Download Inequalites Questions for CAT<\/a><\/p>\n Enroll for CAT 2022 Crash Course<\/a><\/p>\n Question 1:\u00a0<\/b>If n is a positive integer such that $(\\sqrt[7]{10})(\\sqrt[7]{10})^{2}…(\\sqrt[7]{10})^{n}>999$, then the smallest value of n is<\/p>\n 1)\u00a0Answer:\u00a06<\/b><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n $(\\sqrt[7]{10})(\\sqrt[7]{10})^{2}…(\\sqrt[7]{10})^{n}>999$<\/p>\n $(\\sqrt[7]{10})^{1+2+…+n}>999$<\/p>\n $10^{\\frac{1+2+…+n}{7}}>999$<\/p>\n For minimum value of n,<\/p>\n $\\frac{1+2+…+n}{7}=3$<\/p>\n 1 + 2 + … + n = 21<\/p>\n We can see that if n = 6, 1 + 2 + 3 + … + 6 = 21.<\/p>\n Question 2:\u00a0<\/b>The number of distinct pairs of integers (m,n), satisfying $\\mid1+mn\\mid<\\mid m+n\\mid<5$ is:<\/p>\n 2)\u00a0Answer:\u00a012<\/b><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n Let us break this up into 2 inequations [ Let us assume x as m and y as n ]<\/p>\n | 1 + mn | < | m + n |<\/p>\n | m + n | < 5<\/p>\n Looking at these expressions, we can clearly tell that the graphs will be symmetrical about the origin.<\/p>\n Let us try out with the first quadrant and extend the results to the other quadrants.<\/p>\n We will also consider the +X and +Y axes along with the quadrant.<\/p>\n So, the first inequality becomes,<\/p>\n 1 + mn < m + n<\/p>\n 1 + mn – m – n < 0<\/p>\n 1 – m + mn – n < 0<\/p>\n (1-m) + n(m-1) < 0<\/p>\n (1-m)(1-n)\u00a0< 0<\/p>\n (m – 1)(n – 1) < 0<\/p>\n Let us try to plot the graph.<\/p>\n If we consider only mn < 0, then we get<\/p>\n But, we have (m – 1)(n – 1) < 0, so we need to shift the graphs by one unit towards positive x and positive y.<\/p>\n So, we have,<\/p>\n But, we are only considering the first quadrant and the +X and +Y axes. Hence, if we extend, we get the following region.<\/p>\n So, if we look for only integer values, we get<\/p>\n (0,2), (0,3),…….<\/p>\n (0,-2), (0, -3),……<\/p>\n (2,0), (3,0), ……<\/p>\n (-2,0), (-3,0), …….<\/p>\n Now, let us consider the other inequation as well, in which |x + y| < 5<\/p>\n Since one of the values is always zero, the modulus of the other value is less than or equal to 4.<\/p>\n Hence, we get<\/p>\n (0,2), (0,3), (0,4)<\/p>\n (0,-2), (0, -3), (0, -4)<\/p>\n (2,0), (3,0), (4,0)<\/p>\n (-2,0), (-3,0), (-4,0)<\/p>\n Hence, a total of 12 values.<\/p>\n Question 3:\u00a0<\/b>For a real number x the condition $\\mid3x-20\\mid+\\mid3x-40\\mid=20$ necessarily holds if<\/p>\n a)\u00a0$10<x<15$<\/p>\n b)\u00a0$9<x<14$<\/p>\n c)\u00a0$7<x<12$<\/p>\n d)\u00a0$6<x<11$<\/p>\n 3)\u00a0Answer\u00a0(C)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n Case 1 : $x\\ge\\frac{40}{3}$ Question 4:\u00a0<\/b>$f(x) = \\frac{x^2 + 2x – 15}{x^2 – 7x – 18}$ is negative if and only if<\/p>\n a)\u00a0-5 < x < -2 or 3 < x < 9<\/p>\n b)\u00a0x < -5 or -2 < x < 3<\/p>\n c)\u00a0-2 < x < 3 or x > 9<\/p>\n d)\u00a0x < -5 or 3 < x < 9<\/p>\n 4)\u00a0Answer\u00a0(A)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n $f(x) = \\frac{x^2 + 2x – 15}{x^2 – 7x – 18}$<0<\/p>\n $\\frac{\\left(x+5\\right)\\left(x-3\\right)}{\\left(x-9\\right)\\left(x+2\\right)}<0$<\/p>\n We have four inflection points -5, -2, 3, and 9.