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Progression - Online Quiz

Following quiz provides Multiple Choice Questions (MCQs) related to Progression. You will have to read all the given answers and click over the correct answer. If you are not sure about the answer then you can check the answer using Show Answer button. You can use Next Quiz button to check new set of questions in the quiz.

Q 1 - 105 th term of the A.P. 4, 9/2, 5, 11/2, 6 ....is

Answer : A

Explanation

 Here a = 4, d = (9/2-4) = 1/2
T₁0₅ = a+(105-1)*d=4+104*1/2=4+52=56.

Q 2 - What number of 3-digit numbers is distinguishable by 6?

Answer : B

Explanation

Requisite numbers are 102, 108, 114, 120,..., 996.
This is an A.P. in which a = 102 and d = (108-102) = 6
a + (n-1) d = 996    ⇒   102+ (n-1)*6 = 996   ⇒ (n-1)*6 =894    ⇒ (n-1) = 149    ⇒ n = 150

Q 3 - What number between 200 and 600 are distinct by 4, 5, 6?

Answer : B

Explanation

LCM of 4, 5, 6 = 2*2*5*3 = 60
So, each one must be divisible by 60.
Requisite numbers are 240, 300, 360, 420, 480 and 540.
Their no. is 6.

Q 4 - The total of the all odd number somewhere around 100 and 200 is:

Answer : D

Explanation

 Requisite sum = 101 +103 + 105+...+199.
This is an A.P. in which a = 101, d =2 and L= 199.
A + (n-1) d =199 ⇒ 101 + (n-1) *2 = 199
⇒ (n-1) * 2 = 98 ⇒ (n-1) = 49 ⇒ n = 50.
∴ Sum = n/2 * (a+L) = 50/2 * (101 +199) = (50*150) =7500.

Q 5 - In the event that a, a-2,3a are in A.P., Then a=?

Answer : B

Explanation

Since a, (a-2), 3a are in A.P. , we have
(a-2) - a = 3a - (a-2) ⇒ 2a +2 = -2 ⇒ 2a = -4 ⇒ a = -2.

Q 6 - In the event that the fourth and ninth term of a G.P. is 54 and 13122 separately, there its second term is:

Answer : A

Explanation

Let its 1st term be a and common ratio r. Then,
ar³ = 54 and ar⁸ = 13122
∴ ar⁸/ ar³ = 13122/54 ⇒ r⁵ =243 = 3⁵ =r = 3
∴ a* 3³ = 54 ⇒ a*27 = 54 = > a =2
2nd term = ar = (2*3) =6

Q 7 - A man needs to pay Rs 975 in yearly portion every portion being not exactly the prior one by Rs 5. The measure of first portions is Rs 100. In what time, the whole sum will be paid?

Answer : C

Explanation

Let the requisite time be n years.
Then, a =100 and d=-5
Let the number of terms be n. Then,
n/2 *[2a+ (n-1)d]  = 975
⇒ n/2 [200+ (n-1)*(-5) = 975   ⇒ n (205-5n) = 1950
⇒ 5n² -205n +1950= 0 ⇒ n²- 41n+ 390 =0
⇒ n² -26n-15n+ 390 = 0 ⇒ n (n-26) -15(n-26) = 0
⇒ (n-15) (n-26) = 0 ⇒ n= 15.   [∵n ≠ 26]

Q 8 - A clock hums 1 time at 1o, clock, 2 times at 2o, clock, 3 times at 3o, clock etc. What will be the aggregate quantities of hums in a day?

Answer : C

Explanation

Total number of buzzes =2 (1+2+3+..........+12).
This is an A.P. in which a=1, d=1, n = 12 and L = 12.
(1+2+3+ .........+12) = 12/2* (1+12) = 78.
∴ Total number of buzzes = (2 * 78) = 156.

Q 9 - (14² +15² +......+30²) =?

Answer : D

Explanation

We know that (1²+2²+3²+... +a²) = {n(n+1)(2n+1)}/6
Given Exp. = (1²+2²+...+13²+14²+??+30²)-(1²+2²+...+13²)
= (30*31*61)/6 - (13*14*27)/6 = (9455- 819) = 8636

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