SSC CHSL Tier 1 2023 Quantitative Aptitude Questions with Solutions

Explanation

To find the year with the highest difference in the enrolment of boys and girls for football, we calculate the absolute difference for each year:

For 2018: |42 - 37| = 5 For 2019: |44 - 34| = 10 For 2020: |40 - 32| = 8 For 2021: |43 - 25| = 18 For 2022: |41 - 32| = 9

The difference of 18 in the year 2021 is the highest among all the years.

Explanation

Using the standard algebraic identity: a³ + b³ + c³ - 3abc = (a + b + c)(a² + b² + c² - ab - bc - ca)

This can be rewritten as: (a³ + b³ + c³ - 3abc) / (a² + b² + c² - ab - bc - ca) = a + b + c

By setting a = 3.321, b = 2.681, and c = 1.245, the given expression simplifies directly to: a + b + c = 3.321 + 2.681 + 1.245 = 7.247

Explanation

Using the distance-speed relationship, the distance d travelled when two different speeds v₁ and v₂ result in a time difference Δt is given by the formula:

d = (v₁ × v₂) / (v₂ - v₁) × Δt

Here, the two speeds are v₁ = 18 km/h and v₂ = 24 km/h. The difference in time taken by the two men is Δt = 20 minutes. Expressing this time in hours gives:

Δt = 20/60 = ⅓ hour

Substituting these values into the formula:

d = (18 × 24) / (24 - 18) × ⅓ d = 432 / 6 × ⅓ d = 72 × ⅓ d = 24 km

Explanation

Using the rule of alligation on the concentration of milk in each vessel:

In vessel A, the fraction of milk is 5/(5 + 3) = 5/8. In vessel B, the fraction of milk is 2/(2 + 3) = 2/5. In the required mixture C, the fraction of milk is 1/(1 + 1) = 1/2.

By the rule of alligation, the ratio in which the mixtures from A and B must be mixed is: (Fraction of milk in C - Fraction of milk in B) : (Fraction of milk in A - Fraction of milk in C)

Substituting the values: | 1/2 - 2/5 | : | 5/8 - 1/2 | = 1/10 : 1/8

To simplify the ratio, multiply both terms by their common denominator 40: (1/10 × 40) : (1/8 × 40) = 4 : 5

Explanation

Using the fundamental trigonometric identity sin² A + cos² A = 1, we can write sin² A = 1 - cos² A.

Substituting this into the numerator of the given expression:

Numerator = 3 - sin² A + cos² A Numerator = 3 - (1 - cos² A) + cos² A Numerator = 3 - 1 + cos² A + cos² A Numerator = 2 + 2 cos² A

Now, substituting this simplified numerator back into the original expression:

(2 + 2 cos² A) / (2 + 2 cos² A) = 1

Explanation

Let P be the external point from which the two tangents PA and PB are drawn to a circle with center O and radius OA = 6 cm.

The angle between the two tangents is angle APB = 120°. Since the line joining the external point to the center of the circle bisects the angle between the tangents, the angle APO is 120° / 2 = 60°.

The radius OA is perpendicular to the tangent PA at the point of contact, making triangle OAP a right-angled triangle at A.

In the right-angled triangle OAP, the trigonometric ratio for the tangent of angle APO is: tan(60°) = OA / AP

Substituting tan(60°) = √3 and OA = 6 cm into the equation: √3 = 6 / AP AP = 6 / √3 AP = 2√3 cm

Explanation

The exterior angle and interior angle at vertex A form a linear pair, which means they sum to 180°. Therefore, the interior angle A is: Angle A = 180° - 115° = 65°

Since the angles A and B are equal, the interior angle B is: Angle B = 65°

The sum of all three interior angles in a triangle is 180°: Angle A + Angle B + Angle C = 180° 65° + 65° + Angle C = 180° 130° + Angle C = 180° Angle C = 180° - 130° = 50°

Explanation

The volume of a cuboid is V = l × b × h. Since only the length and breadth are increased while the height is assumed to remain constant, the percentage increase in volume is the successive percentage increase of the length and breadth.

