RRB NTPC CBT 1 2021 Mathematics Questions with Solutions

Explanation

Amount with John after receiving ₹1,000 = ₹19,000 + ₹1,000 = ₹20,000

Amount with Joseph after giving ₹1,000 = ₹26,000 - ₹1,000 = ₹25,000

Ratio of John's amount to Joseph's amount = 20,000 : 25,000 = 4 : 5

Explanation

Evaluate the square roots in the expression: √144 = 12 √121 = 11 √484 = 22

Substitute these values into the given equation: (12 / 6) × (11 / 8) × (132 / 22)

Simplify each fraction: 2 × (11 / 8) × 6

Multiply the terms and reduce to lowest terms: (2 × 6 × 11) / 8 = 132 / 8 = 33 / 2

Explanation

Using the standard trigonometric values: sin 30° = ½ cos 60° = ½ cot 30° = √3

Substitute these into the expression: 2 sin² 30° + cos² 60° - ¾ cot² 30° = 2(½)² + (½)² - ¾(√3)² = 2(¼) + ¼ - ¾(3) = ²⁄₄ + ¼ - ⁹⁄₄ = (2 + 1 - 9) / 4 = -6 / 4 = -3/2

Explanation

The lateral surface area (area of four side walls) of a cube with edge length a is given by the formula 4a².

4a² = 36 a² = 9 a = 3 cm

Explanation

The total number of observations is n = 8, which is even. For an even dataset arranged in ascending order, the median is the mean of the (n/2)th term and the ((n/2) + 1)th term, corresponding to the 4th and 5th observations.

Median = (4th term + 5th term) / 2

10 = [(2p + 3) + (3p + 2)] / 2 20 = 5p + 5 5p = 15 p = 3

Explanation

Checking each student's score percentages across all six subjects:

Student C scored 90% in Maths, 60% in Chemistry, 70% in Physics, 70% in Geography, 90% in History, and 70% in Computer Science, all of which are 60% or above.

Student D scored 80% in Maths, 65% in Chemistry, 80% in Physics, 80% in Geography, 60% in History, and 60% in Computer Science, all of which are 60% or above.

All other students scored below 60% in at least one subject: A scored 50% in Chemistry, B scored 40% in Geography, E scored 50% in History, F scored 40% in History, and G scored 35% in Chemistry and 50% in Physics. Exactly 2 students (C and D) meet the criteria.

Explanation

To find the aggregate marks obtained by student C, calculate the absolute marks in each subject from the given percentage and maximum marks:

Maths: 90% of 150 = 135 Chemistry: 60% of 130 = 78 Physics: 70% of 120 = 84 Geography: 70% of 100 = 70 History: 90% of 60 = 54 Computer Science: 70% of 40 = 28

Summing these marks gives the total aggregate: 135 + 78 + 84 + 70 + 54 + 28 = 449

Explanation

Total Maximum Marks = 150 + 130 + 120 + 100 + 60 + 40 = 600

Marks obtained by Student G in each subject: Maths = 65% of 150 = 97.5 Chemistry = 35% of 130 = 45.5 Physics = 50% of 120 = 60 Geography = 77% of 100 = 77 History = 80% of 60 = 48 Computer Science = 80% of 40 = 32

Total marks obtained by G = 97.5 + 45.5 + 60 + 77 + 48 + 32 = 360

Overall Percentage = (Total Marks Obtained / Total Maximum Marks) × 100 = (360 / 600) × 100 = 60%

Explanation

Average percentage marks in Physics = (Sum of percentage marks of all 7 students in Physics) ÷ (Total number of students)

Sum of percentage marks in Physics = 90 + 80 + 70 + 80 + 85 + 65 + 50 = 520%

Average percentage = 520 ÷ 7 = 74.28% ≈ 74%

Explanation

The base of the triangle lies along the x-axis between (-1, 0) and (1, 0), giving a length of b = 2 units. The third vertex is at (0, 1), so the height is h = 1 unit.

Area = ½ × base × height = ½ × 2 × 1 = 1 sq. unit.

