The Number System is one of the key topics in the CAT Quant<\/strong> Section. If you’re not very strong in this topic, it is essential that you know the basics of the CAT Number System. Also, do check out all the Number System questions from the CAT Previous Papers<\/strong><\/span><\/a> with detailed video solutions. This article will look into some important Number System questions for the CAT Exam. If you want to practice these important Number System questions for CAT, you can download the PDF, which is completely Free.<\/p>\n Download Number System Questions for CAT PDF<\/a><\/p>\n CAT Online Coaching<\/a><\/p>\n Question 1:\u00a0<\/b>Suppose n is an integer such that the sum of digits on n is 2, and $10^{10} < n < 10^{11}$. The number of different values of n is<\/p>\n a)\u00a011<\/p>\n b)\u00a010<\/p>\n c)\u00a09<\/p>\n d)\u00a08<\/p>\n 1)\u00a0Answer\u00a0(A)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n The sum of digits should be 2. The \u00a0possibilities are 1000000001,1000000010,10000000100,..these 10 cases . Also additional 1 case where 20000000000. Hence option A .<\/p>\n Question 2:\u00a0<\/b>Each family in a locality has at most two adults, and no family has fewer than 3 children. a)\u00a04<\/p>\n b)\u00a05<\/p>\n c)\u00a02<\/p>\n d)\u00a03<\/p>\n 2)\u00a0Answer\u00a0(D)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n Consider 3 family. Let 1st one have 2 A , 1B and 2 G , 2nd one have 2 A , 2B and 1 G, 3rd one have 2 A , 2B and 1 G . SO total A-6 , B – 5 , g – 4 , F – 3 . Hence minimum is 3 .<\/p>\n Question 3:\u00a0<\/b>The remainder, when $(15^{23} + 23^{23})$ is divided by 19, is<\/p>\n a)\u00a04<\/p>\n b)\u00a015<\/p>\n c)\u00a00<\/p>\n d)\u00a018<\/p>\n 3)\u00a0Answer\u00a0(C)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n The remainder when $15^{23}$ is divided by 19 equals $(-4)^{23}$ Question 4:\u00a0<\/b>In NutsAndBolts 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 4)\u00a0Answer\u00a0(C)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/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 Question 5:\u00a0<\/b>The total number of integers pairs (x, y) satisfying the equation x + y = xy is<\/p>\n a)\u00a00<\/p>\n b)\u00a01<\/p>\n c)\u00a02<\/p>\n d)\u00a0None of the above<\/p>\n 5)\u00a0Answer\u00a0(C)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n xy = x + y<\/p>\n => xy – x – y = 0<\/p>\n => xy – x – y + 1 = 1<\/p>\n => (x – 1) (y – 1) = 1<\/p>\n => Both x – 1 and y – 1 have to be equal to 1 or -1.<\/p>\n So, values taken by (x,y) are (2,2) and (0,0).<\/p>\n => 2 solutions<\/p>\n Checkout: CAT Free Practice Questions and Videos<\/a><\/em><\/p>\n Question 6:\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 6)\u00a0Answer\u00a0(A)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/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 Question 7:\u00a0<\/b>What are the values of x and y that satisfy both the equations? a)\u00a0x = 2, y = 5<\/p>\n b)\u00a0x = 2.5, y = 6<\/p>\n c)\u00a0x =3, y = 5<\/p>\n d)\u00a0x = 3,y = 4<\/p>\n e)\u00a0x = 5,y = 2<\/p>\n 7)\u00a0Answer\u00a0(E)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n $2^{0.7x} * 3^{-1.25y} = 8\\sqrt{6}\/27$ =>\u00a0$2^{0.7x} * 3^{-1.25y}$ = $2^{3.5} * 3^{-2.5}$<\/p>\n => 0.7x = 3.5 => x = 5<\/p>\n => -1.25y = -2.5 => y = 2<\/p>\n $4^{0.3x} * 9^{0.2y} = 8*81^{1\/5}$ =>\u00a0$2^{0.6x} * 3^{0.4y}$ = $2^3 * 3^{0.8}$<\/p>\n => 0.6x = 3 => x = 5<\/p>\n => 0.4y = 0.8 => y = 2<\/p>\n => (5,2) is the solution.