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Number System Questions for CTET mcq
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Number System mcq for Students
The number 49532 rounded to the nearest thousand is
a. 41000
b. 50000
c. 49000
d. 49500
Explanation: To round 49532 to the nearest thousand, focus on the hundreds digit. Numbers are rounded up when this digit is 5 or greater, and rounded down if less than 5.
In 49532, the hundreds digit plays a critical role. Based on its value, the thousands digit may need adjustment, and all digits to the right are converted to zero. This method is useful for estimating large numbers, simplifying calculations, and representing quantities in a more manageable form.
Rounding is widely used in finance, statistics, and daily approximations to simplify reporting while retaining the scale of the number.
Analogously, think of rounding like estimating the number of people in a stadium: if slightly above a round figure, rounding up gives a simpler representation of the crowd size.
The concept emphasizes checking adjacent digits, following standard rounding rules, and adjusting the number accordingly to maintain overall magnitude and practicality.
Option b – 50000
The difference between 5671 and the number formed by reversing its digits is
a. 4906
b. 3916
c. 7436
d. 3906
Explanation: Reversing the digits of a number alters the place values of each digit, producing a different numerical value. The difference between the original and reversed numbers highlights the impact of digit positions.
The process involves identifying the original number and its reversed counterpart, then performing subtraction. Aligning digits by place value is crucial, borrowing as needed from higher digits to ensure accuracy.
This technique demonstrates the significance of positional value in the number system. It also tests arithmetic skills and careful attention to the impact of each digit’s location.
Imagine flipping a number like reflecting it in a mirror; its composition changes, but each digit still contributes according to its position.
The concept reinforces place value understanding and careful calculation while showing how the arrangement of digits affects overall magnitude.
Option d – 3906
The total of place values of 6 in 63606 is
a. 60606
b. 6606
c. 6066
d. 18
Explanation: Each digit in a number has a place value determined by its position. To find the total place value of a digit appearing multiple times, calculate the value contributed by each occurrence and sum them.
In 63606, the digit 6 appears in different positions, each affecting the total differently. Understanding this helps with comprehension of how numbers are constructed and emphasizes the importance of positional notation.
This approach is useful in checking calculations, solving arithmetic problems, and analyzing number composition.
For example, if you think of each digit as a weighted block, its weight depends on its position, and the total effect is the sum of all contributions.
The method illustrates how repeated digits contribute cumulatively to the number’s overall value.
Option a – 60606
Adding 19 thousand, 19 hundred, and 19 units gives
a. 21090
b. 20919
c. 19919
d. 191919
Explanation: When adding quantities of different magnitudes, each component must be expressed in its numeric value before combining. Thousands, hundreds, and units contribute differently to the total sum.
19 thousand, 19 hundred, and 19 units are first converted to 19000, 1900, and 19. Summing these maintains proper place value alignment, ensuring accuracy.
This exercise reinforces understanding of magnitude, addition with multiple terms, and careful attention to place values.
It is similar to combining weights of objects: each object has a different scale, and the total is obtained by summing all correctly.
The concept emphasizes careful conversion of terms to numeric form and the cumulative addition of values across different scales.
Option b – 20919
Which statement is correct?
a. Predecessor of the successor of 1000 is 1000
b. Predecessor of the predecessor of 1000 is 999
c. Successor of the predecessor of 1000 is 1001
d. Successor of the predecessor of 1000 is 1002
Explanation: Understanding predecessor and successor involves knowing that the predecessor of a number is one less, and the successor is one more. Combinations of these operations can create sequences of numbers.
For instance, applying predecessor then successor or successor then predecessor must be considered step by step to evaluate the correctness of statements.
This requires careful attention to arithmetic sequence rules and understanding of basic number operations.
Analogously, it’s like taking a step forward and then back: the final position depends on the sequence of movements, not just individual steps.
The exercise reinforces logical reasoning with numbers and the ability to track changes in value systematically.
Option a – Predecessor of the successor of 1000 is 1000
A shop has 239 toys. If 70 more are purchased and 152 are sold, how many toys remain?
a. 239+70+152
b. 239-70+152
c. 239+70-152
d. 239-70-152
Explanation: To find remaining toys, the total inventory must first account for new purchases and then deduct items sold. Operations must follow the correct sequence to maintain accuracy.
