RikoteMaster/aqua-rat-mcqa huggingface dataset:
Question: Joe needs to paint all the airplane hangars at the airport, so he buys 360 gallons of paint to do the job. During the first week, he uses 1/4 of all the paint. During the second week, he uses 1/3 of the remaining paint. How many gallons of paint has Joe used?
A.18
B.180
C.175
D.216
Answer: B.180
Explanation: Total paint initially = 360 gallons
Paint used in the first week = (1/4)*360 = 90 gallons.
Remaning paint = 270 gallons
Paint used in the second week = (1/3)*270 = 90 gallons
Total paint used = 180 gallons.
Option B
Question: Find the number that fits somewhere into the middle of the series. Some of the items involve both numbers and letters
Look at this series: C99, F102, __, L108, O111, ... What number should fill the blank?
A.I105
B.J106
C.K107
D.L109
Answer: A.I105
Explanation: A
I105
In this series, the letters progress by 3, and the numbers increase by 3.
Question: Which of the following numbers is between 1⁄3 and 2⁄5?
A..45
B..35
C..29
D..22
Answer: B..35
Explanation: 1/3 = .333...
2/5 = .4
The only number between these two is 0.35.
The answer is B.
Question: The average of 10 numbers is calculated as 18. It is discovered later on that while calculating the average, one number namely 36 was wrongly read as 26. The correct average is?
A.A)16
B.B)18
C.C)19
D.D)22
Answer: C.C)19
Explanation: Explanation:
10 * 18 + 36 – 26 = 190=>190/10 = 19
C)
Question: A coin is weighted so that the probability of heads on any flip is 0.2, while the probability of tails is 0.8. If the coin is flipped 5 times independently, which of the following represents the probability that tails will appear no more than twice?
A.(0.2)^5 + 5(0.2)^4(0.8) + 10(0.2)^3(0.8)^2
B.(0.3)^5 + 4(0.3)^4(0.8) + 6(0.3)^3(0.8)^2
C.(0.3)^5 + 3(0.3)^4(0.8) + 2(0.3)^3(0.8)^2
D.(0.3)^5 + 2(0.3)^4(0.8) + (0.3)^3(0.8)^2
Answer: A.(0.2)^5 + 5(0.2)^4(0.8) + 10(0.2)^3(0.8)^2
Explanation: Probability of Head, P(H) = 0.2
Probability of Tail, P(T) = 0.8
Tail will appear NO more than twice
i.e. favourable cases
2 Tails and 3 Heads, Probability = 5C2*(0.2)^3*(0.8)^2
1 Tail and 4 Heads, Probability = 5C1*(0.2)^4*(0.8)^2
0 Tail and 5 Heads, Probability = (0.2)^5
Required Probability = Sum of all Favourable cases = (0.2)^5 + 5(0.2)^4(0.8) + 10(0.2)^3(0.8)^2
Answer: option A
Question: If circles x and y have the same area and circle x has a circumference of 14π, half of the radius of circle y is:
A.16
B.8
C.4
D.3.5
Answer: D.3.5
Explanation: x be radius of circle x
y be radius of circle y
given:
pi*x^2 = pi*y^2
also,
2*pi*x = 14*pi
x =7
thus y =7
y/2 = 3.5
Ans: D
Question: If n is an integer number, and m=115n, then m must be divisible by which of the following?
A.11
B.15
C.23
D.25
Answer: C.23
Explanation: m=115n
m=5∗23n
So, m must be divisible by 5 & 23 both...
Among the given options only (C) satisfies.
Question: Ram draws a card randomly among card 1-23 and keep it back. Then Sam draws a card among those. What is the probability that Sam has drawn a card greater than ram.
A.11/27
B.11/24
C.11/23
D.11/21
Answer: C.11/23
Explanation: If Ram draws 1, then Sam can draw anything from 2 to 23 = 22 ways
If Ram draws 2, then Sam can draw anything from 3 to 23 = 21 ways
. . . .
. . . .
If Ram draws 23, Sam has no option = 0 ways.
Total required ways = 22 + 21 + 20 + . . . . + 0 = 22×23222×232 = 253
Total ways of drawing two cards = 23×23
Required probability = 253/529=11/23
Answer:C
Question: 3.5 can be expressed in terms of percentage as?
