SSC CGL Tier 1 2012 Quantitative Aptitude Questions with Solutions

Explanation

Given x = 2 + √3. The reciprocal 1/x is found by rationalizing the denominator: 1/x = (1)/(2 + √(3)) × 2 - √3{2 - √3} = 2 - √3{4 - 3} = 2 - √3.

The sum of the variable and its reciprocal is: x + ¹⁄ₓ = (2 + √3) + (2 - √3) = 4.

The expression to be evaluated is (x⁶ + x⁴ + x² + 1)/(x³), which simplifies to: x³ + x + ¹⁄ₓ + (1)/(x³) = (x³ + (1)/(x³)) + (x + ¹⁄ₓ).

Using the algebraic identity x³ + (1)/(x³) = (x + ¹⁄ₓ)³ - 3(x + ¹⁄ₓ), we substitute x + ¹⁄ₓ = 4: x³ + (1)/(x³) = 4³ - 3(4) = 64 - 12 = 52.

Substituting this back into the simplified expression: 52 + 4 = 56.

Explanation

The given equation is 50% of (x - y) = 30% of (x + y). This can be written as:

0.50(x - y) = 0.30(x + y)

Dividing both sides by 10, we get:

5(x - y) = 3(x + y)

Expanding the terms:

5x - 5y = 3x + 3y

Rearranging the equation to group x and y terms:

5x - 3x = 3y + 5y 2x = 8y x = 4y

The question asks what percent of x is y. This is calculated as (y / x) × 100. Substituting x = 4y into this expression:

Percentage = (y)/(4y) × 100 = ¼ × 100 = 25%

Explanation

The simple interest B owes to A is calculated using the formula SI = (P × R × T)/(100). Substituting the values P = 5000, R = 6, and T = 2, we get SI = (5000 × 6 × 2)/(100) = 600.

The compound interest B receives from C is calculated using the formula CI = P (1 + R/100)^T - P. For a rate of 10% over 2 years, the effective compound interest rate is 10 + 10 + (10 × 10)/(100) = 21%. Therefore, CI = ²¹⁄₁₀₀ × 5000 = 1050.

The profit made by B is the difference between the compound interest earned and the simple interest paid. Profit = 1050 - 600 = 450.

Explanation

When two objects start at the same time and move towards each other, the time of travel until they meet is identical for both. Since speed v is defined as distance d divided by time t, if time is constant, speed is directly proportional to the distance covered (v ∝ d).

One train covers a distance of d₁ = 110 km. Given the total distance between the stations is 200 km, the second train covers the remaining distance d₂ = 200 - 110 = 90 km.

The ratio of their speeds is equal to the ratio of the distances they traveled: Ratio of speeds = v₁{v₂} = d₁{d₂} = ¹¹⁰⁄₉₀ = ¹¹⁄₉

Explanation

The least number N which when divided by x, y, and z leaves a common remainder r in each case is determined by the formula N = LCM(x, y, z) + r.

To find the Least Common Multiple (LCM) of 25, 40, and 60, we use prime factorization: 25 = 5² 40 = 2³ × 5 60 = 2² × 3 × 5

The LCM is the product of the highest power of every prime factor present in these numbers: LCM(25, 40, 60) = 2³ × 3 × 5² LCM = 8 × 3 × 25 = 600

Now, substitute the LCM and the required remainder r = 7 into the formula: N = 600 + 7 = 607

Explanation

The least number divisible by all natural numbers from 1 to 10 is the Least Common Multiple (LCM) of the set {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}. To find the LCM, we determine the highest power of each prime factor present in the prime factorization of these numbers. The prime factors involved are 2, 3, 5, and 7. The highest power of 2 is 2³ = 8 (found in 8), the highest power of 3 is 3² = 9 (found in 9), the highest power of 5 is 5¹ = 5 (found in 5 and 10), and the highest power of 7 is 7¹ = 7 (found in 7). The LCM is the product of these highest powers: LCM = 2³ × 3² × 5¹ × 7¹ = 8 × 9 × 5 × 7. Multiplying these gives 40 × 63 = 2520.

Explanation

The expression is solved by evaluating each squared term within the brackets first. For the first part, (3² - 2²)² = (9 - 4)² = 5² = 25. For the second part, (5² - 4²)² = (25 - 16)² = 9² = 81. For the third part, (6² - 5²)² = (36 - 25)² = 11² = 121. Adding these values together gives the final result: 25 + 81 + 121 = 227.

Explanation

To find the value of the expression 2 × (2.11 + 2.23 + 2.16), the addition within the parentheses is performed first according to the standard order of operations. Adding the three decimal values gives 2.11 + 2.23 + 2.16 = 6.50. Multiplying this sum by the factor of 2 results in 2 × 6.50 = 13.

Explanation

The total work remains constant and can be expressed using the man-days formula M₁ D₁ = M₂ D₂, where M represents the number of men and D represents the number of days. Initially, we have M₁ = 40 men and D₁ = 15 days.

The total work is 40 × 15 = 600 man-days.

The problem states the work must be completed 3 days earlier than the original schedule, making the new duration D₂ = 15 - 3 = 12 days. To find the total number of men M₂ required to finish the work in 12 days, we set up the equation: M₂ × 12 = 600 M₂ = 600 / 12 = 50 men.

The number of additional people required is the difference between the new total workforce and the original workforce: 50 - 40 = 10.

Explanation

The daughter's current age is 10 years. Since the mother's age is 3 times the daughter's age, the mother is currently 10 × 3 = 30 years old. The father is 7 years older than the mother, making his current age 30 + 7 = 37 years. To determine the father's age at the time of the daughter's birth, we subtract the daughter's age from the father's current age, resulting in 37 - 10 = 27 years.

Explanation

Let the second number be y. Based on the problem, the first number is 2y and the first number is also half of the third, which means the third number is twice the first, or 4y. The average of these three numbers is 56, so their sum is 56 × 3 = 168.

We can write the equation for the sum as 2y + y + 4y = 168. Simplifying this gives 7y = 168, and dividing both sides by 7 yields y = 24. Substituting the value of y back into our expressions for the three numbers, the first number is 2 × 24 = 48, the second number is 24, and the third number is 4 × 24 = 96. The numbers in order are 48, 24, and 96.