RRB NTPC2016MathematicsPage 1 of 6

RRB NTPC CBT 1 2016 Mathematics Questions with Solutions

Questions from every shift of the year, grouped by section. Questions 1–50 of 252, each with the correct answer and a step-by-step explanation.

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1Mathematics
Car B is twice as fast as car A. If car A covers 90 km in 2 hours, what is the speed of car B?
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Answer: (A) 90 kmph

Explanation

Speed = Distance / Time.

Speed of car A = 90 km / 2 hours = 45 km/h.

Since car B is twice as fast as car A: Speed of car B = 2 × 45 km/h = 90 km/h.

2Mathematics
The ratio of speeds of X and Y is 5:7. Y covers a distance of 420 km in 3 hours. What is the speed of X?
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Answer: (C) 100 kmph

Explanation

Speed of Y = Distance ÷ Time = 420 ÷ 3 = 140 km/h.

Given the ratio of speeds of X and Y is 5:7, Speed of X = (5 ÷ 7) × Speed of Y = (5 ÷ 7) × 140 = 100 km/h.

3Mathematics
The standard deviation of the set {10, 11, 9, 11, 9} is
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Answer: (B) 2 / √5

Explanation

The mean (x̅) of the given set is (10 + 11 + 9 + 11 + 9) / 5 = 50 / 5 = 10.

The sum of squared deviations from the mean is Σ(x - x̅)² = (10 - 10)² + (11 - 10)² + (9 - 10)² + (11 - 10)² + (9 - 10)² = 0 + 1 + 1 + 1 + 1 = 4.

Using the standard deviation formula σ = √(Σ(x - x̅)² / N), where N = 5: σ = √(4 / 5) = 2 / √5.

4Mathematics
How many times does the number 5 occur in the range of numbers from 1 to 100?
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Answer: (C) 20

Explanation

To count the total occurrences of the digit 5 in the numbers from 1 to 100, analyze its appearance by place value:

In the unit's place, 5 occurs 10 times: 5, 15, 25, 35, 45, 55, 65, 75, 85, and 95. In the ten's place, 5 occurs 10 times: 50, 51, 52, 53, 54, 55, 56, 57, 58, and 59.

Adding the occurrences across both positions gives 10 + 10 = 20 times. Note that the number 55 contains two 5s, contributing once to the unit's place count and once to the ten's place count.

5Mathematics
If the price of two and a half kg of vegetables is 120 rupees, then what will be the price of half a kg of vegetables?
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Answer: (C) 24

Explanation

Two and a half kg (2.5 kg) consists of five half-kg (0.5 kg) portions, since 2.5 ÷ 0.5 = 5.

Therefore, the price of half a kg of vegetables = 120 ÷ 5 = ₹24.

6Mathematics
The sum of three consecutive even numbers is 42. Find the middle number.
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Answer: (D) 14

Explanation

For an odd number of consecutive terms in an arithmetic progression, the middle term is equal to the average of the set.

Middle number = Sum ÷ Number of terms = 42 ÷ 3 = 14

7Mathematics
Solve: (x + 2y)(2x - y)
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Answer: (B) 2x² + 3xy - 2y²

Explanation

Expanding the given product using the distributive law:

(x + 2y)(2x - y) = x(2x - y) + 2y(2x - y) = 2x² - xy + 4xy - 2y² = 2x² + 3xy - 2y²

Combining the like terms -xy and 4xy yields 3xy.

8Mathematics
The value of (sec²θ + 2tanθcotθ - tan²θ) is
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Answer: (D) 3

Explanation

Using the fundamental trigonometric identities sec²θ - tan²θ = 1 and tanθ × cotθ = 1:

(sec²θ + 2tanθcotθ - tan²θ) = (sec²θ - tan²θ) + 2(tanθ × cotθ) = 1 + 2(1) = 3

9Mathematics
9 women can complete a work in 50 days, if 6 more women are added then in how many days will this work be completed?
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Answer: (A) 30

Explanation

Using the inverse proportion relation M₁ × D₁ = M₂ × D₂:

Initial work capacity = 9 × 50 = 450 woman-days.

