mathewhe/aqua_rat huggingface dataset:
Question: In a group of 6 boys&4 girls a Committee of 4 persons is to be formed. In how many different ways can it be done so that the committee has at least 1girl?
(A) 110
(B) 120
(C) 140
(D) 195
(E) 210
Rationale: The committee of 4 persons is to be so formed that it has at least 1 woman The different ways that we can choose to form such a committee are:
(i) lw. 3 m in t 6X5X4' x 6C3 = 4x — — 80 3X2X1
x 6c2 =42:: x 26:: = 90 (ii) 2w. 2 m in °C2 (iii) 3w. 1 m in 4C3 x 6C1 = 4 x 6 = 24 (iv) 4w in 6C4 = 1 Total no. of different ways in which a committee of 4 persons can be formed so that it has at least one woman. = 80 + 90 + 24 + 1 = 195
D
Answer: (D) 195
Question: What is the probability of getting at least one six in a single throw of three unbiased dice?
(A) 1/6
(B) 125/216
(C) 1/36
(D) 81/216
(E) 91/216
Rationale: Find the number of cases in which none of the digits show a '6'.
i.e. all three dice show a number other than '6', 5 * 5 *5 = 125 cases.
Total possible outcomes when three dice are thrown = 216.
The number of outcomes in which at least one die shows a '6' = Total possible outcomes when three dice are thrown - Number of outcomes in which none of them show '6'.
= 216 - 125 = 91.
The required probability = 91/216.
ANSWER:E
Answer: (E) 91/216
Question: If a ring of diameter 4 inches moves at the rate of x inches/min and another ring of diameter 12 inches moves at the rate of y inches/min, and if they both cover a distance at the same time then find the value of x in terms of y?
(A) 6y/2
(B) 3y/4
(C) 5y/2
(D) 2y/4
(E) 1y/3
Rationale: t=S1/V1=S2/V2
or 4/x=12/y
or x=4y/12=1y/3
Answer is E
Answer: (E) 1y/3
Question: Out of 19 students studying in a class, 7 are from Maharashtra, 5 are from Karnataka and 3 are from Goa. Four students are to be selected at random. What are the chances that at least one is from Karnataka?
(A) 12â„13
(B) 11â„13
(C) 10â„15
(D) 23/31
(E) None of these
Rationale: Total possible ways of selecting 4 students out of 19 students = 19C4 = 3876
The no. of ways of selecting 4 students in which no student belongs to Karnataka = 14C4=1001
âˆ´ Hence no. of ways of selecting at least one student from Karnataka = 19C4 â€“ 14C4 = 2875
âˆ´ Probability = 2875â„3876 = 23/31
Answer D
Answer: (D) 23/31
Question: For the real value of x what is the maximum value of |x+2|+|x-1|-|2x-4|=?
(A) 2
(B) 3
(C) 4
(D) 5
(E) 6
Rationale: value of expression will be maximum if mod(2x-4) will give minimum value
so 2x-4=0
x=2
maximum value=mod(2+2)+mod(2-1)-0=5
ANSWER:D
Answer: (D) 5
Question: In order to obtain an income of Rs. 650 from 10% stock at Rs. 96, one must make an investment of:
(A) 3100
(B) 6240
(C) 6500
(D) 9600
(E) 9800
Rationale: To obtain Rs. 10, investment = Rs. 96.
To obtain Rs. 650, investment = Rs. (96X650)/10 =6240 Answer: B
Answer: (B) 6240
Question: Ram professes to sell his goods at the cost price but he made use of 900 grms instead of a kg, what is the gain percent?
(A) 11 1/5%
(B) 11 1/4%
(C) 11 1/9%
(D) 11 8/9%
(E) 11 3/9%
Rationale: 900 --- 100
100 --- ? => 11 1/9%
Answer: C
Answer: (C) 11 1/9%
Question: -45*29+108=?
(A) 1197
(B) 1897
(C) -1197
(D) -1900
(E) None of these
Rationale: => -45*(30-1)+108;
=> -(45*30)+45+108;
=> -1350+153=-1197.
