mathewhe/aqua_rat huggingface dataset:
Question: From a Group of 8 People, Including Sachin and Rahul, 3 people are to be selected at random to work on a certain project. What is the probability that 3 people selected will include Sachin but not Rahul
(A) 5/56
(B) 9/56
(C) 15/56
(D) 21/56
(E) 25/56
Rationale: Number of ways of selecting 3 people out of 8 people = 8C3
In the three members Sachin will always be there in the team. At this step we have vacancy for 2 more members and 7 members are available. Lina cannot be there in the team. So 2 members have to be selected and the number of available members = 7 - Rahul = 6
Number of ways to form a 3 member team that includes Sachin and excludes Rahul = 6C2
Probability = 6C2/8C3 = 15/56
Answer: C
Answer: (C) 15/56
Question: A man purchased a grinder for Rs. 120 and sold it at a profit of 10%. The selling price is:
(A) Rs. 736
(B) Rs. 812
(C) Rs. 132
(D) None of these
(E) Cannot be determined
Rationale: s.p.=110/100 × 120 = 132
Answer: C
Answer: (C) Rs. 132
Question: What percent of 5 letter combinations that can be made from the letters of the word VERMONT and in which each letter can be used only once are the combinations in which the first letter is a vowel and the last letter is a consonant?
A.
B.
C.
D.
E.
(A) 20%
(B) 23.8%
(C) 25%
(D) 30.2%
(E) 35%
Rationale: TOTAL number of 5 letter combinations = 7P5 = 7!/2! = 3*4*5*6*7
There are 2 vowels: e,o and 5 consonants: v,r,m,n,t
If ABCDE is the required combination, A has 2 options, E has 5 options, B has 5, C has 4, and D has 3
Number of required combinations = 2*5*5*4*3
Answer = 2*5*5*4*3 / 3*4*5*6*7 = 5/21 = 23.8%
ANSWER: B
Answer: (B) 23.8%
Question: If each data point in a certain set Q is increased by a factor of 6, what happens with the set's standard deviation?
(A) Doesn't Change
(B) Decreases by Factor 6
(C) Increases by 6
(D) Increases by Factor 6
(E) Can't be determined
Rationale: Standard deviation questions will always be conceptual in nature, because the GMAT does not require that you be able to calculate standard deviation.Veritasteaches a concept known as themapping strategyfor data sufficiency, which helps make a problem like this easily understandable. The mapping strategy is just looking at data sufficiency with sets of numbers like you would look at a map. Moving left and right on the map does not change the map's scale (which represents standard deviation in this analogy), just as adding or subtracting the same value in all numbers of a set does not change that set's standard deviation. For example, the set: 2, 5, 8, 11 has the same standard deviation of that same set after adding 5 to each number: 7, 10, 13, 16. Multiplying and dividing the entire set by a number other than 1, however, does affect standard deviation, and it does so proportionately. Multiplying by a number with an absolute value greater than 1 is like zooming in on a map...the standard deviation increases. Consider that same set multiplied by 2: 4, 10, 16, 22....this has a larger standard deviation than 2, 5, 8, 11. The opposite goes for division - dividing each number in a set of numbers by a number with an absolute value greater than 1 will reduce standard deviation, like zooming out on a map reduces the map's scale. Here because each number was multiplied by a factor of 6, the standard deviation will increase proportionately, giving us answer choice D.
Answer: (D) Increases by Factor 6
Question: if m>0 and b is m percent of a, then, in terms of m, a is what percent of b?
(A) a)100m
(B) b)1/100m
(C) c)1/m
(D) d)10/m
(E) e)10^4/m
Rationale: Given (m*a)/100=b
to find (a/b)*100
Rearranging first equation
a/b*100 = 100*100/m
=10^4/m
Answer E
Answer: (E) e)10^4/m
Question: The length of a rectangular plot is thrice its breadth. If the area of the rectangular plot is 867 sq m, then what is the breadth of the rectangular plot?
(A) 1
(B) 17
(C) 88
(D) 12
(E) 15
Rationale: Let the breadth of the plot be b m.
Length of the plot = 3 b m
(3b)(b) = 867
3b2 = 867
b2 = 289 = 172 (b > 0)
b = 17 m.
Answer: B
Answer: (B) 17
Question: On flipping 3 coins together what is the outcomes for at least one head?
