niti

A brass ring of diameter 10.00 cm at 22.2°C is heated and slipped over an aluminum rod of diameter 10.01 cm at 22.2°C. Assume the average coefficients of linear expansion are constant. (b) What if the aluminum rod were 9.16 cm in diameter?

According to the formula HCF× LCM=1st no. ×2nd no. Let the no. Be n So, 145×2175=725× n => 315375=725×n As the no. Changes its side , it also changes its sign,minus to plus, plus tu minus,× to ÷ , ÷ to × So,315375÷725=n => 435=n or the second no.

Let the no. S be n, n+1, n+2, n+3, n+4 So n+n+1+n+2+n+3+n+4=135 5n+1+2+3+4=135 5n+10=135 5n=135-10 5n=125 n=125÷5 n=25 25 is the answer

This is the ans of ques 6

5. Let the no. be n As given, n-18=30-n n=30+18-n n+n=30+18 2n=48 n=48÷2 n=24 Now you can check the answer

7. Let the no. be n 1/4of n +1/2of n=12 By taking the LCM of 2and 4, we get 4 So, 1n+2n= 12×4 3n=48 N=48÷3 n=16 16 is the answer