Cracking CAT: Understanding the CAT Percentile Calculation and Its Importance
During class 8-th to class 12-th, a student’s academic goals were to score a high percentage. Percentage was simply the marks scored as a percent of total marks of the paper. And we all understand percentage marks scored very clearly. A slightly more ambitious goal was getting a higher Rank. To be amongst the top rankers i.e. rank 1, rank 2, rank 3, …. Thus, the goal shifted from an absolute percentage of marks scored to a relative one, performing better than the others. And when one entered the field of competitive entrance exams, like CAT, a new term assaulted us – ‘percentile’. Apparently to get admission to the old IIMs one needed 99.5 percentile and above; to get admission to a decently good b-school one needed atleast 95 percentile. And thus began the chase of percentile. Not everyone understood CAT percentile calculation thoroughly, but the chase was on. Come [...]
Time, Speed, Distance
Two towns A and B are $180 \mathrm{kms}$ apart. Car 1 and car 2 start travelling from $A$ and $B$, respectively towards each other. Their speeds are in the ratio $1: 2$ and they start at 7:00 arm. and 8:00 a.m., respectively. After meeting at $\mathrm{C}$, they return to their starting positions and again start travelling towards each other. In order to meet car 1 for the second time at $\mathrm{C}$, car 2 halts at $\mathrm{C}$. For how much in hrs does it halt? TITA type i.e. Type In The Answer type 2hrs Let them meet at time $T$. this. Reaches C again at 7+3(T-7) Reaches D again at 8+3(T-8) Difference is 2 hrs
Number System, Primes
Find the sum of the numbers which are co-prime to 840 and are less than it. A.40320 B.80640 C.120960 D.60480 B. 80640 $$ \begin{aligned} 840= & 2^3 \times 3 \times 5 \times 7 \\ \phi(840) & =840 \times\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{5}\right)\left(1-\frac{1}{7}\right) \\ & =840 \times \frac{1}{2} \times \frac{2}{3} \times \frac{4}{5} \times \frac{6}{7} \\ & =192 \end{aligned} $$ Sum of all numbers co-prime to n and less than $n=\frac{n}{2} \times \phi(n)$. 420*192= 80640
The CAT Maze and the GEM Quandary: A Sideways Glance
There's something about the Indian Institutes of Management (IIMs) that reminds one of an exclusive, members-only club. A place where the doors aren't just casually open to any passerby, but where entry is a game of wits, stamina, and perhaps a touch of masochism. Enter the CAT MBA entrance – not merely a key to this club, but a rite of passage that tests just how badly you want in. Now, amidst the sea of hopeful faces each year, there's a group that's become a bit of a talking point at dinner parties: the GEMs (General Engineering Males). Picture this: here are individuals, often lauded for their academic achievements, caught in a whirlwind of self-doubt and speculation. The chatter on the grapevine suggests these chaps find the uphill battle to IIMs especially steep. But one has to wonder, is this narrative as it seems? The GEM Dilemma On the [...]
