IBPS RRB Clerk2022Numerical AbilityPage 6 of 7

IBPS RRB Clerk Prelims 2022 Numerical Ability Questions with Solutions

Questions from every shift of the year, grouped by section. Questions 251–300 of 308, each with the correct answer and a step-by-step explanation.

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251Numerical Ability
If the radius and height of a cylinder is increased by 25% and 30% respectively, then find out the percent change in its lateral surface area?
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Answer: (D) 62.50%

Explanation

Lateral surface area of a cylinder is given by LSA = 2 × π × r × h. The net percentage change in the product of two variables increased by a% and b% is given by a + b + (a × b) / 100. Substituting a = 25% and b = 30%:

Net Change = 25 + 30 + (25 × 30) / 100 = 55 + 7.5 = 62.5%

252Numerical Ability
Sum of ages of Mohit and Riya five years hence will be 60 years. If the ratio of ages of Riya and Mohit 5 years ago was 3:5 respectively. Find out the present age of Riya?
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Answer: (D) 20 years

Explanation

The sum of present ages of Mohit and Riya is 60 - 2 × 5 = 50 years. Five years ago, the sum of their ages was 50 - 2 × 5 = 40 years. Given the ratio of Riya's age to Mohit's age five years ago was 3:5, Riya's age five years ago was (3 / 8) × 40 = 15 years. Therefore, Riya's present age is 15 + 5 = 20 years.

253Numerical Ability
Find out the compound interest on a sum of Rs. 36000 at the rate of 40% per annum for nine months, if the interest is compounded quarterly?
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Answer: (A) 11916 Rs.

Explanation

With interest compounded quarterly, the rate per quarter is r = 40% / 4 = 10%, and nine months equal n = 3 quarters. Total amount A = P × (1 + r / 100)ⁿ = 36000 × (1.1)³ = 36000 × 1.331 = 47916. Compound interest is A - P = 47916 - 36000 = 11916.

254Numerical Ability
Priya alone and Shushma alone can complete same work in 8 days and (x + 2) days respectively. If they work together, then they can complete the same work in days. Find out the value of x ?
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Answer: (B) 10

Explanation

The sentence states that Priya and Sushma working together can complete the work in days, omitting the required numerical time period. Without this value, x cannot be calculated.

255Numerical Ability
The ratio of upstream speed of a boat and speed of current is 4:1. If in downstream the boat can cover 60 km in 2.5 hours, then find out the speed of current?
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Answer: (B) 4 kmph

Explanation

Downstream speed is 60 / 2.5 = 24 km/h. Let speed of boat in still water be b and current speed be c. Upstream speed is b - c. Given (b - c) / c = 4 / 1, which gives b = 5c. Downstream speed is b + c = 5c + c = 6c. Setting 6c = 24 gives c = 4 km/h.

256Numerical Ability
If x% of 3y = y% of 2z , then find the value of (x + z)/(z) ?
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Answer: (C) 1⅔

Explanation

The relation x% of 3y = y% of 2z simplifies as (x / 100) × 3y = (y / 100) × 2z, which reduces to 3x = 2z or x / z = 2 / 3. Expressing (x + z) / z as (x / z) + 1 yields (2 / 3) + 1 = 5 / 3 = 1⅔.

257Numerical Ability
Ram bought 54 bananas for Rs. 220, and he sold these bananas for Rs. 50 per dozen, find out his total profit (in Rs.).
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Answer: (A) 5 Rs

Explanation

Ram bought 54 bananas for 220. Since 12 bananas make 1 dozen, 54 bananas equal 54 / 12 = 4.5 dozens. The total selling price at 50 per dozen is 4.5 × 50 = 225. Total profit is 225 - 220 = 5.

