# Algebra - Progressions - Previous Year CAT/MBA Questions

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**XAT 2023 QADI | Algebra - Progressions omet Question**

Suppose Haruka has a special key ∆ in her calculator called delta key:

Rule 1: If the display shows a one-digit number, pressing delta key ∆ replaces the displayed number with twice its value.

Rule 2: If the display shows a two-digit number, pressing delta key ∆ replaces the displayed number with the sum of the two digits.

Suppose Haruka enters the value 1 and then presses delta key ∆ repeatedly.

After pressing the ∆ key for 68 times, what will be the displayed number?

- (a)
7

- (b)
4

- (c)
10

- (d)
2

- (e)
8

Answer: Option A

**Explanation** :

Initially the number = 1

Output when ∆ is pressed 1 time = 2 [Single digit number 1 is doubled]

Output when ∆ is pressed 2 times = 4 [Single digit number 2 is doubled]

Output when ∆ is pressed 3 times = 8 [Single digit number 4 is doubled]

Output when ∆ is pressed 4 times = 16 [Single digit number 8 is doubled]

Output when ∆ is pressed 5 times = 7 [Digits of 2-digit number 16 are added]

Output when ∆ is pressed 6 times = 14 [Single digit number 7 is doubled]

Output when ∆ is pressed 7 times = 5 [Digits of 2-digit number 14 are added]

Output when ∆ is pressed 8 times = 10 [Single digit number 5 is doubled]

Output when ∆ is pressed 9 times = 1 [Digits of 2-digit number 10 are added]

After 9 steps we get back the same input 1.

Hence, when the input is 1, we will get the same output at the end of 9 or 18 or ... or 63 steps.

∴ Output at the end of 68 steps will be same as the output at the end of 5 steps i.e., 7.

Hence, option (a).

Workspace:

**XAT 2023 QADI | Algebra - Progressions omet Question**

Consider a_{n + 1} = $\frac{1}{1+{\displaystyle \frac{1}{{a}_{n}}}}$ for n = 1, 2, ...., 2008, 2009 where a_{1} = 1. Find the value of a_{1} a_{2 }+ a_{2} a_{3 }+ a_{3} a_{4 }+ ... + a_{2008} a_{2009 }

- (a)
_{$\frac{2009}{1000}$} - (b)
_{$\frac{2009}{2008}$} - (c)
_{$\frac{2008}{2009}$} - (d)
_{$\frac{6000}{2009}$} - (e)
_{$\frac{2008}{6000}$}

Answer: Option C

**Explanation** :

Given, a_{n + 1} = $\frac{1}{1+{\displaystyle \frac{1}{{a}_{n}}}}$

⇒ a_{1} = 1

⇒ a_{2} = $\frac{1}{1+{\displaystyle \frac{1}{{a}_{1}}}}$ = $\frac{1}{2}$

⇒ a_{3} = $\frac{1}{1+{\displaystyle \frac{1}{{a}_{2}}}}$ = $\frac{1}{3}$

...

⇒ a_{n} = $\frac{1}{n}$

Now: a_{1} a_{2 }+ a_{2} a_{3 }+ a_{3} a_{4 }+ ... + a_{2008} a_{2009} = $\frac{1}{1\times 2}$ + $\frac{1}{2\times 3}$ + $\frac{1}{3\times 4}$ + ... + $\frac{1}{2008\times 2009}$

= $\left(\frac{1}{1}-\frac{1}{2}\right)$ + $\left(\frac{1}{2}-\frac{1}{3}\right)$ + $\left(\frac{1}{3}-\frac{1}{4}\right)$ + ... + $\left(\frac{1}{2008}-\frac{1}{2009}\right)$

= $\frac{1}{1}-\frac{1}{2009}$ = $\frac{2008}{2009}$

Hence, option (c).

Workspace:

**XAT 2022 QADI | Algebra - Progressions omet Question**

A marble is dropped from a height of 3 metres onto the ground. After the hitting theground, it bounces and reaches 80% of the height from which it was dropped. This repeats multiple times. Each time it bounces, the marble reaches 80% of the height previously reached. Eventually, the marble comes to rest on the ground.

What is the maximum distance that the marble travels from the time it was dropped until it comes to rest?

- (a)
15 m

- (b)
27 m

- (c)
24 m

- (d)
12 m

- (e)
30 m

Answer: Option B

**Explanation** :

Given the ball falls from a height of 3 meters.

The ball reaches a height which 0.8 times the original height every time.

Hence this is in the form of a geometric progression. We need to count distance when the ball flies upward

and downward.

Hence considering every time the ball flies upward to a series with terms :

h1, h2,..........................

