RikoteMaster/aqua-rat-mcqa huggingface dataset:
Question: Seven children â€” A, B, C, D, E, F, and G â€” are going to sit in seven chairs in a row. The children C & F have to sit next to each other, and the others can sit in any order in any remaining chairs. How many possible configurations are there for the children?
A.600
B.720
C.1440
D.4320
Answer: C.1440
Explanation: There are six different pairs of â€œnext to each otherâ€ chairs. For each pair, children C & F could be in either order, so thatâ€™s 6*2 = 12 possibilities for these two.
Now, consider the other five children. For any configuration of C & F, the remaining five children could be seated in any order among the five remaining seats. Five items in any order â€” thatâ€™s a permutation of the 5 items â€”- 5P5 = 5! = 120. For any single configuration of C & F, there are 120 ways that the other children could be seated in the remaining seats.
Finally, weâ€™ll combine with the Fundamental Counting Principle. We have 12 ways for the first two, and 120 ways for the remaining five. Thatâ€™s a total number of configurations of 12*120 = 1440. Answer = C
Question: If 222 = 6
333 = 12
444 = 20
Then, 555 = ??
A.30
B.35
C.40
D.45
Answer: A.30
Explanation: 222=2*2+2=6
333=3*3+3=12
444=4*4+4=20
555=5*5+5=30
ANSWER:A
Question: A room is 15m long, 4m broad and 3m height. Find the cost of white washing its four walls at 50p per m2 ?
A.28
B.57
C.58
D.83
Answer: B.57
Explanation: Explanation:
2 *3(15 + 4) = 114
114 * 1/2 = Rs.57
Answer: Option B
Question: If a2 + b2 + c2 = 281 and ab + bc + ca =4, then a + b + c is
A.16
B.18
C.22
D.17
Answer: D.17
Explanation: by formula ,
(a + b + c)^2= a^2 + b^2 + c^2 + 2(ab + bc + ca),
since, a^2 + b^2 + c^2 = 281 and ab + bc + ca = 4,
(a + b + c)^2 = 281 + 2(4)
= 289
= 17^2
therefore: a + b + c = 17
ANSWER:D
Question: Of all the students in a certain dormitory, 1/2 are first-year students and the rest are second-year students. If 4/5 of the first-year students have not declared a major and if the fraction of second-year students who have declared a major is 2 times the fraction of first-year students who have declared a major, what fraction of all the students in the dormitory are second-year students who have not declared a major?
A.1/15
B.1/5
C.4/15
D.3/10
Answer: D.3/10
Explanation: tot students=x
1st year student=x/2----> non majaor=4/5(x/2)-----> maj=1/5(x/2)
2nd year student=x/2---->maj=2(1/5(x/2))=2/10(x)--->non major=x/2-2/10(x)=3/10(x)
hence 3/10
D
Question: How many integers between 400 and 1000 are there such that their unit digit is odd?
A.250
B.150
C.300
D.400
Answer: C.300
Explanation: There are 600 numbers from 401 to 1000 (inclusive).
Half of the numbers are odd, so there are 300 odd numbers.
The answer is C.
Question: Divide $450 among A,B in the ratio 1:4. How many $ that A get?
A.$90
B.$500
C.$150
D.$250
Answer: A.$90
Explanation: Sum of ratio terms = 1+4 = 5
A = 450*1/5 = $90
Answer is A
Question: If 16,666 – n is divisible by 11, and 0 < n < 11, what is n?
A.1
B.3
C.5
D.7
Answer: A.1
Explanation: 16,666 / 11 = 1515 with a remainder of 1.
We need to subtract the remainder to get a multiple of 11.
The answer is A.
Question: If 3f+ 1 < 2f < 9 - f, which of the following must be true?
I. -1 < f < 3
II. f < -1
III. f < 3
A. II only
B. III only
C. I and II only
D. II and III only
Answer: A. II only
Explanation: Just try plugging f in a number like 0 or 1, and you'll quickly find that the inequality doesn't hold.