258Numerical Ability
Pinky and Santosh started a business with investments of Rs. 45000 and Rs. 55000 respectively. Pinky left the business after 9 months, after one- year Santosh gets Rs. 8800 as his profit. find out the total profit (in Rs.).
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Answer: (B) 14200

Explanation

The profit ratio of Pinky to Santosh is (45000 × 9) : (55000 × 12) = 405 : 660 = 27 : 44. Santosh's share of 44 parts corresponds to 8800, so 1 part = 8800 / 44 = 200. Total profit corresponds to 27 + 44 = 71 parts, which equals 71 × 200 = 14200.

259Numerical Ability
When numerator of a fraction is increased by 5 and denominator of the fraction is doubled, then fraction becomes ¾ . Find the original fraction?
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Answer: (E) Cannot be determined.

Explanation

Let the fraction be n / d. The condition gives (n + 5) / (2d) = 3 / 4, which simplifies to 2n - 3d = -10. Because there are infinitely many pairs of (n, d) satisfying a single linear equation in two variables, the original fraction cannot be uniquely determined.

260Numerical Ability
The population of Delhi in 2020 is 14,58,000. Find population of Delhi in 2017, if population of Delhi decreased by 10% in every year?
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Answer: (A) 20 lacs

Explanation

Let the population in 2017 be P. With a 10% annual decrease for 3 years, population in 2020 is P × (1 - 0.10)³ = P × 0.729. Setting P × 0.729 = 14,58,000 gives P = 14,58,000 / 0.729 = 20,00,000, which is 20 lacs.

261Numerical Ability
Train Y crosses train X in 180 seconds while running in same direction and train X crosses a pole in 25 seconds. If length and speed of train Y is 400 meters and 90 km/hr. respectively, then find speed (in m/sec) of train X.
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Answer: (A) 20

Explanation

Let the speed of train X be S m/s and its length be L meters. Since train X crosses a pole in 25 seconds, L = 25 × S. The speed of train Y is 90 × (5/18) = 25 m/s, and its length is 400 meters. When running in the same direction, relative speed is (25 - S) m/s. Using the relative speed formula, (400 + 25 × S) / (25 - S) = 180. Solving gives 400 + 25 × S = 4500 - 180 × S, which simplifies to 205 × S = 4100, yielding S = 20 m/s.

262Numerical Ability
What should come in place of question mark (? ) in the following questions. √(121) +√(961) -√(289) = ?²
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Answer: (B) 5

Explanation

Evaluating each square root gives √121 = 11, √961 = 31, and √289 = 17. Substituting these values into the expression yields 11 + 31 - 17 = 25. Taking the square root of 25 gives 5.

263Numerical Ability
What should come in place of question mark (? ) in the following questions. 20% of240 + 18% of200 = ?
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Answer: (E) 84

Explanation

Calculate 20% of 240 = 0.20 × 240 = 48. Calculate 18% of 200 = 0.18 × 200 = 36. Adding the two results together gives 48 + 36 = 84.

264Numerical Ability
What should come in place of question mark (? ) in the following questions. ¹²¹⁄₇ = (?)/(144)
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Answer: (B) 132

Explanation

Cross-multiplying the proportion 121 / ? = ? / 144 gives ?² = 121 × 144. Taking the square root on both sides yields ? = √(121 × 144) = 11 × 12 = 132.

265Numerical Ability
What should come in place of question mark (? ) in the following questions. 725÷ √(625) +⅖× 600 = ?
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Answer: (A) 269

Explanation

First evaluate √625 = 25. Then perform the division 725 ÷ 25 = 29. Next, calculate (2/5) × 600 = 240. Adding these two terms together gives 29 + 240 = 269.

266Numerical Ability
What should come in place of question mark (? ) in the following questions. [12× (1.9 + 2.1)] - 12 = ?²
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Answer: (C) 6

Explanation

Inside the brackets, 1.9 + 2.1 = 4. Multiplying by 12 gives 12 × 4 = 48. Subtracting 12 yields 48 - 12 = 36. Taking the square root gives √36 = 6.

