Enroll to XAT 2023 Crash Course<\/a><\/p>\n Question 1:\u00a0<\/b>Let $0 \\leq a \\leq x \\leq 100$ and $f(x) = \\mid x – a \\mid + \\mid x – 100 \\mid + \\mid x – a – 50\\mid$. Then the maximum value of f(x) becomes 100 when a is equal to<\/p>\n a)\u00a025<\/p>\n b)\u00a0100<\/p>\n c)\u00a050<\/p>\n d)\u00a00<\/p>\n 1)\u00a0Answer\u00a0(C)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n x>=a, so |x-a| = x-a<\/p>\n x<100, so |x-100| = 100-x<\/p>\n f(x) = (x-a) + (100-x) + |x-a-50| =100<\/p>\n or, |x-a-50| = a<\/p>\n From the graph we can can see that when x=a then<\/p>\n |x-a-50|=a<\/p>\n or, a= 50<\/p>\n Similarly when x=a+100<\/p>\n |x-a-50|=a<\/p>\n or, a= 50<\/p>\n So value of a is 50 when f(x) is 100.<\/p>\n Question 2:\u00a0<\/b>Which of the following equations best describes the graph given below? a)\u00a0$\\mid x+y\\mid-\\mid x-y\\mid=6$<\/p>\n b)\u00a0$\\mid x-y\\mid+\\mid x+y\\mid=10$<\/p>\n c)\u00a0$\\mid x\\mid-\\mid y\\mid=6$<\/p>\n d)\u00a0$\\mid x+y\\mid-\\mid x-y\\mid=0$<\/p>\n 2)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n In these type of questions Question 3:\u00a0<\/b>Which of the following graphs represent the function of x? Choose the correct answer from the options given below :<\/p>\n a)\u00a0(A), (B) and (C) only<\/p>\n b)\u00a0(A), (B) and (D) only<\/p>\n c)\u00a0(A) and (D) only<\/p>\n d)\u00a0(A) and (C) only<\/p>\n 3)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n Graph A and D represent the function of x, because a vertical line is drawn in (a) meets the graph at only one point i.e., for one x, in the domain there exist only one f(x) in the codomain.<\/p>\n Question 4:\u00a0<\/b>The figure below shows the graph of a function f(x). How many solutions does the equation f(f(x)) = 15 have?<\/p>\n a)\u00a05<\/p>\n b)\u00a06<\/p>\n c)\u00a07<\/p>\n d)\u00a08<\/p>\n e)\u00a0cannot be determined from the given graph<\/p>\n 4)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n It has been given that f(f(x)) = 15. When f(x) = 4, x can take 4 values Question 5:\u00a0<\/b>Find the equation of the graph shown below.<\/p>\n a)\u00a0y = 3x – 4<\/p>\n b)\u00a0y = $2x^{2}$ – 40<\/p>\n c)\u00a0x = $2y^{2}$ – 40<\/p>\n d)\u00a0y = $2x^{2}$ + 3x – 19<\/p>\n e)\u00a0x = $2y^{2}$ + 3x – 19<\/p>\n 5)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n When x\u00a0 = -3, y = -10 Hence, option D is the correct answer.<\/p>\n Take XAT 2023 Mock Tests<\/a><\/p>\n Question 6:\u00a0<\/b>In the X-Y plane, the area of the region bounded by the graph of |x+y| + |x-y| = 4 is<\/p>\n a)\u00a08<\/p>\n b)\u00a012<\/p>\n c)\u00a016<\/p>\n d)\u00a020<\/p>\n 6)\u00a0Answer\u00a0(C)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n If the moduli are removed, the equations formed are<\/p>\n x+y+x-y = 4 => x=2<\/p>\n x+y-x+y = 4 => y =2<\/p>\n -x-y+x-y = 4 => y=-2<\/p>\n -x-y-x+y = 4 => x=-2<\/p>\n The area enclosed by these equations is a square with vertices at (2,2), (-2,2), (-2,-2), (2,-2) as shown in figure.<\/p>\n The required area = 4*4 = 16<\/p>\n Question 7:\u00a0<\/b>The graph of y – x (on the y axis) against y + x (on the x axis) is as shown below. (All graphs in this question are drawn to scale and the same scale and the same scale has been used on each axis.)<\/p>\n Which of the following shows the graph of y against x?<\/p>\n a)<\/p>\n b)<\/p>\n c)<\/p>\n d)<\/p>\n 7)\u00a0Answer\u00a0(D)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n For a normal graph with y and x-axis, the equation of the line passing through the origin is y =mx where m is the slope of the line.<\/p>\n m is +ve if the angle made by the line with the x-axis is <\u00a0$90^{\\circ\\ }$<\/p>\n $\\therefore\\ $ The equation of the line in the given graph would be y-x = k( y+\u00a0x) since the axes are y-x and y+x and the line is passing through the origin.<\/p>\n k > 1 because the angle is greater than 45$^{\\circ\\ }$<\/p>\n $y=\\frac{k\\left(x+1\\right)}{1-k}$<\/p>\n Since k>1<\/p>\n Therefore y<0 for x>-1 and y>0 for x<-1<\/p>\n Option d correctly satisfy this condition<\/p>\n Question 8:\u00a0<\/b>Study the following graph and answer the question that follows. What is the total ratio of the revenue of the company in 2016 and 2018 to that of the total expenditure in 2015 and 2017 respectively?