divelab/Sys2Bench huggingface dataset:
Question: How many positive integers E between 200 and 300 (both inclusive) are not divisible by 2, 3 or 5?
Options:
A)3
B)16
C)75
D)24
E)26
Rationale: 1) I figured there are 101 integers (300 - 200 + 1 = 101). Since the set begins with an even and ends with an even, there are 51 evens.
2) Question says integers are not divisible by 2, leaving all of the odds (101 - 51 = 50 integers).
3) Question says integers are not divisible by 5, removing all the integers ending in 5 (already took out those ending in 0). Take out 10 integers (2?5, ? = 0 to 9), leaving us with 40 integers.
4) Now the painstaking part. We have to remove the remaining numbers that are multiples of 3. Those are 201, 207, 213, 219, 231, 237, 243, 249, 261, 267, 273, 279, 291, and 297...a total of 14 numbers. 26 numbers left!
6) Answer choice E .
Answer: E
Question: Find the simple interest on Rs. 3000 at 6 1/4% per year for the period from 4th Feb., 2005 to 18th April, 2005.
Options:
A)36.5
B)33.5
C)35.5
D)37.5
E)None of them
Rationale: Time = (24+31+18)days = 73 days = 73/365 years = 1/5 years.
P = Rs.3000 and R = 6 ¼ %p.a = 25/4%p.a
Therefore, Simple Interest. = Rs.(3,000*(25/4)*(1/5)*(1/100))= Rs.37.50.
Answer is D.
Answer: D
Question: A company organized a recruiting process for 3 vacant positions of assistant manager for its product launches. The company's efforts yielded 9 eligible candidates. How many sets of 3 candidates can the company choose to fill the 3 assistant manager positions?
Options:
A) 84
B) 94
C) 545
D) 315
E) 220
Rationale: 9*8*7/3*2*1 = 84
A
Answer: A
Question: r, b, and c are positive integers. If r, b, and c are assembled into the six-digit number rbcrbc, which one of the following must be a factor of rbcrbc?
Options:
A) 16
B) 13
C) 5
D) 3
E) none of the above
Rationale: Plug in some values and check -
rbcrbc = 123123
Not divisible by 16 and 5
let rbcrbc = 125125
Not divisible by 3
Only option (B) and (E) is left in both the cases...
Check once more to marke (B) as correct answer
let rbcrbc = 135135
Again divisible by 13
So, mark answer as (B) 13
Answer: B
Question: What are the maximum value and minimum values of f(x) = |2sin(2x - pi/3) - 5| + 3
Options:
A)max 8-min 2
B)max 9-min 1
C)max 10-min 6
D)max 11-min 5
E)max 12-min 3
Rationale: -1 <= sin(2x - pi/3) <= 1 : range of a sine function
-2 <= 2sin(2x - pi/3) <= 2 : multiply all terms of the double inequality by 2
-2 - 5 <= 2sin(2x - pi/3) - 5 <= 2 - 5 : add -5 to all terms of the inequality.
-7 <= 2sin(2x - pi/3) - 5 <= -3
3 <= |2sin(2x - pi/3) - 5| <= 7 : change the above using absolute value.
3 + 3 <= |2sin(2x - pi/3) - 5| + 3 <= 7 + 3 : add 3 to all terms of the double inequality.
The maximum value of f(x) is equal to 10 and the minimum value of f(x) is equal to 6.
correct answer C
Answer: C
Question: The interest on Rs. 750 for 2 years is the same as the true discount on Rs. 960 due 2 years hence. If the rate of interest is the same in both cases, it is:
Options:
A)18%
B)14%
C)94%
D)11%
E)54%
Rationale: S.I. on Rs. 750 = T.D. on Rs. 960.
This means P.W. of Rs. 960 due 2 years hence is Rs. 750.
T.D. = Rs. (960 - 750) = Rs. 210.
Thus, S.I. on R.s 750 for 2 years is Rs. 210.