Starting with 239 toys, adding 70 increases the total stock. Subtracting 152 sold items gives the remaining quantity. Aligning addition and subtraction correctly prevents calculation errors.
This scenario illustrates real-life applications of sequential arithmetic operations and reinforces the concept of NET change in quantities.
Think of it as managing items in a basket: adding new items first and then removing some ensures the final count is accurate.
It demonstrates the importance of order of operations and careful arithmetic when dealing with real-world inventory problems.
Option c – 239+70-152
In the product 3759 × 9573, the sum of the digits in the tens and units place is
a. 7
b. 9
c. 16
d. 0
Explanation: Multiplying large numbers produces digits in various places. The sum of specific digits, such as tens and units, focuses on analyzing only the last two digits of the product.
The process involves understanding place value after multiplication, either through long multiplication or modular arithmetic techniques. This highlights the positional significance of digits within a product.
It’s similar to examining only the last two letters of a long word to understand a pattern without analyzing the entire text.
The exercise reinforces attention to detail and selective analysis of numbers within larger computations.
Option a – 7
If 567567567 is divided by 567, the quotient is
a. 111
b. 10101
c. 1001001
d. 3
Explanation: Dividing a large number by a smaller number requires understanding repetition patterns and place value to simplify calculations. Large numbers can often be broken into repeated blocks for easier division.
The approach may involve recognizing repeated sequences, estimating contribution from each block, and summing partial quotients. It emphasizes efficiency in arithmetic and recognizing numerical patterns.
This is akin to dividing a repeated pattern of objects into equal groups to determine how many full groups can be formed.
The concept reinforces both division skills and pattern recognition in large numbers.
Option c – 1001001
Total of place values of 2 in 2424 is
a. 4
b. 220
c. 2002
d. 2020
Explanation: To find the total of a repeated digit’s place values, determine the value contributed by each occurrence based on its position and sum them.
In 2424, the digit 2 appears in thousands and tens places. Each contributes differently, illustrating the significance of place value in positional notation.
Understanding this principle is crucial for analyzing numbers and performing arithmetic operations accurately.
Think of each 2 as a separate weighted block; their combined weight gives the total contribution.
The problem reinforces comprehension of place value and the cumulative effect of repeated digits in numbers.
Option d – 2020
What must be added to 140 × 101 to get 14414?
a. 264
b. 274
c. 278
d. 364
Explanation: To find the required addition, first compute the product of 140 × 101, then determine the difference between this product and the target value.
The method involves multiplication, careful attention to place values, and subtraction to identify the missing component. This ensures accurate calculation without directly computing the final result.
It is like determining how much more paint is needed to fill a container to a desired level after a partial amount is already added.
This reinforces understanding of multiplication, subtraction, and the concept of missing quantities in problem-solving.
Option b – 274
When 121012 is divided by 12, the remainder is
a. 0
b. 2
c. 3
d. 4
Explanation: Finding the remainder when a number is divided by another involves modular arithmetic, where the focus is on the part left after subtracting multiples of the divisor.
For large numbers, breaking them into smaller chunks according to place values simplifies calculations. Each chunk contributes to the total remainder according to its position. This method is efficient for mental math or checking division results.
It is similar to distributing items into equal boxes and observing what is left over that cannot fill a complete box.
The concept reinforces modular reasoning, division rules, and the significance of place value in evaluating remainders.
Option d – 4
Difference between the place value and face value of 3 in 12345 is
a. 305
b. 0
c. 295
d. 297
Explanation: A digit’s place value is determined by multiplying it by the value of its position, while the face value is the digit itself. The difference highlights how positional notation impacts numerical value.
In 12345, identify the place of 3, compute its place value, and then subtract the face value to understand the discrepancy. This deepens understanding of how digits contribute differently depending on location.
Analogously, think of a worker’s role in an organization: the title (face value) is fixed, but the influence or responsibility (place value) depends on position.
This emphasizes place value comprehension and arithmetic reasoning with multi-digit numbers.
Option d – 297
11 ones + 11 tens + 11 hundreds equals
a. 111111
b. 144
c. 1221
d. 12321
Explanation: Each component represents a different scale in the number system: ones, tens, hundreds. To combine them, convert each to its numeric equivalent and sum all values while maintaining place value integrity.
11 ones = 11, 11 tens = 110, 11 hundreds = 1100. Summing these shows how different magnitude components contribute cumulatively to the total.
This is similar to combining coins of different denominations to get the total amount.
