divelab/Sys2Bench huggingface dataset:
Question: In a LAB experiment, a 100 ml compound of element X and element Y needs to be increased to 210 ml by adding some quantity of element X, and some quantity of element Y. If the original compound contains 30% of element X, how much (in ml) of element Y must be added so that element X will be exactly one third of the resulting mixture ?
Options:
A)40 ml
B)70 ml
C)20 ml
D)20/3 ml
E)10 ml
Rationale: There are 0.3*100=30 ml of element X in the initial solution.
We need X to be 1/3rd of 210 ml in the resulting solution, so 70 ml. Thus out of 110 ml of solution added, 40 must be X (in order it to become 70 ml) ans the remaining 70 must be Y.
Answer: B
Answer: B
Question: The sum of two consecutive integers is 51. Find the numbers.
Options:
A)17, 18
B)7, 8
C)5, 6
D)1, 2
E)25,26
Rationale: n + (n + 1) = 51
2n + 1 = 51
2n = 50
n = 25
Answer: E
Answer: E
Question: An investment of $2000 was made in a certain account and earned interest that was compounded annually. The annual interest rate was fixed for the duration of the investment, and after 12 years the $1000 increased to $8000 by earning interest. In how many years after the initial investment was made the $1000 have increased to $16000 by earning interest at that rate?
Options:
A)18
B)12
C)20
D)24
E)30
Rationale: In 12 years the investment quadrupled (from $2,000 to $8,000).
Thus, at the same rate compounded annually, it would need additional 12/2=6 years to double (from $8,000 to $16,000).
Therefore, 12+6=18 years are needed $2,000 to increase to $16,000.
Answer: A.
Answer: A
Question: There are some pigeons and hares in a zoo. If heads are counted, there are 200. If legs are counted, there are 580. The number of hares in the zoo is?
Options:
A)100
B)120
C)90
D)95
E)80
Rationale: 200*2 = 400
580=>180
1----2
?----180 = 90
ANSWER C
Answer: C
Question: A box contains 3 blue marbles, 4 red, 6 green marbles and 2 yellow marbles. If three marbles are drawn what is the probability that one is yellow and two are red?
Options:
A)12/429
B)12/429
C)12/455
D)12/927
E)12/292
Rationale: Given that there are three blue marbles, four red marbles, six green marbles and two yellow marbles.
When three marbles are drawn, the probability that one is yellow and two are red
= (²C₁)(⁴C₂)/¹⁵C₃
= (2 * 4 * 3 * 3 * 2)/(1 * 2 * 15 * 14 * 13)
=12/455
Answer: C
Answer: C
Question: A certain account pays 1.5 percent compound interest every 3 months. A person invested an initial amount and did not invest any more money in the account after that. If after exactly 5 years, the amount of money in the account was T dollars, which of the following is an expression for the original number of dollars invested in the account?
Options:
A)[(1.015)^4]T
B)[(1.015)^15]T
C)[(1.015)^20]T
D)[T/{(1.015)^15}]
E)[T/{(1.015)^20}]
Rationale: The original formula of "annual compound interest rate" is:
t = number of years
i = annual interest rate
T = future value (FV)
P = Present value (PV)
After t year:
FV = PV[1 + i]^t
If interest rate is quarterly, then
new interest rate/period = i/4
new number of periods = t*4
The new formula is:
FV = PV[1 + i/4]^4t
Apply to this question:
T = PV[1 + 0.015]^20
Because i/4 = 1.5%
number of periods = 4*5 = 20
So, PV = T/[1.015]^20
--Or--
PV = T*[1.015]^-20
ANSWER:E
Answer: E
Question: A, B and C enter into a partnership and their shares are in the ratio 1/2 : 1/3 : 1/4. After 2 months, A withdraws half of his capital and after 10 months, a profit of Rs. 378 is divided among them. What is B's share ?
Options:
A)130
B)145
C)140
D)144
E)None of these
Rationale: Explanation :
Ratio of initial investments = 1/2 : 1/3 : 1/4 = 6 : 4 : 3.