Rate = 100 x 210	%	= 14%
750 x 2
Answer: B
Answer: B
Question: A man purchased a grinder for Rs. 320 and sold it at a profit of 10%. The selling price is:
Options:
A)Rs. 736
B)Rs. 352
C)Rs. 684
D)None of these
E)Cannot be determined
Rationale: s.p.=110/100 × 320 = 352
Answer: B
Answer: B
Question: Stacy was 5 days late returning items to the library. Stacy had checked out 40 books, 2 movies, and 3 magazines. Books are fined $0.25 per day, movies are fined $3.00 per day, and magazines are fined $0.50 per day. What is the average fine per item per day Stacy paid?
Options:
A)$0.25
B)$0.30
C)$0.35
D)$0.40
E)none of the above
Rationale: Explanation:
40 2 3
$0.25 $3.00 $0.50
$10.00 + $6.00 + $1.50 = $17.50/45 = $0.39
E
Answer: E
Question: A hare and a jackal are running a race. Three leaps of the hare are equal to four leaps of the jackal. For every three leaps of the hare, the jackal takes five leaps. Find the ratio of the speed of the hare to the speed of the jackal.
Options:
A)6:7
B)4:5
C)5:8
D)25:64
E)3:5
Rationale: The hare takes 3 leaps and the jackal takes 5 leaps.
1 hare leap = 4/3 jackal leaps
Thus the hare's 3 leaps = 3*(4/3) = 4 jackal leaps.
The ratio of their speeds is 4:5.
The answer is B.
Answer: B
Question: The length of a rectangular floor is more than its breadth by 300%. If Rs. 423 is required to paint the floor at the rate of Rs. 3 per sq m, then what would be the length of the floor?
Options:
A)24
B)25
C)26.6m
D)26
E)27
Rationale: Let the length and the breadth of the floor be l m and b m respectively.
l = b + 400% of b = l + 4b = 5b
Area of the floor = 423/3 = 141 sq m
l b = 141 i.e., l * l/5 = 141
l2 = 705 => l = 26.6m.
ANSWER:C
Answer: C
Question: What will be the output of the following code statements?
Integer a = 20, b = 30, c = 8 print (a × b / c) – c
Options:
A)65
B)67
C)70
D)26
E)81
Rationale: Explanation:
Apply BODMAS rule
20 × 30 = 600/8-8 =75 –8 =67
Answer:B
Answer: B
Question: A machine fills 150 bottles of water every 8 minutes. How many minutes it takes this machine to fill 675 bottles?
Options:
A)42 minutes
B)72 minutes
C)62 minutes
D)32 minutes
E)42 minutes
Rationale: 8 minutes are needed to fill 150 bottles. How many groups of 150 bottles are there in 675 bottles?
675 / 150 = 4.5 = 4 and 1/2
For each of these groups 8 minutes are needed. For 4 groups and 1/2
8 * 4 + 4 = 32 + 4 = 36 minutes. (4 is for 1/2 a group that needs half time)
We can also find the final answer as follows
4.5 x 8 = 32 minutes correct answer D
Answer: D
Question: The number of arrangements that can be made with the letters of the word MEADOWS so that the vowels occupy the even places?
Options:
A)117
B)144
C)177
D)155
E)121
Rationale: The word MEADOWS has 7 letters of which 3 are vowels.
-V-V-V-
As the vowels have to occupy even places, they can be arranged in the 3 even places in 3! i.e., 6 ways. While the consonants can be arranged among themselves in the remaining 4 places in 4! i.e., 24 ways.
Hence the total ways are 24 * 6 = 144.
Answer:B
Answer: B
Question: 18 buckets of water fill a tank when the capacity of each bucket is 60 litres. How many buckets will be needed to fill the same tank, if the capacity of each bucket is 5 litres?
Options:
A)216
B)200
C)212
D)220
E)210
Rationale: Capacity of the tank = (18 Ã— 60) litre
Number of buckets required of capacity of each bucket is 17 litre = 18Ã—60/5=18Ã—12=216
Answer is A
Answer: A
Question: It takes 4 workers a total of 8 hours to build a giant machine with each worker working at the same rate. If four workers start to build the machine at 11:00 AM, and one extra worker per hour is added beginning at 3:00 PM, at what time will the machine be complete?
Options:
A)5:43
B)5:56
C)6:09
D)6:22
E)6:35
Rationale: 4 workers build 1/8 of a machine in one hour.
