RikoteMaster/aqua-rat-mcqa huggingface dataset:
Question: 36 identical chairs must be arranged in rows with the same number of chairs in each row. Each row must contain at least three chairs and there must be at least three rows. A row is parallel to the front of the room. How many different arrangements are possible?
A.2
B.4
C.5
D.6
Answer: C.5
Explanation: Three conditions have to be satisfied.
1. The number of students per row has to be at least 3.
2. Number of row has to be at least 3.
3. Equal number of students has to be seated in a row.
The following arrangements satisfy all 3 conditions.
Arrangement 1: 3 students to a row; 12 rows.
Arrangement 2: 4 students to a row; 9 rows.
Arrangement 3: 6 students to a row; 6 rows.
Arrangement 4: 9 students to a row; 4 rows.
Arrangement 5: 12 students to a row; 3 rows.
You will observe that the number of students in a row is a factor of 36.
So, an alternative and faster approach is to list down factors of 36: 1, 2, 3, 4, 6, 9, 12, 18, and 36.
And then start from 3 and quickly find out if the number of rows is at least 2.
Both the conditions are satisfied for the following factors : 3, 4, 6, 9, and 12. i.e., 5 arrangements.
Option C is answer
Question: A and B start a business with Rs.16000 and Rs.32000 respectively. Hoe should they share their profits at the end of one year?
A.1:2
B.3:4
C.2:5
D.3:7
Answer: A.1:2
Explanation: They should share the profits in the ratio of their investments.
The ratio of the investments made by A and B =
16000 : 32000 => 1:2
ANSWER:A
Question: How many numbers up to 300 and 700 are divisible by 2, 3 and 7 both together?
A.9
B.10
C.11
D.12
Answer: A.9
Explanation: (700 – 300)/42 = 9 22/42
=> 9 Numbers
ANSWER:A
Question: In how many ways can 5 letters be posted in 3 post boxes, if any number of letters can be posted in all of the three post boxes?
A.152
B.3^5
C.125
D.217
Answer: B.3^5
Explanation: The first letter can be posted in any of the 3 post boxes. Therefore, we have 3 possibilities.
Similarly, the second, the third, the fourth and the fifth letter can each be posted in any of the 3 post boxes.
Each of the 5 letters has 3 possibilities because we can post any number of letters in all of the boxes.
Therefore, the total number of ways the 5 letters can be posted in 3 boxes is
3 * 3 * 3 * 3 * 3 = 3^5
Ans: B
Question: Two trains are running in opposite directions with the same speed. If the length of each train is 120 m and they cross each other in 12 sec, then the speed of each train is?
A.30
B.31
C.36
D.25
Answer: C.36
Explanation: Let the speed of each train be x m/sec.
Then, relative speed of the two trains = 2x m/sec.
So, 2x = (120 + 120)/12 => x = 10
Speed of each train = 10 m/sec.
= 10 * 18/5 =- 36 km/hr.
Answer: Option C
Question: There are k-2 members in a certain band, including Jim and Ellen. Two members are to be selected to attend the Grammy awards ceremony. If there are 15 possible combinations in which Jim and Ellen are not selected, what is the value of k?
A.8
B.9
C.10
D.11
Answer: C.10
Explanation: There are k-2 members in the band, and k-4 members without Jim and Ellen.
(k-4)C2 = 15
(k-4)(k-5)/2 = 15
(k-4)(k-5) = 30 = 6*5
k = 10
The answer is C.
Question: Find the cost of fencing around a circular field of diameter 28 m at the rate of Rs.1.50 a meter?
A.177
B.132
C.176
D.278
Answer: B.132
Explanation: 2 * 22/7 * 14 = 88
88 * 1 1/2
=Rs.132
Answer:B
Question: Mala and Usha shared their water from a bottle. If mala and usha both drank for the same amount of time, but usha only drank 2/10 of the total bottle, what was the ratio of mala and usha speed?
