divelab/Sys2Bench huggingface dataset:
Question: If 7^r is a factor of the product of the integers from 1 to 100, inclusive, what is the largest value of r?
Options:
A)12
B)13
C)14
D)15
E)16
Rationale: So the question just means that we have to find all the multiples r of 7 between 1 to 100
So there are 14 multiples of 7 ( 7 - 98)
But 49 and 98 contain two 7's as factors
so we have to add 14 + 2 = 16
E
Answer: E
Question: How many integers between 100 and 190, inclusive, cannot be evenly divided by 3 nor 5?
Options:
A)42
B)44
C)46
D)48
E)50
Rationale: The total numbers between 100 and 190, inclusive, is 91.
3*34=102 and 3*63=189 so the number of multiples of 3 is 30.
5*20=100 and 5*38=190 so the number of multiples of 5 is 19.
However, the multiples of 15 have been counted twice.
15*7=105 and 15*12=180 so the number of multiples of 15 is 6.
Thus the total number is 91-30-19+6=48.
The answer is D.
Answer: D
Question: If selling price is doubled, the profit triples. Find the profit percent:
Options:
A)227
B)100
C)278
D)266
E)181
Rationale: Let C.P. be Rs. x and S.P. be Rx. y.
Then, 3(y - x) = (2y - x) => y = 2x
Profit = (y - x) = Rs. (2x - x) = Rs. x
Profit % = x/x * 100 = 100%.
Answer: B
Answer: B
Question: Each of 435 bags contains at least one of the following three items: raisins, almonds, and peanuts. The number of bags that contain only raisins is 10 times the number of bags that contain only peanuts. The number of bags that contain only almonds is 20 times the number of bags that contain only raisins and peanuts. The number of bags that contain only peanuts is one-fifth the number of bags that contain only almonds. 210 bags contain almonds. How many bags contain only one kind of item?
Options:
A)256
B)260
C)316
D)320
E)350
Rationale: there are total of 435 bags and 210 bags contain almonds.
From the diagram 20y=5x --> y=x/4. Now, Total=435={Almonds}+10x+y+x --> 435=210+10x+x/4+x --> x=20 --> # of bags that contain only one kind of item is the sum of yellow segments: 10x+x+5x=16x=320.
Answer: D.
Answer: D
Question: If the price of a certain bond on May 1st was 2/3 the price of the bond on June 1st and the price of the bond on July 1st was 25% greater than the price of the bond on May 1st. then the price of the bond on June 1st st was what percent of the average (arithmetic mean) price of the bond on May 1st and July 1st?
Options:
A)50%
B)75%
C)120%
D)133 1/3%
E)150%
Rationale: The price on June 1st = 12 (assume);
The price on May 1st = 2/3*12 = 8;
The price on July 1st = 8*1.25 = 10.
The average price of the bond on May 1st and July 1st = (8+10)/2 = 9.
The price of the bond on June 1st (12) is 4/3 times (134%) the average price of the bond on May 1st and July 1st.
Answer: D.
Answer: D
Question: I. x2 + 3x - 18 = 0,
II. y2 + y - 30 = 0 to solve both the equations to find the values of x and y?
Options:
A)If x < y
B)If x > y
C)If x ≤ y
D)If x ≥ y
E)If x = y or the relationship between x and y cannot be established.
Rationale: I. x2 + 6x - 3x - 18 = 0
=>(x + 6)(x - 3) = 0 => x = -6, 3
II. y2 + 6y - 5y - 30 = 0
=>(y + 6)(y - 5) = 0 => y = -6, 5
No relationship can be established between x and y.
ANSWER:E
Answer: E
Question: In a jar there are 15 white balls, 25 red balls, 10 blue balls and 20 green balls. How many balls must be taken out in order to make sure we took out 8 of the same color?
Options:
A)a)	8
B)b)	23
C)c)	29
D)d)	32
E)e)	53
Rationale: Thats 29 in C.
