation has no solution.

Figure it Out (Page 181)

1. Write 5 equations whose solution is x = – 2.
Solution:
(i) x + 2 = 0
(ii) 2x + 4 = 0
(iii) x + 5 = 3
(iv) 3x + 6 = 0
(v) x – 1 = -3

2. Find the value of each unknown:

Solution:
(a) 2y = 60
y = $\frac{60}{2}$ = 30.

(b) –8 = 5x – 3
5x – 3 = -8
5x = -8 + 3
5x = -5
x = $\frac{5}{5}$ = 1.

(c) –53w = –15
53w = 15
w = $\frac{15}{53}$.

(d) 13 – z = 8
13 – z – 8 = 0
5 – z = 0
z = 5.

(e) k + 8 = 12 – k
k + k = 12 – 8
2k = 4
k = $\frac{4}{2}$ = 2.

(f) 7m = m – 3
7m – m = -3
6m = -3
m = $\frac{-3}{6}$ = $\frac{-1}{2}$.

(g) 3n = 10 + n
3n – n = 10
2n = 10
n = $\frac{10}{2}$ = 5.

3. I am a 3-digit number. My hundred’s digit is 3 less than my ten’s digit. My ten’s digit is 3 less than my unit’s digit. The sum of all the three digits is 15. Who am I?
Solution:
Let the unit’s digit be x.
Ten’s digit = x − 3
Hundred’s digit = (x − 3) − 3 = x − 6
Since the sum of all three digits = 15
∴ (x − 6) + (x − 3) + x = 15
x – 6 + x – 3 + x = 15
3x – 9 = 15
3x = 15 + 9
3x = 24
x = $\frac{24}{3}$ = 8.
Unit’s digit = x = 8.
Ten’s digit = x – 3 = 8 – 3 = 5.
Hundred’s digit = x – 6 = 8 – 6 = 2.
Therefore, the number is 258.

4. The weight of a brick is 1 kg more than half its weight. What is the weight of the brick?
Solution:
Let the weight of the brick be x kg.
According to the problem,
x = $\frac{x}{2}$ + 1
x – $\frac{x}{2}$ = 1
$\frac{2x-x}{2}$ = 1
$\frac{x}{2}$ = 1
x = 2.
Therefore, the weight of the brick is 2 kg.

5. One quarter of a number increased by 9 gives the same number. What is the number?
Solution:
Let the number be y.
According to the problem,
$\frac{y}{4}$ + 9 = y
$\frac{y}{4}$ – y = -9
$\frac{y-4y}{4}$ = -9
$\frac{-3y}{4}$ = -9
$\frac{3y}{4}$ = 9
y = 9 × $\frac{4}{3}$ = 12.
Therefore, the number is 12.

6. Given 4k + 1 = 13, find the values of:
(a) 8k + 2
(b) 4k
(c) k
(d) 4k – 1
(e) – k – 2
Solution:
4k + 1 = 13
4k = 13 – 1
4k = 12
k = $\frac{12}{4}$ = 3.

(a) 8k + 2 = 8(3) + 2 = 24 + 2 = 26.

(b) 4k = 4(3) = 12.

(c) k = 3

(d) 4k – 1 = 4(3) – 1 = 12 – 1 = 11.

(e) -k – 2 = -(3) – 2 = -3 – 2 = -5.

Page 181 – 182

Q. The following are some equations along with the steps used to solve them to find the value of the letter−number. Go through each solution and decide whether the steps are correct. If there is a mistake, describe the mistake, correct it and solve the equation.

Solution:
(1) Mistake: +6 was moved to the other side as +6 instead of −6.
Correct solution:
4x + 6 = 10
4x = 10 − 6
4x = 4
x = $\frac{4}{4}$ = 1.

(2) Mistake: Instead of dividing 8 by 2, 2 is to be divided by 8.
Correct solution:
7 – 8z = 5
8z = 7 – 5
8z = 2
z = $\frac{2}{8}$ = $\frac{1}{4}$.

