XAT 2019

In the given statement, the expression becomes a whole number only when the powers of all the prime numbers are also whole numbers. 

Let us first simplify the expression a bit by expressing all terms in terms of prime numbers. 

 $$\sqrt[3]{7^a\times 35^{b+1} \times 20^{c+2}}$$

$$\Rightarrow \sqrt[3]{7^a\times 5^{b+1} \times 7^{b+1} \times 2^{2(c+2)} \times 5^{c+2}}$$

$$\Rightarrow \sqrt[3]{2^{2c+4} \times 5^{b+c+3} \times 7^{a+b+1}}$$

$$\Rightarrow 2^{\frac{2c+4}{3}} 5^{\frac{b+c+3}{3}} 7^{\frac{a+b+1}{3}} $$

Now, from the given options, we can put in values of the variables and check the exponents of all the numbers. 

Option A : a = 2, b = 1, c = 1 : 

In this case, we can see that exponent of 5 ie $$\frac{b+c+3}{3} = \frac{5}{3} $$ is not a whole number. 

Option B : a = 1, b = 2, c = 2

In this case, we can see that exponent of 2 ie $$\frac{2c+4}{3} = \frac{8}{3} $$ is not a whole number. 

Option C : a = 2, b = 1, c = 2

In this case, we can see that exponent of 2 ie $$\frac{2c+4}{3} = \frac{8}{3} $$ is not a whole number. 

Option D : a = 3, b = 1, c = 1

In this case, we can see that exponent of 5 ie $$\frac{b+c+3}{3} = \frac{5}{3} $$ is not a whole number. 

Option E : a = 3, b = 2, c = 1

In this case, we can see that all exponents are whole numbers. 

Thus, option E is the correct option.

Let the radius of the cylinder be $$r$$ and height be $$h$$ (as shown).

Let the dimensions of the cone be $$r_c$$ and $$h_c$$ (as shown).

It is given that $$h_c = \frac{h}{2}$$

Area of base of cone $$A_{c}=\pi \times r_{c}^{2}$$

Area of base of cylinder $$A=\pi \times r^{2}$$

Given, $$A_{c}=\frac{1}{5} \times A $$

$$ \Rightarrow \pi \times r_{c}^{2} =\frac{1}{5} \times \pi \times r^{2} $$

$$  \Rightarrow r_{c}^{2} = \frac{1}{5} \times r^{2} $$

$$  \Rightarrow \frac{r^{2}}{r_{c}^{2}} = 5 $$

Now we know that the volume of cylinder = total volume of cones 

Let the number of cones be n. 

So, volume of cylinder = n x volume of  each cone

$$  \Rightarrow V = n \times V_{c} $$

$$  \Rightarrow \pi \times r^{2} \times h = n \times \frac{1}{3} \times \pi \times r_{c}^{2} \times h_c $$

$$  \Rightarrow r^{2} \times h = n \times \frac{1}{3} \times r_{c}^{2} \times h_c $$

$$  \Rightarrow \frac{r^{2}}{r_{c}^{2}} \times \frac{h}{h_c} \times 3 = n $$

$$  \Rightarrow 5 \times 2 \times 3 = n $$

$$  \Rightarrow 30 = n $$

Thus, 30 conical ingots can be made.

Note: For this question, discrepancy is found in question/answer. Full Marks is being awarded to all candidates.

Let CP of object be a. 

It is given that SP = $$(1+\frac{4}{100} )$$ x CP 

$$ \Rightarrow $$ SP = 1.04 x CP

It is given that MP = $$(1+\frac{x}{100} )$$ x CP

It is also given that SP = $$(1-\frac{\frac {2x}{3}}{100} )$$ x MP 

$$ \Rightarrow $$ SP = $$(1-\frac{2x}{300} )$$ x $$(1+\frac{x}{100} )$$ x CP 

$$ \Rightarrow (1+\frac{4}{100})$$ x CP  = $$(1-\frac{2x}{300} )$$ x $$(1+\frac{x}{100} )$$ x CP 

$$ \Rightarrow (1+\frac{4}{100})$$   = $$(1-\frac{2x}{300} )$$ x $$(1+\frac{x}{100} )$$ 

$$ \Rightarrow \frac{104}{100}$$   = $$(1-\frac{2x}{300} )$$ x $$(1+\frac{x}{100} )$$ 

$$ \Rightarrow \frac{104}{100}$$   = $$(\frac{300-2x}{300} )$$ x $$(\frac{x+100}{100} )$$ 

$$ \Rightarrow \frac{104}{100} \times 300 \times 100$$   = $$(300-2x)$$ x $$(x+100)$$ 

$$ \Rightarrow 31200 $$   = $$(300-2x)$$ x $$(x+100)$$ 

Now, we look at the options. 