Answer: C
Question: Alice and Heather are 20 miles apart and walk towards each other along the same route. Alice walks at constant rate that is 1 mile per hour faster than heather's constant rate of 5 miles/hour. If Heather starts her journey 24 minutes after Alice, how far from the original destination has Heather walked when the two meet?
Options:
A)4 miles
B)8 miles
C)10 miles
D)6 miles
E)12 mile
Rationale: Original distance between S and H = 20 miles.
Speed of A = 5+1 = 6 mph, Speed of H = 5 mph.
Time traveled by H = t hours ---> time traveled by A = t+24/60 = t+2/5 hours.
Now, the total distances traveled by A and H = 20 miles ---> 6*(t+2/5)+5*t=20 ---> t= 8/5 hours. Thus H has traveled for 8/5 hours giving you a total distance for H = 5*8/5 = 8 miles.
B is thus the correct answer.
P.S.: based on the wording of the question, you should calculatehow far from theoriginal destination has Heather walkedwhen the two meet. 'Original destination' for H does not make any sense. Original destination for H was situated at a distance of 20 miles.
Answer: B
Question: Let p be a prime number greater than 2, and let N = 32p. How many odd divisors does N have?
Options:
A)1
B)2
C)3
D)4
E)5
Rationale: The answer is A, there is exactly one. Since is the largest (and only) even prime number, any p larger than 2 is odd. Then since 32 = 2*2*2*2, all of which are even, this means that p must be the only odd prime factor.
Answer: A
Question: Given two fair dice, what is the probability Z that the sum of their numbers is 4 if exactly one die shows a 3?
Options:
A)a) 2/11
B)b) 1/18
C)c) 3/11
D)d) 2/39
E)e) 1/5
Rationale: If exactly one die shows 3, there are 10 such combinations (given that we have 2 fair dice).
1:3, 2:3, 4:3, 5:3, 6:3,
3:1, 3:2, 3:4, 3:5, 3:6
The sum of numbers would be 4 for two events: 1:3, 3:1.
Z=2/11 would be the answer if 'at least one of the die shows a 3'.A
Answer: A
Question: If x and y are integers such that |y + 3| ≤ 3 and 2y – 3x + 6 = 0, what is the least possible value W of the product xy?
Options:
A)	-12
B)	-3
C)	0
D)	2
E)	None of the above
Rationale: How to deal with inequalities involving absolute values? First example shows us the so callednumber case
In this case we have |y + 3| ≤ 3 which is generalized |something| ≤ some number. First we solve as if there were no absolute value brackets:
y + 3 ≤ 3
y ≤ 0
So y is 0 or negative
Second scenario - remove the absolute value brackets. Put a negative sign around the other side of the inequality,ANDflip the sign:
y + 3 >= -3
y >= -6
Therefore we have a possible range for y: -6=<y<=0
Ok, so far so good, we're half way through. What about x?
Here's the formula: 2y – 3x + 6 = 0, rewrite it as 2y + 6 = 3x.
You can say that 2y + 6 is a multiple of 3 (=3x). So all values which must be integer must also satisfy this constraint. I'm just saying that, so it's easier to evaluate all the possible numbers W(-6, -3, 0). If you plug in y = 0, x will be 2 and xy = 0 as the lowest possible value.
Hence, Answer Choice C is the one to go.
Answer: C
Question: One-third of 1206 is what percent of 200?
Options:
A)3
B)201
C)300
D)None of these
E)cannot be determined
Rationale: Answer
Let one-third of 1206 is N% of 200.
∵ 1206/3 = (N x 200)/100
∴ N = (402 x 100) / 200 = 201
Correct Option: B
Answer: B
Question: In a clock (Conventional clock with numbers 1 to 12 in order)is cut into 3 pieces such that the sum of each piece in AP with a common difference of 1. What is the count of numbers in each piece? and what are the numbers?