A.0.35%
B.3.5%
C.35%
D.350%
Answer: D.350%
Explanation: 3.5 = 35/10 = (35/10 * 100) = 350%
ANSWER:D
Question: 4 dice are thrown simultaneously on the board. Find the probability show the same face.
A.1/121
B.1/213
C.1/216
D.2/211
Answer: C.1/216
Explanation: The total number of elementary events associated to the random experiments of throwing four dice simultaneously is:
=6×6×6×6=64=6×6×6×6=64
n(S)=64n(S)=64
Let XX be the event that all dice show the same face.
X={(1,1,1,1,),(2,2,2,2),(3,3,3,3),(4,4,4,4),(5,5,5,5),(6,6,6,6)}X={(1,1,1,1,),(2,2,2,2),(3,3,3,3),(4,4,4,4),(5,5,5,5),(6,6,6,6)}
n(X)=6n(X)=6
Hence required probability,
=n(X)n(S)=664=n(X)n(S)=664
=1/216
C
Question: The length of the bridge, which a train 130 metres long and travelling at 45 km/hr can cross in 30 seconds, is:
A.2878
B.277
C.245
D.299
Answer: C.245
Explanation: Speed = [45 X 5/18] m/sec = [25/2] m/sec Time = 30 sec Let the length of bridge be x metres. Then, (130 + x)/30 = 25/2 => 2(130 + x) = 750 => x = 245 m.Answer: C
Question: Tom can hit a target 3 times in 6 shots, Marry can hit the target 2 times in 6 shots & mike can hit 4 times in 4 shots. What is the probability that atleast 2shots hit the target ?
A.3/4
B.4/5
C.2/3
D.5/7
Answer: C.2/3
Explanation: Ashwini hits the target definitely, hence required probability that atleast 2 shots hit the target is given by
Karan hits tha target and Raja not hit the target.
or
Karan does not hit the target and Raja hits the target.
or.
Karan hits the target and Raja hits the target
= 2/6 x 3/6 + 4/6 x 3/6 + 2/6 x 3/6
= 24/36 = 2/3
C
Question: A took 15 seconds to cross a rectangular field diagonally walking at the rate of 52 m/min and B took the same time to cross the same field along its sides walking at the rate of 68m/min. The area of the field is?
A.33
B.60
C.88
D.27
Answer: B.60
Explanation: Explanation:
length of the diagonal = 52*(15/60) =13m
sum of length and breadth = 68*(15/60) = 17m
\inline {\color{Black}\sqrt{l^{2}+b^{2}}=13 \; or\; l+b=17 }
area =lb = 1/2[(2lb)] = 1/2[(l+b)² - (l²+b²)] = 1/2[17² -169]
=1/2*120 = 60 sq meter
Answer: B) 60
Question: 3,14,39,_?
A.81
B.82
C.83
D.84
Answer: D.84
Explanation: n^3 + n^2 + n
Soln : 1^3+1^2+1 = 3
2^3+2^2+2 = 8+4+2 = 14
3^3+3^2+3 = 27+9+3 = 39
4^3+4^2+4 = 64+16+4 = 84
ANSWER:D
Question: In a certain lottery, the probability that a number between 12 and 20, inclusive, is drawn is 1/6 . If the probability that a number 12 or larger is drawn is 1/2 , what is the probability that a number less than or equal to 20 is drawn?
A. 1/18
B. 1/6
C. 2/3
D. 1/2
Answer: C. 2/3
Explanation: You can simply use sets concept in this question. The formula
Total = n(A) + n(B) - n(A and B) is applicable here too.
Set 1: Number 12 or larger
Set 2: Number 20 or smaller
1 = P(Set 1) + P(Set 2) - P(Set 1 and Set 2) (combined probability is 1 because every number will be either12 or moreOR20 or lessOR both)
1/2 + P(Set 2) - 1/6 = 2/3
P(Set 2) = 2/3
Answer (C)
Question: In how many ways can the word DAILY arranged?
A.720
B.100
C.620
D.120
Answer: D.120
Explanation: DAILY-there are 5 letters.
so we have 5 possible arrangements.