A.1:4
B.2:5
C.4:1
D.3:5
Answer: C.4:1
Explanation: The answer is suppose to be C. 4:1. It's from the GMATPrep
Answer:option C
Question: The velocity of a falling object in a vacuum is directly proportional to the amount of time the object has been falling. If after 5 seconds an object is falling at a speed of 84 miles per hour, how fast will it be falling after 12 seconds?
A.204 miles per hour
B.212 miles per hour
C.216 miles per hour
D.1080 miles per hour
Answer: A.204 miles per hour
Explanation: Since Velocity is proportional to Time
Velocity =k*Timewhere k is the constant of proportionality
Time= 5 seconds
Velocity = 5k = 84 miles per Hour
i.e. k = 17
i.e. The relation between Velocity and Time becomes
Velocity =17*Time
Time= 12 seconds
Velocity = 17*12 = 204 miles per hour
Answer: Option A
Question: One farmer takes 12 hours to plow his field, but another farmer takes 8 hours to plow the same field with an equal tractor. How long would it take them together to plow the field?
A.12.5 hours
B.4 hours
C.4.8 hours
D.5 hours
Answer: C.4.8 hours
Explanation: First we must find the least common denominator.
1/12 + 1/8 =
2/24+3/24=5/24
They can plow the field 5/24 per hour. Now to find the actual time, Lets use T for time as a variable.
1/T = 5/24
T/1 = 24/5 = 4.8 hours
Answer is C
Question: Set #1 = {J, K, L, M, E}
Set #2 = {K, L, M, N, O, P}
There are these two sets of letters, and you are going to pick exactly one letter from each set. What is the probability of picking at least one vowel?
A.1/6
B.1/8
C.1/3
D.1/4
Answer: C.1/3
Explanation: At least questions are best solved by taking the opposite scenario and subtracting it from 1. Probability of choosing no vowel from set 1 is 4/5 and set 2 is 5/6. Multiply these to get 2/3. Therefore, probability of picking at least one vowel = 1-2/3=1/3.
ANSWER : C
Question: Bobby bought 2 shares, and which he sold for $106 each. If he had a profit of 120% on the sale of one of the shares but a loss of 47% on the sale of the other share, then on the sale of both shares Bobby had...
A.a profit of 10
B.a profit of 8
C.a loss of 38
D.a loss of 10
Answer: C.a loss of 38
Explanation: Cost of the first stock: 106/2.2=48.18, so profit from it 106-48.18=57.18
Cost of the second stock: 106/0.53=200, so loss from it 106-200=-94
Overall loss 57.18-94=-36.82
Answer: C.
Question: If 1st October is Sunday, then 1st November will be
A.Saturday
B.Thursday
C.Wednesday
D.Tuesday
Answer: C.Wednesday
Explanation: Explanation :
Given that 1st October is Sunday
Number of days in October = 31
31 days = 3 odd days
(As we can reduce multiples of 7 from odd days which will not change anything)
Hence 1st November = (Sunday + 3 odd days) = Wednesday
Answer : Option C
Question: In how many ways can a group of 5 men and 2 women be made out of a total of 7 men and 3 women?
A.1
B.61
C.63
D.64
Answer: C.63
Explanation: We need to select 5 men from 7 men and 2 women from 3 women.
Number of ways to do this
= 7C5 × 3C2
= 7C2 × 3C1 [∵ nCr = nC(n-r)]
= 21 * 3 =63
C
Question: What is the probability of getting exactly three heads on five flips of a fair coin?
A.1/32
B.3/32
C.1/4
D.5/16
Answer: D.5/16
Explanation: 5 Flips of a fair coin to get = HHHTT = no. of ways this can be achieved = 5!/3!x2! = 10
Probability to get any of the above 10 arrangements (HHHTT) = (1/2)^5 = 1/32
Total probability = 1/32 x 10 = 5/16
Answer:D
Question: The average of first seven multiples of 8 is:
A.9
B.16
C.15
D.32
Answer: D.32
Explanation: Explanation :
(8(1+2+3+4+5+6+7)/7
= 8x28/7
= 32
Answer : D
Question: A rectangular field has area equal to 150 sq m and perimeter 50 m. Its length and breadth must be?