Options:
A)25
B)26
C)27
D)28
E)29
Rationale: Sum of the 12 numbers=1+2+3+......+12=12*13/2=78
So we have to divide it into 3 slots such that the sum is 78 and sum of each slot is in AP, thus the 3 slots are 25,26,27. So the numbers can be:
3+4+5+6+7=25
8+9+10=27
11+12+1+2=26
ANSWER:B
Answer: B
Question: In how many different ways can the letters of the word ‘PREPTY’ be arranged?
Options:
A)120
B)230
C)330
D)340
E)360
Rationale: Number of ways = 61/21 6X5X4X3X2X1 - 360
E
Answer: E
Question: Mr. loyd wants to fence his Square shaped land of 150Sqft each side.If a pole is laid every 30ft how many poles do he need?
Options:
A)20
B)30
C)40
D)50
E)60
Rationale: If each side is 120 feet..then total perimeter is 150*4=600
poles every 20 feet hence no of poles= 600/20=30
ANSWER:B
Answer: B
Question: Find the number of ways in which 8 different beads can be arranged to form a necklace.
Options:
A)30
B)20
C)2520
D)60
E)80
Rationale: Since in forming a necklace clockwise and anticlockwise arrangements are not different, therefore,8 beads can be arranged to form a necklace in(8-1)! /2ways
= 7*6*5*4*3*2/2 = 2520 ways
Answer is C
Answer: C
Question: Double of Quarter of 1 percent written as a decimal is :
Options:
A)0.003
B)0.0005
C)0.25
D)0.005
E)None of these
Rationale: Explanation :
Solution: (2)*(1/4)*1% = 2*(1/4 * 1/100) = 1/400 = 2*0.0025 = 0.005.
Answer : D
Answer: D
Question: A cubical block of metal weighs 6 pounds. How much will another cube of the same metal weigh if its sides are twice as long?
Options:
A)48
B)12
C)36
D)60
E)None of these
Rationale: Explanation:
If you double the sides of a cube, the ratio of the surface areas of the old and new cubes will be 1: 4. The ratio of the volumes of the old and new cubes will be 1: 8.
Weight is proportional to volume. So, If the first weighs 6 pounds, the second weighs 6x8 pounds =48.
Answer: A
Answer: A
Question: What will come in place of the x in the following Number series? 11, 14, 19, 22, 27, 30, 35, 38, 43 , 46,51 ,x
Options:
A)23
B)26
C)54
D)35
E)45
Rationale: (C)
The pattern is + 3, + 5, + 3, + 5, …………
So the missing term is = 51 + 3 = 54.
Answer: C
Question: John has 10pairs of dark blue socks and 10pairs of black socks. He keeps them all in the same bag. If he picks out 3socks at random, then what is the probability thathe will get a matching pair?
Options:
A)0
B)1
C)4
D)5
E)6
Rationale: If he draws any combination of 3 socks he will definitely have the matching pair of either colour.
B)
Answer: B
Question: 105 candies are distributed to children with the same number of candies for each child. What can’t be the range which includes the number of children?
Options:
A)1~10.
B)10~20.
C)20~30.
D)30~40.
E)40~50.
Rationale: 105 = 3*5*7
There are factors in each range except for 40~50
The answer is E.
Answer: E
Question: A, B and C invested Rs.6300, Rs.4200 and Rs.10500 respectively, in a partnership business. Find the share of A in profit of Rs.12600 after a year?
Options:
A)3630
B)2881
C)2887
D)9977
E)3780
Rationale: 6300:4200:10500
3:2:5
3/10 * 12600 = 3780.Answer: E
Answer: E
Question: A box contains 5 green, 4 yellow and 3 white marbles, 3 marbles are drawn at random. What is the probability that they are not of the same colour?