267Numerical Ability
What should come in place of question mark (? ) in the following questions. (2343 + 11) + (126 + 3) = ?
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Answer: (D) 255

Explanation

Performing the first division gives 2343 ÷ 11 = 213. Performing the second division gives 126 ÷ 3 = 42. Summing these two values gives 213 + 42 = 255.

268Numerical Ability
What should come in place of question mark (? ) in the following questions. 14²⁄₇% of350 - ⅔× ? = 30
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Answer: (B) 30

Explanation

Convert 14(2/7)% to its fractional equivalent 1/7. Calculate (1/7) × 350 = 50. Substituting into the equation gives 50 - (2/3) × ? = 30. Rearranging gives (2/3) × ? = 20, which yields ? = 20 × (3/2) = 30.

269Numerical Ability
What should come in place of question mark (? ) in the following questions. (42× 12)/(36× 7) +√(121) = ?
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Answer: (A) 13

Explanation

Simplify the fraction (42 × 12) / (36 × 7) = (42 / 7) × (12 / 36) = 6 × (1/3) = 2. Evaluate √121 = 11. Adding the two results together gives 2 + 11 = 13.

270Numerical Ability
What should come in place of question mark (? ) in the following questions. 20% of10% of900 + ⁸⁴⁄₁₂ = ?²
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Answer: (E) 5

Explanation

First evaluate 10% of 900 = 90, then 20% of 90 = 18. Next, calculate 84 ÷ 12 = 7. Adding the two terms gives 18 + 7 = 25. Taking the square root yields √25 = 5.

271Numerical Ability
What should come in place of question mark (? ) in the following questions. 5555÷ 11÷ 5 = 100 + ?
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Answer: (A) 1

Explanation

Applying the order of operations for division, the expression on the left-hand side is evaluated sequentially from left to right. First, calculate 5555 ÷ 11 = 505. Next, divide this result by 5 to obtain 505 ÷ 5 = 101. Equating this to the right-hand side gives 101 = 100 + ?. Subtracting 100 from both sides gives ? = 101 - 100 = 1.

272Numerical Ability
What should come in place of question mark (? ) in the following questions. 2⅓ +2⅚ - 1½ = ? - 6⅙
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Answer: (B) 9⅚

Explanation

Convert all mixed fractions to improper fractions with a common denominator of 6. The left-hand side terms become 2 ⅓ = 7/3 = 14/6, 2 ⅚ = 17/6, and 1 ½ = 3/2 = 9/6. Combining these gives 14/6 + 17/6 - 9/6 = 22/6. The term on the right-hand side is -6 ⅙ = -37/6. Transposing -37/6 to the left-hand side yields ? = 22/6 + 37/6 = 59/6. Converting 59/6 back into a mixed fraction gives 9 ⅚.

273Numerical Ability
What should come in place of question mark (? ) in the following questions. 1024÷ 32 = 2½× ?
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Answer: (D) 10

Explanation

First, calculate the value of the left-hand side by dividing 1024 by 32, which equals 32. Next, rewrite the mixed fraction on the right-hand side, 2 ½, as an improper fraction 5/2, or decimal 2.5. The equation then becomes 32 = (5/2) × ?. To solve for ?, multiply both sides by the reciprocal 2/5, giving ? = 32 × (2/5) = 64/5 = 12.8.

274Numerical Ability
What should come in place of question mark (? ) in the following questions. 1231 + 1312 + 2113 - 3211 = ?
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Answer: (C) 1445

Explanation

To simplify the expression, perform the additions sequentially from left to right before applying subtraction. Adding the first two terms gives 1231 + 1312 = 2543. Adding the third term yields 2543 + 2113 = 4656. Finally, subtracting the last term gives 4656 - 3211 = 1445. Therefore, the value that replaces the question mark is 1445.