<\/p>\n For x<-5, all four terms (x+5), (x-3), (x-9), (x+2) will be negative. Hence, the overall expression will be positive. Similarly, when x>9, all four terms will be positive.<\/p>\n When x belongs to (-2,3), two terms are negative and two are positive. Hence, the overall expression is positive again.<\/p>\n We are left with the range (-5,-2) and (3,9) where the expression will be negative.<\/p>\n Question 5:\u00a0<\/b>The number of integers n that satisfy the inequalities $\\mid n – 60 \\mid < \\mid n – 100 \\mid < \\mid n – 20 \\mid$ is<\/p>\n a)\u00a021<\/p>\n b)\u00a019<\/p>\n c)\u00a018<\/p>\n d)\u00a020<\/p>\n 5)\u00a0Answer\u00a0(B)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n We have\u00a0$\\mid n – 60 \\mid < \\mid n – 100 \\mid < \\mid n – 20 \\mid$.<\/p>\n Now, the difference inside the modulus signified the distance of n from 60, 100, and 20 on the number line. This means that when the absolute difference from a number is larger, n would be further away from that number.<\/p>\n Example: The absolute difference of n and 60 is less than that of the absolute difference between n and 20. Hence, n cannot be $\\le\\ 40$, as then it would be closer to 20 than 60, and closer on the number line would indicate lesser value of absolute difference.\u00a0Thus we have the condition that n>40.<\/p>\n The absolute difference of n and 100 is less than that of the absolute difference between n and 20. Hence, n cannot be $\\le\\ 60$, as then it would be closer to 20 than 100. Thus we have the condition that n>60.<\/p>\n The absolute difference of n and 60 is less than that of the absolute difference between n and 100. Hence, n cannot be $\\ge80$, as then it would be closer to 100 than 60. Thus we have the condition that n<80.<\/p>\n The number which satisfies the conditions are 61, 62, 63, 64……79. Thus, a total of 19 numbers.<\/p>\n Alternatively<\/p>\n as per the given condition :\u00a0$\\mid n – 60 \\mid < \\mid n – 100 \\mid < \\mid n – 20 \\mid$.<\/p>\n Dividing the range of n into 4 segments. (n < 20, 20<n<60, 60<n<100, n > 100 )<\/p>\n 1) For n < 20.<\/p>\n |n-20| = 20-n,\u00a0|n-60| = 60- n,\u00a0|n-100| = 100-n<\/p>\n considering the inequality part :$\\left|n-100\\right|<\\ \\left|n\\ -20\\right|$<\/p>\n 100 -n < 20 -n,<\/p>\n No value of n satisfies this condition.<\/p>\n 2)\u00a0For 20 < n < 60.<\/p>\n |n-20| = n-20, |n-60| = 60- n, |n-100| = 100-n.<\/p>\n 60- n < 100 – n and 100 – n < n – 20<\/p>\n For 100 -n < n – 20.<\/p>\n 120 < 2n and n > 60. But for the considered range n is less than 60.<\/p>\n 3)\u00a0For 60 < n < 100<\/p>\n |n-20| = n-20, |n-60| = n-60, |n-100| = 100-n<\/p>\n n-60 < 100-n and 100-n < n-20.<\/p>\n For the first part 2n < 160 and for the second part 120 < 2n.<\/p>\n n takes values from 61 …………….79.<\/p>\n A total of 19 values<\/p>\n 4)\u00a0For n > 100<\/p>\n |n-20| = n-20, |n-60| = n-60, |n-100| = n-100<\/p>\n n-60 < n – 100.<\/p>\n No value of n in the given range\u00a0satisfies the given inequality.<\/p>\n Hence a total of 19 values satisfy the inequality.