Let their initial investments be 6x, 2x and 3x respectively.
A : B : C = (6x * 2 + 3x * 10) : (4x * 12) : (3x * 12) = 42 : 48 : 36 = 7 : 8 : 6.
B's share = Rs. (378 * 8/21) = Rs. 144.
Answer : D
Answer: D
Question: What is the greatest prime factor of 10!9! + 9!8! ?
Options:
A)7
B)11
C)13
D)17
E)19
Rationale: 10!9! + 9!8! = 9!8!(10*9+1) = 9!*8!*91 = 9!*8!*7*13
The answer is C.
Answer: C
Question: A test has 150 questions. Each question has 5 options, but only 1 option is correct. If test-takers mark the correct option, they are awarded 1 point. However, if an answer is incorrectly marked, the test-taker loses 0.25 points. No points are awarded or deducted if a question is not attempted. A certain group of test-takers attempted different numbers of questions, but each test-taker still received the same net score of 40. What is the maximum possible number of such test-takers?
Options:
A)19
B)21
C)23
D)25
E)27
Rationale: A correct answers get you 1 point, an incorrect answer gets you MINUS 1/4 point and a skipped question gets you 0 points. Since there are 200 total questions, there are a variety of ways to get a total of 40 points.
Let C be the number of correct answers and let I be the number of incorrect answers.
To get 40 points, a test taker must have at least 40 correct answers. Then C=> 40.
For every correct question above 40, the test taker has 4 incorrect answers.
Then, the I = 4*(C - 40).
Also, I + C <= 150. Thus 5C <= 310 and so C <= 62.
Then 40 <= C <= 62 and C can have 23 possible values.
The answer is C.
Answer: C
Question: A man buys Rs. 20 shares paying 9% dividend. The man wants to have an interest of 12% on his money. The market value of each share is:
Options:
A)33
B)15
C)66
D)77
E)81
Rationale: Dividend on Rs. 20 = Rs. 9	x 20 = Rs.	9	.
100	5
Rs. 12 is an income on Rs. 100.
Rs.	9	is an income on Rs. 100	x	9 = Rs. 15.
Answer:B
Answer: B
Question: the present worth of a sum due sometimes hence is Rs. 5760 and the banker's gain is Rs.10 . what is the true discount?
Options:
A)480
B)420
C)240
D)120
E)380
Rationale: TD= square root of PW * BG = square root of 5760 * 10 = square root of 57600= Rs. 240
correct answer : (C)
Answer: C
Question: David gets on the elevator at the 11th floor of a building and rides up at the rate of 57 floors per minute. At the same time, Albert gets on an elevator at the 51st floor of the same building and rides down at the rate of 63 floors per minute. If they continue travelling at these rates, then at which floor will their paths cross ?
Options:
A)19
B)28
C)37
D)53
E)30
Rationale: Suppose their paths cross after x minutes.
Then, 11 + 57x = 51 - 63x 120x = 40
x =	1/3
Number of floors covered by David in (1/3) min. =(1/3)x 57
= 19.3
So, their paths cross at (11 +19) i.e., 30th floor.
Answer is E.
Answer: E
Question: Rectangular tile each of size 70cm by 30cm must be laid horizontally on a rectangular floor of size 900cm by 600cm,such that the tiles do not overlap and they are placed with edges jutting against each other on all edges. A tile can be placed in any orientation so long as its edges are parallel to the edges of floor. No tile should overshoot any edge of the floor. The maximum number of tiles that can be accommodated on the floor is:
Options:
A)50
B)70
C)80
D)90
E)100
Rationale: Area of tile = 90*30 = 2700
Area of floor = 900*300 = 270000
No of tiles = 270000/2700 = 100
So, the no of tile =100
ANSWER:E
Answer: E
Question: Calculate 1307 x 1307 =?
Options:
A)1601249
B)1607249
C)1701249
D)1708249
E)none
Rationale: Answer
1307 x 1307 = (1307)2
= (1300 + 7 )2
= (1300)2 + (7)2 + 2 x 1300 x 7
= 1690000 + 49 +18200
= 1708249
Option: D
Answer: D
Question: A train 960 m long passed a pole in 96 sec. How long will it take to pass a platform 2600 m long?
