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15 Mar 2018, 02:24
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[GMAT math practice question]
X is proportional to Y. If Y is increased by 30%, by approximately what percent is X^2 increased?
A. 30%
B. 40%
C. 50%
D. 60%
E. 70%
=>
X = kY for some k.
Since (1.3kY)^2 = 1.69k^2Y^2 = 1.69(kY)^2 = 1.69X^2, we have (1.69X^2 – X^2) / X^2 = 0.69, which is approximately 70%.
Therefore, the answer is E.
Answer: E
X is proportional to Y. If Y is increased by 30%, by approximately what percent is X^2 increased?
A. 30%
B. 40%
C. 50%
D. 60%
E. 70%
=>
X = kY for some k.
Since (1.3kY)^2 = 1.69k^2Y^2 = 1.69(kY)^2 = 1.69X^2, we have (1.69X^2 – X^2) / X^2 = 0.69, which is approximately 70%.
Therefore, the answer is E.
Answer: E
16 Mar 2018, 00:55
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[GMAT math practice question]
Out of 75 students, 17 students enrolled in a Physics class, 28 students enrolled in a Chemistry class, and 39 students enrolled in a Biology class. 5 students enrolled in both the Physics and Chemistry classes, 7 students enrolled in both the Chemistry and Biology classes, and 6 students enrolled in both the Biology and Physics classes. If 4 students enrolled in Physics, Chemistry, and Biology, how many students did not enroll in any of the three science classes?
A. 2
B. 3
C. 4
D. 5
E. 6
=>
777.png [ 10.71 KiB | Viewed 2127 times ]
The total number of students is given by
a + b + c + d + e + f + g + h = 75.
Of these,
a + d + f + g = 17 are enrolled in Physics,
b + d + e + g = 28 are enrolled in Chemistry, and
c + e + f + g = 39 are enrolled in Biology.
Adding these three equations gives ( a + b + c ) + 2( d + e + f ) + 3g = 84.
We also know that
d + g = 5 study both Physics and Chemistry,
e + g = 7 study both Chemistry and Biology, and
f + g = 6 study both Physics and Biology.
Adding these three questions yields ( d + e + f ) + 3g = 18.
Since 4 students are enrolled in all three classes, we have g = 4,
Plugging this value for g into the above equation yields
(d + e + f) + 3g = 18
(d + e + f) + 12 = 18
d + e + f = 6.
So,
( a + b + c ) + 2( d + e + f ) + 3g = 84 yields
(a + b + c) + 2(6) + 3(4) = 84
(a + b + c) + 12 + 12 = 84
a + b + c = 60.
Finally, using the equation, a + b + c + d + e + f + g + h = 75, we see that
(a + b + c) + (d + e + f) + g + h = 75
60 + 6 + 4 + h = 75
h = 5.
Therefore, the answer is D.
Out of 75 students, 17 students enrolled in a Physics class, 28 students enrolled in a Chemistry class, and 39 students enrolled in a Biology class. 5 students enrolled in both the Physics and Chemistry classes, 7 students enrolled in both the Chemistry and Biology classes, and 6 students enrolled in both the Biology and Physics classes. If 4 students enrolled in Physics, Chemistry, and Biology, how many students did not enroll in any of the three science classes?
A. 2
B. 3
C. 4
D. 5
E. 6
=>
Attachment:
777.png [ 10.71 KiB | Viewed 2127 times ]
The total number of students is given by
a + b + c + d + e + f + g + h = 75.
Of these,
a + d + f + g = 17 are enrolled in Physics,
b + d + e + g = 28 are enrolled in Chemistry, and
c + e + f + g = 39 are enrolled in Biology.
Adding these three equations gives ( a + b + c ) + 2( d + e + f ) + 3g = 84.
We also know that
d + g = 5 study both Physics and Chemistry,
e + g = 7 study both Chemistry and Biology, and
f + g = 6 study both Physics and Biology.
Adding these three questions yields ( d + e + f ) + 3g = 18.
Since 4 students are enrolled in all three classes, we have g = 4,
Plugging this value for g into the above equation yields
(d + e + f) + 3g = 18
(d + e + f) + 12 = 18
d + e + f = 6.
So,
( a + b + c ) + 2( d + e + f ) + 3g = 84 yields
(a + b + c) + 2(6) + 3(4) = 84
(a + b + c) + 12 + 12 = 84
a + b + c = 60.
Finally, using the equation, a + b + c + d + e + f + g + h = 75, we see that
(a + b + c) + (d + e + f) + g + h = 75
60 + 6 + 4 + h = 75
h = 5.
Therefore, the answer is D.
18 Mar 2018, 23:20
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[GMAT math practice question]
What is the minimum distance between the point (3,4) and points on the circle x2+y2=1?
A. 1
B. 2
C. 3
D. 4
E. 5
19.png [ 6.84 KiB | Viewed 2114 times ]
=>
Since the distance between (0,0) the center of the circle x^2+y^2=1 and (3,4) is 5, and the radius of the circle is 1, the minimum distance between a point on the circle and (3,4) is 5 – 1 = 4.
Therefore, D is the answer.
Answer: D
What is the minimum distance between the point (3,4) and points on the circle x2+y2=1?
A. 1
B. 2
C. 3
D. 4
E. 5
Attachment:
19.png [ 6.84 KiB | Viewed 2114 times ]
=>
Since the distance between (0,0) the center of the circle x^2+y^2=1 and (3,4) is 5, and the radius of the circle is 1, the minimum distance between a point on the circle and (3,4) is 5 – 1 = 4.
Therefore, D is the answer.
Answer: D
