SSC CHSL Tier 1 2020 Quantitative Aptitude Questions with Solutions

Explanation

The semi-perimeter of the triangle with sides a = 25, b = 39, and c = 34 is: s = (25 + 39 + 34) / 2 = 49 units

The area of the triangle (A) is calculated using Heron's formula: A = √(s(s - a)(s - b)(s - c)) A = √(49 × (49 - 25) × (49 - 39) × (49 - 34)) A = √(49 × 24 × 10 × 15) A = √(49 × 3600) A = 7 × 60 = 420 square units

The area of the square exceeds the area of the triangle by 21 units: Area of square = 420 + 21 = 441 square units

If the side of the square is x, then: x² = 441 x = √441 = 21 units

Explanation

For the six-digit number 59a44b to be divisible by 36, it must be divisible by both 4 and 9.

According to the divisibility rule of 4, the number formed by the last two digits, 4b, must be divisible by 4. Since b is a single digit, the possible values for b are 0, 4, and 8.

According to the divisibility rule of 9, the sum of the digits of 59a44b must be divisible by 9: 5 + 9 + a + 4 + 4 + b = 22 + a + b

Since a and b are single digits, the maximum possible value of a + b is 9 + 9 = 18. This means the total sum of the digits, 22 + a + b, can range from 22 to 40. The multiples of 9 in this range are 27 and 36.

To maximize the value of a + b, we set the sum of the digits to the largest multiple of 9: 22 + a + b = 36 a + b = 14

For a + b to equal 14, we test the allowed values of b. Setting b = 8 yields a = 6, which are both valid single-digit integers.

Explanation

For the number 59a44b to be divisible by 36, it must be divisible by both 4 and 9, since 4 and 9 are coprime factors of 36.

For divisibility by 4, the number formed by the last two digits, 4b, must be divisible by 4. This restricts the possible values of b to 0, 4, or 8.

For divisibility by 9, the sum of all digits of the number must be a multiple of 9. Sum of digits = 5 + 9 + a + 4 + 4 + b = 22 + a + b

Since a and b are single-digit non-negative integers, the maximum possible value of a + b is 18. This means the sum 22 + a + b can only be 27 or 36. If 22 + a + b = 27, then a + b = 5. If 22 + a + b = 36, then a + b = 14.

To obtain the maximum value of a + b, we select 14. We can achieve this maximum sum by choosing b = 8 and a = 6, both of which are valid single-digit digits.

Explanation

Let the side length of the cube be x. Since the sphere is inscribed inside the cube and touches all its faces, the diameter of the sphere is equal to the side length of the cube. Thus, the radius of the sphere r is x/2.

The volume of the cube is x³, and the volume of the sphere is (4/3) π r³ = (4/3) π (x/2)³ = (π/6) x³. The ratio a of the volume of the cube to the volume of the sphere is: a = x³ / ((π/6) x³) = 6/π

The surface area of the cube is 6 x², and the surface area of the sphere is 4 π r² = 4 π (x/2)² = π x². The ratio b of the surface area of the sphere to the surface area of the cube is: b = (π x²) / (6 x²) = π/6

The product ab is: ab = (6/π) × (π/6) = 1

Explanation

Sum of the serial numbers from 1 to 10 is calculated using the arithmetic progression sum formula n(n + 1) / 2: 10 × 11 / 2 = 55

Since the average of the 10 new numbers formed after adding each number to its serial number is 55, the total sum of these new numbers is: 10 × 55 = 550

The sum of the original 10 numbers is obtained by subtracting the sum of the serial numbers from the sum of the new numbers: 550 - 55 = 495

The sum of the 8 known original numbers is: 24 + 45 + 35 + 59 + 83 + 46 + 29 + 74 = 395

Therefore, the sum of the missing numbers a and b is: a + b = 495 - 395 = 100

The average of the two missing numbers is: (a + b) / 2 = 100 / 2 = 50

Explanation

The given equations are: a + b = 8 a + a²b + b + ab² = 128

By grouping the terms in the second equation, we get: (a + b) + ab(a + b) = 128 (a + b)(1 + ab) = 128

Substituting the value of a + b = 8: 8(1 + ab) = 128 1 + ab = 16 ab = 15

Using the algebraic identity for the sum of cubes: a³ + b³ = (a + b)³ - 3ab(a + b)

Substituting the values of (a + b) and ab: a³ + b³ = 8³ - 3 × 15 × 8 a³ + b³ = 512 - 360 a³ + b³ = 152

Explanation

Let T be the stipulated time in minutes. Since distance D is constant, we can equate the distances traveled in both cases using the formula Distance = Speed × Time:

Speed₁ × Time₁ = Speed₂ × Time₂

Using the given speeds and times: 60 × (T + 20)/60 = 75 × (T - 44)/60

Simplifying the equation: 60 × (T + 20) = 75 × (T - 44)

Dividing both sides by 15: 4 × (T + 20) = 5 × (T - 44) 4T + 80 = 5T - 220 T = 300 minutes

Thus, the stipulated time is 300 minutes, which is equal to 300/60 = 5 hours.

