divelab/Sys2Bench huggingface dataset:
Question: The side of a square is increased by 25% then how much % does its area increases?
Options:
A)56.28
B)56.25
C)56.27
D)56.29
E)56.21
Rationale: a = 100 a2 = 10000
a = 125 a2 = 15625
----------------
10000 --------- 5625
100 -------? => 56.25 %
Answer:B
Answer: B
Question: Excluding stoppages, the speed of a train is 45 kmph and including stoppages it is 35 kmph. Of how many minutes does the train stop per hour?
Options:
A)73
B)83
C)12
D)13
E)28
Rationale: Explanation:
T = 10/45 * 60 = 13
Answer: Option D
Answer: D
Question: A rectangular field has a length 10 meters more than it is width. If the area of the field is 171, what is the length (in meters) of the rectangular field?
Options:
A)17
B)19
C)21
D)23
E)25
Rationale: area = l * w = (L) * (L-10) = 171
trial and error:
20*10 = 200 (too high)
19*9 = 171
The length is 19 meters.
The answer is B.
Answer: B
Question: Two goods trains each 500 m long are running in opposite directions on parallel tracks. Their speeds are 30 km/hr and 90 km/hr respectively. Find the time taken by the slower train to pass the driver of the faster one?
Options:
A)48
B)30
C)26
D)23
E)12
Rationale: Relative speed = 90 + 30 = 120 km/hr.
120 * 5/18 = 100/3 m/sec.
Distance covered = 500 + 500 = 1000 m.
Required time = 1000 * 3/100 = 30 sec.
Answer:B
Answer: B
Question: Dames school has 2000 boys and 5000 girls. What is the percentage increase from boys to total attendance and girls to total attendance?
Options:
A)40%
B)10%
C).4%
D)50%
E)12%
Rationale: Ratio of boys to total attendance (2/7)
Ratio of girls to total attendance (5/7)
percentage increase is (difference/initial quantity)*100
(2/7)/(5/7 )* 100= 40%
Correct answer is A
Answer: A
Question: What is the minimum number of people to join into this group so that when they are grouped into 5, none will remain ungrouped ? Without newcomers when they are grouped into 6, 4 of them remain ungrouped and when they are grouped into 9, 7 of then reamin ungrouped.
Options:
A)2
B)4
C)6
D)8
E)10
Rationale: P = 6a+4 = 9b+7
6a= 9b+3 --> min (a,b) = (2,1) so that number of people in this group is
P= 6*2+4 = 16
5k = 16 + x so that none will be ungrouped --> min( k) = 4 then
5*4 = 20 16 + x
X= 4 newcomers should join to this group
ANSWER: B
Answer: B
Question: Look at this series: 22, 21, 23, 22, 24, 23, ... What number should come next?
Options:
A)30
B)19
C)25
D)5
E)20
Rationale: In this simple alternating subtraction and addition series; 1 is subtracted, then 2 is added, and so on.
so 23+2=25
Option C
Answer: C
Question: A shopkeeper sold an article offering a discount of 5% and earned a profit of 31.1%. What would have been the percentage of profit earned if no discount had been offered?
Options:
A)38
B)27.675
C)30
D)Data inadequate
E)None of these
Rationale: Giving no discount to customer implies selling the product on printed price. Suppose the cost price of the article is 100.
Then printed price = 100Ã—(100+31.1)/(100âˆ’5)
= 138
Hence, required % Profit = 138 â€“ 100 = 38%
Answer A
Answer: A
Question: Alfred buys an old scooter for Rs. 4700 and spends Rs. 800 on its repairs. If he sells the scooter for Rs. 5800, his gain percent is:
Options:
A)12/11 %
B)10 %
C)5 5/11%
D)6/7 %
E)10/11 %
Rationale: Explanation:
Cost Price (C.P.) = Rs. (4700 + 800) = Rs. 5500.
Selling Price (S.P.) = Rs. 5800.
Gain = (S.P.) - (C.P.) = Rs.(5800 - 5500) = Rs. 300.
