Online Maths Training For GRE , GMAT At Your Convenience: 2012

Tuesday, May 8, 2012

GRE/ GMAT : Sample Questions with Solution

1. There are 750 male and female participants in a meeting. Half the female participants and one-quarter of the male participants are Democrats. One-third of all the participants are Democrats. How many of the Democrats are female?

(A) 75

(B) 100

(D) 175

(E) 225

Let m be the number of male participants and f be the number of female participants in the meeting. The total number of participants is given as 750. Hence, we have

m + f = 750

Now, we have that half the female participants and one-quarter of the male participants are Democrats. Let d equal the number of the Democrats. Then we have the equation

f/2 + m/4 = d

Now, we have that one-third of the total participants are Democrats. Hence, we have the equation d = 750/3 = 250

Solving the three equations yields the solution f = 250, m = 500, and d = 250. The number of female democratic participants equals half the female participants equals 250/2 = 125. The answer is (C).

2. A prize of $200 is given to anyone who solves a hacker puzzle independently. The probability that Tom will win the prize is 0.6, and the probability that John will win the prize is 0.7. What is the probability that both will win the prize?

(A) 0.35

(B) 0.36

(D) 0.58

(E) 0.88

Let P(A) = The probability of Tom solving the problem = 0.6, and let P(B) = The probability of John solving the problem = 0.7. Now, since events A and B are independent (Tom’s performance is independent of John’s performance and vice versa), we have

P(A and B) = P(A) × P(B) = 0.6 × 0.7 = 0.42

The answer is (C).

3. In how many ways can 3 boys and 2 girls be selected from a group of 6 boys and 5 girls?

(A) 10

(B) 20

(D) 100

(E) 200

We have two independent actions to do:

1) Select 3 boys from 6 boys.

2) Select 2 girls from 5 girls.

Selection is a combination problem since selection does not include ordering. Hence, by Model 2, the number of ways is

(6C3 ways for boys) . (5C2 ways for girls) = (6!/3!.3!) . (5!/2!.3!)

= 20 .10 = 200

The answer is (E).

4. In how many ways can a committee of 5 members be formed from 4 women and 6 men such that at least 1 woman is a member of the committee?

(A) 112

(B) 156

(D) 246

(E) 252

Forming members of committee is a selection action and therefore this is a combination problem. Whether you select A first and B next or vice versa, it will only be said that A and B are members of the committee.

The number of ways of forming the committee of 5 from 4 + 6 = 10 people is 10C5. The number of ways of forming a committee with no women (5 members to choose from 6 men) is 6C5. Hence, the number of ways of forming the combinations is

10C5 - 6C5 = 10!/5!.5! - 6!/5! = 252 - 6 = 246

The answer is (D).

5. Which one of the following products has the greatest value?

(A) 6.00 × 0.20

(B) 6.01 × 0.19

(D) 6.03 × 0.17

(E) 6.04 × 0.16

Each answer-choice has two factors. The first factor of each answer-choice varies from 6.00 to 6.04, and the second factor varies from 0.16 to 0.20. The percentage change in the first factor is very small (0.67%) compared to the large (almost 25%) change in the second factor. Hence, we can approximate the first factor with 6.00, and the answer-choice that has the greatest second factor [choice (A)] is the biggest. Hence, the answer is (A).

Wednesday, April 11, 2012

GRE / GMAT : Sample Questions with Solution

A: The quantity in Column A is greater.
B: The quantity in Column B is greater.
C: The two quantities are equal.
D: The relationship cannot be determined from the information given.

1. Column A x<0 Column B

x²-x⁵ 0

Sol: If x=-1 then x²-x⁵=(-1)²- (-1)⁵ = 1- -1=1+1 =2 and column A is larger. If x=-2 , then x²-x⁵

=(-2)²–(-2)⁵=4—32 =4+32 =36 and column A is larger. Finally, if x=-1/2 then x²- x⁵

= (-1/2)²- (-1/2)⁵ = ¼ - -1/32 = 9/32 and column A is still larger.

This covers the three types of negative numbers, so we can confidently conclude the answer is A.