So you can immediately cross out any answers with I or III, leaving just (A) II only.
Question: ((1^m)/(5^m))((1^16)/(4^16)) = 1/(2(10)^31)
What is m?
A.17
B.18
C.34
D.31
Answer: D.31
Explanation: ((1^m)/(5^m))((1^16)/(4^16)) = 1/(2(10)^31)
((1/5)^m ) * ((1/2)^32) = 1/ (2 * (2*5)^31))
2^36 will cancel out,since 1 can be written as 1^35 ,So (1/5)^m= (1/5)^31
((1/5)^m ) * ((1/2)^32) = 1/[(2^32)*(5^31)]
So, m=31
Answer D
Question: Two trains 150 m and 160 m long run at the speed of 60 km/hr and 40 km/hr respectively in opposite directions on parallel tracks. The time which they take to cross each other is?
A.10.7
B.10.9
C.10.26
D.11.16
Answer: D.11.16
Explanation: Relative speed = 60 + 40 = 100 km/hr.
= 100 * 5/18 = 250/9 m/sec.
Distance covered in crossing each other = 150 + 160 = 310 m.
Required time = 310 * 9/250 = 11.16 sec.
Answer:D
Question: How many positive integers less than 60 are equal to the product of a positive multiple of 5 and an even number?
A.4
B.5
C.9
D.10
Answer: B.5
Explanation: Number of positive multiples of 5 less than 60 = 11
5*1=5
5*2=10
5*3=15
5*4=20
5*5=25
5*6=30
5*7=35
5*8=40
5*9=45
5*10=50
5*11=55
Only 5 of the above are a product of a positive multiple of 5 and an even number - 10,20,30,40,50
Answer B
Question: In what time will a train 120 m long cross an electric pole, it its speed be 185 km/hr?
A.2.3 sec
B.2.9 sec
C.2.7 sec
D.8.7 sec
Answer: A.2.3 sec
Explanation: Speed = 185 * 5/18 = 51 m/sec
Time taken = 120/51 = 2.3 sec.
Answer: A
Question: How many positive three-digit integers are divisible by both 2 and 7?
A.40
B.46
C.52
D.64
Answer: D.64
Explanation: A number to be divisible by both 2 and 7 should be divisible by the least common multiple of 2 and 7 so by 14.
Multiples of 14 between 100 and 999, inclusive is
(last-first)/multiple+1=(994-112)/14+1=63+1=64
Answer: D.
Question: A bank offers an interest of 5% per annum compounded annually on all its deposits. If $10,000 is deposited, what will be the ratio of the interest earned in the 4th year to the interest earned in the 5th year?
A.1:5
B.625:3125
C.100:105
D.100^4:105^4
Answer: C.100:105
Explanation: The interest earned in the 1st year = $500
The interest earned in the 2nd year = $500*1.05
The interest earned in the 3rd year = $500*1.05^2
The interest earned in the 4th year = $500*1.05^3
The interest earned in the 5th year = $500*1.05^4
(500*1.05^3)/(500*1.05^4) = 1/1.05=100/105.
Answer: C.
Question: There are 300 employees in a room. 99% are managers. How many managers must leave the room to bring down the percentage of managers to 98%?
A.200
B.175
C.150
D.125
Answer: C.150
Explanation: There are 297 managers and 3 others. The 3 others would compose 2% of the total number of people if there were 150 people in the room. Thus 150 managers must leave.
The answer is C.
Question: The area of a square garden is Q square feet and the perimeter is P feet. If Q=P+21, what is the perimeter of the garden in feet?
A.24
B.28
C.32
D.36
Answer: B.28
Explanation: Let x be the length of one side of the square garden.
x^2 = 4x + 21
x^2 - 4x - 21 = 0
(x-7)(x+3) = 0
x = 7, -3
P = 4(7) = 28
The answer is B.