1 worker builds 1/32 of a machine in one hour.
In the first 4 hours, 4 workers build 4*(1/32)*4=16/32 of a machine
From 3:00 to 4:00, 5 workers build another 5/32. The total is 21/32.
From 4:00 to 5:00, 6 workers build another 6/32. The total is 27/32.
To build another 5/32, 7 workers need (5/7)*60 minutes which is about 43 minutes.
The machine is complete at about 5:43.
The answer is A.
Answer: A
Question: The cost price of 20 books is the same as the selling price of ’X’ books. If the profit is 40%, then the value of X is ?
Options:
A)25
B)18
C)16
D)8.9
E)14
Rationale: let S.P. of X books = Rs 100 = C.P. of 20 books
Therefore, C.P. of one book = Rs 8
Profit = 40 %
Therefore, S.P of one book = 11.2
S.P of X articles = 11.2 X
11.2 X = 100 => X = 8.9
ANSWER:D
Answer: D
Question: What is the value of 4^5+4^7?
Options:
A)4^12
B)4^35
C)17(4^5)
D)8^12
E)7(4^5)
Rationale: 4^5+4^7
=4^5(1+4^2)
=4^5 * 17
Answer C
Answer: C
Question: A shipment of 1500 heads of cabbage, each of which was approximately the same size was purchased for $600.The day the shipment arrived 2/3 of the heads were sold, each at 25% above the cost per head.The following day the rest were sold at a price per head equal to 10% less than the price each head sold for the day before.what was the gross profit W on this shipment?
Options:
A)a) $100
B)b) $115
C)c) $125
D)d) $130
E)e) $135
Rationale: I recommend one should use fractions to solve this one rather than converting it into decimals
Sol:
1500 heads -> $600
1 head -> $600/1500
1 head -> $(2/5)
25% more of (2/5) -> 125/100 * 2/5 = $(1/2)
He sold 2/3*1500 = 1000 heads for $(1/2) per head
Total revenue by selling 1000 heads = 1000 * 1/2 = $500
Heads left: 500
Cost per head: 90% of the previous price: 90/100 * 1/2 = $(9/20)
Total revenue by selling 500 heads = 9/20 * 500 = 225
Total revenue after selling 1500 cabbage heads - 225+500 = $725
Money spent on the purchase: $600
Profit W= 725-600 = $125
Ans:C
Answer: C
Question: In the equation x^2+bâ€¢x+c=0, b and c are constants, and x is a variable. If m is a root of x^2+bâ€¢x+c=0, and (x+n) is a factor of x^2+bâ€¢x+c=0, then b=
Options:
A)m:n
B)m-n
C)m+n
D)n-m
E)m
Rationale: the equation is x^2+bx+c=0..
this equation will have two roots...
it is given that m is a root ..
secondly (x+n) is a factor, which means -n is another root..
now we are asked what is b..
in x^2+bx+c=0, the sum of roots = -b/a...
here a=1, so sum of roots=-b..
roots are m and -n, so m+(-n)=-b..
m-n=-b, or b=n-m...
ANSWER D
Answer: D
Question: A woman sells an article at 5% loss. Had she sold it for Rs.60 more she could have gained 10%. What is the cost price of the article?
Options:
A)Rs.200/-
B)Rs.400/-
C)Rs.500/-
D)Rs.600/-
E)Rs.650/-
Rationale: Loss =====> 5 %
Profit ====> 10 %
~
----------------
15 % ---------> Rs.60/-
100 % --------> Rs.400/-
Option 'B'
Answer: B
Question: how many internal diagonals does a pentagon (five sided polygon) have?
Options:
A)5
B)8
C)9
D)10
E)12
Rationale: Number of diagonals in any polygon can be found using this formula: n(n-3)/2
Here n = 5
No. of diagonals = 5(5 - 3)/2
= 5
Ans A
Answer: A
Question: A can do a job in 3 days less time than B. A works at it alone for 4 days and then B takes over and completes it. If altogether 14 days were required to finish the job, how many days would each of them take alone to finish it?
Options:
A)17 days, 20 days
B)12 days, 15 days
C)13 days, 16 days
D)14 days, 11 days
E)None of these
Rationale: Let B can finish the work in x days.