(3) Mistake: Subtracting 2 from 6 is incorrect.
Correct solution:
2v − 4 = 6
2v = 6 + 4
2v = 10
v = $\frac{10}{2}$ = 5.

(4) Mistake: No change in the sign of 3z and 2, even after moving them to the other side.
Correct solution:
5z + 2 = 3z − 4
5z − 3z = −4 − 2
2z = −6
z = $\frac{-6}{2}$ = -3.

(5) Mistake: 26 + 4w cannot be added to get 30.
Correct solution:
15w – 4w = 26
11w = 26
w = $\frac{26}{11}$.

(6) Mistake: Moving 3 to the other side to divide -12 is incorrect.
Correct solution:
3x + 1 = – 12
3x = -12 – 1
3x = -13
x = $\frac{-13}{3}$.

(7) Mistake: Moving +2 to the other side is incorrect.
Correct solution:
4 (4q+ 2) = 50
16q + 8 = 50
16q = 50 – 8
16q = 42
q = $\frac{42}{16}$ = $\frac{21}{8}$.

(8) Mistake: Multiplying (-2) with (-4x) gives 8x not -8x.
Correct solution:
– 2(3 – 4x) = 14
-6 + 8x = 14
8x = 14 + 6
8x = 20
x = $\frac{20}{8}$ = $\frac{5}{2}$.

(9) Mistake: Not dividing 5y by 3 is incorrect.
Correct solution:
3 (7y + 4) = 9 + 5y
7y + 4 = $\frac{9}{3}$ + $\frac{5y}{3}$
7y – $\frac{5y}{3}$ = 3 – 4
$\frac{21y-5y}{3}$ = -1
$\frac{16y}{3}$ = – 1
y = -1 × $\frac{3}{16}$ = $\frac{-3}{16}$.

Page 181 – 182

1. Fill in the blanks with integers.
(a) 5 × ___ – 8 = 37
(b) 37 – (33 – ____ ) = 35
(c) – 3 × (– 11 + ____ ) = 45
Solution:
(a) 5 × ☐ – 8 = 37
5 × ☐ = 37 + 8
5 × ☐ = 45
☐ = $\frac{45}{5}$ = 9.

(b) 37 – (33 – ☐ ) = 35
37 – 33 + ☐ = 35
4 + ☐ = 35
☐ = 35 – 4
☐ = 31.

(c) – 3 × (– 11 + ☐ ) = 45
(– 11 + ☐ ) = $\frac{45}{-3}$
– 11 + ☐ = -15
☐ = -15 + 11
☐ = -4.

2. Ranju is a daily wage labourer. She earns ₹ 750 a day. Her employer pays her in 50 and 100 rupee notes. If Ranju gets an equal number of 50 and 100 rupee notes, how many notes of each does she have?
Solution:
Let the number of ₹50 and ₹100 notes be x.
Total money = ₹750
According to the question,
50x + 100x = 750
150x = 750
x = $\frac{750}{150}$ = 5.
So,
Number of ₹50 notes = 5
Number of ₹100 notes = 5

3. In the given picture, each black blob hides an equal number of blue dots. If there are 25 dots in total, how many dots are covered by one blob? Write an equation to describe this problem.

Solution:
Let the number of dots covered by each blob be x.
Total number of dots covered by 3 blobs = 3x
Total dots = 25
Dots left uncovered = 4
So, the equation is 3x + 4 = 25
Now,
3x + 4 = 25
3x = 25 – 4
3x = 21
x = $\frac{21}{3}$ = 7.
Therefore, each black blob covers 7 dots.

4. Here are machines that take an input, perform an operation on it, and send out the result as an output.

(a)

Find the inputs in the following cases:

(b)

Find the inputs in the following cases:

Solution:
(a)
(i) Let the input be x.
According to the question,
(x + 3) × 4 + (- 5) = 43
(x + 3) × 4 = 43 + 5
(x + 3) × 4 = 48
(x + 3) = $\frac{48}{4}$
x + 3 = 12
x = 12 – 3 = 9.