Explanation

Using the Principle of Inclusion-Exclusion for two sets:

n(C ∪ F) = n(C) + n(F) - n(C ∩ F)

Substituting the given values: n(C ∪ F) = 75 + 60 - 35 = 100

Explanation

To convert 3.42 hours into hours, minutes, and seconds:

Whole hours = 3 hours

Fractional hours to minutes = 0.42 × 60 = 25.2 minutes

Fractional minutes to seconds = 0.2 × 60 = 12 seconds

Combining these gives 3 hours, 25 minutes, and 12 seconds.

Explanation

The speed S of the engine with N wagons attached is given by S = S₀ - k√N, where S₀ = 50 km/h is the speed without wagons and k is a constant.

Using the given condition for N = 25 wagons and S = 35 km/h: 50 - k√25 = 35 5k = 15 k = 3

To keep the speed from falling below 11 km/h: S ≥ 11 50 - 3√N ≥ 11 3√N ≤ 39 √N ≤ 13 N ≤ 13² = 169

Explanation

Let the two positive numbers be x and y, where x > y.

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

Substituting the given values: 81 = (x - y) × 27 x - y = 3

Adding the sum and difference of the two numbers gives the greater number: 2x = (x + y) + (x - y) = 27 + 3 = 30 x = 15

Explanation

Let P(x₁, y₁) be a point on the circle x² + y² = a², so x₁² + y₁² = a².

The equation of the chord of contact of tangents drawn from P(x₁, y₁) to the circle x² + y² = b² is given by x x₁ + y y₁ = b².

Since this line touches the circle x² + y² = c², the perpendicular distance from the center (0, 0) to the line x x₁ + y y₁ - b² = 0 equals the radius c: |0 + 0 - b²| / √(x₁² + y₁²) = c b² / a = c b² = a¹ c¹

Comparing b² = a¹ c¹ with the given relation bᵐ = aⁿ cᵖ for m, n, p ∈ N gives m = 2, n = 1, and p = 1.

Substituting these values into the required expression: m + n + p + 10 = 2 + 1 + 1 + 10 = 14

Explanation

Let the center of the circle be (h, k). Since the center lies on the line y = x - 1, we have k = h - 1, so the center is (h, h - 1).

The distance from the center (h, h - 1) to the given point (7, 3) is equal to the radius, 3 units: (h - 7)² + (h - 1 - 3)² = 3² (h - 7)² + (h - 4)² = 9 h² - 14h + 49 + h² - 8h + 16 = 9 2h² - 22h + 56 = 0 h² - 11h + 28 = 0 (h - 4)(h - 7) = 0

Taking h = 4 gives k = 4 - 1 = 3, so the center is (4, 3).

The standard equation of the circle is: (x - 4)² + (y - 3)² = 3² x² - 8x + 16 + y² - 6y + 9 = 9 x² + y² - 8x - 6y + 16 = 0

Explanation

If a, b, c, and x are in proportion, then a : b = c : x, where x is the fourth proportional.

Using the relation x = (b × c) / a: x = (5 × 15) / 3 = 25

Explanation

Using the standard formula, if A is r% more than B, then B is less than A by [r / (100 + r)] × 100%.

Here r = 50%.

Percentage less = [50 / (100 + 50)] × 100% = (50 / 150) × 100% = 33 ⅓%

Explanation

The length of the tangent L from a point P(x₁, y₁) to a circle S₂ = x² + y² + 2gx + 2fy + c = 0 is given by the formula L = √(S₂ at P).

For any point P(x₁, y₁) lying on the first circle x² + y² + 2011x + 2012y + 2013 = 0, the coordinates satisfy: x₁² + y₁² + 2011x₁ + 2012y₁ = -2013

The length of the tangent from P(x₁, y₁) to the second circle x² + y² + 2011x + 2012y + 2014 = 0 is: L = √(x₁² + y₁² + 2011x₁ + 2012y₁ + 2014)

Substituting x₁² + y₁² + 2011x₁ + 2012y₁ = -2013 into this formula gives: L = √(-2013 + 2014) = √1 = 1

Explanation

Speed = Distance / Time

Speed = 92.5 / 2.5 = 925 / 25 = 37 km/h

Since the truck travels at a constant speed, the distance covered in 1 hour is equal to its speed, which is 37 km.

