RRB NTPC CBT 1 2021 Mathematics Questions with Solutions

Explanation

P requires a total of 10 × 6 = 60 hours to complete the task, while Q requires 9 × 10 = 90 hours.

Taking the total work as LCM(60, 90) = 180 units: P's rate = 180 ÷ 60 = 3 units/hour Q's rate = 180 ÷ 90 = 2 units/hour

Combined rate of P and Q = 3 + 2 = 5 units/hour.

Working together for 6 hours a day, they complete 5 × 6 = 30 units of work per day.

Time required to complete the task = 180 ÷ 30 = 6 days.

Explanation

Effective compound interest rate for 2 years at 10% p.a. compounded annually = 10 + 10 + (10 × 10) / 100 = 21%.

Principal = (630 / 21) × 100 = ₹3000.

Simple interest for 3 years at 10% p.a. = (Principal × Rate × Time) / 100 = (3000 × 10 × 3) / 100 = ₹900.

Explanation

Using the algebraic identity (a + b)² + (a - b)² = 2(a² + b²), let a = 759 and b = 253.

Substituting these into the given expression: [(759 + 253)² + (759 - 253)²] / (759 × 759 + 253 × 253) = 2(759² + 253²) / (759² + 253²) = 2

The terms in the numerator and denominator cancel out, leaving 2 as the result regardless of the specific values of a and b.

Explanation

Let the filling rates of pipes A and B be A and B units per hour respectively.

When pipe B is used to fill the tank alongside A, the combined filling rate is A + B, taking time t. When pipe B is used to empty the tank while A fills it, the net filling rate is A - B, taking time 5t.

Since the volume of the tank is constant: (A + B) × t = (A - B) × 5t A + B = 5A - 5B 4A = 6B A / B = 6 / 4 = 3 / 2

Thus, the ratio of the rates of A and B is 3:2.

Explanation

For a quadratic equation ax² + bx + c = 0 whose roots are in the ratio p : q, the relation between the coefficients is given by:

b² / (a × c) = (p + q)² / (p × q)

From the given equation 12x² + mx + 6 = 0, the coefficients are a = 12, b = m, and c = 6, with p = 2 and q = 3.

Substituting these values into the formula:

m² / (12 × 6) = (2 + 3)² / (2 × 3) m² / 72 = 25 / 6 m² = (25 × 72) / 6 m² = 300 m = √300 = 10√3

Explanation

Using the algebraic identity (a - b)² = a² - 2ab + b²:

1498 × 1498 = (1500 - 2)² = 1500² - (2 × 1500 × 2) + 2² = 22,50,000 - 6,000 + 4 = 22,44,004

Explanation

Using the algebraic identity a² - b² = (a + b)(a - b):

Let a = 515.86 and b = 314.87.

Notice that a - b = 515.86 - 314.87 = 200.99.

The given expression is (a² - b²) / 200.99 = (a² - b²) / (a - b).

Simplifying the expression: (a² - b²) / (a - b) = (a + b)(a - b) / (a - b) = a + b

Substituting the values: a + b = 515.86 + 314.87 = 830.73

Explanation

Perpendicular from the centre of a circle to a chord bisects the chord. Half of chord AB is 3 cm and half of chord CD is 4 cm.

Let r be the radius of the circle, and let x and y be the perpendicular distances from centre O to chords AB and CD respectively. Since the centre lies in the interior of quadrilateral ABCD (between the two chords), the total distance between them is x + y = 7 cm.

Applying Pythagoras theorem to the right triangles formed by the radius, half-chord, and distance from the centre:

r² = x² + 3² = x² + 9 r² = y² + 4² = y² + 16

Equating both expressions for r²: x² + 9 = y² + 16 x² - y² = 7 (x + y)(x - y) = 7

Substituting x + y = 7: 7(x - y) = 7 x - y = 1

Solving the equations x + y = 7 and x - y = 1 gives x = 4 cm and y = 3 cm.

Substituting x = 4 back into the radius equation: r² = 4² + 9 = 25 r = √25 = 5 cm

Explanation

Let the total work be 1 unit.

