Q2 of 38 Page 190

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,


Hence


x+(x+1)+(x+2)-5 = 2(x+1)+11


⇒ 3x+3-5=2x+2+11


⇒ 3x-2=2x+13


⇒ 3x-2x=13+2


⇒ x=15


⇒ x+1=15+1=16


⇒ x+2=15+2=17


Hence the three consecutive numbers are


15, 16 and 17


More from this chapter

All 38 →
1

Shibani has given some marbles to Murad and me. If the number of marbles of Murad is 2 less than 7/3 part of marbles of mine then calculate the possible number of marbles of Murad.

If Murad got 40 marbles, then let's find the number of marbles Shibani has given to me…


 1

Sima has written a number. If 2 is added to twice the number, it becomes 5 less than thrice the number. Let’s write the numbers that Sima has written.

 3

Let’s find a number such that one-fourth of the number is 1 less than one-third of the number.

 4

Let’s find a fraction whose denominator is 2 more than the numerator and when 3 is added to the numerator and 3 subtracted from the denominator, the fraction becomes equal to .