A football team gains
1^st day = 3 points
2^nd day = 4 points = 3 + 4
A football team loses 5 points =  (-5)
Therefore 3rd day = (-5) points (loses)
Total Points scored
((3+4) - 5) = 2
Tree Diagram

Numerical expression for the answer l0
= 8 + 2 or 20 - 10 or 5 \(\times\)2 or 40/4 etc.

i) Algebraic Expression = p + q
ii) Algebraic Expression = I - m
iii) Algebraic Expression
= (a \(\times\) b) - c (or) (ab) - c
iv) Algebraic Expression
= (a + b) - (c + d)
v) Algebraic Expression
= 8 / a + (b / 4 + 3) .
(or) ((8 / a) + (b / 4) + 3))

(a) Equilateral Triangle :
ABC (AB = BC = CA = 4 cm)
(b) Isoceles Triangles :
ABC,AEF
(c) Scalene Triangles :
AEB, AED, ADF, AFC, ABD, ADC, ABF, AEC
(d) Acute Triangles :
ABC, AEF, ABF, AEC
(e) Obtuse Triangles :
AEB, AFC
(f) Right Triangles :
ADB, ADC, ADE, ADF

(i) 60°, 60°, 60° 
The sum of the three angles of a triangle = 180° 
Sum of three angles = 60°  + 60° + 60° = 180°  degrees
Yes, a triangle can be formed.
Type: Acute angled triangle.
(ii) 90°, 55°, 35° 
The sum of three angles of a triangle = 90° + 55° + 35° = 180° 
Yes, a triangle can be formed.
Type: Right angled triangle.
(iii) 60°, 40°, 42°
The sum of three angles of a triangle = 180° 
Sum of the angles = 60°  + 40° + 42°  = 142° 
No a triangle cannot be formed.
(iv) 60°, 90°, 90°
The sum of three angles of a triangle = 180° 
Sum of the angles = 60°  + 90° + 90° 
= 240° \(\neq\) 180° 
No a triangle cannot be formed.
(v) 70°, 60°, 50°
The sum of three angles of a triangle = 180° 
Sum of the angles = 70° + 60° + 50°  = 180°  = 180° .
Yes, a triangle can be formed.
Type: Acute angled triangle.
(vi) 100°, 50°, 30°
The sum of three angles of a triangle = 180° 
Sum of the angles = 100° + 50° + 30° 
= 180° = 180° 
Yes, a triangle cannot be formed.
Type: Obtuse angled triangle

(i) C.P. = Rs. 110
M.P. = Rs. 130
Discount = Rs. 5
S.P. = M.P. - Discount
= Rs. 130 - Rs. 5
=Rs. 125
Profit = S.P. - C.P.
=Rs. 125 - Rs. 110
=Rs. 15
(ii) C.P. = Rs. 110
M.P. = Rs. 130
Discount = Rs. 20
S.P. = M.P. - Discount
= Rs. 130 - Rs. 20
= Rs.110
C.P. = S.P. \(\Rightarrow\) No profit or No loss.
(iii) M.P. = Rs. 130
Discount = Rs. 15
S.P. = M.P. - Discount
= Rs. 130 - Rs15
=Rs. 115
Profit = Rs. 30
Profit = S.P. - C.P.
Rs. 30 = Rs. 115 - C.P.
C.P. = Rs. 115 - Rs. 30
= Rs. 85
(iv) M.P. = Rs.130
Loss = Rs. 25
S.P. = M.P. - Discount
= Rs. 130 - Rs. 0
= Rs. 130
Loss = C.P. - S.P.
Rs. 25 = C.P. - Rs. 130
C.P. = Rs. 25 + Rs. 130
= Rs. 155
(v) M.P. = Rs. 125
Discount = Rs. 0
S.P. = M.P. - Discount
= Rs.125 - Rs. 0
= Rs. 125
No profit / No loss
C.P. = S.P.
C.P. = Rs. 125
(vi) S.P. = Rs. 350
Discount = Rs. 50
Profit = Rs. 100
M.P. = S.P.+ Discount
= Rs. 350+ Rs. 50 = Rs. 400
Profit = S.P. - C.P.
Rs 100 = Rs. 350 - C.P.
C.P. = Rs. 350 - Rs. 100
= Rs. 250

