** Sum of the reciprocals**

**The sum of two numbers is ten. Their product is twenty.Can you find the sum of the reciprocals of the two numbers?**

**Solution: **

**Given :**

** a+b =10 ——-(1)**

** ab=20——–(2)**

** we need (1/a) +(1/b) =?**

** It is simply obtained by dividing equation (1) by (2)**

** We get (1/a) +(1/b) =(10/20) => 1/2.**

**Find out the sum**

**What is the sum of all numbers between 100 and 1000 which are divisible by 14 ?**

**Solution: **

** Numbers divisible by 14 in range 100 and 1000 are 112 are**

** 112, 126,140,…….994.**

** S= 112+126+……..+994.**

** S= 14(8+9+……+71)**

** = [(14)(8+71)(71-8+1)/2]**

** =7(79)(64)**

** =35392.**

**Read out the Figure**

**A London monument is marked as follows:**

**MDCLXVI**

**What year does it represent ?**

** Solution: **

**The given number is represented in Roman Numerals,**

** In Roman Numerals Notation , we have **

** M=1000 D=500 C=100**

** L=50 X=10 VI=6**

** If we add all these together, the result is 1666.**

**Beetles and Spiders **

** Naval collected 8 spiders and beetles into a little box. When he counted the legs he found that there were altogether 54. How many beetles and how many spiders did he collect?**

**Solution:**

** Note: Spiders have 8 legs and beetles have 6 legs.**

** Use trail and error method.**

** Assuming that only beetles are with the Naval , the no of legs will be 8 x 6 = 48 which is less than mentioned number of legs. Therefore 6 legs .We will reduce the no of beetles to 5 and number of 3 spiders.**

** No. of Spiders = 3 and no of beetles = 5.**

**Value of “S”**

** ** **S434S0 What number must be substituted with ‘S’ to make it divisible by 36?**** **

** Solution: **

** If S434S0 is to be divisible by 36, then its also divisible by 4 and 9. To be divisible by 4, S S must be an even number.**

** To be divisible by 4 and 9, sum of digits i.e 2S+11 is a multiple of 9.The digit ‘8’ is the only number that meets these two conditions.when we substitute ‘S’ with ‘*’ we get the number as 843480**

**Missing Terms **

**48,60,58,72,68,104 …….**

**Here is a sequence. Can you find the two missing terms?**

**Solution: **

** It is twin sequence, in which even placed numbers forms one sequence and odd placed numbers forms other sequence.**

** 48 , 58,68 , ?**

** Next number is obatined by adding 10 to previous number. Therefore, 68+10 =78**

** 60,72,104,?**

** Next number is obatined by adding 12to previous number. Therefore, 104+12=116**

**Therefore, next numbers are 78,116.**

**Angle of Hands**

**The time is 2:15 P.M.What is the angle between the hour and minute hands?**

**Solution :**

**θ= ( 11M/2 – 30 ) **

** Using the above formula with H=2 and M=15 , we get **

** = ( 165/2 – 60)**

** =22.5°**

**Average speed**

** It was a long drive. I drove 60 kilometres at 30 kilometres per hour and then an additional 60 kilometres at 50 kilometres per hour.**

** Solution: **

** Time required for the first sixty kilometres = 120 minutes.**

** Time required for the second sixty kilometres =72 minutes.**

** Total time required =120+72=192 minutes.**

** Average Speed= (Total Distance Covered)/(Total Time taken)**

** = 120/192**

** =0.625 Km/minute**

** =0.625 x 60 / hour**

** =37.5 Km/hr**

** Tell the Time**

** Can you tell at what time between 7 and 8 O’clock, the two hands of a clock, will be in a straight line?**

**Solution: **

** Put θ=180° and find the time.**

**A problem of Age **

**Today was Lakshmi’s birthday.She turned 24.Lakshmi is twice as old as Ramu was when Lakshmi was old as old as Ramu now. How old is Ramu now?**

Solution:

**Age of Lakshmi now =24.**

**Laskshmiwa was x years when 24-x years ago.**

** 24-x years ago Ramu was x-(24-x) years old.**

** Today she is x years old.**

** 24= 2(2x-24)**

** Therefore, x=8.**