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## GMAT Test Prep : Quantitative Problem Solving Test III

 Formats Worksheet / Test Paper Quiz Review

 Choose the best answer from the choices given. All numbers used are real numbers.

1. Larry is driving at 70 miles per hour. How many minutes does it take him to travel 28 miles ?
• 2½
• 3½
• 4
• 6
• 24
Speed = Distance / Time.
So, Time = Distance / Speed = (28/70) hr = 0.4 hr = 0.4 x 60 minutes = 24 minutes.

2. If a/7 = 6 and 6/b = 5, then (a + 6) / (b + 6) =
• 5/6
• 6/5
• 20/3
• 288/41
• 11
Cross-multiplication gives a = 42 and b = 6/5 = 1.2
Thus, (a + 6) / (b + 6) = (42 + 6) / (1.2 + 6) = 48/7.2 = 2/0.3 = 20/3.

3. A ship sails 3 nautical miles due south, then 12 nautical miles due west, and then another 12 nautical miles due north. How far, in nautical miles, is the ship from its starting point ?
• 15
• 18
• 21
• 24
• 27
Effectively, the ship sails 12 nautical miles due west and 9 nautical miles due north. These form the sides of a right-angled triangle, whose
hypotenuse is to be determined. By the Pythagorean Theorem,
Hypotenuse = (122 + 92)½= (144 + 81)½ = √225 = 15.

4. The product of the integers p, q, r, sand t is 2310.
If 3 < r < s < t < 231, what is the value of rs − 3t ?
• −24
• −6
• 2
• 6
• 24
The prime factors of 2310 are 2, 3, 5, 7 and 11. Their product is 2310.
Since r = 5, s = 7 and t = 11, the value of
rs − 3t = (5 x 7) − (3 x 11) = 35 − 33 = 2.

5. √(2 + 3 + 5 + 7 + 31) =
• 4 √3
• 8 √3
• 16 √3
• √5 + √43
• √17 + √31
2 + 3 + 5 + 7 + 31 = 48.
So, √48 = √(16 x 3) = √16 x √3 = 4 √3.

6. Four carpenters can individually complete a particular task in 3, 4, 5, and 7 hours, respectively. What is the maximum fraction of the task that can be completed in forty-five minutes if three of the carpenters work together at their respective rates ?
• 11/30
• 47/80
• 3/5
• 11/15
• 5/6
In one hour, the four carpenters can individually complete one-third, one-fourth, one-fifth and one-seventh of the task. For completion of the maximum fraction of the task, the rates of the three quickest carpenters must be added. Thus,
(1/3) + (1/4) + (1/5) = (20 + 15 + 12)/60 = 47/60.
In 1 hour, three carpenters together can complete (47/60)th of the task.
In ¾ hr (i.e., 45 minutes), they can complete (47/80)th of the task.
Note that (47/60) x ¾ = 47/80.

7. The distance between Diane's home and her office is 13 miles. She drives 2 miles from her office to the grocery store, next 3 miles to her son's college, and then 5 miles to her husband's office. If she has to then drive s miles to reach home, what are the possible values of s ?
• 3 < s < 16
• 3 < s < 23
• 10 < s < 13
• 10 < s < 18
• 12 < s < 15
Answer: 3 < s < 23
Diane needs to drive the minimum distance to reach home if the store, college, and her husband's office are on exactly the same route as her drive from her office to home. In that case, the minimum value of s is 3 (i.e., 13 − 2 − 3 − 5).
On the other hand, Diane needs to drive the maximum distance to reach home if the store, college and her husband's office are on a route essentially away from her drive from office to her home. In that case, she retraces her way to her office and then to home. Thus, the maximum value of s is 23 (i.e., 13 + 2 + 3 + 5).

8. The value of (251/3 x 251/3) / 251/2 is
• 51/6
• 51/3
• 5
• 5 (51/3)
• 5 (51/2)
(251/3 x 251/3) / 251/2 = 252/3/251/2 = 252/3 − 1/2 = 251/6 = (52)1/6 = 51/3

9. If 3n is a factor of 1518, the largest possible value of n is
• 5
• 6
• 9
• 15
• 18
1518 = (5 x 3)18 = 518 x 318
If 3n is a factor of 1518, then the maximum possible value of n is 18.