5!=5*4*3*2*1=120 was
Answer is option D
Question: The value of x + x(xx) when x = 2 is:
(a) 10, (b) 16, (c) 18, (d) 36, (e) 64
A.10
B.2
C.8
D.6
Answer: A.10
Explanation: x + x(xx)
Put the value of x = 2 in the above expression we get,
2 + 2(22)
= 2 + 2(2 × 2)
= 2 + 2(4)
= 2 + 8
= 10
Answer is A.
Question: A cylinder of height h is 6/7 of water. When all of the water is poured into an empty cylinder whose radius is 25 percent larger than that of the original cylinder, the new cylinder is 3/5 full. The height of the new cylinder is what percent of h?
A.25%
B.50%
C.60%
D.70%
Answer: D.70%
Explanation: Basically we can disregard the radius is 25% information, as we are only asked about the height of the original and the new cylinder.
This is becausethe new cylinder is 3/5 fullmeans the same as that it's height is 3/5.
Original cylinder 6/7
New cylinder 3/5
So 3/5/6/7 = 3/5*7/6= 0.70 or 70%.
Answer D
Question: A certain sum of money at simple interest amounted Rs.1240 in 6 years at 4% per annum, find the sum?
A.998
B.1000
C.890
D.646
Answer: B.1000
Explanation: 1240 = P [1 + (6*4)/100]
P = 1000
Answer: B
Question: An aeroplane covers a certain distance at a speed of 1500 kmph in 2 hours. To cover the same distance in 1 2/3 hours, it must travel at a speed of:
A.2200 kmph
B.1800 kmph
C.1900 kmph
D.2100 kmph
Answer: B.1800 kmph
Explanation: Explanation: Distance = (1500 x 2) = 3000 km.
Speed = Distance/Time
Speed = 3000/(5/3) km/hr. [We can write 1 2/3 hours as 5/3 hours]
Required speed = 3000 x 3/5 km/hr = 1800 km/hr.
Answer: Option B
Question: The total weight of six people of equal weights inside a car is 12% of the total weight of the car (inclusive of the six people inside). If three people get down from the car, what will be the ratio of the weight of people inside the car to the weight of the empty car?
A.44
B.45
C.46
D.47
Answer: A.44
Explanation: Ans. Assume total weight of the car along with people is 100kgs.
Then Total weight of 6 people is 12% of 100 = 12 kgs. So empty car weight = 88 kgs
That is, the weight of a single person = 12/6 = 2 kgs.
If three people get down from the car, then there is a six kgs reduction in the total weight and the weight of the remaining people is only 6 kgs.
The required ratio = 6 : 88 = 3 : 44
Answer A
Question: If the cost price of 12 pens is equal to the selling price of 7 pens, the gain percent is:
A.80%
B.90%
C.71%
D.40%
Answer: C.71%
Explanation: Explanation:
Let C.P. of each pen be Re. 1.
Then, C.P. of 7 pens = Rs. 7; S.P. of 7 pens = Rs. 12.
Gain % = 5/7 * 100 = 71%
Answer:C
Question: John would make the 3-letter codes with 26 alphabets in condition that the middle letter must be vowel and the first letter and the third letter must be different from each other and both are consonant. How many cases of the codes are there?
A.1,980
B.2,020
C.2,100
D.2,200
Answer: C.2,100
Explanation: No of Vowels: 5 (A, E, I ,O ,U)
No of Consonants: 21 (26-5)
Now,
The Middle letter can be filled up in 5 ways.
The First letter can be filled in 21 ways ( As only consonants are allowed)
As one of the consonants is used in the first letter, the third letter can be filled up in 20 ways, so that the letter must be different from each other.
So total number of ways = 21 x 5 x 20 = 2100
Answer: C
Question: The sum of the two numbers is 15 and their product is 35. What is the sum of the reciprocals of these numbers?
A.15/35
B.1/35
C.35/8
D.7/32
Answer: A.15/35
Explanation: Let the numbers be a and b. Then, a+b=15 and ab =35.
a+b/ab=15/35; (1/b+1/a)=15/35
Sum of reciprocals of given numbers=15/35.
