CAT 2017 Question Paper Slot 1

x - y - z = 25 and $$x\leq40,y\leq12$$, $$z\leq12$$
If x = 40 then y + z = 15. Now since both y and z are natural numbers less than 12, so y can range from 3 to 12 giving us a total of 10 solutions.Similarly, if x = 39, then y + z = 14. Now y can range from 2 to 12 giving us a total of 11 solutions.
If x = 38, then y + z = 13. Now y can range from 1 to 12 giving us a total of 12 solutions.
If x = 37 then y + z = 12 which will give 11 solutions.
Similarly on proceeding in the same manner the number of solutions will be 10, 9, 8, 7 and so on till 1.
Hence, required number of solutions will be (1 + 2 + 3 + 4 . . . . + 12) + 10 + 11
= 12*13/2 + 21
78 + 21 = 99

$$(n - 5) (n - 10) - 3(n - 2)\leq0$$
=> $$ n^2 - 15n + 50 - 3n + 6 \leq 0$$
=> $$n^2 - 18n + 56 \leq 0$$
=> $$(n - 4)(n - 14) \leq 0$$
=> Thus, n can take values from 4 to 14. Hence, the required number of values are 14 - 4 + 1 = 11.

$$f_{1}(x)=x^{2}+11x+n$$ and $$f_{2}(x) = x$$
$$f_{1}(x)=f_{2}(x)$$
=> $$x^{2}+11x+n = x$$
=> $$ x^2 + 10x + n = 0 $$
=> For this equation to have distinct real roots, b$$^2$$-4ac>0
$$ 10^2 > 4n$$
=> n < 100/4 
=> n < 25
Thus, largest integral value that n can take is 24.

For the value of given expression to be minimum, the values of $$a, b, c$$ and $$d$$ should be as close as possible. 30/4 = 7.5. Since each one of these are integers so values must be 8, 8, 7, 7. On putting these values in the given expression, we get
 $$(8 - 8)^{2} + (8 - 7)^{2} + (8 - 7)^{2}$$
=> 1 + 1 = 2

The total number of given points are 11. (10 on circumference and 1 is the center)
So total possible triangles = 11C3 = 165.
However, AOB, COD, EOF, GOH, JOK lie on a straight line. Hence, these 5 triangles are not possible. Thus, the required number of triangles = 165 - 5 = 160

The graph of the given function is as shown below.

We can see that the shortest distance of the point (1/2, 1) will be 1 unit.

The seventh term of an AP = a + 6d. Third term will be a + 2d and second term will be a + 16d. We are given that
$$ (a + 6d)^2 = (a + 2d)(a + 16d)$$
=> $$ a^2 $$ + $$36d^2$$ + 12ad = $$ a^2 + 18ad + 32d^2 $$
=> $$4d^2 = 6ad$$
=> $$ d:a = 3:2$$
We have been asked about a:d. Hence, it would be 2:3

We have been given that a + b + c + d = 7
Total ways of distributing 7 things among 4 people so that each one gets at least one = $$^{n-1}C_{r-1}$$ = 6C3 = 20
Now we need to subtract the cases where any one person got more than 3 erasers. Any person cannot get more than 4 erasers since each child has to get at least 1. Any of the 4 childs can get 4 erasers. Thus, there are 4 cases. On subtracting these cases from the total cases we get the required answer. Hence, the required value is 20 - 4 = 16

$$f(3) = \frac{15 + 2}{9 - 5} = \frac{17}{4}$$
$$f(f(3)) = \frac{5*17/4 + 2}{3*17/4 - 5} = \frac{93/4}{31/4} = \frac{93}{31} = 3$$
$$g(f(f(3))) = 3^2 - 3*2 - 1 = 2$$

$$a_{1}$$ = 3 and $$a_{2}$$ = 7. Hence, the common difference of the AP is 4. If we assume, k=3n 
We have been given that the sum up to 3n terms of this AP is 1830. Hence, $$1830 = \frac{k}{2}[2*3 + (k - 1)*4$$
=> 1830*2 = k(6 + 4k - 4)
=> 3660 = 2k + 4k$$^2$$
=> $$2k^2 + k - 1830 = 0$$
=> (k - 30)(2k + 61) = 0
=> k = 30 or k = -61/2
Since k is the number of terms so k cannot be negative. Hence, must be 30
So, 3n = 30
n = 10
Sum of the first '10' terms of the given AP = 5*(6 + 9*4) = 42*5 = 210
m($$a_1$$+ $$a_{2}$$ +...+ $$a_n$$) > 1830
=> 210m > 1830
=> m > 8.71
Hence, smallest integral value of 'm' is 9.