Options:
A)13⁄44
B)41⁄44
C)13⁄55
D)52⁄55
E)None of these
Rationale: Total no. of ways of drawing 3 marbles
= 12C3 = 12×11×10/1×2×3=220
Total no. of ways of drawing marbles, which are of same colour = 5C3 + 4C3 + 3C3 = 10 + 4 + 1 = 15
\Probability of same colour = 15⁄220 = 3⁄44
\ Probability of not same colour = 1 - 3⁄44 =
Answer B
Answer: B
Question: The product of all the prime numbers less than 12 is closest to which of the following powers of 10 ?
Options:
A) 10^9
B) 10^3
C) 10^7
D) 10^6
E) 10^5
Rationale: P = 2*3*5*7*11
~ 10^ 3
Answer : B
Answer: B
Question: In a class of 20 students, 3 students did not borrow any books from the library, 9 students each borrowed 1 book, 4 students each borrowed 2 books, and the rest borrowed at least 3 books. If the average number of books per student was 2, what is the maximum number of books any single student could have borrowed?
Options:
A)13
B)14
C)15
D)16
E)17
Rationale: The total number of books the students borrowed is 20*2=40.
The students who borrowed zero, one, or two books borrowed 9*1+4*2=17 books.
The 4 students who borrowed at least three books borrowed 40-17=23 books.
If 3 of these students borrowed exactly 3 books, then the maximum that one student could have borrowed is 23-9= 14 books.
The answer is B.
Answer: B
Question: 9773 + x = 13200, then x is ?
Options:
A)3427
B)3237
C)3337
D)2337
E)None of these
Rationale: Answer
x = 13200 - 9773 = 3427
Option: A
Answer: A
Question: 8 coins are tossed. Whats is the probability that one and only one turns up head ?
Options:
A)1/256
B)1/32
C)1/64
D)1/128
E)1/189
Rationale: Explanation :
Possible outcomes(Sample space) when 8 coins are tossed are: (H, H, H, H, H, H, H, H), (H, H, H, H, H, H, H, T), (H, H, H, H, H, H, T, H), (H, H, H, H, H, H, T, T), ....... (T, T, T, T, T, T, T, T) = 28
Required outcomes: (H, T, T, T, T, T, T, T), (T, H, T, T, T, T, T, T), (T, T, H, T, T, T, T, T), (T, T, T, H, T, T, T, T), (T, T, T, T, H, T, T, T), (T, T, T, T, T, H, T, T), (T, T, T, T, T, T, H, T), (T, T, T, T, T, T, T, H) = 8
So the required probability = 8/28 = 1/32
Answer : B
Answer: B
Question: The average age of A, B and C is 29 years. If the average age of A and C is 32 years, what is the age of B in years ?
Options:
A)19
B)35
C)20
D)32
E)23
Rationale: Age of B = Age of (A + B + C) â€“ Age of (A + C) = 29 Ã— 3 â€“ 32 Ã— 2 = 87 â€“ 64 = 23 years
E
Answer: E
Question: The S.I. on a certain sum of money for 4 years at 12% per annum is half the C.I. on Rs. 6000 for 2 years at 15% per annum. The sum placed on S.I. is?
Options:
A)2019.625
B)2017.625
C)2013.625
D)2015.625
E)2011.625
Rationale: C.I. = [6000 * (1 + 15/100)2 - 6000]= (6000 * 23/20 * 23/20 - 6000) = Rs. 1935.Sum = (967.5 * 100)/(4 * 12)= Rs.2015.625Answer:D
Answer: D
Question: $300 is divided amongst A,B and C so that A may get 2/3 as much as B and C together, B may get 6/9 as much as A and C together, then the share of A is
Options:
A)$100
B)$150
C)$125
D)$120
E)$250
Rationale: A:(B+C) = 2:3
A's share = 300*2/5 = $120
Answer is D
Answer: D
Question: An equal number of desks and bookcases are to be placed along a library wall that is 15 meters long. Each desk is 2 meters long, and each bookshelf is 1.5 meters long. If the maximum possible number of desks and bookcases are to be placed along the wall, then the space along the wall that is left over will be how many meters W long?