<\/p>\n Checkout: CAT Free Practice Questions and Videos<\/strong><\/a><\/em><\/p>\n Question 6:\u00a0<\/b>if x and y are positive real numbers satisfying $x+y=102$, then the minimum possible valus of $2601(1+\\frac{1}{x})(1+\\frac{1}{y})$ is<\/p>\n 6)\u00a0Answer:\u00a02704<\/b><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n Now we have\u00a0$2601\\left(1+\\frac{1}{x}\\right)\\left(1+\\frac{1}{y}\\right)=2601\\left(\\frac{xy+y+x+1}{xy}\\right)$<\/p>\n Now we know that x+y=102. Substituting it in the above equation<\/p>\n $2601\\left(\\frac{xy+y+x+1}{xy}\\right)=2601\\left(\\frac{103}{xy}+1\\right)$<\/p>\n Maximum value of xy\u00a0 can be found out by AM>= GM relationship<\/p>\n $\\ \\frac{\\ x+y}{2}\\ge\\ \\sqrt{xy}\\ or\\ \\ \\sqrt{\\ xy}\\le\\ 51\\ or\\ xy\\le\\ 2601$<\/p>\n Hence the maximum value of “xy” is 2601. Substituting in the above equation we get<\/p>\n $2601\\left(\\ \\frac{\\ 103+2601}{2601}\\right)=2704$<\/p>\n Question 7:\u00a0<\/b>For real x, the maximum possible value of $\\frac{x}{\\sqrt{1+x^{4}}}$ is<\/p>\n a)\u00a0$\\frac{1}{2}$<\/p>\n b)\u00a0$1$<\/p>\n c)\u00a0$\\frac{1}{\\sqrt{3}}$<\/p>\n d)\u00a0$\\frac{1}{\\sqrt{2}}$<\/p>\n 7)\u00a0Answer\u00a0(D)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n Now\u00a0$\\frac{x}{\\sqrt{\\ 1+x^4}}=\\ \\frac{\\ 1}{\\sqrt{\\ \\ \\frac{\\ 1+x^4}{x^2}}}=\\frac{1}{\\sqrt{\\ \\frac{1}{x^2}+x^2}}$<\/p>\n Applying A.M>= G.M.<\/p>\n $\\frac{\\left(\\frac{1}{x^2}+x^2\\right)}{2}\\ge\\ 1\\ or\\ \\ \\frac{1}{x^2}+x^2\\ge\\ 2$ Substituting we get the maximum possible value of the equation as\u00a0$\\frac{1}{\\sqrt{\\ 2}}$<\/p>\n Question 8:\u00a0<\/b>The number of pairs of integers $(x,y)$ satisfying $x\\geq y\\geq-20$ and $2x+5y=99$<\/p>\n 8)\u00a0Answer:\u00a017<\/b><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n We have 2x + 5y = 99 or\u00a0$x=\\frac{\\left(99-5y\\right)}{2}$<\/p>\n Now $x\\ge\\ y\\ \\ge\\ -20$ ;\u00a0So\u00a0$\\frac{\\left(99-5y\\right)}{2}\\ge\\ y\\ ;\\ 99\\ge7y\\ or\\ y\\le\\ \\approx\\ 14$<\/p>\n So $-20\\le y\\le14$. Now for this range of “y”, we have to find all the integral values of “x”. As the coefficient of “x” is 2,<\/p>\n then (99 –\u00a05y) must be even, which will happen when “y” is odd. However, there are only 17 odd values of “y” be -20 and 14.<\/p>\n Hence the number of possible values is 17.<\/p>\n Question 9:\u00a0<\/b>Among 100 students, $x_1$ have birthdays in January, $X_2$ have birthdays in February, and so on. If $x_0=max(x_1,x_2,….,x_{12})$, then the smallest possible value of $x_0$ is<\/p>\n a)\u00a08<\/p>\n b)\u00a09<\/p>\n c)\u00a010<\/p>\n d)\u00a012<\/p>\n 9)\u00a0Answer\u00a0(B)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n $x_0=max(x_1,x_2,….,x_{12})$<\/p>\n $x_0$ will be minimum if x1,x2…x12 are close to each other<\/p>\n 100\/12=8.33<\/p>\n .’.\u00a0max$(x_1,x_2,….,x_{12})$ will be minimum if\u00a0$(x_1,x_2,….,x_{12})$=(9,9,9,9,8,8,8,8,8,8,8,8,)<\/p>\n .’. Option B is correct.<\/p>\n Question 10:\u00a0<\/b>If x is a real number, then $\\sqrt{\\log_{e}{\\frac{4x – x^2}{3}}}$ is a real number if and only if<\/p>\n a)\u00a0$1 \\leq x \\leq 3$<\/p>\n b)\u00a0$1 \\leq x \\leq 2$<\/p>\n c)\u00a0$-1 \\leq x \\leq 3$<\/p>\n d)\u00a0$-3 \\leq x \\leq 3$<\/p>\n 10)\u00a0Answer\u00a0(A)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n $\\sqrt{\\log_{e}{\\frac{4x – x^2}{3}}}$ will be real if\u00a0$\\log_e\\ \\frac{\\ 4x-x^2}{3}\\ \\ge\\ 0$<\/p>\n $\\frac{\\ 4x-x^2}{3}\\ >=\\ 1$<\/p>\n $\\ 4x-x^2-3\\ >=\\ 0$<\/p>\n $\\ x^2-4x+3\\ =<\\ 0$<\/p>\n 1=< x=< 3<\/p>\n Question 11:\u00a0<\/b>The smallest integer $n$ such that $n^3-11n^2+32n-28>0$ is<\/p>\n 11)\u00a0Answer:\u00a08<\/b><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n We can see that at n = 2, $n^3-11n^2+32n-28=0$ i.e. (n-2) is a factor of $n^3-11n^2+32n-28$<\/p>\n $\\dfrac{n^3-11n^2+32n-28}{n-2}=n^2-9n+14$<\/p>\n We can further factorize\u00a0n^2-9n+14 as (n-2)(n-7).