Algebra - Progressions - Previous Year CAT/MBA Questions
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Each question is followed by two statements, A and B. Answer each question using the following instructions
Choose 1 if the question can be answered by using one of the statements alone but not by using the other statement alone.
Choose 2 if the question can be answered by using either of the statements alone.
Choose 3 if the question can be answered by using both statements together but not by either statement alone.
Choose 4 if the question cannot be answered on the basis of the two statements.
Is
A. −3 ≤ a ≤ 3
B. One of the roots of the equation 4x2 − 4x + 1 = 0 is a
- (a)
1
- (b)
2
- (c)
3
- (d)
4
Answer: Option A
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Text Explanation :
R.H.S. and L.H.S. are infinite GP's with common ratio = .
∴ We need to find whether
...(i)
This statement (i) is always true for a < 1.
Using statement A alone:
It does not give any conclusion regarding value of a i.e. whether it is less than 1 or not.
So, statement A alone is not sufficient to answer the question.
Using statement B alone:
4a2 − 4a + 1 = 0 …(ii)
∴ a = 1/2
So, statement B alone is sufficient to answer the question.
Hence, option (a).
Workspace:
The 288th term of the series a, b, b, c, c, c, d, d, d, d, e, e, e, e, e, f, f, f, f, f, f... is
- (a)
u
- (b)
v
- (c)
w
- (d)
x
Answer: Option D
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Text Explanation :
In the series a, b, b, c, c, c, d, d, d, d, e, e, e, e, e,...
the first letter of the alphabet is written once, the second is written twice, and the nth letter is written n times.
∴ The number of letters written upto the nth letter is equal to the sum of the first n natural numbers given by, n(n + 1)/2
For n = 23, n(n + 1)/2 = 276 and for n = 24, n(n + 1)/2 = 300
This means the series contains 276 letters in all for the letter corresponding to n = 23 and 300 letters in all for the letter corresponding to n = 24.
∴ The letter corresponding to n = 24 will be the letter occupying the 277th to the 300th place in the series.
But, n = 24 corresponds to letter x.
∴ The 288th letter in the series is x.
Hence, option (d).
Workspace:
If the product of n positive real numbers is unity, then their sum is necessarily
- (a)
a multiple of n
- (b)
equal to n + 1/n
- (c)
never less than n
- (d)
a positive integer
Answer: Option C
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Text Explanation :
Let a1, a2, a3,…, an be n positive real numbers.
Now, a1 × a2 × a3 × … × an = 1
We know that, A.M. ≥ G.M.
a1 + a2 + a3 + … + an ≥ n × (a1 × a2 × a3 × … × an)1/n
∴ a1 + a2 + a3 + … + an ≥ n × (1)1/n
∴ a1 + a2 + a3 + … + an ≥ n
Hence, option (c).
Workspace:
If log3 2, log3 (2x − 5), log3 (2x − 7/2) are in arithmetic progression, then the value of x is equal to
- (a)
5
- (b)
4
- (c)
2
- (d)
3
Answer: Option D
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Text Explanation :
log3 2, log3 (2x − 5), log3 (2x − 7/2) are in A.P.
∴ 2 × log3 (2x − 5) = log3 2 + log3 (2x − 7/2)
∴ log3 (2x − 5)2 = log3 [2 × (2x − 7/2)]
Let 2x = a, then we have,
(a − 5)2 = 2 × (a − 7/2)
∴ a2 − 10a + 25 = 2a − 7
∴ a2 − 12a + 32 = 0
∴ a2 − 8a − 4a + 32 = 0
(a − 8)(a − 4) = 0
a = 8 or 4
2x = 8 or 2x = 4
x = 3 and x = 2
x = 2 cannot be the answer as (2x − 5) would become negative and logarithms of negative numbers are not defined.
∴ x = 3
Hence, option (d).
Workspace:
In a certain examination paper, there are n questions. For j = 1, 2, ..., n, there are 2(n − j) students who answered j or more questions wrongly. If the total number of wrong answers is 4095, then the value of n is
- (a)
12
- (b)
11
- (c)
10
- (d)
9
Answer: Option A
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Text Explanation :
For j = n, 2(n - n) = 1 student has answered wrongly
j = n - 1, 2[n - (n - 1)] = 2 students have answered wrongly
.
.
j = 1, 2(n - 1) students have answered wrongly
∴ 1 + 2 + ... + 2(n – 1) = 4095
The series is in G.P. with common ratio r = 2
∴ 1 × (2n − 1)/(2 − 1) = 4095
∴ 2n − 1 = 4095
∴ 2n = 4096
∴ n = 12
Hence, option (a).
