{"id":38225,"date":"2019-11-26T18:20:20","date_gmt":"2019-11-26T12:50:20","guid":{"rendered":"https:\/\/cracku.in\/blog\/?p=38225"},"modified":"2019-11-26T18:25:56","modified_gmt":"2019-11-26T12:55:56","slug":"quant-questions-for-xat","status":"publish","type":"post","link":"https:\/\/cracku.in\/blog\/quant-questions-for-xat\/","title":{"rendered":"Quant Questions for XAT"},"content":{"rendered":"

Quant Questions for XAT<\/strong><\/span><\/h2>\n

Download important Quant Questions for XAT PDF based on previously asked questions in CAT exam. Practice Quant Questions PDF for XAT exam.<\/p>\n

Download Quant Questions for XAT<\/a><\/p>\n

Get 5 XAT mocks for Rs. 299. Enroll here<\/a><\/p>\n

Question 1:\u00a0<\/b>Two diagonals of a parallelogram intersect each other at coordinates (17.5, 23.5). Two adjacent points of the parallelogram are (5.5, 7.5) and (13.5, 16). Find the lengths of the diagonals.<\/p>\n

a)\u00a015 and 30<\/p>\n

b)\u00a015 and 40<\/p>\n

c)\u00a017 and 30<\/p>\n

d)\u00a017 and 40<\/p>\n

e)\u00a0Multiple solutions are possible<\/p>\n

Question 2:\u00a0<\/b>If the last 6 digits of [(M)! – (N)!] are 999000, which of the following option is not possible for (M) \u00d7 (M – N)? Both (M) and (N) are positive integers and M > N. (M)! is factorial M.<\/p>\n

a)\u00a0150<\/p>\n

b)\u00a0180<\/p>\n

c)\u00a0200<\/p>\n

d)\u00a0225<\/p>\n

e)\u00a0234<\/p>\n

Question 3:\u00a0<\/b>A circular road is constructed outside a square field. The perimeter of the square field is 200 ft. If the width of the road is 7\u221a2 ft. and cost of construction is Rs. 100 per sq.ft. Find the lowest possible cost to construct 50% of the total road.<\/p>\n

a)\u00a0Rs. 70,400<\/p>\n

b)\u00a0Rs. 125,400<\/p>\n

c)\u00a0Rs. 140,800<\/p>\n

d)\u00a0Rs. 235,400<\/p>\n

e)\u00a0None of the above<\/p>\n

Question 4:\u00a0<\/b>Akhtar plans to cover a rectangular floor of dimensions 9.5 meters and 11.5 meters using tiles. Two types of square shaped tiles are available in the market. A tile with side 1 meter costs Rs. 100 and a tile with side 0.5 meters costs Rs. 30. The tiles can be cut if required. What will be the minimum cost of covering the entire floor with tiles?<\/p>\n

a)\u00a010930<\/p>\n

b)\u00a010900<\/p>\n

c)\u00a011000<\/p>\n

d)\u00a010950<\/p>\n

e)\u00a010430<\/p>\n

Question 5:\u00a0<\/b>Consider the set of numbers {1, 3, $3^{2}$, $3^{2}$,\u2026…,$3^{100}$}. The ratio of the last number and the sum of the remaining numbers is closest to:<\/p>\n

a)\u00a01<\/p>\n

b)\u00a02<\/p>\n

c)\u00a03<\/p>\n

d)\u00a050<\/p>\n

e)\u00a099<\/p>\n

XAT Solved Previous papers<\/a><\/p>\n

Take XAT Mock Test<\/a><\/p>\n

Question 6:\u00a0<\/b>If a, b and c are 3 consecutive integers between -10 to +10 (both inclusive), how many integer values are possible for the expression?
\n$\\frac{a^3+b^3+c^3+3abc}{(a+b+c)^2}$=?<\/p>\n

a)\u00a00<\/p>\n

b)\u00a01<\/p>\n

c)\u00a02<\/p>\n

d)\u00a03<\/p>\n

e)\u00a04<\/p>\n

Question 7:\u00a0<\/b>The sum of series, (-100) + (-95) + (-90) + \u2026\u2026\u2026\u2026+ 110 + 115 + 120, is:<\/p>\n

a)\u00a00<\/p>\n

b)\u00a0220<\/p>\n

c)\u00a0340<\/p>\n

d)\u00a0450<\/p>\n

e)\u00a0None of the above<\/p>\n

Question 8:\u00a0<\/b>If $x$ and $y$ are real numbers, the least possible value of the expression $4(x – 2)^{2} + 4(y – 3)^{2} – 2(x – 3)^{2}$ is :<\/p>\n

