Maths Mock Test Quiz Tricks PDF

Q - A started a business with an initial
investment of Rs.2400. After 4 months, B joined the business with an initial investment of Rs.1500. After 4 more months, A withdrew Rs.1200 and C joined them with the initial investment of Rs.4500. At the end of the year,
the total profit of the business is Rs.9000, then
find the profit share of A?

Solution:

Q1.
The efficiency of (A + B) = 100/24 = (25/6)%;
The efficiency of (A + B + C) = 100/8 = (25/2)%;
The efficiency of C = 25/2 – 25/6 = (50/6)%;
Hence, C can alone finish this job in = 100/(50/6) = 12 days;

Q2.
Diagonal of square = side* √2; => Side = 6 cms;
Side = Diameter = 6 cms; => radius = 3 cms;
Hence, the area of circle = 9π sq. cms

Q3.
Original price = Rs. 6000;
Price after discount = 6000 – 1200 = Rs. 4800;
Price after raising service contract = 4800 + 480 = Rs. 5280

Q4.
Suppose the received money by A, B, and C is respectively 5x, 6x, and 9x.
5x = 450; => x = 90;
Hence, the total money = 20x = 20*90 = Rs. 1800;

Q5.
Suppose the Cost price of the bag= Rs. x;
Hence, x + 0.15x = 230; => x = Rs. 200;
Selling price after selling on 20% = 200 + 20% of 200 = Rs. 240

📊 त्रिकोणमिति : महत्वपूर्ण सूत्र 📊

🍎 योग सूत्र
➭ Sin(A+B) = SinACosB+CosASinB
➭ Sin(A-B) = SinACosB-CosASinB
➭ Cos(A+B) = CosACosB-SinASinB
➭ Cos(A-B) = CosACosB+SinASinB

🍏 अन्तर सूत्र
➭ tan(A+B) = tanA+tanB/1-tanAtanB
➭ tan(A-B) = tanA-tanB/1+tanAtanB

🍎 C-D सूत्र
➭ SinC+SinD = 2Sin(C+D/2) Cos(C-D/2)
➭ SinC-SinD = 2Cos(C+D/2) Sin(C-D/2)
➭ CosC+CosD = 2Cos(C+D/2) Cos(C-D/2)
➭ CosC-CosD = 2Sin(C+D/2) Sin(D-C/2)
➭ CosC-CosD = -2Sin(C+D/2) Sin(C-D/2)

🍏 रूपांतरण सूत्र
➛ 2SinACosB = Sin(A+B)+Sin(A-B)
➛ 2CosASinB = Sin(A+B)-Sin(A-B)
➛ 2CosACosB = Cos(A+B)+Cos(A-B)
➛ 2SinASinB = Cos(A-B)-Cos(A+B)

🍎 द्विक कोण सूत्र
➛ Sin2A = 2SinACosA
➛ Cos2A = Cos²A-Sin²A = 2Cos²-1 = 1-2Sin²A
➛ tan2A = 2tanA/1-tan²A
➛ Sin2A = 2tanA/1+tan²A
➛ Cos2A = 1-tan²A/1+tan²A

🍏 विशिष्ट सूत्र
➛ Sin(A+B)Sin(A-B) = Sin²A-Sin²B
= Cos²B-Cos²A
➛ Cos(A+B)Cos(A-B) = Cos²A-Sin²B = Cos²B-Sin²A

🍎 त्रिक कोण सूत्र
➛ Sin3A = 3SinA-4Sin³A
➛ Cos3A = 4Cos³A-3CosA
➛ tan3A = 3tanA-tan³A/1-3tan²A

🍏 महत्वपूर्ण सर्वसमिकाएं
➛ Sin²θ+Cos²θ = 1
➭ Sin²θ = 1-Cos²θ
➭ Cos²θ = 1-Sin²θ
➛ 1+tan²θ = Sec²θ
➭ Sec²θ-tan²θ = 1
➭ tan²θ = Sec²θ-1
➛ 1+Cot²θ = Cosec²θ
➭ Cosec²θ-Cot²θ = 1
➭ Cot²θ = Cosec²θ-1

Pythagoras theorem Handwritten Notes.pdf

बीजगणित के सूत्र:

(a+b)² = a²+2ab+b²

(a-b)² = a²-2ab+b²

(a-b)² = (a+b)²-4ab

(a+b)² + (a-b)² = 2(a²+b²)

(a+b)² – (a-b)² = 4ab(a+b)³ = a³+3a²b+3ab²+b³

(a+b)² – (a-b)² = a³+b³+3ab(a+b)

(a-b)³ = a³-3a²b+3ab²-b³

(a-b)³ = a³+b³+3ab(a+b)

(a+b)³ + (a-b)³ = 2(a³+3ab²)

(a+b)³ + (a-b)³ = 2a(a²+3b²)

(a+b)³ – (a-b)³ = 3a²b+2b³

(a+b)³ – (a-b)³ = 2b(3a²+b²)

a²-b² = (a-b)(a+b)

a³+b³ = (a+b)(a²-ab+b²)

a³-b³ = (a-b)(a²+ab+b²)

a³-b³ = (a-b)³ + 3ab(a-b)

(a+b+c)² = a²+b²+c²+2(ab+bc+ca)

(a+b+c)³ = a³+b³+c³+3(a+b)(b+c)(c+a)

a³+b³+c³ = (a+b+c)³ – 3(a+b)(b+c)(c+a)

(a+b+c+d)² = a²+b²+c²+d²+2(ab+ac+ad+bc+bd+cd)

a³+b³+c³-3abc = (a+b+c)(a²+b²+c²-ab-bc-ca)