The concept reinforces place value addition, understanding magnitude, and structured arithmetic operations.
Option c – 1221
Sum of 509 and 3028 is
a. 387
b. 3537
c. 3087
d. 837
Explanation: Adding numbers requires aligning digits by place value and performing column-wise addition from right to left, carrying over values when needed.
The process emphasizes careful attention to hundreds, tens, and units to ensure accurate results. Place value alignment prevents errors and is essential for multi-digit arithmetic.
It is analogous to stacking layers of blocks where each layer must align correctly for the structure to remain stable.
This reinforces addition skills, column alignment, and carrying techniques in multi-digit arithmetic.
Option b – 3537
The remainder when 90707 is divided by 9 is
a. 7
b. 3
c. 5
d. 6
Explanation: Divisibility by 9 is tested by summing the digits of a number; the remainder upon division by 9 equals the remainder of the sum of digits divided by 9.
For 90707, sum the digits and reduce modulo 9 to find the remainder. This method simplifies the calculation without performing full division.
It is similar to checking a long string of numbers by summarizing their individual contributions rather than counting each unit individually.
The concept reinforces modular arithmetic and divisibility rules to efficiently compute remainders.
Option c – 5
Product of place values of 5 and 6 in 70560 is
a. 300
b. 3000
c. 30000
d. 30
Explanation: Each digit’s place value is obtained by multiplying the digit by the value of its position. To find the product, compute the place values of 5 and 6 separately and then multiply them.
This approach highlights how positional notation affects calculations and how digits in different positions contribute differently to arithmetic operations.
Think of it like multiplying the weights of two objects whose individual impact depends on their position on a scale.
The exercise emphasizes understanding place value and multiplicative reasoning with multi-digit numbers.
Option c – 30000
Sum of 10 tens, 11 hundreds, and 12 ones is
a. 1213
b. 111012
c. 101112
d. 1212
Explanation: Each quantity represents a different place value: tens, hundreds, ones. Converting each to numeric form (10 tens = 100, 11 hundreds = 1100, 12 ones = 12) and summing them yields the total.
This ensures accurate calculation by respecting the scale of each component.
It is similar to combining measures of different units (meters, centimeters, millimeters) into a single total by converting them to the same unit.
The method reinforces arithmetic with multiple scales and the concept of cumulative value.
Option d – 1212
What must be subtracted from 3001 × 101 to get 300311?
a. 2790
b. 2090
c. 2970
d. 270
Explanation: To determine the missing subtraction, first analyze the multiplication of 3001 × 101 as a Base. Then, consider how much needs to be removed from this product to reach the target value.
This involves multiplication, understanding place values, and carefully identifying the difference needed to achieve the desired result.
Analogous to adjusting a recipe: after preparing a mixture, you remove a certain quantity to achieve a specific weight or volume.
This reinforces reasoning with products, differences, and unknown quantities in arithmetic problems.
Option a – 2790
Quotient of 3010301 ÷ 43 is
a. 70707
b. 70007
c. 7077
d. 7007
Explanation: Division of large numbers can be simplified by breaking the dividend into smaller manageable chunks. Each chunk is divided by 43, and partial quotients are combined.
This method emphasizes efficiency, estimation, and understanding the effect of place value in large-number division.
It is similar to splitting a large shipment into smaller boxes for easier handling and counting.
The exercise reinforces division skills, partial quotient reasoning, and careful attention to place value alignment.
Option b – 70007
When 3488 is divided by 12 and 2478 is divided by 11, the difference of remainders is
a. 5
b. 6
c. 7
d. 3
Explanation: First, find the remainders from each division separately. Subtract the smaller remainder from the larger to determine the difference.
This process involves modular arithmetic and reinforces understanding of remainders and sequential operations.
Analogously, it is like measuring leftover items from two separate batches and comparing them.
The concept strengthens reasoning about division, remainders, and differences between modular outcomes.
Option a – 5
Difference between place value of 5 and its face value in 35362 is
a. 0
b. 495
c. 4995
d. 5005
Explanation: The place value of a digit depends on its position in a number, while the face value is the digit itself. Calculating the difference highlights the impact of position on overall value.
In 35362, the digit 5 appears in a specific place. Compute its place value, then subtract the face value. This helps in understanding positional significance and how it influences arithmetic results.
It is similar to the difference between a person’s title (face value) and the influence their role carries (place value).