Gain % =(300/5500*100)%=5 5/11%
Answer:C
Answer: C
Question: A and B start a business, with A investing the total capital of Rs.25000, on the condition that B pays A interest @ 10% per annum on his half of the capital. A is a working partner and receives Rs.250 per month from the total profit and any profit remaining is equally shared by both of them. At the end of the year, it was found that the income of A is twice that of B. Find the total profit for the year?
Options:
A)A)22500
B)B)33500
C)C)22510
D)D)33510
E)E)11500
Rationale: Interest received by A from B = 10% of half of Rs.25000 = 10% * 12500 = 1250.
Amount received by A per annum for being a working partner = 250 * 12 = Rs.3000.
Let 'P' be the part of the remaining profit that A receives as his share. Total income of A = (1250 + 3000 + P)
Total income of B = only his share from the remaining profit = 'P', as A and B share the remaining profit equally.
Income of A = Twice the income of B
(1250 + 3000 + P) = 2(P)
P = 4250
Total profit = 2P + 3000
= 2*4250 + 3000 = 11500
Answer:E
Answer: E
Question: In the first round of the elections, the only two candidates got exactly the same number of votes. During the second round, 21,000 votes switched from the first candidate to the second one. The total number of votes remained the same in both rounds, and no other votes switched sides. If, in the second round, the winning candidate got four times as many votes as the other candidate, how many people have voted in each round?
Options:
A)70,000
B)25,000
C)40,000
D)50,000
E)60,000
Rationale: Let A be the first candidate and B be the second one.
Given x+21000 = 4(x-21000)=>x=35000
2x = 70000
IMO option A.
Answer: A
Question: If x and y are positive integers and 3 + x + y + xy = 17, what is the value of x + y?
Options:
A)3
B)5
C)6
D)8
E)9
Rationale: Try each answer choice.
For A: 3 + 3 + xy = 17; xy = 11 (impossible, 11 prime number. 1 + 11 doesn't equal 3)
For B: 3 + 5 + xy = 17; xy = 9 (no combination of xy = 9 and x + y = 5)
For C: 3 + 6 + xy = 17; xy = 8 (x + y = 6; x = 2, y = 4 or x = 4, y = 2)
For D: 3 + 8 + xy = 17; xy = 6 (no combination of xy = 6 and x + y = 8)
For E: 3 + 9 + xy = 17; xy = 5 (impossible, 5 prime number. 1 + 5 doesn't equal 9)
Therefore, answer C.
Answer: C
Question: What will come in place of the x in the following Number series? 18, 22, 30, 46 ,78, X
Options:
A)35
B)56
C)142
D)87
E)98
Rationale: (C)
The pattern is +4, +8, +16, +32, +64
So the missing term is = 142
Answer: C
Question: What approximate value should come in place of question mark (?) in the following equation?
33 1⁄3% of 768.9 + 25% of 161.2 – 68.12 = ?
Options:
A)230
B)225
C)235
D)220
E)240
Rationale: ? = 33 1⁄3% of 768.9 + 25% of 161.2 – 68.12
= 1⁄3 of 768.9 + 1⁄4 of 161.2 – 68.12
= 256.3 + 40.3 – 68.12 ≈ 230
Answer A
Answer: A
Question: The game of blackjack is played with a deck consisting of 13 cards (one of each of the numbers 2-10, one jack, one queen, one king, and one of ace) in each of four suits: clubs, diamonds, hearts, and spades.
In a game of blackjack, if your first card is a spade, what are the odds that it is an ace?
Options:
A)1/52
B)1/26
C)1/13
D)1/4
E)5/13
Rationale: There are 13 spades, the probability that the spade we already have is an ace is therefore 1/13.
Answer: C.
Answer: C
Question: John and Steve are speed walkers in a race. John is 15 meters behind Steve when he begins his final push. John blazes to the finish at a pace of 4.2 m/s, while Steve maintains a blistering 3.7 m/s speed. If John finishes the race 2 meters ahead of Steve, how long was John’s final push?