Then A can finish the work in (x – 3) days.
B’s one day’s work = 1⁄xth work.
A’s one day’s work = 1⁄(x - 3)th work.
A’s 4 days’ work = 4⁄(x - 3)th work.
Remaining work = 1 - 4⁄(x - 3) = x−7 /x−3
The remaining work done by B in 14 – 4 = 10 days.
Now, in 10 days, work done by B = x−7/x−3th work.
∴ in 1 day, work done by B = 110(x−7/x−3) th work.
and 1/10(x−7/x−3)=1/x
⇒ x = 15 days
∴ B → 15 days and A → 12 days
Answer B
Answer: B
Question: The average weight of a group of boys is 20 kg. After a boy of weight 25 kg joins the group, the average weight of the group goes up by 1 kg. Find the number of boys in the group originally ?
Options:
A)7.0
B)5.0
C)6.0
D)4.0
E)8.0
Rationale: Let the number off boys in the group originally be x.
Total weight of the boys = 20x
After the boy weighing 25 kg joins the group, total weight of boys = 20x + 25
So 20x + 25 = 21(x + 1) = > x = 4.
Answer: D
Answer: D
Question: A military garrison in a wild frontier has 600 men who have provisions to last 20 weeks. At the end of 6 weeks they are joined by another 300 men. How long will the provisions last?
Options:
A)13 1/3
B)9 1/3
C)11 3/3
D)12 1/2
E)12 3/4
Rationale: Total provisions=600*20*7=84000
Provisions used for in 6 weeks=6*7*600=25200
remaining provisions=58800
remaining provisions need to be split between 900(600+300) people.
Number of provisions per person= 58800/900
we need to find the answer in terms of weeks=(58800/900)/7=9 1/3
Answer:B
Answer: B
Question: Two trains of length 60 m and 280 m are running towards each other on parallel lines at 42 kmph and 30 kmph respectively. In what time will they be clear of each other from the moment they meet?
Options:
A)10 sec
B)32 sec
C)17 sec
D)20 sec
E)89 sec
Rationale: Relative speed = (42 + 30) * 5/18 = 4 * 5 = 20 mps.
Distance covered in passing each other = 60 + 280 = 400 m.
The time required = d/s = 340/20 = 20 sec.
Answer:C
Answer: C
Question: Tough and Tricky questions: Functions.
Let a be a positive integer. Let n#a equal n^(2a) . Then (2#3) + (3#2) – (3#3) is equal to
Options:
A)64
B)82
C)-584
D)512
E)1522
Rationale: Answer:
2^6 + 3^4 - 3^6 = -584
ANs C
Answer: C
Question: Abby and Bobby type at constant rates of 80 words per minute and 60 words per minute, respectively. Bobby begins typing before Abby and has typed 500 words when Abby begins typing at 1:30 pm. If they continue typing at their respective rates, at what time will Abby have typed exactly 200 more words than Bobby?
Options:
A)1:40 PM
B)1:50 PM
C)2:05 PM
D)2:10 PM
E)2:20 PM
Rationale: Say time needed for Abby to type 200 more words than Bobby is t. In that time she would type 80t words and Bobby would type 60t words.
Now, total words typed by Bobby would be 500+60t and we want that number to be 200 less than 80t: 500+60t=80t-200 --> t=35.
1:30 PM + 35 minutes = 2:05 PM.
Answer: C.
Answer: C
Question: How many of the positive divisors of 120 are also multiples of 4 not including 120?
Options:
A)3.
B)4.
C)5.
D)7.
E)8.
Rationale: Given to find the positive divisors of 120 and condition is that divisors have to be multiple of 4 but not including 120.
Consider factors of 120
4 * 30
6 * 20
15 * 8
24 * 5
60 * 2
Total we get 5 different ways and all are part of 4.
Answer:C is correct option.
Answer: C
Question: Two trains one from Howrah to Patna and the other from Patna to Howrah, start simultaneously. After they meet, the trains reach their destinations after 81 hours and 9 hours respectively. The ratio of their speeds is
Options:
A)1 : 2
B)2 : 3
C)1 : 3
D)4 : 5
E)None
Rationale: Sol.