2. Column A Column B

ab² a²b

Sol: If a=0 both columns are equal zero. If a=1 and b=2 the two columns are unequal.

This is a double case and answer is D.

3. Column A A precious stone was accidentally Column B

dropped and broke into 3 pieces of

equal weight. The value of this

type of stone is always

proportional to the square of its weight.

The value of the 3 broken The value of the original stone

pieces together

Let x be the weight of the full stone. Then the weight of each of the three broken pieces is x/3.

Since we are given that the value of the stone is proportional to the square of its weight, we have that if kx²is the value of the full stone, then the value of each small stone should be k(x/3)², where k is theproportionality constant. Hence, the value of the three pieces together is

k(x/3)²+ k(x/3)²+ k(x/3)²=3 kx²/9

= 3kx²/9

=kx²/3

Hence, Column A equals kx²/3 and Column B equals kx². Hence, Column A is one third of Column B. The answer is (B).

4. Column A a and b are positive. Column B

(a + 6) : (b + 6) = 5 : 6

a +10/b +10 1

Since a and b are positive, a + 6 and b + 6 are positive. From the ratio (a + 6) : (b + 6) = 5 : 6, we get a + 6/b + 6 =5/6 . Since 5/6 < 1, a + 6 < b + 6. Adding 4 to both sides yields a + 10 < b + 10. Since b is positive, b + 10 is positive. Dividing the inequality by b + 10 yields

a +10/b +10 < 1. Hence, Column A < Column B, and

the answer is (B).

5. Column A 12 students from section A and 15 Column B

students from section B failed an

Anthropology exam. Thus, equal

percentage of attendees failed the

exam from either section.

Number of attendees for the Number of attendees for the

exam from section A exam from section B

Given that an equal percent of attendees failed the exam in sections A and B. Let x be the percent. If a students took the exam from section A and b students took the exam from section B, then number of students who failed from the sections would be a(x/100) and b(x/100), respectively. Given that the two equal 12 and 15, respectively, we have a(x/100) = 12 and b(x/100) = 15. Since 12 < 15, a(x/100) < b(x/100). Canceling x/100 from both sides yields a < b. Hence, Column B > Column A, and the answer is (B).

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Wednesday, April 4, 2012

GRE / GMAT : Sample Questions with Solution

1) 3 pumps , working 8 hours a day can empty a tank in 2 days , how many hours a day must 4

pumps work to empty the tank in 1 day?

a) 9 b) 10 c) 11 d) 12

Ans: Let the required number of working hours per day be x

More pumps Less working hours per day (Indirect Proportion )

Less days, More Working Hours per day (Indirect Proportion )

Pumps 4:3

:: 8: x

Days 1:2

4*1*x=3*2*8

x=3*2*8/4

=12

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2) If the cost of x metres of wire is d dollars , then what is the cost of y metres of wire at the same

rate?

a) xy/d b) xd c) yd d)yd/x

Cost of x metres = d

Cost of 1 metre = d/x

Cost of y metres = (d/x)y= dy/x

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3) In a camp, 95 men had provision for 200 days , After 5 days 30 men left the camp. For how

many days will the remaining food last now?

a) 180 b) 285 c) 139 18/19 d) none of these

Ans: Let the remaining food will last for x days.

95 men had provisions for 195 days . 65 men had provision for x days.

Less men ,More days (Indirect Proportion )

65 : 95 :: 195 : x => 65*x=95*195

x=95*195/65

=285

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4) A rope makes 70 rounds of the circumference of a cylinder whose radius of the base is 14cm. How

many times can it go round a cylinder with radius 20 cm ?

a) 40 b)49 c) 100 d)None of these

Ans : Let the required number of rounds be x

More radius , Less rounds (Indirectly Proportional)

20:14 :: 70:x => 20*x=14*70

x = 14*70/20

=49

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5) If 3/5 of a cistern is filled in 1 minute, how much more time will be required to fill the rest of it?

a) 30 sec b) 40 sec c) 36 sec d) 24 sec

Let the required time be x seconds.