10. If p + qr = s and pq + r = t, then (qr) / p =
• (s + t) / 2
• (st) / 2
• (st) / (s + t)
• (st) / (2(s + t))
• (s + t) / (2(st))
Answer: (st) / (s + t)
The equations provided are:
p + qr = s ...(1)
pq + r = t ...(2)
Adding equations (1) and (2), one obtains p = (s + t) / 2
Subtracting equation (2) from equation (1), one obtains qr = (st) / 2
∴ (qr) / p = (st) / (s + t ).

11. A portion of \$8000 is invested at a 5% annual return, while the remainder is invested at a 3% annual return. If the annual income from each of the two portions is the same, then what is the total income from the two investments after one year?
• \$280
• \$300
• \$320
• \$360
• \$480
Let the \$8000 amount be divided into two parts x and y such that x + y = 8000.
The annual return on x is 5% and that on y is 3%; so, 5x / 100 = 3y / 100 or y = 5x / 3.
On eliminating y, one obtains x + 5x / 3 = 8000 or 8x / 3 = 8000.
Thus, x = 3000 and y = 5000.
So, 5x / 100 = 150 and 3y / 100 = 150.
∴ Total income after one year is \$150 + \$150 = \$300.

12. If 80 pounds of wheat, priced at \$10.00 per pound, are combined with 120 pounds of rice, priced at \$5.00 per pound, what is the price (per pound) of the resulting mixture?
• \$5.4
• \$7
• \$7.5
• \$8
• \$14
Total price of the mixture is (80 x 10) + (120 x 5) = \$800 + \$600 = \$1400.
Total weight of the mixture is 200 pounds.
∴ The price of the resulting mixture is 1400 / 200 = \$7 per pound.

13. A student first reduced a certain number by 20% and then increased the resulting number by 25%. If the original number was 1.643 x 1012, then the value of the new number (after reduction and increment) is
• 1.561 x 1012
• 1.587 x 1012
• 1.60 x 1012
• 1.62 x 1012
• 1.643 x 1012
Let y be the original number. After a reduction of 20%, the result is 0.8y.
Now, this number is increased by 25%. The result is 1.25 x 0.8y.
Note that 1.25 = 5/4 and 0.8 = 4/5; so, 1.25 x 0.8y = y.
Since a reduction by 20% and a subsequent increment by 25% gives the original number, the correct answer is 1.643 x 1012.

14. The sum of the digits of a number is called its digital root. If a two-digit number is equal to q times its digital root (where q is an integer), then the maximum possible value of q is
• 6
• 7
• 8
• 9
• 10
Let xy be the two-digit number,
with x being an integer from 1, 2, ..., 9 and y being an integer from 0, 1, ..., 9.
Thus, 10x + y = q(x + y)
So, (q − 1) y = (10 − q) x
The largest possible value that q can take is 10 (since x and y cannot be negative).
If q = 10, then the above equation gives (9) y = (0) x or y = 0 for any x.
Thus, q = 10 = 90/9 = 80/8 = 70/7 = ... = 10/1.

15. If p = 6q and (ap) / (bq) = 6, then a / b =
• 3
• 4
• 6
• 9
• 12
ap = 6b − 6q
Substituting p = 6q and simplifying gives a = 6b or a / b = 6.
Alternatively, given that (ap) / (bq) = 6 = p / q
By the property of equal ratios, we have ((ap) + p) / ((bq) + q) = 6 or a / b = 6.
Note that the property of equal ratios states that
given a set of equal ratios, a / b = c / d = e / f = ... = k (a constant),
the ratio (a + c + e + ...) / (b + c + d + ...) = k = a / b = c / d = e / f = ...

16. x and y are two distinct positive integers divisible by 5. Which of the following is necessarily divisible by 10?
x + y
xy
xy
• 2xy
x2 + y2
2xy is necessarily divisible by 10 because its factors are 2 and 5.
The other quantities are not necessarily divisible by 10 as shown by the examples below.
If x = 15 and y = 10, then x + y = 25.
If x = 15 and y = 10, then xy = 5.
If x = 15 and y = 5, then xy = 75.
If x = 15 and y = 10, then x2 + y2 = 225 + 100 = 325.