Correct Option: C
Answer: (C) -1197
Question: Farm Tax is levied on the 40% of the cultivated Land. The tax department collected total $ 3840 through the farm Tax from the village of Mr. willam. Mr. willam paid only $ 480 as farm tax. The percentage of total land of Mr. willam over the total taxable land of the village is :
(A) 31.25%
(B) 25%
(C) 12.5%
(D) 0.2083%
(E) none
Rationale: Only trick n this question is to ignore 40% information as farm tax is levied uniformly in the village and that includes Mr William's land.
What percentage of tax Mr William paid ? This will be equal to the percentage of total cultivated land he holds over the total cultivated land in the village. That leads to (480/3840)x100 =12.5%in percentage terms.
But the question asks ratio between his total land to total cultivated land. Hence the answer is 12.5%x (100/40) = 31.25% and the answer is not there in the options.
The correct answer is (A).
Answer: (A) 31.25%
Question: You have a flashlight that takes 2 working batteries. You have 8 batteries but only 4 of them work.
What is the fewest number of pairs you need to test to guarantee you can get the flashlight on?
(A) 5
(B) 8
(C) 7
(D) 6
(E) 4
Rationale: If you break the batteries into 3 groups: Two groups of 3 and one group of 2. By doing this you guarantee that one of the groups has 2 working batteries. Both of the groups of 3 have 3 possible combinations of 2 batteries and the group of 2 only has 1 combination. So, 3 + 3 + 1 = 7 tries at most to find two working batteries.
Answer: (C) 7
Question: What is the sum of all the composite numbers up to 20?
(A) 130
(B) 132
(C) 124
(D) 145
(E) 166
Rationale: 4 + 6 + 8 + 9 + 10 + 12 + 14 + 15 + 16 + 18 = 132
ANSWER B
Answer: (B) 132
Question: A certain music store stocks 800 cellos and 600 violas. Of these instruments, there are 120 cello-viola pairs, such that a cello and a viola were both made with wood from the same tree (each tree can make at most one viola and one cello, so there are no pairs other than these 90). If one viola and one cello are chosen at random, what is the probability that the two instruments are made with wood from the same tree?
(A) 3/16,000
(B) 1/8,000
(C) 3/1,600
(D) 1/90
(E) 2/45
Rationale: Solution provided by stanford2012 is correct: 120/800 choosing one of the cellos which has a pair viola, 1/600 choosing the viola which is the pair of chosen cello --> P=120/800*1/600=1/8,000.
Answer: B.
Answer: (B) 1/8,000
Question: 1,000^2 + 1,001^2 +1,002^2 =
(A) 6,030,053
(B) 6,030,054
(C) 6,006,005
(D) 6,030,056
(E) 6,030,057
Rationale: Interesting problem.
I think key is to notice that all the given answer choices differs in last two digits. Therefore, our entire focus should be to figure out how the given terms contribute to last two digits of total.
1000^2 -> 00
1001^1 -> 01
.
.
.
1002^2 -> 5
Total -> *5
Answer C.
Answer: (C) 6,006,005
Question: f two positive numbers are in the ratio 1/8 : 1/5, then by what percent is the second number more than the first?
(A) 50%
(B) 80%
(C) 60%
(D) 84%
(E) 87%
Rationale: Given ratio = 1/8 : 1/5 = 5 : 8
Let first number be 5x and the second number be 8x.
The second number is more than first number by 3x.
Required percentage = 3x/5x * 100 = 60%.
Answer:C
Answer: (C) 60%
Question: How many 3-digit numbers are completely divisible 6 ?
(A) 149
(B) 150
(C) 151
(D) 166
(E) 182
Rationale: 3-digit number divisible by 6 are: 102, 108, 114,... , 996
This is an A.P. in which a = 102, d = 6 and l = 996
Let the number of terms be n. Then tn = 996.
a + (n - 1)d = 996
102 + (n - 1) x 6 = 996
6 x (n - 1) = 894
(n - 1) = 149
n = 150
Number of terms = 150.