The combined daily work rate of A, B, and C is 1/16, and the daily work rate of C alone is 1/24.

Work completed by A, B, and C together in the first 2 days = 2 × (1/16) = 1/8 unit. Remaining work = 1 - 1/8 = 7/8 unit.

Let b be the daily work rate of B. The remaining work is completed by B and C together for 9 days, followed by C alone for 6 days: 9(b + 1/24) + 6 × (1/24) = 7/8 9b + 15/24 = 7/8 9b + 5/8 = 7/8 9b = 2/8 = 1/4 b = 1/36

Since B completes 1/36 of the task per day, B alone can complete the entire task in 36 days.

Explanation

The side lengths of the two given squares are: Side of first square = 48 / 4 = 12 cm Side of second square = 60 / 4 = 15 cm

Their areas are: Area of first square = 12² = 144 cm² Area of second square = 15² = 225 cm²

Area of the third square = 225 - 144 = 81 cm² Side of the third square = √81 = 9 cm Perimeter of the third square = 4 × 9 = 36 cm

Explanation

Using the algebraic identity x² + y² = (x + y)² - 2xy:

First, find the sum of x and y: x + y = (3 + √5) + (3 - √5) = 6

Next, find the product of x and y using the difference of squares formula (a + b)(a - b) = a² - b²: xy = (3 + √5)(3 - √5) = 3² - (√5)² = 9 - 5 = 4

Now, substitute these values into the identity: x² + y² = (x + y)² - 2xy x² + y² = 6² - 2(4) x² + y² = 36 - 8 = 28

Explanation

In a semicircle with diameter AB, angle ADB is 90°. Since CD is perpendicular to hypotenuse AB, by the right triangle altitude theorem, CD² = AC × CB.

Substituting the given values: 9² = 6 × CB 81 = 6 × CB CB = 13.5 cm

The total diameter AB = AC + CB = 6 + 13.5 = 19.5 cm.

Explanation

The prime numbers strictly between 8 and 59 are 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, and 53.

Summing these 12 prime numbers by pairing terms: (11 + 53) + (13 + 47) + (17 + 43) + (19 + 41) + (23 + 37) + (29 + 31) = 64 + 60 + 60 + 60 + 60 + 60 = 64 + 300 = 364

Explanation

Capacity of the cylindrical tank V = 38,500 litres = 38.5 m³ (since 1000 litres = 1 m³).

Using the cylinder volume formula V = πr²h: 38.5 = (22/7) × r² × 4 r² = (38.5 × 7) / 88 = 245 / 80 = 49/16 r = √(49/16) = 7/4 = 1.75 m

Diameter = 2 × r = 2 × 1.75 = 3.5 m

Explanation

Let the dimensions of the rectangular field be 4x and 3x.

Area = length × breadth 12x² = 300 x² = 25 x = 5

Length = 4 × 5 = 20 m Breadth = 3 × 5 = 15 m

Explanation

Let the number of 50 paise, 25 paise, and 10 paise coins be 9x, 5x, and 6x respectively.

Total value in rupees = (9x × 0.50) + (5x × 0.25) + (6x × 0.10) = 4.50x + 1.25x + 0.60x = 6.35x

Setting the total value equal to ₹317.50: 6.35x = 317.50 x = 317.50 ÷ 6.35 = 50

Number of 25 paise coins = 5x = 5 × 50 = 250.

Explanation

Using the relation between HCF, LCM, and two numbers x and y: Product of numbers xy = HCF × LCM = 21 × 315 = 6615

Using the algebraic identity (x + y)² = (x - y)² + 4xy: (x + y)² = 42² + 4 × 6615 (x + y)² = 1764 + 26460 (x + y)² = 28224 x + y = √28224 = 168

Explanation

The width of a circular path between two concentric circles is the difference between their outer and inner radii (w = R - r).