Correct Option: A
Question: Of the 645 speckled trout in a certain fishery that contains only speckled and rainbow trout, the number of males is 45 more than twice the number of females. If the ratio of female speckled trout to male rainbow trout is 4:3 and the ratio of male rainbow trout to all trout is 3:20, how many female rainbow trout are there?
A.192
B.195
C.200
D.205
Answer: D.205
Explanation: It will be easier if you draw a table for this one. Let T be the total trout.
total speckled trout (ST) = 645
If female speckled trout = fST
then male speckled trout = 45 + 2fST
Therefore 45+ 2fST + fST = 645
We solve and get fST = 200
and mST= 445
Ratio of fST to mRT = 4:3
We have fST = 200
Solve ratio to get mRT = 150
Now ratio of mRT to T = 3:20
We have mRT = 150
Solve ratio to get T = 1000
Now unless youve drawn the table it might get a little complex.
From above we have mST and mRT
mST + mRT = 445 + 150 = 595
So T - 595 = total female trouts
1000 - 595 = 405 female trouts
We have from above that fST = 200
So (whew!) fRT = 405 - 200 = 205 answer is D
Question: There are 16 bees in the hive , then 7 more fly. How many bees are there in all?
A.7
B.23
C.12
D.17
Answer: B.23
Explanation: 16+7=23 . Answer is B.
Question: Rs.2500 is divided into two parts such that if one part be put out at 5% simple interest and the other at 6%, the yearly annual income may be Rs.140. How much was lent at 5%?
A.Rs.1500
B.Rs.1300
C.Rs.1200
D.Rs.1000
Answer: D.Rs.1000
Explanation: Explanation:
(x*5*1)/100 + [(2500 - x)*6*1]/100 = 140
X = 1000
ANSWER IS D
Question: If two dice are thrown together, the probability of getting an even number on one die and an odd number on the other is?
A.1/8
B.1/2
C.4/2
D.1/9
Answer: B.1/2
Explanation: The number of exhaustive outcomes is 36.
Let E be the event of getting an even number on one die and an odd number on the other. Let the event of getting either both even or both odd then
= 18/36 = 1/2
P(E) = 1 - 1/2 = 1/2.
Answer: B
Question: Two taps A and B can fill a cistern in 12 minutes and 18 minutes respectively. They are turned on at the same time. If the tap A is turned off after 4 minutes, how long will tap B take to fill the rest of the cistern ?
A.8 minute
B.9 min.
C.10 min.
D.7 min.
Answer: A.8 minute
Explanation: In one min, (A + B) fill the cistern = 1⁄12 + 1⁄18 = 5⁄36th
In 4 min, (A + B) fill the cistern = 5⁄36 × 4 = 5⁄9th
Rest part = 1 - 5⁄9 = 4⁄9th
∵ 1⁄18th part is filled by B in one min.
∴ 4⁄9th part is filled by B in 18 × 4⁄9 = 8 min
Answer A
Question: IF x is an integer greater than 6, then all the following must be divisible by 3 EXCEPT
A.a) 3x^3
B.b)x (x+3)
C.c)9 x^2
D.d)3 x^7
Answer: B.b)x (x+3)
Explanation: *any number multiplied by 3, will be divided by 3
for option b if we put value x=7, 7*10=70 is not divisible by 3
Answer : B
Question: Jordan passed a point on a track at a constant speed of 50 miles per hour. Then, 15 minutes later, Pauline passed the same point at a constant speed of 60 miles per hour. If both runners maintained their speeds , how long after he passed the gas station did Pauline catch up with Jordan?
A.30 min
B.45 min
C.1 hr
D.1 hr 15 min
Answer: D.1 hr 15 min
Explanation: When Pauline is at the point, Jordan is 50/4 miles ahead on the ahead. (the distance covered by him in 15 min)
every hour, pauline runs 10 miles more than Mary. how many hours will it takes him to run 50/4 miles more? the answer is (50/4)/10 = 25/20=1h15min.
Answer D
Question: If two girls starting from same point, walking in the opposite directions with 7km/hr and 3km/hr as average speeds respectively. Then the distance between them after 12 hours is?