Part Filled = 3/5 , Remaining part = (1-3/5) = 2/5

Less part , Less time (Direct proportion)

3/5 : 2/5 :: 60 : x => 3/5*x =2/5*60

= > x=40

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Wednesday, March 21, 2012

GRE / GMAT : Sample Questions with Solution

1. Define x* by the equation x* = ∏/x . Then ((-∏)*)* =

a) -1/∏ b) -1/2 c) -∏ d) 1/∏ e)∏

Sol: It is given that x* = ∏/x . Then ((-∏)*)* =( -∏/∏)* = ∏/-1 = -∏

So answer is C.

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2. 2ab5 is a four-digit number divisible by 25. If the number formed from the two digits ab is a multiple of 13, then ab =

(A) 10

(B) 25

(D) 65

(E) 75

Sol: . We have that the number 2ab5 is divisible by 25. Any number divisible by 25 ends with the last two digits 00, 25, 50, or 75. So, b5 should equal 25 or 75. Hence, b = 2 or 7. Since a is now free to take any digit from 0 through 9, ab can have multiple values.

We also have that ab is divisible by 13. The multiples of 13 are 13, 26, 39, 52, 65, 78, and 91. Among these, the only number ending with 2 or 7 is 52. Hence, ab = 52. The answer is (C).

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3. What is the remainder when 7² . 8² is divided by 6?

(A) 1

(B) 2

(D) 4

(E) 5

Sol: 7². 8²= (7 . 8)²= 56².

The number immediately before 56 that is divisible by 6 is 54. Now, writing 56² as (54 + 2)², we have

56² = (54 + 2)²

= 54² + 2(2)(54) + 2² by the formula (a + b)²= a² + 2ab + b²

= 54[54 + 2(2)] + 2²

= 6 × 9[54 + 2(2)] + 4 here, the remainder is 4

Since the remainder is 4, the answer is (D).

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4. If 42.42 = k(14 + 7/50), then what is the value of k?

(A) 1

(B) 2

(D) 4

(E) 5

The given equation is

42.42 = k(14 + 7/50)

42.42 = k(14 + 14/100)

42.42 = k(14 + 0.14)

42.42 = k(14.14)

42.42/14.14 = k

3 = k

The answer is (C).

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5. If x + y = 7 and x² + y² = 25, then which one of the following equals the value of x³ + y³ ?

(A) 7

(B) 25

(D) 65

(E) 91

We are given the system of equations:

x + y = 7

x² + y² = 25

Solving the top equation for y yields y = 7 – x. Substituting this into the bottom equation yields

X² + (7 – x)² = 25

X² + 49 – 14x + x² = 25

2x² – 14x + 24 = 0

X² – 7x + 12 = 0

(x – 3)(x – 4) = 0

x – 3 = 0 or x – 4 = 0

x = 3 or x = 4

If x = 3, then y = 7 – 3 = 4. If x = 4, then y = 7 – 4 = 3.

In either case, x³ + y³ = 3³ + 4³ = 27 + 64 = 91. The answer is (E).

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Monday, March 5, 2012

GRE / GMAT : Sample Questions with Solution

1. Hose A can fill a tank in 5 minutes, and Hose B can fill the same tank in 6 minutes. How many tanks would Hose B fill in the time Hose A fills 6 tanks?

(A) 3

(B) 4

(D) 5.5

(E) 6

Hose A takes 5 minutes to fill one tank. To fill 6 tanks, it takes 6 × 5 = 30 minutes. Hose B takes 6

minutes to fill one tank. Hence, in the 30 minutes, it would fill 30/6 = 5 tanks. The answer is (C).

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2. A car traveled at 80 mph for the first half (by time) of a trip and at 40 mph for the second half of the trip. What is the average speed of the car during the entire trip?

(A) 20

(B) 40

(D) 60

(E) 80

Let t be the entire time of the trip.

We have that the car traveled at 80 mph for t/2 hours and at 40 mph for the remaining t/2 hours. Remember that Distance = Speed × Time. Hence, the net distance traveled during the two periods equals 80 × t/2 + 40 × t/2. Now, remember that

Average Speed =Net Distance/Time Taken

=80 × t/2 + 40 × t/2/ t

=80 × ½ + 40 × 1/2

=40 + 20 = 60

The answer is (D).

