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 - Which term of the A.P. 2, 7, 12, 17...are 87?

Answer : C

Explanation

 Here a =2, d= (7-2) =5.
Let the nth term be 87. Then,
a + (n-1) d =87 ⇒ 2+ (n-1)*5 = 87
⇒ (n-1)*5= 85 ⇒ (n-1) = 17 ⇒ n = 18 ∴ 18th term is 87.

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 - (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 - Three numbers are in G.P. Their whole is 28 and item is 512. The numbers are:

Answer : C

Explanation

Let the number be a/r, a, ar.
Then, a/r* a*ar = 512 ⇒ a³ =8³ ⇒ a =8
8/r+8+8r =28 ⇒ 8/r +8r =20 ⇒ 2/r+2r =5
⇒ 2r²-5r+2 =0 ⇒ 2r²-4r-r+2 =0
⇒ 2r (r-2)- (r-2)=0   ⇒ (r-2)(2r-1) =0
⇒r = 2 or r = 1/2
Numbers are 4,8, 16.

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 - The estimation of (1³+2³+3³+? ? ..+15³) - (1+2+3+...+15) =?

Answer : C

Explanation

(1³+2³+3³+...+15³) - (1+2+3+...+15)
= (15*16/2)² -1/2 *15*16 = (120)² -120 =(14400- 120) = 14280.

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