(i) C.P. < S.P. \(\Rightarrow\) Profit = S.P. - C.P. = Rs. 60 - Rs. 50 = Rs. 10
(ii) C.P. > S.P. \(\Rightarrow\) Loss = C.P. - S.P. = Rs. 70 - Rs. 60 = Rs. 10
(iii) Profit = S.P. - C.P.
\(\Rightarrow\) Rs. 20 = S.P. - Rs. 100
\(\Rightarrow\) S.P. = Rs 20 + Rs. 100 = Rs. 120
(iv) Loss = C.P. - S.P.
\(\Rightarrow\) Rs. 15 = Rs. 80 - S.P.
\(\Rightarrow\) S.P. = Rs. 80 - Rs. 15 = Rs. 65
(v) Profit = S.P. - C.P.
\(\Rightarrow\) Rs. 25 = Rs. 70 - C.P.
\(\Rightarrow\) C.P. = Rs. 70 - Rs. 25 = Rs. 45
(vi) Loss = C.P. - S.P.
\(\Rightarrow\) Rs. 30 = C.P. - Rs. 100
\(\Rightarrow\) C.P. = Rs. 30 + Rs. 100 = Rs. 130

(i) The bill number is 100.
(ii) The date of the bill is 15.05.2018.
(iii) There were four different items purchased.
(iv) The cost of 1 hair clip is Rs. 15.
(v) The total cost for the ribbon is Rs. 36.

If the journey crosses the midnight, calculate the time duration from starting hours to midnight, then from midnight to arrival time.

Method-1	Method-2
Duration up to midnight	Total journey time = 3 hours 30 minutes+4 hours 30 minutes = 7 hours 60 minutes = 7 hours +1 hour = 8 hours
Duration from midnight to arrival = 4 hours 30 minutes Total journey time = Duration up to midnight + Duration from midnight to arrival = 3 hours 30 minutes + 4 hours 30 minutes = 7 hours 60 minutes = 7 hours + 1 hour = 8 hours

Solution: By Division method
Step 1: Start with the smallest prime factor and go on dividing till all the numbers are divided as given below.
Step 2: LCM = product of all prime factors
= 2 \(\times\) 2 \(\times\) 3 \(\times\) 13 \(\times\) 31 = 4836
Thus, the LCM of 156 and 124 is 4836.
By Prime Factorisation method
Step 1: We write the prime factors of 156 and 124 as given below (use of divisibility test rules will also help).
156 = 2 \(\times\) 78 = 2 \(\times\) 2 \(\times\) 39 = 2 \(\times\) 2 \(\times\) 3 \(\times\) 13
124 = 2 \(\times\) 62 = 2 \(\times\) 2 \(\times\) 31
Step 2: The product of common factors is 2 \(\times\)2 and also the product of the factors that are not common is 3 \(\times\) 13 \(\times\) 31.
Step 3: Now, LCM = product of common factors x product of factors that are not common
= (2 \(\times\) 2) \(\times\) (3 \(\times\) 13 \(\times\) 31) = 4 \(\times\) 1209 = 4836
Thus, LCM of 156 and 124 is 4836.
                          (or)
156 = 2 \(\times\) 78 = 2 \(\times\) 2 \(\times\) 39 = 2 \(\times\) 2 \(\times\) 3 \(\times\) 13;
124 = 2 \(\times\) 62 = 2 \(\times\) 2 \(\times\) 31
The prime factor 2 appears a maximum of 2 times in the prime factorization of 156 and 124, the prime factor 3 appears only 1 time in the prime factorization of 156, the prime factor 13 appears only 1 time in the prime factorization of 156 and the prime factor 31 appears only 1 time in the prime factorization of 124.
Hence, the required LCM = (2 \(\times\) 2) \(\times\) 3 \(\times\) 13 \(\times\) 31 = 4836.