XAT 2023 - Quant

Let the 4-digit locker key be $$abcd$$, where $$9\ge\ a,b,c,d>0$$, and they are all unique.

It is given that $$a\le\ 3\ \&\ \frac{b}{c}=a$$

It is also given that 'd' is the smallest number.

Case (i): $$a=1$$

Not possible, as 'd' is the smallest number, 'a' cannot be 1.

Case (ii): $$a=2$$

Then 'd' can take only one value, i.e. $$d=1$$ 

$$b=2c$$

 $$if\ c=3,\ then\ b=6$$

$$if\ c=4,\ then\ b=8$$

for $$c\ge\ 5$$, 'b' will not be a single-digit number.

Hence, two cases are possible, i.e. (2631, 2841).

Case (iii): $$a=3$$

'd' can take two values, i.e. $$d=1\ or\ d=2\ $$

& $$b=3c$$

If d = 1, then 'c' can take only one value. i.e. $$\ c=2,\ then\ b=6$$

If d = 2, then 'c' cannot take any value.

Hence one case is possible, i.e. (3621)

Amit has to try 3 different combinations. Option (E) is correct.

The first number Haruka entered is 1.

From Rule-1, the series will be 2, 4, 8, 16

From Rule 2, 16 will be followed by 7

Like-wise, the series will be 1, 2, 4, 8, 16, 7, 14, 5, 10, 1, 2, 4,......

The series is repeating itself for every 9 iterations.

Initially, input is 1 and after first iteration, result is 2. 

This implies,

9n = 1

9n + 1 = 2

9n + 2 = 4

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9n + 7 = 5

9n + 8 = 10

In the question, it is given that key is pressed 68 times. 

68 = 9(7) + 5

68 is in the form of 9n + 5

This implies, 7 is the output.

The answer is option A.

Option A) False, as it is mentioned no-one scored exactly 40 marks.

Option B) False, as only 1 student  or only 2 students can score above 40 and group average can be 40.

                 Case i:- 44, 39, 39, 39 & 39.

                 Case ii:- 42, 41, 39, 39 & 39.

Option C) False,  student's scores could be 44, 39, 39, 39 & 39.

Option D) False, student's scores could be 44, 43, 42, 32 & 39.

Option E) True, if scores of all students are more than 40. The average will be more than 40.

From the given information, we get that it is $$8\ inch\times\ 8\ inch$$ square grid.

Total ways of selecting 2 squares out of 64 in $$^{64}C_2$$.

Two squares with a common side can be selected in the following ways.

(i) Horizontal Pairs.

In the first row, R1, we can select 7 pairs of squares with a common side.

They are (R1C1,R1C2),  (R1C2,R1C3),.....(R1C7,R1C8).

It applies to other rows as well.

Hence the total number of squares = $$7\times\ 8=56$$

(ii) Vertical Pairs.

In the first column, C1, we can select 7 pairs of squares with a common side.

They are (R1C1,R2C1), (R2C1,R3C1),.....(R7C1,R8C1).

It applies to other columns as well.

Hence the total number of squares = $$7\times\ 8=56$$

The probability of two painted squares having a common side = $$\frac{56+56}{^{64}C_2}$$ = $$\frac{112}{2016}$$.

Option (A) is correct.

Let 'm' be the Greatest common divisor and A, B, C, D, E, F & G be the 7 unique numbers.

When 'm' divides A, B, ......& G, we get seven unique quotients. Let the quotients be a, b, c, d, e, f, and g.

It is given,

A + B + C + D + E + F + G = 1740

m(a + b + c + d + e + f + g) = 1740

$$\ \ 1740=2^2\times3\times5\times29$$

m(a + b + c + d + e + f + g) = $$\ \ 1740=2^2\times3\times5\times29$$

29 is a prime number and cannot be factorised further.

Therefore, largest possible value m can take is 60.

Least possible sum of 7 unique numbers is 1 + 2 + 3 +...+ 7 = 28

Replacing 7 with 8, we will get 29.

This implies 60(1 + 2 + 3 .....+ 6 + 8) = 1740

Largest possible G.C.D is 60.

The answer is option B.

Let players at slips 1, 2 and 3 be a, b, and c, respectively.

Let the speed of the ball be m.

The trajectory of the ball is not specified in the question. But let's take it as straight line.

The distance between the 1st and 2nd slips is the same as the distance between the 2nd and 3rd slips.

The length of 'ab' = The length of 'bc'. It implies Xb is the median of triangle Xac.

It takes 3 sec, 4 sec and 5 sec for the ball to reach a, b and c, respectively.

Hence, Xa = 3m, Xb = 4m and Xc = 5m.

Using Apollonius's theorem,

$$2\times\ \left(\left(4m\right)^2+\left(ab\right)^2\right)=\left(3m\right)^2+\left(5m\right)^2$$

$$16m^2+\left(ab\right)^2=17m^2$$

$$ab=m$$

Hence, $$ac=2m$$

Using the properties of triangle, the sum of two sides should be greater than the third side,

But, Xa + ac = Xc

3m + 2m = 5m. 

Hence, this arrangement is not possible. 

As this question has ambiguity, XAT officials awarded full marks to all candidates for this question.

$$Let\ \sin^215^{\circ}\ =a\ and\ \sin^275^{\circ}\ =\ b\ $$

Then the given equation becomes,

$$\ \frac{\ a^3+b^3+6ab}{a^2+b^2+5ab}$$ 

=$$\ \frac{\ \left(a+b\right)^3-3ab\left(a+b\right)+6ab}{\left(a+b\right)^2-2ab+5ab}$$     $$\ \therefore\ \ a^3+b^3=\left(a+b\right)^3-3ab\left(a+b\right)$$ , $$\ \therefore\ \ a^2+b^2=\left(a+b\right)^2-2ab$$

Using $$\sin\left(\theta\right)\ =\cos\left(90-\theta\ \ \right)$$

we get, $$\sin^275^{\circ\ }=\cos^215^{\circ\ }$$

now, a+b = $$\sin^215^{\circ\ }+\sin^275^{\circ\ }$$ = $$\sin^215^{\circ\ }+\cos^215^{\circ\ }$$ = 1.