Options:
A)0.5
B)1
C)1.5
D)2
E)3
Rationale: Let x be the number of desks and bookcases that are placed along the library wall.
2x + 1.5x < 15
3.5x < 15
Since x is a non negative integer, the largest number x can be is 4.
When x is 4, the desks and bookcases take up 3.5 * 4 = 14m, leaving 1m of empty space.
Thus, I believe the answer is B) 1
Answer: B
Question: Some persons can do a piece of work in 16 days. Two times the number of these people will do half of that work in?
Options:
A)3
B)4
C)5
D)6
E)8
Rationale: 16/ (2 * 2) = 4 days
Answer:B
Answer: B
Question: 35.62 + 408.926 + 48.2 = ?
Options:
A)622.441
B)492.243
C)492.746
D)622.443
E)None of these
Rationale: Explanation :
35.62 + 408.926 + 48.2 = 492.746. Answer : Option C
Answer: C
Question: The arithmetic mean and standard deviation of a certain normal distribution are 10.5 and 1, respectively. What value is exactly 2 standard deviations less than the mean?
Options:
A)6
B)8.5
C)11.5
D)12
E)12.5
Rationale: Mean = 10.5
Two standard deviations is 1 + 1 = 2
There could be two calues for this.
Mean + two standard deviations = 12.5
Mean - Two standard deviations = 8.5
Answer choice has 8.5 and so B is the answer.
Answer: B
Question: How many digits are in the number 50^8 × 8^3 × 11^2 x 10^2?
Options:
A)22
B)20
C)21
D)18
E)17
Rationale: Whenever you are asked to find the number of digits, try to bring the number in multiples of 10. This way, we can wasily calculate the umber of 0's through the powers of 10
50^8 × 8^3 × 11^2 x 10^2 = (5^2*2)^8*2^9*11^2x 10^2 = 5^16*2^17*11^2 x10^2 = 2*11^2*10^16x10^2 = 242*10^18
Hence we would have 16 trailing 0's and the three digits from 242
Total digits = 3 + 18 = 21
Correct Option: C
Answer: C
Question: McDonald's ran a campaign in which it gave game cards to its customers. These game cards made it possible for customers to win hamburgers, French fries, soft drinks, and other fast-food items, as well as cash prizes. Each card had 10 covered spots that could be uncovered by rubbing them with a coin. Beneath three of these spots were "No Prize" signs. Beneath the other seven spots were names of prizes, two of which were identical.
For example, one card might have two pictures of a hamburger, one picture of a Coke, one of French fires, one of a milk shake, one of 5 Dollar, one of 1000 Dollar and three "No Prize" signs. For this card the customer could win a hamburger.
To win on any card, the customers had to uncover the two matching spots (which showed the potential prize for that card) before uncovering a "No Prize"; any card with a "No Prize" uncovered was automatically void. Assuming that the two matches and the three "No Prize" signs were arranged randomly on the cards, what is the probability of a customer winning?
Options:
A)0.1
B)0.15
C)0.12
D)0.18
E)0.13
Rationale: Explanation :
As per the question there are 10 cover spots out of which
(i) Three spots are there with no prize (identical)
(ii) Two spots of the same sign (Prize)
(iii) Five other spots which are distinct
Let three spots of no prize are (x,x,x), two spots of same sign are b(P,P) and five other spots are (A,B,C,D,E).
Total number of cases without restriction:
=>10!/(3!x2!).
Total number of favourable cases happen in the following ways which shows sequence of can covering are as follows:-
CASE I: Second uncovering is P
=>P P _ _ _ _ _ _ _ _=8C3×5!×1.
CASE II:
=>_ _ P _ _ _ _ _ _ =7C3×5!×2.
CASE III:
=>6C3×5!×3.
CASE IV:
=>5C3×5!×4.
CASE V:
=>4C3×5!×5.
CASE VI:
=>3C3×5!×6.