(ii) Let the input be y.
According to the question,
(y + 3) × 4 + (- 5) = 75
(y + 3) × 4 = 75 + 5
(y + 3) × 4 = 80
(y + 3) = $\frac{80}{4}$
y + 3 = 20
y = 20 – 3
y = 17.

(b)
(i) Let the input be a.
According to the question,
(a × 3) – (a + 3) = 63
3a – a – 3 = 63
2a = 63 + 3
2a = 66
a = $\frac{66}{2}$ = 33.

(ii) Let the input be b.
According to the question,
(b × 3) – (b + 3) = 227
3b – b – 3 = 227
2b – 3 = 227
2b = 227 + 3
2b = 230
b = $\frac{230}{2}$ = 115.

5. What are the inputs to these machines?

Solution:
(i) Let the input be x.
According to the question,
(x ÷ 3) ÷ 3 = 5
(x ÷ 3) = 5 × 3
x ÷ 3 = 15
x = 15 × 3
x = 45.

(ii) Let the input be y.
According to the question,
{y + (-4)} + (-4) = (-11)
{y + (-4)} = (-11) – (-4)
y – 4 = -11 + 4
y = -11 + 4 + 4
y = -3.

6. A taxi driver charges a fixed fee of ₹800 per day plus ₹20 for each kilometer traveled. If the total cost for a taxi ride is ₹2200, determine the number of kilometres traveled.
Solution:
Let the number of kilometres traveled be x.
Fixed fee = ₹800
Charge per kilometre = ₹20
Total cost = ₹2200
Fixed fee + charge per kilometre = Taxi charge
∴ 800 + 20x = 2200
20x = 2200 – 800
20x = 1400
x = $\frac{1400}{20}$ = 70.
Therefore, the taxi traveled 70 kilometres.

7. The sum of two numbers is 76. One number is three times the other number. What are the numbers?
Solution:
Let the smaller number be x, and the other number be 3x.
The sum of the two numbers = 76
According to the question,
x + 3x = 76
4x = 76
x = $\frac{76}{4}$ = 19.
Thus,
First number = x = 19
Second number = 3x = 3 × 19 = 57
Therefore, the two numbers are 19 and 57.

8. The figure shows the diagram for a window with a grill. What is the gap between two rods in the grill?

Solution:
Let the gap between the two rods be x cm.
Number of gaps in the window = 6
Total gaps’ thickness = 6x cm
Height of the window = 34 cm
Thickness of each rod = 2 cm
Total rods’ thickness = 5 × 2 = 10 cm
Thickness of top and bottom margins = 3 cm
Thus,
Top margin + Rods’ thickness + Gaps’ thickness + Bottom margin = Height of the window
3 cm + 10 cm + 6x cm + 3 cm = 34 cm
16 + 6x = 34
6x = 34 – 16
6x = 18
x = $\frac{18}{6}$ = 3.
Therefore, the gap between the two rods is 3 cm.

9. In a restaurant, a fruit juice costs ₹15 less than a chocolate milkshake. If 4 fruit juices and 7 chocolate milkshakes cost ₹600, find the cost of the fruit juice and milkshake.
Solution:
Let the cost of a chocolate milkshake be ₹x.
Cost of a fruit juice = ₹(x − 15).
Total cost of 4 fruit juices and 7 chocolate milkshakes = ₹600.
∴ 4(x − 15) + 7x = 600
4x – 60 + 7x = 600
11x – 60 = 600
11x = 600 + 60
11x = 660
x = $\frac{660}{11}$ = 60.
So,
Cost of chocolate milkshake = ₹x = ₹60
Cost of fruit juice = ₹(x − 15) = 60 − 15 = ₹45

10. Given 28p – 36 = 98, find the value of 14p – 19 and 28p – 38.
Solution:
28p – 36 = 98
28p = 98 + 36
28p = 134
p = $\frac{134}{28}$ = $\frac{67}{14}$.

(i) 14p – 19 = 14 × $\frac{67}{14}$ – 19 = 67 – 19 = 48.

