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MathRevolution wrote:
This question is frequently on GMAT Math lately, which is 2 by 2 question.
Attachment:
GCDS overview of GMAT (20160118).jpg

On the above, suppose all the passengers 100p(if there is %, use 100 since it is per cent. p is an initial for passengers). Passengers who brought a round ticket and a laptop is 40% of the total passengers, which is 40p. However, what 20% means is that x=10 is derived from 40p:xp=80%:20%=4:1. That is, 40p+10p=50p, which means 50% of the passengers have round tickets. Therefore, the answer is C.

Yeah but this is still wrong. Your chart is wrong..
On a certain transatlantic crossing, 40 percent all passengers held round-trip tickets.
40 is the total of Round-trip laptop + non laptop. You diagram illustrates 40 = round trip and laptop only.

Additionally, they also took their laptops aboard the ships. If 20 percent of the passengers with round-trip tickets did not take their laptops aboard the ship, what percent of the ship’s passengers held round-trip tickets?
If 40 have round trip tickets, 20% of 40 is 8.
So 8 have round trip with no laptop.
40-8 = 32 with round trip and laptop.

32+8 = 40 percent of all passangers with round trip tickets.

Your explanation only works if we are given that 40% is the number of roundtrip tickets with laptops, not total round trip tickets..which is what the question asks for and is given in the question.
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1
1
[GMAT math practice question]

(Algebra) What is a/ab+a+1 + b/bc+b+1 + c/ca+c+1?

1) abc = 1
2) a, b, and c are integers

=>

Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.
Visit https://www.mathrevolution.com/gmat/lesson for details.

The first step of the VA (Variable Approach) method is to modify the original condition and the question. We then recheck the question. We should simplify conditions if necessary.

Condition 1)
If we have abc = 1, then we have

a/ab+a+1 + ab/a(bc+b+1) + abc/ab(ca+c+1)
= a/ab+a+1 + ab/abc+ab+a + abc/abca+abc+ab
=a/ab+a+1 + ab/ab+a+1 + 1/ab+a+1 (substituting 1 for abc)
=ab+a+1/ab+a+1 = 1.
Since condition 1) yields a unique solution, it is sufficient.

Condition 2)
Since we don’t have any information about a, b and c, it is not sufficient.

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[GMAT math practice question]

(Function) The function f(x) is defined as x - 10[x/10]. ([x] is the greatest integer less than or equal to x). What is f(7) + f(7^2) + f(7^3) +…+ f(7^2020)?

A. 10036
B. 10039
C. 10040
D. 10100
E. 10107

=>

f(x) = x – 10[x/10] means the remainder of x when x is divided by 10.
For example, f(1234) = 1234 – 10*[1234/10] = 1234 – 10*[123.4] = 1234 – 10*123 = 1234 – 1230 = 4.

Units digit of Powers of base 7 repeats as follows.
f(7^1) = 7 ~ 7^5 ~ 7^9 ~ …
f(7^2) = f(49) = 9 ~ 7^6 ~ 7^10 ~ …
f(7^3) = f(343) = 3 ~ 7^7 ~ 7^11 ~ …
f(7^4) = f(2401) = 1 ~ 7^8 ~ 7^12 ~ …
f(7) + f(7^2) + f(7^3) +…+ f(7^2020)
= f(7) + f(7^2) + f(7^3) + f(7^4) + f(7^5) +…+ f(7^2017) + f(7^2018) + f(7^2019) + f(7^2020)
= (7 + 9 + 3 + 1) + (7 + 9 + 3 + 1) +…+ (7 + 9 + 3 + 1)
= 20 * 2020/4 = 20*505 = 10100

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1
[GMAT math practice question]

(Geometry) We have a square □ABCD, and △AEF is a triangle inscribed in □ABCD. What is the length of BE?

1) □ABCD is a square with AB = 10
2) △AEF is an equilateral triangle.

Attachment: 3.26(ds).png [ 5.63 KiB | Viewed 552 times ]

=>

Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.
Visit https://www.mathrevolution.com/gmat/lesson for details.

Since we have a triangle and a square, we have 3 variables and 1 variable for the triangle and the square, respectively. Since we have 4 variables and 0 equations, E is most likely the answer. So, we should consider conditions 1) & 2) together first. After comparing the number of variables and the number of equations, we can save time by considering conditions 1) & 2) together first.