Profit,Loss,Percentages
The owner of an art shop conducts his business in the following manner: every once in a while he raises his prices by X%, then a while later he reduces all the new prices by X%. After one such updown cycle, the price of a painting decreased by Rs. 441. After a second up-down cycle the painting was sold for Rs. 1,944.81. What was the original price of the painting? A.Rs. 2756.25 B.Rs.2256.25 C.Rs.2500 D.Rs.2000 A. Rs. 2756.25 Let the price of the painting be P ⇒ P−P×(1+$\frac{x}{100}$)(1−$\frac{x}{100}$)=441 ⇒ P× $\frac{x}{100}$ $\frac{x}{100}$=441 ⇒ Again, P× [(1+$\frac{x}{100}$)(1−$\frac{x}{100}$)]^2=1944.81 ⇒ P×(1−$\frac{x}{100}$ $\frac{x}{100}$)^2=1944.81 Dividing both we get, ∴ x/100=52 ⇒ P=4441×25=Rs.2756.25
Time,speed,distance
Three runners A, B and C run a race, with runner A finishing 12 meters ahead of runner B and 18 meters ahead of runner C, while runner B finishes 8 meters ahead of runner C. Each runner travels the entire distance at a constant speed. What was the length of the race? A.36 B.48 C.60 D.72 48 B and C had a distance of 6 m between them When B finished the race by covering another 12 m, he created a gap of 8 m between them i.e. he created another 2 m gap while running 12 m. Since he created a gap of 2 m by running 12 m, he must have created a gap of 8 m by running a total of 12*4 = 48 m
Geometry,Rectangles
A rectangular pool 20 meters wide and 60 meters long is surrounded by a walkway of uniform width. If the total area of the walkway is 516 square meters, how wide, in meters, is the walkway? A.43m B. 4.3m C.3m D.3.5m C. 3m area of the rectangular pool = length * breadth = 60*20 = 1200 m^2 given the area of walkway is 516 m^2, so total area = 1200 + 516 = 1716 m^2 let the width of walkway be x, as the walkway is uniform in width, the length of total area will be = 60 + 2x and breadth of total area will be = 20 + 2x now, 1716 = (60 + 2x)*(20+2x) 1716 = 1200 + 120x + 40x + 4x^2 We get, x = 3 or -43 since width cannot be negative, the width of the walkway is 3m. [...]
Quadratic Equation
Ujakar and Keshab attempted to solve a quadratic equation. Ujakar made a mistake in writing down the constant term. He ended up with the roots (4, 3). Keshab made a mistake in writing down the coefficient of x. He got the roots as (3, 2). What will be the exact roots of the original quadratic equation? A. 6,1 B.-3,-4 C.4,3 D.-4,3 A. 6,1 We know that quadratic equation can be written as $ x^2 -(sum of roots)*x+(product of the roots)=0$ Ujakar ended up with the roots (4, 3) so the equation is $x^2 -(7)*x+(12)=0$ where the constant term is wrong Keshab got the roots as (3, 2) so the equation is $x^2 -(5)*x+(6)=0$ where the coefficient of x is wrong So the correct equation is $x^2 -(7)*x+(6)=0$ The roots of above equations are (6,1)
Percentages
A school has raised 75% of the amount it needs for a new building by receiving an average donation of Rs. 600 from the parents of the students. The people already solicited represents the parents of 60% of the students. If the School is to raise exactly the amount needed for the new building, what should be the average donation from the remaining students to be solicited? A. Rs. 300 B. Rs.250 C. Rs. 400 D. Rs.500 A. Rs. 300 Let the number of parents be x who has been asked for the donations. People already solicited = 60% of x = 0.6x Remaining people = 40% of x = 0.4x Amount collected from the parents solicited= 600 *0.6x = 360x 360x = 75% Remaining amount = 25% = 120x Thus, Average donations from remaining parents = 120x /0.4x = 300 [...]
Maximum,minimum
The maximum value M of $3^x + 5^x – 9^x + 15^x –25^x$ , as x varies over reals, satisfies – A.3<M<5 B.0<M<2 C.5<M<25 D.0<M<9 B. 0<M<2 $\begin{aligned} & M=a+b-a^2+a b-b^2 \quad \frac{a^2+b^2}{2} \geq a b \\ & \mathrm{a}^2+\mathrm{b}^2 \geq 2 \mathrm{ab} \\ & -\left(a^2+b^2\right) \leq-2 a b \\ & \mathrm{M} \leq \mathrm{a}+\mathrm{b}-\mathrm{ab} \\ & \mathrm{M}<\mathrm{1}-(\mathrm{a}-\mathrm{l})(\mathrm{b}-1) \quad \min \text { is zero } \\ & \text { hence } 0<\mathrm{M}<2 \\ & \end{aligned}$