275Numerical Ability
If y² - x² = 72 and y - x = 4, then find the value of (x × y)?
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Answer: (D) 77

Explanation

Using the identity y² - x² = (y - x)(y + x), substituting the given values yields 72 = 4 × (y + x), which gives y + x = 18. Solving the system of linear equations y + x = 18 and y - x = 4 by addition gives 2y = 22, so y = 11. Subtracting the equations gives 2x = 14, so x = 7. The product x × y = 7 × 11 = 77.

276Numerical Ability
The ratio of length to breadth of a rectangle is 7 : 3 and the perimeter of the rectangle is 40 cm. Find the area (in cm²) of rectangle?
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Answer: (E) 84

Explanation

Let the length be 7x and the breadth be 3x. The perimeter of a rectangle is 2 × (length + breadth) = 2 × (7x + 3x) = 20x. Given perimeter is 40 cm, so 20x = 40, which gives x = 2. Length = 7 × 2 = 14 cm and breadth = 3 × 2 = 6 cm. Area of rectangle = length × breadth = 14 × 6 = 84 cm².

277Numerical Ability
A man invested Rs.1460 in each scheme A and B. If scheme A offers simple interest at 10% p.a. for five years and scheme B offers simple interest at 10% p.a. for eight years, then find difference between interest (in Rs.) received from both schemes?
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Answer: (A) 438

Explanation

Simple Interest SI = (P × R × T) / 100. Since the principal P = ₹1460 and annual rate R = 10% are identical in both schemes, the difference in interest arises from the difference in duration ΔT = 8 - 5 = 3 years. Difference in interest = (1460 × 10 × 3) / 100 = 1460 × 0.30 = ₹438.

278Numerical Ability
A man purchased two articles at same cost price and he sold one article at 18% loss and second at 30% profit. If total profit earned is Rs.210, then find the cost price (in Rs.) of each article?
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Answer: (D) 1750

Explanation

Let the cost price of each article be C. The profit on the second article is 0.30 × C and the loss on the first article is 0.18 × C. Net profit = 0.30 × C - 0.18 × C = 0.12 × C. Given net profit is ₹210, so 0.12 × C = 210. C = 210 / 0.12 = 1750. The cost price of each article is ₹1750.

279Numerical Ability
A man purchase three different articles A, B | C and cost prices of these articles is Rs.52, Rs.78 and Rs.108 respectively. If number of each type of article purchase by man is equal and man spends total of Rs.1190, then find the number of each type of article purchased by the man?
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Answer: (E) 5

Explanation

The combined cost of purchasing one unit of each article A, B, and C is 52 + 78 + 108 = ₹238. Let n be the equal number of each type of article purchased. Total cost = 238 × n = 1190. n = 1190 / 238 = 5. The man purchased 5 of each type of article.

280Numerical Ability
The ratio of present age of A to that of B is 5 : 4 respectively. Four years hence, age of A will be three years more than age of B six years hence. Find the present age of B?
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Answer: (B) 20 years

Explanation

Let the present age of A be 5x years and present age of B be 4x years. Four years hence, age of A = 5x + 4. Six years hence, age of B = 4x + 6. According to the given condition, 5x + 4 = (4x + 6) + 3. Simplifying gives 5x + 4 = 4x + 9, which reduces to x = 5. Present age of B = 4x = 4 × 5 = 20 years.

281Numerical Ability
A vessel contains 120 liters mixture of milk and water in the ratio of 5 : 3. If 20 liters milk added in the vessel, then find the ratio of water to milk in resulting mixture?
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Answer: (D) 9 : 19

Explanation

In 120 liters of mixture with milk and water in 5 : 3 ratio: Initial milk = 120 × (5 / 8) = 75 liters. Initial water = 120 × (3 / 8) = 45 liters. Adding 20 liters of milk gives new milk volume = 75 + 20 = 95 liters. Water volume remains 45 liters. Ratio of water to milk in the resulting mixture = 45 : 95 = 9 : 19.