<\/p>\n $n^3-11n^2+32n-28=(n-2)^2(n-7)$<\/p>\n $\\Rightarrow$\u00a0$n^3-11n^2+32n-28>0$<\/p>\n $\\Rightarrow$\u00a0$(n-2)^2(n-7)>0$<\/p>\n Therefore, we can say that n-7>0<\/p>\n Hence, n$_{min}$ = 8<\/p>\n Question 12:\u00a0<\/b>The minimum possible value of the sum of the squares of the roots of the equation $x^2+(a+3)x-(a+5)=0 $ is<\/p>\n a)\u00a01<\/p>\n b)\u00a02<\/p>\n c)\u00a03<\/p>\n d)\u00a04<\/p>\n 12)\u00a0Answer\u00a0(C)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n Let the roots of the equation\u00a0$x^2+(a+3)x-(a+5)=0 $ be equal to $p,q$<\/p>\n Hence, $p+q = -(a+3)$ and $p \\times q = -(a+5)$<\/p>\n Therefore, $p^2+q^2 = a^2+6a+9+2a+10 = a^2+8a+19 = (a+4)^2+3$<\/p>\n As $(a+4)^2$ is always non negative, the least value of the sum of squares is 3<\/p>\n Question 13:\u00a0<\/b>For how many integers n, will the inequality $(n – 5) (n – 10) – 3(n – 2)\\leq0$ be satisfied?<\/p>\n 13)\u00a0Answer:\u00a011<\/b><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n $(n – 5) (n – 10) – 3(n – 2)\\leq0$ Question 14:\u00a0<\/b>If a and b are integers of opposite signs such that $(a + 3)^{2} : b^{2} = 9 : 1$ and $(a -1)^{2}:(b – 1)^{2} = 4:1$, then the ratio $a^{2} : b^{2}$ is<\/p>\n a)\u00a09:4<\/p>\n b)\u00a081:4<\/p>\n c)\u00a01:4<\/p>\n d)\u00a025:4<\/p>\n 14)\u00a0Answer\u00a0(D)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n Since the square root can be positive or negative we will get two cases for each of the equation.<\/p>\n For the first one,<\/p>\n a + 3 = 3b .. i<\/p>\n a + 3 = -3b … ii<\/p>\n For the second one,<\/p>\n a – 1 = 2(b -1) … iii<\/p>\n a – 1 = 2 (1 – b) … iv<\/p>\n we have to solve i and iii, i and iv, ii and iii, ii and iv.<\/p>\n Solving i and iii,<\/p>\n a + 3 = 3b and a = 2b\u00a0 – 1, solving, we get a = 3 and b = 2, which is not what we want.<\/p>\n Solving i and iv<\/p>\n a + 3 = 3b and a = 3 – 2b, solving, we get b = 1.2, which is not possible.<\/p>\n Solving ii and iii<\/p>\n a + 3 = -3b and a = 2b – 1, solving, we get b = 0.4, which is not possible.<\/p>\n Solving ii and iv,<\/p>\n a + 3 = -3b and a = 3 – 2b, solving, we get a = 15 and b = -6 which is what we want.<\/p>\n Thus, $\\frac{a^2}{b^2} = \\frac{25}{4}$<\/p>\n Question 15:\u00a0<\/b>If | r – 6 | = 11 and | 2q – 12 | = 8, what is the minimum possible value of q \/ r?