When 6 more women are added, the total workforce becomes 9 + 6 = 15 women.

Number of days required = (9 × 50) / 15 = 450 / 15 = 30 days.

10Mathematics
If 4x + 5y = 14 and x - 5y = 16 then the value of x and y are
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Answer: (D) 6 and -2

Explanation

Adding the two equations eliminates y: (4x + 5y) + (x - 5y) = 14 + 16 5x = 30 x = 6

Substituting x = 6 into x - 5y = 16: 6 - 5y = 16 -5y = 10 y = -2

11Mathematics
You have 20 big and 16 small diaries and want to make gift packs containing both in each pack. What is the maximum number of gift packs you can make without any left over?
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Answer: (B) 4

Explanation

To find the maximum number of gift packs that can be made such that each pack contains both types of diaries with equal distribution and none left over, calculate the Highest Common Factor (HCF) of the two quantities, 20 and 16.

Prime factorization of 20 = 2² × 5 Prime factorization of 16 = 2⁴

HCF(20, 16) = 2² = 4

Thus, 4 gift packs can be made, with each pack containing 5 big diaries and 4 small diaries.

12Mathematics
If L:M = 3:5 and M:N = 2:3, then N:L = ?
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Answer: (B) 5:2

Explanation

To find the ratio N:L, express N/L as the product of (N/M) and (M/L):

N/L = (N/M) × (M/L)

Given L:M = 3:5, M/L = 5/3. Given M:N = 2:3, N/M = 3:2.

N/L = (3/2) × (5/3) = 5/2

13Mathematics
10 dozen apples, 15 dozen mangoes, and 20 dozen oranges were kept for sale. If 1/2, 1/3, and 1/4 of them were sold respectively, then how many total fruits are left?
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Answer: (D) 360

Explanation

Total remaining dozen of each fruit: Remaining apples = 10 × (1 - ½) = 5 dozen Remaining mangoes = 15 × (1 - ⅓) = 10 dozen Remaining oranges = 20 × (1 - ¼) = 15 dozen

Total remaining fruits in dozen = 5 + 10 + 15 = 30 dozen Total remaining fruits = 30 × 12 = 360

14Mathematics
The value of tan(-405°) is
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Answer: (B) -1

Explanation

Using the trigonometric identity tan(-θ) = -tan(θ) and the periodicity property tan(360° + θ) = tan(θ):

tan(-405°) = -tan(405°) = -tan(360° + 45°) = -tan(45°) = -1

15Mathematics
What is the highest common factor of 360 and 450?
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Answer: (A) 90

Explanation

To find the highest common factor (HCF) of 360 and 450, express both numbers in terms of their prime factorizations:

360 = 2³ × 3² × 5¹ 450 = 2¹ × 3² × 5²

Taking the product of the lowest powers of all common prime factors: HCF = 2¹ × 3² × 5¹ = 2 × 9 × 5 = 90

Alternatively, using Euclid's division algorithm: 450 = 360 × 1 + 90 360 = 90 × 4 + 0

Since the remainder becomes 0, the divisor 90 is the HCF.

16Mathematics
On a certain principal, simple interest amounts to Rs. 2,000 in two years at the rate of 10% p.a. What will be the difference in the amount of interest if the same is compounded on annual basis?
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Answer: (C) Rs. 100

Explanation

For two years at a simple interest rate of 10% per annum, the total simple interest rate is 20%. Since the simple interest is Rs. 2,000, the principal P = 2000 × (100 / 20) = Rs. 10,000.

The difference (D) between compound interest and simple interest for 2 years is given by the formula:

D = P × (R / 100)²

Substituting P = 10,000 and R = 10:

D = 10,000 × (10 / 100)² = 10,000 × (1 / 100) = Rs. 100

17Mathematics
A prime number
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Answer: (C) Has only 1 and itself as divisors.

Explanation

By definition, a prime number is a natural number greater than 1 that has exactly two distinct positive integer divisors: 1 and the number itself. For example, numbers like 2, 3, 5, 7, and 11 can only be divided evenly by 1 and themselves.