Answer: B
Answer: (B) 150
Question: The average age of 20 students of a class is 20 years. Out of these, the average age of 9 students is 11 years and that of the other 10 students is 24 years. The age of the 20th student is?
(A) 65
(B) 61
(C) 59
(D) 63
(E) 64
Rationale: Age of the 20th student = 20*20 - (9*11+10*24) = 400-339 = 61 years
Answer is B
Answer: (B) 61
Question: A batsman scored 110 runs which included 3 boundaries and 8 sixes. What percent of his total score did he make by running between the wickets?
(A) 45
(B) 500/11
(C) 600/11
(D) 55
(E) 35
Rationale: Explanation :
Number of runs made by running,
=>110−(3×4+8×6).
=>110−(60).
=>50.
Hence, the required percentage is:-
=>(50/100×100).
=>500/11.
Answer : B
Answer: (B) 500/11
Question: A train passes a station platform in 36 seconds and a man standing on the platform in 20 seconds. If the speed of the train is 54 km/hr, what is the length of the platform?
(A) 276 m
(B) 279 m
(C) 240 m
(D) 207 m
(E) 202 m
Rationale: Speed = [54 * 5/18] m/sec
= 15 m/sec.
Length of the train = (15 * 20) m = 300 m.
Let the length of the platform be x meters.
Then, x + 300 / 36 = 15
x + 300 = 540
x =240 m.
Answer: C
Answer: (C) 240 m
Question: A sells a bicycle to B at a profit of 80% and B sells it to C at a loss of 40%. Find the resultant profit or loss.
(A) -4%
(B) 5%
(C) -5%
(D) 8%
(E) -7%
Rationale: The resultant profit or loss = 80 - 40 - (80*40)/100 = 8%
Profit = 8%
Answer is D
Answer: (D) 8%
Question: Which number need to add to 8679921 to get a number exactly divisible by 330?
(A) 69
(B) 62
(C) 61
(D) 63
(E) 64
Rationale: 8679921 / 330 = 26302 and reminder = 261.
330-261 = 69
So, the next number divisible by 339 is 69 places in front of 8679921
Which means 69 + 8679921 =8679990
69 should be added to 8679921
A
Answer: (A) 69
Question: How many integers between 324,700 and 478,600 have tens digit 1 and units digit 3?
(A) 10,300
(B) 10,030
(C) 1,353
(D) 1,352
(E) 1,539
Rationale: There is one number in hundred with 1 in the tens digit and 3 in the units digit:13, 113, 213, 313, ...
The difference between 324,700 and 478,600 is 478,600-324,700=153,900 - one number per each hundred gives 153,900/100=1,539 numbers.
Answer: E.
Answer: (E) 1,539
Question: If Rs. 762 be divided into three parts, proportional to 1/2:2/3:3/4, then the first part is?
(A) A)426
(B) B)526
(C) C)207
(D) D)199
(E) E)297
Rationale: Given ratio
= 1/2:2/3:3/4 = 6:8:9
1st part = 762 * 6/23
= Rs. 199.
Answer:D
Answer: (D) D)199
Question: The total car population in a Mumbai is 90,000 out of which 50,000 are made by Honda. Out of every 100 Honda cars, 90 are red in color, but only 60% of the total car population is red. Find the % of non-Honda cars that are red
(A) 12.5%
(B) 19.5%
(C) 22.5%
(D) 21.5%
(E) 20.5%
Rationale: Total Honda brand red cars = 90/100 *50000 = 45000
Total red cars = 60/100*90000 = 54000
No. of Non-Honda cars = 90000-50000 = 40000
No. of Non-Honda red cars = 54000 - 45000 = 9000
% of Non-Honda red cars = 9000/40000*100 = 22.5%
Answer : C
Answer: (C) 22.5%
Question: If a committee of 3 people is to be selected from among 5 married couples so that the committee does not include two people who are married to each other, how many such committees are possible?
(A) 20
(B) 40
(C) 50
(D) 80
(E) 120
Rationale: total people=10.. ways to select 3 out of them=10c3=120...
it includes comb including couple..
ways in which couple are included =8c1*5=40..
so ans reqd 120-40=80...
