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Question Number 21772 Answers: 2 Comments: 0

integrate ∫2^(4x) dx

$${integrate} \\ $$$$\int\mathrm{2}^{\mathrm{4}{x}} {dx} \\ $$

Question Number 21771 Answers: 0 Comments: 0

The result of 11 chess matches (as win, lose or draw) are to be forecast. Out of all possible forecasts, the number of ways in which 8 correct and 3 incorrect results can be forecast is

$$\mathrm{The}\:\mathrm{result}\:\mathrm{of}\:\mathrm{11}\:\mathrm{chess}\:\mathrm{matches}\:\left(\mathrm{as}\:\mathrm{win},\right. \\ $$$$\left.\mathrm{lose}\:\mathrm{or}\:\mathrm{draw}\right)\:\mathrm{are}\:\mathrm{to}\:\mathrm{be}\:\mathrm{forecast}.\:\mathrm{Out}\:\mathrm{of} \\ $$$$\mathrm{all}\:\mathrm{possible}\:\mathrm{forecasts},\:\mathrm{the}\:\mathrm{number}\:\mathrm{of} \\ $$$$\mathrm{ways}\:\mathrm{in}\:\mathrm{which}\:\mathrm{8}\:\mathrm{correct}\:\mathrm{and}\:\mathrm{3}\:\mathrm{incorrect} \\ $$$$\mathrm{results}\:\mathrm{can}\:\mathrm{be}\:\mathrm{forecast}\:\mathrm{is} \\ $$

Question Number 21768 Answers: 0 Comments: 2

Six cards and six envelopes are numbered 1, 2, 3, 4, 5, 6 and cards are to be placed in envelopes so that each envelope contains exactly one card and no card is placed in the envelope bearing the same number and moreover the card numbered 1 is always placed in envelope numbered 2. Then the number of ways it can be done is

$$\mathrm{Six}\:\mathrm{cards}\:\mathrm{and}\:\mathrm{six}\:\mathrm{envelopes}\:\mathrm{are}\:\mathrm{numbered} \\ $$$$\mathrm{1},\:\mathrm{2},\:\mathrm{3},\:\mathrm{4},\:\mathrm{5},\:\mathrm{6}\:\mathrm{and}\:\mathrm{cards}\:\mathrm{are}\:\mathrm{to}\:\mathrm{be}\:\mathrm{placed} \\ $$$$\mathrm{in}\:\mathrm{envelopes}\:\mathrm{so}\:\mathrm{that}\:\mathrm{each}\:\mathrm{envelope} \\ $$$$\mathrm{contains}\:\mathrm{exactly}\:\mathrm{one}\:\mathrm{card}\:\mathrm{and}\:\mathrm{no}\:\mathrm{card} \\ $$$$\mathrm{is}\:\mathrm{placed}\:\mathrm{in}\:\mathrm{the}\:\mathrm{envelope}\:\mathrm{bearing}\:\mathrm{the} \\ $$$$\mathrm{same}\:\mathrm{number}\:\mathrm{and}\:\mathrm{moreover}\:\mathrm{the}\:\mathrm{card} \\ $$$$\mathrm{numbered}\:\mathrm{1}\:\mathrm{is}\:\mathrm{always}\:\mathrm{placed}\:\mathrm{in}\:\mathrm{envelope} \\ $$$$\mathrm{numbered}\:\mathrm{2}.\:\mathrm{Then}\:\mathrm{the}\:\mathrm{number}\:\mathrm{of}\:\mathrm{ways} \\ $$$$\mathrm{it}\:\mathrm{can}\:\mathrm{be}\:\mathrm{done}\:\mathrm{is} \\ $$

