# Arithmetic Aptitude :: progressions

Home > Arithmetic Aptitude > progressions > General Questions

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The speed of a car increases by 2 kms after every one hour. If the distance travelling in the first one hour was 35 kms. what was the total distance travelled in 12 hours?

A456 kms

B482 kms

C552 kms

D556 kms

| | | Asked In Wipro |

Explanation:

Given, Speed of car increases by 2kms after every one hrs.
=> speed of the car is in AP.
First find the no. of term in the series of speeds.
=> a+(n-1)d= 57
=> 35 + (n-1)*2=57.
=> 33 + 2n = 57.
=> n= 24/2 = 12.
Total distance traveled = (n/2)[2a +(n -1)d].
=> (12/2)[2 * 35 + 11 * 2]
=> 6*[70+22]= 92 * 6 = 552 kms.

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If a,b,c be in GP and p,q be respectively AM between a,b and b,c then

A2/b=1/p+1/q

B2/b=1/p-1/q

C2=a/p-c/q

DNone of the above

| | | Asked In Wipro |

Explanation:

p=(a+b)/2; q=(b+c)/2
=> 1/p+1/q = 2*((1/(a+b))+(1/(b+c))
=> 2*(a+2b+c)/(a+b)(b+c)
=> 2*(a+2ar+a(r^2))/(a+ar)(ar+a(r^2)) because b=ar; c= a(r^2).
where r = common ratio.

On Solving, we get
=> 1/p + 1/q = 2/ar = 2/b.

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If the sum of the roots of the equation ax*2 + bx + c=0 is equal to the sum of the squares of their reciprocals then a/c, b/a, c/b are in

AAP

BGP

CHP

DNone of these

| | | Asked In Wipro |

Explanation:

Suppose the roots are r and s. Then (x-r)(x-s) = 0, so
=> x^2 - (r+s) x + (rs) = 0
and hence r+s = -b/a and rs = c/a.
We are saying that
=> r + s = (1/r^2 + 1/s^2)
=> (rs)^2 (r+s) = (rs)^2 (1/r^2 + 1/s^2) = s^2 + r^2
=> (rs)^2 (r+s) = (r+s)^2 - 2rs
=> (c/a)^2 (-b/a) = (-b/a)^2 - 2(c/a)
Multiplying through by a^3 we get
-bc^2 = ab^2 - 2(a^2)c
Divide through by abc to get
-c/a = b/c - 2a/b
or a/b - c/a = b/c - a/b
which means c/a, a/b and b/c are in AP.
Hence their reciprocals, a/c, b/a and c/b are in harmonic progression (HP).

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After striking the floor , a rubber ball rebounds to 4/5th of the height from which it has fallen. Find the total distance that it travels before coming to rest if it has been gently dropped from a height of 120 m.

A540 m.

B960 m.

C1080 m.

D1020 m.

| | | Asked In Wipro |

Explanation:

It means the ball looses 1/5 of its height i.e 20% from 1 to 4/5 then from 4/5 to (4/5 - (4/5 * 1/5)) = 16/25
So it forms a G.P Series of..Multiplied by 2 because it goes up and comes down
(120 +2*(120* 4/5 + 120*16/25 +120* 64/125 + ...+ till Infinity )
=> 120 + 2*120 * (4/5 )/ (1 - 4/5)
=>120 + 960 =1080m
Total distance travelled will be 1080m

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A ball dropped from H height and moves 80% of height each time. Total distance covered is

A4H

B5H

C7H

D9H

| | | Asked In Accenture |

Explanation:

First time distance= H
Second time = 80H/100 = 4H/5
similarly third time 80% of 4H/5 = H(4^2)/(5^2)
and so on..
This will lead to infinite terms of geometric progression
i.e H+2*4H/5+2*16H/25..
So, Sum = H+ 2*4H/(5(1-4/5)) = 9H

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Two consecutive numbers are removed from the progression 1, 2, 3, ...n. The arithmetic mean of the remaining numbers is 26 1/4. The value of n is

A60

B81

C50

DCannot be determined

| | | Asked In TCS |

Explanation:

As the final average is 105/4, initial number of pages should be 2 more than a four multiple. So in the given options, we will check option C.
Total = n(n+1)/2=(50 * 51)/2 = 1275
Final total = 48 * 105/4 = 1260
So sum of the pages = 15. The page numbers are 7, 8

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In a group of five families, every family is expected to have a certain number of children, such that the number of children forms an arithmetic progression with a common difference of one, starting with two children in the first family. Despite the objection of their parents, every child in a family has as many pets to look after as the number of offsprings in the family. What is the total number of pets in the entire group of five families.

A99

B9

C55

D90

| | | Asked In TCS |

Explanation:

As the number of children are in arithmetic progression starting with 2, the five families have 2, 3, 4, 5, 6 kids respectively.
As each children has kept the pets equal to the number of kids in the family, Each family has n^2 pets.
So total = 2^2+3^2+4^2 +5^2 +6^2 = 90

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If sum of three numbers in A.P is 33 and sum of their squares is 491, then what are the three numbers ?

A5,11,17

B7,11,15

C9,11,13

D3,11,19

| | | Asked In iGate |

Explanation:

Let 'a' be the first number of the series and 'd' be the common difference.
Then, according to the question,
=> a + (a + d ) + ( a + 2d ) = 3*(a + d) = 33.
=> a + d = 11, or second term = 11. Then, first term = 11-d.
Then, ( 11- d )^2 + 11^2 + ( 11 + d )^2 = 491.
=> 2(d^2) = 491 - ( 3 * 121).
=> 491 - 363 = 128.
=> d^2 = 64, d = 8, a = 3.
So, the numbers are 3, 11 and 19.

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If a, b and c are forming increasing terms of G.P., r is the common ratio then find the minimum value of (c-b), given that (log a+log b+log c)/log 6 = 6.
Note that r can be any real no.

A36

B24

C12

D18

| | | Asked In TCS |

Explanation:

(log a + log b + log c)/log 6 = 6
(log abc )/log 6 = 6
log6 abc=6
abc = 6^6
From gp series a/r * a * ar =6^6
a^3= 6^6
a=36
so, a= 36/r, b=36, c=36r
c-b = 36r-36
=36(r-1)
Given r can be real number.

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Three distinct single-digit numbers A,B and C are in geometric progression.
If abs(x) for real x is the absolute value of x (x if x is positive or zero and -x if x is negative ). Then the number of different possible values of abs(A+B-C) is:

A3

B5

C4

D6

| | | Asked In TCS |

Explanation:

As given numbers are in GP,
So, numbers will be GP as 1, 1*3,1*3^2
abs(x) = x (as x is positive no.)
abs(1+3-9) = |1+3-9| = 5

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