Note 1

Take Note:

Take a note while surfing.





Note With Ink

Give your Note a Colorful Tag.




Easy to Access

Stay on same information and in Sync wherever you are.

Note 2

Take Note:

Organize your information,It may take Shape.





Think With Ink

Differ your Content by Color.




Easy to Access

Easy to pull up your content from anywhere anytime.

Note 3

Take Note:

Don't Let information to miss,Because it take shape





Note With Ink

Simple an Easy Way to take a note.




Easy to Access

Get the same in next visit.

Arithmetic Aptitude :: progressions

Home > Arithmetic Aptitude > progressions > General Questions

1. 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?

View Answer | Discuss in Forum | Workspace | Asked In |

Answer: Option C

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.

Workspace

Tags:

Wipro 

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

View Answer | Discuss in Forum | Workspace | Asked In |

Answer: Option A

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.

Workspace

Tags:

Wipro 

3. 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

View Answer | Discuss in Forum | Workspace | Asked In |

Answer: Option C

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).

Workspace

Tags:

Wipro 

4. 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.

View Answer | Discuss in Forum | Workspace | Asked In |

Answer: Option C

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

Workspace

Tags:

Wipro 

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

View Answer | Discuss in Forum | Workspace | Asked In |

Answer: Option D

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

Workspace

Tags:

Accenture 

6. 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

View Answer | Discuss in Forum | Workspace | Asked In |

Answer: Option C

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

Workspace

Tags:

TCS 

7. 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.

View Answer | Discuss in Forum | Workspace | Asked In |

Answer: Option D

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

Workspace

Tags:

TCS 

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

View Answer | Discuss in Forum | Workspace | Asked In |

Answer: Option D

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.

Workspace

Tags:

iGate