Now, calculate the distance D: D = 60 × (300 + 20)/60 = 320 km

Next, find the average speed S required to cover this distance in the stipulated time: S = D / T = 320 / 5 = 64 km/hr

The sum of the numerical values of D and S is: D + S = 320 + 64 = 384

Explanation

Since CosA, SinA, and CotA are in geometric progression, the square of the middle term is equal to the product of the first and third terms:

Sin²A = CosA × CotA

Substituting CotA = CosA / SinA:

Sin²A = Cos²A / SinA Sin³A = Cos²A

Now, express Tan⁶A in terms of SinA and CosA:

Tan⁶A = Sin⁶A / Cos⁶A = (Sin³A)² / Cos⁶A

Substituting Sin³A = Cos²A into the numerator:

Tan⁶A = (Cos²A)² / Cos⁶A = Cos⁴A / Cos⁶A = 1 / Cos²A = Sec²A

Substituting this result into the given expression:

Tan⁶A - Tan²A = Sec²A - Tan²A

Using the fundamental trigonometric identity Sec²A - Tan²A = 1:

Tan⁶A - Tan²A = 1

Explanation

The surface area of a cube with edge length a is given by 6a². Since the surface area is proportional to the square of the edge length (a²), a 10% increase in the edge length results in a successive percentage increase in the surface area.

Using the successive percentage change formula: Net percentage increase = x + y + xy/100

Substituting x = 10 and y = 10: Net percentage increase = 10 + 10 + (10 × 10)/100 = 20 + 1 = 21%

Explanation

According to the BODMAS rule, operations are performed in the order of Brackets, Of, Division, Multiplication, Addition, and Subtraction.

First, evaluate the expression inside the brackets: 6.3 + 9 = 15.3

Next, evaluate the "of" operator: 5 of 15.3 = 5 × 15.3 = 76.5

Then, perform the multiplication: 7 × 0.5 = 3.5

Now, substitute these values back into the expression: 45 - 76.5 + 3.5

Finally, perform the addition and subtraction: 45 + 3.5 - 76.5 = 48.5 - 76.5 = -28

Explanation

Let the cost price (CP) of each item be ₹10. Total cost price of 40 items = 40 × 10 = ₹400.

Since the items are marked up by 20% over the cost price, the marked price (MP) of each item is: MP = 10 × 1.2 = ₹12.

Now, calculate the selling price (SP) of the sold items: SP of 10 items sold at a 10% discount: SP₁ = 10 × 12 × 0.90 = ₹108.

SP of 20 items sold at a 15% discount: SP₂ = 20 × 12 × 0.85 = ₹204.

Total SP of the first 30 items: SPₜotal = 108 + 204 = ₹312.

To earn some profit on the entire transaction, the total selling price of all 40 items must be strictly greater than the total cost price (₹400).

Let the selling price of the remaining 10 items be SPᵣem. 312 + SPᵣem > 400 SPᵣem > 88.

The total MP of the remaining 10 items is: MPᵣem = 10 × 12 = ₹120.

Let d% be the discount percentage offered on these remaining items: 120 × (1 - d/100) > 88 1 - d/100 > 88 / 120 1 - d/100 > 11 / 15 d / 100 < 4 / 15 d < 400 / 15 % = 26.67%

Thus, the maximum discount he can offer on the remaining items while still getting some profit is just under 26.67%.

Explanation

Using the algebraic identity for sum of cubes: If x + 1/x = k, then x³ + 1/x³ = k³ - 3k

Given that x + 1/x = 5/2, we substitute k = 5/2 into the formula: x³ + 1/x³ = (5/2)³ - 3(5/2) x³ + 1/x³ = 125/8 - 15/2 x³ + 1/x³ = 125/8 - 60/8 x³ + 1/x³ = 65/8

Now, substituting this value into the given expression: (4/13) (x³ + 1/x³) = (4/13) × (65/8)

Simplifying the fraction: (4/8) × (65/13) = ½ × 5 = 5/2

Explanation

Using the profit formula: Profit = Selling Price - Cost Price

Let x be the number of litres of duplicate oil added to the mixture.

The merchant purchased 50 litres of genuine oil at ₹236 per litre, and the cost price of the duplicate oil is ₹180 per litre. Total Cost Price = (50 × 236) + (180 × x) = 11,800 + 180x

The total volume of the mixture is 50 + x litres, which is sold at ₹250 per litre. Total Selling Price = 250 × (50 + x) = 12,500 + 250x

Profit = Selling Price - Cost Price Profit = (12,500 + 250x) - (11,800 + 180x) = 700 + 70x

To earn a minimum profit of ₹2,100: 700 + 70x ≥ 2,100 70x ≥ 1,400 x ≥ 20

Thus, the minimum volume of duplicate oil needed is 20 litres.