A.11
B.99
C.89
D.10
Answer: D.10
Explanation: lb = 150
2(l + b) = 50 => l + b = 25
l – b = 5
l = 15 b = 10
Answer:D
Question: A and B can complete a job in 30 and 20 days respectively. They start working together and B leaves 5 days before the work is finished. In how many days is the total work finished?
A.15
B.6
C.5
D.4
Answer: A.15
Explanation: A and B can do the job in 30 and 20 days individually, so together they will take 30*20/(30 + 20) = 600/50 = 12 days.
The work that A can do in 5 days is 1/6th. So, A left when 5/6th of the work was done or A left after (5/6)*12 = 10 days
Total time taken = 10 + 5 = 15 days.
ANSWER:A
Question: A man can row 6 kmph in still water. When the river is running at 1.2 kmph, it takes him 1 hour to row to a place and black. What is the total distance traveled by the man?
A.5.98
B.8.98
C.5.76
D.5.87
Answer: C.5.76
Explanation: M = 6
S = 1.2
DS = 7.2
US = 4.8
x/7.2 + x/4.8 = 1
x = 2.88
D = 2.88 * 2 = 5.76
Answer:C
Question: The length of rectangle is thrice its breadth and its perimeter is 80 m, find the area of the rectangle?
A.432
B.300
C.252
D.992
Answer: B.300
Explanation: 2(3x + x) = 80
l = 30 b = 10
lb = 30 * 10 = 300
Answer:B
Question: The average (arithmetic mean) of six numbers is 9. If 3 is subtracted from each of four of the numbers, what is the new average?
A.1.5
B.2
C.7
D.4
Answer: C.7
Explanation: Sum of 6 numbers = 6*9 =54
If 3 is subtracted from each of four of the numbers , we subtract 3*4=12 from the total sum
Sum of 6 number after subtracting 3 from each of four of the numbers = 54 - 12 = 42
New average = 42/6 = 7
Answer C
Question: The speed of a boat in upstream is 50 kmph and the speed of the boat downstream is 80 kmph. Find the speed of the boat in still water and the speed of the stream?
A.10 kmph.
B.17 kmph.
C.15 kmph.
D.11 kmph.
Answer: C.15 kmph.
Explanation: Speed of the boat in still water = (50+80)/2 = 65 kmph. Speed of the stream = (80-50)/2
= 15 kmph.
Answer:C
Question: A and B run a 1 km race. If A gives B a start of 50m, A wins by 14 seconds and if A gives B a start of 22 seconds, B wins by 20 meters. Find the time taken by A to run 1 km.
A.222
B.100
C.287
D.2976
Answer: B.100
Explanation: To solve these type of questions, always keep in your mind that, the ratio of the speeds of two contestents never change.
A gives B a start of 50 m means, A runs 1000 m and B runs only 950. By the time A reaches the target, B has to take 22 seconds to reach the target.
ab=1000950−14b=9801000−22bab=1000950−14b=9801000−22b
50,000 -1100b = 46550 -686b
Solving we get b = 25/3
Now Assume A's speed = x
1000950−14(25/3)=x25/31000950−14(25/3)=x25/3
x = 10
So x takes 1000/10 = 100 seconds.
Answer:B
Question: The value of 1 + 1/(4 x 3) + 1/(4 x 32) +1/(4 x 33) up to four places of decimals is ?
A.1.1202
B.1.1203
C.1.1204
D.None of these
Answer: B.1.1203
Explanation: Answer
Given expression = (108 + 9 + 3 +1) / 108
= 121 / 108
= 1.1203
Correct Option: B
Question: A number when divided by 342 gives a remainder 47. When the same number ift divided by 19, what would be the remainder ?
A.9
B.7
C.6
D.5
Answer: A.9
Explanation: On dividing the given number by 342, let k be the quotient and 47 as remainder.
Then, number – 342k + 47 = (19 x 18k + 19 x 2 + 9) = 19 (18k + 2) + 9.
The given number when divided by 19, gives (18k + 2) as quotient and 9 as remainder.