Let us name the trains as A and B. Then,
(A's speed) : (B's speed) = √b : √a
= √9 : √81
= 3 : 9. => 1 : 3
Answer C
Answer: C
Question: In a group of cows and hens, the number of legs are 20 more than twice the number of heads. The number of cows is
Options:
A)5
B)7
C)10
D)12
E)14
Rationale: Explanation:
Let the number of cows be x and the number of hens be y.
Then, 4x + 2y = 2 (x + y) + 20
4x + 2y = 2x + 2y + 20
2x = 20
x = 10
Answer: C
Answer: C
Question: A letter lock consists of three rings each marked with six different letters. The number of distinct unsuccessful attempts to open the lock is at the most?
Options:
A)172
B)268
C)215
D)257
E)652
Rationale: Since each ring consists of six different letters, the total number of attempts possible with the three rings is = 6 * 6 * 6 = 216. Of these attempts, one of them is a successful attempt.
Maximum number of unsuccessful attempts
= 216 - 1 =215.
Answer: C
Answer: C
Question: A paint store mixes 3/4 pint of red paint and 2/3 pint of white paint to make a new paint color called Perfect Pink. How many pints of red paint would be needed to make 40 pints of Perfect Pink paint?
Options:
A)9
B)16
C)21
D)25 1/3
E)28 1/2
Rationale: 3/4 pint is required to make 3/4 + 2/3 = 17/12 pint of perfect pink
so 17/12 pint requires 3/4 pint of red..
1 pint will require 3/4 * 12/17=9/17..
40 pints will require 9/17 * 40= 21 pints..
C
Answer: C
Question: The proportion of the ages of Pooja and Charlie is 2 : 5, After 8 years, their ages will be in the proportion of 1 : 2. Then what is the variance in their current ages?
Options:
A)36 years
B)28 years
C)24 years
D)19 years
E)30 years
Rationale: C
24 years
Let Pooja’s age = 2A
and Charlie’s age = 5A
(2A + 8)/(5A+8)= (1/2)
=> 2 (2A + 8) = 5A + 8
=> A = 8
Pooja’s age = 16 years
and Charlie’s age = 40 years
variance of their ages = 24 years.
Answer: C
Question: The temperature of a certain cup of coffee 10 minutes after it was poured was 120 degrees Fahrenheit. If the temperature F of the coffee t minutes after it was poured can be determined by the formula F=120(2^-at) + 60, where F is in degrees Fahrenheit and a is a constant. Then the temperature of the coffee 50 minutes after it was poured was how many degrees Fahrenheit?
Options:
A)63.75
B)75
C)80
D)85
E)90
Rationale: Answer: B
The temperature of coffee 10 minutes after it was poured (120 F) will help in solving the constant “a”.
120 = 120(2 ^ 10a) + 60
2 ^ -1 = 2 ^ 10a
a = -1/10
The temperature of coffee 50 minutes after it was poured is:
F = 120 ( 2 ^ -50/10) + 60
F = 120 * 1/ 32 + 60
F = 15/4 + 60
F = 255/4=63.75
A
Answer: A
Question: A man can row 9 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?
Options:
A)5.98
B)9.98
C)9.06
D)9.87
E)9.82
Rationale: M = 9
S = 1.2
DS = 7.2
US = 10.8
x/10.8 + x/7.8 = 1
x = 4.53
D = 4.53 * 2 = 9.06
Answer:C
Answer: C
Question: Two employees X and Y are paid a total of Rs. 560 per week by their employer. If X is paid 120 percent of the sum paid to Y, how much is Y paid per week?
Options:
A)s.254
B)s.284
C)s.290
D)s.304
E)s.380
Rationale: Let the amount paid to X per week = x
and the amount paid to Y per week = y
Then x + y = 560
But x = 120% of y = 120y/100 = 12y/10
∴12y/10 + y = 560
⇒ y[12/10 + 1] = 560
⇒ 22y/10 = 560
⇒ 22y = 5600
⇒ y = 5600/22 = Rs.254 (Approx)
A)
Answer: A
Question: Which of the following options is the possible total number of people in a hall given that the number of people is a multiple of both 7 and 13.?
