The supplier receives his orders from the five buyers once every 2 weeks, once every 6 weeks, once every 8 weeks, once every 4 weeks, and once every 3 weeks.The number of occasions where all the five buyers place the order on the same day is :The LCM of the 5-time frames during which the 5 buyers place their orders :Hence the LCM is :(2, 6, 8, 4, 3).= 24 weeks.Once every 24 weeks, all five of them place the order simultaneously.A year has 53 weeks in total :Hence all five of them place the orders after 24 weeks, 48 weeks.

Initially considering the number of members = a The number of ladoos each member is required to make as per the original plan = b. Given : a*b = 2400. Given that 10 members were absent and each member had to make an additional 12 ladoos : (a-10)*(b+12) = 2400. ab - 10b+12a-120 = 2400. Since a*b = 2400. Hence 12a-10b = 120. Substituting b = 2400/a.&nbsp; &nbsp; $12a\ -\ 10\cdot\left(\dfrac{2400}{a}\right)\ =\ 120$ =&nbsp;$\ 12a^2-24000-\ 120a\ =\ 0$ The roots are a = 50, a = -40. Hence a = 50, b = 48. The number of people who took part in making ladoos = a - 10 = 40

<img src="https://lightyellow-stork-291267.hostingersite.com/wp-content/uploads/2025/12/image_Nk4852e.png">Given:&nbsp;BD= 6 cm, AC= 9 cm, DC= 5 cm, BP=8 cm, and DP = 5 cm.Since AD is the angular bisector applying the angular bisector theorem we have :$\frac{AB}{BD}=\ \frac{AC}{CD}$Hence : Considering AB = x cm.$\frac{9}{5}=\ \frac{x}{6}$x = 10.8 cm.Now since BD is the angular bisector for angle PBA we have :Applying the internal angle bisector theorem :$\frac{PB}{PD}=\ \frac{BA}{AD}$Considering AD = y cm.$\frac{8}{5}\ =\ \frac{10.8}{y}$y = 6.75 cm.AP = AD + DP.= 6.75 + 5 = 11.75 cm

The two types of apples sold A and B are bought in the ratio of 5: 8. Considering the weights to be 5x and 8x. The cost price of A is 20 percent higher than that of B. Considering the cost price of B = y, A = 6y/5. The total cost price of A =&nbsp;$\left(5x\right)\cdot\left(\frac{6y}{5}\right)$ The total cost price of B =&nbsp;$\left(8x\right)\cdot\left(y\right)$ THe total cost price = 8xy + 6xy = 14xy 14xy = 2800. xy = 200. THe cost price of A = 1200. THe cost price of B = 1600. A is sold a profit of 15 percent. 15 percent of 1200 = 180. B is sold at a profit of 10 percent. 10 percent of 1600 = 160. The total profit is 180 + 160 = 340

Given&nbsp;the ball falls from a height of 3 meters. The ball reaches a height which 0.8 times the original height every time. Hence this is in the form of a geometric progression. We need to count distance when the ball flies upward and downward. Hence considering every time the ball flies upward to a series with terms : h1, h2,.......................... Every time the ball falls down to be d1, d2 ,............... h1 = (0.8)*3, h2 = (0.8)*(0.8)*3 ,......................... d1 = 3, d2 =3*(0.8), d3 = 3*(0.8)*(0.8)................ h1+h2 ........ = Sum of an infinite geometric progression. = 3*0.8(1+0.8+$0.64+...............$) THe sum of an infinite GP with r less than 1 is :&nbsp;$\frac{a}{1-r}$ = $2.4\cdot\left(\frac{1}{1-0.8}\right)\ =\ 12\ meters$ The sum of d1+d2++........................ = 3 + (h1+h2+..................) = 15. The total distance = 15 + 12 = 27 meters.