Answer is A.
Question: (540 - ?) × 1/7 + 12 = 25
A.A)227
B.B)449
C.C)32
D.D)600
Answer: B.B)449
Explanation: Explanation:
=> (540 - ?) × 1/7 + 12 = 25
=> (540 - ?) × 1/7 = 25 - 12 = 13
=> 540 - ? = 13 × 7 = 91
=> ? = 540 - 91 = 449
Answer: Option B
Question: A housewife wishes to purchase three articles A, B and C from a sum of Rs. 200. The unit prices of the articles A, B and C are Rs. 20, Rs. 35 and Rs. 25 respectively. If she spends the entire amount by purchasing 5 numbers of articles of type C, what is the ratio of the number of articles purchased of type A to that of, type B?
A.1 : 2
B.2 : 1
C.1 : 1
D.None of these
Answer: B.2 : 1
Explanation: Explanation:
After spending Rs. 125 (25 * 5) for articles of type C, the housewife is left with Rs. 75 (200 – 125). Since this amount has to be spent in totality, she must have purchased 2 articles of type A (equivalent to Rs. 40) and 1 article of type B (equivalent to Rs. 35). Thus, the required ratio is 2 : 1.
ANSWER B
Question: What is the value of (12 + 22 + 32 + 42 + ----- + 102)
A.2237
B.269
C.268
D.385
Answer: D.385
Explanation: Explanation:
(12 + 22 + ….. + n2) = (1/6) n(n + 1) (2n + 1)
Here, n = 10
Therefore,
(12 + 22 + ….. + 102) = (1/6) 10 (10 + 1) ( 2 x 10 + 1)
= (1/6) x 10 x 11 x 21
= 385
ANSWER: D
Question: A money lender lent a total of $1900 to two of his customers. He charged at the rate of 5%p.a. to one of them and 6% p.a. to the other. If he earned an average of 5.67% on $1900, how much did he lend to each of the two customers?
A.700; 1100
B.1273; 627
C.1000; 800
D.1200; 800
Answer: B.1273; 627
Explanation: The method given above is the one given in the book.
However the method I used was
(1*x*5)/100 + (1900-x)*6/100 = 1900*5.67/100
Simplifying we get x=627
B
Question: There are three numbers. 5/7th of the first number is equal to 48% of the second number. The second number is 1/9th of the third number. If the third number is 1125, then find 25% of the first number?
A.168
B.84
C.42
D.21
Answer: D.21
Explanation: Let the first number and the second number be F and S respectively.
5/2 F = 48/100 S ----> (1)
S = 1/9 * 1125 = 125
(1) => 5/7 F = 48/100 * 125
=> F = 84
25% of F = 1/4 * 84 = 21
ANSWER:D
Question: A reduction of 25% in the price of oil enables a house wife to obtain 5kgs more for Rs.1000, what is the reduced price for kg?
A.s.50
B.s.46
C.s.49
D.s.41
Answer: A.s.50
Explanation: 1000*(25/100) = 250 ---- 5
? ---- 1
=> Rs.50
Answer: A
Question: A customer using a certain telephone calling plan pays a fee of $25 per month, and then receives a discount of 40% on the regular charge for all calls made to country W. If calls to country W are regularly charged at $1.60 per minute for the first 3 minutes, and $0.80 per minute for each minute thereafter, what is the maximum the customer could have saved over regular prices if he was charged for 1 hour of calls made to country W in a certain month?
A.$8.75
B.$12
C.$13.40
D.$17.40
Answer: C.$13.40
Explanation: Discount = 40% of spending
We want to maximise discount so we will need to maximise spending. We do that by assuming that 60 calls were made of 1 min each because the first minute is the most expensive.
Max discount = (40/100)* 60 * 1.6 = (40/100)*96
This would be slightly less than 40. About 38 to 39.
Saving = (Slightly less than 40) - 25 = Slightly less than 15 charged in W country
Answer (C)
Question: For any positive integer n, the sum of the first n positive integers equals [n(n+1)]/2.