Every time the ball falls down to be

d1, d2 ,...............

h1 = (0.8)*3, h2 = (0.8)*(0.8)*3 ,.........................

d1 = 3, d2 = 3*(0.8), d3 = 3*(0.8)*(0.8)................

h1 + h2 ........ = Sum of an infinite geometric progression. = 3 * 0.8(1 + 0.8 + 0.64 + ......)

The sum of an infinite GP with r less than 1 is:

$\frac{a}{1-r}$

= 2.4 ∙ $\left(\frac{1}{1-0.8}\right)$ = 12 meters

The sum of d1 + d2 + + ........................

= 3 + (h1 + h2 + ..................) = 15.

The total distance = 15 + 12 = 27 meters

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**XAT 2022 QADI | Algebra - Progressions omet Question**

Shireen draws a circle in her courtyard. She then measures the circle’s circumference and its diameter with her measuring tape and records them as two integers, A and B respectively. She finds that A and B are coprimes, that is, their greatest common divisor is 1. She also finds their ratio, A : B, to be: 3.141614161416… (repeating endlessly).

What is A - B?

- (a)
21414

- (b)
7138

- (c)
21417

- (d)
21413

- (e)
15

Answer: Option B

**Explanation** :

We have

3.1416141614161416......

= 3 + 0.1416 + 0.00001416 + 0.000000001416.....

= 3 + $\frac{1416}{{10}^{4}}$ + $\frac{1416}{{10}^{8}}$ + $\frac{1416}{{10}^{12}}$ + ....

Now excluding 3 we get a series with infinite Geometric progression such that first term is $\frac{1416}{{10}^{4}}$ and common ratio is $\frac{1}{{10}^{4}}$

Therefore we get sum as $\frac{a}{1-r}$

= $\frac{{\displaystyle \frac{1416}{{10}^{4}}}}{1-{\displaystyle \frac{1}{{10}^{4}}}}$

we get sum as $\frac{1416}{{10}^{4}-1}$

= $\frac{1416}{9999}$

Now adding 3 we get value as :

$\frac{31413}{9999}$

Now adding 3 we get value A : B as :

$\frac{10471}{3333}$

So A - B will be = 7138

Hence, option (b).

Workspace:

**XAT 2022 QADI | Algebra - Progressions omet Question**

If both the sequences x, a1, a2, y and x, b1, b2, z are in A.P. and it is given that y > x and z < x, then which of the following values can $\left\{\frac{(a1-a2)}{(b1-b2)}\right\}$ possibly take?

- (a)
2

- (b)
5

- (c)
-3

- (d)
1

- (e)
0

Answer: Option C

**Explanation** :

The two given sequences in AP are:

x, a1, a2, y and x, b1, b2, z.

Additionally, it is given that : y > x and z < x.

Hence the common difference is not zero for both the series :

Since y > x the common difference is positive for the first series. (Considering the common difference to be d1)

Similarly z < x the common difference is negative for the given series. (Considering the common difference to be d2)

Now for the given value:

$\frac{(a1-a2)}{(b1-b2)}$

The value of a1 - a2 is negative and b1 - b2 is positive.

Hence the value of $\frac{(a1-a2)}{(b1-b2)}$ takes a negative value.

The only possible option is -3.

The answer is option C.

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**XAT 2020 QADI | Algebra - Progressions omet Question**

A man is laying stones, from start to end, along the two sides of a 200-meter-walkway. The stones are to be laid 5 meters apart from each other. When he begins, all the stones are present at the start of the walkway. He places the first stone on each side at the walkway’s start. For all the other stones, the man lays the stones first along one of the walkway’s sides, then along the other side in an exactly similar fashion. However, he can carry only one stone at a time. To lay each stone, the man walks to the spot, lays the stone, and then walks back to pick another. After laying all the stones, the man walks back to the start, which marks the end of his work. What is the total distance that the man walks in executing this work? Assume that the width of the walkway is negligible.

- (a)
16,200 meters

- (b)
8,200 meters

- (c)
8,050 meters

- (d)
16,400 meters

- (e)
4,100 meters

Answer: Option D

**Explanation** :

On one side:

To place 1^{st} rock, he had to travel 10 m.

For 2nd rock he had to travel 20 m…,

Similarly, till last rock he had to travel 400 m.

Total sum would be 10 + 20 + 30 + ... + 400 = 40/2 × 410 = 8200 m.

Similarly, on the other side he will travel 8200 m.

Total 8200 + 8200 = 16400.

Hence, option (d).

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**XAT 2018 QADI | Algebra - Progressions omet Question**

An antique store has a collection of eight clocks. At a particular moment, the displayed times on seven of the eight clocks were as follows: 1:55 pm, 2:03 pm, 2:11 pm, 2:24 pm, 2:45 pm, 3:19 pm and 4:14 pm. If the displayed times of all eight clocks form a mathematical series, then what was the displayed time on the remaining clock?

