The value of $$\sqrt{2023\sqrt{2022\sqrt{2021 \times 2019 + 1} + 1} + 1}$$ is
NMTC kaprekar ( sub junior level) 7th and 8th oct 7 2023
For the following questions answer them individually
Solution
The innermost quantity is $$2021 \times 2019 + 1 = 2020^2$$, so its square root is $$2020$$. The next radical becomes $$\sqrt{2022 \times 2020 + 1} = \sqrt{2021^2} = 2021$$. Therefore, the complete expression is $$\sqrt{2023 \times 2021 + 1} = \sqrt{2022^2} = 2022$$.
In the adjoining figure, $$(\triangle ABC)$$ is an isosceles right-angled triangle. $$(BE)$$ is perpendicular to $$(AD)$$. If $$(BE) = 1\text{ cm}$$, then the area in $$\text{cm}^2$$ of quadrilateral $$(ABCD)$$ is
Solution
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 $$\frac{1}{2}\left|u(1-u)-(u+1)-(1-u)(u+1)\right| = 1$$.
If $$a+b+c=0$$, then the value of $$\frac{(a^2+b^2+c^2)^2}{a^4+b^4+c^4}$$ is
Solution
Let $$S=a^2+b^2+c^2$$. Since $$a+b+c=0$$, we have $$ab+bc+ca=-\frac{S}{2}$$ and $$a^2b^2+b^2c^2+c^2a^2=(ab+bc+ca)^2=\frac{S^2}{4}$$. Thus $$a^4+b^4+c^4=S^2-2\left(\frac{S^2}{4}\right)=\frac{S^2}{2}$$, so the required ratio is $$2$$.
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Ram, Rahim, Robin and Ria are four children, the product of whose ages is $$5040$$. Ram is older than Rahim by one year, older than Robin by two years and older than Ria by three years. The age of Ram (in years) is
Solution
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.
A fruit-seller sells apples, oranges and pineapples. In his stock, $$20\%$$ are pineapples and $$60\%$$ are oranges. There are $$40$$ apples. If half the oranges are replaced by pineapples, how many pineapples are there in the shop now?
Solution
Apples form the remaining $$20\%$$ of the stock, and this 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.
A carpenter can repair $$4$$ tables in $$5$$ hours. The time (in hours) for him to repair, (at the same rate) $$7$$ tables is
Solution
The time required for one table is $$\frac{5}{4}$$ hours. Therefore, the time for $$7$$ tables is $$7 \times \frac{5}{4}=\frac{35}{4}=8\frac{3}{4}$$ hours.
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In the adjoining figure, $$AB=AC$$, $$\angle C=80^\circ$$ and $$BC=BD=DE$$. Then the measure in degrees of $$\angle ADE$$ is
Solution
Since $$AB=AC$$ and $$\angle C=80^\circ$$, we get $$\angle B=80^\circ$$. In $$\triangle BCD$$, $$BC=BD$$, so $$\angle BDC=\angle BCD=80^\circ$$ and hence $$\angle CBD=20^\circ$$. Therefore, $$\angle DBE=80^\circ-20^\circ=60^\circ$$, and because $$BD=DE$$, triangle $$BDE$$ is equilateral. Thus $$\angle ADE=180^\circ-80^\circ-60^\circ=40^\circ$$.
In the adjoining figure, $$ABCD$$ is a square and $$BE=AB$$. If the measure of $$\angle DFC$$ is $$x^\circ$$, then the value of $$2x$$ in degrees is
Solution
Since $$AB=BC=BE$$, triangle $$BEC$$ is isosceles. Also, $$\angle EBC=20^\circ+45^\circ=65^\circ$$, so $$\angle BCE=\angle BEC=\frac{180^\circ-65^\circ}{2}=57.5^\circ$$. 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 $$2x=205^\circ$$.
Two real numbers $$a$$ and $$b$$, where $$a>b$$, are given such that their sum is equal to $$4$$ times their difference. The value of $$\frac{2ab}{3(a^2-b^2)}$$ is
Solution
From $$a+b=4(a-b)$$, we get $$3a=5b$$, so we may write $$a=5k$$ and $$b=3k$$. Substitution gives $$\frac{2ab}{3(a^2-b^2)}=\frac{2 \times 5k \times 3k}{3(25k^2-9k^2)}=\frac{30}{48}=\frac{5}{8}$$.
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The average of three numbers is $$x$$. Two of the three numbers are $$y$$ and $$z$$. Then the third number is
Solution
An average of $$x$$ for three numbers means their total is $$3x$$. Subtracting the two known numbers gives the third number as $$3x-y-z$$.
From a natural number $$3$$ is subtracted, then the result is divided by $$4$$ and the outcome is increased by $$4$$. The whole result is then divided by $$5$$, and the final resulting number is $$2$$. Then the natural number taken in the beginning is
Solution
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.
For all permissible natural numbers $$n$$, the number $$\frac{9n^2-64}{n-1-\frac{1}{1-\frac{n}{n+4}}}$$ is
Solution
We have $$1-\frac{n}{n+4}=\frac{4}{n+4}$$, so the denominator becomes $$n-1-\frac{n+4}{4}=\frac{3n-8}{4}$$. 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 $$n$$.