The exercise reinforces comprehension of place value, subtraction, and positional reasoning in multi-digit numbers.
Option c – 4995
Sum of the largest 4-digit number and the smallest 3-digit number is
a. 7000
b. 9899
c. 10099
d. 10999
Explanation: The largest 4-digit number is at the upper limit of four digits, while the smallest 3-digit number is the minimal three-digit value. Adding them involves understanding numerical limits and performing standard addition.
This illustrates how numbers of different magnitudes combine and highlights the significance of digit positions in addition.
Think of it as combining a large container with a small one; the total is determined by summing their capacities.
The concept reinforces addition skills and understanding of numerical ranges in different digit classes.
Option c – 10099
10 ones + 10 tens + 10 thousands equals
a. 11100
b. 101010
c. 10110
d. 11011
Explanation: Each component has a specific place value: 10 ones = 10, 10 tens = 100, and 10 thousands = 10000. Summing them requires converting each to numeric value and adding while respecting place values.
This approach emphasizes careful attention to magnitude and proper alignment of terms for accurate addition.
It is analogous to combining coins of different denominations to get the total value correctly.
The problem reinforces understanding of place value addition and the cumulative effect of multiple scales.
Option c – 10110
Twenty-six and twenty-six hundredths is written as
a. 2626
b. 26.26
c. 262.6
d. 2.626
Explanation: Decimal representation combines the whole part with fractional parts. “Twenty-six and twenty-six hundredths” expresses 26 + 26/100. Converting words to numeric form involves correctly positioning the decimal point.
This ensures precise representation of fractions, important in measurements, Money, and scientific calculations.
It is like writing ingredients in a recipe where fractions must be accurately placed relative to whole units.
The concept reinforces understanding of decimals and converting verbal statements into numeric form.
Option b – 26.26
What must be subtracted from 1109 × 505 to get 505050?
a. 49495
b. 55005
c. 54995
d. 59495
Explanation: To find the missing subtraction, first analyze the product 1109 × 505 as a Base. Then determine the difference required to achieve the target number.
This involves multiplication, understanding place value, and careful calculation of differences to maintain accuracy.
It is similar to adjusting the weight of a mixture by removing a specific quantity to reach a desired value.
The exercise reinforces reasoning with products, differences, and missing quantities in arithmetic problems.
Option c – 54995
Sum of place values of 5 in 6251, 6521, and 5621 is
a. 550
b. 15
c. 5550
d. 5050
Explanation: Each 5 in a number contributes differently depending on its position. Calculate the place value of 5 in each number and then sum them to find the total contribution.
This illustrates the importance of positional notation and how repeated digits influence overall numeric value.
Think of each 5 as a weighted object; summing them gives the total weight contributed by all instances.
The problem reinforces arithmetic with place values and understanding cumulative effects of repeated digits.
Option c – 5550
Which of these statements is correct?
a. The product of three odd numbers is an even number
b. The difference of an even and an odd number can be even
c. Sum of two odd numbers and one even number is even
d. Sum of three odd numbers is even
Explanation: Understanding sums and products of odd and even numbers helps identify correct statements. Rules: the product of odd numbers is odd, the sum of two odd numbers is even, and combining odd and even numbers produces specific results.
Evaluating each statement systematically ensures correct identification based on arithmetic properties.
It is similar to checking rules in a game: each action has predictable outcomes that must be verified logically.
The exercise strengthens reasoning with number properties and arithmetic logic.
Option c – Sum of two odd numbers and one even number is even
What must be subtracted from 102 × 201 to get 19999?
a. 503
b. 602
c. 103
d. 401
Explanation: Determine the product of 102 × 201, then calculate the amount needed to subtract to reach the target value. This involves multiplication, place value understanding, and subtraction.
This approach emphasizes identifying missing components in arithmetic sequences and ensures proper alignment of digits for accuracy.
It is like adjusting a recipe: after preparing an initial mixture, a certain amount must be removed to achieve the desired portion.
The concept reinforces reasoning with products, differences, and unknown quantities in calculations.
Option a – 503
Sum of quotient and remainder when 110111 is divided by 11 is
a. 11011
b. 11001
c. 10101
d. 10011
Explanation: Division of a number produces a quotient and a remainder. Adding these requires understanding how division distributes values between full multiples and leftover parts.
Breaking the number into place value components simplifies the calculation, ensuring accuracy.