Question: In a forest 10 rabbits were caught, tagged with electronic markers, then released. A month later, 10 rabbits were captured in the same forest. Of these 10 rabbits, it was found that 2 had been tagged with the electronic markers. If the percentage of tagged rabbits in the second sample approximates the percentage of tagged rabbits in the forest, and if no rabbits had either left or entered the forest over the preceding month, what is the approximate number of rabbits in the forest?
A.150
B.70
C.50
D.1,500
Answer: C.50
Explanation: Given that 10 rabbits were caught and tagged in the first sample, and 10 were caught in the second sample of which 2 had tags, then 2/10 approximates the fraction of the population of tagged rabbits in the forest.
Proportion: A/x = C/D
A = rabbits tagged first sample
C/D = the fraction of the population of tagged rabbits in the forest
x = the population of rabbits in the forest
Proportion: 10/x = 1/5
x = (1)(10)(5) = 50. There are approximately 50 rabbits in the forest.
Answer is C
Question: A and B started a business with initial investments in the ratio 14 : 15 and their annual profits were in the ratio 7 : 6. If A invested the money for 10 months, for how many months did B invest his money?
A.2
B.4
C.6
D.8
Answer: D.8
Explanation: Sol.
Suppose A invested Rs. 14x for 10 months and B invested Rs. 15x for y months. Then,
14x * 10 / 15x * y = 7/6 ⇒ y = 840 / 105 = 8
Hence, B invested the money for 8 months.
Answer D
Question: Sum of 3 consecutive even no.'s is 34 more than the 1st no. of the series. Find the middle no.?
A.8
B.12
C.14
D.16
Answer: D.16
Explanation: Let the numbers be x, x+2 and x+4 then
x + x+2 + x+4 = x+34
3x + 6 = x + 34
2x = 28
\inline \therefore x = 14
\inline \therefore Middle number = x + 2 = 14+2 = 16
D
Question: A, B and C enter into partnership. A invests some money at the beginning, B invests double the amount after 6 months, and C invests thrice the amount after 8 months. If the annual gain be Rs.18000. A's share is?
A.6028
B.6020
C.6000
D.6098
Answer: C.6000
Explanation: x* 12 : 2x* 6: 3x* 4
1:1:1
1/3 * 18000 = 6000
Answer: C
Question: If A, B, and C invested some amount in the ratio 3:2:5 in a buisiness. If the total profit incurrec after a year was Rs.12100, find the share of B.
A.Rs.6050
B.Rs.3630
C.Rs.2420
D.Rs.3360
Answer: C.Rs.2420
Explanation: Total profit after a year =12100 Ratio of profit share =3x+2x+5x=10x 10x = 12100. Therefore x=1210 B's share = 2*1210 =Rs.2420 Answer : C
Question: If an amount of Rs 13,640 is divided equally among
62 students, approximately how much amount will
each student get?
A.Rs 206
B.Rs 210
C.Rs 220
D.Rs 218
Answer: C.Rs 220
Explanation: Amount received by each student = 13640/62 â‰ˆ Rs
220
Answer C
Question: Three friends sai, vani and ram divide $1020 amongs them in such a way that if $10, $20 and $15 are removed from the sums that sai, vani and ram received respectively, then the share of the sums that they got will be in the ratio of 8 : 4 : 2. How much did ram receive?
A.$147
B.$142
C.$242
D.$122
Answer: B.$142
Explanation: Explanatory Answer
Let the sums of money received by A, B and C be x, y and z respectively.
Then x - 10 : y - 20 : z -15 is 8a : 4a : 2a
When $10, $20 and $15 are removed, we are removing a total of $45 from $1020.
Therefore, 8a + 4a + 2a = 1020 - 45 = 975
i.e., 14a = 975
or a = 70.
We know that z - 2 = 70a = 2 * 70 =140
Therefore, z = 140 + 2 = $142
The correct choice is (B)
Question: A pupil's marks were wrongly entered as 83 instead of 63. Due to that the average marks for the class got increased by half (1/2). The number of pupils in the class is:
A.38
B.39
C.40
D.46
Answer: C.40
Explanation: Let there be x pupils in the class.