Options:
A)13 seconds
B)17 seconds
C)26 seconds
D)34 seconds
E)51 seconds
Rationale: Let t be the time that John spent for his final push.
Thus, per the question,
4.2t = 3.7t+15+2 ---> 0.5t = 17 ---> t = 34 seconds.
D is the correct answer.
Answer: D
Question: If a coin is tossed 3 times what is the probability that it will land either heads 3 times or tails 3 times?
Options:
A)1/4
B)1/16
C)1/2
D)1/8
E)1/32
Rationale: The probability of 3 heads is (1/2)^3 = 1/8
The probability of 3 tails is (1/2)^3 = 1/8
The probability of 3 heads or 3 tails is 1/8 + 1/8 = 1/4
The answer is A.
Answer: A
Question: The number of sequences in which 7 players can throw a ball, so that the youngest player may not be the last is ?
Options:
A)2389
B)2778
C)4320
D)1987
E)1212
Rationale: Explanation:
x Not younger_______ â†‘
The last ball can be thrown by any of the remaining 6 players. The first 6 players can throw the ball in â¶Pâ‚† ways.
The required number of ways = 6(6!) = 4320
Answer:C
Answer: C
Question: The average of runs of a cricket player of 20 innings was 32. How many runs must he make in his next innings so as to increase his average of runs by 8 ?
Options:
A)96
B)200
C)128
D)116
E)122
Rationale: Average = total runs / no.of innings = 32
So, total = Average x no.of innings = 32 * 20 = 640
Now increase in avg = 4 runs. So, new avg = 32+8 = 40 runs
Total runs = new avg x new no. of innings = 40 * 21 = 840
Runs made in the 11th inning = 840-640 = 200
ANSWER:B
Answer: B
Question: 3 people, harry, john , and Cassie, have speeds of 3 mph, 4 mph and 6 mph respectively. They run a race in which john gives harry a head start of 2 hrs. If both john and Cassie overtake harry at the same time, what head start did Cassie give harry?
Options:
A)3 hours
B)4 hours
C)5 hours
D)9 miles
E)10 miles
Rationale: Question clearly states that A started first. B started after 2 hours and we need to determine When did C start.
As per the question, we know that All three meet at the same point.
So, we can determine when did A and B meet. The point they will meet would be the same at which C will be present at that time.
So,
A and B will meet after t = 6/(4-3) = 6 hours.
In these 6 hours, B would have covered 4*6=24 miles.
=> C would also have covered 24 miles and that would be in 24/6=4 hours.
Thus, C started 2 hours after B or 4 hours after A. hence, answer is B.
Answer: B
Question: A college has classes from 10:00 AM until 1:45 PM. In this duration, there are five periods. If 5 minutes are provided between each period to leave one class and enter the next class, how many minutes long is each period?
Options:
A)38 min
B)41 min
C)40 min
D)42 min
E)52 min
Rationale: The total time is 225 minutes.
There are four breaks of five minutes between the classes for a total of 20 minutes.
The total class time is 205 minutes.
205 / 5 classes = 41 minutes per class
The answer is B.
Answer: B
Question: If x and y are odd integers, which of the following must always be a non-integer
Options:
A)xy
B)xy/4
C)x/y
D)y/x
E)-x^2y
Rationale: as X and Y are odd, multiplication of x and y would result in odd integer
(Odd Integer)/4 will always be non integer
B
Answer: B
Question: What is the unit digit in {(6374)^1793 x (625)^317 x (341^491)}?
Options:
A)0
B)2
C)3
D)5
E)6
Rationale: Unit digit in (6374)^1793 = Unit digit in (4)^1793
= Unit digit in [(42)^896 x 4]
= Unit digit in (6 x 4) = 4
Unit digit in (625)^317 = Unit digit in (5)^317 = 5
Unit digit in (341)^491 = Unit digit in (1)^491 = 1
Required digit = Unit digit in (4 x 5 x 1) = 0.
