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The equation 6 × 5 4 ÷ 2 3 + 7)^0 1)^2 = 6 is missing two opening
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28 Feb 2024, 20:26
28 Feb 2024, 20:26
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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.
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.
| First parenthesis immediately before | Second parenthesis immediately before | |
| 2 | ||
| 3 | ||
| 4 | ||
| 5 | ||
| 6 |
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Official Answer
First parenthesis immediately before: 5
Second parenthesis immediately before: 3
Re: The equation 6 × 5 4 ÷ 2 3 + 7)^0 1)^2 = 6 is missing two opening
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20 Mar 2024, 05:14
20 Mar 2024, 05:14
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Expert Reply
chetan2u wrote:
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.
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.
Hit and Trial in a question like this will take a long time. We need to look for some logic. On the RHS we have 6 and the LHS ends with a square. We can certainly not have an opening bracket at the beginning since no integer squared will give 6.
Then notice that the first term is 6 and then there is a multiplication sign. So it made sense that 6 is multiplied by 1 and hence the entire thing squared is the square of 1
6 × (5 – 4 ÷ 2 – 3 + 7)^0– 1)^2 = 6
The highlighted is likely all 1. This means that 5 – 4 ÷ 2 – 3 + 7)^0 should be 2.
\(5 – 4 ÷ 2 – 3 + 7)^0 = 2\)
The first number is 5 so we need some small numbers to be subtracted out of it to give 2. We have a 0 as exponent. It should take care of some of the numbers. 4 ÷ 2 = 2 so we should not subtract out another 3. Hence (3+7) should be made 1 by the exponent.
\(5 – 4 ÷ 2 – (3 + 7)^0 = 5 - 2 - 1 = 2\) Valid
Hence the parentheses are before 5 and before 3. ANSWER
The equation 6 × 5 4 ÷ 2 3 + 7)^0 1)^2 = 6 is missing two opening
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28 Feb 2024, 20:26
28 Feb 2024, 20:26
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Expert Reply
A new type of question in DI. Purely Logical.
\( 6×5–4÷2–3+7)^0–1)^2=6\)
Now, If the entire term is under SQUARE, then the right side has to be a square integer. But 6 is not a square.
Thus, (\( 6×5–4÷2–3+7)^0–1)^2=6\) is not possible.
Next shift the first parenthesis to one term right to get \( 6×(5–4÷2–3+7)^0–1)^2=6\)
\( (5–4÷2–3+7)^0–1)^2=1\)
This would be ok as 1 is a square.
We get two solutions
(1) \( (5–4÷2–3+7)^0–1)^2=1........5-4÷2–3+7)^0–1=1.........5-4÷2–3+7)^0=2\)
Now, power of 0 will make entire thing as 1. Also 4÷2 should be together, otherwise we are likely to get a fraction.
\(5-2-3+7)^0=2\)
Let us shift '(' towards right.
\(5-2-3+7)^0=2\) => \((5-2-3+7)^0=2\) => \(1=2\)....No
\(5-2-3+7)^0=2\) => \(5-(2-3+7)^0=2\) => \(5-1=2\)....No
\(5-2-3+7)^0=2\) => \(5-2-(3+7)^0=2\) => \(5-2-1=2\)....Yes
Thus the equation is \( 6×(5–4÷2–(3+7)^0–1)^2=6\)
Answer
First parenthesis - Before 5
Second parenthesis - Before 3
Sans8
\( 6×5–4÷2–3+7)^0–1)^2=6\)
Now, If the entire term is under SQUARE, then the right side has to be a square integer. But 6 is not a square.
Thus, (\( 6×5–4÷2–3+7)^0–1)^2=6\) is not possible.
Next shift the first parenthesis to one term right to get \( 6×(5–4÷2–3+7)^0–1)^2=6\)
\( (5–4÷2–3+7)^0–1)^2=1\)
This would be ok as 1 is a square.
We get two solutions
(1) \( (5–4÷2–3+7)^0–1)^2=1........5-4÷2–3+7)^0–1=1.........5-4÷2–3+7)^0=2\)
Now, power of 0 will make entire thing as 1. Also 4÷2 should be together, otherwise we are likely to get a fraction.
\(5-2-3+7)^0=2\)
Let us shift '(' towards right.
\(5-2-3+7)^0=2\) => \((5-2-3+7)^0=2\) => \(1=2\)....No
\(5-2-3+7)^0=2\) => \(5-(2-3+7)^0=2\) => \(5-1=2\)....No
\(5-2-3+7)^0=2\) => \(5-2-(3+7)^0=2\) => \(5-2-1=2\)....Yes
Thus the equation is \( 6×(5–4÷2–(3+7)^0–1)^2=6\)
Answer
First parenthesis - Before 5
Second parenthesis - Before 3
Sans8
