Q. 23.8( 6 Votes )
Let’s write down three consecutive integers such that 5 less than the sum of the number is equal to 11 more than twice the second number. Let’s find the three consecutive integers.
Let the three consecutive numbers be x, (x+1), (x+2).
Now according to the given criteria,
5 less than the sum of the number = x+(x+1)+(x+2)-5…..(i)
11 more than twice the second number = 2(x+1)+11…….(ii)
So from question equation (i) and equation (ii) are equal,
x+(x+1)+(x+2)-5 = 2(x+1)+11
Hence the three consecutive numbers are
15, 16 and 17
Rate this question :
Let’s find the roots of the equations (i.e. let’s solve the equations).
14 (x-2) + 3 (x+5) = 3(x+8) + 5
West Bengal - Mathematics
Divide 56 into parts, so that thrice of the first part becomes 48 more than one third of the second part.West Bengal - Mathematics
Salem uncle of our village used of this savings to buy a house after his retirement from a government job. One day after falling in trouble, he sold his house and got 5% more than his cost price. If he would have taken Rs. 3450 rupees more then he would get 8% more than his cost price. Let’s find the amount of money Salem uncle used to buy his house, and his entire savings.West Bengal - Mathematics