18Mathematics
Ram lent Rs. 6,000/- to Shiva for 3 years and Rs. 8,000/- to Krishna for 5 years at the same rate of simple interest per annum. He got a total interest of Rs. 5,220 from both. Find the rate of interest per annum.
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Answer: (D) 9%

Explanation

Simple Interest formula is SI = (P × R × T) / 100, where P is principal, R is rate of interest per annum, and T is time in years.

Interest from Shiva = (6000 × R × 3) / 100 = 180R Interest from Krishna = (8000 × R × 5) / 100 = 400R

Total Interest = 180R + 400R = 580R

Given total interest is Rs. 5,220: 580R = 5220 R = 5220 / 580 = 9%

19Mathematics
P + Q = 2(P - Q) If Q=10, what is the value of P?
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Answer: (D) 30

Explanation

Substitute Q = 10 into the given equation: P + 10 = 2(P - 10)

Expanding the right side: P + 10 = 2P - 20

Rearranging terms to solve for P: 20 + 10 = 2P - P P = 30

20Mathematics
What is the area of the square, if the length of its diagonal is 13√2 units?
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Answer: (B) 169 sq. units

Explanation

The area of a square with diagonal d is given by the formula Area = d² / 2.

Substituting the given diagonal d = 13√2:

Area = (13√2)² / 2 = (169 × 2) / 2 = 169 sq. units.

21Mathematics
Which of the following fractions is the least of all?
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Answer: (A) 6/5

Explanation

Converting the given fractions into their decimal equivalents: 6/5 = 1.20 5/4 = 1.25 4/3 = 1.33 3/2 = 1.50

Comparing these decimal values, 1.20 is smaller than 1.25, 1.33, and 1.50, which makes 6/5 the smallest fraction.

22Mathematics
If a person has sold an article for Rs. 1,300 and made a profit of 30%, find the cost price of the article.
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Answer: (A) Rs. 1,000

Explanation

Cost Price = Selling Price / (1 + Profit%/100)

Substituting the given values: Cost Price = 1300 / (1 + 30/100) = 1300 / 1.3 = 1000

23Mathematics
Pushpa bought 2 dozen pens for ₹ 480 and sold them in packets of 4 pens each for ₹ 100 per packet. Her profit percentage is:
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Answer: (D) 25%

Explanation

Total pens = 2 dozen = 2 × 12 = 24 pens. Cost price (CP) = ₹ 480.

Number of packets made = 24 / 4 = 6 packets. Total selling price (SP) = 6 × ₹ 100 = ₹ 600.

Profit = SP - CP = 600 - 480 = ₹ 120.

Profit percentage = (Profit / CP) × 100 = (120 / 480) × 100 = 25%.

24Mathematics
Find the range of the data: 11, 13, 9, 17, 13, 19, 10, 11
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Answer: (B) 10

Explanation

Range = Maximum value - Minimum value

From the given data set: Maximum value = 19 Minimum value = 9

Range = 19 - 9 = 10

25Mathematics
Expand: (s + 2)³
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Answer: (C) s³ + 6s² + 12s + 8

Explanation

Using the standard algebraic identity (a + b)³ = a³ + 3a²b + 3ab² + b³:

Substitute a = s and b = 2: (s + 2)³ = s³ + 3(s²)(2) + 3(s)(2²) + 2³ = s³ + 6s² + 12s + 8

26Mathematics
Vikram and Vivek can finish a work in 50 days. They worked together for 20 days and then left. How much work has been done by them?
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Answer: (D) 2/5

Explanation

Fraction of work done = (Number of days worked) / (Total days required)

Substituting the given values: Work done = 20 / 50 = 2/5

27Mathematics
An 18 feet high electric pole is situated at a distance of 10 feet from a 6 feet high tree. What will be the length of the shadow of the tree?
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Answer: (D) 5 feet

Explanation

Using the property of similar triangles, the ratio of the height of the tree to the height of the electric pole is equal to the ratio of the shadow length to the total distance from the pole to the tip of the shadow.

