SSC CHSL Tier 1 2019 Quantitative Aptitude Questions with Solutions

Explanation

Using the standard algebraic identity for the sum of cubes: a³ + b³ = (a + b)(a² - ab + b²)

The term a² - ab + b² can be rewritten by completing the square: a² - ab + b² = (a + b)² - 3ab

Substituting this expression back into the main identity gives: a³ + b³ = (a + b)[(a + b)² - 3ab]

Now, substitute the given values, a³ + b³ = 110 and a + b = 5, into the equation: 110 = 5 × [(a + b)² - 3ab]

Dividing both sides by 5 yields: (a + b)² - 3ab = 110 ÷ 5 = 22

Explanation

Let the number of students in class B be N. The number of students in class A is 40% more than B, so it is 1.4N. Total students = 1.4N + N = 2.4N = 96 N = 40 So, the number of students in class A is 56, and in class B is 40. The ratio of the number of students in class A to class B is 56 : 40 = 7 : 5.

Let the average weight of the students in class A be w kg. Since the average weight of class B is 50% more than class A, it is 1.5w kg.

Using the weighted average relation: Overall Average = (Ratio of students in A × Average weight of A + Ratio of students in B × Average weight of B) / Total ratio sum

Substituting the values: 58 = (7 × w + 5 × 1.5w) / (7 + 5) 58 = (7w + 7.5w) / 12 58 = 14.5w / 12 w = (58 × 12) / 14.5 w = 4 × 12 = 48 kg

The average weight of the students in class B is: 1.5w = 1.5 × 48 = 72 kg

Explanation

Correct Answer: D

The algebraic identity for the given expression is: a³ + b³ + c³ - 3abc = (a + b + c)(a² + b² + c² - (ab + bc + ca))

First, we find the value of (ab + bc + ca) using the square of a trinomial: (a + b + c)² = a² + b² + c² + 2(ab + bc + ca)

Substituting the given values into the equation: 5² = 33 + 2(ab + bc + ca) 25 = 33 + 2(ab + bc + ca) 2(ab + bc + ca) = -8 ab + bc + ca = -4

Now, substituting these values back into the algebraic identity: a³ + b³ + c³ - 3abc = 5 × (33 - (-4)) = 5 × (33 + 4) = 5 × 37 = 185

Explanation

Using the algebraic identity for the difference of two cubes: a³ - b³ = (a - b)(a² + ab + b²)

The given expression is: 40√5 x³ - 3√3 y³ = (2√5 x)³ - (√3 y)³

Let a = 2√5 x and b = √3 y. Applying the identity: (2√5 x)³ - (√3 y)³ = (2√5 x - √3 y)((2√5 x)² + (2√5 x)(√3 y) + (√3 y)²) = (2√5 x - √3 y)(20x² + 2√15 xy + 3y²)

Comparing this with the given equation (2√5 x - √3 y)(Ax² + Bxy + Cy²), we get: A = 20 B = 2√15 C = 3

Substitute these values into the required expression: √(B² + C² - A) = √((2√15)² + 3² - 20) = √(60 + 9 - 20) = √49 = 7

Explanation

For compounding 10-monthly over a total duration of 2½ years (which is 30 months), the number of compounding periods is: n = 30 / 10 = 3

The rate of interest per 10-month period is: R = 15% × (10 / 12) = 12.5%

Converting this rate to a fraction gives: 12.5% = 1/8

The amount after 3 compounding periods is calculated using the formula: A = P × (1 + R)ⁿ A = 4096 × (1 + 1/8)³ A = 4096 × (9/8)³ A = 4096 × 729 / 512 A = 8 × 729 A = 5832

The compound interest is the difference between the amount and the principal: CI = A - P CI = 5832 - 4096 CI = 1736

Explanation

When two trains run in opposite directions, their relative speed is the sum of their individual speeds.

Relative speed = 84 + 52 = 136 km/h

Convert this relative speed from km/h to m/s: Relative speed = 136 × 5/18 = 340/9 m/s

The total distance covered when the two trains completely cross each other is the sum of their lengths. Let the length of train x be L. Since the length of train y is two-third that of x, the length of train y is 2L/3.

Total distance = L + 2L/3 = 5L/3

Using the formula Distance = Speed × Time: 5L/3 = (340/9) × 12 5L/3 = 1360/3 5L = 1360 L = 272 m

The length of train x is 272 m.

Explanation

Given the equation: x² + 1 = 3x

Dividing by x on both sides: x + 1/x = 3

We need to find the value of: (x⁴ + x⁻²) / (x² + 5x + 1) = (x⁴ + 1/x²) / (x² + 5x + 1)

Dividing both the numerator and the denominator by x: Numerator = (x⁴ + 1/x²) / x = x³ + 1/x³ Denominator = (x² + 5x + 1) / x = x + 5 + 1/x

Using the algebraic identity x³ + 1/x³ = (x + 1/x)³ - 3(x + 1/x): Numerator = 3³ - 3(3) = 27 - 9 = 18

For the denominator: Denominator = (x + 1/x) + 5 = 3 + 5 = 8

Thus, the required value is: 18 / 8 = 9/4 = 2¼

Explanation

Let the initial number of successful candidates be 5x and failed candidates be 2x, making the total number of candidates 7x.

If the number of failures had been 14 more, it implies that 14 candidates who initially succeeded instead failed, keeping the total number of candidates constant.

The new number of successful candidates is 5x - 14, and the new number of failed candidates is 2x + 14.

According to the given condition, the new ratio of success to failure is 9 : 5: (5x - 14) / (2x + 14) = 9 / 5

Cross-multiplying the terms: 5 × (5x - 14) = 9 × (2x + 14) 25x - 70 = 18x + 126 7x = 196

Since the total number of candidates who appeared for the examination is represented by 7x, the total number of candidates is 196.

Explanation

According to the Intersecting Chords Theorem, when two chords AB and CD of a circle intersect at an internal point E, the products of the segments of the chords are equal:

AE × BE = CE × DE

Given that the total length of chord CD is 18 cm and the segment DE is 6 cm, the remaining segment CE is:

CE = CD - DE = 18 - 6 = 12 cm

Using the values of AE, CE, and DE in the intersecting chords relation:

18 × BE = 12 × 6

18 × BE = 72

BE = 4 cm

Explanation

In the right-angled triangle ABC, angle A = 90°. Since BL and CM are medians, L is the midpoint of AC and M is the midpoint of AB, meaning AL = AC/2 and AM = AB/2.

Applying Pythagoras' theorem in the right-angled triangle BAL: BL² = AB² + AL² BL² = AB² + (AC/2)² BL² = AB² + AC²/4 4BL² = 4AB² + AC²

Applying Pythagoras' theorem in the right-angled triangle CAM: CM² = AC² + AM² CM² = AC² + (AB/2)² CM² = AC² + AB²/4 4CM² = 4AC² + AB²

Adding these two equations: 4BL² + 4CM² = 5AB² + 5AC² 4(BL² + CM²) = 5(AB² + AC²)

Since ABC is a right-angled triangle with angle A = 90°, by Pythagoras' theorem, BC² = AB² + AC². Substituting this into the previous equation gives: 4(BL² + CM²) = 5BC²

Explanation

The relation between the volume (V) and the curved surface area (CSA) of a cylinder with radius r is given by the formula:

V = (CSA × r) / 2

Substituting the given values of CSA = 440 cm² and r = 7 cm into the formula:

V = (440 × 7) / 2 V = 220 × 7 V = 1540 cm³