The difference between the outer and inner circumferences is given by: C₂ - C₁ = 2πR - 2πr = 2π(R - r) = 2πw

Converting the mixed fractions into improper fractions: Inner circumference C₁ = 56 4/7 = 396/7 m Outer circumference C₂ = 94 2/7 = 660/7 m

Substituting the values into the formula: 660/7 - 396/7 = 2 × (22/7) × w 264/7 = (44/7) × w w = 264 / 44 = 6 m

Explanation

Percentage Growth = [(Value in 2002 - Value in 2000) / Value in 2000] × 100

For Accountants: Increase in number = 154 - 115 = 39 Percentage Growth = (39 / 115) × 100 = 33.91%

Since both Accountants and Programmers saw the same absolute increase of 39 employees, Accountants achieved a higher percentage growth due to their lower base value in 2000 (115 compared to 134). This gives Accountants the highest percentage growth of 33.91% among the choices listed.

Explanation

The total number of employees in the year 2000 is 19 + 115 + 134 + 116 + 32 + 96 = 512.

Half of the total employees in 2000 (50%) equals 512 / 2 = 256.

The sum of Programmers, Technicians, and Analysts in the year 2000 is 134 + 116 + 32 = 282.

Since 282 is greater than 256, the combined total of these three categories was indeed more than 50% of the total employees in 2000.

Explanation

Total strength of employees in 2000 = 19 + 115 + 134 + 116 + 32 + 96 = 512. Total strength of employees in 2002 = 19 + 154 + 173 + 120 + 133 + 149 = 748.

Increase in total strength = 748 − 512 = 236.

Percentage increase = (Increase / Initial value) × 100 = (236 / 512) × 100 ≈ 46.09% ≈ 46%.

Explanation

Percentage increase of Accountants from 2000 to 2002 = [(154 - 115) / 115] × 100 = (39 / 115) × 100 ≈ 33.91%

Percentage increase of Technicians from 2000 to 2002 = [(120 - 116) / 116] × 100 = (4 / 116) × 100 ≈ 3.45%

Difference in percentage increase = 33.91% - 3.45% = 30.46% ≈ 31%

Explanation

Using the BODMAS rule, evaluate the expression step-by-step from the innermost bracket outward:

Innermost parenthesis: (20 × 1/5) = 4

Division and multiplication: 4 ÷ 1/8 × 1/16 = 4 × 8 × 1/16 = 2

Inside parenthesis: 2 - 8 = -6

Multiplying by 2: 2 × (-6) = -12

Square brackets: 20 - (-12) = 20 + 12 = 32

Final addition: 25 + 32 = 57

Explanation

Let b be the number of buffalos.

Total heads H = 20 + 10 + 15 + b = 45 + b

Pigeons have 2 feet each, whereas cows, dogs, and buffalos have 4 feet each. Total feet F = (20 × 2) + (10 × 4) + (15 × 4) + (b × 4) = 140 + 4b

Given that the total number of feet is 125 more than the total heads: F - H = 125 (140 + 4b) - (45 + b) = 125 95 + 3b = 125 3b = 30 b = 10

Explanation

The slope m of a line passing through two points (x₁, y₁) and (x₂, y₂) is given by: m = (y₂ - y₁) / (x₂ - x₁)

Substituting (3, 4) and (4, 5): m = (5 - 4) / (4 - 3) = 1

The angle of inclination θ of the line is related to its slope by: m = tan θ tan θ = 1 θ = 45°

Explanation

The first five prime numbers are 2, 3, 5, 7, and 11.

Mean = (Sum of observations) / (Number of observations) Mean = (2 + 3 + 5 + 7 + 11) / 5 = 28 / 5 = 5.6

Explanation

When interest is compounded half-yearly, the rate per period is halved and the number of periods is doubled: Rate per half-year (r) = 20% / 2 = 10% Number of compounding periods (n) = 1 × 2 = 2

Amount (A) = Principal × (1 + r/100)ⁿ A = 1000 × (1 + 10/100)² A = 1000 × (11/10)² = 1000 × 121/100 = ₹1,210

Compound Interest = Amount - Principal = ₹1,210 - ₹1,000 = ₹210.