(ii) 28p – 38 = 28 × $\frac{67}{14}$ – 38 = 2 × 67 – 38 = 134 – 38 = 96.

11. The steps to solve three equations are shown below. Identify and correct any mistakes.

Solution:
(a) Mistake: 66 can’t be divided directly by 6.
Correct solution:
6x + 9 = 66
6x = 66 – 9
6x = 57
x = $\frac{57}{6}$.

(b) No correction needed.

(c) Mistake: No change in the sign of (-5) even after moving to the other side.
Correct solution:
4x – 5 = 9x + 8
4x – 9x = 8 + 5
-5x =13
x = $\frac{-13}{5}$.

12. Find the measures of the angles of these triangles.

Solution:
(i) Top angle = y°
Base angles = (y + 15)°
y° + (y + 15)° + (y + 15)° = 180°………….(Sum of angles of triangle)
y + y + 15 + y + 15 = 180
3y + 30 = 180
3y = 180 – 30
3y = 150
y = $\frac{150}{3}$ = 50°
Therefore,
Top angle = y = 50°
Base angles = (y + 15)° = (50 + 15)° = 65°

(ii) Top angle = x°
First base angle = (x – 10)°
Second base angle = (x + 10)°
x° + (x – 10)° + (x + 10)° = 180° ……………… (Sum of angles of triangle)
x + x – 10 + x + 10 = 180
3x = 180
x = $\frac{180}{3}$ = 60°
Therefore,
Top angle = x° = 60°
First base angle = (x – 10)° = (60 – 10)° = 50°
Second base angle = (x + 10)° = (60 + 10)° = 70°

13. Write 4 equations whose solution is u = 6.
Solution:
(i) u + 4 = 10
(ii) 2u = 12
(iii) u − 3 = 3
(iv) 3u = 18

14. The Bakhśhāli Manuscript (300 CE) mentions the following problem. The amount given to the first person is not known. The second person is given twice as much as the first. The third person is given thrice as much as the second; and the fourth person four times as much as the third. The total amount distributed is 132. What is the amount given to the first person?
Solution:
Let the amount given to the first person be x.
Second person = 2x
Third person = 3 × 2x = 6x
Fourth person = 4 × 6x = 24x
Total amount distributed = 132
∴ x + 2x + 6x + 24x = 132
33x = 132
x = $\frac{132}{33}$ = 4
So, the amount given to the first person is 4.

15. The height of a giraffe is two and a half metres more than half its height. How tall is the giraffe?
Solution:
Let the height of the giraffe be x metres.
According to the question,
Height = half its height + 2.5 metres
∴ x = $\frac{x}{2}$ + 2.5
x – $\frac{x}{2}$ = 2.5
$\frac{2x-x}{2}$ = 2.5
$\frac{x}{2}$ = 2.5
x = 2 × 2.5 = 5.
Thus, the giraffe is 5 metres tall.

16. Two separate figures are given below. Each figure shows the first few positions in a sequence of arrangements made with sticks. Identify the pattern and answer the following questions for each figure:
(a)  How many squares are in position number 11 of the sequence?
(b)  How many sticks are needed to make the arrangement in position number 11 of the sequence?
(c)  Can an arrangement in this sequence be made using exactly 85 sticks? If yes, which position number will it correspond to?
(d)  Can an arrangement in this sequence be made using exactly 150 sticks? If yes, which position number will it correspond to?

Solution:
(A) First pattern:
(a) The sequence of squares in the pattern: 1, 2, 3….
So, the number of squares in position n = n.
Squares in position number 11 = 11 squares.

(b) The sequence of sticks in the pattern: 6, 9, 12…..
Number of sticks in the nth position = 3(n + 1)
So, the number of sticks in the 11th position = 3(11 + 1) = 3(12) = 36 sticks

(c) Given number of sticks = 85
Number of sticks in the nth position = 3(n + 1)
So, 3(n + 1) = 85
(n + 1) = $\frac{85}{3}$
n = $\frac{85}{3}$ – 1
n = $\frac{85-3}{3}$ = $\frac{82}{3}$ = 27.3
Thus, no arrangement is possible with 150 sticks.