A.120
B.121
C.122
D.123
Answer: A.120
Explanation: Explanation:
Total distance = Distance traveled by person A+ Distance traveled by person B
= (7Ã— 12) + (3Ã— 12)
= 84+ 36
= 120 Km
Answer:A
Question: Perimeter of the backwheel is 9 feet and front wheel is 7 feet at a certain distance. The front wheel gets 10 revolutions more than the back wheel. What is the distance?
A.305
B.310
C.315
D.320
Answer: C.315
Explanation: If distance=d
Number of revolutions by back wheel=d/9 & front wheel=d/7
(d/7)-(d/9)=10 ,2d=630 ,d=315
ANSWER:C
Question: Angelo and Isabella are both salespersons. In any given week, Angelo makes $550 in base salary plus 8 percent of the portion of his sales above $3,000 for that week. Isabella makes 10 percent of her total sales for any given week. For what amount of weekly sales would Angelo and Isabella earn the same amount of money?
A. 15,500
B. 24,500
C. 25,500
D. 26,500
Answer: A. 15,500
Explanation: OFFICIAL SOLUTION:
The problem asks for the amount of weekly sales it takes for Angelo and Isabella to earn the same amount of money. You can write an equation that sets Angelo’s and Isabella’s weekly earnings equal to each other, with x representing weekly sales. Weekly earnings for each salesperson equal base salary plus commission. So Angelo’s earnings are 550 + (0.08)(x – 3,000), and Isabella’s are 0.10x. Set up the equation and solve:
550 + (0.08)(x – 3,000) = 0.10x
Distribute the 0.08:
550 + 0.08x – 240 = 0.10x
Combine terms and subtract 0.08x from both sides:
310 = 0.02x
Divide both sides by 0.02:
15,500 = x
Your answer is A.
Question: John was thrice as old as Tom 6 yrs back. He will be 5/3 times as old as Tom 6 years hence. How old is Tom today?
A.10
B.11
C.12
D.16
Answer: C.12
Explanation: Sol. RMBflSk -6 = 3 (Ravi - 6) ...(j)
and Rupesh + 6 = | (Ravi + 6) ...(ii)
Solving both the equations we get,
Ravi = 12 years.
C
Question: A car covers a distance of 624 km in 6 ½ hours. Find its speed?
A.104 kmph
B.776 kmph
C.886 kmph
D.887 kmph
Answer: A.104 kmph
Explanation: 624/6 = 104 kmph
Answer: A
Question: If (8x)/(11y/z -1) = 3x then which of the following is true?
A.z = -3y
B.z = -y
C.z = -x
D.z = 3y
Answer: D.z = 3y
Explanation: (8x)/(11y/z - 1) = 3x
=> 8z/(11y - z) = 3
=> 8z = 33y - 3z
=> 11z = 33y
=> z = 3y
Hence Ans is D
Question: Which of the following could be the sides of an obtuse angled triangle? all values are degree only
I. 51, 29, 100
II. 32, 33,4 4
III. 52, 53, 55
A.I and III only
B.I only
C.III only
D.I and II only
Answer: B.I only
Explanation: obtuse angled triangle means one angle more than 90deg and total 180 deg.. I
HENCE, (B)
Question: Excluding stoppages, the speed of a bus is 5454 kmph and including stoppages, it is 4545kmph. For how many minutes does the bus stop per hour?
A.12
B.11
C.10
D.9
Answer: C.10
Explanation: Explanation:
Speed of the bus excluding stoppages =54=54 kmph
Speed of the bus including stoppages =45=45 kmph
Loss in speed when including stoppages =54âˆ’45=9 kmph=54âˆ’45=9 kmph
â‡’â‡’ In 11 hour, bus covers 99 km less due to stoppages.
Hence, time in which the bus stops per hour
= Time taken to cover 99 km
=distancespeed=954 hour=16 hour =distancespeed=954 hour=16 hour =606 min=10 min
ANSWER IS C
Question: For a certain set of numbers, if x is in the set, then x - 3 is also in the set. If the number 4 is in the set , which of the following must also be in the set ?
a) 4
b) 1
c) -5
A.a only,
B.b only,
C.c only,
D.a and b only
Answer: B.b only,
Explanation: Notice that the question asks which of the following MUST be in the set, not COULD be in the set.