a)\u00a0– 8<\/p>\n

b)\u00a0– 4<\/p>\n

c)\u00a0– 2<\/p>\n

d)\u00a00<\/p>\n

e)\u00a02<\/p>\n

Question 9:\u00a0<\/b>If $N = (11^{p + 7})(7^{q – 2})(5^{r + 1})(3^{s})$ is a perfect cube, where $p, q, r$ and $s$ are positive integers, then the smallest value of $p + q + r + s$ is :<\/p>\n

a)\u00a05<\/p>\n

b)\u00a06<\/p>\n

c)\u00a07<\/p>\n

d)\u00a08<\/p>\n

e)\u00a09<\/p>\n

Question 10:\u00a0<\/b>Find the value of the expression: $10 + 10^3 + 10^6 + 10^9$<\/p>\n

a)\u00a01010101010<\/p>\n

b)\u00a01001000010<\/p>\n

c)\u00a01001000110<\/p>\n

d)\u00a01001001010<\/p>\n

e)\u00a0100010001010<\/p>\n

XAT Decision making practice questions<\/a><\/p>\n

Question 11:\u00a0<\/b>A girl travels along a straight line, from point A to B at a constant speed, $V_1$ meters\/sec for T seconds. Next, she travels from point B to C along a straight line, at a constant speed of $V_2$ meters\/sec for another T seconds. BC makes an angle 105\u00b0 with AB. If CA makes an angle 30\u00b0 with BC, how much time will she take to travel back from point C to A at a constant speed of $V_2$ meters\/sec, if she travels along a straight line from C to A?<\/p>\n

 <\/p>\n

a)\u00a0$0.53(\\sqrt{3}-1)T$<\/p>\n

b)\u00a0$T$<\/p>\n

c)\u00a0$0.5(\\sqrt{3}+1)T$<\/p>\n

d)\u00a0$\\sqrt{3}$<\/p>\n

e)\u00a0None of the above<\/p>\n

Question 12:\u00a0<\/b>Abdul, Bimal, Charlie and Dilbar can finish a task in 10, 12, 15 and 18 days respectively. They can either choose to work or remain absent on a particular day. If 50 percent of the total work gets completed after 3 days, then, which of the following options is possible?<\/p>\n

a)\u00a0Each of them worked for exactly 2 days.<\/p>\n

b)\u00a0Bimal and Dilbar worked for 1 day each, Charlie worked for 2 days and Abdul worked for all 3 days.<\/p>\n

c)\u00a0Abdul and Charlie worked for 2 days each, Dilbar worked for 1 day and Bimal worked for all 3 days.<\/p>\n

d)\u00a0Abdul and Dilbar worked for 2 days each, Charlie worked for 1 day and Bimal worked for all 3 days.<\/p>\n

e)\u00a0Abdul and Charlie worked for 1 day each, Bimal worked for 2 days and Dilbar worked for all 3 days.<\/p>\n

Question 13:\u00a0<\/b>A, B, C, D and E are five employees working in a company. In two successive years, each of them got hikes in his salary as follows:
\nA : p% and (p+1)%,
\nB : (p+2)% and (p-1)%,
\nC : (p+3)% and (p-2)%,
\nD : (p+4)% and (p-3)%,
\nE : (p+5)% and (p-4)%.
\nIf all of them have the same salary at the end of two years, who got the least hike in his salary?<\/p>\n

a)\u00a0E<\/p>\n

b)\u00a0D<\/p>\n

c)\u00a0C<\/p>\n

d)\u00a0A<\/p>\n

e)\u00a0B<\/p>\n

Question 14:\u00a0<\/b>Let P be the point of intersection of the lines
\n3x + 4y = 2a and 7x + 2y = 2018
\nand Q the point of intersection of the lines
\n3x + 4y = 2018 and 5x + 3y = 1
\nIf the line through P and Q has slope 2, the value of a is:<\/p>\n

a)\u00a04035<\/p>\n

b)\u00a01\/2<\/p>\n

c)\u00a03026<\/p>\n

d)\u00a01<\/p>\n

e)\u00a01009<\/p>\n

Question 15:\u00a0<\/b>Consider the function f(x) = (x + 4)(x + 6)(x + 8) \u22ef (x + 98). The number of integers x for which f(x) < 0 is:<\/p>\n