ANS:D
Answer: (D) 80
Question: 97 students take part in a test and the median score is 85. Which of the following descriptions must be correct?
I. . At most 49 students’ score are equal to or more than 85..
II. At least 49 students’ score are equal to or more than 85.
III. At least 49 students’ score are less than 85.
(A) I
(B) II
(C) III
(D) III
(E) IIII
Rationale: If we order the scores increasingly, then on the 49th place we have the score of 85. There must be 48 scores (those on the right of the 85 score) greater or equal to 85. So, I is necessarily correct.
The scores on the left of the 49th place, all have to be equal or less than 85. Since we can have the extreme case of all the scores 85, neither II, nor III is correct. The question asks for descriptions thatmustbe correct.
Answer B.
Answer: (B) II
Question: What is the probability of drawing a queen from a deck of 52 cards?
(A) 4/13
(B) 1/13
(C) 4
(D) 1
(E) 2/13
Rationale: Total number of cards,n(s)=52
Total number of queen cards,n(e)=4
p(e)=n(e)/n(s)=4/52=1/13
Option B is answer
Answer: (B) 1/13
Question: Find the 2nd least number of four digits which is exactly divisible by 12, 15 and 18?
(A) A)1080
(B) B)6480
(C) C)3240
(D) D)9720
(E) E)1240
Rationale: The largest five digit numbers are
1080,1240,3240,6480,9720
1080 is divisible by 12,15, and 18 -----(1st least no)
1240 is not divisible by 18, 12 and 15
3240 is divisible by 12,15, and 18 -----(2nd least no)
ANSWER:C
Answer: (C) C)3240
Question: Sam invested Rs. 6000 @ 10% per annum for one year. If the interest is compounded half-yearly, then the amount received by Sam at the end of the year will be?
(A) 5300.0
(B) 24580.0
(C) 16537.5
(D) 7120.0
(E) 6615.0
Rationale: P = Rs. 6000 ; R = 10% p.a. = 5% per half-year; T = 1 year
= 2 half-year
Amount = [6000 * (1 + 5/100)2]
= (6000 * 21/20 * 21/20)
= Rs.6615.00
Answer:E
Answer: (E) 6615.0
Question: If the numerator of a fraction W is tripled, and the denominator of a fraction is doubled, the resulting fraction will reflect an increase of what percent?
(A) 16 1⁄6%
(B) 25%
(C) 33 1⁄3%
(D) 50%
(E) 66 2⁄3%
Rationale: This question can be dealt with conceptually or by TESTing VALUES.
We're told that the numerator of a fraction W is tripled and the denominator of that same fraction is doubled. We're asked for the resulting increase, in percentage terms, of that fraction.
Let's TEST 2/1
If we triple the numerator and double the denominator, we end up with....
(2x3)/(1x2) = 6/2 = 3
Since we're increasing 2 to 3, we're increasing that fraction by (3-2)/2 = 1/2 of itself = 50%
D
Answer: (D) 50%
Question: The first half of a 400-mile trip was travelled in x hours and the second half in y hours. Which of the following is equal to the average speed, in miles per hour, for the entire trip?
(A) 200/(x + y)
(B) 400/(x + y)
(C) 50/x + 50/y
(D) 200/x + 200/y
(E) 100/x + 100/y
Rationale: Time taken to cover first half i.e 200 miles of a 400 mile trip = x
time taken to cover the second half i.e 200 miles of a 400 mile trip = y
Total time taken = x+y
Average speed for the entire trip =total distance /total time
=400/(x+y)
Answer B
Answer: (B) 400/(x + y)
Question: Which of the following will come in place of the question mark (?) in the following letter series?
BXF, DVI, FTL, HRO,?
(A) JPL
(B) KPL
(C) JOL
(D) JOK
(E) None of these
Rationale: firstword B then D so theres a diff of 1 word ,so after F then H then J...then second word of BXF is X then -1 word so V ,then with J it will be P ...so ans is JPL...
