NMTC KAPREKAR Sub-Junior 2023 Solved Question Paper PDF

The innermost quantity is $$2021 \times 2019 + 1$$ which can be written as $$(2020+1)(2020-1) + 1$$

Using the property that $$(a+b)(a-b) = a^2 - b^2$$ we get 

$$(2020+1)(2020-1) + 1 = 2020^2 -1^2 +1=2020^2$$

Hence, the square root of the innermost quantity is $$2020$$. The next radical becomes $$\sqrt{2022 \times 2020 + 1} $$

We can again rewrite this using the same property mentioned above to get $$(2021+1)(2021-1) + 1 = 2021^2-1^2 +1 = 2021^2$$

$$\sqrt{2021^2} = 2021$$. Now we can again rewrite this in the same way to get 

$$\sqrt{2023 \times 2021 + 1} = \sqrt{(2022+1)(2022-1) +1} = \sqrt{2022^2-1^2+1}=2022$$.

Place $$A = (0,0)$$, let $$AD$$ be horizontal and write $$B = (u,1)$$ because $$BE = 1$$. Since $$\triangle ABC$$ is right isosceles at $$B$$, a rotation of $$\overrightarrow{BA}$$ through $$90^\circ$$ gives $$C = (u+1,1-u)$$, and hence $$D = (u+1,0)$$. The shoelace formula gives the area as $$\frac{1}{2}\left|u(1-u)-(u+1)-(1-u)(u+1)\right| = 1$$.

Alternative solution

Lets extend CD to F such that FD = BE and $$\angle BFC = 90\degree$$. 

$$\triangle BAC$$ is a right angled isosceles triangle. 

$$ \angle BAC = \angle BCA = 45\degree$$

Lets assume $$\angle CAD = \theta$$

In $$\triangle BAE$$ we get $$ \angle ABE = 180\degree - 90\degree - (45\degree + \theta) = 45 -\theta$$

Now we know that $$\angle EBF = 90\degree$$ (By construction)

$$\angle ABC = 90\degree$$

Equating the two we get

$$\angle ABE + \angle EBC = \angle EBC + \angle CBF$$

$$ \implies \angle ABE = \angle CBF$$

Now, in $$\triangle ABE, \triangle CBF$$

$$AB = BC, \angle ABE = \angle CBF , \angle BFC = \angle BEA$$. Thus, the triangles are congruent to each other.

This implies $$ BE = BF = 1$$

Since all the angles are $$90\degree$$ and the adjacent sides are equal, BEDF is a square.

Now, we need to find the area of ABCD = Area of BAE + Area of BEDC.

Since triangle BAE is congruent to triangle BCF, their areas are equal.

Now we can replace 

Area of ABCD = Area of BCF + Area of BEDC = Area of square BEDF = 1

Let $$S=a^2+b^2+c^2$$. Since $$a+b+c=0$$, 

$$(a+b+c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca)$$

$$ a^2 + b^2 + c^2 = -2(ab + bc + ca)$$

We have $$ab+bc+ca=-\dfrac{S}{2}$$ 

$$a^2b^2+b^2c^2+c^2a^2=(ab+bc+ca)^2=\dfrac{S^2}{4}$$. 

Now we also know that $$(a^2+b^2+c^2)^2 = a^4+b^4+c^4 +2(a^2b^2+b^2c^2+c^2a^2)$$

Thus $$a^4+b^4+c^4=S^2-2\left(\dfrac{S^2}{4}\right)=\dfrac{S^2}{2}$$

Hence the required ratio is $$ \dfrac{S^2}{\dfrac{S^2}{2}} = 2$$.

If Ram is $$r$$ years old, the four ages are $$r$$, $$r-1$$, $$r-2$$ and $$r-3$$. Since $$5040=10 \times 9 \times 8 \times 7$$, the four consecutive ages are $$10,9,8,7$$. Therefore, Ram is $$10$$ years old.

Pineapples form $$20\%$$ and oranges form $$60\%$$ of the stock which means that apples form the remaining $$20\%$$ of the stock.

