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 - In the event that 2x (x+10) and (3x +2) are in A.P. at that point x=?

Answer : C

Explanation

Since 2x, (x+10) and (3x+2) are in A.P. we have
(x+10)-2x = (3x+2)-(x+10) ⇒ 10-x =2x-8 ⇒ 3x= 18 ⇒ x= 6

Q 3 - Aggregate number of whole numbers somewhere around 100 and 200 which are distinct by both 9 and 6 is:

Answer : B

Explanation

LCM of 9 and 6 = 18
So, each one must be multiple of 18.
Requisite no. is 108,126, 144, 162, 180, 198.
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 - In the event that a ≠b, Then which of the accompanying proclamations is valid?

Answer : C

Explanation

For any two unequal numbers a and b, we have (A.M).> (g.m)
∴   (a+b)/2 > √ab

Q 8 - A few buys National Savings Certificates each year whose worth surpasses the earlier years buy by Rs 400. Following 8 years, she finds that she has obtained declarations whose aggregate face worth is Rs 48000. What is the face estimation of the Certificates acquired by her in the first years?

Answer : D

Explanation

Let the required value be Rs a.
Also d= 400, n = 8 and Sn = 48000.
Sn   = n/2 [2a + (n -1) d] ⇒ 8/2 *[2a + 7 *400] = 48000
⇒ 2a + 2800 = 12000⇒ 2a=9200⇒ a = 4600

Q 9 - In the event that (1² + 2² + 3²+ ......... + x²) = x(x+1) (2x+1)/6, then (1² + 3²+ 5² + ...... + 19²) =?

Answer : A

Explanation

(1²+ 3²+5²+..........+19²) = (1²+2²+3²+4²+5²+........+18²+19²) - (2²+4²+6²+.........+18²)
 = {19*(19+1) (38+1)/6} - (1*2²+2²*2²+2²*3²+2²*4²+???. +2²*9²)
= 2470 -2²*{1²+2²+3²+... +9²} = 2470 - (4*9*10*19)/6= (2470-1140) =   1330

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