What is the sum of all the even integers between 99 and 161?
A.1090
B.2060
C.3080
D.4030
Answer: D.4030
Explanation: 100 + 102 +...+ 160 = 31*100 + (2+4...+60) = 31*100 + 2(1+2+...+30) =
31*100 + 2(30)(31)/2 = 130*31 = 4030
The answer is D.
Question: How many alphabets need to be there in a language if one were to make 1 million distinct 3 digit initials using the alphabets of the language?
A.321
B.65
C.120
D.100
Answer: D.100
Explanation: 1 million distinct 3 digit initials are needed.
Let the number of required alphabets in the language be ‘n’.
Therefore, using ‘n’ alphabets we can form n * n * n = n3 distinct 3 digit initials.
Note distinct initials is different from initials where the digits are different.
For instance, AAA and BBB are acceptable combinations in the case of distinct initials while they are not permitted when the digits of the initials need to be different.
This n3 different initials = 1 million
i.e. n3 = 106 (1 million = 106)
=> n3 = (102)3 => n = 102 = 100
Hence, the language needs to have a minimum of 100 alphabets to achieve the objective.
Ans: D
Question: The sum of all solutions for x in the equation x^2 – 8x + 21 = |x – 5|+ 4 is equal to:
A.–7
B.7
C.20
D.12
Answer: C.20
Explanation: x^2 - 8x + 17 = |x-5|
RHS can be -ve or +ve
x^2 - 9x + 22 = 0
x^2 - 7x + 12 = 0
x= 11,4,3,2
We test all 3 values in original equation, all ok.
Thus, Sum = 11 + 4 +3 + 2= 20
Ans (C)
Question: If log10 2 = 0.3010, the value of log10 80 is:
A.2.65
B.1.903
C.3.05
D.6.487
Answer: B.1.903
Explanation: Explanation:
log10 80	= log10 (8 x 10)
= log10 8 + log10 10
= log10 (23 ) + 1
= 3 log10 2 + 1
= (3 x 0.3010) + 1
= 1.9030.
answer B
Question: In a partnership between A, B and C. A's capital is Rs.5000. If his share of a profit of Rs.800 is Rs.200 and C's share is Rs.130, what is B's capital?
A.3377
B.2788
C.27999
D.11750
Answer: D.11750
Explanation: 200 + 130 = 330
800 - 330 = 470
200 ---- 5000
470 ---- ? => 11750
Answer: D
Question: John is traveling to a meeting that is 24 miles away. He needs to be there in 30 minutes. How fast does he need to go to make it to the meeting on time?
A.25 mph
B.37 mph
C.48 mph
D.49 mph
Answer: C.48 mph
Explanation: Well 48mph. C. Time*rate=distance --> 0.5*rate=24 --> rate=48
Question: What is the units digit of (493) (915) (381) (756) (22)
A.0
B.1
C.4
D.5
Answer: B.1
Explanation: Just multiply the digits in the units place for each term and you will get the answer. It should be 0. you got a 5 as a unit digit and an even number term. so the multiplication of this will definitely yield a 0. Answer has to be 0.
I also tried it using the calculator and the answer is 1.
IMO B.
Question: A train covers a distance of 12 km in 10 min. If it takes 6 sec to pass a telegraph post, then the length of the train is?
A.186 m
B.176 m
C.120 m
D.178 m
Answer: C.120 m
Explanation: Speed = (12/10 * 60) km/hr = (72 * 5/18) m/sec
= 20 m/sec.
Length of the train = 20 * 6
= 120 m.
Answer:C
Question: Tomy started a software business by investing Rs.50,000. After 6months, Nanda joined her with a capital of $80,000. After 3years, they earned a profit of $24,500. What was Simran's share in the profit?
A.$8,500
B.$9,500
C.$10,500
D.$11,500
Answer: C.$10,500
Explanation: Tomy : Nanda = (50000 x 36) : (80000 x 30) = 3 : 4.
Tomy 's share = $24500 x	3/7	= $10,500.
C
Question: If n = 7^11 – 7, what is the units digit of n?