<\/p>\n a)\u00a0143:105<\/p>\n b)\u00a0130:103<\/p>\n c)\u00a0115:114<\/p>\n d)\u00a0105:143<\/p>\n 8)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n Total revenue of the company in 2016 and 2018 = 340 + 375 = 715<\/p>\n Total expenditure of the company in 2015 and 2017 = 250 + 275 = 525<\/p>\n Required Ratio = 715 :\u00a0525<\/p>\n i.e; 143 : 105<\/p>\n Hence, option A is correct.<\/p>\n Question 9:\u00a0<\/b>Study the following graph and answer question that follows. The expenditure of the company in 2016 is what percentage more then the average revenue (per year) in 2014, 2015 and 2018 (correct to the nearest integer)?<\/p>\n a)\u00a03%<\/p>\n b)\u00a05%<\/p>\n c)\u00a06%<\/p>\n d)\u00a04%<\/p>\n 9)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n Expenditure of the company in 2016 = 320<\/p>\n Average revenue in 2014,2015 and 2018<\/p>\n $\\therefore\\frac{\\left(revenue\\ in\\ 2014+revenue\\ in\\ 2015+revenue\\ in\\ 2018\\right)}{3}$<\/p>\n $\\therefore\\frac{\\left(260+300+375\\right)}{3}$<\/p>\n = 311.67<\/p>\n Required percentage =\u00a0$\\frac{\\left(320-311.67\\right)}{311.67}$<\/p>\n $=2.67\\simeq\\ 3\\%$<\/p>\n Hence, option A is correct<\/p>\n Question 10:\u00a0<\/b>The following bar graph shows the amount(in Lakh Rs.) invested by a Company in purchasing raw material over the years and the values (in Lakh Rs.) of finished goods sold by the Company over the years. a)\u00a027 : 56<\/p>\n b)\u00a056 : 27<\/p>\n c)\u00a064 : 37<\/p>\n d)\u00a037 : 64<\/p>\n 10)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n Total amount invested for purchasing raw material from 2013 to 2015 = 250 + 350 + 325 = 925<\/p>\n The total sales of finished goods in 2014, 2016 and 2017 = 475 + 600 + 525 = 1600<\/p>\n Required ratio = 925 : 1600<\/p>\n = 37 : 64<\/p>\n Hence, the correct answer is Option D<\/p>\n![](\"https:\/\/cracku.in\/media\/uploads\/Screenshot_20221212_134752.png\")
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\n![](\"https:\/\/cracku.in\/media\/uploads\/image_4OkcBv9.png\")
<\/p>\n
\nwe go through the options :
\nNow when x=6 or -6 the value of y is 0
\nso let us put value of y=0 in all options and check which option satisfies x=6 and -6
\nPutting x=6 and y=0 in option A
\nwe get |6+0|-|6-0| = 0
\nBut RHS =6 so this option is discarded
\nNow putting in option B
\n|6-0|+|6+0| =12
\nBut RHS is 10
\nso this option is also discarded .
\nNow option C we get
\n|6|-0 =6
\nso we can say graph can be of |x|-|y|=6
\nNow satisfying option D
\nwe get |6+0|-|6-0|=0
\nSo we can say 6,0 and -6,0 satisfies option D as well
\nNow you have to eliminate between C and D
\nIf you observe |x+y|-|x-y|=0 will pass through origin
\nand the graph here is not passing through origin so you can eliminate option D
\nTherefore the graph will be |x|-|y|=6<\/p>\n
\n![](\"https:\/\/cracku.in\/media\/uploads\/364944-a_FH8V4So.png\")
\n![](\"https:\/\/cracku.in\/media\/uploads\/364944-b_wzsv3V8.png\")
\n![](\"https:\/\/cracku.in\/media\/uploads\/364944-c.png\")
\n![](\"https:\/\/cracku.in\/media\/uploads\/364944-d.png\")
<\/p>\n![](\"https:\/\/cracku.in\/media\/uploads\/1_jsRA9rD.png\")
<\/figure>\n
\nFrom the graph, we can see that the value of f(4) = 15 and f(12) = 15
\nTherefore, f(x) can be 4 or 12.<\/p>\n![](\"https:\/\/cracku.in\/media\/uploads\/c1_oKaQumS.PNG\")
<\/figure>\n
\nWhen f(x) = 12, x can take 3 values.
\nTherefore, there are 4+3 = 7 solutions in total.
\nTherefore, option\u00a0C is the right answer.<\/p>\n![](\"https:\/\/cracku.in\/media\/uploads\/Review.png\")
<\/figure>\n
\nThis is satisfied only in option D.<\/p>\n![](\"https:\/\/cracku.in\/media\/uploads\/Screen%20Shot%202016-11-01%20at%202.09.20%20pm.png\")
<\/p>\n![](\"https:\/\/cracku.in\/media\/uploads\/2015\/12\/25\/capture_RYCAxo9.PNG\")
<\/figure>\n![](\"https:\/\/cracku.in\/media\/uploads\/2015\/12\/25\/capture-a.PNG\")
<\/figure>\n![](\"https:\/\/cracku.in\/media\/uploads\/2015\/12\/25\/capture-b.PNG\")
<\/figure>\n![](\"https:\/\/cracku.in\/media\/uploads\/2015\/12\/25\/capture-c.PNG\")
<\/figure>\n![](\"https:\/\/cracku.in\/media\/uploads\/2015\/12\/25\/capture-d.PNG\")
<\/figure>\n
\n![](\"https:\/\/cracku.in\/media\/uploads\/12M%2022_Ksa4fi4.png\")
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\n![](\"https:\/\/cracku.in\/media\/uploads\/12M%2022_M3HqO03.png\")
<\/p>\n
\n![](\"https:\/\/media-cdn.cracku.in\/uploads\/Screenshot_462_4jr7D9o.png\")
\nThe ratio of total amount invested for purchasing raw material from 2013 to 2015 to the total sales of finished goods in 2014, 2016 and 2017 is:<\/p>\n