The number of applies equals $$40$$, so the total stock is $$\frac{40}{0.2}=200$$. Initially there are $$40$$ pineapples and $$120$$ oranges. Replacing half the oranges adds $$60$$ pineapples, giving $$40+60=100$$ pineapples.

The time required for one table is $$\dfrac{5}{4}$$ hours. Therefore, the time for $$7$$ tables is $$7 \times \dfrac{5}{4}=\dfrac{35}{4}=8\dfrac{3}{4}$$ hours.

Since $$AB=AC$$ and $$\angle C=80^\circ$$, we get $$\angle B=80^\circ$$. 

In $$\triangle BCD$$, $$BC=BD$$, so $$\angle BDC=\angle BCD=80^\circ$$ 

Applying angle sum property to the triangle we get $$\angle CBD= 180^\circ -80^\circ-80^\circ= 20^\circ$$. 

Therefore, $$\angle DBE=80^\circ-20^\circ=60^\circ$$, and because $$BD=DE$$, triangle $$BDE$$ is equilateral. 

Thus $$\angle ADE=180^\circ-80^\circ-60^\circ=40^\circ$$.

Since $$AB=BC=BE$$, triangle $$BEC$$ is isosceles. 

Also, $$\angle EBC=20^\circ+45^\circ=65^\circ$$

$$\angle BCE=\angle BEC=\dfrac{180^\circ-65^\circ}{2}=57.5^\circ$$. 

Hence $$\angle FCD=90^\circ-57.5^\circ=32.5^\circ$$ and $$\angle FDC=45^\circ$$. 

Therefore, $$x=180^\circ-45^\circ-32.5^\circ=102.5^\circ$$, giving $$2x=205^\circ$$.

From $$a+b=4(a-b)$$, we get $$3a=5b$$, so we may write $$a=5k$$ and $$b=3k$$. 

Substituting in the equation we get

$$\dfrac{2ab}{3(a^2-b^2)}=\dfrac{2 \times 5k \times 3k}{3(25k^2-9k^2)}=\dfrac{30}{48}=\dfrac{5}{8}$$.

An average of $$x$$ for three numbers means their total is $$3x$$. Subtracting the two known numbers gives the third number as $$3x-y-z$$.

Let the number be $$n$$. The operations give $$\frac{\frac{n-3}{4}+4}{5}=2$$, so $$\frac{n-3}{4}=6$$ and $$n=27$$. Since $$27=3^3$$, the number is a perfect cube.

We have $$1-\dfrac{n}{n+4}=\dfrac{4}{n+4}$$, so the denominator becomes $$n-1-\dfrac{n+4}{4}=\dfrac{3n-8}{4}$$. 

Also, $$9n^2-64=(3n-8)(3n+8)$$. The expression therefore simplifies to $$4(3n+8)$$, which is a natural number divisible by $$4$$ for every permissible natural number $$n$$.

$$\sqrt{x+5}+\sqrt{3x+4}=\sqrt{12x+1}$$

Squaring once gives

$$x+5+3x+4 +2\sqrt{(x+5)(3x+4)}=12x+1$$

$$\implies \sqrt{(x+5)(3x+4)}=4x-4$$. 

Since the left side is a positive square root, the right side must also be non-negative which means $$x \geq 1$$. Squaring again gives $$13x^2-51x-4=0$$, whose roots are $$x=4$$ and $$x=-\dfrac{1}{13}$$. Only $$x=4$$ satisfies the required condition and the original equation, so there is exactly one solution.

