divelab/Sys2Bench huggingface dataset:
Question: If x is an odd negative integer and y is an even integer, which of the following statements must be true?
I. (x - 2y) is odd
II. xy^2 is an even negative integer
III. (y^2 - x) is an odd negative integer
Options:
A)I only
B)II only
C)I and II
D)I
E)II and III
Rationale: If x is an odd negative integer and y is an even integer, which of the following statements must be true?
I. (x - 2y) is odd
x - is always negative and odd
2y - can be 0, when y=0, or always even integer (positivie or negative)
odd integer +/- even integer = always odd
I statement is always true
II. xy^2 is an even negative integer
y can be 0 => xy^2 = 0 (non-negative even) => II statement is not true
III. (y^2 - x) is an odd negative integer
y can be 0 => -x will be positive => III statement is not true
basing on above, only I statement will always be true
answer is D
Answer: D
Question: What is the remainder if 7^8 is divided by 100?
Options:
A)7
B)1
C)9
D)10
E)11
Rationale: 7^8 can be written as (7^4)^2
If we divide 7^4 by 100 the reminder is 1
So, (7^4)^2 by 100, the reminder is 1^2 = 1
Answer : B
Answer: B
Question: If it takes a machine 2⁄3 minute to produce one item, how many items will it produce in 1 hours?
Options:
A)1⁄3
B)4⁄3
C)90
D)120
E)180
Rationale: 1 item takes 2/3 min so it takes 60 min to produce x
2x/3=60 the x=90
Answer:C
Answer: C
Question: Two identical trains AA and BB running in opposite direction at same speed tale 2 min to cross each other completely. The number of bogies of AA are increased from 12 to 16. How much more time would they now require to cross each other?
Options:
A)40 Sec
B)50 sec
C)60 sec
D)20 sec
E)25 sec
Rationale: Total initial bogies is 12+12=2412+12=24
Additional bogies =16âˆ’12=4
24 bogies is 12+12=24
additional bogies = 16 - 12 = 4
24 bogies take 2 min
2*60/24 *4=20 sec
ANSWER D
Answer: D
Question: In how many Q ways can a four-letter password be chosen, using the letters A, B, C, D, E, and/or F, such that at least one letter is repeated within the password?
Options:
A)720
B)864
C)900
D)936
E)1,296
Rationale: Total number of four letter passwords = 6*6*6*6= 1296 ------(1)
Total number of passwords in which no letter repeats = 6C4*4!= 15*24=360------(2)
therefore required value Q= (1)-(2)= 1296-360=936.D
Answer: D
Question: When a person aged 39 is added to a group of n people, the average age increases by 2. When a person aged 15 is added instead, the average age decreases by 1. What is the value of W?
Options:
A) 7
B) 8
C) 9
D) 10
E) 11
Rationale: A simple and elegant solution.
As addition of 39, shifts mean by 2, and addition of 15, shifts mean by 1 to the other side, we have the mean lying between 3915, and in a ratio of 2:1
39-15 = 24
24 divide by 3 is 8.
Meaning mean of the n terms is 15+8 = 39-16 = 23
Now, from first statement, When a person aged 39 is added to a group of n people, the average age increases by 2.
W*23 +39 = 25*(W+1)
W = 7
Ans. (A)
Answer: A
Question: An item is being sold for $10 each. However, if a customer will “buy at least 3” they have a promo discount of 12%. Also, if a customer will “buy at least 10” items they will deduct an additional 8% to their “buy at least 3” promo price. If Sam buys 10 pcs of that item how much should he pay?
Options:
A)$92.00
B)$88.00
C)$87.04
D)$80.96
E)$80.00
Rationale: Without any discount Sam should pay 10*10=$100. Now, the overall discount would be slightly less than 20%, thus he must pay slightly more than $80. Only answer choice D fits.
Answer: D.
Answer: D
Question: The labeled price of a table is Rs. 6,000. The shopkeeper sold it by giving 5% discount on the labeled price and earned a profit of 15%. What approximately is the cost price of the table?