The net percentage change for two successive increases of a% and b% is calculated using the formula: Net percentage increase = a + b + ab/100

Substituting a = 7 and b = 6 into the formula: Net percentage increase = 7 + 6 + (7 × 6)/100 = 13 + 42/100 = 13.42%

Explanation

Using the fundamental trigonometric identity: sec² θ = 1 + tan² θ

Substituting the given value tan θ = 4/5 into the identity: sec² θ = 1 + (4/5)² sec² θ = 1 + 16/25 sec² θ = (25 + 16)/25 sec² θ = 41/25

Taking the square root of both sides gives: sec θ = √41/5

Explanation

The total cost of the items purchased is the sum of the individual prices of the bag, the exercise book, and the pen: Total Cost = 560 + 90 + 45 = 695

The merchant offers a 10% discount on the total purchase: Discount = 10% of 695 = 0.10 × 695 = 69.50

Explanation

The relation between the linear scale and the area scale of a map is given by the formula: Area scale = (Linear scale)²

The given linear scale is 0.2 cm to 1 km, which means: Linear scale factor = 0.2 cm/km

Squaring this scale factor gives the area scale factor: Area scale factor = (0.2)² = 0.04 cm²/km²

To find the map area that represents an actual area of 22500 km²: Map area = Actual area × Area scale factor Map area = 22500 × 0.04 = 900 cm²

This corresponds to the numerical value of 900.

Explanation

In an equilateral triangle PQR, all three sides are equal, so PQ = QR = PR.

The perpendicular PM drawn from P to QR acts as the altitude and also bisects the base QR. Thus, QM = QR / 2 = PQ / 2.

In the right-angled triangle PMQ, applying Pythagoras' theorem: PQ² = PM² + QM²

Substituting QM = PQ / 2 into this equation: PQ² = PM² + (PQ / 2)² PQ² = PM² + PQ² / 4 PQ² - PQ² / 4 = PM² ¾ PQ² = PM² 3PQ² = 4PM²

Explanation

In quadrilateral OSPT, the radius is perpendicular to the tangent at the point of contact, so ∠OSP = 90° and ∠OTP = 90°. Since the sum of interior angles in a quadrilateral is 360°, the opposite angles at the centre O and the external point P are supplementary: ∠SOT + ∠SPT = 180° ∠SOT = 180° - 130° = 50°

In triangle OST, OS and OT are radii of the circle, making it an isosceles triangle where OS = OT. Consequently, the angles opposite to these sides are equal: ∠OST = ∠OTS

Using the angle sum property of triangle OST: ∠OST + ∠OTS + ∠SOT = 180° 2 × ∠OST + 50° = 180° 2 × ∠OST = 130° ∠OST = 65°

Explanation

The overall profit percentage can be calculated using the product of the price factor and the weight factor, where the actual weight delivered is 550 grams instead of 600 grams.

Price factor = 42 / 40 = 21 / 20

Weight factor = Claimed weight / Actual weight = 600 / 550 = 12 / 11

Overall profit factor = Price factor × Weight factor = (21 / 20) × (12 / 11) = 63 / 55

Profit percentage = (Overall profit factor - 1) × 100 Profit percentage = (63 / 55 - 1) × 100 = (8 / 55) × 100 = 160 / 11 % ≈ 14.54%

Explanation

Let the marked price of one pen be ₹1.

The retailer buys 50 pens at the marked price of 44 pens. Therefore, the cost price of 50 pens is: CP = ₹44

The retailer sells these 50 pens at a discount of 1% on their marked price. The marked price of 50 pens is ₹50, so the selling price of 50 pens is: SP = 50 × (1 - 0.01) = 50 × 0.99 = ₹49.5

The profit made by the retailer is: Profit = SP - CP = 49.5 - 44 = ₹5.5

The profit percentage is calculated as: Profit percent = (Profit / CP) × 100 Profit percent = (5.5 / 44) × 100 = (1 / 8) × 100 = 12.5%

Explanation

Let the total salary of Prakash for the month be S.

The total percentage of salary spent on children's education and donation to the trust is the sum of the individual percentages: Total percentage spent = 27% + 4% = 31%

According to the question, the total amount spent on these two heads is ₹3,038. Therefore, we can write the relation: 31% of S = 3,038

S × 31/100 = 3,038

S = (3,038 × 100) ÷ 31

S = 98 × 100

S = 9,800

Thus, Prakash's total salary for that month is ₹9,800.