a)\u00a023<\/p>\n

b)\u00a026<\/p>\n

c)\u00a024<\/p>\n

d)\u00a048<\/p>\n

e)\u00a049<\/p>\n

XAT Quant formulas PDF<\/a><\/p>\n

Take XAT Mock Test<\/a><\/p>\n

Answers & Solutions:<\/strong><\/span><\/p>\n

1)\u00a0Answer\u00a0(D)<\/strong><\/p>\n

<\/figure>\n

Using distance formula,<\/p>\n

$CX = \\sqrt{(17.5 – 5.5)^2 + (23.5 – 7.5)^2} = \\sqrt{12^2 + 16^2}$<\/p>\n

= $\\sqrt{144 + 256} = \\sqrt{400} = 20$<\/p>\n

=> $AC = 2 \\times CX = 40$<\/p>\n

$BX = \\sqrt{(17.5 – 13.5)^2 + (23.5 – 16)^2} = \\sqrt{4^2 + 7.5^2}$<\/p>\n

= $\\sqrt{16 + 56.25} = \\sqrt{72.25} = 8.5$<\/p>\n

=> $BD = 2 \\times BX = 17$<\/p>\n

2)\u00a0Answer\u00a0(B)<\/strong><\/p>\n

3)\u00a0Answer\u00a0(B)<\/strong><\/p>\n

<\/figure>\n

Perimeter of square ABCD = 200 ft<\/p>\n

=> AB = $\\frac{200}{4} = 50$ ft<\/p>\n

=> $DB = \\sqrt{50^2 + 50^2} = 50 \\sqrt{2}$ ft<\/p>\n

=> $BO = r = \\frac{50 \\sqrt{2}}{2} = 25 \\sqrt{2}$ ft<\/p>\n

Width of the road = BX = $7 \\sqrt{2}$ ft<\/p>\n

=> $BX = R = 25 \\sqrt{2} + 7 \\sqrt{2} = 32 \\sqrt{2}$<\/p>\n

Area of bigger circle = $\\pi R^2 = \\pi (32 \\sqrt{2})^2 = 2048 \\pi$ sq. ft<\/p>\n

Area of smaller circle = $\\pi r^2 = \\pi (25 \\sqrt{2})^2 = 1250 \\pi$ sq. ft<\/p>\n

=> Area of road = $2048 \\pi – 1250 \\pi = 798 \\times \\frac{22}{7} = 2508$ sq. ft<\/p>\n

But we have to calculate cost of construction of 50% road.<\/p>\n

Required Construction = $\\frac{2508}{2} = 1254$ sq. ft<\/p>\n

$\\therefore$ Cost of 1254 ft = $1254 \\times 100 = Rs. 1,25,400$<\/p>\n

4)\u00a0Answer\u00a0(A)<\/strong><\/p>\n

Area of rectangular floor = $9.5 \\times 11.5$<\/p>\n

= $109.25 m^2$<\/p>\n

Now, cost of covering $109 m^2$ (with 1×1 tiles) = $109 \\times 100$ = Rs. 10,900<\/p>\n

Cost of covering $0.25 m^2$ (with 0.5 m square tile) = Rs. 30<\/p>\n

$\\therefore$ Total cost = $10,900 + 30 = Rs. 10,930$<\/p>\n

5)\u00a0Answer\u00a0(B)<\/strong><\/p>\n

Set\u00a0: \u00a0{$1, 3, 3^{2}, 3^{2},\u2026…,3^{100}$}<\/p>\n

Clearly, this set is a G.P. with common ratio, $r = 3$<\/p>\n

Sum of G.P. = $\\frac{a (r^n – 1)}{r – 1}$<\/p>\n

Number of terms = 101<\/p>\n

Last term = $3^{100}$<\/p>\n

Sum of remaining terms = $\\frac{1 (3^{100} – 1)}{3 – 1}$<\/p>\n

= $\\frac{3^{100} – 1}{2}$<\/p>\n

$\\therefore$ Required ratio = $\\frac{3^{100}}{\\frac{3^{100} – 1}{2}}$<\/p>\n

= $\\frac{3^{100} \\times 2}{3^{100} – 1}$<\/p>\n

$\\approx \\frac{3^{100} \\times 2}{3^{100}} = 2$<\/p>\n

Download XAT General Knowledge PDFs<\/a><\/p>\n

Practice Online Gk tests for XAT<\/a><\/p>\n

6)\u00a0Answer\u00a0(C)<\/strong><\/p>\n

Since a,b,c are consecutive integers<\/p>\n

=> $a = b-1$ and $c = b+1$<\/p>\n

Expression\u00a0:\u00a0$\\frac{a^3+b^3+c^3+3abc}{(a+b+c)^2}$<\/p>\n

= $\\frac{(b – 1)^3 + b^3 + (b + 1)^3 + 3 (b – 1) b (b + 1)}{(b – 1 + b + b + 1)^2}$<\/p>\n