<\/p>\n a)\u00a0-2\/5<\/p>\n b)\u00a02\/17<\/p>\n c)\u00a010\/17<\/p>\n d)\u00a0None of these<\/p>\n 15)\u00a0Answer\u00a0(D)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n | r-6 | = 11 => r = -5 or 17<\/p>\n | 2q – 12 | = 8 => q = 10 or 2<\/p>\n So, the minimum possible value of q\/r = 10\/(-5) = -2<\/p>\n Question 16:\u00a0<\/b>The number of positive integer valued pairs (x, y), satisfying 4x – 17 y = 1 and x < 1000 is:<\/p>\n a)\u00a059<\/p>\n b)\u00a057<\/p>\n c)\u00a055<\/p>\n d)\u00a058<\/p>\n 16)\u00a0Answer\u00a0(A)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n y = $\\frac{4x-1}{17}$<\/p>\n The integral values of x for which y is an integer are 13, 30, 47,……<\/p>\n The values are in the form 17n + 13, where $ n \\geq 0$<\/p>\n 17n + 13 < 1000<\/p>\n => 17n < 987<\/p>\n => n < 58.05<\/p>\n => n can take values from 0 to 58 => Number of values = 59<\/p>\n Question 17:\u00a0<\/b>Which of the following values of x do not satisfy the inequality $(x^2 – 3x + 2 > 0)$ at all?<\/p>\n a)\u00a0$1\\leq x \\leq 2$<\/p>\n b)\u00a0$-1\\geq x \\geq -2$<\/p>\n c)\u00a0$0 \\leq x \\leq 2$<\/p>\n d)\u00a0$0\\geq x \\geq -2$<\/p>\n 17)\u00a0Answer\u00a0(A)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n After solving given equation, we will have inequality resolved to:<\/p>\n (x-1)(x-2)>0<\/p>\n Or we can say range of x will be as follows:<\/p>\n x<1; \u00a0x>2<\/p>\n Hence, option A has a set of values which don’t lie in the possible range of x.<\/p>\n So the\u00a0answer will be A.<\/p>\n Question 18:\u00a0<\/b>The number of integers n satisfying -n+2\u00a0\u2265\u00a00 and 2n \u2265\u00a04 is<\/p>\n a)\u00a00<\/p>\n b)\u00a01<\/p>\n c)\u00a02<\/p>\n d)\u00a03<\/p>\n 18)\u00a0Answer\u00a0(B)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n -n+2 >= 0 Question 19:\u00a0<\/b>If pqr = 1, the value of the expression $1\/(1+p+q^{-1}) + 1\/(1+q+r^{-1}) + 1\/(1+r+p^{-1})$<\/p>\n a)\u00a0p+q+r<\/p>\n b)\u00a01\/(p+q+r)<\/p>\n c)\u00a01<\/p>\n d)\u00a0$p^{-1} + q^{-1} + r^{-1}$<\/p>\n 19)\u00a0Answer\u00a0(C)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n Let p = q = r = 1<\/p>\n So, the value of the expression becomes 1\/3 + 1\/3 + 1\/3 = 1<\/p>\n If we substitute these values, options a), b) and d) do not satisfy.<\/p>\n Option c) is the answer.<\/p>\n Question 20:\u00a0<\/b>What values of x satisfy $x^{2\/3} + x^{1\/3} – 2 <= 0$?<\/p>\n a)\u00a0$-8 \\leq x \\leq 1$<\/p>\n b)\u00a0$-1 \\leq x \\leq 8$<\/p>\n c)\u00a0$1 \\leq x \\leq 8$<\/p>\n d)\u00a0$1 \\leq x \\leq 18$<\/p>\n e)\u00a0$-8 \\leq x \\leq 8$<\/p>\n 20)\u00a0Answer\u00a0(A)<\/strong><\/p>\n![](\"https:\/\/media-cdn.cracku.in\/uploads\/image_RtkDUBh.png\")
<\/p>\n![](\"https:\/\/media-cdn.cracku.in\/uploads\/image_1AGMDmy.png\")
<\/p>\n![](\"https:\/\/media-cdn.cracku.in\/uploads\/image_vxsnVhw.png\")
<\/p>\n
\nwe get 3x-20 +3x-40 =20
\n6x=80
\nx=$\\frac{80}{6}$=$\\frac{40}{3}$=13.33
\nCase 2 :$\\frac{20}{3}\\le\\ x<\\frac{40}{3}\\ $
\nwe get 3x-20+40-3x =20
\nwe get 20=20
\nSo we get x$\\in\\ \\left[\\frac{20}{3},\\frac{40}{3}\\right]$
\nCase 3 $x<\\frac{20}{3}$
\nwe get 20-3x+40-3x =20
\n40=6x
\nx=$\\frac{20}{3}$
\nbut this is not possible
\nso we get from case 1,2 and 3
\n$\\frac{20}{3}\\le\\ x\\le\\frac{40}{3}$
\nNow looking at options
\nwe can say only option C satisfies for all x .
\nHence 7<x<12.<\/p>\n
\n=> $ n^2 – 15n + 50 – 3n + 6 \\leq 0$
\n=> $n^2 – 18n + 56 \\leq 0$
\n=> $(n – 4)(n – 14) \\leq 0$
\n=> Thus, n can take values from 4 to 14. Hence, the required number of values are 14 – 4 +\u00a01 = 11.<\/p>\n
\nor n<=2
\nand 2n>=4
\nor n>=2
\nSo we can take only one value of n i.e. 2<\/p>\n