<\/p>\n Question 8:\u00a0<\/b>The sum of four consecutive two-digit odd numbers, when divided by 10, becomes a perfect square. Which of the following can possibly be one of these four numbers?<\/p>\n a)\u00a021<\/p>\n b)\u00a025<\/p>\n c)\u00a041<\/p>\n d)\u00a067<\/p>\n e)\u00a073<\/p>\n 8)\u00a0Answer\u00a0(C)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n Maximum sum of the four numbers <= 384=99+97+95+93 Question 9:\u00a0<\/b>The number of employees in Obelix Menhir Co. is a prime number and is less than 300. The ratio of the number of employees who are graduates and above, to that of employees who are not, can possibly be:<\/p>\n a)\u00a0101:88<\/p>\n b)\u00a087:100<\/p>\n c)\u00a0110:111<\/p>\n d)\u00a085:98<\/p>\n e)\u00a097:84<\/p>\n 9)\u00a0Answer\u00a0(E)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n The addition of numerator and denominatpr should give a prime no. Only option E gives that.<\/p>\n 3 is a factor of 189 and 183 => A and D eliminated<\/p>\n 17 is a factor of 187 and 221 => B and C eliminated<\/p>\n 181 is prime.<\/p>\n Checkout: CAT Free Practice Questions and Videos<\/strong><\/a><\/em><\/p>\n Question 10:\u00a0<\/b>If R = $(30^{65}-29^{65})\/(30^{64}+29^{64})$ ,then<\/p>\n a)\u00a0$0<R\\leq0.1$<\/p>\n b)\u00a0$0.1<R\\leq0.5$<\/p>\n c)\u00a0$0.5<R\\leq1.0$<\/p>\n d)\u00a0$R>1.0$<\/p>\n 10)\u00a0Answer\u00a0(D)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n $\\frac{(30^{65}-29^{65})}{(30^{64}+29^{64})} = ((30-29)*\\frac{(30^{64}+30^{63}*29+….+29^{64})}{(30^{64}+29^{64})}$ , which is greater than 1 . Hence option D.<\/p>\n Question 11:\u00a0<\/b>If x = $(16^3 + 17^3+ 18^3+ 19^3 )$, then x divided by 70 leaves a remainder of<\/p>\n a)\u00a00<\/p>\n b)\u00a01<\/p>\n c)\u00a069<\/p>\n d)\u00a035<\/p>\n 11)\u00a0Answer\u00a0(A)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n We know that x = $16^3 + 17^3 + 18^3 + 19^3 = (16^3 + 19^3) + (17^3 + 18^3)$<\/p>\n = $(16 + 19)(16^2 – 16 * 19 + 19^2) + (17 + 18)(17^2 – 17 * 18 + 18^2)$ = 35 \u00d7 odd + 35 \u00d7 odd = 35 \u00d7 even = 35 \u00d7 (2k)<\/p>\n =>\u00a0x = 70k<\/p>\n => Remainder when divided by 70 is 0.<\/p>\n Question 12:\u00a0<\/b>Let $n!=1*2*3* …*n$ for integer $n \\geq 1$. a)\u00a010<\/p>\n b)\u00a00<\/p>\n c)\u00a07<\/p>\n d)\u00a01<\/p>\n
\nConsidering all the families together, there are adults than boys, more boys than girls, and more girls than families.
\nThen the minimum possible number of families in the locality is<\/p>\n
\nThe remainder when $23^{23}$ is divided by 19 equals $4^{23}$
\nSo, the sum of the two equals$(-4)^{23}+(4)^{23}=0$<\/p>\n
\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
\n$2^{0.7x} * 3^{-1.25y} = 8\\sqrt{6}\/27$
\n$4^{0.3x} * 9^{0.2y} = 8*81^{1\/5}$<\/p>\n
\n384\/10 = 38.4
\nSo, the perfect square is a number less than 38.4
\nThe possibilities are 36, 25, 16 and 9
\nFor the sum to be 360, the numbers can be 87, 89, 91 and 93
\nThe sum of four consecutive odd numbers cannot be 250
\nFor the sum to be 160, the numbers can be 37,39,41 and 43
\nThe sum of 4 consecutive odd numbers cannot be 90
\nSo, from the options, the answer is 41.<\/p>\n
\nIf $p = 1!+(2*2!)+(3*3!)+… +(10*10!)$, then $p+2$ when divided by 11! leaves a remainder of<\/p>\n