Options:
A)Rs. 4956
B)Rs. 6195
C)Rs. 6425
D)Rs. 7200
E)None of these
Rationale: Explanation:
Labeled price = Rs. 6,000
By giving 5% discount on labeled price, the selling price is
= 95 / 100 * 6,000 = Rs. 5700
By earning 15% profit on the selling of price Rs. 5700 the cost price is
= 100 / 115 * 5700 = Rs. 4956
Therefore, approximate cost is 4956
ANSWER A
Answer: A
Question: A certain amount earns simple interest of 1750 after 7 years. Had the interest been 2% more, how much more interest would it have earned?
Options:
A)35
B)350
C)245
D)Cannot be determined
E)None of these
Rationale: Let p and r be the principal amount and rate of interest respectively.
Then, p×r×7/100=1750
or, pr = 25000
Now, SI = p×(r+2)×7/100=1750
We have to find the value of
p×(r+2)×7/100−p×r×7/100= M - 1750
M = SI when the rate of interest is 2% more. When we solve this equation, we find that we have two variables and one equation. Therefore, can’t be determined the correct answer.
Answer D
Answer: D
Question: A and B walk around a circular track. A and B walk at a speed of 2 rounds per hour and 3 rounds per hour respectively. If they start at 8 a.m. from the same point in opposite directions, how many times shall they cross each other before 10 a.m.?
Options:
A)5
B)6
C)7
D)8
E)10
Rationale: Relative speed = Speed of A + Speed of B (? they walk in opposite directions)
= 2 + 3 = 5 rounds per hour
=> They cross each other 5 times in 1 hour
Time duration from 8 am to 10 am = 2 hour
Hence they cross each other 10 times before 10 am
Answer is E
Answer: E
Question: When the scores of the top four students are deleted, the average score of the remaining 66 students falls by 6 marks. Assume that it is possible for two or more students to have the same net score. What is the minimum score possible for the fourth ranking student if no student got a net score of more than 352?
Options:
A)280
B)320
C)300
D)340
E)350
Rationale: Explanation :
Sum of score of top four students = 1356
Maximum possible score possible for top three students = 352*3 = 1056
Hence the minimum possible score for the 4th student = 300
Answer : C
Answer: C
Question: A and B can do a work in 60 days; B and C can do it in 120 days; A and C can do it in 80 days. In what time A alone can do the work?
Options:
A)100
B)90
C)80
D)70
E)60
Rationale: (A+B)’s one day’s work = 1/60
(B+C)’s one day’s work = 1/120
(A+C)’s one day’s work = 1/80
Adding we get, 2(A+B+C)’ s one day’s work= 1/60 + 1/120 + 1/80
= 28/720
= 14/365
(A+B+C)’ s one day’s work = 14*1/365*2 = 14/720
So, A’s one day’s work = 14/720 - 1/120 = 1/90
A alone can do the work in 90 days.
ANSWER:B
Answer: B
Question: In a 500 m race, the ratio of the speeds of two contestants A and B is 3 : 4. A has a start of 200 m. Then, A wins by:
Options:
A)60 m
B)100 m
C)43m
D)20m
E)23m
Rationale: To reach the winning post A will have to cover a distance of (500 - 200)m, i.e., 300 m.
While A covers 3 m, B covers 4 m.
While A covers 300 m, B covers 4 x 300/3	m	= 400 m.
Thus, when A reaches the winning post, B covers 400 m and therefore remains 100 m behind.
A wins by 100 m.
Answer:	B
Answer: B
Question: IN Common Wealth Games an athlete runs 280 meters in 24 seconds. Hisspeed is ?
Options:
A)25 km/hr
B)42.42 km/hr
C)30 km/hr
D)32 km/hr
E)None of these
Rationale: Explanation :
(280/1000)/(24/3600)=42.42 km/hr
Answer : B
Answer: B
Question: In a throw of dice what is the probability of geÆ«ng number greater than 1
Options:
A)1/2
B)1/3
C)1/5
D)5/6
E)None of these
Rationale: Explanation:
Number greater than 1 is 2, 3, 4, 5 & 6, so only 5 number
Total cases of dice = [1,2,3,4,5,6]
So probability = 5/6
Answer: E
Answer: E
Question: 4^5^6^7^8^9^10 find unit digit
Options:
A)2
B)3
C)4
D)5
E)6
Rationale: guys first find out unit digit of4^5ie is 4 then
4^6 unit digit its 6
then 6^7^8^9^10=6 bcz 6 unit digit alway 6
ANSWER:E
Answer: E
Question: Pipe A can fill a tank in 16 minutes and pipe B cam empty it in 24 minutes. If both the pipes are opened together after how many minutes should pipe B be closed, so that the tank is filled in 30 minutes?
