Arranging the equation, we know that there are no solutions when the lines are parallel:for that the condition had to&nbsp;$\frac{p}{3}=-\frac{4}{k}$ ≠ $\frac{2}{a}$Checking through options:&nbsp;Option A: using the first and last terms in our relation, we see that ap must not be equal to 6 for the lines to be parallel. This option puts no conditions on that and thus is not relevant.&nbsp;Option C: This question is the opposite of what we want; if this is true, the lines can never be parallel.&nbsp;Option D: Using the first and second terms of the relation, we see that we want kp = -12, or kp-12 = 0. Hence, this statement is not what we want.&nbsp;Option B: Using the second and third terms, we see that we do not want k equals to -2a, or we do not want k+2a to be equal to 0Therefore, B is a condition that is necessary for the lines to be parallel and have no solution.&nbsp;Therefore, Option&nbsp;B is the correct answer.&nbsp;

It is given that&nbsp;for some real numbers a and b, the system of equations&nbsp;x+y=4 and&nbsp;(a + 5)x + ($b^2$ − 15)y= 8b has infinitely many solutions for x and y.Hence, we can say that&nbsp;=&gt;&nbsp;&nbsp;$\frac{a + 5}{1}$ = $\frac{b^2 - 15}{1}$ = $\frac{8b}{4}$​This equation can be used to find the value of a, and b.Firstly,&nbsp;we will determine the value of b.&nbsp;=&gt;&nbsp;&nbsp;&nbsp;$\frac{b^2 - 15}{1}$ = $\frac{8b}{4}$​ =&gt; $b^2 - 2b - 15$ = 0=&gt; (b - 5)(b + 3) = 0Hence, the values of b are 5, and -3, respectively.&nbsp;The value of a can be expressed in terms of b, which is&nbsp;a + 5 = $b^2 − 15$&nbsp;=&gt;&nbsp;a&nbsp;= $b^2 − 20$When&nbsp;b=5, a = $5^2$ − 20 = 5When&nbsp;b=−3, a = $3^2$ − 20 = −11The maximum value of&nbsp;ab = (−3)⋅(−11) = 33

Let us assume the initial stock of all the fruits is S.Let us take we have 'b' and 'a' mangoes initially.Stock of Mangoes = 40% of S = 2S/5The total number of fruits sold are Mangoes Sold + Apples Sold + Bananas Sold= 2S/10 + 96 + 4a/10 = S/2 (Given)=&gt; S/5 + 96 + 2a/5 = S/2=&gt; S =&nbsp;$\frac{(4a+960)}{3}$=&gt; $\frac{4a}{3} + 320$'a' has to be a multiple of 3 for the above term to be an integer.But 'a' has to be a multiple of 5 for 4a/10 to be an integer.=&gt; The smallest value of 'a' satisfying both conditions is 15.=&gt;&nbsp;$\frac{4a}{3}$ + 320 = $\frac{4(15)}{3}$ + 320 = 340

xy + yz = 19y(x + z) = 19Observe that 19 is prime and can’t be factorized further. Since x, y and z are all naturalnumbers, the value of (x + z) can’t be 1. (It should at least be 2). Therefore,&nbsp;y = 1&nbsp;and ( x + z) = 19yz + xz = 51z(y + x) = 51z(x + 1) = 5151 = 17 * 3 Case 1: z = 17, x + 1 = 3x = 2, y = 1, z = 17xyz = 2 * 1 * 17 = 34 Case 2: z = 3, x + 1 = 17x = 16, y = 1, z = 3xyz = 16 * 1 * 3 = 48Hence the minimum value of x * y * z = 34Hence, the answer is '34'

In the question, it is given that&nbsp;the equation&nbsp;∣x+a∣+∣x−1∣=2∣x+a∣+∣x−1∣=2&nbsp;has an infinite number of solutions for any value of x. This is possible when x in |x+a| and x in |x-1| cancels out.Case 1:x + a &lt; 0, x - 1&nbsp;≥&nbsp;0- a - x + x - 1 = 2a = -3Case 2:x + a&nbsp;≥&nbsp;0 and x - 1 &lt; 0x + a - x + 1 = 2a = 1Largest value of a is 1.The answer is option C.

a(a + b + 1) = 14 …… (1)b(a + b + 1) = 28 …… (2)$\frac{a}{b} = \frac{1}{2}$b = 2aSubstituting in (1), we geta(3a + 1) = 14$3a^2 + a - 14 = 0$$3a^2 - 6a + 7a - 14 = 0$$3a(a - 2) + 7(a - 2) = 0$Given, a and b are natural numbers.Therefore, a = 2 and b = 42a + b = $2(2) + 4 = 8$

Let the number of questions attempted be x+y out of which x are correct and y are incorrect and the number of questions unattempted be z.It is given,x + y + z = 75 ...... (1)3x - y + z = 97 ...... (2)(2)-(1) -&gt; x - y = 11(1)+(2) -&gt; 2x + z = 86z &gt; x + yz &gt; 75 - zz &gt; 37.5Minimum possible value of z is 382x + 38 = 862x = 48x = 24The maximum number of correct answers is 24.

Let $\dfrac{x}{y}$ be $t$Therefore, $c = 16t + \dfrac{49}{t}$Applying AM ≥ GM $\frac{16t + \frac{49}{t}}{2} \ge (16t \times \frac{49}{t})^{\frac12}$​$16t + \dfrac{49}{t} \ge 56$When $t$ is positive then $c$ is greater than equal to $56$.When $t$ is negative then $c$ is less than equal to $-56$.Therefore $c \in (-\infty, -56] \cup [56, \infty]$

Let the number of 100 cheques, 250 cheques and 500 cheques be x, y and z respectively.We need to find the maximum value of z.x + y + z = 100 ...... (1)100x + 250y + 500z = 152502x + 5y + 10z = 305 ...... (2)2x + 2y + 2z = 200 ....... (1)(2) - (1), we get3y + 8z = 105At z = 12, x = 3Therefore, maximum value z can take is 12.

