Inequalities<\/b>\u00a0is an important topic in the CAT<\/strong> Quant<\/strong> (Algebra) section. <\/span>If you find these questions a bit tough, make sure you solve more CAT Inequalities questions. Learn all the important formulas<\/strong><\/span> and tricks on how to answer questions<\/strong><\/span><\/a><\/span> on<\/strong><\/span> Inequalities<\/strong><\/span><\/a>. You can check out these CAT Inequalities<\/span> questions from the CAT Previous year papers<\/strong><\/a>.<\/span> Practice a good number of sums in the CAT Inequalities<\/strong> so that you can answer these questions with ease in the exam. In this post, we will look into some important Inequalities Questions for CAT quants. These are a good source of practice for CAT 2022 preparation; If you want to practice these questions, you can download this Important CAT Inequalities Questions<\/strong> and answers PDF<\/strong> along with the video solutions below, which is completely Free.<\/p>\n Download Inequalities Questions for CAT<\/a><\/p>\n CAT Online Coaching<\/a><\/p>\n Question 1:\u00a0<\/b>Let a, b, c, d be four integers such that a+b+c+d = 4m+1 where m is a positive integer. Given m, which one of the following is necessarily true?<\/p>\n a)\u00a0The minimum possible value of $a^2 + b^2 + c^2 + d^2$ is $4m^2-2m+1$<\/p>\n b)\u00a0The minimum possible value of $a^2 + b^2 + c^2 + d^2$ is $4m^2+2m+1$<\/p>\n c)\u00a0The maximum possible value of $a^2 + b^2 + c^2 + d^2$ is $4m^2-2m+1$<\/p>\n d)\u00a0The maximum possible value of $a^2 + b^2 + c^2 + d^2$ is $4m^2+2m+1$<\/p>\n 1)\u00a0Answer\u00a0(B)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n Taking lowest possible positive value of m i.e. 1 . Such that a+b+c+d=5 , so atleast one of them must be grater than 1 ,<\/p>\n take a=b=c=1 and d=2<\/p>\n we get $a^2 + b^2 + c^2 + d^2 = 7$ which is equal to $4m^2+2m+1$ for other values it is greater than $4m^2+2m+1$ . so option B<\/p>\n Question 2:\u00a0<\/b>Given that $-1 \\leq v \\leq 1, -2 \\leq u \\leq -0.5$ and $-2 \\leq z \\leq -0.5$ and $w = vz \/u$ , then which of the following is necessarily true?<\/p>\n a)\u00a0$-0.5 \\leq w \\leq 2$<\/p>\n b)\u00a0$-4 \\leq w \\leq 4$<\/p>\n c)\u00a0$-4 \\leq w \\leq 2$<\/p>\n d)\u00a0$-2 \\leq w \\leq -0.5$<\/p>\n 2)\u00a0Answer\u00a0(B)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n We know $w = vz \/u$ so taking max value of u and min value of v and z to get min value of w which is -4.<\/p>\n Similarly taking min value of u and max value of v and z to get max value of w which is 4<\/p>\n Take v = 1, z = -2 and u = -0.5, we get w = 4<\/p>\n Take v = -1, z = -2 and \u00a0u = -0.5, we get w = -4<\/p>\n Question 3:\u00a0<\/b>If x, y, z are distinct positive real numbers the $(x^2(y+z) + y^2(x+z) + z^2(x+y))\/xyz$ would always be<\/p>\n a)\u00a0Less than 6<\/p>\n b)\u00a0greater than 8<\/p>\n c)\u00a0greater than 6<\/p>\n d)\u00a0Less than 8<\/p>\n 3)\u00a0Answer\u00a0(C)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n For the given expression value of x,y,z are distinct positive integers . So the value of expression will always be greater than value when all the 3 variables are equal . substitute x=y=z we get minimum value of 6 .