Number System<\/strong>\u00a0is one of the most important topics in the CAT Quant Section<\/strong>. You can check out these<\/span>\u00a0 CAT Number System Questions from <\/strong>the CAT previous year papers.<\/strong> This article will look into some very important Number System questions PDF(with solutions<\/strong>) for CAT. If you want to practice these questions, you can download this CAT Number System Questions PDF (most important) along with the solutions below, which is completely Free.<\/p>\n Download Number System Questions for CAT<\/a><\/p>\n Question 1:\u00a0<\/b>For a 4-digit number, the sum of its digits in the thousands, hundreds and tens places is 14, the sum of its digits in the hundreds, tens and units places is 15, and the tens place digit is 4 more than the units place digit. Then the highest possible 4-digit number satisfying the above conditions is<\/p>\n 1)\u00a0Answer:\u00a04195<\/b><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n Given the 4 digit number :<\/p>\n Considering the number in thousands digit is a number in the hundredth digit is b, number in tens digit is c, number in the units digit is d.<\/p>\n Let the number be abcd.<\/p>\n Given that a+b+c = 14. (1)<\/p>\n b+c+d = 15. (2)<\/p>\n c = d+4. (3).<\/p>\n In order to find the maximum number which satisfies the condition, we need to have abcd such that a is maximum which is the digit in thousands place\u00a0in order to maximize the value of the number. b, c, and d are less than 9 each as they are single-digit numbers.<\/p>\n Substituting (3) in (2) we have b+d+4+d = 15, b+2*d = 11.\u00a0 (4)<\/p>\n Subtracting (2) and (1) : (2) – (1) = d = a+1.\u00a0 \u00a0(5)<\/p>\n Since c cannot be greater than 9 considering c to be the maximum value\u00a09 the value of d is 5.<\/p>\n If d = 5, using d = a+1, a = 4.<\/p>\n Hence the maximum value of a = 4 when c = 9, d = 5.<\/p>\n Substituting b+2*d = 11. b = 1.<\/p>\n The highest four-digit number satisfying the condition\u00a0is 4195<\/p>\n Question 2:\u00a0<\/b>For all possible integers n satisfying $2.25\\leq2+2^{n+2}\\leq202$, then the number of integer values of $3+3^{n+1}$ is:<\/p>\n 2)\u00a0Answer:\u00a07<\/b><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n $2.25\\leq2+2^{n+2}\\leq202$<\/p>\n $2.25-2\\le2+2^{n+2}-2\\le202-2$<\/p>\n $0.25\\le2^{n+2}\\le200$<\/p>\n $\\log_20.25\\le n+2\\le\\log_2200$<\/p>\n $-2\\le n+2\\le7.xx$<\/p>\n $-4\\le n\\le7.xx-2$<\/p>\n $-4\\le n\\le5.xx$<\/p>\n Possible integers = -4, -3, -2, -1, 0, 1, 2, 3, 4, 5<\/p>\n If we see the second expression that is provided, i.e<\/p>\n $3+3^{n+1}$, it can be implied that n should be at least -1 for this expression to be an integer.<\/p>\n So, n = -1, 0, 1, 2, 3, 4, 5.<\/p>\n Hence, there are a total of 7 values.<\/p>\n Question 3:\u00a0<\/b>How many 4-digit numbers, each greater than 1000 and each having all four digits\u00a0distinct, are there with 7 coming before 3?<\/p>\n 3)\u00a0Answer:\u00a0315<\/b><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n Here there are two cases possible<\/p>\n Case 1: When 7 is at the left extreme<\/p>\n In that case 3 can occupy any of the three remaining places and the remaining two places can be taken by (0,1,2,4,5,6,8,9)<\/p>\n So total ways 3(8)(7)= 168<\/p>\n Case 2: When 7 is not at the extremes<\/p>\n Here there are 3 cases possible. And the remaining two places can be filled in 7(7) ways.(Remember 0 can’t come on the extreme left)<\/p>\n Hence in total 3(7)(7)=147 ways<\/p>\n Total ways 168+147=315 ways<\/p>\n Question 4:\u00a0<\/b>How many pairs(a, b) of positive integers are there such that $a\\leq b$ and $ab=4^{2017}$ ?<\/p>\n a)\u00a02018<\/p>\n b)\u00a02019<\/p>\n c)\u00a02017<\/p>\n d)\u00a02020<\/p>\n 4)\u00a0Answer\u00a0(A)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n $ab\\ =\\ 4^{2017}=2^{4034}$<\/p>\n The total number of factors = 4035.