Workspace:
The infinite sum equals
- (a)
- (b)
- (c)
- (d)
Answer: Option C
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Text Explanation :
Let S = (i)
Dividing (i) by 7, we get,
(ii)
Subtracting (ii) from (i)
(iii)
Dividing (iii) by 7
(iv)
Subtracting (iv) from (iii)
Hence, option (c).
Workspace:
On a straight road XY, 100 metres in length, 5 stones are kept beginning from the end X. The distance between two adjacent stones is 2 metres. A man is asked to collect the stones one at a time and put at the end Y. What is the distance covered by him?
- (a)
460 metres
- (b)
540 metres
- (c)
860 metres
- (d)
920 metres
Answer: Option C
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Text Explanation :
The first stone is kept at 'X' and the subsequent stones are equidistant from each other, i.e. 2 metres.
∴ The total distance covered by the man to keep all the stones at Y = 100 + 2 × (98 + 96 + 94 + 92) = 100 + 2 × (380) = 860 metres
Hence, option (c).
Workspace:
Let S = 2x + 5x2 + 9x3 + 14x4 + 20x5 ... infinity (x < 1)
The coefficient of nth term = The sum is
- (a)
- (b)
- (c)
- (d)
None of these
Answer: Option A
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Text Explanation :
S = 2x + 5x2 + 9x3 + 14x4 + 20x5 + ... …(i)
Multiplying both sides by x, we get
xS = 2x2 + 5x3 + 9x4 + 14x5 + 20x6 + ... ...(ii)
Subtracting (ii) from (i),
(1 – x)S = 2x + 3x2 + 4x3 + 5x4 + 6x5 + ... ...(iii)
Again multiplying both the sides by x, we get
x(1 – x)S = 2x2 + 3x3 + 4x4 + 5x5 + 6x6 + ... ...(iv)
Subtracting (iv) from (iii),
(1 – x)S – x(1 – x)S = 2x + x2 + x3 + x4 + x5 + ...
(1 – x)2S = 2x + x2 + x3 + x4 + x5 +...
(1 – x)2S = x + (x + x2 + x3 + ... + ∞)
(1 - x)2S = x + ∵ S∞ = for r < 1
∴ (1 - x)2S =
∴ S =
Hence, option (a).
Workspace:
Sum of first n natural numbers = S(n)
Sum given by student = 575
S(10) = 55
S(20) = 210
S(30) = 465
S(40) = 820
∴ The student stopped counting somewhere between 30 and 40.
Consider S(35) = 630
The student stopped somewhere before 35.
∴ S(31) = 496, S(32) = 528, S(33) = 561 and S(34) = 595
But the student gave 575 as the sum, so the student missed on the number 20.
Hence, option 4.
A student finds the sum 1 + 2 + 3 + ... as his patience runs out. He found the sum as 575. When the teacher declared the result wrong, the student realized that he missed a number. What was the number the student missed?
- (a)
16
- (b)
18
- (c)
14
- (d)
20
Answer: Option D
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Text Explanation :
Sum of first n natural numbers = S(n)
Sum given by student = 575
S(10) = = 55
S(20) = = 210
S(30) = = 465
S(40) = = 820
∴ The student stopped counting somewhere between 30 and 40.
Consider S(35) = = 630
The student stopped somewhere before 35.
∴ S(31) = 496, S(32) = 528, S(33) = 561 and S(34) = 595
But the student gave 575 as the sum, so the student missed on the number 20.
Hence, option (d).
Workspace:
Two men X and Y started working for a certain company at similar jobs on January 1, 1950. X asked for an initial salary of Rs. 300 with an annual increment of Rs. 30. Y asked for an initial salary of Rs. 200 with a rise of Rs. 15 every six months. Assume that the arrangements remained unaltered till December, 1959. Salary is paid on the last day of the month. What is the total amount paid to them as salary during the period?
- (a)
Rs. 93,300
- (b)
Rs. 93,200
- (c)
Rs. 93,100
- (d)
None of these
Answer: Option A
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Text Explanation :
X’s salary = [3600] + [3600 + 30 × 12] + [3600 + 30 × 12 × 2] + … + [3600 + 30 × 12 × 9]
= [3600 × 10] + [30 × 12(1 + 2 + 3 + … + 9)]
= Rs. 52,200
Y’s salary = [1200 × 20] + [15 × 6(1 + 2 + 3 + … + 19)]
= Rs. 41,100
∴ Sum of X and Y’s salary = Rs. 52,200 + Rs. 41,100 = Rs. 93,300
Hence, option (a).
Alternatively,
X starts at a salary of 300 and ends at a salary = 300 + 30 × 9 = 570
∴ X's average salary =
Y starts at a salary of 200 and ends at a salary = 200 + 15 × 19 = 485
∴ Y's average salary =
∴ X and Y’s total salary = (435 + 342.5) × 12 × 10 = Rs. 93,300
Hence, option (a).
