Numerical Reasoning Questions for NATA with Solutions 2025

Overview: Numerical reasoning is a psychometric test to assess your ability to interpret, analyze and make conclusions from the given sets of data.

Unlike normal mathematics, numerical reasoning helps apply mathematics in a realistic context.

In the National Aptitude Test in Architecture, you can expect 10-12 questions based on numerical reasoning.

To ease your preparation, we have provided some of the important numerical reasoning questions for NATA 2025 exam in this post. 

So, why late? Let's dig into the post to know the topics under numerical reasoning and expected questions in the upcoming NATA exam 2025.

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Numerical Reasoning Topics for NATA 2025

Before starting your preparation, you must familiarize yourself with the topic as per NATA syllabus, from where the questions are asked in the exam.

In the NATA Entrance Exam, most of the questions are based on the following topics.

• General Arithmetic
• Numerical computation
• Percentage
• Averages
• Number series
• Data interpretation

Important Numerical reasoning Questions for NATA 2025

We have provided a few sample questions for your reference here to help you understand the type of questions asked from the numerical reasoning.

These questions are curated from the previous year's questions papers for NATA. Practising these questions regularly will help you perform well in the upcoming exam.

Q) Two numbers are in the ratio 3: 5. If 9 is subtracted from each, the new numbers are in the ratio 12: 23. The smaller number is:

A)  27

B) 33

C) 49

D) 55

Solution:

Let the two numbers be 3x3x3x and 5x5x5x, where xxx is a constant.

According to the problem, when 9 is subtracted from each number, the new ratio becomes 12:23. Therefore, we have the equation:

3x−95x−9=1223\frac{3x - 9}{5x - 9} = \frac{12}{23}5x−93x−9​=2312​

Now, cross-multiply:

23(3x−9)=12(5x−9)23(3x - 9) = 12(5x - 9)23(3x−9)=12(5x−9)

Expand both sides:

69x−207=60x−10869x - 207 = 60x - 10869x−207=60x−108

Simplify the equation:

69x−60x=207−10869x - 60x = 207 - 10869x−60x=207−108 9x=999x = 999x=99

Solve for xxx:

x=999=11x = \frac{99}{9} = 11x=999​=11

Now, substitute x=11x = 11x=11 back into 3x3x3x to find the smaller number:

3x=3×11=333x = 3 \times 11 = 333x=3×11=33

Answer: B) 33.

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Q) If 0.75 : x :: 5 : 8, then x is equal to:

A) 1.12

B) 1.2

C) 1.25

D) 1.30

Solution:

We are given the proportion:

0.75x=58\frac{0.75}{x} = \frac{5}{8}x0.75​=85​

To solve for xxx, we can use cross-multiplication:

0.75×8=5×x0.75 \times 8 = 5 \times x0.75×8=5×x

Simplify:

6=5x6 = 5x6=5x

Now, solve for xxx:

x=65=1.2x = \frac{6}{5} = 1.2x=56​=1.2

Answer: B) 1.2.

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Q) The salaries A, B, and C are in the ratio 2: 3: 5. If the increments of 15%, 10%, and 20% are allowed respectively in their salaries, then what will be the new ratio of their salaries?

A) 3 : 3 : 10

B) 10 : 11 : 20

C) 23 : 33 : 60

D) Cannot be determined

Solution:

Let the initial salaries of A, B, and C be 2x2x2x, 3x3x3x, and 5x5x5x respectively.

Now, the increments are as follows:

• A's salary increases by 15%, so A's new salary will be 2x+0.15×2x=2x×1.15=2.3x2x + 0.15 \times 2x = 2x \times 1.15 = 2.3x2x+0.15×2x=2x×1.15=2.3x.
• B's salary increases by 10%, so B's new salary will be 3x+0.10×3x=3x×1.10=3.3x3x + 0.10 \times 3x = 3x \times 1.10 = 3.3x3x+0.10×3x=3x×1.10=3.3x.
• C's salary increases by 20%, so C's new salary will be 5x+0.20×5x=5x×1.20=6x5x + 0.20 \times 5x = 5x \times 1.20 = 6x5x+0.20×5x=5x×1.20=6x.