Since 1 is in the set, then so must be 4-3=1. Similarly, since 1 is in the set, then so must be 1-3=-2. Could 4 and -1 be in the set? Certainly, but we don't know that for sure.
Answer: B.
Regarding your question: we don't know which is the source integer in the set, if it's 1, then 4 won't be in the set but if the source integer is say 7, then 4 will be in the set. So, 4 may or may not be in the set.
Question: 5^100 is divided by 18. Then what is the remainder?
A.11
B.14
C.13
D.16
Answer: C.13
Explanation: Here N = 18=2×32
ϕ(18)=18(1−12)(1−13) = 6
So 56 when divided by 18, remainder is 1.
So we can write the given expression 5100=(56)16×54 = (1)16×54 = 52×52=7×7=49
Now 49 when divided by 18, remainder is 13.
C
Question: If two letters are chosen at random (without replacement) from the word EQUAL, what is the probability that none of the letters are vowels?
A.1/5
B.2/5
C.1/10
D.3/10
Answer: C.1/10
Explanation: The number of ways to choose two letters is 5C2 = 10
The number of ways to choose two consonants is 2C2 = 1.
P(no vowels) = 1/10
The answer is C.
Question: If nn is a positive integer and (n+1)(n+3)(n+1)(n+3) is odd, then (n+2)(n+4)(n+2)(n+4) must be a multiple of which one of the following?
A.3
B.4
C.5
D.8
Answer: D.8
Explanation: (n+1)(n+3)(n+1)(n+3) is odd only when both (n+1)(n+1) and (n+3)(n+3) are odd. This is possible only when nn is even.
Hence, n=2mn=2m, where mm is a positive integer. Then,
(n+2)(n+4)=(2m+2)(2m+4)=2(m+1)2(m+2)=4(m+1)(m+2)(n+2)(n+4)=(2m+2)(2m+4)=2(m+1)2(m+2)=4(m+1)(m+2)
=4 * (product of two consecutive positive integers, one which must be even)=4 * (product of two consecutive positive integers, one which must be even) =4 * (an even number), and this equals a number that is at least a multiple of 8=4 * (an even number), and this equals a number that is at least a multiple of 8
Hence, the answer is (D).
Question: A car covers 4 successive 3km stretches at speed of 10kmph, 20kmph, 30kmph & 60kmph resp. Its average speed is
A.20 km/h
B.25 km/h
C.30 km/h
D.35 km/h
Answer: A.20 km/h
Explanation: Average speed = total distance / total time
total distance = 4 * 3 = 12 km
total time = 3/10 + 3/20 + 3/30 + 3/60
= 36/60 hr
speed =12/36 * 60 = 20 kmph
Answer : A.
Question: A reading list for a certain course consists of 10 books, of which 4 are poems and the rest are novels. Each student is required to read a selection of 4 books from the list, including at most 2 novels. How many selections of 4 books satisfy the requirements?
A.90
B.115
C.130
D.144
Answer: B.115
Explanation: Case 1 PPNN 2 Poems out of 4 and 2 Novels out of 6 i.e. 4C2 * 6C2 = 90
or
Case 2 PPPN 3 Poems out of 4 and 1 Novels out of 6 i.e 4C3 * 6C1 = 24
or
Case 3 PPPP 4 Poems out of 4 i.e. 4C4 =1
Total no. of selections case 1 + case 2+ case 3 = 90 + 24 +1 = 115
So answer must be B. 115
ANSWER:B
Question: The sum an integer N and its reciprocal is equal to 78/15. What is the value of N?
A.3
B.4
C.5
D.6
Answer: C.5
Explanation: Write equation in N as follows
N + 1/N = 78/15
Multiply all terms by N, obtain a quadratic equation and solve to obtain N = 5.
correct answer C
Question: An accurate clock shows 5 o'clock in the morning. Through how may degrees will the hour hand rotate when the clock shows 11 o'clock in the morning?