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3.A father distributed his total wealth to his two sons. The elder son received 3/5 of the amount. The younger son received $30,000. How much wealth did the father have?

(A) 15,000

(B) 45,000

(D) 75,000

(E) 89,000

Suppose x and y are the amounts received in dollars by the elder and the younger son, respectively.

Then the amount the father had is x + y.

The elder son received 3/5 of the amount. Expressing this as an equation yields

x = (3/5)(x + y)

x = (3/5)x + (3/5)y

(2/5)x = (3/5)y

x = (3/2)y

Hence, x + y, the amount father had, equals 3y/2 + y = 5y/2 = (5/2)(30,000) [Given that the younger son received 30,000 dollars] = 75,000, and the answer is (D).

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4. John has $42. He purchased fifty mangoes and thirty oranges with the whole amount. He then chose to return six mangoes for nine oranges as both quantities are equally priced. What is the price of each Mango?

(A) 0.4

(B) 0.45

(D) 0.55

(E) 0.6

Since 6 mangoes are returnable for 9 oranges, if each mango costs m and each orange costs n, then 6m = 9n, or 2m = 3n. Solving for n yields, n = 2m/3. Now, since 50 mangoes and 30 oranges together cost 42 dollars,

50m + 30n = 42

50m + 30(2m/3) = 42

m(50 + 30 × 2/3) = 42

m(50 + 20) = 42

70m = 42

m = 42/70 = 6/10 = 0.6

The answer is (E).

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5. A ship is sinking and 120 more tons of water would suffice to sink it. Water seeps in at a constant rate of 2 tons a minute while pumps remove it at a rate of 1.75 tons a minute. How much time in minutes has the ship to reach the shore before is sinks?

(A) 480

(B) 560

(D) 680

(E) 720

We have that water enters the ship at 2 tons per minute and the pumps remove the water at 1.75 tons per minute. Hence, the effective rate at which water is entering the ship is 2 – 1.75 = 0.25 tons per minute.

Since it takes an additional 120 tons of water to sink the ship, the time left is (120 tons)/(0.25 tons per minute) = 120/0.25 = 480 minutes. The answer is (A).

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Tuesday, February 14, 2012

GRE : Sample Questions with Solution

A: The quantity in Column A is greater.
B: The quantity in Column B is greater.
C: The two quantities are equal.
D: The relationship cannot be determined from the information given.

1. Column A Column B

15² + 17² + 19² (15 + 17 + 19)²

We know that the square of a sum of n positive numbers is always greater than the sum of the squares of the n numbers. For example, 25 = (2 + 3)² > 2² + 3² = 13. So, (15 + 17 + 19)² is greater than 15² + 17² + 19².

Hence, Column B is greater than Column A, and the answer is (B).

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2. Column A The average temperature in New Column B

Orland from January through

August is 36°C. The minimum

and the maximum temperatures

between September and December

are 26°C and 36°C, respectively.

The average temperature for the 36°C

Year

We are given that the average temperature from January through August is 36°C.

The average of a set of numbers always lies between the smallest number and the greatest number in the set. The minimum and the maximum temperatures between September and December are 26°C and 36°C, respectively. Hence, the average temperature of the period lies between 26°C and 36°C. So, the average temperature for the period September through December is less than 36°C.

So, the overall temperature for the year is less than 36°C. Hence, Column A is less than Column B, and the answer is (B).

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3.Column A Park, Jack, and Galvin distributed Column B

prize money of 120 dollars among

themselves. Park received 3/10 of

what Jack and Galvin together

received. Jack received 3/11 of

what Park and Galvin together

received.

The amount received by Park The amount received by Jack

Let the amounts received by Park, Jack, and Galvin be P, J, and G, respectively.

Since the prize money of $120 was distributed to Park, Jack, and Galvin, the amount that Jack and Galvin together received equals 120 – (the amount received by Park) = 120 – P.