Explanation

Using the algebraic identity x/y + y/x = (x² + y²) / (xy) with x = 1 + √3 and y = 1 - √3:

Numerator: (1 + √3)² + (1 - √3)² = 2(1² + (√3)²) = 2(1 + 3) = 8 Denominator: (1 + √3)(1 - √3) = 1² - (√3)² = 1 - 3 = -2

Dividing the numerator by the denominator: 8 / (-2) = -4

Explanation

For any two numbers, the product of the numbers equals the product of their HCF and LCM:

First Number × Second Number = HCF × LCM

Substituting the given values: 12 × Second Number = 3 × 60 12 × Second Number = 180 Second Number = 180 ÷ 12 = 15

Explanation

Using the combined work formula, Total Time = (A × B) / (A + B), where A = 10 days and B = 20 days.

Substituting the values: Total Time = (10 × 20) / (10 + 20) = 200 / 30 = 20 / 3 = 6⅔ days.

Explanation

HCF of fractions = (HCF of Numerators) / (LCM of Denominators)

Simplifying 6/15 to its lowest terms gives 2/5. The reduced fractions are 9/10, 3/25, 2/5, and 12/5.

HCF of numerators (9, 3, 2, 12) = 1 LCM of denominators (10, 25, 5, 5) = 50

Required HCF = 1/50

Explanation

Following the BODMAS/PEMDAS rule (Brackets, Orders, Division/Multiplication, Addition/Subtraction):

Evaluate the innermost parentheses: (30 - 3 × 9) = 30 - 27 = 3

Evaluate the square brackets: [20 + 7 × 3] = 20 + 21 = 41

Perform the final subtraction: 81 - 41 = 40

Explanation

For a period of two years, the difference (D) between compound interest and simple interest is given by the formula:

D = P × (R / 100)²

where P is the principal sum and R is the annual rate of interest.

Substituting D = 1 and R = 5 into the formula:

1 = P × (5 / 100)² 1 = P × (1 / 20)² 1 = P / 400 P = 400

Explanation

Let the total work be the LCM of 10, 15, and 20, which is 60 units.

Efficiency of A + B = 60 / 10 = 6 units/day Efficiency of B + C = 60 / 15 = 4 units/day Efficiency of A + C = 60 / 20 = 3 units/day

Adding these efficiencies gives 2 × (A + B + C) = 6 + 4 + 3 = 13 units/day. Therefore, the combined efficiency of A, B, and C = 13 / 2 units/day.

Time required for A, B, and C working together = Total Work / Combined Efficiency = 60 ÷ (13 / 2) = 120 / 13 = 9 3/13 days.

Explanation

Gain = Selling Price - Cost Price = ₹1,200 - ₹1,000 = ₹200

Gain Percentage = (Gain / Cost Price) × 100 = (200 / 1,000) × 100 = 20%

Explanation

Express all bases as powers of 3: 27 = 3³ ⇒ 27¹.⁵ = (3³)¹.⁵ = 3⁴.⁵ 9 = 3² ⇒ 9¹.²⁵ = (3²)¹.²⁵ = 3².⁵

Substitute these into the equation: 3⁴.⁵ × 3².⁵ ÷ 3⁵ = 3ⁿ

Using the laws of exponents (aˣ × aʸ = aˣ⁺ʸ and aˣ ÷ aʸ = aˣ⁻ʸ): 3⁴.⁵ ⁺ ².⁵ ⁻ ⁵ = 3ⁿ 3² = 3ⁿ

Comparing the exponents on both sides gives n = 2.

Explanation

Let the side of the square be s and the radius of the circle be r.

Perimeter of square = Circumference of circle 4s = 2πr s = πr / 2

Area of square = s² = (πr / 2)² = π²r² / 4 Area of circle = πr²

Ratio of area of square to area of circle = (π²r² / 4) / (πr²) = π / 4 = π : 4

Explanation

Since the circle x² + y² = a² passes through the point (0, 4), substituting x = 0 and y = 4 into the equation gives:

0² + 4² = a² 16 = a² a = 4

Explanation

Using trigonometric identities, 1 + tan²θ = sec²θ and 1 + cot²θ = cosec²θ.

