SSC CGL Tier 1 2017 Quantitative Aptitude Questions with Solutions

Explanation

The difference between the amounts for two different time periods in simple interest represents the interest earned during that interval. The time difference is 3.5 - 2 = 1.5 years. The interest for this 1.5 year period is 969 - 918 = 51.

To find the annual interest, divide the interest by the number of years: Interest per year = (51)/(1.5) = 34.

The principal P is calculated by subtracting the interest of 2 years from the amount after 2 years: P = 918 - (2 × 34) = 918 - 68 = 850.

The rate of interest R is determined using the formula R = (SI × 100)/(P × T). Using the interest for one year (T = 1): R = (34 × 100)/(850 × 1) = ³⁴⁰⁰⁄₈₅₀ = 4%.

Explanation

The given equations can be simplified by expressing all terms as powers of 4. Since 256 = 4⁴, the first equation 4⁽ˣ⁺ʸ⁾ = 256 becomes 4⁽ˣ⁺ʸ⁾ = 4⁴. Comparing the exponents, we get x + y = 4.

The second equation (256)⁽ˣ⁻ʸ⁾ = 4 can be rewritten as (4⁴)⁽ˣ⁻ʸ⁾ = 4¹. Using the power of a power rule (aᵐ)ⁿ = aᵐⁿ , this becomes 4⁴⁽ˣ⁻ʸ⁾ = 4¹. Comparing the exponents gives 4(x - y) = 1, which simplifies to x - y = 1/4.

To find x, add the two linear equations: (x + y) + (x - y) = 4 + 1/4. This yields 2x = 17/4, so x = 17/8. To find y, subtract the second equation from the first: (x + y) - (x - y) = 4 - 1/4. This yields 2y = 15/4, so y = 15/8.

Explanation

If the expression px³ - qx² - 7x - 6 is completely divisible by x² - x - 6, the quotient must be a linear expression of the form ax + b. By comparing the constant terms of the dividend -6 and the divisor -6, the constant term of the quotient b must be 1. By comparing the coefficients of the x³ terms, the leading coefficient of the quotient a must be p. Thus, the relationship can be expressed as:

px³ - qx² - 7x - 6 = (x² - x - 6)(px + 1)

Expanding the right-hand side, we get: px³ + x² - px² - x - 6px - 6 px³ + (1 - p)x² - (1 + 6p)x - 6

By equating the coefficients of x on both sides: -(1 + 6p) = -7 1 + 6p = 7 6p = 6 p = 1

By equating the coefficients of x² on both sides: -q = 1 - p Substituting p = 1 into the equation: -q = 1 - 1 q = 0

Explanation

If the polynomial f(x) = px³ - 2x² - qx + 18 is completely divisible by x² - 9, it must be divisible by the individual factors of x² - 9, which are (x - 3) and (x + 3). According to the Factor Theorem, if (x - a) is a factor of f(x), then f(a) = 0.

Substituting x = 3 into the expression: p(3)³ - 2(3)² - q(3) + 18 = 0 27p - 18 - 3q + 18 = 0 27p - 3q = 0 27p = 3q 9p = q

By rearranging the equation 9p = q, we find the ratio of p to q as: p/q = 1/9

Substituting x = -3 yields the same result: p(-3)³ - 2(-3)² - q(-3) + 18 = 0 -27p - 18 + 3q + 18 = 0 -27p + 3q = 0 3q = 27p q = 9p

Both substitutions confirm the relationship q = 9p, leading to the ratio p:q = 1:9.

Explanation

To calculate the value of x⁵ + (1)/(x⁵), we use the algebraic relation x⁵ + (1)/(x⁵) = (x² + (1)/(x²))(x³ + (1)/(x³)) - (x + ¹⁄ₓ).

Given the base value x + ¹⁄ₓ = 5, we first find the intermediate square and cube values: x² + (1)/(x²) = (x + ¹⁄ₓ)² - 2 = 5² - 2 = 23 x³ + (1)/(x³) = (x + ¹⁄ₓ)³ - 3(x + ¹⁄ₓ) = 5³ - 3(5) = 125 - 15 = 110

Substituting these values into the fifth power formula: x⁵ + (1)/(x⁵) = (23 × 110) - 5 x⁵ + (1)/(x⁵) = 2530 - 5 = 2525

Explanation

In a right-angled triangle ABC with ∠ ABC = 90°, let BP be the altitude drawn to the hypotenuse AC. This altitude divides the triangle into two smaller right triangles, △ ABP and △ BCP, both of which are similar to the original △ ABC and to each other.

By the property of similarity between △ ABP and △ BCP, the corresponding angles are equal. Specifically, ∠ BAP = ∠ PBC and ∠ ABP = ∠ BCP. Since the problem states that ∠ BAP = 50°, it directly follows from this geometric property that ∠ PBC = 50°.

This can also be verified through angle summation. In △ ABP, since ∠ APB = 90° and ∠ BAP = 50°, the third angle is ∠ ABP = 180° - (90° + 50°) = 40°. Because ∠ ABC is a right angle, ∠ PBC = ∠ ABC - ∠ ABP = 90° - 40° = 50°.

Explanation

The angle formed by the intersection of the external bisectors of two angles of a triangle is given by the formula ∠ QOR = 90° - (∠ P)/(2), where ∠ P is the internal angle at the third vertex. In triangle PQR, the point O is the excenter formed by the intersection of external bisectors of ∠ Q and ∠ R.

Substituting the given value ∠ QOR = 50° into the formula:

50° = 90° - (∠ QPR)/(2)

Rearranging the equation to solve for ∠ QPR:

(∠ QPR)/(2) = 90° - 50°

(∠ QPR)/(2) = 40°

∠ QPR = 40° × 2 = 80°

Explanation

The expression (cosec⁴ A - cot² A) - (cot⁴ A + cosec² A) can be rearranged by grouping the fourth-power terms and the second-power terms: (cosec⁴ A - cot⁴ A) - (cosec² A + cot² A)

Using the algebraic identity x² - y² = (x - y)(x + y), we can factorize the first part of the expression: cosec⁴ A - cot⁴ A = (cosec² A - cot² A)(cosec² A + cot² A)

Applying the fundamental trigonometric identity cosec² A - cot² A = 1, the factored term becomes: 1 · (cosec² A + cot² A) = cosec² A + cot² A

Substituting this back into the rearranged expression: (cosec² A + cot² A) - (cosec² A + cot² A) = 0