Variables, Integers
If a, b and c are 3 consecutive integers between -10 to +10 (both inclusive), how many integer values are possible for the expression - $\frac{a^3+b^3+c^3+3 a b c}{(a+b+c)^2}$ A. 0 B. 1 C. 2 D. 3 C. 2 Since $a, b, c$ are consecutive integers $$ \Rightarrow a=b-1 \text { and } c=b+1 $$ Expression: $\frac{a^3+b^3+c^3+3 a b c}{(a+b+c)^2}$ $$ \begin{aligned} & =\frac{(b-1)^3+b^3+(b+1)^3+3(b-1) b(b+1)}{(b-1+b+b+1)^2} \\ & =\frac{b^3+3 b+b^3+b^3+3 b+3 b^3-3 b}{9 b^2} \\ & =\frac{6 b^3+3 b}{9 b^2}=\frac{2 b^2+1}{3 b} \end{aligned} $$ Putting different values of $b$ from - 10 to 10 , we can verify that only - 1 and 1 satisfies to get integer values for the expression.
Geometry
In the figure given below, the circle has a chord AB of length 12 cm, which makes an angle of 60 degree at the center of the circle, O. ABCD, as shown in the diagram, is a rectangle. OQ is the perpendicular bisector of AB, intersecting the chord AB at P, the arc AB at M and CD at Q. OM = MQ. The area of the region enclosed by the line segments AQ and QB,and the arc BMA is closest to(in cm square) - A. 215 B. 137 C. 35 D. 69 D. 69 In triangle OAP, since AP=6, OA will be 12 Area of AQBMA=Area of Triangle ABQ- (Area of minor arc AMB-Area of OAB) Length of PQ=MQ+PM = 12+(12- 6$\sqrt{3}$)=24- 6$\sqrt{3}$ Area of triangle ABQ= $\frac{1}{2}$*12*24- 6$\sqrt{3}$= 81.64 Area of minor arc AMB-Area of OAB=$\frac{60}{360}$*$\Pi$*144 - $\frac{1}{2}$*12*6$\sqrt{3}$ =13.07 Area of AQBMA= [...]
Venn Diagrams
A survey was conducted among 300 room air conditioner owners. It was found that 125 people had carrier aircon, 145 people had voltas and 90 people had fedders lloyd air conditioners with them. Thirty two of them had exactly two out of the three brands of air conditioners. If 10 of the owners having only carrier aircon now buy voltas also; and 5 of the owners who had only voltas and fedders lloyd now buy carrier aircon also, then how many have at least two out of the three brands of air conditioners? A. 19 B. 61 C. 68 D. 56 61
Roots of quadratic
Q. \(\alpha\) and \(\beta\) are the roots of the equation \({ax}^{2} + bx + c = 0\). What is the condition for which \(\beta = 5 \alpha\)? A. \({5b}^{2}=24ac\) B. \({3b}^{2}=12ac\) C. \({3b}^{2}=16ac\) D. \({5b}^{2}=36ac\) Use properties of roots. D. \({5b}^{2}=36ac\) For roots we know, \(\alpha+\beta=\frac{-b}{a} \) and \(\alpha\beta=\frac{c}{a} \) and we know \(\beta =5\alpha\), substituting this in the above relations we get \(\Rightarrow 6\alpha=\frac{-b}{a} \) [∵ \(5\alpha+\alpha=6\alpha\)] \(36{\alpha}^{2}=\frac{{-b}^{2}}{{a}^{2}} .........\text{(i)}\) \(\Rightarrow\alpha\beta=\frac{c}{a} \) \({5\alpha}^{2}=\frac{c}{a}\) [∵ \(\alpha\times5\beta={5\alpha}^{2}\)] \({\alpha}^{2}=\frac{c}{5a}.........\text{(ii)}\) ⇒ combining (i) and (ii) we get, \(36\times\frac{c}{5a}=\frac{{b}^{2}}{{a}{2}}\) \(36ac=5{b}^{2}\)
GP which is also an AP
Q. If the second, third and first terms of a geometric progression (GP) form an arithmetic progression (AP), then find the first term of the GP, given that the sum to infinite terms of the GP is 36. A. 9 B. 54 C. 27 D. 18 Think of common properties of an AP. B. 54 Let us take terms of GP as \(\Rightarrow a, ar, {ar}^{2}\) then the terms in AP will be \(\Rightarrow ar, {ar}^{2}, a\) In that case, \({ar}^{2} = \frac{a+ar}{2}\) \({2ar}^{2}=a+ar\) on simplifying the quadratic we get r = 1 OR r = -1/2 r can not be 1 as that will make each term of the GP same which will not make it an infinite GP, so we get r = -1/2 ⇒ Also given that sum to infinite terms of the GP = 36 \(\frac{a}{1-r}=36\) plugging r=-1/2 we [...]