A.140º
B.160º
C.180º
D.200º
Answer: C.180º
Explanation: Angle traced by the hour hand in 6=(360x12)x6º = 180º.
answer :C
Question: How many 5 digit numbers can be formed which are divisible by 3 using the numerals 0, 1, 2, 3, 4, 5 (WITHOUT REPETITION)
A.216
B.3152
C.240
D.600
Answer: A.216
Explanation: Divisibility rule for 3: Sum of all digits is a multiple of 3
Selections for the above to be valid
1,2,3,4,5
No of possible ways = 5! = 120
0,1,2,4,5
No of possible ways = 4*4! = 96
Total = 120+96 = 216
ANS:A
Question: Monica planned her birthday party. She prepared 5 muffins for each of her guests and kept aside two additional muffins in case someone will want extra. After the party, it turned out that one of the guests didn't come but every one of the guests that did come ate six muffins and 3 muffins remained. How many guests did Monica plan on?
A.3.
B.4.
C.5.
D.6.
Answer: C.5.
Explanation: Let x be the number of guests.
Number of muffins prepared = 5x + 2
Number of muffins eaten + Number of muffins remaining = Number of muffins prepared
6(x - 1) + 3 = 5x + 2
6x - 3 = 5x + 2
x = 5
Answer: C
Question: Two pipes P and Q can fill a cistern in 10 and 15 minutes respectively. Both are opened together, but at the end of 3 minutes the first is turned off. How much longer will the cistern take to fill?
A.1/2
B.1/4
C.2/4
D.1/4
Answer: A.1/2
Explanation: 3/10 + x/15
= 1
x= 10 1/2
Answer: A
Question: In a marketing survey, 60 people were asked to rank three flavors of ice cream, chocolate, vanilla, and strawberry, in order of their preference. All 60 people responded, and no two flavors were ranked equally by any of the people surveyed. If 3/5 of the people ranked vanilla last, 1/10 of them ranked vanilla before chocolate, and 1/3 of them ranked vanilla before strawberry, how many people ranked vanilla first?
A.2 people
B.6
C.14
D.16
Answer: A.2 people
Explanation: FIRST OFFUnderstand what you are trying to find - In this case, you are trying to find how many people rank Vanilla first. For someone to rank Vanilla first, they have to rank them before Chocolate or Strawberry. This should be the very first hint that the correct answer cannot be bigger than 1/10 * 60 = 6.
So if you are time pressured on the real test, at least you narrowed down the selections to two choices - not bad. But what now?
Solve the easier problem. From 60 people, we are already told that 3/5 of them rank Vanilla last, leaving us with 2/5 of the people not ranking vanilla last - 24 people. So what are those combinations that vanilla are not last?
A) C, V, S
B) S, V, C
C) C, S, V (vanilla first)
D) S, C, V (vanilla first)
What are we ultimately hunting for here? The number of people that rank vanilla first:C+D
Now let's represent the given info into three equations
1) A+B+C+D = 24(2/5 of the people do not rank vanilla last)
2) A+C+D = 6(1/10 rank vanilla before chocolate)
3) B+C+D = 20(1/3 rank vanilla before strawberry)
Subtract equation 1 from 2, leaving us with
B = 24 - 6 = 18
Sub in B into equation 3 and solve for C+D
18 + C + D = 20
C+D = 20 - 18 = 2
There you go! 2 people out of the 60 ranked Vanilla as first A
Question: Water consists of hydrogen and oxygen, and the approximate ratio, by mass, of hydrogen to oxygen is 2:16. Approximately how many grams of oxygen are there in 135 grams of water?
A. 16
B. 72
C. 112
D. 120
Answer: D. 120
Explanation: Since the ratio, by mass, of hydrogen to oxygen is 2:16, then oxygen is 16/(2+16)=8/9 of mass of water.
Therefore there are 135*8/9=120 grams of oxygen in 135 grams of water.
Answer: D.
Question: If (x/y)=(7/5), find the value (x^2+y^2)/(x^2-y^2)
A.37/12
B.59/11
C.51/77
D.41/11
Answer: A.37/12
Explanation: = (x^2+y^2)/(x^2-y^2) = ( x^2 /y^2+ 1)/ ( x^2 /y^2-1) = [(7/5)^2+1] / [(7/5)^2-1]
= [(49/25)+1] / [(49/25)-1] = 37/12
Answer is A.