Question Number 21766 Answers: 1 Comments: 0

There are 3 apartments A, B and C for rent in a building. Each apartment will accept either 3 or 4 occupants. The number of ways of renting the apartments to 10 students

$$\mathrm{There}\:\mathrm{are}\:\mathrm{3}\:\mathrm{apartments}\:{A},\:{B}\:\mathrm{and}\:{C}\:\mathrm{for} \\ $$$$\mathrm{rent}\:\mathrm{in}\:\mathrm{a}\:\mathrm{building}.\:\mathrm{Each}\:\mathrm{apartment}\:\mathrm{will} \\ $$$$\mathrm{accept}\:\mathrm{either}\:\mathrm{3}\:\mathrm{or}\:\mathrm{4}\:\mathrm{occupants}.\:\mathrm{The} \\ $$$$\mathrm{number}\:\mathrm{of}\:\mathrm{ways}\:\mathrm{of}\:\mathrm{renting}\:\mathrm{the} \\ $$$$\mathrm{apartments}\:\mathrm{to}\:\mathrm{10}\:\mathrm{students} \\ $$

Question Number 21760 Answers: 1 Comments: 0

There are m points on the line AB and n points on the line AC, excluding the point A. Triangles are formed joining these points (i) When point A is not included, (ii) When point A is included. The ratio of the number of such triangles is

$$\mathrm{There}\:\mathrm{are}\:{m}\:\mathrm{points}\:\mathrm{on}\:\mathrm{the}\:\mathrm{line}\:{AB}\:\mathrm{and} \\ $$$${n}\:\mathrm{points}\:\mathrm{on}\:\mathrm{the}\:\mathrm{line}\:{AC},\:\mathrm{excluding}\:\mathrm{the} \\ $$$$\mathrm{point}\:{A}.\:\mathrm{Triangles}\:\mathrm{are}\:\mathrm{formed}\:\mathrm{joining} \\ $$$$\mathrm{these}\:\mathrm{points} \\ $$$$\left(\mathrm{i}\right)\:\mathrm{When}\:\mathrm{point}\:{A}\:\mathrm{is}\:\mathrm{not}\:\mathrm{included}, \\ $$$$\left(\mathrm{ii}\right)\:\mathrm{When}\:\mathrm{point}\:{A}\:\mathrm{is}\:\mathrm{included}. \\ $$$$\mathrm{The}\:\mathrm{ratio}\:\mathrm{of}\:\mathrm{the}\:\mathrm{number}\:\mathrm{of}\:\mathrm{such}\:\mathrm{triangles} \\ $$$$\mathrm{is} \\ $$

Question Number 21759 Answers: 1 Comments: 0

A block of mass 1 kg is pushed against a rough vertical wall with a force of 20 N, coefficient of static friction being (1/4). Another horizontal force of 10 N is applied on the block in a direction parallel to the wall. Will the block move? If yes, with what acceleration? If no, find the frictional force exerted by wall on the block. (g = 10 m/s^2 )

$$\mathrm{A}\:\mathrm{block}\:\mathrm{of}\:\mathrm{mass}\:\mathrm{1}\:\mathrm{kg}\:\mathrm{is}\:\mathrm{pushed}\:\mathrm{against} \\ $$$$\mathrm{a}\:\mathrm{rough}\:\mathrm{vertical}\:\mathrm{wall}\:\mathrm{with}\:\mathrm{a}\:\mathrm{force}\:\mathrm{of}\:\mathrm{20} \\ $$$$\mathrm{N},\:\mathrm{coefficient}\:\mathrm{of}\:\mathrm{static}\:\mathrm{friction}\:\mathrm{being}\:\frac{\mathrm{1}}{\mathrm{4}}. \\ $$$$\mathrm{Another}\:\mathrm{horizontal}\:\mathrm{force}\:\mathrm{of}\:\mathrm{10}\:\mathrm{N}\:\mathrm{is} \\ $$$$\mathrm{applied}\:\mathrm{on}\:\mathrm{the}\:\mathrm{block}\:\mathrm{in}\:\mathrm{a}\:\mathrm{direction} \\ $$$$\mathrm{parallel}\:\mathrm{to}\:\mathrm{the}\:\mathrm{wall}.\:\mathrm{Will}\:\mathrm{the}\:\mathrm{block}\:\mathrm{move}? \\ $$$$\mathrm{If}\:\mathrm{yes},\:\mathrm{with}\:\mathrm{what}\:\mathrm{acceleration}?\:\mathrm{If}\:\mathrm{no}, \\ $$$$\mathrm{find}\:\mathrm{the}\:\mathrm{frictional}\:\mathrm{force}\:\mathrm{exerted}\:\mathrm{by}\:\mathrm{wall} \\ $$$$\mathrm{on}\:\mathrm{the}\:\mathrm{block}.\:\left({g}\:=\:\mathrm{10}\:\mathrm{m}/\mathrm{s}^{\mathrm{2}} \right) \\ $$

