Distance from A to D = 7 m
Distance from A to D (via B) = 4 + 5 = 9 m
Distance from A to D (via C) = 2 + 6 = 8 m
Therefore, the route directly from A to D requires minimum cable of length 7 m.

Since we do not know the number of students who get down at the first stop, let us take the number as x. The number of students get down at the second bus stop is 2x.
x + 2x + 6 + 3 = 60
(1 + 2) x + 9 = 60
3x + 9 = 60
3x + 9 – 9 = 60 − 9 [Subtract 9 on both sides]
3x = 51
\(\frac { 3x }{ 3 } =\frac { 51 }{ 3 } \) [Divide by 3 on both sides]
Therefore, x = 17.
Thus, the number of students got down in the first bus stop is 17.

Profit of 1 kg mangoes = Rs. 15
Profit of 5 kg mangoes = Rs. 15 x 5
= Rs. 75
Loss of 1 kg apples = Rs. 30
Loss of 3 kg apples = 30 x 3
= Rs. 90
Loss = Rs. 90 − Rs. 75
= Rs. 15

Lift in the ground floor denotes 0.
It goes 5 floors down = -5
It moves up to 10 floors = +10
Position of the lift = 10 - 5 = 5
Lift will be in the 5m floor

Let the height of the parallelogram = h cm
Then the base of the parallelogram = 3h cm
Area of the parallelogram = 192 sq. cm
b x h = 192
3h x h = 192
3h² = 192
h² = 64
h x h = 8 x 8
h = 8 cm
base = 3h = 3 x 8 = 24 cm
Therefore, base of the parallelogram is 24 cm and height is 8 cm.

The amount Sita has Rs. 225
The amount spent for stationery on credit = Rs. 400
The due amount to be paid = Rs. 225 – Rs. 400 = – Rs.175
Therefore, Sita has to pay Rs. 175

\(\frac{\mathrm{P}-\mathrm{Q}}{\mathrm{P}+\mathrm{Q}}=\frac{-15-5}{-15+5}=\frac{-20}{-10}=2\)

Let a, b be two integers.
The product of two integers = -135
a x b = -135
One number,a = -15
b = ?
Other number \(=\frac{135}{-15}=9\)
∴ The other number = 9

In diagram (i)
Consider DF as height (h) and AB as base (b)
Area of the parallelogram = 1470 sq.cm
b x h = 1470
49 x h = 1470
h = 30 
height, DF = 30 cm

In diagram (ii),
Consider BE as height (h) and AD as base (b) 
Area of the parallelogram = 1470 sq.cm
b x h = 1470
35 x h = 1470
height BE = 42 cm
Ans: DF = 30 cm and BE = 42 cm

To cover 100 km quantity of petrol required = 2 litres
\(\therefore\) To cover 1km petrol required = 2litres
To cover 1km petrol required = \(\cfrac { 2 }{ 100 } =\cfrac { 1 }{ 50 } litres\)
\(\therefore\) To cover 250km petrol required = \(\cfrac { 1 }{ 50 } \times 250=5litres\)
5 litres of petrol required to cover 250 km

Given : Area = 100 sq.cm, d₁ = 8 cm, d₂ = ?
Area of rhombus = 100 sq.cm
\(\frac{1}{2} \times\left(d_{1} \times d_{2}\right)=100 \)
\(\frac{1}{2} \times\left(8 \times d_{2}\right)=100 \)
8 x d₂ = 100 x 2
\(\mathrm{d}_{2}=\frac{100 \times 2}{8} \Rightarrow \mathrm{d}_{2}=25 \mathrm{~cm}\)

(i) 7x, -8x, -12x are like terms containing same algebraic variable 'x'.
(ii) 5y, 12 -9y are like terms containing same algebraic variable 'y'.
(ili) 6z, z and 11 z are like terms containing same algebraic variables 'z'.

h = 14 m,b = 14 + 8 = 22 m
Area of ground (parallelogram) = b x h sq.units
= 22 x 14 = 308 m²
Cost of levelling the ground = Rs. 15  per sq.m
Cost of levelling the ground = 308 x 15 = Rs. 4620

Marks given for one correct answer = 4
Marks given for 10 correct answers = 10 x 4
= 40
Kalaivizhi’s final score = 20
Marks reduced for incorrect answers = 40 − 20
= 20
Therefore, number of questions answered incorrectly = 20 ÷ 2 = 10

Let x be the required number of days. The decrease in number of workers lead to the increase in number of days. (Therefore, both are in inverse proportion)
For inverse proportion x₁y₁ = x₂ y₂

Hence 60 x 7 = 42 \(\times\) x
42 \(\times\) x = 60 x 7
x = \(\frac { 60\times 7 }{ 42 } \)
x = 10
In 10 days 42 workers can spin a bale of cotton.

In the given algebraic expression, x,y,z are the variables.
Let us group the like terms.
100x + 99y − 98z + 10x + 10y + 10z − x − y + z
= (100x + 10x − x) + (99y + 10y − y) + ( −98z + 10z + z)
= (100 + 10 − 1)x + (99 + 10 − 1)y + ( −98 + 10 + 1)z
= (110 − 1)x + (109 − 1)y + ( −98 + 11)z
= 109x + 108y + ( −87)z
= 109x + 108y − 87z.