Download important IIFT Number System Questions PDF based on previously asked questions in IIFT and other MBA exams. Practice Number System questions and answers for IIFT exam.<\/p>\n
Download Number System Questions for IIFT PDF<\/a><\/p>\n Get 50% Off on IIFT & Other MBA Test Series\u00a0<\/a><\/p>\n Download IIFT Previous Papers PDF<\/a><\/p>\n Practice IIFT Mock Tests<\/a><\/p>\n Question 1:\u00a0<\/b>Let N = 1421 * 1423 * 1425. What is the remainder when N is divided by 12?<\/p>\n a)\u00a00<\/p>\n b)\u00a09<\/p>\n c)\u00a03<\/p>\n d)\u00a06<\/p>\n Question 2:\u00a0<\/b>When $2^{256}$ is divided by 17, the remainder would be<\/p>\n a)\u00a01<\/p>\n b)\u00a016<\/p>\n c)\u00a014<\/p>\n d)\u00a0None of these<\/p>\n Question 3:\u00a0<\/b>Number S is obtained by squaring the sum of digits of a two-digit number D. If difference between S and D is 27, then the two-digit number D is<\/p>\n a)\u00a024<\/p>\n b)\u00a054<\/p>\n c)\u00a034<\/p>\n d)\u00a045<\/p>\n Question 4:\u00a0<\/b>How many 3 – digit even number can you form such that if one of the digits is 5, the following digit must be 7?<\/p>\n a)\u00a05<\/p>\n b)\u00a0405<\/p>\n c)\u00a0365<\/p>\n d)\u00a0495<\/p>\n Question 5:\u00a0<\/b>To decide whether a number of n digits is divisible by 7, we can define a process by which its magnitude is reduced as follows: $(i_{1}, i_{2}, i_{3}$,…..\u00a0are the digits of the number, starting from the most significant digit). $i_{1} i_{2} … i_{n} => i_{1}.3^{n-1} + i_{2}.3^{n-2} + … + i_{n}.3^0$. a)\u00a02<\/p>\n b)\u00a03<\/p>\n c)\u00a04<\/p>\n d)\u00a01<\/p>\n IIFT Free Mock Test<\/a><\/p>\n Enroll for CAT\/MBA Courses<\/a><\/p>\n Question 6:\u00a0<\/b>If m and n are integers divisible by 5, which of the following is not necessarily true?<\/p>\n a)\u00a0m – n is divisible by 5<\/p>\n b)\u00a0m2 – n2 is divisible by 25<\/p>\n c)\u00a0m + n is divisible by 10<\/p>\n d)\u00a0None of these<\/p>\n Question 7:\u00a0<\/b>A certain number, when divided by 899, leaves a remainder 63. Find the remainder when the same number is divided by 29.<\/p>\n a)\u00a05<\/p>\n b)\u00a04<\/p>\n c)\u00a01<\/p>\n d)\u00a0Cannot be determined<\/p>\n Question 8:\u00a0<\/b>If a, b, c and d are four different positive integers selected from 1 to 25, then the highest possible value of ((a + b) + (c +d ))\/((a + b) + (c – d)) would be:<\/p>\n a)\u00a047<\/p>\n b)\u00a049<\/p>\n c)\u00a051<\/p>\n d)\u00a096<\/p>\n e)\u00a0None of the above<\/p>\n Question 9:\u00a0<\/b>Two numbers, $297_{B}$ and $792_{B}$ , belong to base B number system. If the first number is a factor of the second number then the value of B is:<\/p>\n a)\u00a011<\/p>\n b)\u00a012<\/p>\n c)\u00a015<\/p>\n d)\u00a017<\/p>\n e)\u00a019<\/p>\n Question 10:\u00a0<\/b>In a Green view apartment, the houses of a row are numbered consecutively from 1 to 49. Assuming that there is a value of \u2018x\u2019 such that the sum of the numbers of the houses preceding the house numbered \u2018x\u2019 is equal to the sum of the numbers of the houses following it. Then what will be the value of \u2018x\u2019?<\/p>\n a)\u00a021<\/p>\n b)\u00a030<\/p>\n c)\u00a035<\/p>\n d)\u00a042<\/p>\n Top 500+ Free Questions for IIFT<\/a><\/p>\n IIFT Previous year question\u00a0 answer PDF<\/a><\/p>\n Answers & Solutions:<\/strong><\/span><\/p>\n 1)\u00a0Answer\u00a0(C)<\/strong><\/p>\n The numbers 1421, 1423 and 1425 when divided by 12 give remainder 5, 7 and 9 respectively.