(A) {HHH, HHT, HTH, THH, TTH, THT, TTT}
(B) {HHH, HHT, HTH, THH, TTH, THT, HTT}
(C) {HHH, HHT, HTH, TTT, TTH, THT, HTT}
(D) {TTT, HHT, HTH, THH, TTH, THT, HTT}
(E) None
Rationale: Sample space, S = {HHH, HHT, HTH, THH, TTH, THT, HTT, TTT}
Favorable outcomes for at least one heads
= favorable outcomes for exactly one heads and favorable outcomes for more than one heads
= {HHH, HHT, HTH, THH, TTH, THT, HTT}
B
Answer: (B) {HHH, HHT, HTH, THH, TTH, THT, HTT}
Question: One pipe can fill a tank four times as fast as another pipe. If together the two pipes can fill the tank in 40 minutes, then the slower pipe alone will be able to fill the tank in?
(A) 100min
(B) 150min
(C) 200min
(D) 250min
(E) 300min
Rationale: Let the slower pipe alone fill the tank in x minutes
Then, faster pipe will fill it in x/4 minutes
1/x + 4/x = 1/40
5/x = 1/40
x = 200 min
Answer is C
Answer: (C) 200min
Question: The average age of 36 students in a group is 14 years. when teacher's age is included to it, the average increases by one.What is the teacher's age in years?
(A) 31
(B) 36
(C) 51
(D) 53
(E) 57
Rationale: age of the teacher=(37*15-36*14)
=51 years.
ANSWER:C
Answer: (C) 51
Question: A car traveling at a certain constant speed takes 5 seconds longer to travel 1 km than it would take to travel 1 km at 90 km/hour. At what speed, in km/hr, is the car traveling?
(A) 70
(B) 80
(C) 74
(D) 75
(E) 78
Rationale: Time to cover 1 kilometer at 80 kilometers per hour is 1/90 hours = 3,600/90 seconds = 40 seconds;
Time to cover 1 kilometer at regular speed is 40 + 5 = 45 seconds = 45/3,600 hours = 1/80 hours;
So, we get that to cover 1 kilometer 1/80 hours is needed --> regular speed 80 kilometers per hour (rate is a reciprocal of time or rate=distance/time).
Answer: B.
Answer: (B) 80
Question: John weighs twice as much as Susan. Susan's weight is 60% of Sam's weight. Mark weighs 50% of Lynn's weight. Lynn weighs 140% of John's weight. Which of these 5 people weighs the least?
(A) John
(B) Susan
(C) Sam
(D) Mark
(E) Lynn
Rationale: John weighs twice as much as Susan --> J=2S --> John cannot weigh the least;
Susan's weight is 60% of Sam's weight --> S=0.6*Sam --> Sam cannot weigh the least;
Mark weighs 50% of Lynn's weight --> M=0.5*L --> Lynn cannot weigh the least;
Lynn weighs 140% of John's weight --> L=1.9*J --> John cannot weigh the least (we already know that);
As you can see the answer is either Susan (S) or Mark (M). To compare their weights we should express their weights in regards of the same person: M=0.5*L=0.5*(1.4*J)=0.7*J and from J=2S, S=0.5*J --> Susan weights the least: Susan's weight is 50% of that of John and Mark's weight is more than 50%of that of John.
Answer: B.
Answer: (B) Susan
Question: The sum of ages of 5 girls born at the intervals of 3 years each is 40 years. What is the age of the youngest girl
(A) years
(B) years
(C) years
(D) years
(E) years
Rationale: Let the ages of children be x, (x + 3), (x + 6), (x + 9) and (x + 12) years.
Then, x + (x + 3) + (x + 6) + (x + 9) + (x + 12) = 40
5x = 10
x = 2.
Age of the youngest child = x = 2 years.
A)
Answer: (A) years
Question: If x^3 < y^2 > z, which of the statements could be true?
I. x < y < z
II. x < z < y
III. y < x < z
(A) I only
(B) III only
(C) I and II only
(D) II and III only
(E) I, II and III
Rationale: Given : x^3 < y^2 > z
I. x < y < z is true for x = 3, y = 6, z = 7
II. x < z < y is true for x = 3 , y = 5 , z = 4
III. y < x < z is not true
Answer: option C
Answer: (C) I and II only
Question: Wendy begins sanding a kitchen floor by herself and works for 6 hours. She is then joined by Bruce, and together the two of them finish sanding the floor in 2 hours. If Bruce can sand the floor by himself in 20 hours, how long would it take Wendy to sand the floor by herself?