Clocks, Calenders
Three clocks were set to true time. First runs with the exact time. Second slows one minute/day. Third gains one minute/day. After how many days will they show true time again? A. 360 B. 180 C. 720 D. 480 E. None of these A faulty watch will show correct time when it is either 12 hrs ahead or 12 hrs behind. This is so because we are using a 12hr watch just to check the time and ignoring the am/pm aspect or date aspect. C. 720 The second watch gains 1min in 1 day It will gain 12hrs * 60 min = 720 min in 720 days Similarly, the third watch will be 12hrs ahead in 720 days. Thus in 720 days, all three watches will be showing the same time.
Weights, mixtures
Grapes have 80% water (by weight). When kept in the sun to dry, only some of the water evaporates and I get raisins, having 10 % water(by weight). If I need 1kg of raisins, how many kgs of grapes should I dry? TITA type i.e. Type In The Answer type In such cases, keep in mind that portion having no water i.e pulp must be equal before and after drying. 4.5 kg pulp in grapes= dried portion of raisins Let weight of grapes needed in kgs= x 20% of x = 90% of 1 $\frac{20 x}{100}$ = $\frac{90*1}{100}$ x= 9/2 kgs =4.5kgs
Profit, Market price
Of all the articles manufactured by a company, 5% are defective. If the manufacturing cost of 100 articles is Rs. 1900, what should be the market price of the article so that the company gets 40% profit, even if it gives the articles to the distributor for 30% less price than the market price? A. Rs. 40 B. Rs. 22 C. Rs. 28 D. Rs. 34 E. Rs. 24 Rs. 40 Total manufacturing cost = 1900 Profit = 40% of 1900 =190 *4=760 Total selling price to the distributer= 1900+760 = 2660 2660= $\frac{70}{100}$* market price market price= $\frac{26600}{7}$ market price of 1 article= $\frac{26600}{7*95}$ = 40 Rs.
Clocks and Calendar
A watch which gains uniformly, is 5 min slow at 8 o'clock in the morning on Sunday and it is 5 min 48 sec. fast at 8 p.m on following Sunday. when was it correct? A. 7 pm on Wednesday B. 20 min past 7 pm on Wednesday C. 15 min past 7 pm on Wednesday D. 8 pm on Wednesday E. None of these 20 min past 7 pm on Wednesday Consider 1st Sunday at 8 am and 2nd Sunday at 8 pm Time Passed= 7 days 12 hrs = 180 hrs 5 min 48 s = 54/5 min 5 min gain = $\frac{180 * 5}{54/5}$ = 250/3 = 83 hrs 20 mins = 3 days 11 hrs 20 mins Adding to Sunday 8 am = Wednesday, 7 pm 20 min
Man and log in stream
A man swims from A to B and back in 4.5 hours. A block of wood when allowed to go with the stream from A to B takes 6 hours. What is the ratio of the speed of the man in still water to that of the stream? A. 2 : 1 B. 3 : 1 C. 4 : 1 D. 5: 1 B. 3 : 1 $$ \begin{align} \frac{d}{b+s}+\frac{d}{b-s} & =\frac{9}{2} \\ \\ \frac{d}{s} & =6 \end{align} $$ Substituting d = 6s $$ \frac{6 s}{b+s}+\frac{6 s}{1-s}=\frac{9}{2} $$ Rather than solve, use options - (A) $\frac{6}{2+1}+\frac{6}{2-1}=8 \neq 9 / 2$. (B) $\frac{6}{3+1}+\frac{6}{3-1}=4.5$