The equation becomes, $$=\ \frac{\ 1-3ab+6ab}{1-2ab+5ab}\ =\ \frac{\ 1+3ab}{1+3ab}\ =1\ $$

Let 'a' and 'b' be the speeds of Rahim and the speed boat in still water, respectively

Let 'c' be the speed of the stream.

Speed boat starts after 1 minute of Rahim's departure and crosses him after 4 minutes.

It implies that the speed boat covered the same distance in 4 minutes that Rahim took 5 minutes.

Hence, the ratio of upstream speeds of Rahim and Speed boat = $$\frac{a-c}{b-c}=\frac{4}{5}$$

Using Statement 1, we get $$b=30\ kmph$$. But, it's not sufficient to get the value of 'a'.

Using Statement 2, we get $$a-c=\frac{30}{3}=10\ kmph$$. But it's also not sufficient to get the value of 'a'.

using both statements, we get

$$\ \frac{\ a-c\ }{b-c}=\frac{4}{5}$$

$$\ \frac{\ 10\ }{30-c}=\frac{4}{5}$$

$$\ \ 50\ =120-4c$$

$$\ \ c\ =17.5$$

$$\ \ a=17.5+10=27.5\ kmph$$

Let 'm' be the number of matches Guava played till now. They won '0.4m' matches. 

After playing another 'n' matches and wining all of them, their winning percentage will improve to 50.

i.e, 

$$\ \frac{\ 0.4m+n}{m+n}=0.5$$  

 $$m=5n$$

Playing 15 more matches and wining all of them, their winning percentage will improve from 50 to 60.

i.e $$\ \frac{\ 0.4m+n+15}{m+n+15}=0.6$$

Solving, we get $$6+0.4n\ =\ 0.2m$$

Substituting $$m=5n$$,

we get $$6+0.4n\ =\ n$$ 

$$n=10\ \&\ m\ =50$$

Hence, Guava club played 50 matches so far. 

The answer is option A.

Given that, $$AD\bot\ BC$$

                   $$BE=3$$

                   $$BC=5$$

Using Pythagoras' theorem,

    $$AD^2+BD^2=AB^2$$ .....(1)

    $$AD^2+DE^2=AE^2$$......(2)

    $$AD^2+DC^2=AC^2$$......(3)

(1) - (2) gives

$$BD^2+DE^2=AB^2-AE^2$$

$$x^2-\left(3-x\right)^2=AB^2-AE^2$$

$$AB^2-AE^2=6x-9$$

$$AB^2-AE^2+6CD=6x-9+6\left(5-x\right)$$       ($$CD=\left(5-x\right)$$)

$$AB^2-AE^2+6CD=6x-9+30-6x$$

$$AB^2-AE^2+6CD=21$$

Let the amount Jose borrowed be 'x'. The rate of interest is 10%

Interest occurred on 'x' is = $$\ \frac{\ x\times1\times\ \ 0.1\ }{5}+\ \frac{\ 2x\ \times2\times\ \ 0.1}{5}\ +\ \ \frac{2x\times\ 3\times\ 0.1\ }{5}$$

                                         = $$\ \frac{\ 1.1x\ }{5}$$

The invested amount doubled at the end of the fourth year, i.e. 2x.

Jose's profit after paying principal and interest amounts to his friend at the end of the fourth year is Rs 97500.

i.e., $$\ 2x-x-\frac{\ 1.1x\ }{5}=97500$$

       $$\frac{\ 3.9x}{5}=97500$$

       $$\ \ \ \ x=125000$$

The amount that Jose borrowed is Rs 1,25,000.

Option (D) is correct.

Let the amount invested by Jack and Sristi be $$'x'$$.

Jack's amount after four months will become $$\frac{x}{2^4}$$ and after six months it will become $$\frac{x}{2^6}$$.

Sristi's amount after four months will become $$x-\left(4\times\ 15000\right)$$ and after six months it will become $$x-\left(6\times\ 15000\right)$$

From the given information,

$$\ \frac{\ \ \frac{\ x}{2^4}}{x-\left(4\times\ 15000\right)}=\ \frac{\ \ \frac{\ x}{2^6}}{x-\left(6\times\ 15000\right)}$$

$$\ \ \left(\ x-\left(4\times\ 15000\right)\right)=\ 2^2\times\ \left(x-\left(6\times\ 15000\right)\right)$$

$$3x\ =\ 300000$$

$$x\ =\ 100000$$.

The amount invested by Jack and Sristi is 100000.

Option 'A' is correct.

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.

From the graph, we can say that the graph is similar to y=log(x).

y = log(x) graph is given below-

The correct option is C

We will check the options and figure out which option is more suitable.

Option A is not correct since it is not evident from the figure that there is a direct correlation between life satisfaction and affect balance.

Options B, C, and D are not correct since we cannot definitely conclude the correlation between the two mentioned factors.

Option E is correct because from the figure, and the fish selling distribution we can conclude the randomness than the others.

The correct option is E

In this case, we need to calculate for which state (fish/economic) is the maximum. To get the answer, we need to check the maximum fish-selling state and the corresponding economic status.

From the figure, we can say that Roviana has the maximum ratio.

We get the above table after filling in the data from the given question.

This question has ambiguity and was deleted.

From the given table, we can say that the 81+ age group has the highest female-to-male ratio.

The correct option is E

This question was deleted for ambiguous phrasing.