282Numerical Ability
A car cover 160 km in five hours at the speed of x km/h and same distance cover by another car in 3 hour 20 minutes at the speed of y km/hr. Find the value of (x : y)?
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Answer: (B) 2 : 3

Explanation

Time taken by first car = 5 hours. Time taken by second car = 3 hours 20 minutes = 3 + 20 / 60 = 10 / 3 hours. Ratio of time taken = 5 : (10 / 3) = 15 : 10 = 3 : 2. Since distance covered is identical, speed is inversely proportional to time, making x : y = 2 : 3.

283Numerical Ability
A boat takes five hours to cover certain distance in upstream and takes two hours to cover the same distance in downstream. If the speed of the stream is 3 km/hr, then find the distance?
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Answer: (D) 20 km

Explanation

Let speed of boat in still water be v km/hr. Stream speed s = 3 km/hr. Upstream speed = v - 3 and downstream speed = v + 3. Since distance = speed × time, equating upstream and downstream distances gives: 5 × (v - 3) = 2 × (v + 3) 5v - 15 = 2v + 6 3v = 21, so v = 7 km/hr. Distance = 5 × (7 - 3) = 20 km.

284Numerical Ability
A is 40% more efficient than B. If B takes 6 days more than A to complete a work, then in how many days 'A' alone can complete the whole work?
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Answer: (B) 15 days

Explanation

Efficiency ratio of A to B = 140 : 100 = 7 : 5. Since efficiency is inversely proportional to time taken, time ratio of A to B = 5 : 7. Let time taken by A be 5x days and B be 7x days. Given time difference 7x - 5x = 2x = 6 days, so x = 3. Time taken by A alone = 5x = 5 × 3 = 15 days.

285Numerical Ability
What will come in the place of question (?) mark in following number series - 2, 4, 7, 12, 19, ?
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Answer: (B) 30

Explanation

To find the missing term in the number series, calculate the difference between consecutive terms: 4 - 2 = 2 7 - 4 = 3 12 - 7 = 5 19 - 12 = 7 The differences form a sequence of consecutive prime numbers starting from 2, which are 2, 3, 5, and 7. The next prime number following 7 is 11. Adding this next prime difference of 11 to the preceding term 19 gives the missing number: 19 + 11 = 30.

286Numerical Ability
What will come in the place of question (?) mark in following number series - 67, 98, 129, 160, ?, 222
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Answer: (C) 191

Explanation

To determine the missing term in the given number series, compute the common difference between successive terms: 98 - 67 = 31 129 - 98 = 31 160 - 129 = 31 This establishes that the series is an arithmetic progression with a constant common difference of 31. Adding 31 to 160 gives the missing value: 160 + 31 = 191. Verifying with the subsequent term, 191 + 31 = 222, confirms the pattern.

287Numerical Ability
What will come in the place of question (?) mark in following number series - 10, 18, 45, 109, ?, 450
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Answer: (C) 234

Explanation

To solve the number series, examine the differences between adjacent terms: 18 - 10 = 8 = 2³ 45 - 18 = 27 = 3³ 109 - 45 = 64 = 4³ The differences are consecutive perfect cubes starting from 2³. The next term in the pattern requires adding 5³, which equals 125, to 109. Calculating 109 + 125 gives 234. Verifying with the final term, 234 + 6³ = 234 + 216 = 450, confirms that 234 is correct.

288Numerical Ability
What will come in the place of question (?) mark in following number series - 12, 16, 25, 41, ?, 102
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Answer: (A) 66

Explanation

To find the missing term in the given sequence, evaluate the difference between consecutive terms: 16 - 12 = 4 = 2² 25 - 16 = 9 = 3² 41 - 25 = 16 = 4² The sequence of differences consists of perfect squares of consecutive integers starting from 2². Following this pattern, the next difference must be 5² = 25. Adding 25 to 41 yields 41 + 25 = 66. Verifying the subsequent term gives 66 + 6² = 66 + 36 = 102.