<\/p>\n $(x^2(y+z) + y^2(x+z) + z^2(x+y))\/xyz$ = x\/z + x\/y + y\/z + y\/x + z\/y + z\/x<\/p>\n Applying AM greater than or equal to GM, we get minimum sum = 6<\/p>\n Question 4:\u00a0<\/b>The number of solutions of the equation 2x + y = 40 where both x and y are positive integers and x <= y is:<\/p>\n a)\u00a07<\/p>\n b)\u00a013<\/p>\n c)\u00a014<\/p>\n d)\u00a018<\/p>\n e)\u00a020<\/p>\n 4)\u00a0Answer\u00a0(B)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n y = 38 => x = 1<\/p>\n y = 36 => x = 2<\/p>\n …<\/p>\n …<\/p>\n y = 14 => x = 13<\/p>\n y = 12 => x = 14 => Cases from here are not valid as x > y.<\/p>\n Hence, there are 13 solutions.<\/p>\n Question 5:\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 5)\u00a0Answer\u00a0(A)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n Try to solve this type of questions using the options.<\/p>\n Subsitute 0 first => We ger -2 <=0, which is correct. Hence, 0 must be in the solution set.<\/p>\n Substitute 8 => 4 + 2 – 2 <=0 => 6 <= 0, which is false. Hence, 8 must not be in the solution set.<\/p>\n => Option 1 is the answer.<\/p>\n Question 6:\u00a0<\/b>If x > 5 and y < -1, then which of the following statements is true?<\/p>\n a)\u00a0(x + 4y) > 1<\/p>\n b)\u00a0x > -4y<\/p>\n c)\u00a0-4x < 5y<\/p>\n d)\u00a0None of these<\/p>\n 6)\u00a0Answer\u00a0(D)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n Substitute x=6 and y=-6 ,<\/p>\n x+4y = -18<\/p>\n x = 6, -4y = 24<\/p>\n -4x = -24, 5y = -30<\/p>\n So\u00a0none of the options out of a,b or c satisfies .<\/p>\n Question 7:\u00a0<\/b>x and y are real numbers satisfying the conditions 2 < x < 3 and – 8 < y < -7. Which of the following expressions will have the least value?<\/p>\n a)\u00a0$x^2y$<\/p>\n b)\u00a0$xy^2$<\/p>\n c)\u00a0$5xy$<\/p>\n d)\u00a0None of these<\/p>\n 7)\u00a0Answer\u00a0(C)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n $xy^2$ will have it’s least value when y=-7 and x=2 and equals 98. $x^2y$ will have it’s least value when y=-8 and x=3 and equals -72. $5xy$ will have it’s least value when y=-8 and x=3 and equals -120 So, of the three expressions, the least possible value is that of 5xy a)\u00a04<\/p>\n b)\u00a05<\/p>\n c)\u00a08<\/p>\n d)\u00a0None of these<\/p>\n 8)\u00a0Answer\u00a0(D)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n $n^3 – 7n^2 + 11n – 5 = (n-1)(n^2 – 6n +5) = (n-1)(n-1)(n-5)$ Question 9:\u00a0<\/b>If a, b, c and d are four positive real numbers such that abcd = 1, what is the minimum value of (1 + a)(1+b)(1+c)(1+d)?<\/p>\n a)\u00a04<\/p>\n b)\u00a01<\/p>\n c)\u00a016<\/p>\n d)\u00a018<\/p>\n 9)\u00a0Answer\u00a0(C)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n Since the product is constant,<\/p>\n (a+b+c+d)\/4 >= $(abcd)^{1\/4}$<\/p>\n We know that abcd = 1.