<\/p>\n 5*7*9 mod 12 = 11 * 9 mod 12 = 99 mod 12 = 3<\/p>\n 2)\u00a0Answer\u00a0(A)<\/strong><\/p>\n $2^4 = 16 = -1$ (mod $17$) 3)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Consider the options:<\/p>\n 24: (Square of sum of digits – the number) = 36 – 24 = 12<\/p>\n 54: (Square of sum of digits – the number) = 81 – 54 = 27<\/p>\n 34: (Square of sum of digits – the number) = 49 – 34 = 15<\/p>\n 45: (Square of sum of digits – the number) = 81 – 45 = 36<\/p>\n So, option b) is the correct answer.<\/p>\n 4)\u00a0Answer\u00a0(C)<\/strong><\/p>\n For a number to be even, its unit digit should be 0,2,4,6,8 5)\u00a0Answer\u00a0(A)<\/strong><\/p>\n For 203 : Quant Formula For IIFT PDF<\/a><\/p>\n IIFT Free Mock Test<\/a><\/p>\n 6)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Let’s say m=5k and n=5t <\/p>\n 7)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Let’s say N is our number 8)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Expression\u00a0: $\\frac{a + b + c + d}{a + b + c – d}$<\/p>\n To maximize the above expression, we have to minimize the denominator<\/p>\n Minimum value of the denominator = 1<\/p>\n So we can make $a + b + c = 26$ and $d = 25$ \u00a0 (as maximizing d will give denominator the least value).<\/p>\n So required maximum value = $\\frac{a + b + c + d}{a + b + c – d}$<\/p>\n = $\\frac{26 + 25}{26 – 25} = 51$<\/p>\n 9)\u00a0Answer\u00a0(E)<\/strong><\/p>\n In Base B, $297_B = 2B^2 + 9B + 7$<\/p>\n and $792_B = 7B^2 + 9B + 2$<\/p>\n It is given that $297_{B}$ is a factor of $792_{B}$<\/p>\n => $\\frac{7B^2 + 9B + 2}{2B^2 + 9B + 7}$ must be an integer<\/p>\n => $\\frac{(2B^2 + 9B + 7) + (5B^2 – 5)}{2B^2 + 9B + 7}$<\/p>\n => $\\frac{5B^2 – 5}{2B^2 + 9B + 7} + 1 = k$<\/p>\n => $5B^2 – 5 = (2B^2 + 9B + 7) k$ \u00a0 \u00a0 \u00a0(where $k$ is factor)<\/p>\n Put $k = 1$<\/p>\n => $5B^2 – 5 = 2B^2 + 9B + 7$<\/p>\n => $B^2 – 3B – 4 = 0$<\/p>\n => $(B – 4) (B + 1) = 0$<\/p>\n => $B = 4 , -1$<\/p>\n Since, B is a base,so B must be greater than 9. Hence, it is not possible<\/p>\n Put $k = 2$<\/p>\n => $5B^2 – 5 = 4B^2 + 18B + 14$<\/p>\n => $B^2 – 18B – 19 = 0$<\/p>\n => $(B – 19) (B + 1) = 0$<\/p>\n => $B = 19 , -1$<\/p>\n $\\therefore B = 19$<\/p>\n 10)\u00a0Answer\u00a0(C)<\/strong><\/p>\n It is given that sum of the first $(x – 1)$ numbers is equal to sum of the numbers from $(x + 1)$ to 49
\ne.g. $259 => 2.3^2 + 5.3^1 + 9.3^0 = 18 + 15 + 9 = 42$
\nUltimately the resulting number will be seven after repeating the above process a certain number of times. After how many such stages, does the number 203 reduce to 7?<\/p>\n
\nSo, $2^{256} = (-1)^{64} $(mod $17$)
\n$= 1$ (mod $17$)
\nHence, the answer is 1. Option a).<\/p>\n
\nCase 1: One of the digit is 5
\nHence according to question, 5 can’t come in middle and at unit’s place, so numbers will be 570,572,574,576,578.
\nCase 2: No digit is 5
\nHence the hundreds place can be filled in 8 ways (except 0,5) and tens place can be filled in 9 ways (except 5).
\nNumber of ways = 8 * 9 * 5 = 360
\nHence total number of ways = 360 + 5 = 365<\/p>\n
\nfirst step =\u00a0$2\\times 3^2 + 0 \\times 3^1 + 3 \\times 3^0$ = 21
\nsecond step = $2 \\times 3^1 + 1 \\times 3^0$ = 7
\nSo two steps needed to reduce it to 7<\/p>\n
\nSo m-n = 5(k-t) will be divisible by 5.
\n$m^2 – n^2 = 25(k^2 – t^2)$ will be divisible by 5.
\n$m+n = 5(k+t)$ will be divisible by 5 but not necessarily with 10.<\/p>\n
\nN = (899K + 63) or N = ($29 \\times 31$K) + 63
\nSo when it is divided by 29, remainder will be $\\frac{63}{29}$ = 5<\/p>\n
\nor, sum of $(x – 1)$ numbers = sum of first 49 numbers – sum of first $x$ numbers
\n$\\dfrac{x(x – 1)}{2} = \\dfrac{49 * 50}{2} – \\dfrac{x(x + 1)}{2}$
\nOn solving, we get $x$ = 35
\nHence, option C is the correct answer.<\/p>\n