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?

**Answer: **552 kms

**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.

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

**Answer: **2/b=1/p+1/q

**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.

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

**Answer: **HP

**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).

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.

**Answer: **1080 m.

**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

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

**Answer: **9H

**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

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

**Answer: **50

**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

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.

**Answer: **90

**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

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

**Answer: **3,11,19

**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.