Gmat Problem Solving 1000 - Section 1

For all 1000 question 

http://www.quantfunda.in/2013/12/blog-post.html

Aim of this part is to put all the 1000 questions solved called to be must do 1000 sums for Problem Solving section of GMAT.

Section 1          Section 2             Section 3              Section 4               Section 5

Section 6          Section 7             Section 8              Section 9               Section 10

Section 11        Section 12           Section 13            Section 14             Section 15

1. The 180 students in a group are to be seated in rows so that there is an equal number of students in each row. Each of the following could be the number of rows EXCEPT

(A) 4

(B) 20

(C) 30

(D) 40

(E) 90

Solution:
180 Students to be seated in rows having equal number of students. Hence 180 should be integer multiple of number of rows.
180 is integer multiple of 4,20,30 and 90 but not 40. Hence answer is D.

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 2. A parking garage rents parking spaces for $10 per week or $30 per month. How much does a person save in a year by renting by the month rather than by the week?
(A) $140
(B) $160
(C) $220
(D) $240
(E) $260

Solution:
Rent to be paid per week: $10
Approx Rent to be paid for a year on weekly base: $10 × 52 = $520
Rent to be paid per month: $30
Approx Rent to be paid for a year on monthly base: $30 × 12 = $360
Amount that can be saved = $520 - $360 = $160
Hence Answer is B.

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3.If y = 5x^2 - 2x then y =
(A) 24
(B) 27
(C) 39
(D) 51
(E) 219

Solution:
Simply substitute x = 3 in equation.
y = 5 * 3^2 - 2 * 3
y = 5*9 - 6
y = 39

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4. Of the following, which is the best approximation to √ (0.0026)?
(A) 0.05
(B) 0.06
(C) 0.16
(D) 0.5
(E) 0.6

Solution:
√(0.0026)
= √(0.0026) * 100 ÷ 100
= √(0.0026) * √(10000) ÷ 100
= √(26) ÷ 100
= 5.099 ÷ 100
= 0.05099
Hence option A.

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5. At a certain diner, a hamburger and coleslaw cost $3.59, and a hamburger and french fries cost $4.40. If french fries cost twice as much as coleslaw, how much do french fries cost?
(A) $0.30
(B) $0.45
(C) $0.60
(D) $0.75
(E) $0.90

Solution:
Hamburger (H), coleslaw(C) and french fries (F)
Then
H + C = 3.59 … (i)
H + F = 4.40 … (ii)
And F = 2C
Putting this in (ii)
We get H + 2C = 4.40 … (iii)
Subtract (i) from (iii)
We get C = 0.81
Hence cost of a French fries is = 2C = $1.62
This answer does not match with any of the option. Please correct me if I am wrong. Put your views in comments.

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6. If∠XYZ in the figure above is a right angle, what is the value of x?


(A) 155
(B) 145
(C) 135
(D) 125
(E) 110

Solution:
Angle XYZ = 90°
So angle XYB = 90° - 55° = 35°
Hence x° = 180° - 35° = 145°

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Experssion : (a÷b)÷c


7. In the expression above, a, b, and c are different numbers and each is one of the numbers 2, 3, or 5. What is the least possible value of the expression?
(A) 1/30
(B) 2/15
(C) 1/6
(D) 3/10
(E) 5/6

Solution:
(a÷b)÷c
= a ÷ (bc)
for least possible value a should be least number that is a = 2 , b= 3 and c = 5
hence least value = 2 / 15

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8. A certain culture of bacteria quadruples every hour. If a container with these bacteria was half full at 10:00 a.m., at what time was it one-eighth full?
(A) 9:00 a.m.
(B) 7:00 a.m.
(C) 6:00 a.m.
(D) 4:00 a.m.
(E) 2:00 a.m.

Solution:
A bacterium quadruples every hour means its quantity is multiplied by 4 times every hour. It was half full at 10.00 am hence it was one-eight at 9.00 am.

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9. Al, Lew, and Karen pooled their funds to buy a gift for a friend. Al contributed $2 less than 1/3 of the cost of the gift and Lew contributed $2 more than 1/4 of the cost. If Karen contributed the remaining $15, what was the cost of the gift?
(A) $24
(B) $33
(C) $36
(D) $43
(E) $45

Solution:
Let the cost of the give be $x
Contribution of A1 = (x/3) – 2
Contribution of Lew = (x/4) + 2
Contribution of Karen = $15
So we can say that
(x/3) – 2 + (x/4) + 2 + 15 = x
on solving equation we get that x = $36

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10. What is the total number of integers between 100 and 200 that are divisible by 3?
(A) 33
(B) 32
(C) 31
(D) 30
(E) 29

Solution:
Highest number divisible by 3 lesser than 100 is 99
Smallest number divisible by 3 greater than 200 is 201
Number of digits divisible by 3 between 100 and 200 is
= (201/3) – (99/3)
= 66 -33
= 33

Or

From 100 to 130 there are 10 numbers divisible by 3 (3's Multiplication table).
So, from 100 to 190 there are 30!.
From 190 to 200 there are 3 more (192,195 and 198).
Tihs way, you have 3*10+3 = 33.

