XAT 2023 Question Paper

Let printed price be 100.

Rajnish bought it at 25% discount, C.P of Rajnish = 75% of 100 = 75.

He sold it at 10% discount on printed price, S.P of Rajnish = 90% of 100 = 90

Hence, Rajnish, profit% = $$\ \frac{\ 90-75}{75}\times\ 100=20%$$.

The answer is option D.

Given equation, $$A = |x + 3| + | x - 2 | - | 2x -8|$$.

Case (i):- $$x+3\ge0,\ x-2\ge\ 0\ \&\ 2x-8\ge\ 0$$

then, $$A = x + 3 +  x - 2  -  2x +8 =9$$.

The maximum value of |A| = 9

Case (ii):- $$x+3\ge0,\ x-2\ge\ 0\ \&\ 2x-8< 0$$

$$x\ge-3,\ x\ge\ 2\ \&\ x<4$$

then $$A = x + 3 + x - 2 + 2x -8 = 4x-7$$.

The range of x is [2,4). Hence the value of A varies from [1,9).

The maximum value of |A| < 9

Case (iii):- $$x+3\ge0,\ x-2 < 0\ \&\ 2x-8< 0$$

$$x\ge-3,\ x < 2\ \&\ x<4$$

then $$A = x + 3 - x + 2 + 2x -8 = 2x-3$$.

The range of x is [-3, 2). Hence the value of A varies from [-9,1).

The maximum value of |A| = 9

Case (iv):- $$x+3 < 0,\ x-2 < 0\ \&\ 2x-8< 0$$

$$x < 3,\ x < 2\ \&\ x<4$$

then $$A = - x - 3 - x + 2 + 2x -8 = -9$$.

The maximum value of |A| = 9

From the above cases, The maximum value of |A| = 9. Option (B) is correct.

The length of AB = $$\left(b+4c\right)-\left(b-2c\right)=6c$$  (X-coordinates of A&B are same).

The altitude of triangle ABC, CD = a-(-2a)=3a.

Area of triangle ABC = $$\ \frac{\ AB\ \times\ CD}{2}$$ = $$\ \frac{6c\ \times\ 3a\ }{2}=9ac$$

Option (D) is correct.

The given equation is,

x (x - p) - y (y + p) = 7p

$$x^2-px-y^2-py=7p$$

$$x^2-y^2-px-py=7p$$

$$\left(x+y\right)\left(x-y\right)-p\left(x+y\right)=7p$$

$$\left(x-y-p\right)\left(x+y\right)=7p$$

As '7' & 'p' both are prime numbers

$$\left(x-y-p\right)\left(x+y\right)$$ can be expressed as $$\left(7\times\ p\right)\ or\ \left(7p\times\ 1\right)$$

Case (i) - $$\left(x+y\right)\ \times\ \left(x-y-p\right)=7\times\ p$$

                $$x+y+x-y-p=7+ p$$

                $$2x-p=7+p$$

                $$x=\frac{7}{2}+p$$

But it's given that 'x' is a positive integer. This case is not possible.

Case (ii) - $$\left(x+y\right)\ \times\ \left(x-y-p\right)=7p\times\ 1$$

                 $$x+y+x-y-p=7p+1$$

                 $$2x-p=7p+1$$   

                 $$x=\frac{1}{2}+4p$$

But it's given that 'x' is a positive integer. This case is not possible.

The given equation is not possible with given conditions.

Option (E) is correct. 

Given that $$a_1=1$$ & $$a_{n+1} =\frac{1}{1+\frac{1}{a_{n}}}$$

$$a_2=\ \frac{\ 1}{1+\ \frac{\ 1}{1}}=\ \ \frac{1}{2},\ a_3=\ \frac{\ 1}{1+\ \frac{\ 1}{\left(\frac{1}{2}\right)}}=\frac{1}{3},....$$

This implies, $$\ a_n=\ \frac{1}{n}$$.

Required value$$=a_1a_2+a_2a_3+........+a_{2008}a_{2009}$$

= $$\frac{1}{1}\times\ \frac{1}{2}+\frac{1}{2}\times\ \frac{1}{3}+.....+\frac{1}{2008}\times\ \frac{1}{2009}$$

= $$\left(\frac{1}{1}-\ \frac{1}{2}\right)+\left(\frac{1}{2}-\ \frac{1}{3}\right)+.....+\left(\frac{1}{2008}-\ \frac{1}{2009}\right)$$

= $$1-\ \frac{1}{2009}$$ 

= $$\frac{2008}{2009}$$

The answer is option C.

The shortest route ant takes to travel from the bottom corner to the diagonally opposite top corner is  shown below.

The route goes through lateral faces (vertical faces) of the cubical room.

The length of route, r = $$\sqrt{\ \left(2+2\right)^2+2^2}$$ = $$\sqrt{\ 20}$$ = $$2\sqrt{5}$$

Let the positive integers Raju & Sarita chose be 'r' & 's'.

After doing the given operations, the final numbers they get are $$\frac{2r-20}{5}\ \&\frac{2s-20}{5}$$

Adding results we get, $$\frac{2r-20}{5}\ +\frac{2s-20}{5}\ =\ 16$$

$$2r-20\ +2s-20\ =\ 80$$

$$2r+2s\ =\ 80+40+40=120$$

$$r+s\ =60$$

For maximum value of |r-s|, one of 'r' & 's' has to be maximum and other has to be minimum.

As r & s are positive integers minimum value they can take is "1".

If $$r=1,\ then\ s=59$$

$$\left|r-s\right|=\left|1-59\right|=58$$.

If $$s=1,\ then\ r=59$$

$$\left|r-s\right|=\left|59-1\right|=58$$

Hence maximum value of $$\left|r-s\right|$$ is 58. Option B is correct.

Given that, CPD is an equilateral triangle

$$\angle\ CPD=\angle\ PDC=\angle\ DCP=60^{\circ\ }$$

As AQ is parallel to PC $$\angle\ CPD=\angle\ QAP=60^{\circ\ }$$

As BC is parallel to AD $$\angle\ QAP=\angle\ AQB=60^{\circ\ }$$ ( alternate interior angles).

Given, Area of equilateral triangle CPD is $$4\sqrt{\ 3}$$

This implies, $$\ \frac{\ \sqrt{3}a^2}{4}=4\sqrt{3}$$

Solving, we get $$a=\ 4$$.

From figure, length of $$AB$$ = Height of $$\triangle\ CPD=\frac{\sqrt{\ 3}a}{2}=2\sqrt{\ 3}$$.

From $$\triangle\ ABQ,\ BQ=\ \frac{\ AB}{\tan\ 60^{\circ\ }}=\frac{2\sqrt{\ 3}}{\sqrt{\ 3}}=2$$

Area of $$\triangle\ ABQ=\frac{1}{2}\times\ AB\times\ BQ=\ \ \frac{\ 2\times\ 2\sqrt{\ 3}}{2}=2\sqrt{\ 3}\ $$

Option A is correct.

The ratio of contents in the bigger jar was 85:5:10 (water, lime and sugar respectively).

Let volume of water be $$85x$$

Percentage of water in bigger jar = $$\ \frac{\ 85x}{85x+5x+10x}\times100$$ = $$85$$%

Option E is correct.

From the given information we get the following table,

The possible maximum marks are shown below.

Therefore, sixth highest possible score is 94.

Option B is correct.