Given f(1), f(2), f(3), f(4), find f(5)
Q. f(x) is a four -degree polynomial and the coefficient of \({x}^{4}\) is 1. If f(1) = 5, f(2) = 10, f(3) = 15 and f(4) = 20 then, what is the value of f(5) ? A. 25 B. 30 C. 49 D. 34 E. 36 C. 49 f(x) = c(x-1)(x-2)(x-3)(x-4) + 5x As leading coefficient = 1 ,c = 1 \(\Rightarrow\) f(x) = (x-1)(x-2)(x-3)(x-4) + 5x \(\Rightarrow\) f(5) = 4×3×2×1 + 5×5 = 49
Probability of customer getting his choice color
Q A small store has five units of a new phone model in stock: two white, two black, and one red. Three customers arrive at the shop to buy a unit each. Each one has a pre-determined choice of the colour and will not buy a unit of any other colour. All the three customers are equally likely to have chosen any of the three colour. What is the probability that the store will be able to satisfy all the three customers? A. \(\frac{4}{5}\) B. \(\frac{7}{9}\) C. \(\frac{2}{3}\) D. \(\frac{8}{9}\) E. \(\frac{1}{3}\) C. \(\frac{2}{3}\) White = 2 Black = 2 Red = 1 All three customers has three choices i.e. W,B,R \(\Rightarrow\) Total combination possible is 3 × 3 × 3 = 27 \(\Rightarrow\) Combination not possible are given below WWW - 1 BBB - 1 RRR - 1 RRB - \(\frac{3!}{2!}\) = 3 [...]
Rate at which shadow increases
Q. A 1.5 m tall man walks at a uniform speed of 60 m/min away from a lamp post of height 6 m. Find the rate at which his shadow increases A. \(\frac{15m}{min}\) B. \(\frac{24m}{min}\) C. \(\frac{30m}{min}\) D. \(\frac{20m}{min}\) E. \(\frac{25m}{min}\) D. \(\frac{20 m}{min}\) \(\Rightarrow\) \(\frac{6}{1.5}\) = \(\frac{a+b}{b}\) = \(\frac{4}{1}\) = a + b = 4b = a = 3b \(\Rightarrow\) Time to travel dist "a" = \(\frac{a}{60}\)min \(\frac{a}{60}\)min \(\rightarrow\) b 1 min \(\rightarrow\) ? = \(\frac{1×b}{a/60}\) = 20 m/ min
Arranging letters
Q. The letters D, G, I, I and T can be used to form 5-letter strings such as DIGIT or DGIIT. Using these letters, how many 5-letters strings can be formed in which the two occurrences of the letter I are separated by at least one other letter? A. 12 B. 18 C. 24 D. 36 E. 48 D. 36 D, I, I, G, T Total possible arrangements of the 5 letters is given below 5!/2! = 60 Cases where the "I"s are together = 4! = 24 So, 60 - 24 = 36 OR Arrange D, G, T first and then place "I"s. D, G, T can be arranged in 3! = 6 ways 4 spaces are created. The "I"s need two places, so total ways = ⁴C₂ = 6 ways Total ways = 6×6 = 36
Whoever wins 5 matches wins the series
Q. England and Australia play a test series until one team wins 5 matches. No match ends in a draw. In how many ways can the series be won? A. 126 B. 252 C. 144 D. 231 E. 280 B. 252 The minimum number of matches that the series can have is 5 matches. And this case is the easiest. The team winning the series, will win all 5 matches in a row i.e. W W W W W, just 1 way. If the series ends after 6 matches, the 6th match i.e. the last match will be won by the team winning the series ... that is how the series ends. And amongst the first 5 matches, 4 matches will be won by the team winning the series and 1 match will be won by the team losing the series. {W, W, [...]