<\/p>\n Therefore, a+b+c+d >= 4<\/p>\n $(a+1)(b+1)(c+1)(d+1)$ Question 10:\u00a0<\/b>Let x and y be two positive numbers such that $x + y = 1.$ a)\u00a012<\/p>\n b)\u00a020<\/p>\n c)\u00a012.5<\/p>\n d)\u00a013.3<\/p>\n 10)\u00a0Answer\u00a0(C)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n Approach 1:<\/p>\n The given expression is symmetric in x and y and the limiting condition (x+y=1) is also symmetric in x and y.<\/p>\n =>This means that the expression attains the minimum value when x = y Approach 2:<\/p>\n $(x+1\/x)^2$ +\u00a0$(y+1\/y)^2$ =\u00a0$(x+1\/x+y+1\/y)^2$ – $2*(x+1\/x)(y+1\/y)$<\/p>\n Let x+1\/x and y+1\/y be two terms. Thus\u00a0(x+1\/x+y+1\/y)\/2 would be their Arithmetic Mean(AM) and $\\sqrt{(x+1\/x)(y+1\/y)}$ would be their Geometric Mean (GM).<\/p>\n Therefore, we can express the above equation as\u00a0$(x+1\/x)^2$ +\u00a0$(y+1\/y)^2$ = $4AM^2$ – $GM^2$. As AM >= GM, the minimum value of expression would be attained when AM = GM.<\/p>\n When AM = GM, both terms are equal. That is x+1\/x = y\u00a0+1\/y.<\/p>\n Substituting y=1-x we get<\/p>\n x+1\/x = (1-x) + 1\/(1-x)<\/p>\n On solving we get 2x-1 = (2x-1)\/ x(1-x)<\/p>\n So either 2x-1 = 0 or x(1-x) = 1<\/p>\n x(1-x) = x * y<\/p>\n As x and y are positive numbers whose sum = 1, 0<= x, y\u00a0<=1. Hence, their product cannot be 1.<\/p>\n Thus, 2x-1 = 0 or x=1\/2<\/p>\n => y = 1\/2<\/p>\n So, the value = $(x+\\frac{1}{x})^2+(y+\\frac{1}{y})^2$ =\u00a0$(2+\\frac{1}{2})^2+(2+\\frac{1}{2})^2$ = 12.5<\/p>\n Question 11:\u00a0<\/b>If x>2 and y>-1,then which of the following statements is necessarily true?<\/p>\n a)\u00a0xy>-2<\/p>\n b)\u00a0-x<2y<\/p>\n c)\u00a0xy<-2<\/p>\n d)\u00a0-x>2y<\/p>\n 11)\u00a0Answer\u00a0(B)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n This kind of questions must be solved using the counter example method.<\/p>\n x = 100 and y = -1\/2 rules out option a) a)\u00a05\/3<\/p>\n b)\u00a0$\\sqrt{19}$<\/p>\n c)\u00a013\/3<\/p>\n d)\u00a0None of these<\/p>\n 12)\u00a0Answer\u00a0(C)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n The given equations are \u00a0x + y + z = 5 — (1) , xy + yz + zx = 3 — (2) =>$b^2-4ac>0$<\/p>\n => $ ( x – 5)^2 – 4(x ^2 – 5x +3 ) \\geq 0$ Question 13:\u00a0<\/b>If $x^2 + 5y^2 + z^2 = 2y(2x+z)$, then which of the following statements is(are) necessarily true? a)\u00a0Only A<\/p>\n b)\u00a0B and C<\/p>\n c)\u00a0A and B<\/p>\n d)\u00a0None of these<\/p>\n 13)\u00a0Answer\u00a0(C)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n The equation is not satisfied for only x = 2y.<\/p>\n Using statements B and C, i.e., x = 2z and 2x = z, we see that the equation is not satisfied.<\/p>\n Using statements A and B, i.e., x = 2y and x = 2z, i.e., z = y = x\/2, the equation is satisfied.<\/p>\n Option c) is the correct answer.