Workspace:
All the page numbers from a book are added, beginning at page 1. However, one page number was mistakenly added twice. The sum obtained was 1000. Which page number was added twice?
- (a)
44
- (b)
45
- (c)
10
- (d)
12
Answer: Option C
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Text Explanation :
First we have to find the total number of pages in the book. For this we need to find the sum of the first few natural numbers, such that we reach to a number just short of 1000.
By trial and error, we can find that,
Sum of the first 44 natural numbers = = 90
∵ 990 + 10 = 1000
∴ The page number added twice was 10.
Hence, option (c).
Workspace:
For a Fibonacci sequence, from the third term onwards, each term in the sequence is the sum of the previous two terms in that sequence. If the difference in squares of seventh and sixth terms of this sequence is 517, what is the tenth term of this sequence?
- (a)
147
- (b)
76
- (c)
123
- (d)
Cannot be determined
Answer: Option C
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Text Explanation :
Let the nth term of the series be Fn.
∵ (F7)2 – (F6)2 = 517
∴ (F7 + F6)(F7 – F6) = 517
∴ (F7 + F6) (F7 – F6) = 11 × 47
∴ F8 × (F6 + F5 – F6) = 11 × 47
∴ F8 × F5 = 11 × 47
∴ F8 = 47 and F5 = 11
∴F8 = F7 + F6 = 2F6 + F5 = 3F5 + 2F4
∴ F4 = 7
Now the series F4 onwards is: 7, 11, 18, 29, 47, 76, 123
∴ The 10th term will be 123.
Hence, option (c).
Workspace:
If a1 = 1 and an+1 = 2an + 5, n = 1, 2 ... , then a100 is equal to
- (a)
(5 × 299 – 6)
- (b)
(5 × 299 + 6)
- (c)
(6 × 299 + 5)
- (d)
(6 × 299 – 5)
Answer: Option D
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Text Explanation :
a1 = 1 = 6 – 5
a2 = 7 = 12 – 5 = 6 × 2 – 5
a3 = 19 = 24 – 5 = 6 × 22 – 5
a4 = 43 = 48 – 5 = 6 × 23 – 5
and so on.
Thus, a100 = 6 × 299 – 5
Hence, option (d).
Workspace:
What is the value of the following expression?
+ + + ... +
- (a)
- (b)
- (c)
- (d)
Answer: Option C
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Text Explanation :
Hence, option (c).
Workspace:
Directions: Answer the questions based on the following information.
There are 50 integers a1, a2 … a50, not all of them necessarily different. Let the greatest integer of these 50 integers be referred to as G, and the smallest integer be referred to as L. The integers a1 through a24 form sequence S1, and the rest form sequence S2. Each member of S1 is less than or equal to each member of S2.
All values in S1 are changed in sign, while those in S2 remain unchanged. Which of the following statements is true?
- (a)
Every member of S1 is greater than or equal to every member of S2.
- (b)
G is in S1.
- (c)
If all numbers originally in S1 and S2 had the same sign, then after the change of sign, the largest number of S1 and S2 is in S1.
- (d)
None of the above
Answer: Option D
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Text Explanation :
The ideal approach is to pick up the options one by one.
Option (a) – Let S1 and S2 be two sequences of positive numbers. After change of sign, S1 will consist of negative numbers while S2 remains unchanged.
Clearly, the members of S1 would be less than that of S2.
Hence, option (a) is not correct.
Option (b) - Let S1 and S2 be two sequences of positive numbers. After change of sign, S1 will consist of negative numbers while S2 remains unchanged.
Clearly, G would remain in S2 itself. Hence, option (b) is not correct.
Option (c) – If S1 and S2 had same sign, say positive initially, then the largest number of S1 and S2 would be in S2. Then after the change of sign, every member of S1 will be negative and therefore, less than every member of S2. This implies that the largest number would remain in S2.
Hence, option (c) is not correct.
Workspace:
Elements of S1 are in ascending order, and those of S2 are in descending order. a24 and a25 are interchanged. Then which of the following statements is true?
- (a)
S1 continues to be in ascending order.
- (b)
S2 continues to be in descending order.
- (c)
S1 continues to be in ascending order and S2 in descending order.
- (d)
None of the above
Answer: Option A
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Text Explanation :
The elements of S1 are in the order : a1 < a2 < a3 < a4 < . . . < a24
The elements of S2 are in the order: a25 > a26 > . . . > a49 > a50
Even if a24 and a25 are interchanged, the elements of S1 continues to be in ascending order. However, nothing can be concluded about the elements of S2.
Workspace:
Every element of S1 is made greater than or equal to every element of S2 by adding to each element of S1 an integer x. Then x cannot be less than
- (a)
210
- (b)
the smallest value of S2
- (c)
the largest value of S2
- (d)
(G – L)
Answer: Option D
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Text Explanation :
Since every element of S1 is less than or equal to each member of S2, L will be in S1 and G in S2.