Total increase in marks = (x x 1/2) = x/2
x /2 = (83 - 63)
x /2= 20
x= 40
Answer is C
Question: There are two circles of different radii. The are of a square is 784 sq cm and its side is twice the radius of the larger circle. The radius of the larger circle is seven - third that of the smaller circle. Find the circumference of the smaller circle?
A.16
B.15
C.12
D.87
Answer: C.12
Explanation: Let the radii of the larger and the smaller circles be l cm and s cm respectively. Let the side of the square be a cm.
a2 = 784 = (4)(196) = (22).(142)
a = (2)(14) = 28
a = 2l, l = a/2 = 14
l = (7/3)s
Therefore s = (3/7)(l) = 6 Circumference of the smaller circle = 2∏s
= 12∏ cm.
Answer:C
Question: A train running at the speed of 90 km/hr crosses a pole in 6 seconds. What is the length of the train?
A.150
B.2766
C.187
D.267
Answer: A.150
Explanation: Speed=(60 * 5/18) m/sec = (50/3) m/sec Length of the train = (Speed x Time) = (50/3 * 9) m = 150 m.
Answer: A
Question: A and B invests Rs.8000 and Rs.9000 in a business. After 4 months, A withdraws half of his capital and 2 months later, B withdraws one-third of his capital. In what ratio should they share the profits at the end of the year?
A.32:47
B.32:49
C.32:45
D.32:46
Answer: C.32:45
Explanation: A : B
(8000*4)+(4000*8) : (9000*6)+(6000*6)
64000 : 90000
32 : 45
Answer: C
Question: In a graduate physics course, 60 percent of the students are male and 30 percent of the students are married. If two-sevenths of the male students are married, what fraction of the female students is single?
A.2/7
B.17/30
C.1/2
D.27/40
Answer: D.27/40
Explanation: let assume there are 100 students of which 60 are male and 40 are females
if 30 are married then 70 will be single.
now its given that two-sevenths of the male students are married that means 2/7 of 60 = 17 males are married
if 30 is the total number of students who are married and out of that 17 are males then the remaining 13 will be females who are married.
total females = 40
married females = 13
then single females = 40-13 = 27
we need to find the fraction of female students who are single i.e single female students / total female student
= 27/40 [D]
Question: In how many ways, a committee of 5 members can be selected from 7 men and 5 ladies, consisting of 3 men and 2 ladies?
A.75
B.152
C.175
D.350
Answer: D.350
Explanation: (3 men out 7) and (2 ladies out of 5) are to be chosen
Required number of ways = (7C3 * 5C2) = 350
Answer is D
Question: Compound X contains elements A and B at an approximate ratio, by weight, of 2:10. Approximately how many grams of element B are there in 300 grams of compound X?
A.54
B.162
C.250
D.270
Answer: C.250
Explanation: total number of fractions =2+10=12
element B constitutes =10 out of 12 parts of X
So in 300 gms of X have 300*10/12=250 gms of B
and 300-250=50 gms of A.
cross check:- A/B = 50/250 =2/10 (as given)
Ans C
Question: The sum of the present ages of two persons A and B is 60. If the age of A is twice that of B, find the sum of their ages 7 years hence?
A.58
B.74
C.62
D.70
Answer: B.74
Explanation: Explanation:
A + B = 60, A = 2B
2B + B = 60 => B = 20 then A = 40.
7 years, their ages will be 47 and 27.
Sum of their ages = 47 + 27 = 74.
B)
Question: A trader bought a car at 20% discount on its original price. He sold it at a 40% increase on the price he bought it. What percent of profit did he make on the original price?
A.10%
B.11%
C.12%
D.15%
Answer: C.12%
Explanation: Original price = 100
CP = 80
S = 80*(140/100) = 112
100 - 112 = 12%
ANSWER:C
Question: Two hoses are pouring water into an empty pool. Hose 1 alone would fill up the pool in 6 hours. Hose 2 alone would fill up the pool in 3 hours. How long would it take for both hoses to fill up two-thirds of the pool?
