$\log_4mn=\log_2(m+n)$&nbsp;$\sqrt{\ mn}=(m+n)$Squarring on both sides$m^2+n^2+mn\ =\ 0$Since m, n are positive real numbers, no value of m and n satisfy the above equations.

Let the number of eggs bought = 9x+2 number of eggs left after throwing away 2 = 9x number of eggs kept in fridge = 5x number of eggs brought to his mothers' house = 4x number of eggs left after cooking 2 which are kept in fridge&nbsp;= 4x-2 Given, 4x-2 &lt;=5 =&gt; x$\le\frac{7}{4}$ Hence the max value of x is 1 Max number of eggs bought = 11

Let the total amount be 3x Case 1:&nbsp; Smaller amount = x, rate of interest = 10 Larger amount = 2x, rate of interest = 5 Total amount received at the end of two years( smaller amount) =&nbsp;$x\left(1+\frac{10}{100}\right)^2\ =\ 1.21x$. CI = 0.21x Total amount received at the end of two years( larger amount) =&nbsp;$2x\left(1+\frac{5}{100}\right)^2\ =\ 2.205x$&nbsp; &nbsp;CI = 0.205x Given, 0.21x + 0.205x = 830 =&gt; x = 2000 2x= 4000 Case 2: Smaller amount = x, rate of interest = 20 Larger amount = 2x, rate of interest = 10 Total amount received at the end of two years( smaller amount) = $x\left(1+\frac{20}{100}\right)^2\ =\ 1.44x$. CI = 0.44x Total amount received at the end of two years( larger amount) = $2x\left(1+\frac{10}{100}\right)^2\ =\ 2.42x$ CI = 0.42x Given, 0.44x+0.42x = 830 =&gt; x = 965.11 which is not valid since it should be greater than 1000

Number of white phones = 2 Number of black phones = 2 Number of red phones = 1 customer 1 will have 3 choices customer 2 will have 3 choices customer 3 will have 3 choices Hence total choices = 3 x 3 x 3 = 27 The cases not possible = BBB, RRR,WWW, RRB,RBR,BRR, RRW,RWR, WRR Possible cases = 18 Probability = 18/27 = 2/3

The difference between the hour and minute hand of a clock is given by&nbsp;$\left|30H-5.5m\right|$. Here H is the current hour and m represents the number of completed minutes in the current hour.In the given time frame of 2: 30 to 3: 00 pm.At 2 : 30 pm the angle =&nbsp;$\left|30\cdot2-5.5 \cdot30 \right| = 105$ degreesAt 3: 00 pm the angle =&nbsp;$\left|30\cdot3-5.5 \cdot0 \right| = 90$ degreesThe function of&nbsp;$\left|30\cdot H-5.5 \cdot m\right| =$ constantly increases as the value of m increases from 31, 32................ 59.Because of the modulus function, the net value of the function&nbsp;remains positiveBetween 2: 30 to 2: 59 the angle is constantly increasing. The minimum value is 2: 30 which is equal to 105 degrees which is greater than the 90 degrees when the time is 3: 00.Hence 90 degrees is the minimum angle.

Let us assume there are three buses, each carrying 30 passengers.Now it is given that one-third of the buses from City A to City B stop at City C, while the rest go non-stop to City B. This means that one bus stops at C while two buses go directly to B. This means that $2\times 30=60$ passengers directly reach to B.For the buses that stops at C,&nbsp;One-third of the passengers continue to City B, that is 10 passengers in each bus continue to city B, while the rest alight at City C. Since there is only one bus which stops at city C, this means that $1\times 20=20$ passengers alight at C, while $1\times 10=10$ passengers travel from city C to city B.<img src="https://lightyellow-stork-291267.hostingersite.com/wp-content/uploads/2025/12/image_ElywAjQ.png"> We are asked what proportion of the passengers going to City B from City A travel by a bus stopping at City C. So, in total, we can see the total number of passengers who are travelling to City B is 70, while the number of passengers travelling by the buses that stop at City&nbsp;C is 10.So, the proportion of the passengers going to City B from City A who travel by a bus stopping at City C =&nbsp;$\dfrac{10}{70}=\dfrac{1}{7}$

Based on the information, the following figure can be obtained.<img src="https://lightyellow-stork-291267.hostingersite.com/wp-content/uploads/2025/12/Screenshot-2025-05-08-10.36.19-AM.png">Given, CE=0.5, BC = 1.3 and ED=2.5Triangle CEB is a right-angled triangle =&gt; EB = 1.2$\triangle$ ECB is similar to&nbsp;$\triangle$ EDAEB/EC = AE/ED&nbsp;=&gt; AE = 6Hence total distance travelled = AB + BC + CD = 7.2 + 1.3 + 3 = 11.5km

$\frac{f\left(x\right)}{g\left(x\right)}=\frac{\left(x^2+1\right)}{x^2-1}\cdot\frac{\left(x-1\right)}{x+1}=\frac{\left(x^2+1\right)}{\left(x+1\right)^2}$This function is definitely greater than 0$let\ y=\frac{\left(x^2+1\right)}{\left(x+1\right)^2}$=&gt;&nbsp;$x^2\left(y-1\right)+2yx+\left(y-1\right)=0$ which is quadratic in xDisctiminant should be greater than 0$4y^2-4\left(y-1\right)^2\ge0$=&gt; y&gt;=1/2When x =1, f(x)/g(x) = 1/3Hence either the value should be greater than 1/2 or should be equal to 1/3

Let the speed of the stream be x $\frac{12}{5-x}=\frac{12}{5+x}+1$ The value of x satisfying the above equation is 1 Now, $\frac{N}{5+1}+\frac{N}{5-1}\ge2$ $\frac{2N+3N}{12}\ge2$ =&gt;&nbsp;$N\ge4.8$

Sum of 3rd row = sum of 2nd column=&gt; 2x+4y = y+2z-1&nbsp;=&gt; 2x+3y-2z= -1 ------- (A)Sum of diagonals are also equal=&gt; 3x+4y+z-1 = y+z+2x+y+z=&gt; x+2y-z=1 -----(B)Solving A and B we get y= 3Putting it in A, we get x-z = -5 ----- (C)Sum of 1st row = sum of 2nd column5x+5y+z = 3x+4y+2z=&gt; 2x +y - z =0Since y=3, 2x-z = -3 ------ (D)Solving C and D we get x=2 and z=7Hence N = 36

xat 2021 Complete Paper Solution | QADI

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