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The four consecutive odd numbers which add up to 120 are 27, 29, 31 and 33.
It can be found out as follows:
Let 'x' be the smallest odd number.
The next odd numbers will be 'x+2', 'x+4', 'x+6'.
The addition of these four numbers will give 120.
Therefore,
x+(x+2)+(x+4)+(x+6)=120
4x+12=120
x+3=30
Therefore, x=27.
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Find two consecutive even whole numbers whose product is 120?
10 and 12
What are two consecutive integers of 242?
120 and 122.
What are the 13 consecutive composite numbers in the integers 1-200?
114 115 116 117 118 119 120 121 122 123 124 125 126
The sum of 2 numbers is 120 their difference is 0 What are the two numbers?
60
Find 4 consecutive even integers where the product of the two smaller numbers is 72 less than the product of the two larger numbers?
((x+6)(x+8))-((X + 2)(X+4)) = 72(x^2 + 14x + 48)-(x^2 + 6x + 8) = 728x + 40 = 728x = 32x = 4check:((10)(12))-((6)(8)) = 72(120)-(48) = 72120 = 12012,10,8,6The numbers are 6, 8, 10, and 12.Here is how we find them:(n + 6)(n + 4) - (n + 2)n = 72; whence,(n2 + 10n + 24) - (n2 + 2n) = 8n + 24 = 72,8n = 48, andn = 6.So, let's try it!(12)(10) - (8)(6) = 120 - 48 = 72.It works!
