NMTC RAMANUJAN Inter Level 2019 Solved Question Paper PDF

For Ram to win exactly at the end of the sixth game, the first four games must leave the score tied, and Ram must then win the last two games. There are $$4$$ possible sequences for the first four games, each with probability $$\left(\frac{3}{5}\right)^2\left(\frac{2}{5}\right)^2$$. Thus the required probability is $$4\left(\frac{3}{5}\right)^4\left(\frac{2}{5}\right)^2=\frac{1296}{15625}$$.

Let the three numbers of volunteers be $$x<y<z$$, so $$x+y+z=30$$. The possible ordered triples are counted by considering the possible values of the smallest number, giving $$61$$ unordered triples. Since the three booths are distinct, each triple can be assigned in $$3!=6$$ ways, giving $$61\times6=366$$. Therefore none of the listed numerical choices is correct.

Put $$t=\cos x$$, where $$-1\le t\le1$$. Then $$f(x)=2t^2+2at+2a-1$$. Checking the minimum of this quadratic on $$[-1,1]$$ shows that the required minimum $$\frac{1}{2}$$ occurs only for $$a=1$$, so there is exactly one such value.

Direct composition gives $$f^2(x)=x$$ wherever the expressions are defined. Hence the iterates alternate between $$f(x)$$ and $$x$$, so an odd iterate gives $$f(x)$$. Therefore $$f^{2019}(\sqrt{2})=f(\sqrt{2})=\sqrt{2}$$.

Translating the centre does not change the area, so consider $$|x|+|y|=2$$. This is a rhombus with diagonals $$4$$ and $$4$$. Its area is $$\frac{1}{2}\times4\times4=8$$.

A line containing two points with rational coordinates has rational slope. If such a line with slope $$m$$ passes through $$(a,2a)$$ and a rational point $$(h,k)$$, then $$k-2a=m(h-a)$$. Rearranging gives $$k-mh=(2-m)a$$, and because $$a$$ is irrational while the left side is rational, we must have $$m=2$$. Thus there is only one possible line.

Counting incidences in the first family gives $$33\times6=9|S|$$, so $$|S|=22$$. Counting incidences in the second family gives $$8n=4|S|=88$$. Hence $$n=11$$.

Rewrite the second equation as $$(2b-c)^2+5bc-7a-8=0$$. Adding this to five times the first equation gives $$10a^2+(2b-c)^2-52a+42=0$$. Therefore $$10a^2-52a+42\le0$$, which factors as $$10(a-1)(a-4.2)\le0$$. Hence $$a\in[1,4.2]$$.

Since $$\frac{1}{\operatorname{cosec}x}=\sin x$$, we have $$-1\le\sin x\le1$$. The greatest-integer value of a number in this interval can only be $$-1$$, $$0$$, or $$1$$. All three values occur, so the range is $$\{-1,0,1\}$$.

Let $$p=xy$$. Using $$x+y=3$$ and the identity for $$x^5+y^5$$ reduces the second equation to $$5p^2-45p+135=0$$. Thus $$p=\frac{9\pm3\sqrt{3}i}{2}$$, which is non-real. Since $$x+y$$ is real but $$xy$$ is non-real, all resulting pairs are non-real, so option d is correct.

Using coordinates for the rectangle and the given division ratios gives the three segments on $$BF$$ in the ratio $$26:29:10$$, which reduces no further. Hence their sum is $$26+29+10=65$$. The published solution for this problem gives the same minimum sum as $$65$$, but the uploaded paper prints the options $$4,8,12,16$$, so the keyed answer cannot be matched to any printed option.

They reach home $$20$$ minutes earlier than usual, so the husband must have met Rita $$40$$ minutes before the normal meeting time. Since the husband travels back along the route, the time saved on his journey is twice the time Rita walked. Therefore Rita walked for $$20$$ minutes.

The area of a regular polygon with $$n$$ sides of side length $$s$$ is $$\frac{ns^2}{4\tan(\pi/n)}$$. Therefore the ratio of the area of the $$200$$-gon to the $$100$$-gon is $$4\frac{\tan(\pi/100)}{2\tan(\pi/200)}$$, which is approximately $$8$$. Hence the closest listed integer is $$8$$.

Put $$x=1$$ to get $$f(4)=f(3)-3$$, so $$f(3)=6$$. Put $$x=4$$ to get $$f(3)=f(6)-3$$, giving $$f(6)=9$$. Finally, putting $$x=3$$ gives $$f(6)=f(5)-3$$, so $$f(5)=12$$.

Let $$t=xy$$. Then $$xy^4+x^4y=xy(x^3+y^3)=t(1-3t)$$ because $$x+y=1$$. Since $$0<t\le\frac14$$, the quadratic $$t-3t^2$$ is maximized at $$t=\frac16$$, giving the maximum value $$\frac{1}{12}$$.