Notice that

$$\dfrac{1}{k(k+2)}=\dfrac{1}{2}\left(\dfrac{1}{k}-\dfrac{1}{k+2}\right).$$

Using this, the given sum becomes

$$\dfrac{1}{2}\left(\dfrac{1}{1}-\dfrac{1}{3}\right)+\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{4}\right)+\cdots+\dfrac{1}{2}\left(\dfrac{1}{n}-\dfrac{1}{n+2}\right).$$

Most of the terms cancel out, leaving only

$$\dfrac{1}{2}\left(1+\dfrac{1}{2}-\dfrac{1}{n+1}-\dfrac{1}{n+2}\right).$$

Therefore,

$$\dfrac{1}{1 \times 3}+\dfrac{1}{2 \times 4}+\cdots+\dfrac{1}{n(n+2)}=\dfrac{3}{4}-\dfrac{2n+3}{2(n+1)(n+2)}.$$

We are given that this sum is $$\dfrac{3553}{4830}.$$
We can substitute the values and check.

For $$n=68$$.

$$\dfrac{3}{4}-\dfrac{2(68)+3}{2(69)(70)}=\dfrac{3}{4}-\dfrac{139}{9660}=\dfrac{3553}{4830}.$$

This matches the given value.

$$\angle FDG = 180\degree - (68\degree + 50\degree) = 62\degree$$ (Straight angle)

AE || DC, 

$$ \angle FIE = \angle FDG = 62\degree$$ (Corresponding angles)

In $$\triangle KIA$$

$$\angle KIA = 180\degree- 62\degree = 118\degree$$

$$\angle AKI = 180\degree - (16\degree + 118\degree) = 46\degree$$

Now $$ x = \angle AKI = 46\degree$$ ( Vertically opposite angles)

Similarly,

$$\angle BLE = 82 \degree$$ (corresponding angles)

$$ \angle BLJ = 180\degree - \angle BLE = 98\degree $$(Straight angle)

$$ \angle BJL = 180\degree - 98\degree - 12\degree = 70\degree$$( Angle sum property of triangle)

$$ y = \angle BJL = 70\degree$$ (Vertically opposite angles)

Thus,

$$x+y = 116\degree$$

Let the cost of one apple be $$a$$, one banana be $$b$$, and one pineapple be $$p$$.

From the first row,

$$a+b=22.$$

From the third row,

$$p+a+2b=54.$$

From the first column,

$$2a+p+b=72.$$

From the first equation,

$$a=22-b.$$

Substitute this into the second equation.

$$p+(22-b)+2b=54.$$

$$p+b=32.$$

So,

$$p=32-b.$$

Now substitute $$a=22-b$$ and $$p=32-b$$ into the third equation.

$$2(22-b)+(32-b)+b=72.$$

$$44-2b+32-b+b=72.$$

$$76-2b=72.$$

$$2b=4.$$

$$b=2.$$

Now,

$$a=22-2=20,$$

and

$$p=32-2=30.$$

The cost of the second row is

$$a+2p+b=20+2(30)+2=82.$$

Hence, the cost of the second row is $$82$$

Since $$16=2^4$$,

$$2^{2^{x^2-1}}=2^4$$

equating the powers we get $$2^{x^2-1}=4=2^2$$. 

Equating the powers once again we get $$x^2-1=2$$, so $$x^2=3$$ and therefore $$x^4=9$$.

Let the left and right widths be $$w_1,w_2$$ and the upper and lower heights be $$h_1,h_2$$.

The areas of the regions are as follows

$$w_1h_1=72$$, $$\dfrac{1}{2}w_2h_1=48$$ and $$w_2h_2=128$$. 

Hence $$w_1h_2=\dfrac{(w_1h_1)(w_2h_2)}{w_2h_1}=\dfrac{72 \times 128}{96}=96$$. 

Therefore, $$A=\dfrac{1}{2}w_1h_2=48$$.

Let the length and breadth of the rectangle be $$l$$ and $$b$$.

We are given that

$$l+b=6.$$

The side of the square is equal to the diagonal of the rectangle.

By the Pythagorean Theorem,

$$\text{Diagonal}^2=l^2+b^2.$$

Hence, the area of the square is $$l^2+b^2.$$

Also, the ratio of the areas of the square and the rectangle is $$\dfrac{5}{2}$$.