Since we are given that Park received 3/10 of what Jack and Galvin together received, we have the equation

P = (3/10)(120 – P)

P = 3/10 ⋅ 120 – 3/10 ⋅ P

P + 3/10 ⋅ P = 3/10 ⋅ 120

13/10 ⋅ P = 3/10 ⋅ 120

P = 3/13 ⋅ 120 = Column A

Similarly, since we are given that Jack received 3/11 of what Park and Galvin together received (120 – J),

we have the equation

J = (3/11)(120 – J)

J = 3/11 ⋅ 120 – 3/11 ⋅ J

J + 3/11 ⋅ J = 3/11 ⋅ 120

14/11 ⋅ J = 3/11 ⋅ 120

J = 3/14 ⋅ 120 = Column B

Since 3/13 ⋅ 120 is greater than 3/14 ⋅ 120, Column A > Column B and the answer is (A).

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4.Column A Jack, Karl, Marc, and Kate are Column B

friends. They collected just

enough money to buy a car. Jack

contributed 1/3 of what his three

friends contributed together. Karl

contributed 1/4 of what his three

friends contributed together. Marc

contributed 2/5 of what his three

friends contributed together.

The amount paid by Jack The amount paid by Marc

Marc contributed 2/5 of what his three friends contributed together, while Jack contributed only 1/3 of what his three friends contributed together. Clearly, Marc must have contributed more than Jack. Let’s see this in detail:

Let the total contribution of the four friends be T, the contribution by Jack be J, and the contribution by Marc be M.

Now, the contribution by the three friends other than Jack is T – J. Jack contributed 1/3 of this. Hence, we have J = (1/3)(T – J), or J = T/4 (by solving for J).

Also, the contribution by the three friends other than Marc is T – M. As given, Marc contributed 2/5 of this.

Hence, we have M = (2/5)(T – M), or M = 2T/7 (by solving for M).

Now, 2T/7 is greater than T/4 and therefore Marc contributed more. The answer is (B).

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5. Column A 720 cents can buy 72 bananas. Column B

If the price of each banana is

reduced by 1 cent, 720 cents can

buy 72 + x bananas.

If the price of each banana is

reduced by 2 cents, 720 cents can

buy 72 + y bananas.

2 x y

Since 72 bananas cost 720 cents, each banana costs 720/72 = 10 cents.

When price is reduced by 1, the new price is 10 – 1 = 9 cents. Hence, 720 cents can buy 80 (= 720/9) bananas. Equating this to 72 + x yields 72 + x = 80, or x = 80 – 72 = 8.

When price is reduced by 2, the new price is 10 – 2 = 8 cents. Hence, 720 cents can buy 90 (= 720/8) bananas. Equating this to 72 + y yields 72 + y = 90, or y = 90 – 72 = 18.

Now, Column A equals 2x = 2(8) = 16, and this is less than 18 (= Column B). The answer is (B).

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Monday, February 13, 2012

GRE : Sample Questions with Solution

1. In how many ways can the letters of the word MAXIMA be arranged such that all vowels are

together?

(A) 12

(B) 18

(D) 36

(E) 72

The base set can be formed as {{A, I, A}, M, X, M}. The unit {A, I, A} arranges itself in 3P3/2!

ways. The 4 units in the base set can be arranged in 4P4/2! ways. Hence, the total number of ways of arranging the letters is 3P3/2! . 4 P4/2!

=3!/2! . 4!/2!

= 3.12 = 36

The answer is (D).

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2. If x + z > y + z, then which of the following must be true?

(I) x – z > y – z

(II) xz > yz

(III) x/z > y/z

(A) I only

(B) II only

(D) I and II only

(E) II and III only

Canceling z from both sides of the inequality x + z > y + z yields x > y. Adding –z to both sides yields x –z > y – z. Hence, I is true.

If z is negative, multiplying the inequality x > y by z would flip the direction of the inequality resulting in the inequality xz < yz. Hence, II may not be true.

If z is negative, dividing the inequality x > y by z would flip the direction of the inequality resulting in the inequality x/z < y/z. Hence, III may not be true.

The answer is (A) since we are asked for statements that MUST be true.

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3. If yz – zx = 3 and zx – xy = 4, then xy – yz =

(A) –7

(B) 1

(D) 4

(E) 7

Adding the two given equations yields

(yz – zx) + (zx – xy) = 3 + 4

yz – zx + zx – xy = 7

yz – xy = 7

xy – yz = –7 multiplying both sides by –1

The answer is (A).