Explanation

Simplify the left-hand side of the equation: 2/5 + 2/10 = 4/10 + 2/10 = 6/10 = 3/5

Equating this to the right-hand side: 3/5 = (N + 1)/(N + 5)

Cross-multiplying to solve for N: 3(N + 5) = 5(N + 1) 3N + 15 = 5N + 5 2N = 10 N = 5

Explanation

The smallest number that leaves a remainder r when divided by x, y, and z is given by the formula LCM(x, y, z) + r.

Prime factorization of the given divisors: 35 = 5 × 7 40 = 2³ × 5 60 = 2² × 3 × 5

LCM(35, 40, 60) = 2³ × 3 × 5 × 7 = 840

Required number = 840 + 7 = 847.

Explanation

Let the number to be subtracted be x. The modified numbers are (510 - x) and (270 - x). Their difference is (510 - x) - (270 - x) = 240.

For 24 to be the GCD of the two numbers, both (510 - x) and (270 - x) must be divisible by 24. Subtracting 6 from both numbers gives: 510 - 6 = 504 = 24 × 21 270 - 6 = 264 = 24 × 11

Since 21 and 11 are coprime (gcd(21, 11) = 1), the greatest common divisor of 504 and 264 is 24.

Explanation

Multiplying numerator and denominator inside the square root by (1 - sin θ) gives:

√((1 - sin θ) / (1 + sin θ)) = √((1 - sin θ)² / (1 - sin² θ)) = √((1 - sin θ)² / cos² θ) = (1 - sin θ) / |cos θ|

Since -1 ≤ sin θ ≤ 1, (1 - sin θ) is always non-negative, so |1 - sin θ| = 1 - sin θ.

For π/2 < θ < 3π/2 (Quadrants II and III), cos θ is negative, which means |cos θ| = -cos θ.

Substituting this gives:

(1 - sin θ) / (-cos θ) = -1/cos θ + sin θ / cos θ = -sec θ + tan θ = tan θ - sec θ

Explanation

A fraction in its lowest terms p/q has a non-terminating repeating decimal expansion if and only if its denominator q contains prime factors other than 2 and 5. In the fraction 100/11, the denominator is 11, which is a prime number distinct from 2 and 5. Dividing 100 by 11 gives 9.090909..., where the sequence 09 repeats infinitely.

Explanation

Simplify the left side of the equation by finding a common denominator: 1/3 + 2/9 = 3/9 + 2/9 = 5/9

Equating this to the right side gives: 5/9 = (N + 1)/(N + 5)

Cross-multiply to solve for N: 5 × (N + 5) = 9 × (N + 1) 5N + 25 = 9N + 9 25 - 9 = 9N - 5N 16 = 4N N = 4

Explanation

The amount under compound interest is given by the formula A = P(1 + r/100)ⁿ, where A is the amount, P is the principal, r is the rate of interest per annum, and n is the time in years.

Substituting the given values: 441 = 400(1 + r/100)² (1 + r/100)² = 441 / 400 1 + r/100 = 21 / 20 r / 100 = (21 / 20) - 1 r / 100 = 1 / 20 r = 5%

Explanation

Earnings are distributed in the ratio of work efficiencies, which is inversely proportional to the time taken by each person.

Ratio of efficiencies of A, B, and C = (1/15) : (1/20) : (1/30)

Multiplying each term by the LCM of denominators (60): Efficiency ratio = 4 : 3 : 2

Total ratio parts = 4 + 3 + 2 = 9

Share of B = (3 / 9) × 270 = 90

Explanation

Speed of the train = 96 × (5/18) = 80/3 m/s.

When the train crosses a post in 12 seconds, it covers its own length. When it crosses the platform in 30 seconds, it covers the combined length of the train and the platform.

Therefore, the train takes 30 - 12 = 18 seconds to cover the length of the platform alone.