= $\\frac{b^3 + 3b + b^3 + b^3 + 3b + 3b^3 – 3b}{9 b^2}$<\/p>\n

= $\\frac{6 b^3 + 3 b}{9 b^2} = \\frac{2 b^2 + 1}{3 b}$<\/p>\n

Putting different values of b from – 10 to 10, we can verify that only – 1 and 1 satisfies to get integer values for the expression.<\/p>\n

Ans – (C)<\/p>\n

7)\u00a0Answer\u00a0(D)<\/strong><\/p>\n

The given series is (-100)+ (-95)+ (-90)+….+110 +\u00a0115 + 120.
\nWe can observe that -100 will cancel out 100, -95 will cancel out 95 and so on. Therefore, the only terms that will be remaining are 105, 110, 115 and 120.
\nSum of the series = 105 + 110 +\u00a0115 +\u00a0120 = 450.
\nTherefore, option D is the right answer.<\/p>\n

8)\u00a0Answer\u00a0(B)<\/strong><\/p>\n

$4(x – 2)^{2} + 4(y – 3)^{2} – 2(x – 3)^{2}$
\n$y$ is an independent variable. The value of $y$ is unaffected by the value of $x$. Therefore, the least value that the expression $4(y-3)^2$ can take is $0$ (at $y=3$).<\/p>\n

Let us expand the remaining terms.
\n$4(x-2)^2-2(x-3)^2$=4*$(x^2-4x+4)-2*(x^2-6x+9)$
\n$=2x^2-4x-2$
\n=$2(x^2-2x-1)$
\n=$2(x^2-2x+1-2)$
\n=$2((x-1)^2-2)$
\nThe least value that the expression $(x-1)^2$ can take is $0$ (at $x$ = $1$)
\nTherefore, the least value that the expression\u00a0$2((x-1)^2-2$ can take is $2*(0-2)=2*(-2) = -4$
\nTherefore, option\u00a0B is the right answer.<\/p>\n

9)\u00a0Answer\u00a0(E)<\/strong><\/p>\n

It has been given that\u00a0$N = (11^{p + 7})(7^{q – 2})(5^{r + 1})(3^{s})$ is a perfect cube. All the factors given are prime. Therefore, the power of each number should be a multiple of 3 or 0.<\/p>\n

$p,q,r$ and $s$ are positive integers. Therefore, only the power of the expressions in which some number is subtracted from these variables or these variables are subtracted from some number can be made $0$.<\/p>\n

$11^{p + 7}$:<\/p>\n

This expression must be made a perfect cube. The nearest perfect cube is $11^9$. Therefore, the least value that $p$ can take is $9-7=2$.<\/p>\n

$7^{q – 2}$<\/p>\n

The least value that $q$ can take is 2. If $q=2$, then the value of the expression $7^{q-2}$ will become $7^0=1$, without preventing the product from becoming a perfect cube.<\/p>\n

$5^{r+1}$:<\/p>\n

The least value that $r$ can take is $2$.<\/p>\n

$3^{s})$:<\/p>\n

The least value that $s$ can take is $3$.<\/p>\n

Therefore, the least value of the expression\u00a0$p + q + r + s$ is $2+2+2+3=9$.
\nTherefore, option E is the right answer.<\/p>\n

10)\u00a0Answer\u00a0(D)<\/strong><\/p>\n

$10 + 10^3 + 10^6 + 10^9 = 10 + 1000 + 1000000 + 1000000000$
\n=$1001001010$
\nTherefore, option\u00a0D is the right answer.<\/p>\n