Conditions 1) & 2)

Attachment: 3.26ds(a).png [ 9.37 KiB | Viewed 552 times ]

Assume BE = x (0 < x < 10).
Then we have BE = DF since the triangle AEF is equilateral.
AE^2 = 100 + x^2 = EF^2 = (10 - x)^2 + (10 - x)^2.
Then we have
100 + x^2 = (10 - x)^2 + (10 - x)^2
100 + x^2 = 2(10 – x)^2 (adding like terms)
100 + x^2 = 2(10 – x)(10 – x)
100 + x^2 = 2(100 – 10x – 10x + x^2) (foiling out the brackets)
100 + x^2 = 2x^2 – 40x + 200 (multiplying 2 through the bracket and combining like terms)
x^2 – 40x + 100 = 0 (bringing all terms to one side)
Thus, we have x = 20 ± 10√3. (using the quadratic formula)
But we have x = 20 - 10√3 since 0 < x < 10.

Since both conditions together yield a unique solution, they are sufficient.

In cases where 3 or more additional equations are required, such as for original conditions with “3 variables”, or “4 variables and 1 equation”, or “5 variables and 2 equations”, conditions 1) and 2) usually supply only one additional equation. Therefore, there is an 80% chance that E is the answer, a 15% chance that C is the answer, and a 5% chance that the answer is A, B or D. Since E (i.e. conditions 1) & 2) are NOT sufficient, when taken together) is most likely to be the answer, it is generally most efficient to begin by checking the sufficiency of conditions 1) and 2) when taken together. Obviously, there may be occasions on which the answer is A, B, C, or D.
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[GMAT math practice question]

(Function) For a positive integer n, f(n) and g(n) are defined as follows

f(n)={0 ,(n is a multiple of 5) 1 , (n is not a multiple of 5)
g(n)={0 ,(n is a multiple of 7) 1 , (n is not a multiple of 7)

Moreover, h(n) is defined as (1 - f(n))(1 - g(n)).
What is the value of h(3) + h(6) + h(9) + … + h(2004) + h(2007)?

A. 13
B. 19
C. 38
D. 57
E. 152

=>

h(n) = 0 when f(n)=1 or g(n) = 1
h(n) = 1 when f(n)=0 and g(n)=0
Thus, we have h(n) = 1 when n is a multiple of both 5 and 7. It means we have h(n) = 1 when n is a multiple of 35.

3, 6, 9, … , 2016, 2019 are multiples of 3.
The function values of h for those numbers is 1 when they are multiples of 105 = 35*3.
2019 = 105*19 + 24
Thus, we have 19 multiples of 105 between 1 and 2007, inclusive.

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1
[GMAT math practice question]

(Function) What is the value of f(2019)?

1) f(3) = 5
2) f(x+2)=f(x - 1)/f(x + 1)

=>

Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.
Visit https://www.mathrevolution.com/gmat/lesson for details.

The first step of the VA (Variable Approach) method is to modify the original condition and the question. We then recheck the question. We should simplify conditions if necessary.

Since we have many variables to determine the function f(x), E is most likely the answer. So, we should consider conditions 1) & 2) together first. After comparing the number of variables and the number of equations, we can save time by considering conditions 1) & 2) together first.

Conditions 1) & 2)
Since f(3) = 5, we have:
f(5) = (f(3) - 1) / (f(3) + 1)
f(5) = (5 - 1) / (5 + 1)
f(5) = 4/6 = 2/3.

Then we have:
f(7) = (f(5) - 1) / (f(5) + 1)
f(7) = ((2/3) - 1) / ((2/3) + 1)
f(7) = -(1/3) / (5/3) = -(1/5).
We have:
f(9) = (f(7) - 1) / (f(7) + 1)
f(9) = (-(1/5) - 1)/(-(1/5) + 1)
f(9) = -(6/5) / (4/5) = -(3/2).

We have:
f(11) = (f(9) - 1) / (f(9) + 1)
f(11) = (-(3/2) - 1)) / (-(3/2) + 1)
f(11) = -(5/2) / -(1/2) = 5.

Since f(3) = f(11), we have f(8k - 5) = 5.

Then we have:
f(8k-3) = 2/3, f(8k-1) = -(1/5) and f(8k+1) = -(3/2).
f(2007) = f(8*251-1) = -(1/5).

Since both conditions together yield a unique solution, they are sufficient.

