XAT 2013

Let 'A' be the point from where all started and D is the destination. At 'x' hours both Ram and Hari reached point C, Hari got off and Ram turned back to pick up Shyam. At the same instant, Shyam was at point B.

Let 'T' is the total amount of time taken by all three to reach point D.

For Hari,

T = $$x+\dfrac{100-25x}{5}$$  ... (1)

For Ram,

T = $$x+\dfrac{20x*5/6}{25}$$ + $$\dfrac{20x*5/6+100-25x}{25}$$  ... (2)

By equating (1) and (2),

$$x+\dfrac{100-25x}{5}$$ = $$x+\dfrac{20x*5/6}{25}$$ + $$\dfrac{20x*5/6+100-25x}{25}$$

$$\Rightarrow$$ $$x = 3$$ hours.

Therefore, time taken by all to reach destination  = $$x+\dfrac{100-25x}{5}$$

$$\Rightarrow$$ 3 + $$\dfrac{100-25*3}{5}$$

$$\Rightarrow$$ 3 + 5 

$$\Rightarrow$$ 8 hours

Hence, we can say that option A is the correct answer. 

Let us draw the diagram according to the information given in the questions. 

Let us assume that 'T' is the width of the road as shown in the diagram. 

Total area covered by road = $$30*T+40*T-T^2$$

Also it is given that the mayor wants that the area of the two roads to be equal to the remaining area of the park.

$$\Rightarrow$$ $$30*T+40*T-T^2$$ = $$\dfrac{30*40}{2}$$

$$\Rightarrow$$ $$T^2-70T+600$$ = $$0$$

$$\Rightarrow$$ $$(T-10)(T-60)$$ = $$0$$

$$\Rightarrow$$ $$T$$ = $$10$$ or $$60$$ m. T $$\neq$$ 60 m as the width of park is 30 m only. 
Therefore, we can say that T = 10 meters. Hence, option A is the correct answer.

It has been given that Arun's task is to complete as many tasks as possible and then to focus on completing tasks based on their priority. 
Arun has 10 days with him. Except 1 task, all other tasks require at least 3 days to be completed. Therefore, Arun can complete a maximum of 3 tasks. 
Arun can complete T1, T2, and T5 or T1, T3, and T4 or  T1, T3, and T5 or T2, T3, and T5 or T1, T4, and T5. 
In all these combinations, Arun would have completed 2 higher priority tasks in the combination T1, T2, and T5. 
Therefore, Arun should complete T1, T2, and T5 and hence, option B is the right answer.

According to the data given in previous question, Arun have 10 days to do a background check, as he have to enter within 10 days of receiving permit.

The number of tasks that can be completed in this case is also 3. 
Arun can complete T1, T2, and T5 or T1, T3, and T4 or  T1, T3, and T5 or T2, T3, and T5 or T1, T4, and T5. 
In this case, the time required to complete T1, T2, and T5 exceeds 10 days. The same is the case with T1, T2, and T3.
Among the given options, both options D and E are possible cases.

T5 is a higher priority task than T3. Therefore, Arun should try to complete T1, T4, and T5.

Therefore, option E is the right answer. 

We know that ABC is a right angled triangle.
=> $$AC=\sqrt{6^2+2^2}$$
$$AC=\sqrt{40}$$
$$AC=2*\sqrt{10}$$
Let the coordinates of A be (0,0)
We know that the radius of the circle, OA = $$\sqrt{50}$$cm
Let OD be the height of the triangle AOC.
$$AC/2=*\sqrt{10}$$

By applying Pythagoras theorem, we get,
=>The height(y) of  the point  O $$= \sqrt{50-10}$$
The height(y) of  the point  O$$=\sqrt{40}$$cm
=> Coordinates of point O = $$(\sqrt{10},\sqrt{40})$$

Area of triangle ABC = $$0.5*6*2$$ = $$6$$ square units.
Let the height of triangle ABC be h.
0.5*h*AC=6 
h*AC = 12
h*$$2*\sqrt{10}$$ = $$12$$
h = $$\frac{6}{\sqrt{10}}$$
X-coordinate of point B =$$\sqrt{6^2-\frac{36}{10}}$$
                                    = $$\sqrt{32.4}$$ cm
Distance between points O and B = $$\sqrt{(\sqrt{10}-\sqrt{32.4})^2+({\sqrt{40}-\frac{6}{\sqrt{10}})^2}}$$
Expanding, we get,
Square of the distance between points O and B = $$26$$ cm.
Therefore, option A is the right answer.

There are 6 cards and 2 out of the 6 cards are kings. 
Number of ways of selecting 2 cards = 6C2 = 15 ways.
Number of ways in which 2 cards can be selected such that both of them are King = 2C2 = 1
Number of ways in which 2 cards can be selected such that exactly one of them is a King = 2C1*4C1 = 8
=> a = (1+8)/15 = 9/15
b = 1-(9/15) = 6/15
a/b = 9/6 = 1.5
1.5 > 1.25
Therefore, option E is the right answer. 

We can draw the towns as shown in the diagram below.

For the minimum length the Government should diagonal roads. The length of road will be same as the length of a diagonal of a square whose side length is 10 km.

Length of one diagonal = $$\sqrt{2}*10$$ = 1.414*10 = 14.14 km

Therefore, total length of both the roads = 2*14.14 $$\approx$$ 28.30 km. 

$$(xxx)_{b}=x^3$$
=> $$xb^2+xb+x = x^3$$
=> $$b^2+b+1=x^2$$
On substituting b=1,and b=2, we get $$x^2$$ as $$3$$, and $$7$$. Since $$3$$ and $$7$$ are not perfect squares, we can infer that no number satisfies the given condition. Therefore, option E is the right answer.

$$x^5-1 = (x-1)(x^4+x^3+x^2+x+1)$$
$$x^5-1=(x-1)f(x)$$
$$x^5 = 1+(x-1)f(x)$$
$$f(x^5) = (1+(x-1)f(x))^4+(1+(x-1)f(x))^3+(1+(x-1)f(x))^2+(1+(x-1)f(x))+1$$
On dividing by $$f(x)$$, every term will leave $$1$$ as the remainder. Therefore, the remainder when $$f(x^5)$$ is divided by $$f(x)$$ is $$1+1+1+1+1=5$$.
Therefore, option A is the right answer. 

Let us evaluate the statements one by one:
I: 103 and 7 are the only prime factors of 1000027
On successively dividing 1000027 by 103 and 7, we get 1387 as the answer. 
1387 is divisible by 19. 
Therefore, statement I is false. 
II: $$\sqrt[6]{6!}>\sqrt[7]{7!}$$
Raising the power to 42 on both sides, we get,
$$[6!]^7>[7!]^6$$
$$6!*[6!]^6 > 7^6*[6!]^6$$
$$7^6$$ is greater than $$6!$$.
Therefore, statement II is false. 
III:It has been given that the person travels one-half of the journey at x kmph. 
Let us assume the distance to be '2d'. 
Let us assume the average speed to be 2d and check for the feasibility.
Let the speed at which the person travels the other half of the journey be y.
d/x + d/y = 2d/2x
d/x + d/y = d/x
=> d/y = 0 or y tends to infinity.
Therefore, such a scenario is not possible and hence, statement III is true.
Only statement III is true. Therefore, option C is the right answer.