So, $$\dfrac{l^2+b^2}{lb}=\dfrac{5}{2},$$

or $$l^2+b^2=\dfrac{5}{2}lb.$$

Now use the identity

$$\left(l+b\right)^2=l^2+b^2+2lb.$$

Substituting the given values,

$$6^2=\dfrac{5}{2}lb+2lb.$$

$$36=\dfrac{9}{2}lb.$$

Multiplying both sides by $$2$$,

$$72=9lb.$$

$$lb=8.$$

Therefore, the area of the square is

$$l^2+b^2=\dfrac{5}{2}\times8=20.$$

Hence, the required area is $${20\text{ cm}^2}.$$

The area is $$\pi r^2 = 25\pi$$ and the circumference is $$ 2\pi r = 10\pi$$. Thus $$25\pi=\frac{x}{100}\times 10\pi$$, which gives $$x=250$$.

Let the three numbers be

$$a,\quad a+d,\quad a+2d,$$

where $$d$$ is the common difference.

We are given that

$$a(a+d)=85,$$

and

$$(a+d)(a+2d)=115.$$

Divide the second equation by the first equation.

$$\dfrac{a+2d}{a}=\dfrac{115}{85}=\dfrac{23}{17}.$$

Cross-multiplying,

$$17(a+2d)=23a.$$

$$17a+34d=23a.$$

$$6a=34d.$$

$$3a=17d.$$

So,

$$a=\dfrac{17d}{3}.$$

Substitute this into

$$a(a+d)=85.$$

$$\dfrac{17d}{3}\left(\dfrac{17d}{3}+d\right)=85.$$

$$\dfrac{17d}{3}\times\dfrac{20d}{3}=85.$$

$$\dfrac{340d^2}{9}=85.$$

$$340d^2=765.$$

$$d^2=\dfrac{9}{4}.$$

Since the numbers are positive,

$$d=\dfrac{3}{2}.$$

The difference between the greatest and the smallest numbers is

$$(a+2d)-a=2d=3.$$

Hence, the required difference is $$3$$

We have $$T_{3n+1}=\dfrac{(3n+1)(3n+2)}{2}$$ and $$9T_n=\dfrac{9n(n+1)}{2}$$. 

$$T_{3n+1}-9T_n =\dfrac{(3n+1)(3n+2)}{2}-\dfrac{9n(n+1)}{2} = \dfrac{(9n^2+9n+2)-9n^2-9n}{2} =1$$

After the first day, $$\frac{5}{8}$$ of the book remains. Reading $$\frac{4}{5}$$ of this remainder leaves $$\frac{1}{5}\times\frac{5}{8}=\frac{1}{8}$$ of the book unread. Since $$\frac{1}{8}$$ equals $$30$$ pages, the book has $$30 \times 8=240$$ pages.

Let the times taken by $$A,$$ $$B,$$ and $$C$$ to complete the race be $$T_A,$$ $$T_B,$$ and $$T_C$$ seconds, respectively.

Since $$B$$ finishes $$60$$ seconds after $$A$$,

$$T_B=T_A+60.$$

Also, $$C$$ finishes $$30$$ seconds after $$B$$, so

$$T_C=T_B+30=T_A+90.$$

When $$A$$ finishes the $$1000\text{ m}$$ race, $$C$$ has covered only

$$1000-375=625\text{ m}.$$

Since both run for the same amount of time until $$A$$ finishes,

$$\dfrac{1000}{s_A}=\dfrac{625}{s_C},$$

where $$s_A$$ and $$s_C$$ are the speeds of $$A$$ and $$C$$.

Thus,

$$s_C=\dfrac{625}{1000}s_A=\dfrac{5}{8}s_A.$$

Now,

$$T_A=\dfrac{1000}{s_A}\quad \text{and}\quad T_C=\dfrac{1000}{s_C}.$$

Therefore,

$$\dfrac{T_A}{T_C}=\dfrac{\dfrac{1000}{s_A}}{\dfrac{1000}{s_C}}=\dfrac{s_C}{s_A}=\dfrac{5}{8}.$$

Hence,

$$T_C=\dfrac{8}{5}T_A.$$

But,

$$T_C=T_A+90.$$

So,

$$T_A+90=\dfrac{8}{5}T_A.$$

Multiplying both sides by $$5$$,

$$5T_A+450=8T_A.$$

$$3T_A=450.$$

$$T_A=150\text{ seconds}.$$

Therefore,

$$T_B=150+60=210\text{ seconds}.$$

Since $$210$$ seconds is equal to $$3$$ minutes $$30$$ seconds.