This is similar to dividing candies into equal boxes and then combining the count of full boxes and remaining pieces.
The concept reinforces division, remainders, and the relationship between quotient and remainder in arithmetic.
Option d – 10011
Ravi has 36 chocolates. He gives one-third to his neighbour, one-sixth to Rehana, and one-fourth to his sister. How many remain?
a. 9
b. 10
c. 6
d. 8
Explanation: Solve by computing fractional portions of the total chocolates given away, then subtracting their sum from the initial quantity.
Stepwise fraction calculations ensure each part is accounted for correctly, respecting the total amount.
This is similar to distributing items among friends and keeping track of what remains after each distribution.
The exercise reinforces fractional operations, sequential subtraction, and careful arithmetic reasoning.
Option a – 9
Difference between place value of 6 and face value of 4 in 56431 is
a. 999
b. 5600
c. 5996
d. 2
Explanation: The place value of a digit is determined by multiplying it by the value of its position, while the face value is the digit itself. To find the difference, compute each value separately and subtract the face value from the place value.
This highlights how positional value changes the contribution of a digit to the overall number and reinforces understanding of the decimal system.
Think of each digit like a worker whose output depends on the position in the workflow; the difference shows the effect of location on overall contribution.
The exercise strengthens reasoning about place value, subtraction, and positional significance in multi-digit numbers.
Option c – 5996
10 thousand + 10 hundred + 10 tens equals
a. 11110
b. 11100
c. 11010
d. 101010
Explanation: Each quantity represents a different magnitude. Convert them to numeric values: 10 thousand = 10000, 10 hundred = 1000, 10 tens = 100. Summing these ensures the total is calculated accurately while respecting place value.
This reinforces structured addition across different scales and understanding of how each component contributes to the final sum.
It is similar to adding different units of measurement, like combining kilograms, grams, and milligrams into a single total.
The concept emphasizes careful addition with varying magnitudes and place value comprehension.
Option b – 11100
Sum of remainders when 12112 ÷ 11 and 13223 ÷ 13 are added is
a. 3
b. 4
c. 5
d. 2
Explanation: To compute the sum of remainders, first find each remainder individually using division or modular arithmetic. Then, sum the results to get the combined remainder value.
This method emphasizes understanding of divisibility, modular operations, and sequential arithmetic.
Think of it like distributing objects into different sized boxes, then summing the leftovers from each distribution.
The exercise reinforces division, remainder calculation, and modular reasoning.
Option a – 3
There are 660 students. Two-thirds are boys. Three-fourths of boys and one-fourth of girls play. Total players are
a. 370
b. 385
c. 395
d. 330
Explanation: Start by calculating the number of boys and girls. Then, find the number of players in each group using the given fractions. Sum these to determine the total players.
This problem demonstrates the application of fractions to real-life scenarios and sequential arithmetic operations for mixed groups.
Analogously, it is like counting how many participants from different teams qualify for a competition based on proportions.
The concept reinforces fractional reasoning, sequential calculations, and understanding distribution within groups.
Option b – 385
If place value of 5 in 15201 + place value of 6 in 2659 = 7 × …, the missing number is
a. 90
b. 900
c. 80
d. 800
Explanation: Compute the place values of 5 and 6 based on their positions in the numbers. Sum these values, then divide by 7 to find the missing factor.
This requires knowledge of place value, multiplication, and division to solve for an unknown in an equation.
Think of it as combining weights of two objects and then determining how many times a standard unit fits into the total.
The exercise reinforces arithmetic reasoning, place value comprehension, and solving equations with unknowns.
Option d – 800
If 112 ones + 12 thousand = 11012 + … tens, the missing number is
a. 111
b. 112
c. 101
d. 110
Explanation: Convert the left-hand side to numeric form: 112 ones = 112, 12 thousand = 12000. Combine to get the total. Then, subtract 11012 to find the number of tens needed to complete the equation.
This reinforces arithmetic operations, place value understanding, and breaking down numbers into units, tens, hundreds, and thousands.
It is like adjusting components in a mixture to reach a desired total by calculating the remaining amount.
The problem strengthens reasoning with place values, addition, and subtraction across different scales.
Option d – 110
If 1001 × 111 = 110000 + 11 × …, the missing number is
a. 121
b. 101
c. 211
d. 111
Explanation: Break the multiplication into manageable parts, representing it as 1001 × 111 = 1001 × 100 + 1001 × 11. Comparing with 110000 + 11 × … helps identify the unknown multiplier.