Options:
A)350
B)352
C)354
D)356
E)358
Rationale: Speed = 960/96 = 10 m/sec.
Required time = (960 + 2600)/10
= 356 sec.
Answer:D
Answer: D
Question: A trained covered x km at 40 kmph and another 4x km at 20 kmph. Find the average speed of the train in covering the entire 3x km.
Options:
A)11.33 kmph
B)12.33 kmph
C)13.33 kmph
D)14.33 kmph
E)23.33 kmph
Rationale: Total time taken = x/40 + 4x/20 hours
= 9x/40 hours
Average speed = 3x/(9x/40) = 13.33 kmph
ANSWER:C
Answer: C
Question: In how many ways the letters of the word REDUCTION be arranged?
Options:
A)344500
B)325479
C)226478
D)362880
E)344900
Rationale: Total ways of arranging the letters = 9! = 362880 ways.
D)
Answer: D
Question: An engineer undertakes a project to build a road 10 km long in 30 days and employs 30 men for the purpose. After 10 days, he finds only 2 km of the road has been completed. Find the (approximate) number of extra men he must employ to finish the work in time.
Options:
A)30
B)40
C)50
D)60
E)70
Rationale: 30 workers working already
Let x be the total men required to finish the task in next 20 days
2 km done hence remaining is 8 km
Also, work has to be completed in next 20 days (30 - 10 = 20)
We know that, proportion of men to distance is direct proportion
and, proportion of men to days is inverse proportion
Hence, X = (30 * 8 * 10) / (2 *20)
thus, X = 60
Thus, more men needed to finish the task = 60-30=30
Answer : A
Answer: A
Question: r = number of flags;n = number of poles;Any number of flags can be accommodated on any single pole.r= 5 n = 3 . If first pole has 2 flags ,third pole has 1 flag, how many ways the remaining can be arranged?
Options:
A)4 ways
B)5 ways
C)7 ways
D)8 ways
E)6 ways
Rationale: 2 flags are remaining and as it is given that 3 poles can have any no. of flags therefore
3p2=3!/(3-2)!
=3*2=6 ways
ANSWER:E
Answer: E
Question: 7528:5306::4673:?
Options:
A)2351
B)2451
C)2551
D)2651
E)2751
Rationale: As there is a difference of 2222. 7528-2222 = 5306. so 4673-2222 = 2451
ANSWER:B
Answer: B
Question: There are two positive numbers in the ratio 5:8. If the larger number exceeds the smaller by 15, then find the smaller number?
Options:
A)25
B)81
C)86
D)16
E)27
Rationale: Let the two positive numbers be 5x and 8x respectively.
8x - 5x = 15
3x = 15 => x = 5
=> Smaller number = 5x = 25.
Answer: A
Answer: A
Question: A can do a piece of work in 4 days. B can do it in 6 days. With the assistance of C they completed the work in 2 days. Find in how many days can C alone do it?
Options:
A)22 days
B)12 days
C)67 days
D)17 days
E)18 days
Rationale: C = 1/2 - 1/6 - 1/4 = 1/12
=> 12 days
Answer:B
Answer: B
Question: There are certain number of hats and gloves in a box.They are of 41 red,23 green,11 orange.Power gone but a woman can differentiate between hats and gloves.How many draws are required to obtain a pair of each color?
Options:
A)66
B)67
C)68
D)69
E)70
Rationale: the worst case senario will be
first take 40 red
then take 22 green
then take 1+1 red+green
then take 2 orange.
so total 40+22+2+2=66
ANSWER:A
Answer: A
Question: A car started running at a speed of 30 km/hr and the speed of the car was increased by 2 km/hr at the end of every hour. Find the total distance covered by the car in the first 10 hours of the journey.
Options:
A)227
B)390
C)287
D)2771
E)0212
Rationale: The total distance covered by the car in the first 10 hours = 30 + 32 + 34 + 36 + 38 + 40 + 42 + 44 + 46 + 48 = sum of 10 terms in AP whose first term is 30 and last term is 48 = 10/2 [30 + 48] = 390 km.
