Aptitude - Questions & Answers

Progression - Online Quiz

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 - The twelfth term of the A.P. 14, 9, 4, - 1, - 6....is:

Answer : D

Explanation

 Here a = 14, d = (9 - 14) = -5.
∴ T₁₂ = a + (12 -1) d = a + 11d = 14 + 11 * (-5) = -41.

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 of products of arrive between the whole numbers 15 and 105, both comprehensive?

Answer : B

Explanation

Requisite no. are 15, 18, 21, 24, 105.
Here a = 15 and d = (18-15) = 3
Let their number be n. Then,
a + (n-1) d =105 ⇒ 15+ (n-1)*3 = 105
⇒ a+ (n-1) d = 105 ⇒ 15+ (n-1)*3 = 105
⇒ (n-1)*3 = 90   ⇒ n-1 =30 ⇒ n = 31

Q 4 - (45+46+47....+113+114+115)?

Answer : D

Explanation

Here a=45, d =1 and L=115.
A+ (n-1) d = 115 ⇒ 45+ (n-1) *1 = 115
(n-1 ) = 70 ⇒ n = 71
Sum = n/2 * (a+L) = 71/2 * (45 +115) = 71/2 *160 = (71 *80) = 5680.

Q 5 - The total of every single common number from 75 to 97 is:

Answer : D

Explanation

Sum = 75+76+77+...+97.
Here a =75, d = (76-75) =1
Let the number of terms be n. Then,
A+ (n-1) d =97⇒ 75 + (n-1)*1 =97 ⇒ (n-1) = 22 ⇒ n= 23.
∴ Sum = 23/2 (75 + 97) = (23/2 *172) = (23 *86 ) = 1978.

Q 6 - Which term of the G.P. 5,10,20,40 ...is 1280?

Answer : B

Explanation

Let T_n = 1280. Then ar (ⁿ⁻ⁱ) = 1280 ⇒ Here a= 5 and r =2
∴ 5*2(ⁿ⁻ⁱ) =1280 ⇒256 = 2(ⁿ⁻ⁱ) =2⁸ ⇒ n-1 =8 ⇒ n= 9

Q 7 - The second term of a geometrical progression is 2/3 and its fifth term is 16/81. Its seventh term is:

Answer : D

Explanation

Let the G.P. be a, ar, ar²...
Then T2 =2/3 and T₅ = 16/81 ⇒ ar = 2/3 ...(i) And ar⁴= 16/81 ...(ii)
∴ar⁴/ar =16/81*3/2 = 8/27 ⇒ r³ = 8/27 ⇒ r= 2/3
Putting r =2/3 in (i), we get =a= 1.
T7 = ar⁶ = 1* (2/3)⁶ = 2⁶/3⁶ = 64/729

Q 8 - A man orchestrates to pay off an obligation of Rs 3600 by 40 yearly portions which are in A.P. At the point when 30 of the portions are paid, he bites the dust abandoning 33% of the obligation unpaid. The estimation of the eighth portion is:

Answer : C

Explanation

Total debt = Rs 3600 to be paid in 40 installments.
Debt paid = Rs (3600 * 2/3) = Rs 2400, paid in 30 Installments.
Let the installments. Be a, a+d, a+2d...
S₃₀ = 2400 ⇒ 30/2* [2a +(30-1) d] = 2400⇒ 2a+29d = 160.......(i)
S ₄₀ = 3600 ⇒ 40/2 *[2a + (40 -1) d] = 3600 ⇒ 2a + 39d = 180......... (ii)
On solving (i) and (ii) , we get d = 2 and a = 51.
T₈= (a+7d) = (51 + 7 *2) = 65
∴ 8th installments= Rs 65.

Q 9 - (1³ +2³ + 3³+...... + 15³) =?

Answer : D

Explanation

We know that (1³+2³+3³+??+n³) = {n (n-1)/2)²
∴ (1³+2³+3³+?..+15³) = (15*16/2)² = (120)² = 14400.³

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