The number of solutions of the equation $$\sqrt{x+5}+\sqrt{3x+4}=\sqrt{12x+1}$$ is
Solution
Squaring once gives $$\sqrt{(x+5)(3x+4)}=4x-4$$, so a valid solution must satisfy $$x \geq 1$$. Squaring again gives $$13x^2-51x-4=0$$, whose roots are $$x=4$$ and $$x=-\frac{1}{13}$$. Only $$x=4$$ satisfies the required condition and the original equation, so there is exactly one solution.
If $$\frac{1}{1 \times 3}+\frac{1}{2 \times 4}+\frac{1}{3 \times 5}+\cdots+\frac{1}{n(n+2)}=\frac{3553}{4830}$$, then $$n$$ is
Solution
Since $$\frac{1}{k(k+2)}=\frac{1}{2}\left(\frac{1}{k}-\frac{1}{k+2}\right)$$, the sum telescopes to $$\frac{3}{4}-\frac{2n+3}{2(n+1)(n+2)}$$. For $$n=68$$, this equals $$\frac{3}{4}-\frac{139}{9660}=\frac{3553}{4830}$$. Hence $$n=68$$.
In the adjoining figure, $$\angle BAE=16^\circ$$ and $$\angle CBG=12^\circ$$. Then the measure of $$x+y$$ in degrees is
Solution
Using the parallel horizontal lines, the line $$AD$$ makes $$112^\circ$$ with the rightward horizontal. Since $$\angle ADF=50^\circ$$, line $$DF$$ makes $$62^\circ$$ with the horizontal, while $$AB$$ makes $$16^\circ$$, so $$x=46^\circ$$. At the other side, the marked $$82^\circ$$ angle and $$\angle CBG=12^\circ$$ give $$y=82^\circ-12^\circ=70^\circ$$. Therefore, $$x+y=46^\circ+70^\circ=116^\circ$$.
There are $$3$$ pineapples, $$6$$ bananas and $$7$$ apples. Each fruit of the same category has the same price. The total amount of the fruits in the first row is Rs. $$44$$, that of the third row is Rs. $$54$$ and that of the first column is Rs. $$72$$. Then the total amount in rupees of the fruits in the second row is
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Solution
Let the prices of an apple, banana and pineapple be $$a$$, $$b$$ and $$p$$. The first row gives $$a+b=22$$, the third row gives $$p+a+2b=54$$, and the first column gives $$2a+p+b=72$$. Solving gives $$a=20$$, $$b=2$$ and $$p=30$$. The second row costs $$a+2p+b=20+60+2=82$$.
If $$2^{2^{x^2-1}}=16$$, then the value of $$x^4$$ is
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Solution
Since $$16=2^4$$, we get $$2^{x^2-1}=4=2^2$$. Thus $$x^2-1=2$$, so $$x^2=3$$ and therefore $$x^4=9$$.
In the adjoining figure, two rectangles and two right-angled triangles are arranged as shown. The numbers shown inside each are their respective areas. Then the area of $$A$$ is
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Solution
Let the left and right widths be $$w_1,w_2$$ and the upper and lower heights be $$h_1,h_2$$. The given regions give $$w_1h_1=72$$, $$\frac{1}{2}w_2h_1=48$$ and $$w_2h_2=128$$. Hence $$w_1h_2=\frac{(w_1h_1)(w_2h_2)}{w_2h_1}=\frac{72 \times 128}{96}=96$$. Therefore, $$A=\frac{1}{2}w_1h_2=48$$.
The sum of the length and breadth of a rectangle is $$6\text{ cm}$$. A square is constructed whose side is equal to the diagonal of the rectangle. If the ratio of the areas of the square and the rectangle is $$\frac{5}{2}$$, then the area of the square in $$\text{cm}^2$$ is
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Solution
Let the rectangle sides be $$l$$ and $$b$$. The square area is $$l^2+b^2$$, and the given ratio gives $$l^2+b^2=\frac{5}{2}lb$$. Since $$(l+b)^2=36=l^2+b^2+2lb$$, we obtain $$36=\frac{9}{2}lb$$ and hence $$lb=8$$. Therefore, the square area is $$\frac{5}{2}\times 8=20$$.
If the area of a circle of radius $$5$$ is numerically $$x\%$$ of its circumference, then $$x$$ is
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Solution
The area is $$25\pi$$ and the circumference is $$10\pi$$. Thus $$25\pi=\frac{x}{100}\times 10\pi$$, which gives $$x=250$$.
There are $$3$$ positive real numbers. The second is greater than the first by the same amount that the third is greater than the second. The product of the two smaller numbers is $$85$$ and that of the two bigger numbers is $$115$$. Then the difference between the smallest and the greatest numbers is
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Solution
Let the three numbers in arithmetic progression be $$u-d,u,u+d$$. Dividing the two product equations gives $$\frac{u+d}{u-d}=\frac{115}{85}=\frac{23}{17}$$, so the numbers may be written as $$17k,20k,23k$$. Since $$17k \times 20k=85$$, we get $$k=\frac{1}{2}$$. The required difference is $$23k-17k=6k=3$$.