Let the cost of an apple, an orange and a mango be a, o, and m respectively.Then it is given that:2a+4o+6m = a+4o+8mor a = 2m.Also, a+4o+8m = 8o + 7m10m-7m = 4o3m = 4o.We can now express the cost of a basket in terms of mangoes only:2a+4o+6m = 4m+3m+6m = 13m.

We need to check for all regions:x &gt;= 0, y &gt;= 0x &gt;= 0, y &lt; 0x &lt; 0, y &gt;= 0x &lt; 0, y &lt; 0However, once we find out the answer for any one of the regions, we do not need to calculate for other regions since the options suggest that there will be a single answer.Let us start with&nbsp;x &gt;= 0, y &gt;= 0,3x + 3y = 72x + 3y = 1Hence, x = 6 and y = -11/3Since y &gt; = 0, this is not satisfying the set of rules.Next, let us test&nbsp;x &gt;= 0, y &lt; 0,3x - y = 72x + 3y = 1Hence, y = -1x = 2.This satisfies both the conditions. Hence, this is the correct point.WE need the value of x + 2yx + 2y = 2 + 2(-1) = 2 - 2 = 0.

Let the number of large shirts be l and the number of small shirts be s.Let the price of a small shirt be x and that of a large shirt be x + 50.Now, s + l = 64l (x+50) = 5000sx = 1800Adding them, we get,lx + sx + 50l = 680064x +50l = 6800Substituting l = (6800 - 64x) / 50, in the original equation, we get$\frac{(6800−64x)}{50}$(x+50)=5000(6800 - 64x)(x +&nbsp;50) = 250000$6800x+340000−64x^2−3200x=250000$$64x^2−3600x−90000=0$Solving, we get, x=&nbsp;$\frac{225±375}8 = \frac{600}{8} or −\frac{150}8$SO, x = 75x +&nbsp;50 = 125Answer = 75 + 125 = 200.Alternate approach: By options.Hint: Each option gives the sum of the costs of one large and one small shirt. We know that large = small + 50Hence, small + small + 50 = option.SMALL = (Option - 50)/2LARGE = Small +&nbsp;50 = (Option + 50)/2Option A and Option D gives us decimal values for SMALL and LARGE, hence we will consider them later.&nbsp;Lets start with&nbsp;Option B.Large = 150 + 50 / 2 = 100Small = 150 - 50 / 2 = 50Now, total shirts = 5000/100 + 1800/50 = 50 + 36 = 86 (X - This is wrong)Option C -&nbsp;Large = 200 + 50 / 2 = 125Small = 200 - 50 / 2 = 75Total shirts = 5000/125 + 1800/75 = 40 + 24 = 64&nbsp;( This is the right answer)

Let tom's age = x=&gt;&nbsp;Dick=3x=&gt; harry = 6xGiven,&nbsp;3x+1 = (x+3x+6x)/3&nbsp;=&gt; x= 3Hence, Harry's age = 18 years

$\text{Two linear equations } ax + by = c \text{ and } dx + ey = f \text{ have a unique solution if } \frac{a}{d}$ ≠ $\frac{b}{e}$$\therefore \frac{k}{4}$ ≠ $\frac{1}{k} \Rightarrow k^{2}$ ≠ $4$$\Rightarrow k$ ≠ $|2|$

CAT Linear Equations

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CAT Linear Equations

Question 1.

<div><p>For some constant real numbers p, k and a, consider the following system of linear equations in x and y:</p> <p>px - 4y = 2</p> <p>3x + ky= a</p> <p>A necessary condition for the system to have no solution for (x, y), is</p></div>

Question 2.

<div><p>For some real numbers a and b, the system of equations x + y = 4 and $(a+5) x+\left(b^2-15\right) y=8 b$ has infinitely many solutions for x and y. Then, the maximum possible value of ab is</p></div>

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<div>For natural numbers x,y, and z, if xy + yz = 19 and yz + xz = 51, then the minimum possible value of xyz is</div>

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<div><p>The largest real value of a for which the equation |x + a| + |x &minus; 1| = 2 has an infinite number of solutions for x is</p></div>

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<div><p>Let a and b be natural numbers. If $a^2+a b+a=14$ and $b^2+a b+b=28$, then (2a + b) equals</p></div>

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<div><p>If $c=\frac{16 x}{y}+\frac{49 y}{x}$ for some non-zero real numbers x and y, then c cannot take the value</p></div>

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<div><p>A basket of 2 apples, 4 oranges and 6 mangoes costs the same as a basket of 1 apple, 4 oranges and 8 mangoes, or a basket of 8 oranges and 7 mangoes. Then the number of mangoes in a basket of mangoes that has the same cost as the other baskets is</p></div>

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<div><p>If 3x+2|y|+y=7 and x+|x|+3y=1 , then x+2y is</p></div>

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<div><p>Dick is thrice as old as Tom and Harry is twice as old as Dick. If Dick's age is 1 year less than the average age of all three, then Harry's age, in years, is</p></div>

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<div><p>Let k be a constant. The equations kx + y = 3 and 4x + ky = 4 have a unique solution if and only if</p></div>

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<div><p>If $a - 6b + 6c = 4$ and $6a + 3b - 3c = 50$, where $a$, $b$ and $c$ are real numbers, the value of $2a + 3b - 3c$ is.</p></div>

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