- (a)
1:53 pm

- (b)
1:58 pm

- (c)
2:18 pm

- (d)
3:08 pm

- (e)
5:08 pm

Answer: Option B

**Explanation** :

Observe the following times.

Thus, we see that “5” is missing in the first difference. Therefore, the missing time is 5 minutes before 2.03, that is, 1.58.

Hence, option (b).

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**XAT 2017 QADI | Algebra - Progressions omet Question**

The sum of the series, (-100) + (-95) + (-90) + … + 110 + 115 + 120, is

- (a)
0

- (b)
220

- (c)
340

- (d)
450

- (e)
None of the above

Answer: Option D

**Explanation** :

(–100) + (-95) + ….. + 120

= [(–100) + (-95) + ….. + 95 + 100] + 105 + 110 + 115 + 120

= 105 + 110 + 115 + 120

= 450

Hence, option (d).

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**IIFT 2017 QA | Algebra - Progressions omet Question**

In a certain sequence the term x_{n} is given by formula ${x}_{n}=5{x}_{n-1}-\frac{3}{4}{x}_{n-\mathrm{2\; for\; n\; \ge \; 2.}}$

What is the value of x_{3} if x_{0} = 4 and x_{1} = 2?

- (a)
67/2

- (b)
37/2

- (c)
123/4

- (d)
None

Answer: Option A

**Explanation** :

*x*_{2} = 5*x*_{1} − (3/4)(*x*_{0}) = 5(2) − (3/4)(4) = 7

∴ *x*_{3} = 5*x*_{2} − (3/4)(*x*_{1}) = 5(7) − (3/4)(2) = 67/2

Hence, option (a).

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**XAT 2016 QADI | Algebra - Progressions omet Question**

Consider the set of numbers {1, 3, 3^{2}, 3^{3},…...,3^{100}}. The ratio of the last number and the sum of the remaining numbers is closest to:

- (a)
1

- (b)
2

- (c)
3

- (d)
50

- (e)
99

Answer: Option B

**Explanation** :

The last number is 3^{100}.

Sum of the remaining numbers = S = 1 + 3 + 3^{2} + … + 3^{99} …(1)

This is a GP with a = 1; r = 3 and n = 100

∴ S = $\frac{1\times ({3}^{100}-1)}{3-1}$

∴ Required ratio = $\hspace{0.17em}\frac{{3}^{100}}{\frac{1\times ({3}^{100}-1)}{2}}$ = $\hspace{0.17em}\frac{2}{1-{\displaystyle \frac{1}{{3}^{100}}}}$

∵ 1/3^{100} is approximately zero, hence it can be ignored.

∴ Ratio = 2

Hence, option (b).

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**IIFT 2016 QA | Algebra - Progressions omet Question**

A child, playing at the balcony of his multi-storied apartment, drops a ball from a height of 350 m. Each time the ball rebounds, it rises 4/5th of the height it has fallen through. The total distance travelled by the ball before it comes to rest is

- (a)
2530 m

- (b)
2800 m

- (c)
3150 m

- (d)
3500 m

Answer: Option C

**Explanation** :

Distance covered by ball when it touches ground for the first time = 350 m.

Distance travelled when it touches for the second time = 2 × (4/5) × 350 m

Distance travelled when it touches for the third time = 2 × (4/5) × (4/5) × 350 m

∴ Total distance = 350 + [2 × (4/5) × 350] + [2 × (4/5) × (4/5) × 350] + …= 350 + [2 × (4/5) × 350][(1 + (4/5) + (4/5)^{2} + …]

The last term above is an infinite G.P. with *a* = 1 and *r* = 4/5

The sum of such an infinite G.P. = *a*/(1 – *r*) = 1/([1 – (4/5)] = 1/(1/5) = 5

∴ Total distance = 350 + [2 × (4/5) × 350](5) = 350 + 8(350) = 9(350) = 3150 m

Hence, option (c).** **

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**IIFT 2016 QA | Algebra - Progressions omet Question**

What is the sum of integers 54 through 196 inclusive?

- (a)
28,820

- (b)
24,535

- (c)
20,250

- (d)
17,875

Answer: Option D

**Explanation** :

Sum of integers from 54 to 196 = (sum of integers from 1 to 196)– (sum of integers from 1 to 53)

Sum from 1 to 196 = (196)(197)/2 = 98 × 197 = 19306

Observe that this value is already less than three out of four options. Hence, it will further decrease when the sum from 1 to 53 is subtracted from it.

Hence, the answer should be less than 19306. Only 17875 satisfies this condition.

Hence, option (d).