Options:
A)11
B)21
C)33
D)51
E)61
Rationale: Explanation:
Let the pipe B be closed after x minutes.
30/16 - x/24 = 1 => x/24 = 30/16 - 1 = 14/16
=> x = 14/16 * 24 = 21.
Answer:B
Answer: B
Question: Simple interest on a sum at 9% per annum for 2 years is Rs.200. The C.I. on the same sum for the same period is?
Options:
A)201.1
B)201.5
C)200.4
D)109.1
E)81.1
Rationale: SI = 100 + 100
CI = 100 + 100 + 1.1 = 201.1
Answer: A
Answer: A
Question: A walks at 30 kmph and 30 hours after his start, B cycles after him at 55 kmph. How far from the start does B catch up with A?
Options:
A)1200km
B)1980km
C)2000km
D)2500km
E)3600km
Rationale: Suppose after x km from the start B catches up with A. Then, the difference in the time taken by A to cover x km and that taken by B to cover x km is 30 hours.
x/30 - x/55 = 30
x = 1980 km
Answer is B
Answer: B
Question: A man's speed with the current is 15 km/hr and the speed of the current is 2.5 km/hr. The man's speed against the current is:
Options:
A)8.5 km/hr
B)10 km/hr
C)12.5 km/hr
D)9 km/hr
E)8 km/hr
Rationale: Man's speed with the current = 15 km/hr
=> speed of the man + speed of the current = 15 km/hr
speed of the current is 2.5 km/hr
Hence, speed of the man = 15 - 2.5 = 12.5 km/hr
man's speed against the current = speed of the man - speed of the current
= 12.5 - 2.5 = 10 km/hr
ANSWER:B
Answer: B
Question: Jack and Jill are marathon runners. Jack can finish a marathon (42 km) in 6 hours and Jill can run a marathon in 4.2 hours.What is the ratio of their average running speed? (Jack: Jill)
Options:
A)7/10
B)15/14
C)4/5
D)5/4
E)Cannot be determined
Rationale: Average Speed of Jack =Distance/ Time
= 42 /6= 7
Average Speed of Jill = 42/(4.2) = 10
Ratio of Average speed of Jack to Jill = 7/10
Answer A
Answer: A
Question: A man can row downstream at 22 kmph and upstream at 10 kmph. Find the speed of the man in still water and the speed of stream respectively?
Options:
A)13, 3
B)12, 6
C)15, 3
D)14, 4
E)16, 6
Rationale: Explanation:
Let the speed of the man in still water and speed of stream be x kmph and y kmph respectively.
Given x + y = 22 --- (1)
and x - y = 10 --- (2)
From (1) & (2) 2x = 32 => x = 16, y = 6.
Answer: Option E
Answer: E
Question: If x=y=2z and xyz=256 then what is the value of x?
Options:
A)12
B)8
C)16
D)6
E)2
Rationale: Given that x=y=z.
so,x=y,
x=2z,
y=2z.
given xyz=256
=>(2z)*(2z)*z=256
=>4z^3=256
=>z=4
then x=2z
=2(4)
=8
ANSWER:B
Answer: B
Question: Harkamal purchased 8kg of grapes at the rate of 70 per kg and 9 kg of mangoes at the rate of 50 per kg. How much amount did he pay to the shopkeeper?
Options:
A)A)1000
B)B)1010
C)C)1065
D)D)1075
E)E)1080
Rationale: Cost of 8 kg grapes = 70 × 8 = 560.
Cost of 9 kg of mangoes = 50 × 9 = 450.
Total cost he has to pay = 560 + 450 = 1010.
B)
Answer: B
Question: There are 1000 students in a school and among them 20% of them attends chess class. 10% of the students who are in the chess class are also enrolled for swimming. No other students are interested in swimming so how many will attend the swimming class if all enrolled attends?
