In this article certain questions have been randomly taken from different topics of the quantitative section of CAT. Each question has been selected keeping in mind certain challenges you need to manage as a test taker. These challenges may range from concept basics to application orientation and test taking strategies.
Consider the sequence of numbers a1,a2,a3…..to infinity where a1 = 81.33, a2 = -19 and
aj = aj-1 – aj-2, for j greater than equal to 3. What is the sum of the first 6002 terms of this sequence?
1. 0
2. 62.33
3. 100.33
4. Indeterminate
Topic : Sequence & Series
Approach :Given that aj = aj-1 – aj-2, for j greater than equal to 3,
a3 = a2-a1
a4 = a3-a2 = a2-a1-a2 = -a1
a5 = a4-a3 = -a1-(a2-a1)= -a2
a6 = a5-a4 = -a2-(-a1) = a1-a2
a7 = a6-a5 = a1-a2+a2= a1
a8 = a7-a6 = a1-(a1-a2) = a2
After a6 there is a repetition in a cyclic way, so we just need to focus on the first 6 terms a1 to a6.
Now a1+a2+a3+a4+a5+a6= a1+a2+(a2-a1)-a1-a2+a1-a2= 0
Thus, sum of first 6000 terms will be 0
Sum of 6002 terms = 6001th term+6002th term = a1+a2= 81.33-19=62.33
Answer : Option 2
Challenge :Understanding the concept of cyclicity in sequences & series
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If the sum of the first 11 terms of an Arithmetic Progression (A.P) equals that of the first 19 terms of the same A.P, then what is the sum of the first 30 terms of the same A.P?
1. 0
2. 1
3. -1
4. None of these
Topic : Sequence & Series
Approach: The sum of n terms of an A.P = n/2 (2a+(n-1)d)
From the given information,
11/2 (2a+10d) = 19/2 (2a+18d)
Rearranging, we get 16a + 232d= 0
Or 2a + 29d= 0
Now, sum of first 30 terms = 30/2 (2a+29d)= 15 *0 =0
Answer: Option 1
Challenge: Understanding the application of summation in an AP
A and B run along a circular track at constant speeds. A completes a lap in 5 seconds less than B. If they start from the same point and run in the same direction, then they meet after ½ a minute. How many times will they meet in a minute if they run in opposite directions?
1. 5 times
2. 10 times
3. Once
4. None of these
Topic: Time-Speed-Distance
Approach: Let the circumference of the track be l metres, and speeds of A and B be Sa m/s and Sb m/s respectively.
Since they meet after 30 seconds, l/(Sa-Sb) =30
Or Sa/l – Sb/l = 1/30…………….I
Also, l/Sb – l/Sa = 5………..II (as A takes 5 seconds less than B to complete l metres)
Let Sa/l = x and Sb/l = y
Then from I and II, we get x-y=1/30 and 1/y -1/x=5
Or xy = (x-y)/5 = 1/150
Now (x+y)2=(x-y)2+4xy= (1/30)2+4(1/150)= 1/36
Or x+y=1/6
Thus Sa/l +Sb/l = 1/6, or (Sa+Sb)/l = 6, Thus they will meet every 6 seconds if they run in opposite directions. So they will meet 10 times in a minute.
Answer : Option 2
Challenge: Understanding the concept of time-speed-distance in circular motion
One step of A is one and a half times as long as one step of B. If B takes twice as many steps per second as A does, and B gives a lead of 400m to A in a 2000m race, then who wins the race and by what distance?
1. A wins by 100m
2. B wins by 100m
3. A wins by 200m
4. B wins by 200m
Topic : Time-Speed-Distance
Approach :Ratio of distance per step of A and B = 3/2 :1 or 3:2
Ratio of Number of steps/second for A and B= 1:2
Thus, ratio of speeds of A and B – (3*1): (2*2) = 3:4
Now B gives A a lead of 400m in a 2000m race, so A has to cover 1600m while B has to cover 2000m
If B wins the race then he will cover 2000m while A in that time will cover 2000*3/4= 1500m
But A already has a lead of 400m, thus A will have covered 1900 m from the start point.
Thus B wins the race by 100m.
Answer : Option 2
Challenge : Application of time-speed-distance to races