Answer: B
Answer: B
Question: A bag contains a certain number of 50 paise coins, 20 paise coins and 10 paise coins inthe ratio 3:4:5. If the total value of all the coins in the bag is Rs.280, find the number of 50 paise coins ?
Options:
A)498
B)488
C)300
D)528
E)418
Rationale: 50*3k + 20*4k + 10*5k = 28000
280k = 28000 => k = 100
50p coins = 3k = 3*100 = 300
ANSWER:C
Answer: C
Question: Fox jeans regularly sell for $15 a pair and Pony jeans regularly sell for $18 a pair. During a sale these regular unit prices are discounted at different rates so that a total of $8.82 is saved by purchasing 5 pairs of jeans: 3 pairs of Fox jeans and 2 pairs of Pony jeans. If the sum of the two discount rates is 22 percent, what is the discount rate on Pony jeans?
Options:
A)9%
B)10%
C)11%
D)12%
E)15%
Rationale: Let x be the discount on Pony jeans. Then 0.22-x is the discount on Fox jeans.
3(0.22-x)(15) + 2x(18) = 8.82
9.9 - 45x + 36x = 8.82
9x = 1.08
x = 0.12
The answer is D.
Answer: D
Question: If 2 < x < 7 and 8 << y < 9, then the lowest possible value for xy is
Options:
A)between 60 and 70
B)between 0 and 60
C)0
D)between 0 and -70
E)less than -70
Rationale: If 2 < x < 7 and 8 << y < 9,
Say y = 8.9 (8<<y<9)
x = 6
So xy =53.4 which is less than 60.
Answer is B.
Answer: B
Question: By looking at a rectangular box, a carpenter estimates that the length of the box is between 2 to 2.1 meters, inclusive, the breadth is between 1 to 1.1 meters, inclusive and the height is between 2 to 2.1 centimeters, inclusive. If the actual length, breadth and height of the box do indeed fall within the respective ranges estimated by the carpenter, which of the following is the closest to the maximum possible magnitude of the percentage error K that the carpenter can make in calculating the volume of the rectangular box?
Options:
A)1%
B)3%
C)10%
D)18%
E)22%
Rationale: the dimensions are between 2-2.1, 1-1.1 and 2-2.1..
so MAX error in Volume=?..
IMPORTANT POINTS TO NOTE
1) Max ERROR would be when actuals and estimate are at the extreme ends..
2) BUT what about MAX % ERROR--
It will be when the base is the lowest...
so the ACTUALS should be lowest or at lower end and the ESTIMATE at the higher end..
SOLUTION
Actuals= 2*2*1=4..
Estimate=2.1*2.1*1.1=4.841..
Max % error K=(4.841-4)/4 *100= 84.1/4 = 21.25%..
21.25% is closest to 22% in the choices..
ans E
Answer: E
Question: Victor gets 92 % marks in examinations. If these are 460 marks, find the maximum marks.
Options:
A)334
B)500
C)376
D)288
E)271
Rationale: Let the maximum marks be m
Then 92 % of m = 460
⇒ 92/100 × m = 460
⇒ m = (460 × 100)/92
⇒ m = 46000/92
⇒ m = 500
Therefore, maximum marks in the examinations are 500.
Answer:B
Answer: B
Question: If y exceeds x by 25%, then x is less than y by?
Options:
A)20%
B)16 3/3%
C)16 2/3%
D)12 2/3%
E)16 1/3%
Rationale: X=100	y=125
125------25
100-------? => 20%
Answer:A
Answer: A
Question: From a group of 3 boys and 3 girls, 4 children are to be randomly selected. What is the probability that equal numbers of boys and girls will be selected?
Options:
A)1/10
B)4/9
C)1/2
D)3/5
E)2/3
Rationale: prob reqd= 1- probab of not having eq numbers
prob of not having eq no's is = prob of having 3 Boys and 1 girl or prob of having 3 girls and 1 boy
=(1* 3/15) +(3/15 * 1)=2/5
reqd prob= 1-2/5=3/5
Answer:D
Answer: D
Question: A, B and C can do a work in 6 days, 8 days and 12 days respectively. In how many days can all three of them working together, complete the work?