<\/p>\n Question 14:\u00a0<\/b>If u, v, w and m are natural numbers such that $u^m + v^m = w^m$, then which one of the following is true?<\/p>\n a)\u00a0m >= min(u, v, w)<\/p>\n b)\u00a0m >= max(u, v, w)<\/p>\n c)\u00a0m < min(u, v, w)<\/p>\n d)\u00a0None of these<\/p>\n 14)\u00a0Answer\u00a0(D)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n Substitute value of u = v = 2, w = 4 and m = 1. Here the condition holds and options A and B are false. Hence, we can eliminate options A and B.<\/p>\n Substitute u = v = 1, w=2 and m= 1. Here m=min(u, v, w). Hence, option C also does not hold. Hence, we can eliminate option C.<\/p>\n Option d) is the correct answer.<\/p>\n Question 15:\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 15)\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 16:\u00a0<\/b>From any two numbers $x$ and $y$, we define $x* y = x + 0.5y – xy$ . Suppose that both $x$ and $y$ are greater than 0.5. Then a)\u00a01 > x > y<\/p>\n b)\u00a0x > 1 > y<\/p>\n c)\u00a01 > y > x<\/p>\n d)\u00a0y > 1 > x<\/p>\n
\nSo $xy^2>98$<\/p>\n
\nSo, $x^2y > -72$<\/p>\n
\nSo, $5xy > -120$<\/p>\n
\nQuestion 8:\u00a0<\/b>$m$ is the smallest positive integer such that for any integer $n \\geq m$, the quantity $n^3 – 7n^2 + 11n – 5$ is positive. What is the value of $m$?<\/p>\n
\nThis is positive for n > 5
\nSo, m = 6<\/p>\n
\n= $1+a+b+c+d+ ab + ac + ad + bc + bd + cd+ abc+ bcd+ cda+ dab+abcd$
\nWe know that $abcd = 1$
\nTherefore, $a = 1\/bcd, b = 1\/acd, c = 1\/bda$ and $d = 1\/abc$
\nAlso, $cd = 1\/ab, bd = 1\/ac, bc = 1\/ad$
\nThe expression can be clubbed together as $1 + abcd + (a + 1\/a) + (b+1\/b) + (c+1\/c) + (d+1\/d) + (ab + 1\/ab) + (ac+ 1\/ac) + (ad + 1\/ad)$
\nFor any positive real number $x$, $x + 1\/x \\geq 2$
\nTherefore, the least value that $(a+1\/a), (b+1\/b)….(ad+1\/ad)$ can take is 2.
\n$(a+1)(b+1)(c+1)(d+1) \\geq 1 + 1 + 2 + 2 + 2+ 2 + 2 + 2 + 2$
\n=> $(a+1)(b+1)(c+1)(d+1) \\geq 16$
\nThe least value that the given expression can take is 16. Therefore, option\u00a0C is the right answer.<\/p>\n
\nThen the minimum value of $(x+\\frac{1}{x})^2+(y+\\frac{1}{y})^2$ is<\/p>\n
\nx=y=1\/2
\nSo, the value = $(x+\\frac{1}{x})^2+(y+\\frac{1}{y})^2$ =\u00a0$(2+\\frac{1}{2})^2+(2+\\frac{1}{2})^2$ =12.5<\/p>\n
\nx = 3 and y = 0 rules out options c) and d)
\nOption b) is correct.
\nQuestion 12:\u00a0<\/b>If x, y and z are real numbers such that x + y + z = 5 and xy + yz + zx = 3, what is the largest value that x can have?<\/p>\n
\nxy + yz + zx = 3
\nx(y + z) + yz = 3
\n=> x ( 5 -x ) +y ( 5 – x – y) = 3
\n=> $ – y^2 – y (5 -x) – x ^2 + 5x = 3$
\n=> $ y^2 + y (x-5) + ( x ^2 – 5x +3) \u00a0= 0 $
\nThe above equation should have real roots for y, => Determinant >= 0<\/p>\n
\n=> \u00a0$ 3x^2 -10x – 13 \\leq 0$
\n=> \u00a0$ -1 \\leq x \\leq \\frac{13}{3}$
\nHence maximum value x can take is $\\frac{13}{3}$, and the corresponding values for y,z are $\\frac{1}{3},\\frac{1}{3}$<\/p>\n
\nA. x = 2y B. x = 2z C. 2x = z<\/p>\n
\n$x* x < y* y$ if<\/p>\n