Number System — Set #3 | Practice

{"setId":"0247776c-e66a-4673-bc76-69247c72ae44","setTitle":"Number System Aptitude Quiz Vol 2 – UPSC CSAT","topicName":"Number System","topicSlug":"number-system","setNumber":3,"totalQuestions":20,"questions":[{"id":"0f1d6c5c-c6e0-4fa5-baba-55474c8782f9","order":1,"statement":"Count the total occurrence of digit 2 in the numbers from 200 to 350.","options":{"A":"140","B":"160","C":"210","D":"135"},"correctOption":"D","solution":"2 in unit's place: 200-299 gives 10 occurrences, 300-350 gives 5, total 15. 2 in tens place: 200-299 gives 10, 300-350 gives 10, total 20. 2 in hundreds place: all 100 numbers from 200-299 have 2 in hundreds place = 100. Total = 15+20+100 = 135."},{"id":"788fa904-7d7c-4105-a68c-c1d43979c507","order":2,"statement":"In a 3-digit prime number 'M2N' the digit at the unit place is 1 more than the cube of the digit at the hundreds place. Find the square of the sum of the digits of the number.","options":{"A":"121","B":"169","C":"196","D":"225"},"correctOption":"B","solution":"The pair of single digits where one is the cube of the other is 2 and 8 (2^3=8). Since the unit digit is one more than the cube of the hundreds digit, hundreds digit = 2, unit digit = 8+1 = 9. The number is 229 (a valid prime). Square of sum of digits = (2+2+9)^2 = 13^2 = 169."},{"id":"a1dfedb2-6a46-469d-a50b-8f799df023af","order":3,"statement":"The number of times the digit 7 will appear while writing the integers from 1 to 700 is","options":{"A":"139","B":"140","C":"141","D":"142"},"correctOption":"C","solution":"Considering 0-100: 7 appears 10 times in unit's place and 10 times in tens place, totaling 20 times per group of 100. From 0-700, there are 7 such groups, giving 7×20=140. Additionally, 7 appears once more in the hundreds place (in 700 itself). Total = 140+1 = 141."},{"id":"872192d4-b84a-4cbf-b5c5-eac17a540fdb","order":4,"statement":"If A00A × CC = 44044, then find the value of (A + C).","options":{"A":"0","B":"5","C":"4","D":"Either (b) or (c)"},"correctOption":"D","solution":"A00A = 1001A and CC = 11C, so 1001A×11C = 44044, giving 11011AC = 44044, so AC = 4. Possible pairs: A=1,C=4 (A+C=5); A=2,C=2 (A+C=4); A=4,C=1 (A+C=5). All three pairs satisfy the original equation, so A+C can be either 5 or 4."},{"id":"a866dd8d-c30f-4187-bdd7-668f59fbcfe5","order":5,"statement":"Find a two-digit number where the sum of its digits is 9. When the digits are reversed, the resulting number is 9 more than the original number. What is the original number?","options":{"A":"45","B":"54","C":"34","D":"43"},"correctOption":"A","solution":"Let the number be 10x+y, with x+y=9. Given (10x+y)+9=(10y+x), which simplifies to 9x-9y=-9, so x-y=-1. Adding this to x+y=9 gives 2x=8, so x=4 and y=5. The number is 45."},{"id":"4ea74933-345c-45b7-9ae7-264d3dded15a","order":6,"statement":"If 124B + 7C83 = 9031, then find the value of (B × C).","options":{"A":"48","B":"72","C":"20","D":"56"},"correctOption":"D","solution":"From the units column: B+3=11 (with a carry), so B=8. From the tens/hundreds analysis: 2+C=9, so C=7. Thus B×C = 8×7 = 56."},{"id":"2728be92-fc8a-4775-b82e-6ec36c568e56","order":7,"statement":"ABC is a three-digit number such that ABC = (AB + BC + CA), where all 'AB', 'BC' and 'CA' are all natural numbers. Find the maximum possible value of (A + 3B + 2C).","options":{"A":"24","B":"34","C":"44","D":"54"},"correctOption":"C","solution":"Setting up 100a+10b+c = 11a+11b+11c gives 89a-b=10c. Testing a=1 with maximum b=9 gives c=8, satisfying the equation. Testing a=2 or higher makes the equation unsatisfiable within single-digit constraints. So the only valid triple is (a,b,c)=(1,9,8). Required value = 1+(3×9)+(2×8) = 1+27+16 = 44."},{"id":"4bed3235-b6e4-43c2-915a-77dd39f8a4d5","order":8,"statement":"How many digits are required to print a book of 1250 pages, with page number starting from 1?","options":{"A":"2639","B":"3893","C":"4694","D":"4458"},"correctOption":"B","solution":"Digits for pages 1-9 = 9×1=9. Digits for pages 10-99 = 90×2=180. Digits for pages 100-999 = 900×3=2700. Digits for pages 1000-1250 = 251×4=1004. Total = 9+180+2700+1004 = 3893."},{"id":"6069e959-dfff-4792-9bf3-6a5d09d6497f","order":9,"statement":"What is the difference between the cube of the greatest single-digit prime number and the cube of smallest 3-digit prime number?","options":{"A":"1025648","B":"1029958","C":"1059348","D":"1025638"},"correctOption":"B","solution":"The greatest single-digit prime is 7, and the smallest 3-digit prime is 101. The difference between their cubes = 101^3 - 7^3 = 1030301 - 343 = 1029958."},{"id":"c64ccbe5-83aa-4a62-b8ee-d4f2d9f6f4f4","order":10,"statement":"A three-digit number is 495 more than number obtained by interchanging the hundred's place and unit place digits. Hundred's place digit is 3 more than ten's place digit while the sum of three digits is 16. Find the three-digit number?","options":{"A":"853","B":"963","C":"736","D":"743"},"correctOption":"A","solution":"Let the number be 100a+10b+c. Given (100a+10b+c)-(100c+10b+a)=495, this simplifies to 99(a-c)=495, so a-c=5. Also a=b+3, and a+b+c=16. Substituting b=a-3 and c=a-5 into a+b+c=16 gives 3a-8=16, so a=8, b=5, c=3. The number is 853."