289Numerical Ability
What will come in the place of question (?) mark in following number series - 15, 20, 25, 30, ?, 40
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Answer: (B) 35

Explanation

To determine the missing number in the sequence, analyze the difference between each pair of consecutive terms: 20 - 15 = 5 25 - 20 = 5 30 - 25 = 5 The given series is an arithmetic progression where each term increases by a fixed constant of 5. Adding 5 to the term 30 yields 30 + 5 = 35. Verifying with the next term, 35 + 5 = 40, confirms that 35 correctly completes the series.

290Numerical Ability
Table given below shows total three types (Sunflowers, lilies and Roses) of flowers used by an event management company in three different months. Read the data carefully and answer the questions. Flowers | January | February | March Sunflowers | 68 | 112 | 96 Lilies | 84 | 64 | 80 Roses | 72 | 88 | 118 Total number of sunflowers used in January | March together are what percentage more than total number of roses used in January | February together?
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Answer: (A) 2.5%

Explanation

To calculate the required percentage excess: Total sunflowers used in January and March = 68 + 96 = 164. Total roses used in January and February = 72 + 88 = 160. The difference between total sunflowers (January and March) and total roses (January and February) is 164 - 160 = 4. Percentage more = [(Total Sunflowers - Total Roses) / Total Roses] × 100 = (4 / 160) × 100 = 2.5%.

291Numerical Ability
Table given below shows total three types (Sunflowers, lilies and Roses) of flowers used by an event management company in three different months. Read the data carefully and answer the questions. Flowers | January | February | March Sunflowers | 68 | 112 | 96 Lilies | 84 | 64 | 80 Roses | 72 | 88 | 118 Find the average number of all three types of flowers used by event management company in March?
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Answer: (E) 98

Explanation

To find the average number of all three types of flowers used by the company in March, sum the values for sunflowers, lilies, and roses in March and divide by 3: Sunflowers in March = 96 Lilies in March = 80 Roses in March = 118 Total flowers in March = 96 + 80 + 118 = 294. Average = Total Flowers / Number of Flower Types = 294 / 3 = 98.

292Numerical Ability
Table given below shows total three types (Sunflowers, lilies and Roses) of flowers used by an event management company in three different months. Read the data carefully and answer the questions. Flowers | January | February | March Sunflowers | 68 | 112 | 96 Lilies | 84 | 64 | 80 Roses | 72 | 88 | 118 Find the ratio of total number of roses used in January to total number of lilies used in January?
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Answer: (C) 6 : 7

Explanation

To find the ratio of total roses used in January to total lilies used in January, extract the required values from the provided table: Number of roses used in January = 72 Number of lilies used in January = 84 Ratio = Roses in January : Lilies in January = 72 : 84. Dividing both terms by their greatest common divisor, 12, simplifies the ratio: 72 ÷ 12 = 6 84 ÷ 12 = 7 The simplified ratio is 6 : 7.

293Numerical Ability
Line graph shows total quantity (in kg) of five different dry fruits sold by a shop. The values are: Almond = 60 kg, Cashew = 80 kg, Walnut = 50 kg, Raisin = 90 kg, Pistachio = 40 kg. Read the data carefully and answer the questions. Find the difference between total Walnut | Raisin sold by the shop and total Cashew | Pistachio sold by the shop?
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Answer: (D) 20 kg

Explanation

Total quantity of Walnut and Raisin sold = 50 kg + 90 kg = 140 kg.

Total quantity of Cashew and Pistachio sold = 80 kg + 40 kg = 120 kg.

Required difference = 140 kg - 120 kg = 20 kg.

294Numerical Ability
What value should come in place of (?) in the following questions? (510)/(?) = √(324) + √(256)
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Answer: (D) 15

Explanation

To solve for the unknown variable x in the algebraic equation 510 / x = √324 + √256, compute the principal square roots of the terms on the right-hand side first: √324 = 18 √256 = 16 Substituting these simplified square root values back into the expression gives: 510 / x = 18 + 16 510 / x = 34 Rearranging the equation to solve for x yields x = 510 / 34 = 15.