Let s be the length of the shadow of the tree.

Height of tree / Height of pole = s / (10 + s) 6 / 18 = s / (10 + s) 1 / 3 = s / (10 + s) 10 + s = 3s 2s = 10 s = 5 feet

28Mathematics
The standard deviation of the set {10, 10, 10, 10, 10} is
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Answer: (A) 0

Explanation

Standard deviation measures the spread or dispersion of data points relative to their mean. It is calculated using the formula σ = √(Σ(xᵢ - μ)² / N), where μ is the mean and N is the total number of observations.

For the given set {10, 10, 10, 10, 10}: Mean μ = (10 + 10 + 10 + 10 + 10) / 5 = 10 Variance σ² = ((10 - 10)² + (10 - 10)² + (10 - 10)² + (10 - 10)² + (10 - 10)²) / 5 = 0 Standard deviation σ = √0 = 0.

29Mathematics
Karan purchased one dozen pens for Rs. 120/- and sold a pack of 3 pens for Rs. 35/- each. What is his gain?
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Answer: (A) 16.67%

Explanation

Cost price of 12 pens = ₹120.

Since there are 4 packs of 3 pens each in a dozen (12 ÷ 3 = 4), and each pack is sold for ₹35: Total selling price = 4 × 35 = ₹140.

Gain = Selling Price - Cost Price = 140 - 120 = ₹20. Gain percentage = (20 / 120) × 100 = 16.67%.

30Mathematics
G is twice as fast as S in doing work. If G can do a work in 30 days less than S, how many days will they take to complete the work together?
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Answer: (B) 20

Explanation

Since efficiency is inversely proportional to time taken, the ratio of efficiencies of G and S is 2 : 1, which means the ratio of time taken by G and S is 1 : 2.

The difference in the time ratio is 2 - 1 = 1 unit, which corresponds to 30 days. Time taken by G = 1 × 30 = 30 days Time taken by S = 2 × 30 = 60 days

Total work = Efficiency × Time = 2 × 30 = 60 units. Combined efficiency of G and S = 2 + 1 = 3 units per day.

Time taken working together = Total work / Combined efficiency = 60 / 3 = 20 days.

31Mathematics
What is the ratio of simple interest earned on certain amount at the rate of 12% p. a. for 9 years and that for 12 years?
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Answer: (C) 3:4

Explanation

Simple Interest SI = (P × R × T) / 100

Since the principal amount P and interest rate R = 12% are the same in both cases, simple interest is directly proportional to time period T.

Ratio of Simple Interest = T₁ : T₂ = 9 : 12 = 3 : 4

32Mathematics
Find the range and mode of the data 17, 18, 28, 19, 16, 18, 17, 29, 18
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Answer: (B) 13 and 18

Explanation

Range is calculated as the difference between the maximum and minimum values in the dataset: Range = Maximum value - Minimum value = 29 - 16 = 13.

Mode is the value that appears most frequently in the dataset. Frequency of each number: 16 appears 1 time 17 appears 2 times 18 appears 3 times 19 appears 1 time 28 appears 1 time 29 appears 1 time

Since 18 has the highest frequency (3 times), the mode is 18.

33Mathematics
K purchased a table for Rs. 11,000 and sold it for Rs. 13,500. What is his gain in percentage?
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Answer: (C) 22.7%

Explanation

Gain Percentage = [(Selling Price - Cost Price) / Cost Price] × 100

Gain = 13,500 - 11,000 = 2,500

Gain Percentage = (2,500 / 11,000) × 100 = 250 / 11 = 22.727...% ≈ 22.7%

34Mathematics
The value of (sin θ +cos θ)² =
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Answer: (C) 1 + 2cos θ sin θ

Explanation

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

(sin θ + cos θ)² = sin²θ + cos²θ + 2 sin θ cos θ

Substituting the fundamental trigonometric identity sin²θ + cos²θ = 1 yields:

(sin θ + cos θ)² = 1 + 2 cos θ sin θ

35Mathematics
How many times does the number 3 occur in unit's place for numbers ranging from 1 to 100?
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Answer: (C) 10

Explanation

In every set of 10 consecutive integers, each digit from 0 to 9 appears exactly once in the unit's place. The range from 1 to 100 contains 10 complete sets of 10 numbers (1–10, 11–20, ..., 91–100).