There are $$\binom84=70$$ subsets. The numbers of subsets whose greatest elements are $$8,7,6,5,4$$ are respectively $$\binom73,\binom63,\binom53,\binom43,\binom33$$. Hence the mean is $$\frac{8\binom73+7\binom63+6\binom53+5\binom43+4\binom33}{70}=\frac{504}{70}=7.2$$.

Adding the repunits produces a number that can be grouped into $$11$$ blocks with the pattern $$123456790$$, followed by the final carry pattern. The digit $$1$$ therefore occurs exactly once in each of the $$11$$ blocks. Hence the required number of occurrences is $$11$$.

The integer side constraints allow a rectangle $$PQRS$$ with side lengths $$40$$ and $$10$$. Its area is therefore $$40\times10=400$$ square units. The construction shown in the figure satisfies the required conditions, and no larger admissible rectangle occurs.

Each blue ball must be adjacent to one green and one white, while green and white balls cannot be adjacent directly. Counting the possible placements of the six blue positions and distributing the remaining green and white balls gives $$\binom62\binom93+\binom92\binom63=1980$$. Hence the number of arrangements is $$1980$$.

To maximize the sum with a fixed number of terms, increase by $$1$$ from $$1$$ to a peak and then decrease by $$1$$ back to $$1$$, with a repeated peak when necessary. The construction reaching $$2019$$ uses $$89$$ terms, with the central values rising through $$44$$ and then descending symmetrically. Therefore the minimum number of terms is $$89$$.

Let the common scale of the three perpendicular distances be $$s$$. Resolving the geometry around the equilateral triangle gives the required area ratio as $$\frac{36}{11}$$. Thus the coprime integers are $$36$$ and $$11$$, whose sum is $$47$$.

By the angle bisector theorem, $$BD:DC=AB:AC=6:8=3:4$$. Since $$BC=7$$, we get $$BD=3$$. Applying the angle bisector theorem again in triangle $$ABD$$ gives $$AF:FD=AB:BD=6:3=2:1$$. Under the numeric-only TITA rule, the answer field is recorded as $$2$$, while the explanation preserves the required ratio.

Since every root of the cubic $$f$$ is a root of the quartic $$h$$, polynomial division gives $$h(x)=(x+10/c)f(x)$$. Comparing coefficients gives $$c=30$$, $$a=-2$$ and $$b=-5$$. Therefore $$h(1)=1+1-5+13+30=40$$.

In triangle $$ABD$$, the triangle inequalities give $$5<BD<25$$. In triangle $$BCD$$, they give $$23<BD<43$$. Therefore the only integer satisfying both sets of inequalities is $$BD=24$$.

Factor the expression as $$f(n)=(n^2+3n+3)(n^2-3n+3)$$. For the product to be prime, the second factor must equal $$1$$, which occurs for $$n=1$$ and $$n=2$$. These give $$f(1)=7$$ and $$f(2)=13$$, so the required sum is $$20$$.

Let $$S_n$$ denote the number of permitted rearrangements of $$n$$ boys when each boy can remain in the same seat or move to an adjacent seat. Then $$S_n=S_{n-1}+S_{n-2}$$, with $$S_2=2$$ and $$S_3=3$$. This gives $$S_{13}=377$$, and excluding the original unchanged arrangement leaves $$377-1=376$$.

Each pair of the $$15$$ polygon vertices determines two possible equilateral triangles in the plane, giving $$2\binom{15}{2}=210$$. However, the $$15$$ equilateral triangles formed entirely by polygon vertices have been counted, so they must be removed. Therefore the required number is $$210-15=195$$.

If the roots are $$\alpha,\beta,\gamma$$, their mean is $$-\frac{a}{3}$$ and their product is $$-c$$. The sum of the coefficients is $$1+a+b+c$$, and the $$y$$ intercept gives $$c=2$$. Thus $$-\frac{a}{3}=-2=1+a+b+2$$, which gives $$a=6$$ and $$b=-11$$.

By Varignon's theorem, the midpoint quadrilateral has sides parallel to the diagonals of $$ABCD$$. Since it is a square, the diagonals $$AC$$ and $$BD$$ are perpendicular and equal. The conditions give $$3q-p=2$$ and $$40=(p-1)^2+(q-1)^2$$, whose first-quadrant solution is $$p=7,q=3$$. Hence $$p+q=10$$.

Rewrite the three conditions as $$(a+1)(b+1)=1225=49\times25$$, $$(b+1)(c+1)=550=25\times22$$, and $$(c+1)(d+1)=352=22\times16$$. Hence $$a+1=49$$, $$b+1=25$$, $$c+1=22$$ and $$d+1=16$$, giving $$a=48,b=24,c=21,d=15$$. Their sum is $$48+24+21+15=108$$.