Length of the platform = Speed × Time = (80/3) × 18 = 480 m.

Explanation

In the initial 120 litres mixture, the quantities of milk and water in the ratio 2:1 are: Milk = 120 × (2/3) = 80 litres Water = 120 × (1/3) = 40 litres

Since only water is added to the mixture, the quantity of milk remains constant at 80 litres.

In the new mixture, the ratio of milk to water is 1:2. New quantity of water = 2 × Milk = 2 × 80 = 160 litres

Amount of water to be added = 160 - 40 = 120 litres

Explanation

Prime factorization of each number gives: 144 = 2⁴ × 3² 180 = 2² × 3² × 5 384 = 2⁷ × 3

To find the LCM, take the highest power of each prime factor: LCM = 2⁷ × 3² × 5 = 128 × 9 × 5 = 5760

Explanation

Translating the point (1, 2) by 2 units along the positive x-axis gives the new point P(3, 2).

For the circle x² + y² = 9, the center is the origin O(0, 0) and the radius r = 3.

The distance from the center O to the point P is: OP = √(3² + 2²) = √13

If α is half the angle between the tangents drawn from P to the circle, then in the right-angled triangle formed by the origin, point of tangency, and P: tan α = r / √(OP² - r²) = 3 / √(13 - 9) = 3 / √4 = 3/2

Thus, α = tan⁻¹(3/2), and the total angle between the tangents is 2α = 2 tan⁻¹(3/2).

Explanation

By the quotient law of exponents, aᵐ ÷ aⁿ = aᵐ⁻ⁿ.

Applying this rule to 7¹ᐟ⁵ ÷ 7¹ᐟ³:

1/5 - 1/3 = (3 - 5) / 15 = -2/15

Therefore, 7¹ᐟ⁵ ÷ 7¹ᐟ³ = 7⁻²ᐟ¹⁵.

Explanation

Percentage less = [r / (100 + r)] × 100%

Substituting r = 20%: Percentage less = [20 / (100 + 20)] × 100% = (20 / 120) × 100% = (1 / 6) × 100% = 50 / 3% = 16 ⅔%

Explanation

Using the reciprocal trigonometric identity: cot A + cot B = (1 / tan A) + (1 / tan B) = (tan A + tan B) / (tan A tan B)

Substituting tan A + tan B = p and cot A + cot B = q: q = p / (tan A tan B) tan A tan B = p / q

Applying the compound angle formula for tangent: tan(A + B) = (tan A + tan B) / (1 - tan A tan B) = p / (1 - p/q) = pq / (q - p)

Taking the reciprocal gives: cot(A + B) = 1 / tan(A + B) = (q - p) / pq

Explanation

Area of a path running around and outside a rectangular plot is given by the formula: Area of path = 2w(L + B + 2w)

Where: Length of plot (L) = 10 m Breadth of plot (B) = 8 m Width of path (w) = 5 m

Substituting the values into the formula: Area of path = 2 × 5 × (10 + 8 + 2 × 5) Area of path = 10 × (18 + 10) Area of path = 10 × 28 = 280 m²

Explanation

For any two positive integers, Product of the two numbers = LCM × HCF.

Substituting the given values: 248 × 868 = 1736 × HCF

HCF = (248 × 868) / 1736 Since 1736 = 2 × 868: HCF = 248 / 2 = 124

Explanation

Evaluating the expression step-by-step from the innermost square root outwards:

√169 = 13

Substituting this value back into the expression: √(248 + √(51 + 13)) = √(248 + √64)

Since √64 = 8: √(248 + 8) = √256 = 16

Explanation

Simplifying the left-hand side by taking the common denominator: [(√5 - 1)/(√5 + 1)] - [(√5 + 1)/(√5 - 1)] = [(√5 - 1)² - (√5 + 1)²] / [(√5 + 1)(√5 - 1)]

Using the algebraic identity (x - y)² - (x + y)² = -4xy: [(6 - 2√5) - (6 + 2√5)] / (5 - 1) = -4√5 / 4 = -√5

Comparing -√5 + 0 with a + b√5 gives: a = 0 and b = -1

Substituting these values into the required expression: (a + b)(a - b) ÷ (a² + b²) = (a² - b²) / (a² + b²) = [0² - (-1)²] / [0² + (-1)²] = -1 / 1 = -1

Explanation

Kiran completes 1/15th of the task in 1 day, so in 5 days he completes 5/15 = 1/3 of the task.