Hence, the total number of favourable cases are :-
=>case (I+II+III+IV+V+VI)/(10!3!X2!).
=>0.10.
Answer : A
Answer: A
Question: A box contains 60 balls, numbered from 1 to 60. If three balls are selected at random and with replacement from the box, what is the probability that the sum of the three numbers on the balls selected from the box will be odd?
Options:
A)1/4
B)3/8
C)1/2
D)5/8
E)3/4
Rationale: I don't think order matters in this case, because 2 + 2 + 1 = 2 + 1 + 2
My answer is: 1/4 A
Answer: A
Question: Shanthi went to the stationers and bought things forh Rs. 25, out of which 30 paise went on sales tax on taxable purchases. If the tax rate was 6%, then what was the cost of the tax free items?
Options:
A)19.7
B)20
C)25
D)24.5
E)32.9
Rationale: amt tax consider as x
6% of x = 30/100
x=30/100*100/6 = 5
cost of tax free items = (25-(5+0.30) = 19.70
ANSWER A
Answer: A
Question: A boy has Rs. 720 in the denominations of one-rupee notes, five-rupee notes and ten-rupee notes. The number of notes of each denomination is equal. What is the total number of notes that he has ?
Options:
A)90
B)110
C)140
D)135
E)120
Rationale: Let number of notes of each denomination be x.
Then x + 5x + 10x = 720
16x = 720
x = 45
Hence, total number of notes = 3x = 135
D
Answer: D
Question: If ax = by
Options:
A)log a/b = x/y
B)log a/ log b = x/y
C)log a/ log b = y/x
D)None
E)Cannot be determined
Rationale: Solution
ax = by	= log ax= log by
‹=› x log a = y log b
‹=› log a / log b= y/x.
Answer C
Answer: C
Question: Fox jeans regularly sell for $15 a pair and Pony jeans regularly sell for $18 a pair. During a sale these regular unit prices are discounted at different rates so that a total of $4 is saved by purchasing 5 pairs of jeans: 3 pairs of Fox jeans and 2 pairs of Pony jeans. If the sum of the two discounts rates is 18 percent, what is the discount rate on Pony jeans?
Options:
A)45%
B)13%
C)11%
D)12%
E)15%
Rationale: You know that Fox jeans costs $15, and Pony jeans costs $18, you also know that 3 pairs of Fox jeans and 2 pairs of Pony jeans were purchased. So
3(15) = 45 - Fox
2(18) = 36 - Pony
The total discount discount is $4 and you are asked to find the percent discount of pony jeans,
so 45(18 - X)/100 + 36(X)/100 = 4 or 45*18 -45*x + 36*x =400 or - 9x = 400-45*18
or x= 410/9 =45%
=45% A
Answer: A
Question: A batsman scored 1 runs in his 18th innings and that makes his average 18. Find his average upto
the 17th innings?
Options:
A)19
B)20
C)21
D)22
E)23
Rationale: Avg = sum of the value/No. of the value
18 = (Sum of the first 17 innings Score +1)/18
(18*18-1)= Sum of the first 17 innings Score
Sum of the first 17 innings Score = 323
Ave upto 17 innings = 323/17=19
ANSWER:A
Answer: A
Question: The perimeter of a semi circle is 140 cm then the radius is?
Options:
A)27
B)28
C)19
D)11
E)12
Rationale: 36/7 r = 140 => r
= 27
Answer: A
Answer: A
Question: According to the directions on the can of frozen orange juice concentrate, 1 can of concentrate is to be mixed with 3 cans of water to make orange juice. How many 12 ounces cans of the concentrate are required to prepare 280 6 ounces servings of orange juice?
Options:
A)a) 35
B)b) 34
C)c) 50
D)d) 67
E)e) 100
Rationale: Its A.
Total juice rquired = 280*6 = 1680 ounce
12 ounce concentate makes = 12 * 4 = 48 ounce juice
Total cans required = 1680/48 = 35.