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[GMAT math practice question]

There are 1 card written ‘1’, 2 cards written ‘2’, …, and n cards written ‘n’. The average of all the numbers written on the cards is 17. How many cards are there? (Use the fact : 1 + 2 + … + n = n(n+1)/2, 1^2 + 2^2 + … + n^2 = n(n+1)(2n+1)/6)

A. 25 B. 125 C. 225 D. 325 E. 425

=>

( 1*1 + 2*2 + 3*3 + … +n*n ) / ( 1 + 2 + 3 + … + n ) = 17
[(1/6)n(n+1)(2n+1)] / [(1/2)n(n+1) ] = (2n+1)/3 = 17.
Thus, we have 2n+1 = 51 or n = 25.

Then the number of cards is
1 + 2 + 3 + … + 25 = (1/2)25*26 = 325

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[GMAT math practice question]

(Geometry) We have a right triangle ABC with ∠A = 90, AD = 6 and CD = 8. AD and BC are perpendicular to each other. If x = AC and y = BD, what is the value of xy?

Attachment: 4.6ps.png [ 8.7 KiB | Viewed 477 times ]

A. 40
B. 42
C. 45
D. 50
E. 56

=>

We have x^2 = 6^2 + 8^2 = 36 + 64 = 100 and x = 10.
Since we have 6^2 = 8y or 8y = 36, we have y = 9/2.

Thus, we have xy = 10*(9/2) = 45.

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[GMAT math practice question]

(Statistics) s is the standard deviation of 3 real numbers x, y, and z, where a = x - y, b = y – z and c = z - x. What is the expression of s^2 in terms of a, b, and c?

A. a^3 + b^3 + c^3
B.(abc)^2
C. 0
D. a^2 + b^2 + c^2
E. 1/9(a^2 + b^2 + c^2)

=>

Assume m = (1/3)(x + y + z), which is the average of x, y, and z.
x – m = x – (1/3)(x + y + z) = (1/3)(3x – x – y - z) = (1/3)(2x – y - z) = (1/3)(x – y - (z - x)) = (1/3)(a - c).
We have y – m = (1/3)(b - a) and z – m = (1/3)(c - b) similarly.

Then, we have a + b + c = (x - y) + (y - z) + (z - x) = 0.

s^2 = (1/3)[(x - m)^2 + (y - m)^2 + (z - m)^2]
= (1/3)(1/3)[(a - c)^2 + (b - a)^2 + (c - b)^2]
= (1/27)[(a - c)^2 + (b - a)^2 + (c - b)^2]
= (1/27)[(a – c)(a – c) + (b – a)(b -a ) + (c – b)(c – b)]
= (1/27)(a^2 – 2ac + c^2 + b^2 – 2ab + a^2 + c^2 – 2bc + b^2)
= (1/27)[2(a^2 + b^2 + c^2) - 2(ab + bc + ca)]
= (1/27)[3(a^2 + b^2 + c^2) - (a + b + c)^2]
= (1/27)[3(a^2 + b^2 + c^2)], since a + b + c = 0
= (1/9)(a^2 + b^2 + c^2)

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[GMAT math practice question]

(Inequalities) What is the value of (x, y) satisfying √x^2+4y, with the integer portion equaling 5.

1) x and y are the numbers of eyes on two dice.
2) x and y are positive integers.

=>

Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.
Visit https://www.mathrevolution.com/gmat/lesson for details.

The first step of the VA (Variable Approach) method is to modify the original condition and the question. We then recheck the question. We should simplify conditions if necessary.

Since the integer part of √x^2+4y, we have
5≤√x^2+4y < 6
=> 25 x^2 + 4y < 36
X = 1, y = 6 and x = 2, y = 6 are possible solutions.

Since both conditions together do not yield a unique solution, they are not sufficient.

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[GMAT math practice question]

(Statistics) The average of x1, x2, x3, x4, x5 is m. What is the average of x1-a, x2-2a, x3-3a, x4-4a, x5-5a?

A. m
B. m - 1/2a
C. m - a
D. m - 3a
E. m - 3/2a

=>

Since m is the average of x1, x2, x3, x4, x5, we have m = (x1 + x2 + x3 + x4 + x5) / 5 or x1 + x2 + x3 + x4 + x5 = 5m.
Then the average of x1-a, x2-2a, x3-3a, x4-4a, x5-5a is
[(x1 - a) + (x2 - 2a) + (x3 - 3a) + (x4 - 4a) + (x5 - 5a)] / 5
= (x1 + x2 + x3 + x4 + x5 - 15a) / 5
= (x1 + x2 + x3 + x4 + x5) / 5 – 3a
= m – 3a

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[GMAT math practice question]

(Algebra) a, b, and c are positive integers with abc + 2ab + 2bc + 2ca + 4a + 4b + 4c = 447. What is the value of a + b + c?