Options:
A)1
B)10
C)100
D)50
E)20
Rationale: 20% of 1000 gives 200. so 200 attends chess and 10% of 200 gives 20.
so 20 enrolled for swimming
Answer: E
Answer: E
Question: Find the odd man out. 1,4,9,16,20
Options:
A)8
B)10
C)25
D)6
E)14
Rationale: Explanation :
1^2=1
2^2=4
3^2=9
4^2=16
5^2=25
Hence, it is clear that 25 should have come instead of 20
Answer : Option C
Answer: C
Question: Havid has d books, which is 3 times as many as Jeff and 1/2 as many as Paula. How many books do the three of them have altogether, in terms of d?
Options:
A) 5/6*d
B) 7/3*d
C) 10/3*d
D) 7/2*d
E) 9/2*d
Rationale: Havid has d books;
Jeff has d/3 books;
Paula has 2d books;
Total = d+d/3+2d=10d/3.
Answer: C.
Answer: C
Question: Mike needs 30% to pass. If he scored 212 marks and falls short by 25 marks, what was the maximum marks he could have got?
Options:
A)343
B)677
C)790
D)867
E)932
Rationale: If Mike had scored 25 marks more, he could have scored 30%
Therefore, Mike required 212 + 25 = 237 marks
Let the maximum marks be m.
Then 30 % of m = 237
(30/100) × m = 237
m = (237 × 100)/30
m = 23700/30
m = 790
Answer:C
Answer: C
Question: Bag A contains red, white and blue marbles such that the red to white marble ratio is 1:3 and the white to blue marble ratio is 2:3. Bag B contains red and white marbles in the ratio of 1:4. Together, the two bags contain 24 white marbles. How many red marbles could be in bag A?
Options:
A)1
B)3
C)4
D)6
E)8
Rationale: 6 is the answer.
Bag A-
R:W:B = 2:6:9
Let W in bag A be 6K
Bab B -
R:W = 1:4
Let W in bag B be 4P
W = 24 = 6K+4P
=> K =2, P=3
Total Red's in bag A will be 2K = 4
C
Answer: C
Question: The timing of a college is from 12 p.m to 4.20 p.m. Five lectures are held in the given duration and a break of 5 minutes after each lecture is given to the students. Find the duration of each lecture.
Options:
A)52 minutes
B)45 minutes
C)30 minutes
D)48 minutes
E)58 minutes
Rationale: Explanation:
Total time a student spends in college = 4 hours 20 minutes = 260 minutes
As there are 5 lectures, the number of breaks between lectures is 4. Total time of the break = 20 minutes
Hence, the duration of each lecture is = (260 – 20)/5
= 48 minutes
ANSWER D
Answer: D
Question: On dividing 12401 by a certain number, we get 76 as quotient and 13 as remainder. What is the divisor?
Options:
A)163
B)785
C)852
D)456
E)852
Rationale: Divisor *quotient+remainder=dividend
divisor=(dividend)-(remainder)/quotient
(12401-13)/76=163
Answer(A)
Answer: A
Question: Three people each took 5 tests. If the ranges of their scores in the 5 practice tests were 15, 25 and 30, what is the minimum possible range in scores of the three test-takers?
Options:
A)25
B)30
C)35
D)15
E)10
Rationale: I simply looked at the 3 different possible scores for each individual test: 15,30,25
We have to find the minimum range:
30-15=15
30-25=5
25-15=10
The find the minimum range, you have to make the set of the 5 scores as small as possible.
Which means that 4 of the 5 scores of each individual person is zero.
5*5 = 25.
Answer:A
Answer: A
Question: There are 30 people in a party. If everyone is to shake hands with one another, how many hand shakes are possible?
Options:
A)435
B)367
C)361
D)297
E)261
Rationale: Explanation:
Total number of persons = n = 30
Shakehands involve only 2 persons = r = 2
Number of shakehands = nCr = 30C2
30C2 = (30 * 29) /(2 * 1) = 435
nCr = (n!) / r! (n – r)!
= 30! / 2! 28!