Question Number 21756 Answers: 0 Comments: 0

Prove that the product for all nth roots of unity is equal to zero, except n=1. Note: U_n ={e^(2kπi/n) ∣ k∈{1, 2, ..., n}} x^n =1

$$\mathrm{Prove}\:\mathrm{that}\:\mathrm{the}\:\mathrm{product}\:\mathrm{for}\:\mathrm{all} \\ $$$${n}\mathrm{th}\:\mathrm{roots}\:\mathrm{of}\:\mathrm{unity}\:\mathrm{is}\:\mathrm{equal}\:\mathrm{to}\:\mathrm{zero}, \\ $$$$\mathrm{except}\:{n}=\mathrm{1}. \\ $$$$\: \\ $$$$\mathrm{Note}: \\ $$$${U}_{{n}} =\left\{{e}^{\mathrm{2}{k}\pi{i}/{n}} \:\mid\:{k}\in\left\{\mathrm{1},\:\mathrm{2},\:...,\:{n}\right\}\right\} \\ $$$${x}^{{n}} =\mathrm{1} \\ $$

Question Number 21754 Answers: 1 Comments: 0

Q. In 2014, country X had 783 miles of paved roads. Starting in 2015, the country has been building 8 miles of new paved roads each year. At this rate, how many miles of paved roads will country X have in 2030?

$$\mathrm{Q}.\:\mathrm{In}\:\mathrm{2014},\:\mathrm{country}\:\mathrm{X}\:\mathrm{had}\:\mathrm{783}\:\mathrm{miles}\:\mathrm{of}\:\mathrm{paved}\:\mathrm{roads}.\:\mathrm{Starting}\:\mathrm{in}\: \\ $$$$\:\:\:\:\:\:\:\mathrm{2015},\:\mathrm{the}\:\mathrm{country}\:\mathrm{has}\:\:\mathrm{been}\:\mathrm{building}\:\:\mathrm{8}\:\:\mathrm{miles}\:\mathrm{of}\:\mathrm{new}\:\mathrm{paved} \\ $$$$\:\:\:\:\:\:\mathrm{roads}\:\mathrm{each}\:\mathrm{year}.\:\mathrm{At}\:\mathrm{this}\:\mathrm{rate},\:\:\mathrm{how}\:\:\mathrm{many}\:\mathrm{miles}\:\mathrm{of}\:\mathrm{paved}\:\mathrm{roads} \\ $$$$\:\:\:\:\:\:\mathrm{will}\:\mathrm{country}\:\mathrm{X}\:\mathrm{have}\:\mathrm{in}\:\mathrm{2030}? \\ $$

Question Number 21750 Answers: 0 Comments: 2

Column II gives five arrangements of two blocks A and B. In each arrangement their masses are 3 kg and 2 kg respectively. Match the entries in column I with arrangements in column II. Column I (A) Force on B w.r.t. A is equal to force on A w.r.t. B (B) Net force on B w.r.t. an inertial frame is 4 N (C) Acceleration of the block A is 2 m/s^2 . (D) Net force on B w.r.t. A is 8 N