The remaining 2/3 of the task is completed by Venkat in 30 days.

Therefore, time taken by Venkat alone to complete the full task = 30 × 3/2 = 45 days.

Explanation

Profit share ratio = Investment × Time duration

Let the capital invested by both P and Q be C. Since P stayed for 4 months and Q for the full 12 months:

Ratio of profit (P : Q) = (C × 4) : (C × 12) = 4 : 12 = 1 : 3

Total profit = ₹6,400 P's share = [1 / (1 + 3)] × ₹6,400 = ¼ × ₹6,400 = ₹1,600

Explanation

Let the capitals of A, B, and C be represented by A, B, and C. Given, 2 × A = 3 × B, which gives A : B = 3 : 2 = 6 : 4. Also, B = 4 × C, which gives B : C = 4 : 1.

Combining the ratios, A : B : C = 6 : 4 : 1. Total parts = 6 + 4 + 1 = 11.

Since profit is shared in the ratio of capital invested: Share of B = (4 / 11) × 2,97,000 = 4 × 27,000 = ₹1,08,000.

Explanation

Interest earned on ₹5,000 in 1 year = ₹5,900 − ₹5,000 = ₹900.

Rate of simple interest R = (Interest × 100) / (P × T) = (900 × 100) / (5,000 × 1) = 18% per annum.

Simple interest on ₹8,000 at 18% for 5 years = (P × R × T) / 100 = (8,000 × 18 × 5) / 100 = ₹7,200.

Total amount = Principal + Simple Interest = 8,000 + 7,200 = ₹15,200.

Explanation

Using the radical property √(a) × √(b) = √(a × b):

√(54) × √(6) = √(54 × 6) = √324 = 18

Alternatively, simplifying √54 as √(9 × 6) = 3√6 gives:

3√6 × √6 = 3 × 6 = 18

Explanation

If a number N leaves a remainder r when divided by a divisor d, then N² leaves a remainder equal to (r²) mod d when divided by the same divisor d.

Here, the given remainder r = 4 and divisor d = 7.

Remainder of N² when divided by 7 = 4² mod 7 = 16 mod 7 = 2.

Explanation

The total number of candidates who appeared for the interview from UP across the five years from 2012 to 2016 is 300 + 400 + 450 + 500 + 430 = 2080. Dividing this total by 5 years gives an average of 2080 / 5 = 416, which is greater than 400.

Explanation

To convert 1.08 hours into hours, minutes, and seconds:

Whole hours = 1 hour

Fractional hours to minutes = 0.08 × 60 = 4.8 minutes

Fractional minutes to seconds = 0.8 × 60 = 48 seconds

This gives 1 hour, 4 minutes, and 48 seconds.

Explanation

For an even number of observations N = 12, the median is the average of the (N/2)th and (N/2 + 1)th terms, which correspond to the 6th and 7th terms in the ordered data.

6th term = 2 7th term = x

Median = (2 + x) / 2 = 2.5 2 + x = 5 x = 3

Explanation

Expressing each number as a product of its prime factors: 6 = 2¹ × 3¹ 72 = 2³ × 3² 120 = 2³ × 3¹ × 5¹

To find the Least Common Multiple (LCM), take the highest power of each prime factor present: LCM = 2³ × 3² × 5¹ = 8 × 9 × 5 = 360

Explanation

Total distance covered by the train to completely clear the tunnel equals the sum of the length of the train and the length of the tunnel: Total Distance = 500 m + 1000 m = 1500 m = 1.5 km

Time taken = 1 minute = 1/60 hour

Speed = Distance ÷ Time = 1.5 ÷ (1/60) = 1.5 × 60 = 90 km/hr