Answer:A
Answer: A
Question: Each digit 1 through 5 is used exactly once to create a 5-digit integer. If the 5 and the 4 cannot be adjacent digits in the integer, how many 5-digit integers are possible?
Options:
A) 48
B) 66
C) 76
D) 78
E) 72
Rationale: Number of arrangements using 5 distinct digits = 5!
Number of arrangements in which 4 and 5 are adjacent - consider 4 and 5 together as one group. Now you have 4 numbers/groups to arrange which can be done in 4! ways. In each of these arrangements, 4 and 5 can be arranged as 45 or 54.
Number of arrangements in which 3 and 4 are not adjacent = 5! - 2*4! = 72
Answer : E
Answer: E
Question: Adding two non multiples of E always results in a multiple of E; which of the following must be the value of E ?
Options:
A)[A] two
B)[B] three
C)[C] five
D)[D] ten
E)[E] hundred
Rationale: NON- multiples of 2 would always be ODD and sum of two odd would be EVEN.
so irrespective of value of non-multiples, sum of two them would always be even and thus multiple of 2..
A
Answer: A
Question: When 20 per cent of a number is added to another number the second number increases to its 140 per cent. What is the ratio between the first and the second number?
Options:
A)2 : 1
B)4 : 3
C)3 : 2
D)Data inadequate
E)None of these
Rationale: Let the first and the second numbers be x and y respect then
y + 20% of x = 140% of y
or, y + 0.2x = 1.4y
or, 0.2x = 0.4y
∴ x : y = 0.4 : 0.2 = 2 : 1
Answer A
Answer: A
Question: A can do a piece of work in 10 days and B can do it in 15 days and C can do it 20 days. They started the work together and A leaves after 2 days and B leaves after 4 days from the beginning. How long will work lost?
Options:
A)8 2/3 days
B)9 2/3 days
C)10 2/3 days
D)10 days
E)11 days
Rationale: Explanation:
2/10 + 4/15 + x/20 = 1
x = 32/3 = 10 2/3
Answer: C
Answer: C
Question: If two students starting from same point, walking in the opposite directions with 6km/hr and 9km/hr as average speeds respectively. Then the distance between them after 4 hours is?
Options:
A)65 km
B)55km
C)15km
D)60km
E)75km
Rationale: Explanation:
Total distance = Distance traveled by person A+ Distance traveled by person B
= (6× 4) + (9× 4)
= 24+ 36
= 60 Km
Answer:D
Answer: D
Question: There are 6 girls and 6 boys. If they are to be seated in a row, what is the probability that no two girls sit together?
Options:
A)1/12!
B)6!5!/12!
C)6!6!/12!
D)2∗6!6!/12!
E)6!7!/12!
Rationale: If girls are alternate, then only no 2 girls can be together.
So among 12 places, girls can be seated in 6 alternate places in 6! ways.
And boys can be seated in the remaining 6 places in 6! ways.
Therefore total number of ways = 6! * 6!
But there could be 2 such arrangements-->
1. When first place is filled by a girl
2. When first place is filled by a boy
Therefore total number of ways = 2*6! * 6!
Also total number of ways to fill 12 places = 12!
Hence probability =2∗6!6!/12!
Answer:D
Answer: D
Question: How many ways can 4 friends be arranged around a circular dinner table?
Options:
A)A)6
B)B)48
C)C)96
D)D)120
E)E)720
Rationale: The number of arrangements of n distinct objects in a row is given by n!n!.
The number of arrangements of n distinct objects in a circle is given by (n−1)!(n−1)!.
The difference between placement in a row and that in a circle is following: if we shift all object by one position, we will get different arrangement in a row but the same relative arrangement in a circle. So, for the number of circular arrangements of n objects we have:
n!n=(n−1)!n!n=(n−1)!.
So, the answer is (4 - 1)! = 3! = 6.
Answer: A.