A.5/12 hours
B.5/8 hours
C.4/3 hours
D.12/7 hours
Answer: C.4/3 hours
Explanation: Hose 1 would fill the pool alone in 6 hours
i.e. 1 Hour work of Hose 1 = 1/6
Hose 2 would fill the pool alone in 3 hours
i.e. 1 Hour work of Hose 2 = 1/3
i.e. 1 Hour work of Hose1 and Hose2 together = (1/6)+(1/3) = (1/2)
i.e. (1/2) work can be finished by Hose1 and Hose2 together in 1 hour
i.e. (2/3) work can be finished by Hose1 and Hose2 together in 1*(2)*(2/3) = 4/3 hour
Answer: Option C
Question: 10^185 - 10^9 = Which of the following best approximates the value of the expression above?
A.10^170
B.10^180
C.10^(185)
D.10^160
Answer: C.10^(185)
Explanation: 10^185 - 10^9 =
Which of the following best approximates the value of the expression above?
Yes, answer 10^185 is correct. Note that we need approximate value of the given expression and as 10^(185) is much larger number than 10^(9) then 10^(9) is pretty much negligible in this case: 10^(185)-10^(9)=~10^(185).
Answer : C
Question: 45 pupil,out of them 12 in debate only and 22 in singing only.then how many in both?
A.9
B.10
C.11
D.12
Answer: C.11
Explanation: The intersection for two = 45 - 12 - 22 = 11 play both games.
ANSWER:C
Question: In the first 10 overs of a cricket game, the run rate was only 4.8. What should be the run rate in the remaining 40 overs to reach the target of 282 runs?
A.4.25
B.5.85
C.6.25
D.7
Answer: B.5.85
Explanation: Required run rate = 282 - (4.8 x 10) =	234
234/40= 5.85
B)
Question: A and B can do a work in 6 days, B and C in 2 days and C and A in 8 days. In how many days will the work be completed, if all three of them work together?
A.12/24
B.48/63
C.45/56
D.48/19
Answer: D.48/19
Explanation: One day work of A and B = 1/6
One day work of B and C = 1/2
One day work of C and A = 1/8
2(A + B + C) = 1/6 + 1/2 + 1/8
2(A + B + C) = 19/24
(A + B + C) = 19/48
Number of days required = 48/19 days.
Answer: D
Question: A series of book was published at seven years intervals when the seventh book was published the total sum of publication year was 13, 1524. first book was published in?
A.1911
B.1912
C.1913
D.1914
Answer: A.1911
Explanation: First of all total sum of publication years is 13524 not 131524.
Let the first year is x.
so the years are x, x+7, x+14, x+21, x+28, x+35 and x+42.
Now Sum of all years = (n/2)(2a+(n-1)d) [n=number of years, a=first year, d=difference of two years]
=(7/2)(2x+(7-1)7)
=7x+147
Again Sum of all years = 13524
So, 7x+147=13524
=> x=1911
ANSWER:A
Question: Every student of a certain school must play an instrument. In last year, 1/2 of the students picked a woodwind instrument, 2/5 of the students picked a brass instrument, and all of the other students took percussion. In this year, 1/2 of the students who play woodwind and 1/4 of the students who play brass instruments left school, other students did not leave, and no fresh student come in. What fraction of all students play either a woodwind or brass instrument?
A.7/13
B.4/5
C.9/13
D.11/13
Answer: D.11/13
Explanation: Lets pick smart numbers. Total number of students: 20
Woodwind (1/2): 10
Brass (2/5): 8
Percussion (1/10): 2
After leaving school
Woodwind: 5
Brass: 6
Percussion: 2
New Total number of students: 13
Woodwind and Brass: 11
Answer 11/13 or D
Question: Two trains 111 meters and 165 meters in length respectively are running in opposite directions, one at the rate of 80 km and the other at the rate of 65 kmph. In what time will they be completely clear of each other from the moment they meet?