Options:
A)2 2/3 days
B)2 8/3 days
C)2 23/3 days
D)8 2/3 days
E)3 2/3 days
Rationale: Work done by all three of them in one day = 1/6 + 1/8 + 1/12 = 3/8.
The number of days required = 8/3
= 2 2/3 days.
Answer: A
Answer: A
Question: A dishonest dealer professes to sell goods at the cost price but uses a weight of 800 grams per kg, what is his percent?
Options:
A)2%
B)25%
C)22%
D)29%
E)45%
Rationale: 800 --- 200
100 --- ? => 25%
Answer:B
Answer: B
Question: The product of all the prime numbers less than 18 is closest to which of the following powers of 10?
Options:
A)10^9
B)10^8
C)10^7
D)10^6
E)10^5
Rationale: OA: D
Answer: D
Question: The H.C.F. of two numbers is 20 and the other two factors of their L.C.M. are 13 and 14. The larger of the two numbers is:
Options:
A)276
B)280
C)322
D)345
E)354
Rationale: Clearly, the numbers are (20 x 13) and (20 x 14).
Larger number = (20 x 14) = 280.
Answer: Option B
Answer: B
Question: Which of the following fractions Q, if written as a decimal, would have a 2 in the thousandths place ?
Options:
A) 3/11
B) 7/9
C) 1/8
D) 4/7
E) 1/6
Rationale: (A) Q=3/11 = 0.2727272727....
Answer: A
Answer: A
Question: Two family reunions are happening at the Park Avenue hotel, the Oates reunion and the Hall reunion. All 100 guests at the hotel attend at least one of the reunions. If 40 people attend the Oates reunion and 70 people attend the Hall reunion, how many people attend both reunions
Options:
A) 2
B) 5
C) 10
D) 16
E) 22
Rationale: No of people in Oates reunion=40
No of people in Hall reunion=70
Attending both=x
All guests attend at least one.
Therefore, 100=40+70-(Both)
Both=10
Answer C
Answer: C
Question: Sony and Johnny caught 75 fishes. Sony caught four times as many as Johnny. How many fishes did Johnny catch?
Options:
A)12
B)16
C)34
D)38
E)15
Rationale: s+j=75
s=4*j
j=15
ANSWER:E
Answer: E
Question: The cash difference between the selling prices of an article at a profit of 4% and 6% is Rs. 3. The ratio of the two selling prices is:
Options:
A)52:59
B)52:53
C)52:51
D)52:52
E)52:58
Rationale: Explanation:
Let C.P. of the article be Rs. x.
Then, required ratio = 104% of x / 106% of x
= 104/106 = 52/53 = 52:53
Answer: B
Answer: B
Question: From a group of 7 men and 6 women, five persons are to be selected to form a committee so that at least 3 men are there in the committee. In how many ways can it be done?
Options:
A)650
B)656
C)756
D)760
E)780
Rationale: From a group of 7 men and 6 women, five persons are to be selected with at least 3 men.
Hence we have the following 3 options.
We can select 5 men ...(option 1)
Number of ways to do this = 7C5
We can select 4 men and 1 woman ...(option 2)
Number of ways to do this = 7C4 × 6C1
We can select 3 men and 2 women ...(option 3)
Number of ways to do this = 7C3 × 6C2
Total number of ways
= 7C5 + (7C4 × 6C1) + (7C3 × 6C2)
= 7C2 + (7C3 × 6C1) + (7C3 × 6C2)[∵ nCr = nC(n - r) ]
=
7
×
6
2
×
1
+
7
×
6
×
5
3
×
2
×
1
×
6
=7×62×1+7×6×53×2×1×6
+
7
×
6
×
5
3
×
2
×
1
×
6
×
5
2
×
1
+7×6×53×2×1×6×52×1
=
21
+
210
+
525
=756
C)
Answer: C
Question: If 5 (A's capital) = 10 (B's capital) = 15 (C's capital). Then the ratio of their capitals is?