(d) Given number of sticks = 150
Number of sticks in the nth position = 3(n + 1)
So, 3(n + 1) = 150
(n + 1) = $\frac{150}{3}$
(n + 1) = 50
n = 50 – 1 = 49.
Thus, an arrangement at position 49 can be made using exactly 150 sticks.

(B) Second pattern:
(a) The sequence of squares in the pattern: 4, 7, 10, 13….
So, the number of squares in position n = 3n + 1.
Squares in position number 11 = 3 × 11 + 1 = 33 + 1 = 34 squares.

(b) The sequence of sticks in the pattern: 13, 22, 31, 40…..
Number of sticks in the nth position = 9n + 4
So, the number of sticks in the 11th position = 9 × 11 + 4 = 99 + 4 = 103 sticks

(c) Given number of sticks = 85
Number of sticks in the nth position = 9n + 4
So, 9n + 4 = 85
9n = 85 – 4
9n = 81
n = $\frac{81}{9}$ = 9.
Thus, an arrangement is possible with 85 sticks.

(d) Given number of sticks = 150
Number of sticks in the nth position = 9n + 4
So, 9n + 4 = 150
9n = 150 – 4
9n = 146
n = $\frac{146}{9}$ = 16.22
Thus, no arrangement is possible with 150 sticks.

17. A number increased by 36 is equal to ten times itself. What is the number?
Solution:
Let the number be x.
According to the question:
x + 36 = 10x
36 = 10x − x
36 = 9x
x = $\frac{36}{9}$ = 4.
Therefore, the number is 4.

18. Solve these equations:

Solution:
(a) 5(r + 2) = 10
5r + 10 = 10
5r = 10 – 10
5r = 0
r = $\frac{0}{5}$ = 0.

(b) – 3(u + 2) = 2(u – 1)
-3u – 6 = 2u – 2
-3u – 2u = -2 + 6
-5u = 4
u = $\frac{4}{-5}$ = $\frac{-4}{5}$.

(c) 2(7 – 2n) = – 6
14 – 4n = -6
-4n = -6 – 14
-4n = -20
4n = 20
n = $\frac{20}{4}$ = 5.

(d) 2(x – 4) = – 16
2x – 8 = -16
2x = -16 + 8
2x = -8
x = $\frac{-8}{2}$ = 4.

(e) 6(x – 1) = 2(x – 1) – 4
6x – 6 = 2x – 2 – 4
6x – 6 = 2x – 6
6x – 2x = -6 + 6
4x = 0
x = $\frac{0}{4}$ = 0.

(f) 3 – 7s = 7 – 3s
-7s + 3s = 7 – 3
-4s = 4
s = $\frac{4}{-4}$ = -1.

(g) 2x + 1 = 6 – (2x – 3)
2x + 1 = 6 – 2x + 3
2x + 1 = 9 – 2x
2x + 2x = 9 – 1
4x = 8
x = $\frac{8}{4}$ = 2.

(h) 10 – 5x = 3(x – 4) – 2(x – 7)
10 – 5x = 3x – 12 – 2x + 14
10 – 5x = x + 2
10 – 2 = x + 5x
8 = 6x
x = $\frac{8}{6}$ = $\frac{4}{3}$.

19. Solve the equations to find a path from Start to the End. Show your work in the given boxes provided and colour your path as you proceed.

Solution:

20. There are some children and donkeys on a beach. Together they have 28 heads and 80 feet. How many donkeys are there? How many children are there?
Solution:
Let the number of children = x
Let the number of donkeys = y
x + y = 28
y = 28 – x ………(1)
Since each child has 2 feet, and each Donkey has 4 feet, so
2x + 4y = 80
2x + 4(28 – x) = 80
2x + 112 – 4x = 80
-2x = 80 – 112
-2x = -32
2x = 32
x = $\frac{32}{2}$ = 16.
Therefore,
Children = x = 16.
Donkeys = y = 28 – 16 = 12.