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4. The cost of production of a certain instrument is directly proportional to the number of units produced. The cost of production for 300 units is $300. What is the cost of production for 270 units?

(A) 270

(B) 300

(D) 370

(E) 395

The cost of production is proportional to the number of units produced. Hence, we have the equation The Cost of Production = k × Quantity, where k is a constant.

We are given that 300 units cost 300 dollars. Putting this in the proportionality equation yields 300 = k × 300. Solving the equation for k yields k = 300/300 = 1. Hence, the Cost of Production of 270 units equals k × 270 = 1 × 270 = 270. The answer is (A).

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5. The list price of a commodity is the price after a 20% discount on the retail price. The festival discount price on the commodity is the price after a 30% discount on the list price. Customers purchase commodities from stores at a festival discount price. What is the effective discount offered by the stores on the commodity on its retail price?

(A) 20%

(B) 30%

(D) 50%

(E) 56%

Let r be the retail price. The list price is the price after a 20% discount on the retail price. Hence, it equals r(1 – 20/100) = r(1 – 0.2) = 0.8r.

The festival discount price is the price after a 30% discount on the list price. Hence, the festival discount price equals (list price)(1 – 30/100) = (0.8r)(1 – 30/100) = (0.8r)(1 – 0.3) = (0.8r)(0.7) = 0.56r. Hence, the total discount offered is (Original Price – Price after discount)/Original Price × 100 =(r – 0.56r)/ r × 100 = 0.44 × 100 = 44%.

The answer is (C).

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Tuesday, January 24, 2012

GRE/GMAT Sample questions with solution

1. AD is the longest side of the right triangle ABD . What is the length of longest side of ΔABC ?

(A) 2

(B) 3

(C)√ 41

(D) 9

(E) 41

In a right triangle, the angle opposite the longest side is the right angle. Since AD is the longest side of the right triangle ABD, ∠B must be a right angle and ΔABC must be a right triangle. Applying The Pythagorean Theorem to the right triangle ABC yields

AC² = AB² + BC²

AC = √(AB² + BC²)

AC = √(4² + 5²)

AC = √(16 + 25)

AC = √41

The answer is (C).

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2. From the figure, which of the following must be true?

(I) x + y = 90

(II) x is 35 units greater than y

(III) x is 35 units less than y

(A) I only

(B) II only

(D) I and II only

(E) I and III only

Angle x is an exterior angle of the triangle and therefore equals the sum of the remote interior angles, 35 and y. That is, x = y + 35. This equation says that x is 35 units greater than y. So, (II) is true and (III) is false. Now, if x is an obtuse angle (x > 90), then x + y is greater than 90. Hence,

x + y need not equal 90. So, (I) is not necessarily true. The answer is (B).

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3. Column A x/15 > y/25 Column B

6y + 5x 10x + 3y

Multiplying the given inequality x/15 > y/25 by 75 yields 5x > 3y.

Now, subtracting 3y and 5x from both columns yields

Column A 5x > 3y Column B

3y 5x

Since we know that 5x > 3y, Column B is greater than Column A and the answer is (B).

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4. Jane gave three-fifths of the amount of money she had to Jack. Jane now has 200 dollars. How much did she give to Jack?

(A) $80

(B) $120

(D) $300

(E) $500

Let the original amount of money Jane had be x. Since she gave 3/5 of her money to Jack, she now has 1 – 3/5 = 2/5 of the original amount. We are given that this 2/5 part equals 200 dollars. Hence, we have the equation 2/5 x = 200. Solving for x yields x = 500. Since she gave 3/5 of this amount to Jack, she gave him $300 ( = 3/5 × 500). The answer is (D).

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5. At Stephen Stores, 3 pounds of cashews costs $8. What is cost in cents of a bag weighing 9 ounces?

(A) 30

(B) 60

(D) 120

(E) 150

This problem can be solved by setting up a proportion. Note that 1 pound has 16 ounces, so 3 pounds has 48 (= 3 × 16) ounces. Now, the proportion, in cents to ounces, is 800/48

=cents/9

Or cents = 9 × 800/48 = 150

The answer is (E).

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