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Q11. Which of the following inequalities is equivalent to 10 – 2x > 18?
(A) x > -14
(B) x > -4
(C) x > 4
(D) x < 4
(E) x < -4

Solution:
10 – 2x > 18
= - 2x > 8
= x < -4 (Division by -2 will change the equality sign)

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12. In 1979 approximately 1/3 of the 37.3 million airline passengers traveling to or from the United States used Kennedy Airport. If the number of such passengers that used Miami Airport was 1/2 the number that used Kennedy Airport and 4 times the number that used Logan Airport, approximately how many millions of these passengers used Logan Airport that year?
(A) 18.6
(B) 9.3
(C) 6.2
(D) 3.1
(E) 1.6

Solution:
Number of people used Kennedy Airport = 1/3 * 37.3 = 12.43
So Number of people used Miami Airport = 1/2 * 12.43 = 6.22
Hence Number of people used Logan Airport = 1/4 * 6.22 = 1.6

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13. A certain basketball team that has played 2/3 of its games has a record of 17 wins and 3 losses. What is the greatest number of the remaining games that the team can lose and still win at least ¾ of all of its games?
(A) 7
(B) 6
(C) 5
(D) 4
(E) 3

Solution:
2/3 of the total games = 20 games
Hence total games = 30 games.
To win ¾ of all games team need to win = ¾ * 30 = 22.5 = 23 games.
Team can lose = 7 games.
Team can lose in last 10 games = 4 games.

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14. Dan and Karen, who live 10 miles apart meet at a cafe that is directly north of Dan’s house and directly east of Karen’s house. If the cafe is 2 miles closer to Dan’s house than to Karen’s house, how many miles is the cafe from Karen’s house?
(A) 6
(B) 7
(C) 8
(D) 9
(E) 10

Solution:

http://4.bp.blogspot.com/_PULTbNiVGWg/S1w2cYYsJhI/AAAAAAAAAWA/MTqPRaWn2qE/s1600-h/S1Q14.bmp

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15. If n is an integer and n = (2.3.5.7.11.13) / 77 K then which of the following could be the value of k?
(A) 22
(B) 26
(C) 35
(D) 54
(E) 60

n = (2.3.5.7.11.13) / 77 K
i.e. (2.3.5.13)/K
to n be integer , number will be 2, 3, 5,13 or multiplication of any of these numbers.
Hence answer is 26 as it is multiplication of 13*2.

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16. There were 36.000 hardback copies of a certain novel sold before the paperback version was issued. From the time the first paperback copy was sold until the last copy of the novel was sold. 9 times as many paperback copies as hardback copies were sold. If a total of 441.000 copies of the novel were sold in all, how many paperback copies were sold?
(A) 45.000
(B) 360.000
(C) 364.500
(D) 392.000
(E) 396.900

Solution:


Total Copies sold = 441.000
Total copies sold after first paperback copy was sold = 441.000 – 36.000 = 405.000
Total paperback copes sold = 9 / 10 * 405.000 = 364.500

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17. In the formula w = p ÷ (t √v) , integers p and t are positive constants. If w =2 when v = 1 and if when v = 64, then t =
(A) 1
(B) 2
(C) 3
(D) 4
(E) 16

Solution:

Cant solve for t.. bec p cant be determined . Please put answer in comments.

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18. Last year Mrs. Long received $160 in dividends on her shares of Company X stock, all of which she had held for the entire year. If she had had 12 more shares of the stock last year, she would have received $15 more in total annual dividends. How many shares of the stock did she have last year?
(A) 128
(B) 140
(C) 172
(D) 175
(E) 200

Solution:
Dividend on 12 shares = $15
Dividend on one share = $15 ÷ 12 = $1.25
Dividend of $160 is received on = $160 ÷ $ 1.25 = 128 shares.

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Month	Average Price per Dozen
April May June	$1.26 $1.20 $1.08

19. The table above shows the average (arithmetic mean) price per dozen of the large grade A eggs sold in a certain store during three successive months. If 2/3 as many dozen were sold in April as in May, and twice as many were sold in June as in April, what was the average price per dozen of the eggs sold over the three-month period?

(A) $1.08

(B) $1.10

(C) $1.14

(D) $1.16

(E) $1.18

Solution:

Let the number of dozens of eggs sold in May = x

Then number of dozens of eggs sold in April = 2/3x

Then number of dozens of eggs sold in June = 4/3x

Average price of dozen of eggs = (1.26 *2/3x + 1.2*x+ 1.08 * 4/3x) ÷ (2/3x + x + 4/3x)

= (.84x + 1.2x + 1.44x) ÷ (3x)

= 3.48x ÷ 3x

= 1.16

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20. If y ≠ 3 and 3x/y is a prime integer greater than 2, which of the following must be true?
Ⅰ. x = y
Ⅱ. y = 1
Ⅲ. x and y are prime integers.
(A) None
(B) Ⅰ only
(C) Ⅱonly
(D) Ⅲonly
(E) Ⅰand Ⅲ

Solution: 3X/Y should be 3. it means x=y.

Y =1 may or may not be true. to tell this we have to know the x value. since we do not the x value, Y=1 is uncertain.

looking at the answers B is right.

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If you have better answer or quick way of doing it. Please put in comments. We will introduce it to page.

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