A.6.85
B.7.18
C.7.16
D.7.15
Answer: A.6.85
Explanation: T = (111 + 165)/ (80 + 65) * 18/5
T = 6.85
Answer: A
Question: The market value of a 10.5% stock, in which an income of Rs. 756 is derived by investing Rs. 6000, brokerage being 1/4%, is:
A.83.08
B.114
C.114.75
D.124
Answer: A.83.08
Explanation: Face Value = Rs. 6000.
Dividend = 10.5%.
Annual Income = Rs. 756.
Brokerage per Rs. 100 = Rs. 0.25.
Dividend is always paid on the face value of a share.
Face Value * Dividend / (Market Value + Brokerage per Rs. 100) = Annual income.
= 6000 * 10.5 / 756 = Market Value of Rs. 100 stock + Brokerage per Rs. 100.
= Market Value of Rs. 100 stock + Brokerage per Rs. 100 = Rs. 83.33.
= Market Value of Rs. 100 stock = Rs. 83.33 - Re. 0.25.
= Market Value of Rs. 100 stock = Rs. 83.08
answer : A
Question: A, B and C start a business each investing 20,000. After 5 months A withdrew 10000, B withdrew 15000 and C invests 5000 more. At the end of the year, a total profit of 58000 was recorded. Find the share of B.
A.20,000
B.21,200
C.28,200
D.13,500
Answer: D.13,500
Explanation: Ratio of the capitals of A, B and C
= 20000 Ã— 5 + 10000 Ã— 7 : 20000 Ã— 5 + 5000 Ã— 7 : 20000 Ã— 5 + 25000 Ã— 7
=170000 : 135000 : 275000 = 170 : 135 : 275.
Bâ€™s share = (58000 Ã— 135â„580) = 13500
Answer D
Question: A train of 24 carriages, each of 60 meters length, when an engine also of 60 meters length is running at a speed of 60 kmph. In what time will the train cross a bridge 1.5 km long?
A.1 Â½ mins
B.3 mins
C.3 Â½ mins
D.7 mins
Answer: B.3 mins
Explanation: Explanation:
D = 25 * 60 + 1500 = 3000 m
T = 3000/60 * 18/5 = 180 sec = 3 mins
ANSWER IS B
Question: In a certain country ½ of 5 = 3. Assuming the same proportion, what would be the value of 1/3 of 10 ?
A.3
B.4
C.8
D.7
Answer: B.4
Explanation: B
4
Question: Which of the following Equation is NOT equivalent to 25x^2 = y^2 - 49?Which of the following Equation is NOT equivalent to 25x^2 = y^2 - 4?
A.25x^2 + 4 = y^2
B.75x^2 = 3y^2 - 12
C.25x^2 = (y + 2)(y - 2)
D.5x = y - 7
Answer: D.5x = y - 7
Explanation: 1) Take the original equation and try to manipulate it so that it becomes 4 of the 5 answers
2) Take each answer choice and see if you can manipulate it to make it into the original equation.
With Answer D, we have...
5X = Y - 7
If we square both sides, we end up with....
(5X)^2 = (Y-7)^2
(5X)(5X) = (Y-7)(Y-7)
25X^2 = Y^2 - 14Y + 49
This is NOT a match for the original equation, so it's the answer that is NOT equivalent.D
Question: A certain bus driver is paid a regular rate of $14 per hour for any number of hours that does not exceed 40 hours per week. For any overtime hours worked in excess of 40 hours per week, the bus driver is paid a rate that is 75% higher than his regular rate. If last week the bus driver earned $976 in total compensation, how many total hours did he work that week?
A.54
B.51
C.59
D.55
Answer: C.59
Explanation: For 40 hrs = 40*14=560
Excess = 976-560=416
For extra hours =.75(14)=10.5+14=24.5
Number of extra hrs =416/24.5=18.82=19 approx.