Since the question says that $$x\epsilon[25,30]$$ so 50% of x ie 0.5x cannot be less than 12.50 ie $$\frac {25}{2}$$ and cannot be more than 25 ie $$\frac {50}{2}$$

This eliminates option A.

Putting values of $$x$$ given in the remaining options in the final expression, [here we need to be careful to use the value of x and not the value of 50% of x as given in the expression]

Look carefully in the remaining options : 16,13,15 as 0.5x ie 32,26,30 as possible values of x. 

In the question since the discount rate offered is $$\frac{2x}{3}$$%, then a safer choice would be to check for the option of 30 in the beginning. It is a safer choice because the percentage of discount that we get in the other options are not whole numbers. 

NOTE : THIS IS JUST A SAFE CHOICE AND NEVER MARK AN ANSWER DIRECTLY ON THIS PRESUMPTION WITHOUT CHECKING IT.

Putting x=30 in the expression : 

(300 - (2x30))x(30+100) = (300-60)x(100+30) = 240x130 = 24x13x100= 312x100 = 31200= LHS of expression.

Thus the value of $$x$$ is 30.

Therefore value of 50% of $$x$$ = 0.5x30 = 15 

Amount on which interest will be charged = 19200 - 4800 = 14400

The total amount paid will be equal to the sum of all monthly instalments. Therefore, we have

$$14400*k^{5a} = I (k^{4a} +k^{3a}+k^{2a}+k^{a}+1 )$$   .....(1)

where, k = $$1+\frac{12}{100}$$ & a = $$\frac{1}{12}$$

We know that, $$k^{5a}-1 = (k-1)(k^{4a} +k^{3a}+k^{2a}+k^{a}+1)$$

=>$$k^{4a} +k^{3a}+k^{2a}+k^{a}+1$$ = $$\frac{k^{5a}-1}{k-1}$$ ....(2)

Substituting in equation (1) we get

I = $$14400*k^{5a}*\left[\frac{k-1}{k^{5a}-1}\right]$$ ....(3)

On substituting the values of k and a in equation (3) we get
I $$\approx$$ 2965

Hence, option B.

The stack of apples can be imagined as shown below. For every row, there is an increase of 1 apple from the previous row.

This means that the sum total of the apples present in the stack will be sum of AP given as : 1,2,3,4.... ie the sum of numbers from 1.

This is given by the formula S = $$\frac{n \times (n+1)}{2}$$

Since the initial number of apples was not more than 150, we find the maximum number of rows that was possible. 

Let S be the total initial sum of apples in the cart. 

We know that $$S\leq 150 $$

$$ \Rightarrow \frac{n \times (n+1)}{2} \leq 150 $$

$$ \Rightarrow n \times (n+1) \leq 300 $$

$$ \Rightarrow n^2 + n \leq 300 $$

Since the biggest number with square less than 300 is 17, we try to see if 17 works in the expression by hit and trial method. 

So, 17 x 18 = 306 > 300, therefore the maximum number of rows possible is not 17.

Since 306 is JUST over 300, let us check at 16.

So, 16 x 17 = 272 < 300, therefore the maximum number of rows possible is 16.

Let us tabulate the total number of apples present in the cart based on the number of rows : 

Since the total number of apples from row 15 is less than 126 (which was found by the customer), we can safely say that the total number of rows of apples with the shopkeeper that day was 16. 

Now, if 'n' rows of apples are sold, the total available apples can be calculated as : $$\frac{16 \times 17}{2} -  \frac{n \times (n+1)}{2}$$

Let us tabulate the result obtained in this case : 

From the table we can see that when the customer saw 126 apples, the apples in the top 4 rows had been sold. 

Therefore, the number of rows = 16 - 4 = 12 rows. 

