Given statement
The numbers divisible by 4 and 6 are divisible. by 24.
Example:
Let the number be 12.
12 is divisible by both 4 and 6 but not divisible by 24.
Therefore statement is False.

Tree diagram

Algebraic expression
8b \(\div \) 6

 
Numerical Expression [7 \(\times\) (9+2)] + [(6-3) \(\div \)3].

Quantity of rice to be distributed to 5000 families = 1,00,000 kg
Quantity of rice distributed to each family = 1,00,000 \(\div \) 5,000 = 20 kg
Each family was given 20 kg of rice.

Let

l || m and P || Q
Distance between the lines are 1cm. 
Each of the pairs of lines are same distance.

Given AB = 7 cm

Construction:
1. Draw a line segment AB =7 cm and take a point P anywhere on the line.
2. Place the set square on the line AB.
3. Draw a line PQ through P.
4. The line PQ is perpendicular to the line AB at P. That is PQ f \(\perp\) AB

The sum of two smaller sides = 3 + 4 = 7 cm <8 cm, the third side.
It is less than the third side.
So, a triangle cannot be formed with the given sides.

Instead of checking triangle inequality by all the sides in the triangle, check only with two smaller sides.
Sum of two smaller sides of the triangle = 5+7=12 cm > 10 cm, the third side. It is greater than the third side. So, a triangle can be formed with the given sides.

Given
Valarmathi sold her pearl set
s.p = Rs. 30,000
profit = Rs. 5,000
c.p = s.p - profit
= 30,060 - 5,000
c.p = Rs. 25,000

Somu bought a bike =Rs. 28,000
He spent = 2,000
c.p = Rs. 30,000
s.p = Rs. 30,000
Therefore s.p = c.p
Therefore No profit / No loss

(i) Given
c.p = Rs. 110
m.p = Rs. 130
Discount = NIL
Therefore m.p = s.p
Therefore s.p = 130
Here, s.p > c.p 
Therefore Profit = s.p - c.p 
= 130 - 110
Profit = Rs. 20
(ii) Given
c.p = Rs. 110
m.p = Rs. 130
Discount = Rs. 10
Here, s.p = m.p - Discount
= 130 - 10
s.p = Rs. 120
Here, s.p > c.p
Profit = s.p - c.p
= 120 - 110
Profit = Rs. 10
(iii) Given
c.p = Rs. 110
m.p = Rs. 130
Discount = 30
s.p = m.p - Discount
= 130 - 30
s.p = 100
Here, s.p < c.p
Therefore Loss = c.p - s.p
= 110 - 100
Loss = 10
(iv) Given
c.p = Rs. 110
m.p = Rs. 120
Profit = NIL
Loss = Rs. 10
s.p = c.p - loss
= Rs. 110 - Rs. 10
= Rs. 100
Discount = m.p - s.p
= 120 - 100
=20
(v) Given
m.p = Rs. 120
Discount = Rs. 10
Profit = Rs. 20
Loss = NIL
s.p = m.p - Discount
= 120 - 10
= Rs. 110
Here, s.p = Rs. 110
c.p = s.p - Profit
= Rs. 110 - Rs. 20
c.p = Rs. 90

C.P. = Rs. 1500; Profi t = Rs. 150; M.P. = Rs. 1800
S.P. = M.P. - Discount = Rs. 1800 - Discount
Profit = S.P. - C.P. \(\Rightarrow\) Rs. 150 = Rs. 1800 - Discount - Rs. 1500
Discount = Rs. 1800 - Rs. 1500 -  Rs. 150 = Rs. 150

Discount = Rs. 50
Profit = Rs. 100
M.P. = Rs. 800
S.P. = M.P. - Discount
= Rs. 800 - Rs. 50
= Rs. 750
Profit = S.P. - C.P.
Rs. 100 = Rs. 750 - C.P.
C.P. = Rs. 750 - Rs. 100
= Rs. 650

i) 10:15 hours; quarter past 10
ii) 6:45 hours; quarter to 7
iii) 4:10 hours; 10 minutes past 4
iv) 3:30 hours; half past 3
v) 9:40 hours; 20 minutes to 10;

170 day \(\div \) 7 (why?)
170 days = 24 weeks + 2 days
Required day is the second day after Wednesday.
Therefore 8^th of june is Friday.

1 Student = 150gm of rice
Therefore 40 students = 150gm \(\times\) 40
= 6000 gm
= 6 kg
Therefore 1 student = 15ml of oil
40 students = 40 \(\times\) 15
= 600 ml
= 0.6 ml
Therefore 6 kg of rice and 0:6 I of oil needed.

Geetha brought = 2 l 250 ml
Her friend drank = 0 300 ml
Remaining water in the bottle = 1 950  = 1 l 950 ml

15 km (m, cm, mm)
1 km = 1000 m
1 m = 100 cm
1 cm= 10mm
* 1 km = 1000 m
15 km = 15 \(\times\) 1000 m
= 15000 m
* 1 m = 100 m
15000 m = 15000 \(\times\) 100
= 150000 cm
* 1 cm = 10 mm
Therefore 1500000 em = 10 \(\times\) 1500000
= 15000000 mm
Ans : 15 km = 15000 m, 1500000 cm, 15000000mm
(ii) 12kg (g, mg)
1000 g = 1 kg
1000mg = 1 g
* 1000g = 1 kg
12kg = 12 \(\times\) (1000) g
= 12000 g
* 1g = 1000mg
12000 g = 12000 \(\times\) 1000 mg
= 12000000 mg

We know that, the product of the two numbers = LCM \(\times\) HCF
As the HCF of co-primes is 1,
65 \(\times\) (the other number) = 5005 \(\times\) 1
The other number = 5005 \(\div \) 65 = 77

Now, 18 = 2 \(\times\) 3 \(\times\) 3 and 30 = 2 \(\times\) 3 \(\times\) 5
and their HCF is 2 \(\times\) 3 = 6 and LCM is 2 \(\times\) 3 \(\times\) 3 \(\times\) 5 = 90
Hence, HCF : LCM = 6 : 90 = 1 : 15

Get all the common factors of 62 − 2, 78 − 3 and 109 − 4, i.e., 60, 75 and 105 and see that the common factors will divide them all. The greatest number is the H.C.F of 60, 75 and 105.
60 = 2 \(\times\) 2 \(\times\) 3 \(\times\) 5         75 = 3 \(\times\) 5 \(\times\) 5        105 = 3 \(\times\) 5 \(\times\) 7
Hence, the HCF is 3 \(\times\) 5=15, which is the greatest number that will divide 62, 78, 109 leaving remainders 2, 3 and 4 respectively.