Options:
A)87
B)65
C)35
D)39
E)91
Rationale: the number of people are a multiple of both 7 and 13, so it must be divisible by 7 and 13.
using the elimination method look for a number that is both divisible by 7 and 13,
of all the options only 91 is divisible by 7 and 13.
[ 91/7=13 and 91/13=7]
answer :E
Answer: E
Question: 1480 men have provisions for 14 days. If 460 more men join them, for how many days will the provisions last now?
Options:
A)9.9
B)8.0
C)12.5
D)12.2
E)12.1
Rationale: 1480*13 = 1940*x
x = 9.9
Answer:A
Answer: A
Question: Given a certain number (Q), find Q given that 1 is 5% of 25% of Q?
Options:
A)80
B)60
C)95
D)100
E)110
Rationale: Using elimination method to pick the correct option it will be discovered that option A i.e 80 is the correct option because 25% of 80 is 20 and 5% of 20 is 1.
ANSWER:A
Answer: A
Question: Ravi purchased 20 dozens of toys at the rate of Rs. 375 per dozen. He sold each one of them at the rate of Rs. 33. What was his percentage profit?
Options:
A)23%
B)1.2%
C)5.6%
D)34.1%
E)12.2%
Rationale: C
C.P. of 1 toy = 375/12 = Rs. 31.25
S.P of 1 toy = Rs. 33
Profit = 1.75/31.25 * 100 = 28/5 = 5.6%
Answer: C
Question: The average of six numbers is 3.95. The average of two of them is 3.1, while the average of the other two is 3.85. What is the average of the remaining two numbers?
Options:
A)4.5
B)4.6
C)4.7
D)4.8
E)4.9
Rationale: Sum of the remaining two numbers
= (3.95 × 6) – [(3.1 × 2) + (3.85 × 2)]
= 23.70 – (6.1 + 7.7)
= 23.70 – 13.9 = 9.80
∴ Required average = (9.8⁄2) = 4.9
Answer E
Answer: E
Question: If I walk at 3 kmph, I miss the train by 2 min, if however, I walk at 4 kmph. I reach the station 2 min before the arrival of the train. How far do I walk to reach the station?
Options:
A)4/5 km
B)4/6 km
C)4/9 km
D)4/1 km
E)4/3 km
Rationale: x/3 – x/4 = 4/60
x = 4/5 km
Answer: A
Answer: A
Question: The price of rice falls by 40%. How much rice can be bought now with the money that was sufficient to buy 20 kg of rice previously?
Options:
A)5 kg
B)15 kg
C)25 kg
D)30 kg
E)None
Rationale: Solution: Let Rs. 100 be spend on rice initially for 20 kg.
As the price falls by 40%, new price for 20 kg rice,
= (100 -40% of 100) = 60
New price of rice = 60/20 = Rs. 3 per kg.
Rice can bought now at = 100/3 = 33.33 kg.
Answer: Option E
Answer: E
Question: Rebecca's annual income is $20,and Jimmy's annual income is $10,. By how much must Rebecca's annual income increase so that it constitutes 55% of Rebecca and Jimmy's combined income?
Options:
A)8
B)9
C)7.77
D)10
E)11
Rationale: total rebecca = x+20; total = x+20+10
x+20/x+30 = 55/100 therefore x =7.77
C
Answer: C
Question: How many positive integers arethere from 0 to 1000 that leave a remainder of 3 on division by 7 & a remainder of 2 on division by 4?
Options:
A)30
B)36
C)38
D)42
E)46
Rationale: Number should be of the form 7n + 3 and 4m + 2.
The LCM of 7 and 4 is 28. So, let us see what are the possible remainders when we divide this number by 28.
A number of the form 7n + 3 can be written as 28K + 3 or 28k + 10 or 28 + 17 or 28k + 24.
A number of the form 4m + 2 can be written as 28l + 2, 28l + 6, 28l + 10, 28l + 14, 28l + 18, 28l + 22, 28l + 26.
For a detailed discussion on how we get to this, look at this post.
Within these, the only common term is 28K + 10.
The numbers in this sequence are 10, 38, 66.....990.