Answer: A
Question: A lent Rs. 4000 to B for 2 years and Rs. 2000 to C for 4 years on simple interest at the same rate of interest and received Rs. 2200 in all from both of them as interest. The rate of interest per annum is :
Options:
A)14%
B)15%
C)12%
D)13.75%
E)14.56%
Rationale: Explanation:
Let the rate of interest per annum be R%
Simple Interest for Rs.4000 for 2 years at R% + Simple Interest for Rs.2000 for 4 years at R%
= 2200
(4000×R×2)/100 + (2000×R×4)/100 = 2200
80R + 80R = 2200
160R = 2200
16R = 220
4R = 55
R = 55/4 = 13.75%
Answer: Option D
Answer: D
Question: Insert the missing number. 12, 25, 49, 99, 197, 395, X
Options:
A)579
B)722
C)780
D)812
E)789
Rationale: Each number is twice the previous one with 1 added or subtracted alternatively.
395X2-1=789
Answer: E
Question: The diameter of the driving wheel of a bus in 140 cm. How many revolutions per minute must the wheel make in order to keep a speed of 24 kmph?
Options:
A)21.21
B)22.26
C)90.9
D)24.86
E)25.25
Rationale: distance covered in 1 min=(24*1000)/60=400m
circumference of the wheel=(2*(22/7)*.70)=4.4m
no of revolution per min=400/4.4=90.90
ANSWER:C
Answer: C
Question: How many 6 letter words are formed using the letters of the word POLICY?
Options:
A)129
B)420
C)750
D)680
E)720
Rationale: The number of letters in the given word is 6.
The number of 6 letter words that can be formed using these four letters is 6P6=6*5*4*3*2*1=720
Answer:E
Answer: E
Question: A marketing firm determined that, of 200 households surveyed, 80 used neither Brand W nor Brand B soap, 60 used only Brand W soap, and for every household that used both brands of soap, 3 used only Brand B soap. How many of the 200 households surveyed used both brands of soap?
Options:
A) 15
B) 20
C) 30
D) 40
E) 45
Rationale: SOLUTION FOR SOAP W AND SOAP B
(D) 40
Answer: D
Question: 12 + 13 + 14 + ... 51 + 52 + 53+ 54 = ?
Options:
A)1419
B)1362
C)1363
D)1364
E)1365
Rationale: Sum = 12 + 13 + 14 + ... 51 + 52 + 53 + 54
Sum of n consecutive positive integers starting from 1 is given as n(n+1)/2
Sum of first 54 positive integers = 55*54/2
Sum of first 11 positive integers = 11*12/2
Sum = 12 + 13 + 14 + ... 51 + 52 + 53 + 54 = 55*54/2 - 11*12/2 = 1419
Answer : A
Answer: A
Question: Two trains of equal are running on parallel lines in the same direction at 46 km/hr and 36 km/hr. The faster train passes the slower train in 36 sec. The length of each train is?
Options:
A)50
B)87
C)52
D)76
E)21
Rationale: Let the length of each train be x m.
Then, distance covered = 2x m.
Relative speed = 46 - 36 = 10 km/hr.
= 10 * 5/18 = 25/9 m/sec.
2x/36 = 25/9 => x = 50.
Answer: A
Answer: A
Question: In a function they are distributing noble prize. In how many ways can 3 prizes be distributed among 4 boys when a boy gets any no. of prizes?
Options:
A)55
B)58
C)64
D)65
E)70
Rationale: Ans.(C)
Sol. In this case, repetitions are allowed, so all the three prizes can be given in 4 ways, i.e. (4 x 4 x 4) ways = 43 ways = 64 ways Or nr ways = 43 ways = 64 ways
Answer: C
Question: a = 5^15 - 125^3 and a/x is an integer, where x is a positive integer greater than 1, such that it does NOT have a factor p such that 1 < p < x, then how many different values for x are possible?
Options:
A)None
B)One
C)Two
D)Three
E)Four
Rationale: This is a tricky worded question and I think the answer is should be D not C...