= 435
ANSWER:A
Answer: A
Question: A pump can fill a tank with a water in 2 hours. Because of a leak, it took 2 and 4/5 hours to fill the tank. The leak can drain all the water of the full tank in how many hours?
Options:
A)4
B)5
C)6
D)7
E)8
Rationale: The rate of the pump + leak = 5/14
1/2 - leak's rate = 5/14
leak's rate = 1/2 - 5/14 = 1/7
The leak will empty the tank in 7 hours.
The answer is D.
Answer: D
Question: P alone can complete a piece of work in 6 days. Work done by Q alone in one day is equal to one-third of the work done by P alone in one day. In how many days can the work be completed if P and Q work together?
Options:
A)491/2)
B)4(1/2)
C)4(1/8)
D)7(1/2)
E)4(5/2)
Rationale: Work done by P alone in one day = 1/6th of the total work done by Q alone in one day = 1/3(of that done by P in one day) = 1/3(1/6 of the total) = 1/18 of the total.
Work done by P and Q, working together in one day = 1/6 + 1/18 = 4/18 = 2/9 of the total
They would take 9/2 days = 4 (1/2) days to complete the work working together.
Answer:D
Answer: D
Question: At present George is 10 years older than Jason. After 5 years George will be twice the age of Jason. Find their present age.
Options:
A)2 and 12
B)6 and 16
C)4 and 14
D)10 and 20
E)5 and 15
Rationale: Let the present age be x and x+10
After 5 years ages will be x+5 and x+15
2 (x + 5) = x + 15
x = 5
Hence the present ages are 5 and 15
Answer : E
Answer: E
Question: A cyclist traveled for two days. On the second day the cyclist traveled 6 hours longer and at an average speed 10 mile per hour slower than she traveled on the first day. If during the two days she traveled a total of 280 miles and spent a total of 12 hours traveling, what was her average speed on the second day?
Options:
A)5 mph
B)10 mph
C)20 mph
D)30 mph
E)40 mph
Rationale: SOLUTION:
D = 280 Mi
T = 12 hrs
Đây 1 time = T1
Dây 2 time = T2
T2 - T1 = 4 hrs ----- (I)
T1 + T2 = 12 hrs ----- (II)
Adding I and II, T2 = 8 hrs and T1 = 4 hrs
Dày 1 Rate = R1
Dây 2 Rate = R2
R1 - R2 = 10 mph
í.ẹ. R1 = 10 + R2
280 = 8R2 + 4R1
í.ẹ. 280 = 8R2 + 4 (10 + R2)
í.ẹ. R2 = 20 mph
ANSWER: A
Answer: A
Question: If the two-digit integers A and B are positive and have the same digits, but in reverse order, which of the following could be the sum of A and B?
Options:
A)77
B)78
C)79
D)80
E)81
Rationale: A two-digit number "cd" can be expressed algebraically as 10c+d.
The sum of A + B = (10c+d)+(10d+c)=11(c+d).
The only answer that is a multiple of 11 is 77.
The answer is A.
Answer: A
Question: The value of B+5B(BB)(2B)when B = 3 is:
Options:
A)900
B)810
C)809
D)803
E)813
Rationale: Solution:
B+5B(BB)(2B)
Put the value of B= 3 in the above expression we get,
3+ 15(33)(2 x 3)
= 3+ 15(3 Ã— 3)(2 Ã— 3)
= 3 + 810
=813
Answer : E
Answer: E
Question: The average monthly income of P and Q is Rs. 5050. The average monthly income of Q and R is 6250 and the average monthly income of P and R is Rs. 5250. The monthly income of P is?
Options:
A)A)Rs.4078
B)B)Rs.4000
C)C)Rs.4050
D)D)Rs.4027
E)E)Rs.4020
Rationale: Let P, Q and R represent their respective monthly incomes. Then, we have:
P + Q = (5050 * 2) = 10100 --- (i)
Q + R = (6250 * 2) = 12500 --- (ii)
P + R = (5200 * 2) = 10500 --- (iii)
Adding (i), (ii) and (iii), we get:
2(P + Q + R) = 33100 = P + Q + R = 16550 --- (iv)
Subtracting (ii) from (iv), we get, P = 4050.