Workspace:

**IIFT 2016 QA | Algebra - Progressions omet Question**

A multi-storied office building has a total of 17 rows of parking spaces. There are 20 parking spaces in the first row and 21 parking spaces in the second row. In each subsequent row, there are 2 more parking spaces than in the previous row. The total number of parking spaces in the office building is

- (a)
380

- (b)
464

- (c)
596

- (d)
712

Answer: Option C

**Explanation** :

Observe that the last 16 rows form an A.P. with a = 21, d = 2 and n = 16.

The term before that is 20.

Hence, sum = (19 + 1) + A.P. of 16 terms

Hence, this can even be considered as an A.P. of 17 terms with a = 19 and d = 2

∴ Total parking spaces = 1 + (17/2) × [2(19) + 16(2)] = 1 + 17(19 + 16) = 596

Hence, option (c).

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**IIFT 2016 QA | Algebra - Progressions omet Question**

The sum of 4 + 44 + 444 + …. upto n terms in

- (a)
$\frac{40}{81}({8}^{n}-1)-\frac{5n}{9}$

- (b)
$\frac{40}{81}({8}^{n}-1)-\frac{4n}{9}$

- (c)
$\frac{40}{81}({10}^{n}-1)-\frac{4n}{9}$

- (d)
$\frac{40}{81}({10}^{n}-1)-\frac{5n}{9}$

Answer: Option C

**Explanation** :

A simple way to solve this is to take the value of n as a suitable integer and tally the options with the sum.

e.g. when n = 1, sum = 4

Only the expression in option 3 gives this value.

Hence,option 3.

**Note:**

You can verify option 3 by putting n = 2.

Workspace:

**IIFT 2015 QA | Algebra - Progressions omet Question**

If p, q and r are three unequal numbers such that p, q and r are in A.P., and p, r-q and q-p are in G.P., then p : q : r is equal to:

- (a)
1 : 2 : 3

- (b)
2 : 3 : 4

- (c)
3 : 2 : 1

- (d)
1 : 3 : 4

Answer: Option A

**Explanation** :

Without loss of generality, p < q < r

∴ r – q = q – p = d

p, r – q and q – p are in G.P.

i.e., p, d, d are in G. P.

So, p = d

∴ q = p + d = 2d and r = 3d

∴ p : q : r = 1 : 2 : 3

Hence, option (d).

Workspace:

**IIFT 2015 QA | Algebra - Progressions omet Question**

Seema has joined a new Company after the completion of her B.Tech from a reputed engineering college in Chennai. She saves 10% of her income in each of the first three months of her service and for every subsequent month, her savings are Rs. 50 more than the savings of the immediate previous month. If her joining income was Rs. 3000, her total savings from the start of the service will be Rs.11400 in:

- (a)
6 months

- (b)
12 months

- (c)
18 months

- (d)
24 months

Answer: Option C

**Explanation** :

Seema saves Rs. 900 in first three months.

Let she reach the given amount in X more months.

She would save 300X + 50 + 50 × 2 + 50 × 3 +......... + 50 × X = 11400 − 900

∴ 300X + 50(1 + 2 + 3 +......+ X) = 10500

∴ *X*^{2 }+ 13*X* - 420 = 0

On solving, we get X = 15.

Thus, in 15 + 3 = 18 months her savings will be Rs. 11,400.

Hence, option (c).

Workspace:

**XAT 2015 QA | Algebra - Progressions omet Question**

What is the sum of the following series?

–64, –66, –68,……..….., –100

- (a)
-1458

- (b)
-1558

- (c)
-1568

- (d)
-1664

- (e)
None of the above

Answer: Option B

**Explanation** :

Number of terms = (100 – 64)/2 + 1 = 19

Sum = *n*/2 (*a* + *a _{n}*)

= 19/2 × (–64 –100) = –1558

Hence, option (b).

Workspace:

**XAT 2015 QA | Algebra - Progressions omet Question**

An ascending series of numbers satisfies the following conditions:

- When divided by 3, 4, 5 or 6, the numbers leave a remainder of 2.
- When divided by 11, the numbers leave no remainder.

The 6th number in this series will be:

- (a)
242

- (b)
2882

- (c)
3542

- (d)
4202

- (e)
None of the above

Answer: Option C

**Explanation** :

LCM (3, 4, 5, 6) = 60

From (i), numbers are of the form 60*k* + 2.

From (ii), numbers are of the form 11*m*.

11*m* = 60*k* + 2 = 55*k* + (5*k* + 2)

11 divides (5*k* + 2)

Units digit of (5*k* + 2) is 2 or 7

So, values of (5*k* + 2) are 11 × 2, 11 × 7, 11 × 12, 11 × 17, 11 × 22, 11 × 27, … and so on.

We need to find the 6th number of the ascending series.

∴ 5*k* + 2 = 11 × 27 ⇒ *k* = 59

The required number = 60 × 59 + 2 = 3542

Hence, option (c).

Workspace:

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