We still need to figure out how many numbers are there in this sequence. We are going in steps of 28, so let us see if we can write these numbers in terms of 28p + r.
10 = 28 * 0 + 10
38 = 28 * 1 + 10
66 = 28 * 2 + 10
94 = 28 * 3 + 10
...........................
990 = 28 * 35 + 10
There are 36 numbers in this sequence.
Correct Answer: 36
B
Answer: B
Question: Three 6 faced dice are thrown together. The probability that no two dice show the same number on them is ?
Options:
A)5/6
B)5/9
C)2/9
D)5/7
E)1/9
Rationale: No two dice show same number would mean all the three faces should show different numbers. The first can fall in any one of the six ways. The second die can show a different number in five ways. The third should show a number that is different from the first and second. This can happen in four ways.
Thus 6 * 5 * 4 = 120 favourable cases.
The total cases are 6 * 6 * 6 = 216.
The probability = 120/216 =5/9.
Answer:B
Answer: B
Question: A rectangular circuit board is designed to have a width of W inches, a length of L inches, a perimeter of P inches, and an area of C square inches. Which of the following equations must be true?
Options:
A)2W^2 + PW + 2C = 0
B)2W^2 − PW + 2C = 0
C)2W^2 − PW − 2C = 0
D)W^2 + PW + C = 0
E)W^2 − PW + 2C = 0
Rationale: P = 2(L+W)-----------------1)
C= LW------------------------2)
option A is not possible why ?? because all the terms are positive.
lets try option B , put value of P and A from 1 and 2 we have,
2W^2-2(L+W)W + 2(LW)
2W^2 - 2LW - 2W^2 + 2LW=0.
hence answer is B.
Answer: B
Question: If the average of a sequence of consecutive multiple of 18 is 153, and the greatest term is 270, how many terms R in the sequence are smaller than the average
Options:
A)5
B)6
C)7
D)8
E)9
Rationale: Let the sequence have N terms starting 18*K
--> 18K, 18(K+1), 18(K+2), .... , 18(K+N-1) is the sequence.
The sum of the sequence is
---> 18K + 18(K+1) + 18(K+2) + .... + 18(K+N-1) =18*(K+(K+1)+...+(K+N-1)) = 18*(N*(2K+N-1)/2).
So the average is
153=[18*(N*(2K+N-1)/2)]/N = 18*(2K+N-1)/2 = 9 *(2K+N-1) ---> by cancelling 9 we get 2K+N-1=17 --> 2K+N=18 ---*).
On the other hand, since the greatest term is 270, 18(K+N-1)=270 --> by cancelling 18 we get K+N-1=15 --> K+N=16 ---**).
By *), **) we have K=2, N=14.
Since 153/18 = 8.5, 18*8<153<18*9. So the number of terms R which are smaller than the average is 18*2, 18*3, ..... ,18*8 ---> 7. The answer is C.
Answer: C
Question: The cost of an article is decreased by 10%. If the original cost is $60, find the decrease cost.
Options:
A)33
B)12
C)68
D)36
E)54
Rationale: Original cost = $60
Decrease in it = 10% of $60
= 10/100 Ã— 60
= 600/100
= $6
Therefore, decrease cost = $60 - $6 = $54
Answer:E
Answer: E
Question: A and B started a business with their capitals Rs. 36,000 and 12,000 respectively. Annual profit will be distributed between them in the ratio
Options:
A)2:5
B)1:5
C)2:3
D)1:3
E)3:1
Rationale: 36,000:12,000
Ans. 3:1
ANSWER:E
Answer: E
Question: Which expression is the greatest?
Options:
A)3252/3257
B)3456/3461
C)3591/3596
D)3346/3351
E)3453/3458
Rationale: Let 3252/3257 be the original fraction.
Notice that it is a positive Proper fraction.
Hence adding the same constant to both the numerator and the denominator will increase the fraction as the fraction will move closer to one.
Hence we just have to pick the number that adds the greatest value to both the numerator and the denominator.
Options can be re-written as (x-5)x => 1-(5/x)
A) 1- (5/3257)
B) 1- (5/3461)
C) 1- (5/3596)
D) 1- (5/3351)
E) 1- (5/3458)
ANSWER:C
Answer: C