SSC CHSL Tier 1 2013 Quantitative Aptitude Questions with Solutions

Explanation

Given the equation: a + b + c = 4

We can rewrite this as: a + b = 4 - c

Cubing both sides of the equation: (a + b)³ = (4 - c)³

Applying the algebraic expansion rules: a³ + b³ + 3ab(a + b) = 64 - 48c + 12c² - c³

Substitute a + b = 4 - c into the left side of the equation: a³ + b³ + 3ab(4 - c) = 64 - 48c + 12c² - c³ a³ + b³ + 12ab - 3abc = 64 - 48c + 12c² - c³

Rearranging the terms to match the target expression: a³ + b³ + c³ - 12c² + 48c - 64 = 3abc - 12ab

Explanation

To find the maximum number of solid cubes that can fit inside the box, we calculate how many cubes can be aligned along each of the box's inside dimensions.

The edge length of each cube is 0.2 m, and the dimensions of the box are 1 m by 1 m by 2 m.

Number of cubes along the first dimension = 1 ÷ 0.2 = 5 Number of cubes along the second dimension = 1 ÷ 0.2 = 5 Number of cubes along the third dimension = 2 ÷ 0.2 = 10

The total number of cubes that can be packed in the box is the product of the number of cubes along each dimension: 5 × 5 × 10 = 250

Since each dimension of the box is perfectly divisible by the edge length of the cube, the cubes fit perfectly without leaving any wasted space.

Explanation

To find the number of circles that can be made, divide the total length of the wire by the circumference of a single circle.

First, convert the total length of the wire from meters to centimeters to ensure the units are consistent: Total length of the wire = 66 m = 66 × 100 cm = 6600 cm

The circumference of one circle is given as 1.32 cm.

Now, apply the division: Number of circles = Total length of the wire / Circumference of one circle Number of circles = 6600 / 1.32 Number of circles = (6600 × 100) / 132 Number of circles = 5000

Explanation

In the incircle of Δ ABC, O is the center and the lines AQ and AR are tangents to the circle from the external point A. Since the radius is perpendicular to the tangent at the point of contact: ∠AQO = 90° ∠ARO = 90°

In the quadrilateral AQOR, the sum of all interior angles is 360°: ∠QOR = 360° - (∠AQO + ∠ARO + ∠A) ∠QOR = 360° - (90° + 90° + 60°) = 120°

The angle subtended by the arc QR at the center O is ∠QOR = 120°. The angle subtended by the same arc at any point P on the remaining part of the circle is half of the angle subtended at the center: ∠QPR = ½ ∠QOR ∠QPR = ½ × 120° = 60°

Explanation

Let AB be the diameter of the circle. Since the angle subtended by a diameter at any point on the circle is a right angle, angle ADB = 90°. This means AD is perpendicular to BC.

In the isosceles triangle ABC, it is given that AB = AC. The altitude AD drawn from the vertex A to the base BC bisects the base BC in any isosceles triangle where the two lateral sides are equal. Therefore, BD = CD.

Since BD = 3 cm, the length of CD is also 3 cm.

Explanation

Since the stick and the tower cast shadows at the same time, the angle of elevation of the sun is the same for both. This means the right-angled triangle formed by the stick and its shadow is similar to the triangle formed by the tower and its shadow.

By the property of similar triangles, the ratio of the height to the shadow length is constant:

Height of stick / Length of stick shadow = Height of tower / Length of tower shadow

Substituting the given values:

12 / 8 = Height of tower / 40

Solving for the height of the tower:

Height of tower = (12 × 40) / 8 = 12 × 5 = 60 cm

Explanation

By Thales's theorem, an angle inscribed in a semicircle is a right angle. In a right-angled triangle ABC, right-angled at B, the angle opposite to the hypotenuse CA is 90°. This implies that the hypotenuse CA acts as the diameter of the circumcircle of the triangle. Since the circumcentre of a circle is the centre of that circle, it must lie at the midpoint of its diameter CA.

Explanation

Total selling price of both tables = 2 × ₹ 720 = ₹ 1440.

For the table sold at a 20% gain: Cost price = 720 ÷ 1.20 = ₹ 600.

For the table sold at a 20% loss: Cost price = 720 ÷ 0.80 = ₹ 900.

Total cost price of both tables = 600 + 900 = ₹ 1500.

Total loss = Total cost price - Total selling price = 1500 - 1440 = ₹ 60.

Explanation

The simple interest (SI) on a principal of ₹ 1,000 at a rate of 10% per annum for 4 years is calculated using the formula:

SI = (P × R × T) / 100 SI = (1000 × 10 × 4) / 100 = 400

The compound interest (CI) on the same principal, rate, and time is calculated using the formula:

CI = P × ((1 + R/100)ᵀ - 1) CI = 1000 × ((1 + 10/100)⁴ - 1) CI = 1000 × ((1.1)⁴ - 1) CI = 1000 × (1.4641 - 1) CI = 1000 × 0.4641 = 464.10

The difference between the compound interest and the simple interest is:

Difference = CI - SI Difference = 464.10 - 400 = 64.10

Explanation

The sum of all 15 numbers is calculated by multiplying the average of the 15 numbers by 15: 15 × 7 = 105

The sum of the first 8 numbers is: 8 × 6.5 = 52

The sum of the last 8 numbers is: 8 × 8.5 = 68

When we add the sum of the first 8 numbers and the sum of the last 8 numbers, the middle number (the 8th number) is included twice. Therefore, the middle number is the difference between this combined sum and the sum of all 15 numbers: (52 + 68) - 105 = 120 - 105 = 15

Explanation

Using the given relation √15625 = 125, the decimal point shifts in the square root by half the number of places it shifts in the radicand:

√156.25 = 12.5 √1.5625 = 1.25 √0.015625 = 0.125

Substituting these values into the expression:

12.5 + 1.25 − 0.125 = 13.75 − 0.125 = 13.625

Explanation

Let the three numbers be x, y, and z. The ratio of the second to the third is y : z = 9 : 6 = 3 : 2, and the ratio of the first to the third is x : z = 1 : 4.