Thus, the value of $$ x = 3$$.

To determine any particular length we need to choose a set of two points from the given set of points on the ruler.

In the first case we had to choose 2 points out of 2 points given (ends of the ruler)  which is $$\binom{2}{2} =1$$ ways. 

In the second case we are given 3 points given (end points and one mark in between) out of which we must choose 2 points given which can be done in $$\binom{3}{2} =3$$ ways.

In the second case we are given 4 points (end points and two marks in between)  out of which we must choose 2 points given which can be done in $$\binom{4}{2} =6$$ ways.

Extending this to the last case we are given 6 points (end points and four marks in between) out of which we must choose 2 points given which can be done in $$\binom{6}{2} = \dfrac{6!}{4!2!} = 3\times 5 =15$$ ways.

$$ \angle DBE = 180\degree - 54\degree-108\degree = 18\degree$$(Straight angle)

$$\angle EDB = 180\degree- 60\degree \text{(vertically opposite angle)}  -18\degree =102\degree$$ (Angle sum property of triangle)

$$\angle ADB = 180\degree- 102\degree = 78\degree$$ (Straight angle)

In $$\triangle ADB$$

$$x = 180\degree - 78\degree -54\degree = 48\degree$$ (Using angle sum property)

Given that AB || CD,

$$ \angle BCD = 180\degree - \angle ABC = 180\degree- 142\degree = 38\degree$$ (Interior angles are supplementary)

$$\angle CEJ = \angle BCD = 38\degree$$ (Vertically opposite angles)

$$\angle EPJ = 360\degree - 90\degree- 90\degree - \angle CEJ = 360\degree - 90\degree-90\degree - 38\degree =142\degree$$ (Angle sum property of a quadrilateral)

Since PG = PF, $$\triangle PGF$$ is an isosceles triangle  and $$ \angle PGF = \angle PFG$$

$$\angle PGF = \dfrac{1}{2} (180\degree - 142\degree) = 19\degree$$

$$ x = 180\degree-\angle PGF = 180\degree - 19\degree = 161\degree$$ (Straight angle)

Let us assume the four odd integers as $$2n+1,2n+3,2n+5,2n+7$$. 

We are given that $$2n+1+2n+7=52$$, so $$4n+8 = 52 \implies 4n=44$$ or $$n =11$$

Thus the numbers are $$23,25,27,29$$

The sum of these numbers is 

$$23+25+27+29 =104$$

Let the number be $$N$$ 

We are given that $$\sqrt{N}+2 = N$$

Let us substitute $$y=\sqrt{N}$$ to obtain a quadratic. 

Then we can rewrite the equation as 

$$y+2=y^2 \implies y^2-y-2=0$$

Factorising the equation we get

$$(y-2)(y+1)=0$$. 

Since $$y=\sqrt{N}$$, it is non-negative hence we get

 $$y=2$$ and $$N=y^2 = 4$$.

The side of square $$CEFG$$ is $$CG=4\sqrt{2}$$, so its diagonal $$GE$$ is $$\sqrt{(4\sqrt{2})^2+(4\sqrt{2})^2 }= 8$$cm

In a square, the diagonals bisect each other perpendicularly. 

Thus, $$ CH = EH = \dfrac{8}{2} = 4$$cm

In $$\triangle BHC$$ we can apply Pythagoras theorem to find the length of BC

$$BH = 6+4 = 10, CH = 4 \implies BC = \sqrt{10^2 + 4^2} = \sqrt{116}$$cm

Thus, Area of the square is

$$BC^2 = 116\text{cm}^2$$