ANSWER:A
Answer: (A) JPL
Question: Pakistan, India and Srilanka are engaged in a missile war, such that they fire one after the other, in a sequence, the sequence selected by draw of lots. The rules of war are such that the 3 countries keep on shooting at each other until they hit with the missile. This goes on until there is one last country left unhit, and then the war ends. The chances that Pakistan hits its target is 100%, Srilanka hits the target is 80% and India hits the target is 50%.
Who has the highest chance of winning the war?
(A) India
(B) Pakistan
(C) Srilanka
(D) None of these
(E) Cannot be determined
Rationale: Solution:
India has the highest chances of winning the war (47/90)
Missile Firing order:
Chances of Remaining Unhit - Pakistan:
Chances of Remaining Unhit - Srilanka:
Chances of Remaining Unhit - India
Pakistan - Srilanka - India
1/2 0 1/2
Pakistan - India - Srilanka
1/2 0 1/2
India - Pakistan - Srilanka
1/10 16/45 49/90
Srilanka- Pakistan - India
1/2 0 1/2
Srilanka - India - Pakistan
1/10 16/45 49/90
India - Srilanka - Pakistan
1/10 16/45 49/90
Total chances of being unhit (sum of the probabilities divided by 6):
27/90 16/90 47/90
Answer D
Answer: (D) None of these
Question: A certain list consists of 21 different numbers. If n is in the list and n is 4 times the average(arithmetic mean) of the other 20 numbers in the list, then n is what fraction W of the sum of the 21 numbers in the list?
(A)  1/20
(B)  1/6
(C)  1/5
(D)  4/21
(E)  5/21
Rationale: This is how I used to calculate which I think works pretty well:
if you let the average of the 20 other numbers equal a, can you write this equation for sum of the list (S)
n + 20a = S
the question tells us that
n = 4a
plug this back into the first equation and you get that the sum is 24a
4a + 20a = 24a
therefore fraction W of n to the total would be
4a/24a or 1/6
answer B
Answer: (B)  1/6
Question: A gold mine has 5% depletion rate. i.e. after each year of mining, the miners extract 5/100 of the mine's ore. If the current quantity of the gold reserves of the mine is currently estimated at 2 tons, how much gold would the miners be able extract over the next 4 years?
(A) 0.400 tons
(B) 0.200 tons
(C) 0.175 tons
(D) 0.160 tons
(E) 0.185 tons
Rationale: X is the quantity of gold reserves of the mine
After a year of mining the quantity of gold left would be X1=X-5/100X i.e. X1=95/100X
Same thing for X2, X3, X4, respectively representing the quantity of gold left after 2, 3, 4 years of mining.
So we have:
X1=95/100 X
X2=95/100 X1
X3=95/100 X2
X4=95/100 X3
By multiplying both ends of the equations we get:
X1*X2*X3*X4 = (95/100)^4 X*X1*X2*X3
Simplification leaves us with:
X4 = (95/100)^4 X
X4 is the quantity of gold left after four years of mining. So the cumulative quantity of gold (Q4) which has been extracted after 4 years is: Q4= X-X4=X-(95/100)^4X i.e. Q4 = X (1-(95/100)^4)
In this case, given X was: X=2 tons
So Q4 = 2 * (1-(95/100)^4) = 2 * (1-81450625/10^8) =2 * 18549375/10^8 = 0.1854 tons round to Q4=0.185 ton
Answer E.
Answer: (E) 0.185 tons
Question: In how many ways can seven friends be seated in a row having 35 seats, such that no two friends occupy adjacent seats?
(A) 29P7
(B) 29C7
(C) 28P7
(D) 28C7
(E) None
Rationale: Solution:
First let us consider the 28 unoccupied seats.
They create 29 slots- one on the left of each seat and one on the right of the last one.
We can place the 7 friends in any of these 29 slots i.e. 29P7 ways.
Answer A
Answer: (A) 29P7
Question: A fraction becomes 2/3 when 1 is added to both, its numerator and denominator.
And ,it becomes 1/2 when 1 is subtracted from both the numerator and denominator. Find the fraction.