$$\mathrm{Column}\:\mathrm{II}\:\mathrm{gives}\:\mathrm{five}\:\mathrm{arrangements}\:\mathrm{of} \\ $$$$\mathrm{two}\:\mathrm{blocks}\:{A}\:\mathrm{and}\:{B}.\:\mathrm{In}\:\mathrm{each} \\ $$$$\mathrm{arrangement}\:\mathrm{their}\:\mathrm{masses}\:\mathrm{are}\:\mathrm{3}\:\mathrm{kg}\:\mathrm{and} \\ $$$$\mathrm{2}\:\mathrm{kg}\:\mathrm{respectively}.\:\mathrm{Match}\:\mathrm{the}\:\mathrm{entries}\:\mathrm{in} \\ $$$$\mathrm{column}\:\mathrm{I}\:\mathrm{with}\:\mathrm{arrangements}\:\mathrm{in}\:\mathrm{column} \\ $$$$\mathrm{II}. \\ $$$$\boldsymbol{\mathrm{Column}}\:\boldsymbol{\mathrm{I}} \\ $$$$\left(\mathrm{A}\right)\:\mathrm{Force}\:\mathrm{on}\:{B}\:\mathrm{w}.\mathrm{r}.\mathrm{t}.\:{A}\:\mathrm{is}\:\mathrm{equal}\:\mathrm{to}\:\mathrm{force} \\ $$$$\mathrm{on}\:{A}\:\mathrm{w}.\mathrm{r}.\mathrm{t}.\:{B} \\ $$$$\left(\mathrm{B}\right)\:\mathrm{Net}\:\mathrm{force}\:\mathrm{on}\:{B}\:\mathrm{w}.\mathrm{r}.\mathrm{t}.\:\mathrm{an}\:\mathrm{inertial} \\ $$$$\mathrm{frame}\:\mathrm{is}\:\mathrm{4}\:\mathrm{N} \\ $$$$\left(\mathrm{C}\right)\:\mathrm{Acceleration}\:\mathrm{of}\:\mathrm{the}\:\mathrm{block}\:{A}\:\mathrm{is}\:\mathrm{2} \\ $$$$\mathrm{m}/\mathrm{s}^{\mathrm{2}} . \\ $$$$\left(\mathrm{D}\right)\:\mathrm{Net}\:\mathrm{force}\:\mathrm{on}\:{B}\:\mathrm{w}.\mathrm{r}.\mathrm{t}.\:{A}\:\mathrm{is}\:\mathrm{8}\:\mathrm{N} \\ $$

Question Number 21747 Answers: 0 Comments: 4

Three blocks of masses m_1 , m_2 and m_3 are connected as shown. All the surfaces are frictionless and the string and the pulleys are light. Find the acceleration of m_1 .

$$\mathrm{Three}\:\mathrm{blocks}\:\mathrm{of}\:\mathrm{masses}\:{m}_{\mathrm{1}} ,\:{m}_{\mathrm{2}} \:\mathrm{and}\:{m}_{\mathrm{3}} \\ $$$$\mathrm{are}\:\mathrm{connected}\:\mathrm{as}\:\mathrm{shown}.\:\mathrm{All}\:\mathrm{the}\:\mathrm{surfaces} \\ $$$$\mathrm{are}\:\mathrm{frictionless}\:\mathrm{and}\:\mathrm{the}\:\mathrm{string}\:\mathrm{and}\:\mathrm{the} \\ $$$$\mathrm{pulleys}\:\mathrm{are}\:\mathrm{light}.\:\mathrm{Find}\:\mathrm{the}\:\mathrm{acceleration} \\ $$$$\mathrm{of}\:{m}_{\mathrm{1}} . \\ $$

Question Number 21738 Answers: 1 Comments: 0

What is the last digit from the sum of 1 . 2^1 + 2 . 2^2 + 3 . 2^3 + ... + 50 . 2^(50) ?

$$\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{last}\:\mathrm{digit}\:\mathrm{from}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of} \\ $$$$\mathrm{1}\:.\:\mathrm{2}^{\mathrm{1}} \:+\:\mathrm{2}\:.\:\mathrm{2}^{\mathrm{2}} \:+\:\mathrm{3}\:.\:\mathrm{2}^{\mathrm{3}} \:+\:...\:+\:\mathrm{50}\:.\:\mathrm{2}^{\mathrm{50}} \:? \\ $$

Question Number 21784 Answers: 0 Comments: 0

Call a positive integer n good if there are n integers, positive or negative, and not necessarily distinct, such that their sum and product are both equal to n (e.g. 8 is good since 8=4∙2∙1∙1∙1∙1(−1)(−1)=4+2+1+1+1 +1+(−1)+(−1)). Show that integers of the form 4k + 1 (k ≥ 0) and 4l (l ≥ 2) are good.

