In the next few articles, we are going to cover some interesting questions from Quant section. Before referring to the solutions, please try to solve these five questions in five minutes. Yes, five minutes is the upper limit to solve these five questions.
Q.1 A graph may be defined as a set of points connected by lines called edges. Every edge contains a pair of points. Thus, a triangle is a graph with 3 edges and 3 points. The degree of a point is the number of edges connected to it. For example, a triangle is a graph with three points of degree 2 each. Consider a graph with 12 points. It is possible to reach any point from any point through a sequence of edges. The number of edges, e, in the graph must satisfy the condition (CAT 2003)
(1) 11 ≤ e ≤ 66
(2) 10 ≤ e ≤ 66
(3) 11 ≤ e ≤ 65
(d) 0 ≤ e ≤ 11
Q.2 If the sum of the first 11 terms of an AP equals that of the first 19 terms, what is the sum of the first 30 terms? (CAT 2004)
(1) 0
(2) -1
(3) 1
(4) none of these
Q.3 Ten years ago, the ages of the members of a joint family of eight people added up to 231 years. Three years later, one member died at the age of 60 years and a child was born during the same year. After another three years, one more member died again at 60, and a child was born during the same year. The current average age of this eight member joint family is nearest to (CAT 2007)
(1) 23 years
(2) 22 years
(3) 21 years
(4) 25 years
(5) 24 years
Q.4 Suppose, the seed of any positive integer n is defined as follows:
seed (n) = n, if n < 10 = seed(s(n)),
Otherwise, where s(n) indicates the sum of digits of n.
For example, seed (7) = 7, seed (248) = seed (2 + 4 + 8) = seed (14) = seed (1 + 4) = seed (5) = 5 etc. How many positive integers n, such that n < 500, will have seed (n) = 9? (CAT 2008)
(1) 39
(2) 72
(3) 81
(4) 108
(5) 55
Q.5 The 288th term of the series a, b, b, c, c, c, d, d, d, d, e, e, e, e, e, is (CAT 2003)
(1) u
(2) v
(3) w
(4) x
These questions can be solved using simple logic and conceptual understanding. We’ll go through each of them now.
Q.1 A graph may be defined as a set of points connected by lines called edges. Every edge contains a pair of points. Thus, a triangle is a graph with 3 edges and 3 points. The degree of a point is the number of edges connected to it. For example, a triangle is a graph with three points of degree 2 each. Consider a graph with 12 points. It is possible to reach any point from any point through a sequence of edges. The number of edges, e, in the graph must satisfy the condition (CAT 2003)
Solution:
The problem is written in a difficult manner to make things complicated. Essentially, we are looking at a closed figure with 12 points and 12 edges. Clearly d is not our answer. We need to connect any two points with a line to create pathways through which any point can be connected to any other point. Hence, we need to draw all the diagonals in this image.
Selecting 2 points from 12 can be done in 12C2 ways = 66 ways.
Alternate approach: Number of diagonals in a polygon = n (n-3)/2 = 12(9)/2 = 54. Remember, as these are only diagonals, we will have to add edges = 54 + 12 = 66. Hence, the answer is 11 ≤ e ≤ 66
Q.2 If the sum of the first 11 terms of an AP equals that of the first 19 terms, what is the sum of the first 30 terms? (CAT 2004)
Solution:
Using normal Sn = n/2 (2a + (n-a)d), we get
22a + 110d = 28a + 342d
2a + 29d = 0
30/2 (2a + 29d) = 0; hence sum of 30 terms is zero.
Alternate approach:
If the first 11 terms are giving sum x and the next 8 terms are y, we can basically conclude that these 8 terms add up to zero.
First 11 terms + a b c d e f g h
Which means effect of a b c d is being nullified by e f g h. hence the last 15 terms will be of one sign and they will nullify the positive effect of the first 15 terms (of opposite sign) giving us zero as the sum. Answer is 0.
Q.3 Ten years ago, the ages of the members of a joint family of eight people added up to 231 years. Three years later, one member died at the age of 60 years and a child was born during the same year. After another three years, one more member died again at 60, and a child was born during the same year. The current average age of this eight member joint family is nearest to (CAT 2007)
Solution:
One way is to solve using numbers. So initial age 231, after three years, 255 – 60 = 195. After three years, 219. 219 – 60 = 159 and 159 + 32 = 191/8 = 24
The other way is to look at averages (which are simpler to add and subtract). Initial average = 231/8 = 29
29 + 3 years = 32
A person of age 60 died making his contribution to overall average zero. 60/8 = 7.5
32 – 7.5 = 24.5
24.5 + 3 years = 27.5
A person of age 60 died making his contribution to overall average zero. 60/8 = 7.5
27.5 – 7.5 = 20
20 + 4 years = 24
Answer = 24 years
Q.4 Suppose, the seed of any positive integer n is defined as follows:
seed (n) = n, if n < 10 = seed(s(n)),
Otherwise, where s(n) indicates the sum of digits of n.
For example, seed (7) = 7, seed (248) = seed (2 + 4 + 8) = seed (14) = seed (1 + 4) = seed (5) = 5 etc. How many positive integers n, such that n < 500, will have seed (n) = 9? (CAT 2008)
Solution:
Essentially, we are taking digit sum till we arrive at a single digit. All numbers divisible by 9, also have 9 as their digit sum. So every multiple of 9, will have seed (n) = 9. In first 500 numbers, there are 55 multiples of 9. Hence, the answer is 55.
Q.5 The 288th term of the series a, b, b, c, c, c, d, d, d, d, e, e, e, e, e, is (CAT 2003)
Solution:
How many terms will be there in this series? 26 * 27 / 2 = 351.
Out of these 351 terms, the last 26 terms will be z. 351 – 26 = 325
Out of these 325 terms, the last 25 terms will be y. 325 – 25 = 300
Out of these 300 terms, the last 24 terms will be x. 300 – 24 = 276
Hence, the answer is x.
Hope you are reading our 75 days to CAT series. Do you know that you can submit a topic and ask us to write an article explaining it? That’s how we covered cauchy-schwarz-inequality and chicken mcnugget theorem.
You can also practice CAT level questions in a test environment on our website. Have a look.