Options:
A)5:3:2
B)7:3:2
C)6:3:3
D)6:3:2
E)5:3:2
Rationale: 5A = 10B = 15 C
A:B:C = 1/5:1/10:1/15
= 6:3:2
Answer: D
Answer: D
Question: How many ways are there to arrange the letters in the word Teressa ?
Options:
A)1
B)1260
C)3780
D)7560
E)11340
Rationale: 7 total words, 2E, 2S
= 7! / 2 ! * 2! = 1260. Answer B
Answer: B
Question: 617+6.017+0.617+6.0017=?
Options:
A)6.2963
B)62.965
C)629.6357
D)6296.357
E)None of them
Rationale: 617.00
6.017
0.617
6.0017
------------
629.6357
Answer is C
Answer: C
Question: The expression 10^13−120 is divisible by all of the following integers EXCEPT
Options:
A)2
B)13
C)4
D)8
E)11
Rationale: Yeah. I agree this question has two answers which are (B) and (E). If 10 is power to odd number and equal or more than 5, then the answer choice (B) is correct. B
Answer: B
Question: Two cards are drawn at random from a pack of 52 cards.what is the probability that either both are black or both are queen?
Options:
A)55/228
B)55/228
C)55/221
D)55/212
E)55/2234
Rationale: We have n(s) = \inline {\color{Black}52C_{2}} = = 1326.
Let A = event of getting both black cards
B = event of getting both queens
A∩B = event of getting queen of black cards
n(A) = \inline {\color{Black}26C_{2}} = = 325, n(B)= \inline {\color{Black}4C_{2}} = = 6 and n(A∩B) = \inline {\color{Black}2C_{2}} = 1
P(A) = n(A)/n(S) = 325/1326;
P(B) = n(B)/n(S) = 6/1326 and
P(A∩B) = n(A∩B)/n(S) = 1/1326
P(A∪B) = P(A) + P(B) - P(A∩B) = (325+6-1) / 1326 = 330/1326 = 55/221
Answer: C
Answer: C
Question: A, B and C enter into a partnership. They invest Rs. 80,000, Rs. 80,000 and Rs. 1,20,000 respectively. At the end of the first year, B withdraws Rs. 40,000, while at the end of the second year, C withdraws Rs. 80,000. In what ratio will the profit be shared at the end of 3 years?
Options:
A)2 : 3 : 4
B)3 : 2 : 4
C)6 : 2 : 1
D)6 : 4 : 7
E)none
Rationale: Sol.
A : B : C = (80000 * 36) : (80000 * 12 + 40000 * 24)
: (120000 * 24 + 40000 * 12) =288 : 192 : 336 = 6: 4 : 7
Answer D
Answer: D
Question: . Rs.1170 is divided so that 4 times the first share, thrice the 2nd share and twice the third share amount to the same. What is the value of the third share?
Options:
A)540
B)300
C)890
D)750
E)460
Rationale: A+B+C = 1170
4A = 3B = 2C = x
A:B:C = 1/4:1/3:1/2 = 3:4:6
6/13 * 1170 = Rs.540
ANSWER A
Answer: A
Question: Evaluate : (2.39)(power 2) - (1.61) (power 2) / 2.39 - 1.61
Options:
A)4
B)5
C)6
D)7
E)8
Rationale: Given Expression = a(power 2) - b(power 2) / a-b
= (a + b)(a - b) / (a-b)
= (a + b) = (2.39 + 1.61) = 4.
Answer is A.
Answer: A
Question: If x+|x|+y=4 and x+|y|-y=3 what is x+y=?
Options:
A)1
B)-1
C)3
D)5
E)13
Rationale: If x<0 and y<0, then we'll have x-x+y=7 and x-y-y=6. From the first equation y=7, so we can discard this case since y is not less than 0.
If x>=0 and y<0, then we'll have x+x+y=7 and x-y-y=6. Solving gives x=4>0 and y=-1<0 --> x+y=3. Since in PS questions only one answer choice can be correct, then the answer is C (so, we can stop here and not even consider other two cases).
