Download important CAT Best Number System Questions with Solutions PDF based on previously asked questions in CAT exam. Practice Best Number System Questions with Solutions for CAT exam.<\/p>\n
Download Best Number System Questions for CAT<\/a><\/p>\n 3 Months Crash Course for CAT<\/a><\/p>\n Question 1:\u00a0<\/b>The integers 1, 2, \u2026, 40 are written on a blackboard. The following operation is then repeated 39 times: In each repetition, any two numbers, say a and b, currently on the blackboard are erased and a new number a + b – 1 is written. What will be the number left on the board at the end?<\/p>\n a)\u00a0820<\/p>\n b)\u00a0821<\/p>\n c)\u00a0781<\/p>\n d)\u00a0819<\/p>\n e)\u00a0780<\/p>\n Question 2:\u00a0<\/b>How many three digit positive integers, with digits x, y and z in the hundred’s, ten’s and unit’s place respectively, exist such that x < y, z < y and x $\\neq$ 0 ?<\/p>\n a)\u00a0245<\/p>\n b)\u00a0285<\/p>\n c)\u00a0240<\/p>\n d)\u00a0320<\/p>\n Question 3:\u00a0<\/b>In Nuts And Bolts factory, one machine produces only nuts at the rate of 100 nuts per minute and needs to be cleaned for 5 minutes after production of every 1000 nuts. a)\u00a0130 minutes<\/p>\n b)\u00a0135 minutes<\/p>\n c)\u00a0170 minutes<\/p>\n d)\u00a0180 minutes<\/p>\n Question 4:\u00a0<\/b>If a\/b = 1\/3, b\/c = 2, c\/d = 1\/2 , d\/e = 3 and e\/f = 1\/4, then what is the value of abc\/def ?<\/p>\n a)\u00a03\/8<\/p>\n b)\u00a027\/8<\/p>\n c)\u00a03\/4<\/p>\n d)\u00a027\/4<\/p>\n e)\u00a01\/4<\/p>\n Question 5:\u00a0<\/b>The digits of a three-digit number A are written in the reverse order to form another three-digit number B. If B > A and B-A is perfectly divisible by 7, then which of the following is necessarily true?<\/p>\n a)\u00a0100<A<299<\/p>\n b)\u00a0106<A<305<\/p>\n c)\u00a0112<A<311<\/p>\n d)\u00a0118<A<317<\/p>\n Download CAT Quant Questions PDF<\/a><\/p>\n Take a free mock test for CAT<\/a><\/p>\n Question 6:\u00a0<\/b>A red light flashes three times per minute and a green light flashes five times in 2 min at regular intervals. If both lights start flashing at the same time, how many times do they flash together in each hour?<\/p>\n a)\u00a030<\/p>\n b)\u00a024<\/p>\n c)\u00a020<\/p>\n d)\u00a060<\/p>\n Question 7:\u00a0<\/b>Let $b$ be a positive integer and $a = b^2 – b$. If $b \\geq 4$ , then $a^2 – 2a$ is divisible by<\/p>\n a)\u00a015<\/p>\n b)\u00a020<\/p>\n c)\u00a024<\/p>\n d)\u00a0All of these<\/p>\n Question 8:\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 9:\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 10:\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 Take a free CAT online mock test<\/a><\/p>\n Answers & Solutions:<\/strong><\/span><\/p>\n 1)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Let the first operation be (1+40-1) = 40, the second operation be (2+39-1) = 40 and so on<\/p>\n So, after 20 operations, all the numbers are 40. After 10 more operations, all the numbers are 79<\/p>\n Proceeding this way, the last remaining number will be 781<\/p>\n 2)\u00a0Answer\u00a0(C)<\/strong><\/p>\n x, y and z in the hundred’s, ten’s and unit’s place. So y should start from 2<\/p>\n If y=2 , possible values of x=1 and z = 0,1 .So 2 cases 120,121.<\/p>\n Also if y=3 , possible values of x=1,2 and z=0,1,2.<\/p>\n Here 6 three digit nos. possible .<\/p>\n Similarly for next cases would be 3*4=12,4*5=20,5*6=30,…..,8*9=72 . Adding all we get 240 cases.<\/p>\n 3)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Machine A takes 15 min to produce 1000 nuts with clean time. machine b takes 30 min to make 1500 nuts with clean time . So B is slower. So with B 900 nuts will be made in 180 mins but at last round cleaning time of 10 min no need to count hence 170 mins<\/p>\n 4)\u00a0Answer\u00a0(A)<\/strong><\/p>\n a\/d = a\/b * b\/c * c\/d = 1\/3 * 2 * 1\/2 = 1\/3<\/p>\n Similarly, b\/e and c\/f are 3 and 3\/8 respectively.<\/p>\n b\/e = b\/c*c\/d*d\/e = 3<\/p>\n c\/f = c\/d*d\/e*e\/f = 3\/8<\/p>\n => Value of abc\/def = 1\/3 * 3 * 3\/8 = 3\/8<\/p>\n 5)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Let A = 100x + 10y + z \u00a0and B = 100z + 10y + x .According to given condition\u00a0B – A = 99(z – x) As (B – A) is divisible by 7 . So clearly \u00a0(z – x) should be\u00a0\u00a0divisible by 7.\u00a0\u00a0z and x can have values 8,1 or 9,2 , such that 8-2=9-2=7 and\u00a0\u00a0y can have \u00a0value from 0 to 9. Take a free mock test for CAT<\/a><\/p>\n Download CAT Syllabus PDF<\/a><\/p>\n 6)\u00a0Answer\u00a0(A)<\/strong><\/p>\n A red light flashes three times per minute and a green light flashes five times in 2 min at regular intervals. So\u00a0red light fashes after every 1\/3 min and green light flashes every 2\/5 min. LCM of both the fractions is 2 min .<\/p>\n Hence they flash together after every 2 min. So in an hour they flash together 30 times .<\/p>\n 7)\u00a0Answer\u00a0(C)<\/strong><\/p>\n We know that a=$b^2-b$.<\/p>\n So$a^2-a$ = b($b^3-2b^2-b+2$) . = (b – 2)(b – 1)( b)(b + 1)<\/p>\n The above given is a product of 4 consecutive numbers with the lowest number of the product being 2(given b >= 4)<\/p>\n In any set of four consecutive numbers, one of the numbers would be divisible by 3 and there would be two even numbers with the minimum value of the pair being (2,4).<\/p>\n Thus, for any value of b >=4, $a^2-4$ would be divisible by 3 x 2 x 4 = 24.<\/p>\n Thus, option C is the right choice. Options A and B are definitely wrong as a set of four consecutive numbers need not always include a multiple of 5 eg:(6,7,8,9)<\/p>\n <\/p>\n <\/p>\n 8)\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 9)\u00a0Answer\u00a0(A)<\/strong><\/p>\n $2^4 = 16 = -1$ (mod $17$) 10)\u00a0Answer\u00a0(A)<\/strong><\/p>\n For 203 :
\nAnother machine produces only bolts at the rate of 75 bolts per minute and needs to be cleaned for 10 minutes after production of every 1500 bolts. If both the machines start production at the same time, what is the minimum duration required for producing 9000 pairs of nuts and bolts?<\/p>\n
\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 Lowest possible value of A lowest x,y and z which is \u00a0is 108 and the highest possible value of A is 299.<\/p>\n
\nSo, $2^{256} = (-1)^{64} $(mod $17$)
\n$= 1$ (mod $17$)
\nHence, the answer is 1. Option a).<\/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