11)\u00a0Answer\u00a0(C)<\/strong><\/p>\n

<\/figure>\n

We draw BD perpendiculat to AC.
\nIn right angled triangle BDC,\u00a0 BD \/ BC = sin 30\u00b0
\nor, BD = (V2 * T)\/2 ……(i)
\nIn right angled triangle BDA, BD \/ BA = sin 45\u00b0
\nOr, BD = (V1 * T)\/$\\sqrt{2}$ ……(ii)
\nFrom (i) and (ii), we get
\nV2 \/ V1 = $\\sqrt{2}$
\nTotal distance to be travelled from C to A = CD + DA\u00a0 = $\\sqrt{3}$BD + BD
\n= BD($1 + \\sqrt{3}$)
\nReplacing\u00a0BD = (V2 * T)\/2 in the avove equation,
\nCA = $\\dfrac{\\text{(V2 * T)}}{2} (1 + \\sqrt{3}$)
\nTime taken at speed V2 =\u00a0$0.5(\\sqrt{3}+1)$T
\nHence, option C is the correct answer.<\/p>\n

12)\u00a0Answer\u00a0(E)<\/strong><\/p>\n

Let us assume the amount of work to be finished = LCM of {10, 12, 15, 18} = 180 units.<\/p>\n

The amount of work which\u00a0Abdul can complete in a day = $\\dfrac{180}{10}$ = 18 units.<\/p>\n

The amount of work which\u00a0Bimal can complete in a day = $\\dfrac{180}{12}$ = 15 units.<\/p>\n

The amount of work which\u00a0Charlie\u00a0 can complete in a day = $\\dfrac{180}{15}$ = 12 units.<\/p>\n

The amount of work which\u00a0Dilbar can complete in a day = $\\dfrac{180}{18}$ = 10 units.<\/p>\n

It is given that\u00a050 percent of the total work gets completed after 3 days. Therefore, we can say that 90 units of work was completed in 3 days.<\/p>\n

Let us check options.
\nOption A:\u00a0Each of them worked for exactly 2 days.
\nIn this case amount of work completed = 2*(10+15+12+18) = 110 units.<\/p>\n

Option B:\u00a0Bimal and Dilbar worked for 1 day each, Charlie worked for 2 days and Abdul worked for all 3 days.
\nIn this case amount of work completed = 1*(10+15)+2*(12)+3*(18) = 103 units.<\/p>\n

Option C:\u00a0Abdul and Charlie worked for 2 days each, Dilbar worked for 1 day and Bimal worked for all 3 days.
\nIn this case amount of work completed = 1*(10)+3*(15)+2*(18+12) = 115 units.<\/p>\n

Option D:\u00a0Abdul and Dilbar worked for 2 days each, Charlie worked for 1 day and Bimal worked for all 3 days.
\nIn this case amount of work completed = 1*(12)+3*(15)+2*(18+10) = 113 units.<\/p>\n

Option E:\u00a0Abdul and Charlie worked for 1 day each, Bimal worked for 2 days and Dilbar worked for all 3 days.
\nIn this case amount of work completed = 1*(18+12)+2*(15)+3*(10) = 90 units.<\/p>\n

Therefore, we can say that option E is the correct answer.<\/p>\n

13)\u00a0Answer\u00a0(A)<\/strong><\/p>\n

Let the initial salary of A,B,C,D,E be $a,b,c,d,e$ respectively and let the final salary of everyone be $x$<\/p>\n

Now, $a*(1+ \\frac{p}{100})*(1+ \\frac{p+1}{100}) = x$<\/p>\n

$\\Rightarrow a = \\frac{x}{(1+ \\frac{p}{100})*(1+ \\frac{p+1}{100})}$<\/p>\n

$\\Rightarrow a = \\frac{x*100*100}{(100+p)*(100+p+1)}$<\/p>\n

$\\Rightarrow a = \\frac{x*100*100}{(p+100)*(p+101)}$<\/p>\n

Similarly, $b=\u00a0\\frac{x}{(1+ \\frac{p+2}{100})*(1+ \\frac{p-1}{100})}$<\/p>\n

$\\Rightarrow b = \\frac{x*100*100}{(p+102)*(p+99)}$<\/p>\n

Similarly, $c=\u00a0\\frac{x}{(1+ \\frac{p+3}{100})*(1+ \\frac{p-2}{100})}$<\/p>\n

$\\Rightarrow c = \\frac{x*100*100}{(p+103)*(p+98)}$<\/p>\n

Similarly, $d=\u00a0\\frac{x}{(1+ \\frac{p+4}{100})*(1+ \\frac{p-3}{100})}$<\/p>\n

$\\Rightarrow d = \\frac{x*100*100}{(p+104)*(p+97)}$<\/p>\n

Similarly, $e=\u00a0\\frac{x}{(1+ \\frac{p+5}{100})*(1+ \\frac{p-4}{100})}$<\/p>\n