Answer: C.
Adding both eqn we get 2x +IxI+IyI = 13
Now considering X <0 and y>0
2x-x +y = 13
we get x+y =3
Hence answer should be C
Answer: C
Question: What approximate value should come in place of question mark (?) in the following equation
287.532 + 1894.029 – 757.48 = 443.095 + ?
Options:
A)870
B)790
C)680
D)770
E)890
Rationale: ? = 285 + 1895 – 755 – 745 or, ? = 680
Answer C
Answer: C
Question: If 522x is a 3 digit no. with as a digit x. If the no.is divisible by 6, What is the value of the digit x is?
Options:
A)1
B)6
C)4
D)7
E)9
Rationale: If a number is divisible by 6 , it must be divisible by both 2 and 3
In 522x, to this number be divisible by 2, the value of x must be even. So it can be 2,4 or 6 from given options
552x is divisible by 3, If sum of its digits is a multiple of 3.
5+5+2+x =12+x ,
If put x =2 , 12+2=14 not a multiple of 3
If put x =4 , 12+6=18 is a multiple of 3
If put x =6 , 12+2=14 not a multiple of 3
The value of x is 6.
B
Answer: B
Question: 4 points are marked on a straight line and another 5 points are marked on a second straight line with no points in common. How many triangles can be constructed with vertices from among the above points?
Options:
A)91
B)105
C)196
D)2
E)14
Rationale: assuming that the top line can have either 2 points or 1 point...
4! / 1!3! - 4! / 2! 2! yields 2.
D
Answer: D
Question: 56, 63, 59, 11, 57, 6, 11
Which of the following numbers is lower than the median but greater than the mean of the set of the numbers in the list?
Options:
A)21
B)32
C)49
D)56
E)58
Rationale: Yes, it can be solved with observation even without approximation in 30 sec.
Just observe given numbers: 6,11 ,11, 56, 57, 59,63
Median is very easy to tell - its 56. Now we want a number lower than 55 and greater than mean. How many numbers are lower than 56? DE are already out.
Now, we have 21, 32 and 49 and we want a number among these which is greater than mean. logically whatever be the mean, if one value out of these 5 has to be greater than mean it has to be 49. (Notice, if 21 is more than mean then 49 is also. If 32 is more than mean, 49 is also. )
Hence C it is!
Answer: C
Question: Based on this year's costs, a factory budgets P dollars for planing N new cars next year. If the average cost of building each car were to increase 50% from this year's cost, then the greatest number of cars that the factory could build next year using P dollars would be:
Options:
A)33% less than N
B)67% less than N
C)Equal to N
D)33% greater than N
E)67% greater than N
Rationale: Correct answer is (A).
This year, the price of a car is price1=P/N.
If this price increases by 50% it becomes Price2=P/N*1.5
Then with P dollars, you can build P/Price2 cars i.e. P/(P/N*1.5) i.e. N/1.5 i.e. 0.67*N
Which is 33% less than N.
Answer: A
Question: A and B walk around a circular track. A and B walk at a speed of 4 rounds per hour and 5 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 10.30 a.m.?
Options:
A)10
B)20
C)22
D)9
E)5
Rationale: Relative speed = Speed of A + Speed of B (âˆ´ they walk in opposite directions)
= 4 + 5 = 9 rounds per hour
=> They cross each other 9 times in 1 hour and 4 times in 1/2 hour
Time duration from 8 am to 10.30 am =2.5 hour
Hence they cross each other 22 times before 10.30 am
Answer : C
Answer: C
Question: The population of a bacteria colony doubles every day. If it was started 9 days ago with 2 bacteria and each bacteria lives for 13 days, how large is the colony today?
Options:
A)512
B)1024
C)768
D)2048
E)4096
Rationale: 9 days ago- 2
8 days ago- 4
7 days ago- 8
6 days ago- 16
5 days ago- 32
4 days ago- 64
3 days ago- 128
2 days ago- 256
Yesterday- 512
Today- 1024
Ans:B
Answer: B