P's monthly income = Rs.4050.
Answer: C
Answer: C
Question: Timothy leaves home for school, riding his bicycle at a rate of 8 miles per hour. Fifteen minutes after he leaves, his mother sees Timothy’s math homework lying on his bed and immediately leaves home to bring it to him. If his mother drives at 36 miles per hour, how far (in terms of miles) must she drive before she reaches Timothy?
I think is a 700 level problem but I tag it as 600/700, let me know. Either way I hope in an explanationThanks
Options:
A)1/3
B)3
C)4
D)9
E)36/13
Rationale: In 15 mins, Timothy travels=8/4 miles.
Now, let his mother takes x hours to reach him, traveling at 36mph.
So, 36x=8x+8/4
x=1/13 hrs.
Thus, the distance traveled by his mother to reach= 36*1/13=36/13 miles. Ans E
Answer: E
Question: A teacher can divide her class into groups into groups of 5,13 and 17. What is the smallest possible strength of the class?
Options:
A)1387
B)1768
C)1105
D)1286
E)2812
Rationale: Explanation:
For smallest possible class strength, we consider LCM of the given numbers. So LCM (5, 13, 17) = 1105
Answer: C
Answer: C
Question: is two years older than B who is twice as old as C. If the total of the ages of A, B and C be 37, then how old is B?
Options:
A)7
B)8
C)9
D)10
E)14
Rationale: Let C's age be x years.
Then, B's age = 2x years.
A's age = (2x + 2) years.
(2x + 2) + 2x + x = 37
5x = 35 => x = 7
Hence, B's age = 2x = 14 years.
ANSWER:E
Answer: E
Question: If the cost of 6 erasers and 4 sharpeners is Rs.21/-. What will be the cost of 24 erasers and 16 sharpeners?
Options:
A)Rs.62/-
B)Rs.84/-
C)Rs.64/-
D)Rs.88/-
E)Rs.99/-
Rationale: Explanation:
6 erasers + 4 sharpeners = Rs.21.
Multiply by 4
We get,
24 erasers + 16 sharpeners = Rs.21x4 = Rs.84/-.
Answer: Option B
Answer: B
Question: The contents of two vessels containing copper and tin in the ratio 1:2 and 2:5 are mixed in the ratio 1:4. The resulting mixture will have copper and tin in the ratio?
Options:
A)15:32
B)21:28
C)31:74
D)33:37
E)41:92
Rationale: The ratio of copper and tin the new vessel= (1/3*1/5 + 2/7*4/5) : (2/3*1/5 + 5/7*4/5)
= 31/105 : 74/105 = 31:74
Answer is C
Answer: C
Question: What will come in place of the question mark (?) in the following question?
112.36 + 225.05 + ? = 815.30
Options:
A)477.79
B)477.8
C)477.81
D)477.89
E)477.84
Rationale: Explanation:
815.30 - ( 112.36 + 225.05) = ?
= 477.89
ANSWER:D
Answer: D
Question: A wooden box of dimensions 8 m x 7 m x 6 m is to carry rectangularboxes of dimensions 4 cm x 7 cm x 6 cm. The maximum number ofboxes that can be carried in the wooden box, is
Options:
A)9800000
B)1000000
C)7500000
D)2000000
E)None of these
Rationale: Explanation: Number = (800*700*600)/8*7*6 = 2000000 Answer: D
Answer: D
Question: Two unbiased coins are tossed. What is the probability of getting at most one head?
Options:
A)1/2
B)1/4
C)2/3
D)3/4
E)5/7
Rationale: S = {HH, HT, TH, TT}
E = event of getting at most one head
E = {TT, HT, TH}
P(E) = 3/4
Answer is D
Answer: D
Question: Natasha climbs up a hill, and descends along the same way she went up. It takes her 4 hours to reach the top and 2 hours to come back down. If her average speed along the whole journey is 3 kilometers per hour, what was her average speed (in kilometers per hour) while climbing to the top?
Options:
A)1.75
B)2.25
C)2.5
D)2.75
E)3.25
Rationale: Let the distance to the top be x, so the total distance traveled by Natasha is 2x.