Explanation

Correct Answer: D

The rate of discount is calculated using the formula: Discount % = ((Marked Price - Selling Price) / Marked Price) × 100

Substituting the given values into the formula: Marked Price = ₹850 Selling Price = ₹700

Discount = 850 - 700 = ₹150

Discount % = (150 / 850) × 100 Discount % = (15 / 85) × 100 Discount % = (3 / 17) × 100 Discount % = 300 / 17 Discount % ≈ 17.64%

Explanation

Let √(a/b) = x. Since √(b/a) = 1/x, the given equation can be written as: x = 8/3 - 1/x x + 1/x = 8/3

Squaring both sides of this equation: (x + 1/x)² = (8/3)² x² + 2 + 1/x² = 64/9 x² + 1/x² = 64/9 - 2 x² + 1/x² = 46/9

Substituting x² = a/b back into the equation: a/b + b/a = 46/9

We are given that a - b = 10. Squaring both sides: (a - b)² = 100 a² - 2ab + b² = 100

Dividing the entire equation by ab: (a² - 2ab + b²) / ab = 100 / ab a/b - 2 + b/a = 100 / ab (a/b + b/a) - 2 = 100 / ab

Substituting the value of a/b + b/a = 46/9 into this equation: 46/9 - 2 = 100 / ab 28/9 = 100 / ab ab = 900 / 28 ab = 225 / 7 ab = 32 1/7

Explanation

The algebraic identity for the square of a sum of two terms is derived by multiplying the binomial (a + b) by itself:

(a + b)² = (a + b)(a + b)

By applying the distributive property of multiplication over addition, we expand this product:

(a + b)(a + b) = a(a + b) + b(a + b) = a² + ab + ba + b²

Since multiplication is commutative for real numbers, ab is equal to ba. Substituting this back into the equation gives:

a² + ab + ab + b² = a² + 2ab + b²

Rearranging the terms, we get:

a² + b² + 2ab

This equation holds true for all possible real or complex values of variables a and b, making it a standard mathematical identity.

Explanation

By using the algebraic identity (a - b)² = a² + b² - 2ab, we can square both sides of the given equation to find the required value.

We are given: tan θ - cot θ = 4

Squaring both sides of the equation: (tan θ - cot θ)² = 4² tan²θ + cot²θ - 2 tan θ cot θ = 16

Since tan θ and cot θ are reciprocal trigonometric functions, their product is equal to 1: tan θ cot θ = 1

Substituting this relation into the squared equation: tan²θ + cot²θ - 2(1) = 16 tan²θ + cot²θ - 2 = 16 tan²θ + cot²θ = 16 + 2 tan²θ + cot²θ = 18

Explanation

The formula for compound interest is: CI = P × ((1 + R/100)ⁿ - 1)

Given: Principal P = 25000 Rate R = 10% Time n = 3 years

Substituting these values into the formula: CI = 25000 × ((1 + 10/100)³ - 1) CI = 25000 × (1.1³ - 1) CI = 25000 × (1.331 - 1) CI = 25000 × 0.331 CI = 8275

Explanation

In a rhombus STUV, all sides are equal, which means SV = UV and ST = UT.

In triangle SUV, since SV = UV, the angles opposite to these sides are equal. Therefore, ∠USV = ∠SUV = 44°.

The sum of angles in triangle SUV is 180°: ∠SVU + ∠SUV + ∠USV = 180° ∠SVU + 44° + 44° = 180° ∠SVU + 88° = 180° ∠SVU = 92°

In triangle STU, since ST = UT, the angles opposite to these sides are equal, so ∠TSU = ∠TUS.