A. 17
B. 19
C. 21
D. 23
E. 25

=>

abc + 2ab + 2bc + 2ca + 4a + 4b + 4c + 8 = 455
a(bc + 2b + 2c + 4) + 2(bc + 2b + 2c + 4) = 455 (taking out a common fraction of a from the first 4 terms and a common factor of 2 from the last 4 terms)
= (a + 2)(bc + 2b + 2c + 4) = 455 (taking out a common factor of (bc + 2b + 2c +4))
(a + 2)[b(c + 2) + 2(c + 2)] = 455 (taking out common factors of b and 2)
= (a + 2)(b + 2)(c + 2) = 455 (taking out a common factor of (c + 2))
= 3*5*11.
a, b, and c are interchangeable since the equation (a + 2)(b + 2)(c + 2) = 5*7*13 is symmetric.
Then we have a + b + c = 3 + 5 + 11 = 19.

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[GMAT math practice question]

(Algebra) If the decimal part of √5 is x, what is the decimal part of √45?

A. 3x - 2
B. 3x - 1
C. 3x
D. 3x + 1
E. 3x + 2

=>

Since 2 < √5 < 3, we have x = √5 - 2.
Since 6 < √45 < 7, the decimal part of √45 is √45 – 6 = 3√5 – 6 = 3(√5 - 2) = 3x.

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[GMAT math practice question]

(Algebra) What is the value of x^2(x - y) + y^2(y - x)?

1) x + y = 3.
2) xy = 1.

=>

Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.
Visit https://www.mathrevolution.com/gmat/lesson for details.

The first step of the VA (Variable Approach) method is to modify the original condition and the question. We then recheck the question. We should simplify conditions if necessary.

The expression in the question x^2(x - y) + y^2(y - x) is equivalent to (x - y)^2(x + y) for the following reason:
x^2(x - y) + y^2(y - x)
= x^2(x - y) - y^2(x - y) (taking out a common factor of -1 from the second bracket)
= (x - y)(x^2 - y^2) (taking out a common factor of (x – y))
= (x - y)(x - y)(x + y) (factoring (x^2 – y^2) using a difference of squares)
= (x - y)^2(x + y) (putting like terms together)

(x - y)^2
= x^2 - 2xy + y^2 (foiling (x - y)(x – y))
= x^2 + 2xy + y^2 – 4xy (since 2xy – 4xy = -2xy in the previous equation)
= (x + y)^2 – 4xy (factoring x^2 + 2xy + y^2 using trinomial factoring)
= 3^2- 4*1 = 9 – 4 = 5 when we have x + y = 3 and xy = 1 from both conditions 1) & 2).

Then we have (x - y)^2(x + y) = 5*3 = 15.

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[GMAT math practice question]

(Algebra) What is the coefficient of x^2 in the expression (-3x^2 + 2x + 1)(x - 1)^2?

A. -6
B. -5
C. 4
D. 5
E. 6

=>

Since (-3x^2 + 2x + 1)(x - 1)^2 = (-3x^2 + 2x + 1)(x – 1)(x – 1)
= (-3x^2 + 2x + 1)(x^2 – 2x + 1)
= -3x^4 + 6x^3 – 3x^2 + 2x^3 – 4x^2 + 2x + x^2 – 2x + 1
= -3x^4 + 8x^3 - 6x^2 + 1, and the coefficient of x^2 is -6.

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[GMAT math practice question]

(Number Properties) What is the sum of all possible solutions of the equation (3^n-9)^3+(9^n-3)^3=(3^n+9^n-12)^3, where n is a positive integer?

A. 1
B. 2
C. 3
D. 4
E. 5

=>

Assume a = 3^n - 9 and b = 9^n - 3.

We have the equation a^3 + b^3 = (a + b)^3 or 3ab(a + b) = 0, since (a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3 = a^3 + b^3 + 3ab(a + b).
Then we have a = 0, b = 0 or a + b = 0.

Case 1: a = 0
Since we have 3^n – 9 = 0, we have n = 2.

Case 2: b = 0
Since we have 9^n – 3 = 0, we have n = ½, which is not a solution, since n is a positive integer.

Case 3: a + b = 0
3^n + 9^n – 12 = 0
=> 9^n + 3^n – 12 = 0
=> 3^2n + 3^n – 12 = 0
=> (3^n - 3)(3^n + 4) = 0
=> 3^n = 3 since 3^n + 4 > 0
=> n = 1

Then, the sum of all solutions is 1 + 2 = 3.