A.0
B.1
C.4
D.6
Answer: D.6
Explanation: Always divide the power (incase 11) by 4 and use the remainder as the new power. The question now becomes 7^3 - 7. Now 7^3 has last digit 3. Since 7^11 (or for that matter 7^3) is greater than 7, we subtract 7 from 13 ( the 10+3 --> 10 has come from carry over from the tenth place). thus 13 - 7 = 6 is the answer. Option D
Question: Three persons A, B and C are standing in a queue. There are five persons between A and B and eight persons between B and C. If there be three persons ahead of C and 21 persons behind A, what could be the minimum number of persons in the queue?
A.33
B.37
C.28
D.36
Answer: C.28
Explanation: Explanation:
Three persons A, B, C can be arranged in a queue in six different ways, ie ABC, CBA, BAC, CAB, BCA, ACB. But since there are only 3 persons ahead of C, so C should be in front of the queue. Thus, there are only two possible arrangements, ie CBA and CAB.
We may consider the two cases as under:
Case I:
Clearly, number of persons in the queue = (3+1+8+1+5+1+21=) 40
Case II:
Number of persons between A and C
= (8 - 6) = 2
Clearly number of persons in the queue = (3+1+2+1+21) = 28
Now, 28 < 40. So, 28 is the minimum number of persons in the queue.
Answer: C) 28
Question: By selling a house for Rs.45000, it was found that 1/8 of the outlay was gained, what ought the selling to price to have been in order to have lost 6 p.c?
A.38028
B.38000
C.32977
D.37600
Answer: D.37600
Explanation: CP + CP/8 = 45000
CP = 40000
SP = 40000*(94/100)
= 37600
Answer:D
Question: A number when divided by a divisor leaves a remainder of 21. When twice the original number is divided by the same divisor, the remainder is 11. What is the value of the divisor?
A.12
B.13
C.34
D.37
Answer: D.37
Explanation: Let the number is N, the divisor = D,
I will make the two equations-
N = xD+21
2N = yD+11
where x and y are integers
Solving them: D(y-2x) = 34
as D is also integer and 34 is a prime number, the D should be 34 to satisfy the above equation.
Hence answer is 'D'
Question: A dress on sale in a shop is marked at $D. During the discount sale its price is reduced by 65%. Staff are allowed a further 60% reduction on the discounted price. If a staff member buys the dress what will she have to pay in terms of D ?
A.0.15D
B.0.16D
C.0.65D
D.0.14D
Answer: D.0.14D
Explanation: Effective discount = a + b + ab/100 = - 65 - 60 + (-65)(-60)/100 = -86
Sale price = D * ( 1 - 86/100)
Sale price = .14 * D
Answer (D)
Question: A bus travel 5/7 of its speed covers 42 km in 1 hr 40 min48 sec. What is theactual speed of the bus ?
A.25 km/hr
B.35 km/hr
C.40 km/hr
D.42 km/hr
Answer: B.35 km/hr
Explanation: Time = 1 hr 40 min 48 sec = 1hr +4060hr+483600hr=1+23+175=12675hrdistance = 42 kmspeed=distancetime=42(12675)=42×75126⇒57 of the actual speed = 42×75126⇒actual speed = 42×75126×75=42×1518=7×153=7×5=35 km/hr
B
Question: A sum of money becomes triple itself in 5 years at simple interest. How many years will it become six times at the same rate?
A.12 1/2 years
B.12 1/8 years
C.17 1/2 years
D.12 3/2 years
Answer: A.12 1/2 years
Explanation: 100 ---- 200 ---- 5
200 ---- 5
100 ---- 2 1/2
------------------
600 ---- 12 1/2 years
Answer:A
Question: 4+4
A.8
B.2
C.1
D.5
Answer: C.1
Explanation: C
Question: If sum of three numbers in A.P is 33 and sum of their squares id 491, then what are the three numbers.