17. The co-ordinates of the foot of the perpendicular from the origin to the line x + 2y − 15 = 0 are
• (1, 7)
• (3, 6)
• (6, 3)
• (5, 5)
• (5, 10)
The general equation of a line is y = mx + c (where m is the slope of the line and c is its y-intercept).
The slope of the line x + 2y − 15 = 0 is −½ (obtained on rearranging the equation as y = −½ x + 15/2).
The slope of a line perpendicular to this line is 2 (since the product of the slopes of the two lines must be −1).
For the line perpendicular to the given line, m = 2 and c = 0 (because it passes through the origin).
So, the required equation of the perpendicular is y = 2x.
The intersection point of the lines (y = 2x & x + 2y = 15) is found by solving the equations simultaneously.
Substitute y = 2x in x + 2y = 15 to get x + 4x = 15 or 5x = 15. So, x = 3 and y = 6.
∴ The required co-ordinates are (3,6).
Note: The correct point must lie on the given line. This fact can be used to eliminate a few of the choices.

18. A number x is chosen at random from the set of integers that satisfy |x| < 9.
What is the probability that (4 / x) > x?
• 1 / 5
• 1 / 8
• 7 / 8
• 2 / 17
• 7 / 17
The values of x that satisfy |x| < 9 are −9 < x < 9.
There are 17 integers (−8, −7, ..., −1, 0, 1, ..., 7, 8) in this interval.
If both sides of (4 / x) > x are multiplied by a positive value of x, then x2 < 4.
The only (positive) integer that satisfies −2 < x < 2 is 1.
If both sides of (4 / x) > x are multiplied by a negative value of x, then x2 > 4.
The six integers that satisfy x < −2 are −8, −7, −6, −5, −4 and −3.
Since only 7 out of 17 integers satisfy (4 / x) > x, the probability is 7 / 17.

19. A bag contains 3 black balls and 4 white balls. What is the probability of drawing a black ball and a white ball in succession (in that order)?(Assume that once a ball is drawn from a bag, it is not replaced.)
• 1 / 7
• 2 / 7
• 4 / 7
• 9 / 49
• 12 / 49
There are totally 7 balls in the bag.
The probability of drawing a black ball (first from the bag) is 3/7.
Now, 6 balls remain in the bag, and the probability of drawing a white ball is 4/6 or 2/3.
Since the two events are independent, the probability of drawing a black ball and then a white ball is
(3/7) x (2/3) = 2/7.
Note that the order in which the balls are drawn makes no difference to the final answer in this case.

20. A sum of \$600 is divided among Lando, Jango and Haan. The division is such that for every \$3 that Haan receives, Jango receives \$2, and for every dollar that Jango receives, Lando receives half a dollar. What is Jango's share?
• \$100
• \$200
• \$250
• \$300
• \$350
Haan, Jango and Lando receive the money in the ratio 3 : 2 : 1.
Thus, Jango receives (2/6) x \$600 = \$200.
Alternatively, if the shares of Lando, Jango and Haan are x, y and z dollars respectively, then
x + y + z = \$600.
Since Jango receives \$2 for every \$3 that Haan receives, y = 2z/3
Since Lando receives half a dollar for every dollar Jango receives, x = y/2
Eliminating x and z, one obtains y/2 + y + 3y/2 = \$600
∴ 3y = \$600 or y = \$200.

21. The smallest number, which when decreased by 3 is divisible by 14, 18 and 30 is
• 633
• 663
• 693
• 723
• 753
14 = 2 x 7; 18 = 2 x 3 x 3; 30 = 2 x 3 x 5.
So, Least Common Multiple (LCM) of 14, 18 and 30 = 2 x 3 x 3 x 5 x 7 = 630.
Required number = 630 + 3 = 633.

22. When the integer n is divided by 3, the quotient is p and the remainder is 1, but when it is divided by 5, the quotient is q and the remainder is 7.
The value of 5q − 3p is
• 8
• 6
• −3
• −6
• −8