Total hrs =40+19=59
Answer C 59
Question: 24 oz of juice P and 25 oz of juice F are mixed to make smothies X and Y . The ratio of p to F in smothie X is 4 is to 1 and that in Y is 1is to 5.How many ounces of juice P are contained in the smothie X?
A.5
B.10
C.15
D.20
Answer: D.20
Explanation: Easy way to solve this question is start from the answer and then conform the information provided in the question.
we can start from option D i.e 20 ... as a quantity of juice P in X because it is the only one option that gets divided by 4 is 20 ... since in the X the juice P to F ratio is 4:1
this gives us that quantity of juice P in X = 20 therefore quantity of Juice F will be 5 ... hence ratio = 4:1
This will lead to quantity of juice P in X = 4 and quantity of Juice F = 20 ... hence ratio 1:5
if we calculate total Juice P = 24 and total of juice V = 25
it fits because totals are same as what mentioned in the question ...
thus ans is D
Question: Harkamal purchased 3 kg of grapes at the rate of 70 per kg and 9 kg of mangoes at the rate of 55 per kg. How much amount did he pay to the shopkeeper?
A.A)700
B.B)1065
C.C)1070
D.D)1075
Answer: A.A)700
Explanation: Cost of 3 kg grapes = 70 × 3 = 210.
Cost of 9 kg of mangoes = 55 × 9 = 490.
Total cost he has to pay = 210 + 490 = 700
A
Question: There are 76 lights which are functional and each is controlled by a separate On/Off switch. Two children A and B start playing with the switches. A starts by pressing every third switch till he reaches the end. B, thereafter, presses every fifth switch till he too reaches the end. If all switches were in Off position at the beggining, How many lights are switched On by the end of this operation?
A.28
B.30
C.32
D.34
Answer: A.28
Explanation: Editing my solution:
Number of switches = 76
Number of switches turned on by A: 3, 6, ... 75 = 25
Number of switches turned on by B: 5, 10, ....75 = 15
Few switches are turned on by A and later turned off by B: LCM(3,5) = 15x = 15, 30,....90 = 6.
Subtract the above 6 switches from both A and B as they are turned off.
Number of switches that are turned on = (25 - 6) + (15 - 6) = 28
Answer: A
Question: For positive integers n, the integer part of the nth term of sequence B equals n, while the infinite decimal part of the nth term is constructed in order out of the consecutive positive multiples of n, beginning with 2n. For instance, B_1 = 1.2345678…, while B_2 = 2.4681012… The sum of the first seven terms of sequence B is between:
A.28 and 29
B.29 and 30
C.30 and 31
D.31 and 32
Answer: C.30 and 31
Explanation: First, construct the first seven terms, though only out to a few decimal places, following the given pattern.
B_1 = 1.23…
B_2 = 2.46…
B_3 = 3.69…
B_4 = 4.812…
B_5 = 5.10…
B_6 = 6.12…
B_7 = 7.14…
Now, to add up the first seven terms, you should be strategic about how many decimal places to keep. You can drop the hundredths place and get a good approximation with the tenths place—and if you find the sum too close to a boundary between choices, then you can refine your answer if necessary.
1.2 + 2.4 + 3.6 + 4.8 + 5.1 + 6.1 + 7.1 = 30.3
Including more decimal places would only add a couple of tenths to the sum—not enough to tip the sum over 31.
The correct answer is C.
Question: A jar is filled of liquid which is 3 parts water and 5 parts alcohol. How much of this mixture should be drawn out and replaced such that this mixture may contain half
water and half alcohol?
A.28%
B.10%
C.20%
D.29%
Answer: C.20%
Explanation: Explanation:
Let the jar initially contain 8 litres of mixture:3 litres water and 5 litres alcohol
Let x litres of this mixture is drawn out and is replaced by x litres of water.
Amount of water now in the solution: 3-(3/8)x+x.
Amount of alcohol now in the solution: 5-(5/8)x
Desired ratio: 1/1
=>3-(3/8)x+x=5-(5/8)x
x=8/5 litres
%x=((8/5)/8)*100=20%
ANSWER: C