The total time is 4 + 2 = 6 hours
The average speed = total distance/total time taken = 2x/6 = x/3
The average speed of the complete journey is 3 km/hour
x/3 = 3
x = 9 km
The average speed while climbing = distance/time = 9/4 = 2.25 km/h
The answer is B.
Answer: B
Question: What is the unit’s digit of 213 *212*347*468*778 ?
Options:
A)5
B)6
C)7
D)8
E)9
Rationale: 3 × 2 = 6
6 × 7 = 42
2 × 8 = 16
6 × 8 = 48
So at units place 8 is present
thus ans is 8
ANSWER:D
Answer: D
Question: The product D of the first 10 prime numbers is closest to which of the following?
Options:
A)6.5 x 10^7
B)6.5 x 10^8
C)6.5 x 10^9
D)6.5 x 10^10
E)6.5 x 10^11
Rationale: The first 10 primes are: 2, 3, 5, 7, 11, 13, 17, 19, 23, and 29.
So, we need to find approximate value of 2*3*5*7*11*13*17*19*23*29.
# of different approximations are possible. For example:
2*5=10
3*17=~50 (actually more than 50);
7*13=~100 (actually less than 100);
11*19=~200 (actually more than 200)
23*29=667=D
2*3*5*7*11*13*17*19=~10*50*100*200*667=10^7*667=6.67*10^9.
Answer: C.
Answer: C
Question: On a sum of money ,the simple interest for 2 years is Rs. 660,while the compound interest is Rs. 696.30,the rate of interest being the same in both the cases.The rate of interest is :
Options:
A)10%
B)10.5%
C)12%
D)None of these
E)Cannot be determined
Rationale: Solution
Difference in C.I. and S.I. for 2 years = Rs.(696.30 - 660)= Rs. 36.30.
S.I. for one year = Rs. 330.
∴ S.I. on Rs. 330 for 1 year = Rs. 36.30.
∴ Rate (100x36.30/330x1)% = 11%. Answer D
Answer: D
Question: How many liters of a 40% iodine solution need to be mixed with 35 liters of a 20% iodine solution to create a 34% iodine solution?
Options:
A)35
B)49
C)100
D)105
E)136
Rationale: Solution 1:
Assume the iodine solution to be mixed = x lts.
Iodine = 0.4x lts, Water = 0.6x lts.
Solution 2: 35 liters of a 20% iodine solution
Iodine = 7 lts, Water = 28 lts.
Total iodine = 0.4x + 7
Total water = 0.6x + 28
The resultant is a 35% idoine solution.
Hence (0.4x + 7) / (x + 35) = 34/100
40x + 700 = 34x + 1220
6x = 820
x = 136 lts
Correct Option: E
Answer: E
Question: The duplicate ratio of 3:4 is?
Options:
A)9:15
B)9:10
C)9:16
D)9:13
E)9:11
Rationale: 32: 42 = 9:16
Answer: C
Answer: C
Question: Which of the following is NOT a factor of 5! ?
Options:
A)2
B)3
C)4
D)5
E)6
Rationale: 120=5*4*3*2 and we dont have 6
ANSWER IS E
Answer: E
Question: A number consists of two digits whose sum is 11. If 27 is added to the number, then the digits change their places. What is the number ?
Options:
A)47
B)65
C)83
D)92
E)98
Rationale: Explanation:
Let the ten's digit be x. Then, unit's digit = (11 - x).
So, number = 10x + (11 - x) = 9x + 11.
Therefore (9x + 11) + 27 = 10 (11 - x) + x 9x + 38 = 110 - 9x 18x = 72 x = 4.
Thus, ten's digit = 4 and unit's digit = 7.
Hence, required number = 47.
Answer: Option A
Answer: A
Question: solve the following quadratic equation by factoring.
u^2-5u-14=0
Options:
A)(5,7)
B)(8,9)
C)(3,2)
D)(5,3)
E)(-2,7)
Rationale: We already have zero on one side of the equation, which we need to proceed with this problem. Therefore, all we need to do is actually factor the quadratic.
(u+2)(u-7)=0
u=-2, u=7
Therefore the two solutions are :(-2,7)
Ans: E
Answer: E