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ਜਮਾਤ 10 (ਬੁਨਿਆਦ)
ਕੋਰਸ: ਜਮਾਤ 10 (ਬੁਨਿਆਦ) > Unit 4
Lesson 2: ਲਗਾਤਾਰ ਸੰਪੂਰਨ ਸੰਖਿਆਵਾਂ ਦਾ ਜੋੜਸੰਪੂਰਨ ਸੰਖਿਆਵਾਂ ਦਾ ਜੋੜ ਚੁਣੌਤੀ
ਸੈਲ ਹੇਠਲੀ ਸਮੱਸਿਆ ਦਾ ਹੱਲ ਕਰਦਾ ਹੈ: ਲਗਾਤਾਰ ਤਿੰਨ ਪੂਰਨ ਅੰਕ ਦਾ ਜੋੜ 231 ਹੈ । ਸਭ ਤੋਂ ਵੱਡੀ ਸੰਪੂਰਨ ਸੰਖਿਆ ਕੀ ਹੈ ? ਸੈਲ ਖਾਨ ਅਤੇ ਮੋਂਟਰੇ ਤਕਨਾਲੋਜੀ ਅਤੇ ਸਿੱਖਿਆ ਸੰਸਥਾ ਦੁਆਰਾ ਬਣਾਇਆ।
ਕੀ ਗੱਲਬਾਤ ਵਿੱਚ ਸ਼ਾਮਲ ਹੋਣਾ ਚਾਹੁੰਦੇ ਹੋ?
ਹਾਲੇ ਤੱਕ ਕੋਈ ਪੋਸਟ ਨਹੀਂ ਕੀਤੀ ਗਈ ਹੈ।
ਵੀਡੀਓ ਪ੍ਰਤੀਲਿਪੀ
We're told that the sum
of three consecutive odd integers is 231. What is the largest integer? So let's think about
this a little bit. Let's say that x is the
smallest integer. x is equal to the smallest
odd of these three. It's not the smallest odd
integer of all integers, it's is the smallest odd of
these three, the smallest odd integer. So what's the next
one going to be? Well, if I have one odd integer,
what's going to be the next odd integer? Let's think about this. If x was 3, what's the
next odd integer? It's 5. And then what's the next
one after that? It's 7. And the next one after that? It's 9. So every time we add 2. So if the smallest one is x, the
next smallest odd integer is x plus 2, is equal to the
next smallest odd integer-- I'll write integer here--
odd integer. And then what would
be the next one? Well, we're going to add
2 to this one, right? So it's going to be x plus 4. Think about it. If the smallest is 3, then you
have x plus 2, which is 5. And then you have x plus
4, which is 7. So this will be the largest of
the consecutive odd integer in this group. And they tell us that the sum
of these consecutive odd integers is 231. What is the largest integer? So if I take x, x plus 2 and x
plus 4, and I sum them, they should be equal to 231. So let's do that. So we have x plus x plus 2, plus
x plus 4, and this needs to be equal to 231. And when they ask us what's the
largest one, we're going to have to tell them what
x plus 4 is equal to. So let's just solve
this equation. So let's add our x terms. We
have one x, two x, three x's, so we get 3x plus-- and then
what are our constants? We have a 2 and we have a 4. So 3x plus 6 is equal to 231. Now, let's get rid
of the 6 from the left-hand side of the equation. The best way to do that it is to
subtract 6 from both sides. So let's subtract 6
from both sides. The left-hand side, we're
just left with the 3x. The 6's cancel out. The right-hand side,
231 minus 6 is 225. We have 3x is equal to 225. To isolate the x, let's just
divide both sides by 3. The left-hand side, the 3's
cancel out-- that was the whole point behind dividing by
3-- we get just an x is being equal to-- and 225
divided by 3. Let me do it over here. So 3 goes into 225. It goes into 22 7 times. 7 times 3 is 21. 22 minus 21 is 1. Bring down the 5. 3 goes into 15 5 times. 5 times 3 is 15. Subtract, no remainder. So 225 divided by 3 is 75. So the smallest, the smallest
of the odd integers is 75. So this one is 75. What's x plus 2 going
to be equal to? Well, that's going
to be 2 more, 77. And what's x plus 4 going
to be equal to? Well, that's the largest
of them. x is 75 plus 4 is
going to be 79. And notice, we have three
odd integers. They're consecutive. They're the, you know, they're
the odd integers that come directly after each other. And let's verify that when we
add them up, we get 231. So if we get 75 plus-- let me
just write it like this-- 75 plus 77, plus 79, want
to add them all up. 5 plus 7 is 12. 12 plus 9 is 21. Carry the 2. 2 plus 7 is 9. 9 plus 7 is 16. 16 plus 7 is 23. So there you have it. The three consecutive odd
integers, when you add them up, you got 231. They're consecutive and odd. They ask, what's the largest? The largest is 79.