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MathRevolution wrote:
[GMAT math practice question]

(Algebra) a, b, and c are positive integers with abc + 2ab + 2bc + 2ca + 4a + 4b + 4c = 447. What is the value of a + b + c?

A. 17
B. 19
C. 21
D. 23
E. 25

=>

abc + 2ab + 2bc + 2ca + 4a + 4b + 4c + 8 = 455
a(bc + 2b + 2c + 4) + 2(bc + 2b + 2c + 4) = 455 (taking out a common fraction of a from the first 4 terms and a common factor of 2 from the last 4 terms)
= (a + 2)(bc + 2b + 2c + 4) = 455 (taking out a common factor of (bc + 2b + 2c +4))
(a + 2)[b(c + 2) + 2(c + 2)] = 455 (taking out common factors of b and 2)
= (a + 2)(b + 2)(c + 2) = 455 (taking out a common factor of (c + 2))
= 3*5*11.
a, b, and c are interchangeable since the equation (a + 2)(b + 2)(c + 2) = 5*7*13 is symmetric.
Then we have a + b + c = 3 + 5 + 11 = 19.

@MathRevolution--How will I judge that I should change 447 to 455 ?

Originally posted by ammuseeru on 28 Apr 2020, 22:26.
Last edited by ammuseeru on 03 May 2020, 20:14, edited 1 time in total.
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[GMAT math practice question]

(Algebra) What is the equivalent expression of (1/x+y+z)(1/x+1/y+1/z)(1/xy+yz+zx)(1/xy+1/yz+1/zx)?

A. 1/x+y+z
B. 1/xy+yz+zx
C. 1/xyz
D. (1/xyz)2
E. xy+yz+zx/x+y+z

=>

(1/x+y+z)(1/x+1/y+1/z)(1/xy+yz+zx)(1/xy+1/yz+1/zx)
=(1/x+y+z)(xy+yz+zx/xyz)(1/xy+yz+zx)(x+y+z/xyz)
=(1/xyz)^2

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Re: Overview of GMAT Math Question Types and Patterns on the GMAT  [#permalink]

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[GMAT math practice question]

(Algebra) abc ≠ 0. What is the value of a^2+b^2+c^2?

1) a+b+c=3.
2) a^3+b^3+c^3=27.

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Since we have 3 variables (a, b, and c) and 0 equations, E is most likely the answer. So, we should consider conditions 1) & 2) together first. After comparing the number of variables and the number of equations, we can save time by considering conditions 1) & 2) together first.

Conditions 1) & 2)

If a = 1, b = -1, c = 3, then we have a^2 + b^2 + c^2 = 1 + 1 + 9 = 11.
If a = 2, b = -2, c = 3, then we have a^2 + b^2 + c^2 = 4 + 4 + 9 = 17.

Since both conditions together do not yield a unique solution, they are not sufficient.

In cases where 3 or more additional equations are required, such as for original conditions with “3 variables”, or “4 variables and 1 equation”, or “5 variables and 2 equations”, conditions 1) and 2) usually supply only one additional equation. Therefore, there is an 80% chance that E is the answer, a 15% chance that C is the answer, and a 5% chance that the answer is A, B, or D. Since E (i.e. conditions 1) & 2) are NOT sufficient, when taken together) is most likely to be the answer, it is generally most efficient to begin by checking the sufficiency of conditions 1) and 2) when taken together. Obviously, there may be occasions on which the answer is A, B, C, or D.
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GMAT 1: 760 Q51 V42
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Re: Overview of GMAT Math Question Types and Patterns on the GMAT  [#permalink]

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[GMAT math practice question]

(Geometry) The figure shows a quadrangle ABCD with ∠ADC = 90°, ∠DCB = 90° and AD = 6, AB = 8 and BC = 10. What is the length BD?

Attachment: 5.4ps.png [ 14.35 KiB | Viewed 106 times ]

A. 2√37
B. 10
C. 3√39
D. 14
E. 5√41

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Attachment: 5.4PS(A).png [ 18.47 KiB | Viewed 106 times ]

From the above figure, since we have EC = 6, we have BE = 4. Then we use Pythagorean Theorem to solve for AE, giving us AE = √8^2- 4^2 = √64-16 = √48 = 4√3 and AE = DC = 4√3.
Thus, we have x = √10^2+(4√3)^2 = √100+48 = 2√37.

_________________ Re: Overview of GMAT Math Question Types and Patterns on the GMAT   [#permalink] 06 May 2020, 18:06

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# Overview of GMAT Math Question Types and Patterns on the GMAT  