This illustrates how factoring and distributive properties simplify multiplication of large numbers.
Think of it as splitting a task into two phases: handling the main portion first, then the smaller remainder.
The exercise reinforces multiplication strategies, factoring, and decomposition of products.
Option b – 101
If the difference of remainders from 26679 ÷ 39 and 29405 ÷ 34 is divided by 18, the remainder is
a. 8
b. 9
c. 3
d. 5
Explanation: First, find the remainders of 26679 ÷ 39 and 29405 ÷ 34. Compute their difference, then divide that result by 18 to find the final remainder.
This involves sequential modular arithmetic operations, emphasizing careful computation and order of operations.
It is similar to checking leftover items after distributing two batches of objects, then dividing the difference among a new group.
The exercise reinforces remainder calculation, modular reasoning, and sequential arithmetic operations.
Option a – 8
Difference between place value of 5 in 29503 and face value of 7 in 32071 is
a. 493
b. 2
c. 43
d. 430
Explanation: Identify the place value of 5 and the face value of 7. Compute the difference by subtracting the smaller value from the larger.
This reinforces understanding of how positional significance affects numeric values and emphasizes subtraction based on place value reasoning.
Analogously, it is like comparing a person’s influence (place value) versus a Base trait (face value).
The problem strengthens skills in arithmetic, place value analysis, and logical reasoning.
Option a – 493
If 30028 = 28 ones + 28 thousands + … tens, the missing number is
a. 280
b. 28
c. 128
d. 200
Explanation: Break the number into its components: ones, thousands, and tens. Calculate the tens value needed to complete the total by subtracting known components from the overall number.
This emphasizes understanding of place value, decomposition of numbers, and structured calculation.
It is similar to splitting a total distance into known segments and figuring out the remaining part.
The exercise reinforces arithmetic, place value comprehension, and problem-solving with unknowns.
Option d – 200
If 603 × 28 = 63 × 4 × …, the missing number is
a. 67
b. 21
c. 28
d. 63
Explanation: To simplify 603 × 28, factor it into smaller components: 603 × 28 = (63 × 10 + 3) × 28. The missing number represents a factor that balances the equation.
This emphasizes factoring, distributive property, and manipulating large numbers by breaking them into smaller, manageable parts.
It is like splitting a task among workers to find how much each group should handle for the total to match.
The problem reinforces multiplication strategies, factoring, and understanding relationships between numbers in an equation.
Option a – 67
A number is less than half of 100, lies between 40 and 50, ones digit is one less than tens digit, and sum of digits is 7. Product of digits is
a. 24
b. 12
c. 16
d. 20
Explanation: Analyze the number using the given clues: it is a two-digit number, tens digit is between 4 and 5 tens, ones digit is one less than tens digit, and the sum of digits is 7.
By logically combining these constraints, you can determine the digits without performing blind calculation. This tests understanding of place value, arithmetic reasoning, and logical deduction.
It is similar to solving a puzzle where each clue progressively narrows down the possible solution.
The exercise strengthens problem-solving skills using constraints and digit relationships in two-digit numbers.
Option b – 12
Remainder of (80808 ÷ 108) ÷ remainder of (90909 ÷ 109) gives quotient
a. 12
b. 3
c. 6
d. 8
Explanation: First, compute the remainders of the two divisions. Then, divide the remainder from the first by the remainder of the second to find the quotient.
This sequential modular arithmetic emphasizes order of operations, careful calculation, and understanding of how remainders interact in subsequent divisions.
Analogously, it is like comparing leftover items from two batches and determining how many times one leftover fits into the other.
The problem reinforces modular reasoning, division, and sequential arithmetic logic.
Option d – 8
In a School of 360 students, two-thirds are girls and the rest are boys. Three-fourths of boys play. Number of boys who do not play is
a. 75
b. 25
c. 30
d. 60
Explanation: Begin by calculating the number of boys (total students minus two-thirds girls). Then, determine how many boys are players using the given fraction. Finally, subtract to find the number of boys who are not players.
This illustrates practical application of fractions and sequential operations in group problems.
Think of it as distributing tasks among a group and identifying the leftover portion that does not participate.
The exercise reinforces fractional reasoning, sequential subtraction, and practical problem-solving.
Option c – 30
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