$$\mathrm{Call}\:\mathrm{a}\:\mathrm{positive}\:\mathrm{integer}\:{n}\:\boldsymbol{\mathrm{good}}\:\mathrm{if}\:\mathrm{there} \\ $$$$\mathrm{are}\:{n}\:\mathrm{integers},\:\mathrm{positive}\:\mathrm{or}\:\mathrm{negative},\:\mathrm{and} \\ $$$$\mathrm{not}\:\mathrm{necessarily}\:\mathrm{distinct},\:\mathrm{such}\:\mathrm{that}\:\mathrm{their} \\ $$$$\mathrm{sum}\:\mathrm{and}\:\mathrm{product}\:\mathrm{are}\:\mathrm{both}\:\mathrm{equal}\:\mathrm{to}\:{n} \\ $$$$\left(\mathrm{e}.\mathrm{g}.\:\mathrm{8}\:\mathrm{is}\:\boldsymbol{\mathrm{good}}\:\mathrm{since}\right. \\ $$$$\mathrm{8}=\mathrm{4}\centerdot\mathrm{2}\centerdot\mathrm{1}\centerdot\mathrm{1}\centerdot\mathrm{1}\centerdot\mathrm{1}\left(−\mathrm{1}\right)\left(−\mathrm{1}\right)=\mathrm{4}+\mathrm{2}+\mathrm{1}+\mathrm{1}+\mathrm{1} \\ $$$$\left.+\mathrm{1}+\left(−\mathrm{1}\right)+\left(−\mathrm{1}\right)\right). \\ $$$$\mathrm{Show}\:\mathrm{that}\:\mathrm{integers}\:\mathrm{of}\:\mathrm{the}\:\mathrm{form}\:\mathrm{4}{k}\:+\:\mathrm{1} \\ $$$$\left({k}\:\geqslant\:\mathrm{0}\right)\:\mathrm{and}\:\mathrm{4}{l}\:\left({l}\:\geqslant\:\mathrm{2}\right)\:\mathrm{are}\:\boldsymbol{\mathrm{good}}. \\ $$

Question Number 21782 Answers: 0 Comments: 0

a_1 =1, a_(n+1) =(a_n /(√(a_n +n+1))) Σ_(n=1) ^∞ a_n =?

$${a}_{\mathrm{1}} =\mathrm{1},\:{a}_{{n}+\mathrm{1}} =\frac{{a}_{{n}} }{\sqrt{{a}_{{n}} +{n}+\mathrm{1}}} \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}{a}_{{n}} =? \\ $$

Question Number 21781 Answers: 2 Comments: 0

Which is greater 10^(11) or 11^(10) ?

$$\mathrm{Which}\:\mathrm{is}\:\mathrm{greater}\:\mathrm{10}^{\mathrm{11}} \:\mathrm{or}\:\mathrm{11}^{\mathrm{10}} ? \\ $$

Question Number 21733 Answers: 1 Comments: 0

If p is one of roots from x^2 − 2x + 6 = 0 then p^4 + 16p is equal to ...

$$\mathrm{If}\:{p}\:\mathrm{is}\:\mathrm{one}\:\mathrm{of}\:\:\mathrm{roots}\:\mathrm{from}\:{x}^{\mathrm{2}} \:−\:\mathrm{2}{x}\:+\:\mathrm{6}\:=\:\mathrm{0} \\ $$$$\mathrm{then}\:{p}^{\mathrm{4}} \:+\:\mathrm{16}{p}\:\mathrm{is}\:\mathrm{equal}\:\mathrm{to}\:... \\ $$

Question Number 21732 Answers: 0 Comments: 2

(i) Find the first three terms in the expansion of (2 − x)^6 in ascending power of x. (ii) Find the value of k for which there is no term in x^2 in the expansion (1 + kx)(2 − x)^6

$$\left(\mathrm{i}\right)\:\mathrm{Find}\:\mathrm{the}\:\mathrm{first}\:\mathrm{three}\:\mathrm{terms}\:\mathrm{in}\:\mathrm{the}\:\mathrm{expansion}\:\mathrm{of}\:\left(\mathrm{2}\:−\:\mathrm{x}\right)^{\mathrm{6}} \:\mathrm{in}\:\mathrm{ascending}\:\mathrm{power} \\ $$$$\mathrm{of}\:\mathrm{x}. \\ $$$$\left(\mathrm{ii}\right)\:\mathrm{Find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\mathrm{k}\:\mathrm{for}\:\mathrm{which}\:\mathrm{there}\:\mathrm{is}\:\mathrm{no}\:\mathrm{term}\:\mathrm{in}\:\mathrm{x}^{\mathrm{2}} \:\mathrm{in}\:\mathrm{the}\:\mathrm{expansion} \\ $$$$\left(\mathrm{1}\:+\:\mathrm{kx}\right)\left(\mathrm{2}\:−\:\mathrm{x}\right)^{\mathrm{6}} \\ $$