$\\Rightarrow e = \\frac{x*100*100}{(p+105)*(p+96)}$<\/p>\n

The numerators of the fractions are same, therefore the one with the smallest value of denominator will have the greatest value. If we compare the denominators, we can find out the fraction with the highest value. The person with the highest initial salary got the least raise, as we know that the final salary of all the candidates is same.<\/p>\n

Thus, denominator of a, $a_{den}= (p+100)*(p+101)= p^{2}+201p+100*101$<\/p>\n

Similarly,\u00a0$b_{den}= (p+102)*(p+99)= p^{2}+201p+102*99$<\/p>\n

$c_{den}= (p+103)*(p+98)= p^{2}+201p+103*98$<\/p>\n

$d_{den}= (p+104)*(p+97)= p^{2}+201p+104*97$<\/p>\n

$e_{den}= (p+105)*(p+96)= p^{2}+201p+105*96$<\/p>\n

We see that we need to compare only the last terms of the denominators as the other terms are same.<\/p>\n

Thus, last term of a, $a_{lt}= 100*101= 100.5^{2}-0.5^{2}$<\/p>\n

last term of b, $b_{lt}= 102*99= 100.5^{2}-1.5^{2}$<\/p>\n

last term of c, $c_{lt}= 103*98= 100.5^{2}-2.5^{2}$<\/p>\n

last term of d, $d_{lt}= 104*97=\u00a0100.5^{2}-3.5^{2}$<\/p>\n

last term of e, $d_{lt}= 105*96=\u00a0100.5^{2}-4.5^{2}$<\/p>\n

Thus, we can see that since denominator of $e$ is the smallest, therefore E has the highest initial salary.<\/p>\n

14)\u00a0Answer\u00a0(C)<\/strong><\/p>\n

On solving for x and y from the equations<\/p>\n

3x + 4y = 2018 and 5x + 3y = 1<\/p>\n

we get Q(-550,917)<\/p>\n

Let, P(x,y)<\/p>\n

So, $\\frac{y – 917}{x + 550}$ = 2<\/p>\n

=> y – 2x = 2017 ….(1)<\/p>\n

Considering the equations<\/p>\n

3x + 4y = 2a ……..(2)<\/p>\n

7x + 2y = 2018 …..(3)<\/p>\n

On subtracting equation (2) from (3) we have,<\/p>\n

4x – 2y = 2018 – 2a<\/p>\n

=> 2x – y = 1009 – a<\/p>\n

=> y – 2x = a -1009 …..(4)<\/p>\n

From equation (1) and (4)<\/p>\n

2017 = a – 1009<\/p>\n

=> a = 3026<\/p>\n

Hence, option C.<\/p>\n

15)\u00a0Answer\u00a0(C)<\/strong><\/p>\n

The critical points of the function are -4, -6, -8, … , -98 ( 48 points).<\/p>\n

For all integers less than -98 \u00a0and greater than -4 f(x) > 0 always .<\/p>\n

for x= -5, f(x) < 0<\/p>\n

Similarly, for x= -9, -13, …., -97 (This is an AP with common difference -4)<\/p>\n

Hence, in total there are 24 such integers satisfying f(x)< 0.<\/p>\n

XAT Solved Previous papers<\/a><\/p>\n

Take XAT Mock Test<\/a><\/p>\n

We hope this Quant Questions PDF for XAT with Solutions will be helpful to you.<\/p>\n","protected":false},"excerpt":{"rendered":"

Quant Questions for XAT Download important Quant Questions for XAT PDF based on previously asked questions in CAT exam. Practice Quant Questions PDF for XAT exam. Question 1:\u00a0Two diagonals of a parallelogram intersect each other at coordinates (17.5, 23.5). Two adjacent points of the parallelogram are (5.5, 7.5) and (13.5, 16). Find the lengths of […]<\/p>\n","protected":false},"author":42,"featured_media":38231,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"om_disable_all_campaigns":false,"_mi_skip_tracking":false,"footnotes":""},"categories":[3,169,125,350,362,366],"tags":[214,1425],"class_list":{"0":"post-38225","1":"post","2":"type-post","3":"status-publish","4":"format-standard","5":"has-post-thumbnail","7":"category-cat","8":"category-downloads","9":"category-featured","10":"category-iift","11":"category-snap","12":"category-xat","13":"tag-quant","14":"tag-xat"},"better_featured_image":{"id":38231,"alt_text":"Quant Questions for XAT","caption":"Quant Questions for XAT\n","description":"Quant Questions for 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