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The equation 6 × 5  4 ÷ 2  3 + 7)^0 1)^2 = 6 is missing two opening

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Re: The equation 6 × 5  4 ÷ 2  3 + 7)^0 1)^2 = 6 is missing two opening [#permalink]
Hi chetan2u,

How to think through such questions when faced in exams?
Any specific approach you would like to suggest

Thanks
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Re: The equation 6 × 5  4 ÷ 2  3 + 7)^0 1)^2 = 6 is missing two opening [#permalink]
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Re: The equation 6 × 5  4 ÷ 2  3 + 7)^0 1)^2 = 6 is missing two opening [#permalink]
Where can we practice questions like these? Bunuel bb
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Re: The equation 6 × 5  4 ÷ 2  3 + 7)^0 1)^2 = 6 is missing two opening [#permalink]
PReciSioN wrote:
Where can we practice questions like these? Bunuel bb

­OG book is the best place
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Re: The equation 6 × 5  4 ÷ 2  3 + 7)^0 1)^2 = 6 is missing two opening [#permalink]
­According to general mathematics principles the externalmost bracket is known as the 2nd or 3rd bracket and the internal bracket is called the 1st bracket, I got the answer correct but messed up the order. Can someone please explain the logic behind this?
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Re: The equation 6 × 5  4 ÷ 2  3 + 7)^0 1)^2 = 6 is missing two opening [#permalink]
Thanks for the solution.
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The equation 6 × 5  4 ÷ 2  3 + 7)^0 1)^2 = 6 is missing two opening [#permalink]
srishti_6 wrote:
­According to general mathematics principles the externalmost bracket is known as the 2nd or 3rd bracket and the internal bracket is called the 1st bracket, I got the answer correct but messed up the order. Can someone please explain the logic behind this?

­chetan2u KarishmaB Bunuel Pls explain the logic behind this?­
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The equation 6 × 5  4 ÷ 2  3 + 7)^0 1)^2 = 6 is missing two opening [#permalink]
Sans8 wrote:
srishti_6 wrote:
­According to general mathematics principles the externalmost bracket is known as the 2nd or 3rd bracket and the internal bracket is called the 1st bracket, I got the answer correct but messed up the order. Can someone please explain the logic behind this?

­chetan2u KarishmaB Bunuel Pls explain the logic behind this?­

­
I am not sure what the problem is here. We solve the inner bracket first and the other bracket later.

So first solve highlighted.
6 × (5 – 4 ÷ 2 – (3 + 7)^0– 1)^2 = 6

I get
$$6 × (5 – 4 ÷ 2 – (10)^0– 1)^2 = 6$$

$$6 × (5 – 4 ÷ 2 – 1 – 1)^2 = 6$$

Next solve highlighted using BODMAS/PEMDAS

6 × (5 – 4 ÷ 2 – 1 – 1)^2 = 6

$$6 × (5 – 2 – 1 – 1)^2 = 6$$

$$6 × (1)^2 = 6$$

$$6 = 6$$­
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Re: The equation 6 × 5  4 ÷ 2  3 + 7)^0 1)^2 = 6 is missing two opening [#permalink]

chetan2u wrote:
­
All Data Insight question: TPA [ Official Guide DI Review 2023-24]

The equation $$6 × 5 – 4 ÷ 2 – 3 + 7)^0– 1)^2 = 6$$ is missing two opening parentheses. It can be made true if opening parentheses are placed immediately before each of two of the numbers.

Assume that standard order-of-operations principles are followed, whereby multiplication and division are performed before addition and subtraction, unless parentheses indicate otherwise. Select in the table a number before which the first opening parenthesis could be placed and a number before which the second opening parenthesis could be placed, such that the equation would express a true statement. Make only two selections, one in each column.­

$$­6×5–4÷2–3+7)^0–1)^2=6$$

We can PLUG IN THE ANSWERS.

If opening parentheses are placed before 6,  we get:
$$(­6×5–4÷2–3+7)^0–1)^2=6$$
Taking the square root of both sides, we get:
$$­6×5–4÷2–3+7)^0–1=\sqrt{6}$$­
Not viable.

If opening parentheses are placed before 5,  we get:
$$­6×(5–4÷2–3+7)^0–1)^2=6$$
Dividing both sides by 6, we get:
$$­(5–4÷2–3+7)^0–1)^2=1$$­
Taking the square root of both sides, we get:
$$­5–4÷2–3+7)^0–1=1$$­
$$­5–4÷2–3+7)^0=2$$­
This seems viable.

Before 4 --> $$­5–(4÷2–3+7)^0=2$$­ --> 5-1=2 --> Nope
Before 2 --> $$­5-4÷(2–3+7)^0=2$$­ --> 5-4=2 --> Nope
Before 3 --> $$­5–4÷2–(3+7)^0=2$$­ --> 5-2-1 = 2 --> Success!

­
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Re: The equation 6 × 5  4 ÷ 2  3 + 7)^0 1)^2 = 6 is missing two opening [#permalink]
KarishmaB Thank you for your reply. My sincere apologies I believe my question could have been framed better. My problem is not with which bracket to solve first. In the question stem they are asking that the "first" parenthesis should be placed immediately before which number, and as you mentioned innermost bracket should be solved first, so I inferred that the innermost bracket (which, as per me is the first bracket) which is before 3 should be the answer of the first column. I thus messed up the order of the answers and I am unable to understand the logic.
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Re: The equation 6 × 5  4 ÷ 2  3 + 7)^0 1)^2 = 6 is missing two opening [#permalink]

Sans8 wrote:
srishti_6 wrote:
­According to general mathematics principles the externalmost bracket is known as the 2nd or 3rd bracket and the internal bracket is called the 1st bracket, I got the answer correct but messed up the order. Can someone please explain the logic behind this?

­chetan2u KarishmaB Bunuel Pls explain the logic behind this?­

­Sans8 and  srishti_6,

We are talking of two opening parenthesis, so what would they be like ( (.
Next we look for first opening parenthesis. It would surely be red colored one, (, between ( and (.

I hope it settles the query.
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Re: The equation 6 × 5  4 ÷ 2  3 + 7)^0 1)^2 = 6 is missing two opening [#permalink]
chetan2u Thanks for your response. Understood.
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Re: The equation 6 × 5  4 ÷ 2  3 + 7)^0 1)^2 = 6 is missing two opening [#permalink]

chetan2u wrote:
­
All Data Insight question: TPA [ Official Guide DI Review 2023-24]

The equation $$6 × 5 – 4 ÷ 2 – 3 + 7)^0– 1)^2 = 6$$ is missing two opening parentheses. It can be made true if opening parentheses are placed immediately before each of two of the numbers.

Assume that standard order-of-operations principles are followed, whereby multiplication and division are performed before addition and subtraction, unless parentheses indicate otherwise. Select in the table a number before which the first opening parenthesis could be placed and a number before which the second opening parenthesis could be placed, such that the equation would express a true statement. Make only two selections, one in each column.­

­I was thinking if somehow I can get a 2 from the first parentheses, then from the second parentheses I can get (2-1)^2=1. Finally, we get 6*1=6. I did a bit of trial and error and reached the answer. It was not that difficult, to be honest.
Re: The equation 6 × 5  4 ÷ 2  3 + 7)^0 1)^2 = 6 is missing two opening [#permalink]
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