**Que 6: You have two non-uniform threads, each takes 30 mins to burn if we burn it from one side. How will you measure 45 minutes with those two threads without cutting or breaking them?**

Ans 6: If we burn it from one side, it will take 30 mins. to burn. So if we burn it from both the side it will burn in 15 mins. So, to measure 45 mins. 1st we will burn one thread from one side for 30 mins and immediately after it we will burn 2nd thread from both sides to measure 15 mins. and thus we can measure 45 mins.** **

**Que 7: You have 10 packets of chocolate, each packet consists of 10 chocolates. Each chocolate is 10gm except in one packet which has 11gms chocolate each, all the packets look similar. Use a digital weighing machine once to find out which packet has 11 gm of **chocolates.

Ans 7: As we can use the weighing machine once, we have to make some combination of the chocolates in such a way that one weigh will tell the packet. One such possible combination is to take 1 chocolate from the 1st packet, 2 chocolate from the 2nd packet, and so on.

So, the total chocolates become 550gm. If 1st packet has the 11gm chocolate the measured weight will be 560gm. Likewise, if the nth packet has 11 gm chocolate measured weight will be (550+10n)gm.

**Que 8: If a Pen and a pencil cost Rs 10, if the price of a pen is Rs 9 more than that of a pencil. What is the cost of a pen and pencil? Ans 8: **Suppose the price of the Pen is x and the price of the pencil is y.

By the question x + y = 10 and x – y = 9

Solving both equations x(Pen) will be Rs 9.5 and y(Pencil) will be Rs 0.5.

**Que 9: There are 25 horses. You can conduct a **race among at most 5 to find out their relative speed. At no point, you can find out the actual speed of the horse in a race. Find out how many races are required to get the top 3 fastest horses.

Ans 9: Make 5 groups of 5 horses and run 5 races. Suppose five groups are a,b,c,d, and e and the next alphabet represents its rank in the group(of 5 horses) eg. d3 means horse in group d and has ranked 3rd in his group. [ 5 RACES DONE ]

a1 b1 c1 d1 e1

a2 b2 c2 d2 e2

a3 b3 c3 d3 e3

a4 b4 c4 d4 e4

a5 b5 c5 d5 e5

Now make a race of (a1,b1,c1,d1,e1).[RACE 6 DONE] suppose result is a1>b1>c1>d1>e1

which implies a1 must be FIRST.

b1 and c1 MAY BE(but not must be) 2nd and 3rd.

For the II positions, the horse will be either b1 or a2.

(we have to find the top 3 horses therefore we choose horses b1,b2,a2,a3, and c1 to race among them [RACE 7 DONE].

The only possibilities are :

c1 may be third

b1 may be the second or third

b2 may be third

a2 may be the second or third

a3 may be third

The final result will give ANSWER. suppose result is a2>a3>b1>c1>b2

then answer is a1,a2,a3,b1,c1.

HENCE ANSWER is 7 RACES

**Que 10: Prove that the number between 2 consecutive odd prime numbers(eg 17 19), **is always divisible by 6.

Ans 10: To prove this, we need to prove that the number between two consecutive odd prime numbers is divisible by 2 and 3.

The number between two odd numbers will be even so it will be divisible by 2.

In each consecutive 3 odd number 1 number is always divisible by 3, let’s say two consecutive odd prime numbers are n and n+2, then n-2 and n+4 will be divisible by 3 and hence the number between them n+1 will also be divisible by 3.