},{"id":"cee4be47-c0d8-47ea-9db2-aa8d654e28bf","order":11,"statement":"The average of the largest five double-digit prime numbers is:","options":{"A":"84.2","B":"87.5","C":"79.5","D":"78.2"},"correctOption":"A","solution":"The five largest two-digit prime numbers are 97, 89, 83, 79, 73. Their average = (97+89+83+79+73)/5 = 421/5 = 84.2."},{"id":"a52233a7-8f5b-4e0e-874d-956fff241180","order":12,"statement":"How many integers between 300 and 723 have a sum of digits equal to 7?","options":{"A":"12","B":"14","C":"15","D":"13"},"correctOption":"C","solution":"We need three-digit combinations summing to 7 that form numbers between 300 and 723: (0,0,7), (0,1,6), (0,2,5), (0,3,4), (1,1,5), (1,2,4), (1,3,3), (2,2,3). Counting all valid arrangements within the range: 700; 601,610; 502,520; 304,340,403,430; 511; 412,421; 313,331; 322 — totaling 15 integers."},{"id":"4aa5b300-d7f6-441c-a1a6-5575fd10637e","order":13,"statement":"Total number of digits used for numbering pages in a book starting from 1 is 4545. Find the total number of pages the book contains.","options":{"A":"1385","B":"1375","C":"1413","D":"1434"},"correctOption":"C","solution":"Digits used for pages 1-999 = 9(one-digit)+180(two-digit)+2700(three-digit) = 2889. Remaining digits = 4545-2889 = 1656. Number of 4-digit pages = 1656/4 = 414. Total pages = 999+414 = 1413."},{"id":"3c2f466e-b320-491e-bcf6-e90da86d3853","order":14,"statement":"A printer numbers the pages of a book starting with 1 and uses 4749 digits in all. How many pages does the book have?","options":{"A":"1465","B":"1365","C":"1364","D":"1464"},"correctOption":"D","solution":"Digits used for pages 1-999 = 9+180+2700 = 2889. Remaining digits = 4749-2889 = 1860. Number of 4-digit pages = 1860/4 = 465. Total pages = 999+465 = 1464."},{"id":"eed28a17-93dd-4dd1-bf1e-be5ee92ab663","order":15,"statement":"How many times the digit '5' will appear in integers from 1 to 500?","options":{"A":"101","B":"91","C":"111","D":"100"},"correctOption":"A","solution":"In each group of 100 (e.g., 1-100), digit 5 appears 10 times in the unit's place and 10 times in the ten's place, totaling 20 times per group. From 1-500, there are 5 such groups, giving 5×20=100. Additionally, 5 appears once more in the hundred's place (in 500). Total = 100+1 = 101."},{"id":"26185a94-d55f-4d18-8e20-ed5e88912cab","order":16,"statement":"In a double-digit number, the digit at the unit place is thrice more than the digit at the tens place. If the product of the original number and the sum of the digits of the number is 70. Then find the original number.","options":{"A":"29","B":"13","C":"14","D":"26"},"correctOption":"C","solution":"Let the number be XY with Y=4X (unit digit is 4 times the tens digit, interpreting 'thrice more'). Given (10X+Y)(X+Y)=70. Substituting Y=4X: (10X+4X)(X+4X)=70, giving 14X×5X=70, so 70X^2=70, meaning X=1 and Y=4. The number is 14."},{"id":"6dd58b23-e027-4864-9485-d5f62ca58c89","order":17,"statement":"A two-digit number is such that the product of the digits is 15. When 18 is added to the number, then the digits are reversed. The number is:","options":{"A":"25","B":"35","C":"53","D":"Inadequate data"},"correctOption":"B","solution":"Let the number be 10x+y with xy=15, so y=15/x. Given (10x+y)+18=10y+x. Substituting and solving the resulting quadratic gives x=3, y=5 as the valid solution (satisfying xy=15). The number is 35."},{"id":"25fb81e0-85f6-4f45-a42b-e42ff23911a1","order":18,"statement":"How many prime numbers are there between 50 and 85?","options":{"A":"7","B":"9","C":"10","D":"8"},"correctOption":"D","solution":"The prime numbers between 50 and 85 are: 53, 59, 61, 67, 71, 73, 79, 83 — a total of 8 prime numbers."},{"id":"fd4509cb-2265-4527-b8f3-3f67adfaab8c","order":19,"statement":"Consider the following statements: Statement 1: All the single-digit odd numbers are prime numbers. Statement 2: No single-digit even number is a prime number. Which of the following statement(s) is/are not correct?","options":{"A":"Only 1","B":"Only 2","C":"Both 1 & 2","D":"Neither 1 nor 2"},"correctOption":"C","solution":"Statement 1 is incorrect because 1 and 9 are odd single-digit numbers but not prime. Statement 2 is incorrect because 2 is an even number that is prime. Since both statements are incorrect, the answer is 'Both 1 & 2'."},{"id":"9d42a5e6-bfc5-4bad-9a53-5d235cd8bce7","order":20,"statement":"If A7A × 9 = A266, then find the value of 'A'.","options":{"A":"4","B":"3","C":"6","D":"8"},"correctOption":"A","solution":"Setting up the equation (100A+70+A)×9 = 1000A+266 gives (101A+70)×9 = 1000A+266, so 909A+630 = 1000A+266, giving 91A=364, so A=4."}]}