# Algebra - Progressions - Previous Year CAT/MBA Questions

You can practice all previous year OMET questions from the topic Algebra - Progressions. This will help you understand the type of questions asked in OMET. It would be best if you clear your concepts before you practice previous year OMET questions.

**XAT 2020 QADI | Algebra - Progressions**

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).

Workspace:

**XAT 2018 QADI | Algebra - Progressions**

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).

Workspace:

**XAT 2017 QADI | Algebra - Progressions**

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).

Workspace:

**IIFT 2017 QA | Algebra - Progressions**

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 1.

Workspace:

**XAT 2016 QADI | Algebra - Progressions**

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).

Workspace:

**IIFT 2016 QA | Algebra - Progressions**

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 3.** **

Workspace:

**IIFT 2016 QA | Algebra - Progressions**

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 4.

Workspace:

**IIFT 2016 QA | Algebra - Progressions**

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 3.

Workspace:

**IIFT 2016 QA | Algebra - Progressions**

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**

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 4.

Workspace:

**IIFT 2015 QA | Algebra - Progressions**

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 3.

Workspace:

**XAT 2015 QA | Algebra - Progressions**

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 2.

Workspace:

**XAT 2015 QA | Algebra - Progressions**

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 3.

Workspace:

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