The sum of angles in triangle STU is 180°: ∠STU + ∠TSU + ∠TUS = 180° 92° + 2 ∠TSU = 180° 2 ∠TSU = 88° ∠TSU = 44°

Substitute these values into the required expression: 4 ∠SVU - 3 ∠TSU = 4 × 92° - 3 × 44° = 368° - 132° = 236°

Explanation

Using the order of operations, simplify each term of the equation:

First, evaluate the term (2² + 4)² × (3 - 5): 2² + 4 = 4 + 4 = 8 8² = 64 3 - 5 = -2 64 × (-2) = -128

Next, calculate 20% of 400: 20% × 400 = 0.20 × 400 = 80

Then, calculate 30% of 30: 30% × 30 = 0.30 × 30 = 9

Now, substitute these simplified terms back into the original equation: -128 + 80 + (x/100) × 30 = 9

Simplify the linear equation: -48 + 0.3x = 9 0.3x = 9 + 48 0.3x = 57 x = 57 / 0.3 x = 190

Explanation

Profit sharing in a business is directly proportional to the investment ratio of the partners.

Let the profits of Anam and Bikash be 3x and 2x respectively. The total profit is therefore 3x + 2x = 5x.

Anam's profit is given as 3x = ₹3,840.

Solving for x: x = 3840 ÷ 3 = 1280

The total profit is: 5x = 5 × 1280 = ₹6,400

Explanation

Using the work equivalence formula:

M₁ × D₁ × H₁ = M₂ × D₂ × H₂

where M is the number of people, D is the number of days, and H is the number of hours worked per day.

Given: M₁ = 7, D₁ = 14, H₁ = 4 M₂ = 12, H₂ = 8

Substituting these values into the formula:

7 × 14 × 4 = 12 × D₂ × 8

392 = 96 × D₂

D₂ = 392 / 96

Simplifying the fraction by dividing the numerator and denominator by 8:

D₂ = 49 / 12 = 4 1/12 days

Explanation

The speed of the train can be found using the formula for relative speed when two objects move in the same direction. The relative speed is the length of the train divided by the time taken, which is 130 meters divided by 6.5 seconds, giving 20 m/s (or 72 km/h). Adding the man's speed of 12 km/h to this relative speed gives the speed of the train as 84 km/h.

Explanation

Using the double-angle trigonometric identity: tan(2A) = (2 tan A) / (1 - tan² A)

Substituting the given value of A = 30° into the identity: tan(2 × 30°) = tan(60°)

Since tan(60°) = √3, the value of the given expression is √3.

Explanation

The volume of a cube with side s is given by the formula V = s³.

When the three metal cubes are melted and recast into a single new cube, the total volume of the new cube is the sum of the volumes of the three individual cubes:

V = 15³ + 18³ + 25³ V = 3375 + 5832 + 15625 V = 24832 cm³

Let a be the side of the newly formed cube. The volume of this cube is a³ = 24832.

To find the side length a, we take the cube root of the total volume:

a = ∛24832 a = ∛(64 × 388) a = 4 ∛388 cm

Explanation

To find the vehicle type with the maximum total number, we calculate the sum of each vehicle type across all three cities (B, C, and K) in thousands:

Total Cars = 4.5 + 3.0 + 3.0 = 10.5 thousand Total Buses = 2.5 + 3.0 + 4.5 = 10.0 thousand Total Trucks = 3.4 + 3.5 + 3.5 = 10.4 thousand Total Bikes = 3.2 + 3.3 + 3.4 = 9.9 thousand

Comparing the totals, the number of Cars is the highest at 10.5 thousand.

Explanation

Given the equation: x + 2/x = 1

Multiplying the entire equation by x gives: x² + 2 = x

We need to find the value of the expression: (x² + 7x + 2) / (x² + 13x + 2)

By rearranging the terms in both the numerator and the denominator, we can group x² and 2 together: ((x² + 2) + 7x) / ((x² + 2) + 13x)

Substituting x² + 2 = x into this expression: (x + 7x) / (x + 13x) = 8x / 14x = 8/14 = 4/7

Explanation

The angle between two tangents drawn from an external point to a circle and the angle subtended by the radii at the center are supplementary. This is because the radius is perpendicular to the tangent at the point of contact, making the two angles at the points of contact 90° each. Since the sum of angles in a quadrilateral is 360°, the remaining two opposite angles must sum to 180°.

Using the relation: Angle between tangents + Angle between radii = 180°

Substituting the given values: Angle between tangents + 100° = 180° Angle between tangents = 180° - 100° = 80°