(A) 1/2
(B) 3/4
(C) 2/3
(D) 3/5
(E) 2/5
Rationale: Let the required fraction be x/y. Then,
x+1 / y+1 = 2 / 3 => 3x – 2y = - 1 …(i) and x – 1 / y – 1 = 1 / 2
	2x – y = 1 …(ii)
Solving (i) and (ii), we get : x = 3 , y = 5
therefore, Required fraction= 3 / 5.
Answer is D.
Answer: (D) 3/5
Question: Given that there are 5 basketball players per team, how many ways can you select 3 basketball players from 3 teams if no more than one player can be selected from each team?
(A)  25
(B)  30
(C)  60
(D)  75
(E)  90
Rationale: 3 teams can be selected from 3 teams in 3C3 ways = 1 ways
as the condition is that the 2 players should be from different teams
one player can be chosen from 1st team in 5 ways
similarly another player can be chosen from 2nd team in 5 ways
Total number of ways to choose the player = 5 * 5 * 1 = 25 ways
Correct Answer - A
Answer: (A)  25
Question: At what time will the train reach city X from city Y?
I.The train crosses another train of equal length of 200 metres and running in opposite directions in 15 seconds.
II.The train leaves city Y and 7.15 a.m. for city X situated at a distance of 558 km.
III.The 200 metres long train crosses a signal pole in 10 seconds.
(A) I only
(B) II only
(C) III only
(D) II and III only
(E) All I, II and III are required.
Rationale: Explanation:
From the statement I, we get length of the train is 200 metres (Redundant info while comparing with Statement III).
The rest of the info given in this statement cannot be used for calculating the speed of the train,
because the two trains might run at different speed.
III gives, speed =200/1o m/sec = 20 m/sec = {20 x 18/5} km/hr = 72 km/hr.
II gives, time taken =(558/72) hrs = 31/4 hrs = 7 3/4 hrs = 7 hrs 45 min.
So, the train will reach city X at 3 p.m.
.’. Hence II and III only gives the answer.
Answer D
Answer: (D) II and III only
Question: How to additions fast. For example we have to do a simple addition like 234 + 563+ 985 + 349.
(A) 2131
(B) 7399
(C) 2997
(D) 2769
(E) 2771
Rationale: Instead of doing like this we first add all the digits in the hundred's place. 2 + 5 + 9 + 3 = 19. Put 2 zero's to the right hand side so we get 1900.
Now add the digits in the tenth's place. 3 + 6 + 8 + 4 = 21. Put 1 zero to the right hand side so we get 210.
Now add the digits in the units place 4 + 3 + 5 + 9 = 21.
Now 1900 + 210 + 21 = 2131
Answer:A
Answer: (A) 2131
Question: Of the 11 members of a football team in school, 9 are trained to play the offensive position, 6 can play defensive position, and 4 can play both position. If one member is chosen at random, what is the probability that the member chosen will be someone who play either position ?
(A) a) 3/11
(B) b) 4/11
(C) c) 0
(D) d) 6/11
(E) e) 7/11
Rationale: You know how to solve for Total number of people who can play at one position:
Total = n(A) + n(B) - n(A and B)
Total people who can play at least one position = 9+6-4=11
So all 11 people know how to play at least one position. Hence, if you pick anyone, the probability is 0 that he wouldn't know how to play any position
Answer (C)
Answer: (C) c) 0
Question: Josh spends a total of $6.00 buying N items in the convenience store. If each of the items is either a 5 cents single bubblegum, or a 50 cents bubblegum pack, then N may be which of the following?
(A) 99
(B) 111
(C) 101
(D) 109
(E) 140
Rationale: Let x be 0.05$ single bubblegum and y be the total 0.5$ packs ---> 0.05x+0.5y=6.00 ---> x+10y=120 ...(1)
Also, x+y=N ...(2)
Solving 1 and 2 , we get, N = 120-9y ---> only value satisfied is when y = 1---> N = 120-9=111. B is the correct answer. Other options are not equal to multiples of 9 removed from 120.