295Numerical Ability
What value should come in place of (?) in the following questions? 3 × (?)² = (5)² + (7)² + (17)²
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Answer: (D) 11

Explanation

Evaluating the right-hand side of the equation: 5² + 7² + 17² = 25 + 49 + 289 = 363

Substituting this back into the equation: 3 × (?)² = 363 (?)² = 363 ÷ 3 = 121 ? = √121 = 11

296Numerical Ability
What value should come in place of (?) in the following questions? 2² = 40% of 420 + 44% of 200
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Answer: (D) 16

Explanation

Calculate the individual percentages on the right side of the equation: 40% of 420 = (40 ÷ 100) × 420 = 168 44% of 200 = (44 ÷ 100) × 200 = 88

Adding these two values gives: 168 + 88 = 256

The equation simplifies to: (?)² = 256

Taking the square root of both sides yields: ? = √256 = 16

297Numerical Ability
What value should come in place of (?) in the following questions? 2 ⅓ of 4 ²⁄₇ of 3 ⅓ of 81 = ?
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Answer: (C) 2700

Explanation

Convert each mixed fraction into an improper fraction: 2 ⅓ = 7/3 4 2/7 = 30/7 3 ⅓ = 10/3

Now substitute these values into the product: (7/3) × (30/7) × (10/3) × 81

Multiplying the first two terms gives: (7/3) × (30/7) = 10

Next, multiply 10 by the remaining terms: 10 × (10/3) × 81 = (100 ÷ 3) × 81 = 100 × 27 = 2700

298Numerical Ability
What value should come in place of (?) in the following questions? √(256) × 49 + (16)² - ³√(343) = (?)²
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Answer: (E) 19

Explanation

Evaluate each term step by step: √256 = 16 √49 = 7 16² = 256 ∛343 = 7

Substitute these calculated values back into the expression: (16 × 7) + 256 - 7

Perform multiplication before addition and subtraction: 112 + 256 - 7 = 361

The equation becomes: (?)² = 361

Taking the square root of both sides gives: ? = √361 = 19

299Numerical Ability
What value should come in place of (?) in the following questions? √(256) × √(169) + 3600 ÷ 12 = 800 - ?
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Answer: (C) 292

Explanation

Calculate each component of the expression on the left side: √256 = 16 √169 = 13 16 × 13 = 208 3600 ÷ 12 = 300

Adding these results gives: 208 + 300 = 508

Set up the simplified equation: 508 = 800 - ?

Subtracting 508 from 800 yields the required answer: ? = 800 - 508 = 292

300Numerical Ability
What value should come in place of (?) in the following questions? 37.5 × 14 + 800 - (26)² + 136 = ?
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Answer: (A) 785

Explanation

Follow the order of operations to simplify the left side of the equation: 37.5 × 14 = 525 26² = 676

Substitute these values back into the full expression: 525 + 800 - 676 + 136

Perform addition and subtraction from left to right: 525 + 800 = 1325 1325 - 676 = 649 649 + 136 = 785

Answer key

Tap a number to jump to that question.

  1. 251D
  2. 252D
  3. 253A
  4. 254B
  5. 255B
  6. 256C
  7. 257A
  8. 258B
  9. 259E
  10. 260A
  11. 261A
  12. 262B
  13. 263E
  14. 264B
  15. 265A
  16. 266C
  17. 267D
  18. 268B
  19. 269A
  20. 270E
  21. 271A
  22. 272B
  23. 273D
  24. 274C
  25. 275D
  26. 276E
  27. 277A
  28. 278D
  29. 279E
  30. 280B
  31. 281D
  32. 282B
  33. 283D
  34. 284B
  35. 285B
  36. 286C
  37. 287C
  38. 288A
  39. 289B
  40. 290A
  41. 291E
  42. 292C
  43. 293D
  44. 294D
  45. 295D
  46. 296D
  47. 297C
  48. 298E
  49. 299C
  50. 300A

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