Here is my reason :
The stem says that x is a positive integer such that has no factor grater than 2 and less than x itself . The stem wants to say that X is a PRIME NUMBER . because any prime
Number has no factor grater than 1 and Itself .
On the other hand the stem says that X COULD get how many different number NOT MUST get different number ( this is very important issue )
AS our friends say, if we simplify Numerator more we can obtain : 5^12 ( 5^3-1) = 5^12 (124) = 5^12 (31*2*2) divided by x and we are told that this fraction is
An integer . so, X COULD Be ( not Must be) 5 , 31 ,or 2 !!! so , X could get none different values and answer is A....
Answer: A
Question: If n is an integer greater than 6, which of the following must be divisible by 11 ?
Options:
A)n(n + 5)(n – 6)
B)n(n + 4)(n – 2)
C)n(n + 3)(n – 5)
D)n(n + 1)(n – 4)
E)n(n + 2)(n – 1)
Rationale: If we assume n as being either 7 , 8 or 9 we can easily see why A is the correct answer .. Because in every instance , either of the 3 numbers multiplied will be divisible by 3 and the entire product will meet the requirements..
if n =10 , then n +1 is divisible by 11...
D
Answer: D
Question: Set I consists of integers {3, -8, Y, 19, -6} and set B consists of integers {k, -3, 0, 16, -5, 9}. Number L represents the median of set I, number M represents mode of set B and number Z = L^M. If Y is an integer greater than 21, for what value of k will Z be a divisor of 26?
Options:
A)-2
B)-1
C)0
D)1
E)2
Rationale: I = {3,-8,Y,19,-6}
B = {k,-3,0,16,-5,9}
Y > 21
So L = Median of I = 3
M = Mode of Set B
Z = (3)^M
If Z is a divisor of 26, (3)^M = 1 because 26 does not have 3 as a factor
=> M = 0
Hence k = 0, as M is mode and 0 will be the most frequently occuring number in set B.
Answer - C
Answer: C
Question: What will come in place of the x in the following Number series? 12, 38, 116, 350, 1052, x
Options:
A)2345
B)3245
C)3457
D)5688
E)3158
Rationale: 12
12 × 3 + 2 = 38
38 × 3 + 2 = 116
116 × 3 + 2 = 350
350 × 3 + 2 = 1052
1052 × 3 + 2 = 3158
E
Answer: E
Question: A began a business with Rs. 85,000. He was joined afterwards by B with Ks. 42,500. For how much period does B join, if the profits at the end of the year are divided in the ratio of 3 : 1 ?
Options:
A)2
B)1
C)7
D)8
E)9
Rationale: Explanation:
Suppose B joined for x months . Then, ( 85000 * 12 )/(42500 * x) = 3. or x = (85000 * 12) / (42500 * 3) = 8.
So, B joined for 8 months.
Answer: D) 8 months
Answer: D
Question: 14 is what percent of 70?
Options:
A)0.2%
B)2%
C)5%
D)20%
E)500%
Rationale: 14= x*70/100
x = 20%
Ans; D
Answer: D
Question: How many integers between 100 and 300 are there such that their unit digit is odd?
Options:
A)100
B)120
C)200
D)50
E)150
Rationale: 200 numbers between - 100 and 300
Out of which half would be even, half odd.
Number of odd unit digit number = 100.
correct option is A
Answer: A
Question: A train starts from Delhi at 6.00 a.m. and reaches Meerut at 10 a.m. The other train starts from Meerut at 8 a.m. and reaches Delhi at 11.30 a.m. If the distance between Delhi and Meerut is 200 km, then at what time did the two trains meet each other?
Options:
A)8.56 a.m
B)8.46 a.m
C)7.56 a.m
D)8.30 a.m
E)None of these
Rationale: The Speed of the train starting from Delhi = 200/4 = 50 km/h
The Speed of train starting from Meerut = 200/3.5 = 400/7 km/h
Suppose the two trains meet x hours after 6.00 am
Then x X 50 + (x - 2) x 400/7 = 200
or, 350x + 400x - 800 = 1400
or, 750x = 2200
or, x = 2200/750 = 2h 56 min
Hence, the required time = 8.56 am
ANSWER:A
Answer: A
Question: A person can row at 9 kmph and still water. He takes 4 1/2 hours to row from A to B and back. What is the distance between A and B if the speed of the stream is 1 kmph?