Answer A
Answer: A
Question: Fill in the blank with the correct number in this number series: 20, 65, 27, 61, __, 57, 41, ?
Options:
A)34
B)35
C)38
D)25
E)56
Rationale: This is a simple alternating subtraction and addition series; The first series begins with 20 and adds 7; the second begins with 65 and subtracts 4, and so on.
Answer: A
Answer: A
Question: The Racing magic takes 50 seconds to circle the racing track once. The Charging bull makes 40 rounds of the track in an hour. If they left the starting point together, how many minutes will it take for them to meet at the starting point for the second time?
Options:
A)3
B)6
C)9
D)12
E)15
Rationale: Time taken by Racing magic to make one circle = 50 seconds
Time taken byCharging bullto make one circle = 60 mins / 40 = 1.5 mins = 90 seconds
LCM of 90 and 50 seconds = 450 seconds
Time taken for them to meet at the starting point for the second time = 450 *2 =900 seconds = 15 mins
Answer E
Answer: E
Question: A loan has a variable interest rate that fluctuates between 4% and 9% of the base payment per month. If base payments remain at $500 each month and an additional monthly surcharge of 2% is added to the combined (base + interest), what would be the greatest possible payment due in any given month?
Options:
A)$555.45
B)$555.55
C)$555.90
D)$556.45
E)$556.90
Rationale: The maximum interest in one month is 9%*$500=$45.
The combined payment (base plus interest) is $500 + $45 = $545.
The greatest possible payment is: 1.02*$545=$555.90.
The answer is C.
Answer: C
Question: In the city of San Durango, 60 people own cats, dogs, or rabbits. If 30 people owned cats, 40 owned dogs, 16 owned rabbits, and 12 owned exactly two of the three types of pet, how many people owned all three?
Options:
A)2
B)4
C)8
D)7
E)32
Rationale: You are considering a case when cat dogs and rabbit are not exactly equal to 12 .
The solution shall be 60=30+40+16-12-2x
and hence x shall be 7.
Answer D
Answer: D
Question: A sum of Rs. 12500 amounts to Rs. 15500 in 4 years at the rate of simple interest. What is the rate of interest?
Options:
A)4%
B)5%
C)6%
D)8%
E)14%
Rationale: S.I. = (15500 - 12500) = Rs. 3000/-
Rate = (100 * 3000) / (12500 * 4) = 6%
ANSWER:C
Answer: C
Question: The population of a town increased from 1,75,000 to 2,62,500 in a decade. What is the average percent increase of population per year?
Options:
A)5%
B)4%
C)6%
D)7%
E)2%
Rationale: increase in the population in 10 yrs=
262500 - 175000 = 87500
percent increase in the population in 10 yrs
= 87500/175000*100 = 8750/175=50%
avg percent increase of population per yr
50%/10=5%
ANSWER A
Answer: A
Question: A tightrope approximately 320 m long is suspended between two poles. During a performance, a break occurs in the line. Assume that the line has an equal chance of breaking anywhere along its length. What is the probability that the break happened in the first 50 meters of the rope?
Options:
A)27/32
B)1/2
C)5/32
D)5/27
E). 2/3
Rationale: Simly 50/320 = 5/32
Answer will be (C)
Answer: C
Question: 8.036 divided by 0.04 gives :
Options:
A)200.9
B)2.06
C)20.06
D)100.9
E)200.6
Rationale: = 8.036/0.04
= 803.6/4
= 200.9
Answer is A.
Answer: A
Question: If s,u, and v are positive integers and 2s=2u+2v, which of the following must be true?
i. s=u
ii. u is not equal to v
iii. s > v
Options:
A)None
B)I only
C)II only
D)III only
E)II and III
Rationale: Given: 2s=2u+2v --> s=u+v. Now, since s, u, and v are positive integers then s is more than either u or v, so I is never true and III is always true. As for II: it's not necessarily true, for example 4=2+2. So, we have that only option III must be true.
Answer: D.
Answer: D