<\/p>\n out of these 4035 factors, we can choose two numbers a,b such that a<b in [4035\/2] = 2017.<\/p>\n And since the given number is a perfect square we have one set of two equal factors.<\/p>\n .’.\u00a0many pairs(a, b) of positive integers are there such that $a\\leq b$ and $ab=4^{2017}$ = 2018.<\/p>\n Question 5:\u00a0<\/b>How many of the integers 1, 2, \u2026 , 120, are divisible by none of 2, 5 and 7?<\/p>\n a)\u00a042<\/p>\n b)\u00a041<\/p>\n c)\u00a040<\/p>\n d)\u00a043<\/p>\n 5)\u00a0Answer\u00a0(B)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n The number of multiples of 2 between 1 and 120 = 60<\/p>\n The number of multiples of 5 between 1 and 120 which are not multiples of 2 = 12<\/p>\n The number of multiples of 7 between 1 and 120 which are not multiples of 2 and 5 = 7<\/p>\n Hence,\u00a0number\u00a0of the integers 1, 2, \u2026 , 120, are divisible by none of 2, 5 and 7 = 120 – 60 – 12 – 7 = 41<\/p>\n Checkout: CAT Free Practice Questions and Videos<\/strong><\/a><\/em><\/p>\n Question 6:\u00a0<\/b>Let N, x and y be positive integers such that $N=x+y,2<x<10$ and $14<y<23$. If $N>25$, then how many distinct values are possible for\u00a0N?<\/p>\n 6)\u00a0Answer:\u00a06<\/b><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n Possible values of x = 3,4,5,6,7,8,9<\/p>\n When x = 3, there is no possible value of y<\/p>\n When x = 4, the possible values of y = 22<\/p>\n When x = 5, the possible values of y=21,22<\/p>\n When x = 6, the possible values of y = 20.21,22<\/p>\n When x = 7, the possible values of y = 19,20,21,22<\/p>\n When x = 8, the possible values of y=18,19,20,21,22<\/p>\n When x = 9, the possible values of y=17,18,19,20,21,22<\/p>\n The unique values of N = 26,27,28,29,30,31<\/p>\n Question 7:\u00a0<\/b>How many integers in the set {100, 101, 102, …, 999} have at least one digit\u00a0repeated?<\/p>\n 7)\u00a0Answer:\u00a0252<\/b><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n Total number of numbers from 100 to 999 = 900<\/p>\n The number of three digits numbers with unique digits:<\/p>\n _ _ _<\/p>\n The hundredth’s place can be filled in 9 ways ( Number 0 cannot be selected)<\/p>\n Ten’s place can be filled in 9 ways<\/p>\n One’s place can be filled in 8 ways<\/p>\n Total number of numbers = 9*9*8 = 648<\/p>\n Number of integers in the set {100, 101, 102, …, 999} have at least one digit repeated = 900 – 648 = 252<\/p>\n Question 8:\u00a0<\/b>Let m and n be natural numbers such that n is even and $0.2<\\frac{m}{20},\\frac{n}{m},\\frac{n}{11}<0.5$. Then $m-2n$ equals<\/p>\n a)\u00a03<\/p>\n b)\u00a01<\/p>\n c)\u00a02<\/p>\n d)\u00a04<\/p>\n 8)\u00a0Answer\u00a0(B)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n $0.2<\\frac{n}{11}<0.5$<\/p>\n => 2.2<n<5.5<\/p>\n Since n is an even natural number, the value of n = 4<\/p>\n $0.2<\\frac{m}{20}<0.5$\u00a0 => 4< m<10. Possible values of m = 5,6,7,8,9<\/p>\n Since\u00a0$0.2<\\frac{n}{m}<0.5$, the only possible value of m is 9<\/p>\n Hence m-2n = 9-8 = 1<\/p>\n Question 9:\u00a0<\/b>If a, b and c are positive integers such that ab = 432, bc = 96 and c < 9, then the smallest possible value of a + b + c is<\/p>\n a)\u00a049<\/p>\n b)\u00a056<\/p>\n c)\u00a059<\/p>\n d)\u00a046<\/p>\n 9)\u00a0Answer\u00a0(D)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n Since $c<9$, we can have the following viable combinations for\u00a0$b\\times\\ c\\ =96$ (given our objective is to minimize the sum):<\/p>\n $48\\times\\ 2$ ;\u00a0$32\\times3$ ;\u00a0$24\\times\\ 4$ ;\u00a0$16\\times6$ ;\u00a0$12\\times8$<\/p>\n Similarly, we can factorize\u00a0$a\\times\\ b\\ = 432$ into its factors. On close observation, we notice that\u00a0$18\\times24\\ and\\ 24\\ \\times\\ 4\\ $ corresponding to\u00a0$a\\times b\\ and\\ b\\times\\ c\\ $ respectively\u00a0together render us with the least value of the sum of\u00a0$a+b\\ +\\ c\\ \\ =\\ 18+24+4\\ =46$<\/p>\n Hence, Option D is the correct answer.