A.5,11,17
B.7,11,15
C.9,11,13
D.3,11,19
Answer: D.3,11,19
Explanation: a + (a + d ) + ( a + 2d ) = 3( a + d ) = 33,
a + d = 11, or second term = 11 , first term = 11– d ,
Then ( 11- d ) 2 + 11 2 + ( 11 + d ) 2 = 491
2 d 2 = 491 - ( 3 * 121 ) = 491 - 363 = 128
d 2 = 64, d = 8, a = 3,
ANSWER:D
Question: How many number are immediately preceded and immediately followed by different numbers?
7 7 7 5 7 5 7 5 7 7 7 7 5 7 5 7 5 7 7 7 7 7 5 7 5
A.2
B.3
C.4
D.5
Answer: D.5
Explanation: five
ANSWER:D
Question: An exhibition was conducted for 4 weeks. The number of tickets sold in 2nd work week was increased by 20% and increased by 16% in the 3rd work week but decreased by 20% in the 4th work week. Find the number of tickets sold in the beginning, if 1392 tickets were sold in the last week
A.6
B.35
C.120
D.1250
Answer: D.1250
Explanation: Answer: 1250
Explanation:
let initially A ticket has been sold.
So now in 2nd week 20% increases so
A × 120100120100
In 3rd week 16% increases so
A × 120100120100 × 116100116100
In 4th week 20% decrease so
A × 120100120100 × 116100116100 × 120100120100 = 1392
A = 1250
Answer:D
Question: Let exp(m,n) = m to the power n. If exp(10, m)= n exp(2,2) where m and n are integers then n= ___?
A.24
B.25
C.26
D.27
Answer: B.25
Explanation: 10^m=n*(2^2)
(2*5)^m=n*(2^2)
(2^m)*(5^m)=n*(2^2) now equate both side we get m=2 and (5^m)=n
(5^2)=n=>n=25
ANSWER:B
Question: Difference of 2 numbers is 100. If 5% of one number is 10% of the other number, find the addition of2 numbers?
A.150
B.300
C.210
D.290
Answer: B.300
Explanation: Let the numbers be x and y
5% of x = 10% of y
x = 2y
x-y = 100
2y - y =100
y = 100
x = 2y = 200
x+y = 300
Answer is B
Question: If y = 30p, and p is prime, what is the greatest common factor of y and 24p, in terms of p?
A.p
B.2p
C.6p
D.8p
Answer: C.6p
Explanation: y = 30p = 2*3*5*p
24p = 2^3*3*p
The greatest common factor of 30p and 20p is the product of all the common prime factors, using the lower power of repeated factors. The greatest common factor is 2*3*p = 6p
The answer is C.
Question: A man has some hens and cows. If the number of heads be 48 and the number of feet equals 140, then the number of hens will be
A.22
B.24
C.26
D.20
Answer: C.26
Explanation: Explanation :
Let number of hens = h and number of cows = c
number of heads = 48
=> h + c = 48 ---(Equation 1)
number of feet = 140
=> 2h + 4c = 140
=> h + 2c = 70 ---(Equation 2)
(Equation 2) - (Equation 1) gives
2c - c = 70 - 48
=> c = 22
Substituting the value of c in Equation 1, we get
h + 22 = 48
=> h = 48 - 22 = 26
i.e., number of hens = 26
Answer : C
Question: It takes 8 high school students, working at identical constant individual rates, 10 hours to paint a certain house. At what time will the house be fully painted if 8 students start painting at 9:00 am and one student, working at the same rate, is added per hour starting at 2:00 pm?
A.5:10 pm
B.5:30 pm
C.5:50 pm
D.6:10 pm
Answer: C.5:50 pm
Explanation: Each student paints at a rate of 1/80 of the house per hour.
In 5 hours, the 8 students can paint 40/80 of the house.
From 2 pm to 3 pm, 9 students paint another 9/80 for a total of 49/80.
From 3 pm to 4 pm, 10 students paint another 10/80 for a total of 59/80.
From 4 pm to 5 pm, 11 students paint another 11/80 for a total of 70/80.
12 students can paint the remaining 10/80 of the house in 10/12 of an hour = 50 minutes.
The house is completed at 5:50 pm.
The answer is C.