Question Number 21731 Answers: 0 Comments: 0

A very flexible uniform chain of mass M and length L is suspended vertically so that its lower end just touches the surface of a table. When the upper end of the chain is released, it falls with each link coming to rest the instant it strikes the table. Find the force exerted by the chain on the table at the moment when x part of chain has already rested on the table.

Question Number 21722 Answers: 1 Comments: 0

Question Number 21721 Answers: 1 Comments: 0

∫_0 ^(π/2) sin^2 xcos^3 xdx

$$\int_{\mathrm{0}} ^{\pi/\mathrm{2}} {sin}^{\mathrm{2}} {xcos}^{\mathrm{3}} {xdx} \\ $$

Question Number 21720 Answers: 1 Comments: 0

∫sin^5 θdθ

$$\int{sin}^{\mathrm{5}} \theta{d}\theta \\ $$

Question Number 21713 Answers: 0 Comments: 7

A constant force F = 20 N acts on a block of mass 2 kg which is connected to two blocks of masses m_1 = 1 kg and m_2 = 2 kg. Calculate the accelerations produced in all the three blocks. Assume pulleys are frictionless and weightless.

$$\mathrm{A}\:\mathrm{constant}\:\mathrm{force}\:{F}\:=\:\mathrm{20}\:\mathrm{N}\:\mathrm{acts}\:\mathrm{on}\:\mathrm{a} \\ $$$$\mathrm{block}\:\mathrm{of}\:\mathrm{mass}\:\mathrm{2}\:\mathrm{kg}\:\mathrm{which}\:\mathrm{is}\:\mathrm{connected}\:\mathrm{to} \\ $$$$\mathrm{two}\:\mathrm{blocks}\:\mathrm{of}\:\mathrm{masses}\:{m}_{\mathrm{1}} \:=\:\mathrm{1}\:\mathrm{kg}\:\mathrm{and} \\ $$$${m}_{\mathrm{2}} \:=\:\mathrm{2}\:\mathrm{kg}.\:\mathrm{Calculate}\:\mathrm{the}\:\mathrm{accelerations} \\ $$$$\mathrm{produced}\:\mathrm{in}\:\mathrm{all}\:\mathrm{the}\:\mathrm{three}\:\mathrm{blocks}.\:\mathrm{Assume} \\ $$$$\mathrm{pulleys}\:\mathrm{are}\:\mathrm{frictionless}\:\mathrm{and}\:\mathrm{weightless}. \\ $$

Question Number 21707 Answers: 0 Comments: 3

∫_(π/6) ^(π/3) (1/2)cot^2 2θdθ

$$\int_{\pi/\mathrm{6}} ^{\pi/\mathrm{3}} \frac{\mathrm{1}}{\mathrm{2}}{cot}^{\mathrm{2}} \mathrm{2}\theta{d}\theta \\ $$

Question Number 21702 Answers: 0 Comments: 1

∫_(π/6) ^(π/3) 1/2cot^2 2θdθ

$$\int_{\pi/\mathrm{6}} ^{\pi/\mathrm{3}} \mathrm{1}/\mathrm{2}{cot}^{\mathrm{2}} \mathrm{2}\theta{d}\theta \\ $$

Question Number 21701 Answers: 2 Comments: 1

∫2cot^2 2t

$$\int\mathrm{2}{cot}^{\mathrm{2}} \mathrm{2}{t} \\ $$

Question Number 21686 Answers: 0 Comments: 0

I have 6 friends and during a vacation I met them during several dinners. I found that I dined with all the 6 exactly on 1 day; with every 5 of them on 2 days; with every 4 of them on 3 days; with every 3 of them on 4 days; with every 2 of them on 5 days. Further every friend was present at 7 dinners and every friend was absent at 7 dinners. How many dinners did I have alone?