Therefore, the digit 3 occurs in the unit's place exactly once in each set, giving a total of 10 times: 3, 13, 23, 33, 43, 53, 63, 73, 83, and 93.

36Mathematics
Find the G.C.F. | L.C.M. of 20 and 28.
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Answer: (D) 4,140

Explanation

Prime factorization of the given numbers: 20 = 2² × 5 28 = 2² × 7

The Greatest Common Factor (G.C.F.) is the product of the lowest powers of common prime factors: G.C.F.(20, 28) = 2² = 4

The Least Common Multiple (L.C.M.) is the product of the highest powers of all prime factors involved: L.C.M.(20, 28) = 2² × 5 × 7 = 140

Thus, the G.C.F. and L.C.M. of 20 and 28 are 4 and 140 respectively.

37Mathematics
Ali travels a distance of 300m in 2 minutes and 30 seconds. What is his speed in kmph?
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Answer: (C) 7.2

Explanation

Total distance = 300 m Total time = 2 minutes 30 seconds = 150 seconds

Speed (in m/s) = Distance / Time = 300 / 150 = 2 m/s

To convert speed from m/s to km/h, multiply by 18/5: Speed = 2 × (18/5) = 36/5 = 7.2 km/h

38Mathematics
Twice the difference between two numbers is equal to their sum. If one number is 15, find the other number.
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Answer: (C) 5

Explanation

Let the other number be y.

According to the given condition, twice the difference between the two numbers is equal to their sum: 2(15 - y) = 15 + y 30 - 2y = 15 + y 3y = 15 y = 5

39Mathematics
The sum of three consecutive numbers is 126. Find the highest number.
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Answer: (C) 43

Explanation

Let the three consecutive numbers be x − 1, x, and x + 1.

Sum = (x − 1) + x + (x + 1) = 3x = 126 x = 126 ÷ 3 = 42

The three numbers are 41, 42, and 43. The highest number among them is 43.

40Mathematics
Surya is 25 years older than his son. In 5 years, he will be twice as old as his son. What will be Surya's age after 3 years?
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Answer: (D) 48

Explanation

Let Surya's present age be S and his son's present age be s.

According to the given conditions: S - s = 25 In 5 years, S + 5 = 2(s + 5), which simplifies to S - 2s = 5.

Subtracting the second equation from the first: (S - s) - (S - 2s) = 25 - 5 s = 20

Surya's present age S = 20 + 25 = 45 years. Surya's age after 3 years = 45 + 3 = 48 years.

41Mathematics
If the interior angle of a polygon is 108° , then it is a
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Answer: (C) Pentagon

Explanation

The exterior angle of a regular polygon is E = 180° - 108° = 72°. The number of sides n is given by n = 360° / E = 360° / 72° = 5. A polygon with 5 sides is a pentagon.

42Mathematics
If sin θ = 1 / √2 then (tan θ +cos θ) =
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Answer: (D) (1 + √2) / √2

Explanation

For sin θ = 1 / √2, θ = 45°.

Therefore, cos θ = 1 / √2 and tan θ = 1.

Substituting these values into the expression: tan θ + cos θ = 1 + 1 / √2 = (1 + √2) / √2

43Mathematics
The mean and median of 7, 5, 5, 2, 7, 6, 5, 3, 7, 6
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Answer: (D) 5.3 and 5.5

Explanation

Mean is calculated as the sum of all values divided by the number of observations: Mean = (7 + 5 + 5 + 2 + 7 + 6 + 5 + 3 + 7 + 6) / 10 = 53 / 10 = 5.3

To find the median, arrange the 10 values in ascending order: 2, 3, 5, 5, 5, 6, 6, 7, 7, 7

Since n = 10 is even, the median is the average of the 5th and 6th values: Median = (5 + 6) / 2 = 5.5

44Mathematics
B has 32 pens, 24 pencils and 16 erasers. How many sets of these three items can B make without any left over?
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Answer: (C) 8

Explanation

To find the maximum number of sets that can be formed without any items leftover, calculate the Highest Common Factor (HCF) of 32, 24, and 16.