Answer: (B) 111
Question: Define a* by the equation a* = π-x. Then ((−π)*)* =
(A)  −2π
(B)  -1
(C)  −π
(D)  2π
(E)  4π
Rationale: for a* f(f(−π)) = f(π − (−π)) = f(π + π) = f(2π) = π − 2π = −π=C
Answer: (C)  −π
Question: If two sides of a triangle are 12 and 2, which of the following could be the area of triangle?
1. 35
2. 48
3. 56
(A) a) 1 only
(B) b) 1 and 2 only
(C) c) 1 and 3 only
(D) d) 2 and 3 only
(E) e) 1, 2 and 3 only
Rationale: when two sides of a triangle are known , the maximum area occurs when the angle between those two sides is 90.
lets say base = 12 , height =8 (angle = 90)
maximum area = (1/2)*12*8 = 48 ( this is the same maximum area even when base is 8 and height is 12).
if we fix the base and keep lowering the angle, it will result in a lower height . hence the resulting area will always be < 48.
C,D and E are ruled out.
1, 2 and 3 are possible areas as their areas are less than 48.
Hence the answer is E.
Answer: (E) e) 1, 2 and 3 only
Question: A box contains 3 blue marbles, 4 red, 6 green marbles and 2 yellow marbles. If three marbles are picked at random, what is the probability that they are all blue?
(A) 1/45
(B) 1/29
(C) 1/455
(D) 2/275
(E) 1/495
Rationale: Given that there are three blue marbles, four red marbles, six green marbles and two yellow marbles.
Probability that all the three marbles picked at random are blue = Â³Câ‚ƒ/Â¹âµCâ‚ƒ = (1 * 3 * 2 * 1)/(15 * 14 * 13) = 1/455
Answer: C
Answer: (C) 1/455
Question: At 3:40, the hour hand and the minute hand of a clock form an angle of:
(A) 130
(B) 125
(C) 140
(D) 154
(E) 145
Rationale: Angle traced by hour hand in 12 hrs. = 360Â°.
abgle traced it in 11/3 hrs = (360/12 * 11/3) = 110
angle traced by min hand in 60 min = 360
angle traced by it in 40 min = (360/60*40) = 240
req (240-110)=130d
Answer: (A) 130
Question: If there are 30 red and blue marbles in a jar, and the ratio of red to blue marbles is 2:3, what is the probability that, drawing twice, you will select two red marbles if you return the marbles after each draw?
(A) 1/25
(B) 2/25
(C) 3/25
(D) 4/25
(E) 21/25
Rationale: 0 red and blue marbles in a jar, and the ratio of red to blue marbles is 2:3
So, 5x=30 ---> x=6
Red marbles = 2*x = 2*6 = 12
Blue marbles = 3*x = 3*6 = 18
Prob to select a red marble = 12/30 = 2/5
Prob to select 2 red marbles with replacement = 2/5*2*5 = 4/25
Hence, answer will be D.
Answer: (D) 4/25
Question: A certain car dealership sells economy cars, luxury cars, and sport utility vehicles. The ratio of economy to luxury cars is 3:5. The ratio of economy cars to sport utility vehicles is 4:3. What is the ratio of luxury cars to sport utility vehicles?
(A) 9:8
(B) 8:9
(C) 3:2
(D) 20:9
(E) 1:2
Rationale: The ratio of economy to luxury cars is 3:5 --> E:L = 3:5 = 12:20.
The ratio of economy cars to sport utility vehicles is 4:3 --> E:S = 4:3 = 12:9.
Thus, L:S = 20:9.
Answer: D.
Answer: (D) 20:9
Question: The ratio of kamla and nisha is 6:5 and the sum of their ages is 44.what will be the ratio of their ages after 8 years.
(A) 5:6
(B) 7:8
(C) 8:7
(D) 12:11
(E) 14:13
Rationale: Answer:	let age of kamla is 6x and age of nisha is 5x.
then,6x+5x=44,11x=44,x=4
after 8 year the ratio of their ages=6x+8/5x+8=8:7
answer C
Answer: (C) 8:7