<\/p>\n Question 10:\u00a0<\/b>The mean of all 4-digit even natural numbers of the form ‘aabb’,where $a>0$, is<\/p>\n a)\u00a04466<\/p>\n b)\u00a05050<\/p>\n c)\u00a04864<\/p>\n d)\u00a05544<\/p>\n 10)\u00a0Answer\u00a0(D)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n The four digit even numbers will be of form:<\/p>\n 1100, 1122, 1144 … 1188, 2200, 2222, 2244 … 9900, 9922, 9944, 9966, 9988<\/p>\n Their sum ‘S’ will be (1100+1100+22+1100+44+1100+66+1100+88)+(2200+2200+22+2200+44+…)….+(9900+9900+22+9900+44+9900+66+9900+88)<\/p>\n => S=1100*5+(22+44+66+88)+2200*5+(22+44+66+88)….+9900*5+(22+44+66+88)<\/p>\n => S=5*1100(1+2+3+…9)+9(22+44+66+88)<\/p>\n =>S=5*1100*9*10\/2 + 9*11*20<\/p>\n Total number of numbers are 9*5=45<\/p>\n .’. Mean will be S\/45 = 5*1100+44=5544.<\/p>\n Option D<\/p>\n Question 11:\u00a0<\/b>How many 3-digit numbers are there, for which the product of their digits is more than 2 but less than 7?<\/p>\n 11)\u00a0Answer:\u00a021<\/b><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n Let the number be ‘abc’. Then, $2<a\\times\\ b\\times\\ c<7$. The product can be 3,4,5,6.<\/p>\n We can obtain each of these as products with the combination 1,1, x where x = 3,4,5,6. Each number can be arranged in 3 ways, and we have 4 such numbers: hence, a total of 12<\/strong> numbers fulfilling the criteria.<\/p>\n We can factories 4 as 2*2 and the combination 2,2,1 can be used to form 3<\/strong> more distinct numbers.<\/p>\n We can factorize 6 as 2*3 and the combination 1,2,3 can be used to form 6<\/strong> additional distinct numbers.<\/p>\n Thus a total of 12 + 3 + 6 = 21<\/strong> such numbers can be formed.<\/p>\n Question 12:\u00a0<\/b>The product of two positive numbers is 616. If the ratio of the difference of their cubes to the cube of their difference is 157:3, then the sum of the two numbers is<\/p>\n a)\u00a058<\/p>\n b)\u00a085<\/p>\n c)\u00a050<\/p>\n d)\u00a095<\/p>\n 12)\u00a0Answer\u00a0(C)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n Assume the numbers are a and b, then ab=616<\/p>\n We have,\u00a0$\\ \\ \\frac{\\ a^3-b^3}{\\left(a-b\\right)^3}$ = $\\ \\frac{\\ 157}{3}$<\/p>\n =>\u00a0$\\ 3\\left(a^3-b^3\\right)\\ =\\ 157\\left(a^3-b^3+3ab\\left(b-a\\right)\\right)$<\/p>\n => $154\\left(a^3-b^3\\right)+3*157*ab\\left(b-a\\right)$\u00a0= 0<\/p>\n =>\u00a0$154\\left(a^3-b^3\\right)+3*616*157\\left(b-a\\right)$ = 0\u00a0 \u00a0 \u00a0 \u00a0\u00a0(ab=616)<\/p>\n =>$a^3-b^3+\\left(3\\times\\ 4\\times\\ 157\\left(b-a\\right)\\right)$\u00a0 \u00a0 (154*4=616)<\/p>\n =>\u00a0$\\left(a-b\\right)\\left(a^2+b^2+ab\\right)\\ =\\ 3\\times\\ 4\\times\\ 157\\left(a-b\\right)$<\/p>\n =>\u00a0$a^2+b^2+ab\\ =\\ 3\\times\\ 4\\times\\ 157$<\/p>\n Adding ab=616 on both sides, we get<\/p>\n $a^2+b^2+ab\\ +ab=\\ 3\\times\\ 4\\times\\ 157+616$<\/p>\n =>\u00a0$\\left(a+b\\right)^2=\\ 3\\times\\ 4\\times\\ 157+616$ = 2500<\/p>\n => a+b=50<\/p>\n Question 13:\u00a0<\/b>In a six-digit number, the sixth, that is, the rightmost, digit is the sum of the first three digits, the fifth digit is the sum of first two digits, the third digit is equal to the first digit, the second digit is twice the first digit and the fourth digit is the sum of\u00a0fifth and sixth digits. Then, the largest possible value of the fourth digit is<\/p>\n 13)\u00a0Answer:\u00a07<\/b><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n Let the six-digit number be ABCDEF<\/p>\n F = A+B+C, E= A+B, C=A, B= 2A, D= E+F.<\/p>\n Therefore D = 2A+2B+C = 2A + 4A + A= 7A.<\/p>\n A cannot be 0 as the number is a 6 digit number.<\/p>\n A cannot be 2 as D would become 2 digit number.<\/p>\n Therefore A is 1 and D is 7.<\/p>\n Question 14:\u00a0<\/b>How many factors of $2^4 \\times 3^5 \\times 10^4$ are perfect squares which are greater than 1?<\/p>\n 14)\u00a0Answer:\u00a044<\/b><\/p>\n