For some i (1≤ i ≤ 24), ai = L and for some j (25 ≤ j ≤ 50), aj = G.
Every other element of S1 is greater than ai and every other member of S2 is less than aj.
Therefore, to make every element of S1 greater than or equal to that of S2, we need to add a minimum of (aj –ai) = G – L.
Workspace:
Answer the questions based on the following information.
A series S1 of five positive integers is such that the third term is half the first term and the fifth term is 20 more than the first term. In series S2, the nth term defined as the difference between the (n + 1)th term and the nth term of series S1, is an arithmetic progression with a common difference of 30.
First term of S1 is
- (a)
80
- (b)
90
- (c)
100
- (d)
120
Answer: Option C
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Text Explanation :
First series: (S1) = x, y, z, x + 20
Second series: (S2) = a1, a2, a3, a4
Now a1 = y – x, a2 = - y, a3 = z - and a4 = x + 20 – z
a2 – a1 = 30 gives 3x – 4y = 60 ... (1)
a4 – a3 = 30 gives 3x – 4z = 20 ... (2)
and a4 – a2 = 60 gives x – 2z + 2y = 80 ... (3)
Solving these equations we get the values of x = 100, y = 60, z = 70
∴ S1 is 100, 60, 50, 70, 120
S2 is –40, –10, 20, 50.
Hence, option (c).
Workspace:
Second term of S2 is
- (a)
50
- (b)
60
- (c)
70
- (d)
None of these
Answer: Option D
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Text Explanation :
Consider the solution to first question of this set.
Second term of S2 = -10.
Hence, option (d).
Workspace:
What is the difference between second and fourth terms of S1?
- (a)
10
- (b)
20
- (c)
30
- (d)
60
Answer: Option A
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Text Explanation :
Consider the solution to first question of this set.
Difference between second and fourth term of S1 = 70 - 60 = 10
Hence, option (a).
Workspace:
What is the average value of the terms of series S1?
- (a)
60
- (b)
70
- (c)
80
- (d)
Average is not an integer
Answer: Option C
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Text Explanation :
Consider the solution to first question of this set.
Average of S1 = (100 + 60 + 50 + 70 + 120)/5 = 80.
Hence, option (c).
Workspace:
What is the sum of series S2?
- (a)
10
- (b)
20
- (c)
30
- (d)
40
Answer: Option B
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Text Explanation :
Consider the solution to first question of this set.
Sum of S2 = (-40) + (-10) + 20 + 50 = 20.
Hence, option (b).
Workspace:
If the harmonic mean between two positive numbers is to their geometric mean as 12 : 13; then the numbers could be in the ratio
- (a)
12 : 13
- (b)
1/12 : 1/13
- (c)
4 : 9
- (d)
2 : 3
Answer: Option C
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Text Explanation :
The harmonic mean of two numbers x and y is and the geometric mean is
∴ =
⇒ =
Although this can be simplified to get the answer, the best way to proceed from here would be to look out for the answer choices and figure out which pair of x & y satisfies the above equation.
Option (c) satisfies the above expression
Alternately,
Please note that this sum is a classic example of how you could have gone for intelligent guess work. Since we know that the denominator of the ratio is the geometric mean, which is , the two numbers should be in such a ratio that their product should be a perfect square.
The only pair from the answer choices that supports this is 4 & 9, as = = 6
Hence, option (c).
Workspace:
Fourth term of an arithmetic progression is 8. What is the sum of the first 7 terms of the arithmetic progression?
- (a)
7
- (b)
64
- (c)
56
- (d)
Cannot be determined
Answer: Option C
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Text Explanation :
Middle term of an A.P. is average of all the terms in A.P.
Number of terms = 7
Average of all these 7 terms = Middle term = Fourth term = 8
Therefore, sum of all the terms = 7 × 8 = 56.
Hence, option (c).
Workspace:
Along a road lie an odd number of stones placed at intervals of 10m. These stones have to be assembled around the middle stone. A person can carry only one stone at a time. A man carried out the job starting with the stone in the middle, carrying stones in succession, thereby covering a distance of 4.8 km. Then the number of stones is
- (a)
35
- (b)
15
- (c)
29
- (d)
31
Answer: Option D
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Text Explanation :
As the person covers 4.8 km, he covers 2.4 kms on one side and 2.4 kms on other side.
Let the number of stones be ‘n’ on either side of the road excluding the middle stone.
So total distance covered by him on one side = 20 + 40 + 60 + … + 20n
∴ 2400 = (2 × 20 + (n - 1)20) = 10n (n + 1)
(Here n is the number of stones)
After solving, we get n = 15
∴ Total number of stones = 15 + 15 + 1 = 31
Hence, option (d).
Workspace:
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