Considering the two numbers to a, b : We were given that : $a^3+b^3\ =\ 128$ $\frac{1}{a^3}+\ \frac{1}{b^3}=\ 2$ $\frac{\left(a^3+b^3\right)}{a^3\cdot b^3}=\ 2\ =\ \frac{128}{k}$ k = 64. Hence&nbsp;$a^3\cdot b^3\ =\ 64$ a*b = 4 and&nbsp;$a^2\cdot b^2\ =\ 16$

The age of Nadeem is a two-digit number. When squared yields a three-digit number whose last digit is Y. Y is a prime number. Using statement 1: When a number is squared : The last digit of the number can be : 1, 2, 3, 4, 5, 6, 7, 8, 9. When squared the last digit can be : For a number ending with 1: 1 For a number ending with 2: 4 For a number ending with 3: 9 For a number ending with 4: 6 For a number ending with 5: 5 For a number ending with 6: 6 For a number ending with 7: 9 For a number ending with 8: 4 For a number ending with 9: 1 The only possible prime number is 5. Hence the last digit of X is 5 and Y is 5. Using statement 2 : Y = X/3. This alone cannot be&nbsp;sufficient to determine the possibilities for Y and X. Combining both the statements : Since Y = 5, then the value of X = 15. The age is equal to 15.

<img src="https://lightyellow-stork-291267.hostingersite.com/wp-content/uploads/2025/12/image_dxykAXy.png">Given the distances are :AE = 4 meters , EB = 8 meters and EC = 16 meters.Considering the length of ED = K.Given the angles DAE and angle DCE&nbsp;are complementary.Hence the angles are A and 90 - A.Tan(90-A) = Cot A$\ \tan\ DAE\ =\frac{k}{4}$ and&nbsp;$\ \tan\ DCE\ =\frac{1}{\tan\ DAE}=\ \frac{k}{16}$Hence&nbsp;$\frac{k}{16}=\ \frac{4}{k}$k = 8 meters.The angle DBE is given by&nbsp;$Tan\ DBE\ =\ \frac{k}{8}=\ 1$Hence the angle is equal to 45 degrees.

Using condition W :The magnitudes of the net change in enrollment between 2014 and 2021 is closest among any two universities for TRU and MPU.Going by the color of the lines the net change for different universities&nbsp;is:Univ 1: 0.7Univ 2: 4.4Univ 3: 0.1Univ 4: 0.2Univ 5: 3.3The closest among these are: Univ 3 and Univ 4. They can possibly be : (TRU/MPU)Using condition X :The university LTU must have the same enrollment in consecutive years&nbsp;at least twice :LTU can either be Univ 3 or Univ 1 but since Univ 3 must be among TRU and MPU. LTU is university 1.Using condition Y :The increase in the enrollment of JSU between the years 2015 and 2019 is 50 percent of TRU's total enrollment in 2020.Considering :TRU = Univ 4&nbsp;The enrollment is 5.TRU = Univ 3The enrollment is 0.7For TRU as Univ 3, there is no university whose increase in enrollment between 2015 and 2019 is 50 percent of TRU.Hence TRU =&nbsp;Univ 4.Since the increase in enrollment for JSU is half of TRU. The increase must be half of 5 = 2.5The only possible case is JSU = Univ 2.MPU = Univ 3.LTU = Univ 1PKU = Univ 5.

Using condition W : The magnitudes of the net change in enrollment between 2014 and 2021 is closest among any two universities for TRU and MPU. Going by the color of the lines the net change for different universities is: Univ 1: 0.7 Univ 2: 4.4 Univ 3: 0.1 Univ 4: 0.2 Univ 5: 3.3 The closest among these are: Univ 3 and Univ 4. They can possibly be : (TRU/MPU) Using condition X : The university LTU must have the same enrollment in consecutive years at least twice : LTU can either be Univ 3 or Univ 1 but since Univ 3 must be among TRU and MPU. LTU is university 1. Using condition Y : The increase in the enrollment of JSU between the years 2015 and 2019 is 50 percent of TRU's total enrollment in 2020. Considering : TRU = Univ 4 The enrollment is 5. TRU = Univ 3 The enrollment is 0.7 For TRU as Univ 3, there is no university whose increase in enrollment between 2015 and 2019 is 50 percent of TRU. Hence TRU = Univ 4. Since the increase in enrollment for JSU is half of TRU. The increase must be half of 5 = 2.5 The only possible case is JSU = Univ 2. MPU = Univ 3. LTU = Univ 1 PKU = Univ 5. SInce LTU is univ 1 the university with an enrollment twice that of LTU = 2*(3.2) = 6.4.&nbsp; Hence the only possible case close by is Univ 5 (PKU)

xat 2022 Complete Paper Solution | QADI

Question 1.

Question 2.

Question 3.

Question 4.

Question 5.

Question 6.

Question 7.

Question 8.

Instructions

Question 9.

Question 10.