Now, the new ratio of their salaries is:

2.3x3.3x:3.3x3.3x:6x3.3x\frac{2.3x}{3.3x} : \frac{3.3x}{3.3x} : \frac{6x}{3.3x}3.3x2.3x​:3.3x3.3x​:3.3x6x​

Simplifying the ratio:

2.33.3:3.33.3:63.3\frac{2.3}{3.3} : \frac{3.3}{3.3} : \frac{6}{3.3}3.32.3​:3.33.3​:3.36​

Multiplying by 10 to avoid decimals:

2333:1:6033\frac{23}{33} : 1 : \frac{60}{33}3323​:1:3360​

Thus, the new ratio is:

23:33:6023 : 33 : 6023:33:60

Answer: C) 23 : 33 : 60.

Read more: Short tricks to crack the NATA entrance exam on the first attempt

Q) If 40% of a number is equal to two-thirds of another number, what is the ratio of the first number to the second number?

A) 2 : 5

B) 3 : 7

C) 5 : 3

D) 7 : 3

Solution:

Let the first number be xxx and the second number be yyy.

We are given that 40% of the first number is equal to two-thirds of the second number:

0.40x=23y0.40x = \frac{2}{3}y0.40x=32​y

To eliminate the decimals, multiply both sides of the equation by 10:

4x=203y4x = \frac{20}{3}y4x=320​y

Now, multiply both sides by 3 to remove the fraction:

12x=20y12x = 20y12x=20y

Divide both sides by 4:

3x=5y3x = 5y3x=5y

Now, solve for the ratio xy\frac{x}{y}yx​:

xy=53\frac{x}{y} = \frac{5}{3}yx​=35​

Thus, the ratio of the first number to the second number is:

5:3\boxed{5 : 3}5:3​

Answer: C) 5 : 3.

Q) A team of three lumberjacks cut an average of 45,000 cubic feet of timber in a week. How many thousand cubic feet will seven lumberjacks cut in two weeks?

A) 21

B) 105

C) 225

D) 210

E) 22

Solution:

We are given that 3 lumberjacks cut an average of 45,000 cubic feet of timber in a week.

To find how much one lumberjack cuts in one week:

Timber cut by one lumberjack in a week=45,0003=15,000 cubic feet\text{Timber cut by one lumberjack in a week} = \frac{45,000}{3} = 15,000 \text{ cubic feet}Timber cut by one lumberjack in a week=345,000​=15,000 cubic feet

Now, for 7 lumberjacks working for 2 weeks:

Total timber cut by 7 lumberjacks in 2 weeks=7×15,000×2=210,000 cubic feet\text{Total timber cut by 7 lumberjacks in 2 weeks} = 7 \times 15,000 \times 2 = 210,000 \text{ cubic feet}Total timber cut by 7 lumberjacks in 2 weeks=7×15,000×2=210,000 cubic feet

Since the question asks for the answer in thousands of cubic feet:

210,0001,000=210\frac{210,000}{1,000} = 2101,000210,000​=210

Answer: D) 210.

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Q) A man covers a distance on a scooter. Had he moved 3kmph faster, he would have taken 40 min less. If he had moved 2kmph slower, he would have taken 40min more. The distance is

A) 30 km

B) 40 km

C) 45 km

D) 50 km

Solution:

Let the distance be ddd km, and let the original speed be vvv km/h.

The time taken to cover the distance at speed vvv is dv\frac{d}{v}vd​ hours.

Case 1: Speed increased by 3 km/h

When the speed increases by 3 km/h, the new time taken is dv+3\frac{d}{v + 3}v+3d​ hours. The difference in time is given as 40 minutes, or 23\frac{2}{3}32​ hours. Therefore, we have the equation:

dv−dv+3=23\frac{d}{v} - \frac{d}{v + 3} = \frac{2}{3}vd​−v+3d​=32​

Case 2: Speed decreased by 2 km/h

When the speed decreases by 2 km/h, the new time taken is dv−2\frac{d}{v - 2}v−2d​ hours. The difference in time is given as 40 minutes, or 23\frac{2}{3}32​ hours. Therefore, we have the equation:

dv−2−dv=23\frac{d}{v - 2} - \frac{d}{v} = \frac{2}{3}v−2d​−vd​=32​

Now, we solve these two equations.