The numbers $$1,3,6,10,\ldots$$ are called triangular numbers. The $$n$$th triangular number is $$T_n=\frac{n(n+1)}{2}$$. Then the value of $$T_{3n+1}-9T_n$$ is equal to
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Solution
We have $$T_{3n+1}=\frac{(3n+1)(3n+2)}{2}$$ and $$9T_n=\frac{9n(n+1)}{2}$$. Their difference is $$\frac{9n^2+9n+2-9n^2-9n}{2}=1$$.
I read $$\frac{3}{8}$$ of a book on one day and $$\frac{4}{5}$$ of the remainder on another day. If $$30$$ pages are still unread, then the total number of pages in the book is
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Solution
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.
Three persons $$A,B,C$$ participate in a running race of $$1\text{ km}$$ distance. When $$A$$ and $$B$$ run, $$A$$ wins by $$60$$ seconds. When $$A$$ and $$C$$ run, $$A$$ wins by $$375\text{ m}$$. When $$B$$ and $$C$$ run, $$B$$ wins by $$30$$ seconds. If the time taken by $$B$$ to run $$1\text{ km}$$ is $$x$$ minutes and $$30$$ seconds, then $$x$$ is
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Solution
Let the times taken by $$A,B,C$$ be $$T_A,T_B,T_C$$ seconds. Then $$T_B=T_A+60$$ and $$T_C=T_B+30=T_A+90$$. Since $$C$$ covers only $$625\text{ m}$$ when $$A$$ covers $$1000\text{ m}$$, $$T_C=\frac{8}{5}T_A$$. Hence $$T_A+90=\frac{8}{5}T_A$$, giving $$T_A=150$$ and $$T_B=210$$ seconds, which is $$3$$ minutes $$30$$ seconds.
A ruler with no mark on it can measure its own length $$AB$$. A ruler with only one mark on it can measure $$3$$ lengths $$AB,AC,BC$$. A ruler with two marks on it can measure $$6$$ lengths $$AB,AC,CD,DB,AD,CB$$. Then the number of lengths a ruler with $$4$$ marks on it can measure is
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Solution
Four internal marks together with the two endpoints give $$6$$ points on the ruler. Every pair of points determines one measurable length. Therefore, the number of lengths is $$\binom{6}{2}=15$$.
In the adjoining figure, the value of $$x$$ in degrees is
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Solution
At $$B$$, the line through the upper intersection makes an angle of $$108^\circ$$ with the rightward horizontal line. At the upper intersection, the angle between this line and the line through $$A$$ is $$60^\circ$$. Therefore, the latter line makes $$108^\circ-60^\circ=48^\circ$$ with the horizontal, so $$x=48^\circ$$.
In the adjoining figure, $$AB$$ || $$DC$$. Also, $$PG=PF$$. The measure of angle $$x$$ in degrees is
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Solution
Since $$AB$$ is parallel to $$DC$$ and $$\angle ABC=142^\circ$$, the acute angle between the two intersecting directions at $$C$$ is $$38^\circ$$. The two marked right angles in quadrilateral $$CEPJ$$ then give $$\angle EPJ=142^\circ$$. This is vertically opposite to $$\angle GPF$$, so the vertex angle of isosceles triangle $$GPF$$ is $$142^\circ$$. Each base angle is $$19^\circ$$, and the exterior angle marked $$x$$ is $$180^\circ-19^\circ=161^\circ$$.
The sum of the first and last of four consecutive odd integers is $$52$$. The sum of all of them is
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Solution
Write the four odd integers as $$n,n+2,n+4,n+6$$. Then $$n+(n+6)=52$$, so $$n=23$$. Their sum is $$23+25+27+29=104$$.
The square root of a number plus $$2$$ gives the number itself. Then the number is
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Solution
Let the number be $$N$$ and put $$y=\sqrt{N}$$. Then $$y+2=y^2$$, so $$(y-2)(y+1)=0$$. Since $$y$$ is non-negative, $$y=2$$ and hence $$N=4$$.
$$ABCD$$ and $$CEFG$$ are two squares such that the extension of $$GE$$, a diagonal of $$CEFG$$, passes through $$B$$. Given $$BE=6\text{ cm}$$ and $$CG=4\sqrt{2}\text{ cm}$$, then the area of square $$ABCD$$ in $$\text{cm}^2$$ is
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Solution
The side of square $$CEFG$$ is $$CG=4\sqrt{2}$$, so its diagonal $$GE$$ is $$8$$. Since $$B,E,G$$ are collinear, $$BG=BE+EG=6+8=14$$. A diagonal of a square bisects a right angle, so $$\angle BGC=45^\circ$$. By the cosine rule, $$BC^2=14^2+(4\sqrt{2})^2-2 \times 14 \times 4\sqrt{2}\cos45^\circ=116$$, which is the area of square $$ABCD$$.