$$\mathrm{I}\:\mathrm{have}\:\mathrm{6}\:\mathrm{friends}\:\mathrm{and}\:\mathrm{during}\:\mathrm{a}\:\mathrm{vacation} \\ $$$$\mathrm{I}\:\mathrm{met}\:\mathrm{them}\:\mathrm{during}\:\mathrm{several}\:\mathrm{dinners}.\:\mathrm{I} \\ $$$$\mathrm{found}\:\mathrm{that}\:\mathrm{I}\:\mathrm{dined}\:\mathrm{with}\:\mathrm{all}\:\mathrm{the}\:\mathrm{6}\:\mathrm{exactly} \\ $$$$\mathrm{on}\:\mathrm{1}\:\mathrm{day};\:\mathrm{with}\:\mathrm{every}\:\mathrm{5}\:\mathrm{of}\:\mathrm{them}\:\mathrm{on}\:\mathrm{2}\:\mathrm{days}; \\ $$$$\mathrm{with}\:\mathrm{every}\:\mathrm{4}\:\mathrm{of}\:\mathrm{them}\:\mathrm{on}\:\mathrm{3}\:\mathrm{days};\:\mathrm{with} \\ $$$$\mathrm{every}\:\mathrm{3}\:\mathrm{of}\:\mathrm{them}\:\mathrm{on}\:\mathrm{4}\:\mathrm{days};\:\mathrm{with}\:\mathrm{every}\:\mathrm{2} \\ $$$$\mathrm{of}\:\mathrm{them}\:\mathrm{on}\:\mathrm{5}\:\mathrm{days}.\:\mathrm{Further}\:\mathrm{every}\:\mathrm{friend} \\ $$$$\mathrm{was}\:\mathrm{present}\:\mathrm{at}\:\mathrm{7}\:\mathrm{dinners}\:\mathrm{and}\:\mathrm{every} \\ $$$$\mathrm{friend}\:\mathrm{was}\:\mathrm{absent}\:\mathrm{at}\:\mathrm{7}\:\mathrm{dinners}.\:\mathrm{How} \\ $$$$\mathrm{many}\:\mathrm{dinners}\:\mathrm{did}\:\mathrm{I}\:\mathrm{have}\:\mathrm{alone}? \\ $$

Question Number 21685 Answers: 0 Comments: 0

In a group of ten persons, each person is asked to write the sum of the ages of all the other 9 persons. If all the ten sums form the 9-element set {82, 83, 84, 85, 87, 90, 91, 92} find the individual ages of the persons (assuming them to be whole numbers of years).

$$\mathrm{In}\:\mathrm{a}\:\mathrm{group}\:\mathrm{of}\:\mathrm{ten}\:\mathrm{persons},\:\mathrm{each}\:\mathrm{person} \\ $$$$\mathrm{is}\:\mathrm{asked}\:\mathrm{to}\:\mathrm{write}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{the}\:\mathrm{ages}\:\mathrm{of} \\ $$$$\mathrm{all}\:\mathrm{the}\:\mathrm{other}\:\mathrm{9}\:\mathrm{persons}.\:\mathrm{If}\:\mathrm{all}\:\mathrm{the}\:\mathrm{ten} \\ $$$$\mathrm{sums}\:\mathrm{form}\:\mathrm{the}\:\mathrm{9}-\mathrm{element}\:\mathrm{set}\:\left\{\mathrm{82},\:\mathrm{83},\:\mathrm{84},\right. \\ $$$$\left.\mathrm{85},\:\mathrm{87},\:\mathrm{90},\:\mathrm{91},\:\mathrm{92}\right\}\:\mathrm{find}\:\mathrm{the}\:\mathrm{individual} \\ $$$$\mathrm{ages}\:\mathrm{of}\:\mathrm{the}\:\mathrm{persons}\:\left(\mathrm{assuming}\:\mathrm{them}\:\mathrm{to}\right. \\ $$$$\left.\mathrm{be}\:\mathrm{whole}\:\mathrm{numbers}\:\mathrm{of}\:\mathrm{years}\right). \\ $$

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