Step 1: Solve the first equation.

dv−dv+3=23\frac{d}{v} - \frac{d}{v + 3} = \frac{2}{3}vd​−v+3d​=32​

Multiply both sides by 3v(v+3)3v(v + 3)3v(v+3):

3d(v+3)−3d(v)=2v(v+3)3d(v + 3) - 3d(v) = 2v(v + 3)3d(v+3)−3d(v)=2v(v+3)

Simplify:

3d⋅3=2v(v+3)3d \cdot 3 = 2v(v + 3)3d⋅3=2v(v+3) 9d=2v2+6v9d = 2v^2 + 6v9d=2v2+6v

Step 2: Solve the second equation.

dv−2−dv=23\frac{d}{v - 2} - \frac{d}{v} = \frac{2}{3}v−2d​−vd​=32​

Multiply both sides by 3v(v−2)3v(v - 2)3v(v−2):

3d(v)−3d(v−2)=2v(v−2)3d(v) - 3d(v - 2) = 2v(v - 2)3d(v)−3d(v−2)=2v(v−2)

Simplify:

3d⋅2=2v2−4v3d \cdot 2 = 2v^2 - 4v3d⋅2=2v2−4v 6d=2v2−4v6d = 2v^2 - 4v6d=2v2−4v

Step 3: Solve both equations.

From equation 1:

9d=2v2+6v9d = 2v^2 + 6v9d=2v2+6v

From equation 2:

6d=2v2−4v6d = 2v^2 - 4v6d=2v2−4v

Now, solve for ddd in both equations, set them equal, and find the value of vvv. After solving for vvv, substitute it into one of the original equations to find ddd.

Answer: B) 40 km.

Read more: Important GK questions for NATA exam

Q) 20, 19, 17, ?, 10, 5

A) 15

b) 14

C) 13

D) 12

Solution:

The given number series is: 20, 19, 17, ?, 10, 5.

To find the missing number, let's check the pattern of the differences between consecutive numbers:

• 20 to 19: Decrease by 1.
• 19 to 17: Decrease by 2.
• ? to 10: Decrease by 7.
• 10 to 5: Decrease by 5.

We see that the differences are: -1, -2, ?, -7, -5.

The missing difference between 17 and ? seems to be -3, based on the alternating pattern in the differences.

Thus, the missing number is:

17−3=1417 - 3 = 1417−3=14

Answer: B) 14.

Q) An express train traveled at an average speed of 100 km/hr, stopping for 3 minutes after every 75 km. How long did it take to reach its destination 600 km from the starting point?

A) 6 hrs 30 min

B) 6 hrs 49 min

C) 6 hrs 45 min

D) 6 hrs 21 min

Solution:

Given:

• Distance to travel = 600 km
• Speed of the train = 100 km/h
• The train stops for 3 minutes after every 75 km.

Step 1: Calculate the time taken for travel without any stops.

The time to travel 600 km at a speed of 100 km/h is:

Time=DistanceSpeed=600100=6 hours.\text{Time} = \frac{\text{Distance}}{\text{Speed}} = \frac{600}{100} = 6 \text{ hours}.Time=SpeedDistance​=100600​=6 hours.

Step 2: Calculate the number of stops made.

The train travels 75 km before stopping, and the total distance is 600 km. The number of stops made is:

Number of stops=60075−1=8−1=7 stops.\text{Number of stops} = \frac{600}{75} - 1 = 8 - 1 = 7 \text{ stops}.Number of stops=75600​−1=8−1=7 stops.

(The subtraction of 1 accounts for the fact that the train doesn't stop after the final segment.)

Step 3: Calculate the total time spent stopping.

Each stop lasts 3 minutes, so the total stopping time is:

Total stopping time=7×3=21 minutes.\text{Total stopping time} = 7 \times 3 = 21 \text{ minutes}.Total stopping time=7×3=21 minutes.

Step 4: Calculate the total time taken.

The total time taken is the travel time plus the stopping time:

Total time=6 hours+21 minutes=6 hours21 minutes.\text{Total time} = 6 \text{ hours} + 21 \text{ minutes} = 6 \text{ hours} 21 \text{ minutes}.Total time=6 hours+21 minutes=6 hours21 minutes.

Answer: D) 6 hrs 21 min.

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Q) 12 men can complete work in 18 days. Six days after they started working, 4 more men joined them. How many days will all of them together complete the remaining work? 

A) 10 days

B) 8 days

C) 11 days

D) 9 days

Solution:

Given:

• 12 men can complete the work in 18 days.
• After 6 days, 4 more men join, so there are 16 men working now.

Step 1: Calculate the total work in man-days.

The total work in man-days is:

Total work=12 men×18 days=216 man-days.\text{Total work} = 12 \text{ men} \times 18 \text{ days} = 216 \text{ man-days}.Total work=12 men×18 days=216 man-days.

Step 2: Calculate the work done in the first 6 days.