Options:
A)32
B)28
C)29
D)54
E)20
Rationale: Let the distance between A and B be x km.
Total time = x/(9 + 1) + x/(9 - 1) = 4.5
=> x/10 + x/8 = 9/2 => (4x + 5x)/40 = 9/2 => x
= 20 km.
Answer:E
Answer: E
Question: Jacob is 10 years old. He is 5 times as old as his brother. How old will Jacob be when he is twice as old?
Options:
A)13
B)14
C)15
D)16
E)17
Rationale: j=10;
j=5b;
b=10/5=2;
twice as old so b=2(now)+(2)=4;
jacob is 10+4=14
ANSWER:B
Answer: B
Question: There are ten houses on the street. Three people live in each house. How many people live on the street?
Options:
A)20
B)25
C)30
D)32
E)9
Rationale: Ans is :C, 30 because 10 x 3 = 30.
Answer: C
Question: On the shelf, there are three books of poetry, five novels, and eight reference works. Suppose we picked two books at random. What is the probability that we pick one novel and one reference work?
Options:
A)1/2
B)1/3
C)1/4
D)1/5
E)1/6
Rationale: We could pick one novel and then one reference work, or, we could choose one reference work and then one novel.
P(one novel and one reference work) = 5/16*8/15 + 8/16*5/15 = 1/3
The answer is B.
Answer: B
Question: Sum of the consequent 7 integers equals 1617. Find the no. of prime numbers in that sequence.
Options:
A)1
B)2
C)3
D)4
E)5
Rationale: Let the numbers be x, x+1, x+2, x+3, x+4, x+5, x+6
Since their sum equals 1617, we have-
x+(x+1)+(x+2)+(x+3)+(x+4)+(x+5)+(x+6)=1617
7x+21=1617
7x=1596
x=228
Hence the numbers are: 228,229,230,231,234,235,236
Prime Numbers: 229, 231
ANSWER:B
Answer: B
Question: To gain profit, a milkman adds 1.5 litre of water to a certain amount of milk. He sells the adulterated milk at Rs.28 per litre and pure milk costs Rs.36 per litre. Find the amount of milk.
Options:
A)5.25
B)5.27
C)5.22
D)5.21
E)5.29
Rationale: Explanation:
Let the amount of milk be x.
36x = 28(x + 1.5)
36x = 28x + 42
8x = 42
x = 5.25 litres
ANSWER:A
Answer: A
Question: How many pieces of 85 cm length can be cut from a rod of 25.5 meters long?
Options:
A)50
B)40
C)30
D)20
E)10
Rationale: Number of pieces = 2550/85 = 30
The answer is C.
Answer: C
Question: Matt and Peter can do together a piece of work in 24 days. After they have worked together for 12 days Matt stops and Peter completes the remaining work in 10 days. In how many days Peter complete the work separately.
Options:
A)24
B)25
C)26
D)27
E)28
Rationale: Together they complete the job in 24 days means they complete 12/24 of the job after 12 days.
Peter completes the remaining (12/24) of the job in 10 days which means that the whole job(1) can be completed in X days.
X=12/(12/24)=24
A
Answer: A
Question: The tank contains 16 Fishes. 5 of the fishes are shark, 6 are jelly fishes, and 5 are gold fishes. If two fishes are randomly selected from the tank, one after the other, what is the probability that both will be jelly fishes?
Options:
A)2/8
B)1/8
C)1/6
D)2/7
E)8/7
Rationale: Probability of getting the first Jelly Fish - 6/16
Probability of getting the second Jelly Fish - 5/15
total probability = 6/16*5/15 = 1/8
correct option B
Answer: B