SSC CGL Tier 1 2021 Quantitative Aptitude Questions with Solutions

Explanation

To solve the expression sin⁴30° + cos⁴30° - sin 25°cos 65° - sin 65°cos 25°, we evaluate the components using standard trigonometric values and identities.

Substituting the standard values sin 30° = ½ and cos 30° = √3{2} into the first part of the expression gives: sin⁴30° + cos⁴30° = (½)⁴ + (√3{2})⁴ = ¹⁄₁₆ + ⁹⁄₁₆ = ¹⁰⁄₁₆ = ⅝

The second part of the expression can be simplified by factoring out the negative sign: -(sin 25°cos 65° + cos 25°sin 65°) Applying the compound angle identity sin(A + B) = sin A cos B + cos A sin B where A = 25° and B = 65°: -(sin(25° + 65°)) = -sin 90° = -1

Combining the two parts results in the final calculation: ⅝ - 1 = (5 - 8)/(8) = -⅜

Explanation

The theorem for similar triangles states that the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding altitudes. Let h₁ be the altitude of △ ABC and h₂ be the altitude of △ DEF.

The relationship is expressed as: (Area(△ ABC))/(Area(△ DEF)) = ( (h₁)/(h₂) )²

Substituting the given values into the formula: ¹⁰⁰⁄₈₁ = ( (h₁)/(6.3) )²

Taking the square root of both sides to solve for the linear ratio: √(¹⁰⁰⁄₈₁) = (h₁)/(6.3) ¹⁰⁄₉ = (h₁)/(6.3)

Multiplying both sides by 6.3 to isolate h₁: h₁ = (10 × 6.3)/(9) h₁ = ⁶³⁄₉ h₁ = 7 cm

Explanation

Let the usual speed of Akhil be v km/h. According to the problem, the difference in time for a 150 km journey when speed is reduced by 10 km/h is 30 minutes, or 0.5 hours. Using the relation Time = Distance / Speed, we have:

(150)/(v - 10) - ¹⁵⁰⁄ᵥ = 0.5

Multiplying by 2v(v - 10) to clear the fractions:

300v - 300(v - 10) = v(v - 10) 3000 = v² - 10v v² - 10v - 3000 = 0

Solving this quadratic equation by splitting the middle term:

v² - 60v + 50v - 3000 = 0 (v - 60)(v + 50) = 0

Since speed cannot be negative, v = 60 km/h.

The question asks for the time taken to cover 90 km if he drives 15 km/h slower than his usual speed. The new speed is 60 - 15 = 45 km/h.

Time taken = (Distance)/(New Speed) = ⁹⁰⁄₄₅ = 2 hours