In the first 6 days, 12 men work, so the work done is:

Work done in 6 days=12×6=72 man-days.\text{Work done in 6 days} = 12 \times 6 = 72 \text{ man-days}.Work done in 6 days=12×6=72 man-days.

Step 3: Calculate the remaining work.

The remaining work is:

Remaining work=216−72=144 man-days.\text{Remaining work} = 216 - 72 = 144 \text{ man-days}.Remaining work=216−72=144 man-days.

Step 4: Calculate the number of days needed to complete the remaining work.

Now, 16 men are working on the remaining 144 man-days of work. The time required to complete the remaining work is:

Time=Remaining workNumber of men=14416=9 days.\text{Time} = \frac{\text{Remaining work}}{\text{Number of men}} = \frac{144}{16} = 9 \text{ days}.Time=Number of menRemaining work​=16144​=9 days.

Answer: D) 9 days

Q) A two-digit number is three times the sum of its digits. If 45 is added to it, the digits are reversed. The number is 

A) 23

B) 32

C) 27

D) 72

Solution:

Let the two-digit number be represented as 10a+b10a + b10a+b, where:

• aaa is the tens digit,
• bbb is the ones digit.

First condition:

The number is three times the sum of its digits, so:

10a+b=3(a+b)10a + b = 3(a + b)10a+b=3(a+b)

Expanding the equation:

10a+b=3a+3b10a + b = 3a + 3b10a+b=3a+3b

Simplifying:

10a−3a=3b−b10a - 3a = 3b - b10a−3a=3b−b 7a=2b(Equation 1).7a = 2b \quad \text{(Equation 1)}.7a=2b(Equation 1).

Second condition:

If 45 is added to the number, the digits are reversed. So:

10a+b+45=10b+a10a + b + 45 = 10b + a10a+b+45=10b+a

Simplifying:

10a+b+45=10b+a10a + b + 45 = 10b + a10a+b+45=10b+a 10a−a=10b−b−4510a - a = 10b - b - 4510a−a=10b−b−45 9a=9b−459a = 9b - 459a=9b−45 a=b−5(Equation 2).a = b - 5 \quad \text{(Equation 2)}.a=b−5(Equation 2).

Solving the system of equations:

Substitute a=b−5a = b - 5a=b−5 into Equation 1:

7(b−5)=2b7(b - 5) = 2b7(b−5)=2b

Expanding:

7b−35=2b7b - 35 = 2b7b−35=2b

Simplifying:

7b−2b=357b - 2b = 357b−2b=35 5b=355b = 355b=35 b=7.b = 7.b=7.

Substitute b=7b = 7b=7 into Equation 2:

a=7−5=2.a = 7 - 5 = 2.a=7−5=2.

Thus, the number is:

10a+b=10(2)+7=27.10a + b = 10(2) + 7 = 27.10a+b=10(2)+7=27.

Answer: C) 27.

Read more: Enhance your preparation with the best online coaching for NATA

Q) A square garden has fourteen posts along each side at equal intervals. Find how many posts are there on all four sides.

A) 56

B) 44

C) 52

D) 60

Solution:

Given that a square garden has 14 posts along each side at equal intervals, and we need to calculate the total number of posts on all four sides.

Since the posts are placed along the sides, there will be 14 posts on each of the four sides. However, the posts at the corners are shared by two sides. Therefore, we need to account for these corner posts only once.

• On each side, there are 14 posts, and there are 4 sides, so if we just multiply, we get:

4×14=56 posts.4 \times 14 = 56 \text{ posts}.4×14=56 posts.

• But we have 4 corner posts that are counted twice, so we subtract 4 from 56 to avoid double-counting them:

56−4=52 posts.56 - 4 = 52 \text{ posts}.56−4=52 posts.

Answer: C) 52.

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Conclusion

Numerical Reasoning Questions for NATA are a vital component that tests an aspiring architect's mathematical prowess and logical problem-solving skills. These questions reflect the practical mathematical skills necessary in the field of architecture, ensuring candidates are well-prepared for the challenges they will face in their careers. By focusing on strengthening your mathematical foundation, practicing regularly, and utilizing various resources, you can excel in Numerical Reasoning Questions for NATA.

Remember, consistent practice and strategic preparation are the keys to mastering this section and boosting your chances of success in the exam. With diligence and the right approach, you can enhance your numerical reasoning abilities and pave the way for a promising career in architecture.

Read more: Logical reasoning questions and answers for NATA Exam