Lengths of two intersecting arcs
Q. In the below figure, the two circular curves create 60° and 90° angles with their respective centers. If the length of the bottom curve Y is 10π, find the length of the other curve. A. \(\frac{15π}{\sqrt{2}}\) B. \(\frac{20π\sqrt{2}}{3}\) C. \(\frac{60π}{\sqrt{2}}\) D. \(\frac{20π}{3}\) E. \(15π\) A. \(\frac{15π}{\sqrt{2}}\) Join the endpoints , let X1 and Y1 be the respective centers. Given that length of Y is 10π and angle subtended is 60° \(\Rightarrow \frac{60}{360} \times 2 \pi r = 10 \pi \), where r is the radius. ⇒ r = 30 Triangle AY1B is an equilateral triangle ⇒ AB = 30 units In 45-45-90 triangle AX1B, \(AX_{1}\) = \(BX_{1} = \frac{30}{\sqrt{2}}\) = \(15\sqrt{2}\) Now, length of arc X = \(\frac{90}{360}×2π×15\sqrt{2}\) = \(\frac{15π}{\sqrt{2}}\)
Two circular tracks
Q. Ram and Shyam are running on different circular tracks. In the time that Ram completes 4 rounds of his track, Shyam completes 3 rounds of his track. If Ram and Shyam interchange their tracks, the time taken by both of them to complete 1 round is same. What is the ratio of speeds of Ram and Shyam? A. \(\sqrt{\frac{3}{4}}\) B. \(\sqrt{\frac{7}{9}}\) C. \(\frac{2}{\sqrt{3}}\) D. \(\frac{\sqrt{7}}{4}\) E. Cannot be determined C. \(\frac{2}{\sqrt{3}}\) On first reading, the way forward is not crystal clear but what makes it doable is that time being same, track length is proportional on speeds; and there is a second relation also ... so two variables, two data seems doable. Let the speeds of Ram and Shyam be x and y. Using the first equation and equating the time, \(\frac{4\;×\;\text{length of track 1}}{x}\) = \(\frac{3\;×\;\text{length of track 2}}{y}\) \(\Rightarrow \frac{\text{length of [...]
Simultaneous Equations, Unique Solution
Q. The equations ax - (a + b) y = 1 and (a - b) x + ay = 5 have a unique solution .... A. ... for all value of a and b B. ... only when \(a = b\) C. ... only when \({a}^{2} : {b}^{2} \neq 1 : 2\) D. ... only when a = 0 and b = 0 E. ... only when \(a \neq b\) C. Only when \({a}^{2} : {b}^{2} \neq 1 : 2\) \(\frac{a}{a-b} \neq -\frac{a+b}{a}\) \({a}^{2} \neq [{a}^{2} - {b}^{2}]\) \(2{a}^{2} \neq {b}^{2}\) \(\frac{{a}^{2}}{{b}^{2}} \neq \frac{1}{2}\)
Clocks, time
A watch gains 3 seconds every 2 minutes. It was set right at 12 noon on 1 July 2022. From 12 noon of 1 July 2022 to 12 noon of 30 June 2023 (including both the noons), for how many instances the watch will show the same time as shown by a watch that always shows the correct time? (Assuming all watches show time as per 24-hour format) A. 9 B. 7 C. 8 D. 10 D. 10 From 1 July 2022 to 30 June 2023 = 365 days Gain in 2 mins = 3s gain in 20 mins = 30 s gain in 1 hour= 1.5 min gain in 24 hours= 36 min i.e gain in one day= 36 mins = 36/60 hrs number of days= $\frac{24hrs}{36/60 hrs}$ = 40 After every 40 days clock will show the correct time 365/40 = 9 [...]
Functions
Let $\mathrm{f}(\mathrm{x})=\frac{x^2+1}{x^2-1}$, if $\mathrm{x} \neq 1,-1$, and 1 if $\mathrm{x}=1,-1$. Let $\mathrm{g}(\mathrm{x})=\frac{x+1}{x-1}$ if $\mathrm{x} \neq 1$ and 3 if $x=1$. What is the minimum possible value of $\frac{f(x)}{g(x)}$ ? A. 1 B. -1 C. 1/4 D. 1/3 D. 1/3 $\frac{f(x)}{g(x)}=\frac{(x^2+1)(x-1)}{(x^2-1)(x+1)} =\frac{(x^2+1)}{(x+1)^2}$ i.e x cannot be -1 we observe when x=1(which is possible here), we get f(x)=1 and g(x)=3 Hence, $\frac{f(x)}{g(x)} =\frac{1}{3}$