32 = 8 × 4 24 = 8 × 3 16 = 8 × 2

HCF(32, 24, 16) = 8

Thus, 8 sets can be made, with each set containing 4 pens, 3 pencils, and 2 erasers.

45Mathematics
Expand: (w - 9)²
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Answer: (C) (w² - 18w + 81)

Explanation

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

Substituting a = w and b = 9: (w - 9)² = w² - 2(w)(9) + 9² = w² - 18w + 81

46Mathematics
On a certain principal, simple interest amounts to Rs. 1,000 in 1 year at the rate of 10% p.a. What will be the effective rate of interest if the same is compounded on half yearly basis?
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Answer: (D) 10.25

Explanation

The effective annual rate of interest with half-yearly compounding is calculated using the formula:

Effective Rate = (1 + r/n)ⁿ - 1

Here, nominal annual rate r = 10% = 0.10, and compounding frequency n = 2 per year.

Rate per half-year = 10% / 2 = 5% = 0.05

Effective Rate = (1 + 0.05)² - 1 = 1.05² - 1 = 1.1025 - 1 = 0.1025 = 10.25%

47Mathematics
1S = Rs.67.89 Money changer adds a margin of Rs. 1.11. What will be the cost of 150?
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Answer: (B) Rs. 10,350

Explanation

Total rate per unit = Base rate + Margin = 67.89 + 1.11 = Rs. 69.00

Cost of 150 units = 150 × 69 = Rs. 10,350

48Mathematics
If the area of a circle is 9π sq. cm then its circumference is
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Answer: (B) 6π cm

Explanation

The area of a circle is given by the formula Area = πr², where r is the radius.

Given Area = 9π sq. cm: πr² = 9π r² = 9 r = 3 cm

The circumference of a circle is given by the formula Circumference = 2πr.

Substituting r = 3 cm: Circumference = 2 × π × 3 = 6π cm

49Mathematics
A person carries Rs. 165 / - in the form of currency notes of denominations Rs. 5, Rs. 10 | Rs. 20 in the ratio of 3:2:1. What is the value of currency notes of Rs. 20 denomination?
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Answer: (A) Rs. 60

Explanation

Let the number of Rs. 5, Rs. 10, and Rs. 20 currency notes be 3x, 2x, and 1x respectively.

Total value = (5 × 3x) + (10 × 2x) + (20 × 1x) = 15x + 20x + 20x = 55x

Given total value = 165 55x = 165 x = 3

Number of Rs. 20 notes = 1x = 3 notes. Value of currency notes of Rs. 20 denomination = 3 × 20 = Rs. 60.

50Mathematics
Which of the following fractions is the highest of all?
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Answer: (C) 3/2

Explanation

Converting each fraction to decimal form: 3/2 = 1.50 4/3 = 1.33... 5/4 = 1.25 6/5 = 1.20

Comparing these values, 1.50 is the largest decimal.

Answer key

Tap a number to jump to that question.

  1. 1A
  2. 2C
  3. 3B
  4. 4C
  5. 5C
  6. 6D
  7. 7B
  8. 8D
  9. 9A
  10. 10D
  11. 11B
  12. 12B
  13. 13D
  14. 14B
  15. 15A
  16. 16C
  17. 17C
  18. 18D
  19. 19D
  20. 20B
  21. 21A
  22. 22A
  23. 23D
  24. 24B
  25. 25C
  26. 26D
  27. 27D
  28. 28A
  29. 29A
  30. 30B
  31. 31C
  32. 32B
  33. 33C
  34. 34C
  35. 35C
  36. 36D
  37. 37C
  38. 38C
  39. 39C
  40. 40D
  41. 41C
  42. 42D
  43. 43D
  44. 44C
  45. 45C
  46. 46D
  47. 47B
  48. 48B
  49. 49A
  50. 50C

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