Substituting these into the given expression: (1 + tan²θ) / (1 + cot²θ) = sec²θ / cosec²θ = (1 / cos²θ) / (1 / sin²θ) = sin²θ / cos²θ = tan²θ

Alternatively, expressing cot²θ in terms of tan²θ: (1 + tan²θ) / (1 + 1/tan²θ) = (1 + tan²θ) / ((tan²θ + 1) / tan²θ) = tan²θ

Explanation

For any two positive numbers: Product of two numbers = HCF × LCM

Substituting the given values: 270 = HCF × 90 HCF = 270 / 90 = 3

Explanation

The simple interest formula is: SI = (P × R × T) / 100

where P is the principal sum borrowed, R is the interest rate per annum, and T is the time period in years.

Substituting the given values into the formula: 10 = (P × 10 × 10) / 100 10 = (P × 100) / 100 10 = P P = ₹10

Explanation

The sum of the squares of the first n natural numbers is given by the formula:

S = n(n + 1)(2n + 1) / 6

For the first 10 natural numbers, substitute n = 10:

S = 10 × (10 + 1) × (2 × 10 + 1) / 6 S = 10 × 11 × 21 / 6 S = 2310 / 6 = 385

Explanation

Area of a square = side² Given area = 1225 cm² side = √1225 = 35 cm

Perimeter of a square = 4 × side = 4 × 35 cm = 140 cm

Explanation

In a proportion a : b :: c : d, the product of the means equals the product of the extremes:

b × c = a × d

Given 1.75 : x :: 5 : 8:

x × 5 = 1.75 × 8 5x = 14 x = 14 ÷ 5 = 2.8

Explanation

For a standard quadratic equation ax² + bx + c = 0, the sum of the roots is given by the formula -b/a.

For the given equation x² - 11x + 30 = 0, the coefficients are a = 1, b = -11, and c = 30.

Substituting these values into the formula: Sum of roots = -(-11)/1 = 11

Explanation

The third proportional x to two numbers a and b is given by the formula x = b² / a.

Given a = 4 and b = 9: x = 9² / 4 = 81 / 4 = 20.25

Explanation

The compound ratio of multiple ratios a:b, c:d, and e:f is calculated using the formula (a × c × e) : (b × d × f).

Substituting the given values: Compound ratio = (1 × 3 × 10) : (2 × 10 × 21) = 30 : 420

Dividing both terms by 30 gives 1 : 14.

Explanation

The Highest Common Factor (HCF) of fractions is given by the formula:

HCF of fractions = HCF of numerators / LCM of denominators

For the fractions ½ and ¾: Numerators are 1 and 3, so HCF(1, 3) = 1. Denominators are 2 and 4, so LCM(2, 4) = 4.

Substituting these values gives: HCF(½, ¾) = 1 / 4 = ¼

Explanation

To compare 2¹/², 3¹/³, 4¹/⁴, and 5¹/⁵, raise each number to the power of 60 (the LCM of the root indices 2, 3, 4, and 5):

(2¹/²)⁶⁰ = 2³⁰ = (2³)¹⁰ = 8¹⁰ (3¹/³)⁶⁰ = 3²⁰ = (3²)¹⁰ = 9¹⁰ (4¹/⁴)⁶⁰ = 4¹⁵ = 2³⁰ = 8¹⁰ (5¹/⁵)⁶⁰ = 5¹² = (5³)⁴ = 125⁴

Since 9¹⁰ > 8¹⁰, 3²⁰ > 2³⁰. Expressing 3²⁰ as (3⁵)⁴ = 243⁴ shows that 243⁴ > 125⁴, making 3²⁰ the largest resulting value.

Explanation

Using the BODMAS rule to evaluate the expression step-by-step:

First, solve inside the innermost parenthesis: 17 - 2 = 15

Next, evaluate the division inside the curly brackets: 45 ÷ 15 = 3

Then, perform the multiplication: 5 × ½ × 10 = 25

Finally, substitute these simplified values back into the main expression and perform addition and subtraction from left to right: 24 + 13 - 25 - 3 = 37 - 25 - 3 = 9