To find the combined ratio x : y : z, we make the term for z common in both ratios. Since the ratio of the first to the third has z as 4, we multiply the terms of the ratio of the second to the third by 2 to get y : z = 6 : 4.

Thus, the ratio of the three numbers is x : y : z = 1 : 6 : 4.

The sum of the ratio parts is 1 + 6 + 4 = 11. Since the actual sum of the three numbers is 121, the value of the second number is:

(6 / 11) × 121 = 6 × 11 = 66.

Explanation

Let the two numbers be 3x and 5x, where x is their highest common factor (H.C.F.).

Since 3 and 5 are co-prime numbers, the least common multiple (L.C.M.) of 3x and 5x is: L.C.M. = 3 × 5 × x = 15x

Given that the L.C.M. is 75, we can set up the equation: 15x = 75

Solving for x: x = 75 ÷ 15 x = 5

Since x represents the H.C.F. of the two numbers, the H.C.F. is 5.

Explanation

Let B's salary be 100. Since A's salary is 30% higher than B's salary, A's salary is 130.

The percentage by which B's salary is less than A's salary is calculated using the formula: Percentage difference = (Difference / A's salary) × 100%

Substituting the values into the formula: Percentage difference = ((130 - 100) / 130) × 100% Percentage difference = (30 / 130) × 100% Percentage difference = 300 / 13 % Percentage difference = 23 1/13 %

Explanation

The relation between the number of men and the number of days required to complete a work is inversely proportional, which is expressed using the formula M₁ × D₁ = M₂ × D₂.

Here, the initial number of men M₁ is 8 and the initial number of days D₁ is 40.

When 2 more men join them, the new number of men M₂ becomes 8 + 2 = 10.

Substituting these values into the formula: 8 × 40 = 10 × D₂ 320 = 10 × D₂ D₂ = 32 days

Explanation

The speed of the car is converted from km/h to m/s: 36 × 5/18 = 10 m/s

Initially, the bus is 40 m ahead of the car, and after 20 seconds, it is 60 m behind. This means the car covers the initial gap of 40 m and creates an additional gap of 60 m relative to the bus. Relative distance = 40 + 60 = 100 m Time = 20 seconds

The relative speed of the car with respect to the bus is: Relative speed = Relative distance / Time = 100 / 20 = 5 m/s

Converting this relative speed to km/h: 5 × 18/5 = 18 km/h

Since the car is moving faster than the bus to overtake it, the relative speed is the difference between their speeds: Relative speed = Speed of car - Speed of bus 18 = 36 - Speed of bus Speed of bus = 36 - 18 = 18 km/h

Explanation

Let the quantities of the two varieties of tea mixed be 5 kg and 3 kg, respectively.

Cost Price of 5 kg of the first variety = 5 × 240 = ₹ 1200 Cost Price of 3 kg of the second variety = 3 × 300 = ₹ 900 Total Cost Price of 8 kg of the blend = 1200 + 900 = ₹ 2100

Selling Price of 8 kg of the blended variety at ₹ 315 per kg = 8 × 315 = ₹ 2520

Gain = Selling Price - Cost Price = 2520 - 2100 = ₹ 420

Gain percent = (Gain / Cost Price) × 100 = (420 / 2100) × 100 = 20%

Explanation

Let the initial capitals of A and B be 4x and 5x.

For the first 3 months, A invests 4x and B invests 5x.

After 3 months, A withdraws 1/4 of his capital, so his remaining capital is 4x - (1/4 × 4x) = 3x. B withdraws 1/5 of his capital, so his remaining capital is 5x - (1/5 × 5x) = 4x.

These remaining capitals are invested for the next 7 months of the 10-month period.

The ratio of their equivalent capitals for 1 month is: A's equivalent capital = (4x × 3) + (3x × 7) = 12x + 21x = 33x B's equivalent capital = (5x × 3) + (4x × 7) = 15x + 28x = 43x

The profit-sharing ratio of A to B is 33 : 43.

With a total profit of ₹ 760, the sum of the ratio parts is 33 + 43 = 76.

A's share of the profit is: (33 / 76) × 760 = 33 × 10 = ₹ 330

Explanation

The relationship between the Cost Price (CP) and the Printed Price (MP) given the discount percentage (d%) and profit percentage (p%) is determined by the formula:

CP / MP = (100 - d) / (100 + p)

Substituting the given values of discount (d = 10) and profit (p = 15) into the formula:

CP / MP = (100 - 10) / (100 + 15)

CP / MP = 90 / 115

Simplifying this fraction by dividing both the numerator and the denominator by 5 yields:

CP / MP = 18 / 23

Explanation

Let x be the cost price of the article. Since the profit percentage is numerically equal to the cost price, the profit percentage is x%.

Using the formula for selling price: SP = CP × (1 + Profit%/100)

Substituting SP = 39 and CP = x: 39 = x × (1 + x/100) 39 = x + x²/100 3900 = 100x + x² x² + 100x - 3900 = 0

Factorizing the quadratic equation: x² + 130x - 30x - 3900 = 0 x(x + 130) - 30(x + 130) = 0 (x - 30)(x + 130) = 0

This yields x = 30 or x = -130. Since the cost price cannot be a negative value, x must be 30.

This can also be quickly verified because a cost price of ₹ 30 with a 30% profit results in a profit of: 30% of 30 = 9

Adding this profit to the cost price gives: 30 + 9 = 39