On solving for x and y from the equations 

3x + 4y = 2018 and 5x + 3y = 1

we get Q(-550,917)

Let, P(x,y)

So, $$\frac{y - 917}{x + 550}$$ = 2

=> y - 2x = 2017 ....(1)

Considering the equations

3x + 4y = 2a ........(2)

7x + 2y = 2018 .....(3)

On subtracting equation (2) from (3) we have,

4x - 2y = 2018 - 2a

=> 2x - y = 1009 - a

=> y - 2x = a -1009 .....(4)

From equation (1) and (4)

2017 = a - 1009

=> a = 3026

Hence, option C.

As mentioned in the question,

Lines L1 and L2 intersect at point P as shown in the figure.

On solving for x and y from equations

x + 2y - 18 = 0

2x - y - 1 = 0

We get x = 4 and y =7.

Given, radius = $$\sqrt{20}$$

Using the equation of a circle, we have

$$(4-a)^{2}$$ + $$(7-b)^{2}$$ = 20....(1)

The center of the circle will lie on the line: 2x - y - 1 = 0

a,b will satisfy this equation.

So 2a-b-1=0

b=2a-1

From equation 1...

$$(4-a)^{2}$$ + $$(7-b)^{2}$$ = 20

$$(4-a)^{2}$$ + $$(8-2a)^{2}$$ = 20

5$$(4-a)^{2}$$ = 20

a=6 or a=2

b=11 or b=3

The sum of the coordinates possible=6+11  or 2+3 

i.e. 17 or 5

Option B is one of the solution.

Simplify the expression a bit to remove the root sign in the denominator

$$\dfrac{\log{97-56\sqrt{3}}}{\dfrac{1}{2}\times (\log{7+4\sqrt{3})}}$$

$$ \Rightarrow 2 \times \dfrac{\log{97-56\sqrt{3}}}{\log{7+4\sqrt{3}}} $$ 

To move further, let us see the root of the numerator. 

Assume the root of the numberator to be $$\sqrt{a}-\sqrt{b}$$.

When we square it, we get $$a + b - (2 \times \sqrt{a} \sqrt{b}) = a+b-2\sqrt{ab} $$

comparing the value of terms under root with the terms in the numerator, we get 

$$\sqrt{ab} = 28\sqrt{3} $$ and $$a+b=97$$ 

From solving this, we get to know that $$a=7$$ and $$b=4\sqrt{3}$$ 

Thus the expression can be written as $$ 2 \times2 \times \dfrac{\log{7-4\sqrt{3}}}{\log{7+4\sqrt{3}}} $$

$$ \Rightarrow 4 \times \dfrac{\log{7-4\sqrt{3}}}{\log{7+4\sqrt{3}}} $$ 

Now, let us look at the reciprocal of the term in log in the denominator. 

$$ \frac{1}{7+4\sqrt{\ 3}} = \frac{1}{7+4\sqrt{\ 3}} \times \dfrac{7-4\sqrt{3}}{7-4\sqrt{3}} $$

$$ \Rightarrow \dfrac{7-4\sqrt{3}}{7^{2}-(4\sqrt{3})^{2}} $$

$$ \Rightarrow \dfrac{7-4\sqrt{3}}{49-48} = 7-4\sqrt{3}$$

$$\frac{\log\left(7-4\sqrt{\ 3}\right)}{\log\left(7+4\sqrt{\ 3}\right)}=\frac{\log\left(7-4\sqrt{\ 3}\right)}{\log\left(\frac{1}{7-4\sqrt{\ 3}}\right)}=\frac{\log\left(7-4\sqrt{\ 3}\right)}{-\log\left(7-4\sqrt{\ 3}\right)}=-1$$

Thus the value of the expression can be further simplified as 

$$ \Rightarrow 4 \times (-1) = -4 $$

Hence the correct answer is option C

Construct a perpendicular in the trapezium. Let its height be 'h'

Now, $$sin\angle BAC = \dfrac{h}{AC}$$

and $$sin\angle BAD = \dfrac{h}{AD}$$

Thus, $$\dfrac{sin\angle BAC}{sin\angle BAD} = \dfrac{\frac{h}{AC}}{\frac{h}{AD}} = \dfrac{AD}{AC}$$