(A) 9/80 hours
(B) 3/20 hours
(C) 20/3 hours
(D) 80/9 hours
(E) 10 hours
Rationale: Let Wendy finishes sanding the floor alone in W hours while B be the hours for Bruce.
Thus in 1 hour Wendy finishes 1/w of the work while Bruce finishes 1/B of the work.
If wendy works for 6 hours and is then joined by Bruce to finish the work in 2 more hours,
6/W + 2/W+2/B = 1 (1 denotes the total amount of work)
8/W + 2/B =1 and given B =20 hours.
Thus W = 80/9 hours , D is the correct answer.
Answer: (D) 80/9 hours
Question: An amount of Rs. 100000 is invested in two types of shares. The first yields an interest of 9% p.a and the second, 11% p.a. If the total interest at the end of one year is 9 3/4 %, then the amount invested in each share was?
(A) Rs. 52500; Rs. 47500
(B) Rs. 62500; Rs. 37500
(C) Rs. 72500; Rs. 27500
(D) Rs. 82500; Rs. 17500
(E) Rs. 72500; Rs. 17500
Rationale: Let the sum invested at 9% be Rs. x and that invested at 11% be Rs. (100000 - x). Then,
(x * 9 * 1)/100 + [(100000 - x) * 11 * 1]/100 = (100000 * 39/4 * 1/100)
(9x + 1100000 - 11x)/100 = 39000/4 = 9750
x = 62500
Sum invested at 9% = Rs. 62500
Sum invested at 11% = Rs. (100000 - 62500) = Rs. 37500.
ANSWER:B
Answer: (B) Rs. 62500; Rs. 37500
Question: what is summation of [3^k(28Ck)] (summation range 28 to k=0) where 28Ck is the number of ways of choosing k items from 28 items??
(A) 4^28
(B) 4^27
(C) 4^26
(D) 4^25
(E) 4^24
Rationale: The expansion of (1+x)^n=nC0 x^0+nC1 x^1+nC2 x^2+......
putting the value of x=3 and n=28 we get
(1+3)^28=28C0 3^0+28C1 3^1+28C2 3^2+28C3 3^3.......
which is the given summation we need to find hence,
ans= 4^28
ANSWER:A
Answer: (A) 4^28
Question: If an integer is randomly chosen from the numbers 250 to 449, inclusive, what is the probability that all the digits of the number are even?
(A) 3/5
(B) 17/50
(C) 1/8
(D) 51/199
(E) 1/4
Rationale: Another method is to enumerate the hundreds and tens digits which are acceptable:
26
28
40
42
44
In each case, any of the 5 even digits (0, 2, 4, 6, 8) can take the units place.
So acceptable cases = 5*5 = 25
Total number of integers = 200
Probability = 25/200 = 1/8
ANSWER:C
Answer: (C) 1/8
Question: Difference of two numbers is 1660. If 7.5% of the number is 12.5% of the other number , find the number ?
(A) 1470
(B) 4150
(C) 3012
(D) 4430
(E) 5507
Rationale: Let the numbers be x and y. Then , 7.5 % of x =12.5% of y
X = 125*y/75 = 5*y/3.
Now, x-y =1660
5*y/3 –y =1660
2*y/3= 1660
y =[ (1660*3)/2] =2490.
One number = 2490, Second number =5*y/3 =4150.
ANSWER B 4150
Answer: (B) 4150
Question: A and B together have Sterling 1220. If 4/15 of A's amount is equal to 2/5 of B's amount, how much amount does B have?
(A) Sterling 460
(B) Sterling 488
(C) Sterling 550
(D) Sterling 664
(E) None
Rationale: Explanation: 4/15A = 2/5B
A = (2/5 x 15/4)B
A = 3/2 B
A/B = 3/2
A : B = 3 : 2.
B's share = Sterling (1220 x 2/5) = Sterling 488.
Answer: Option B
Answer: (B) Sterling 488
Question: A box contains nine slips that are each labeled with one number: 1, 2, 3, 5, 8, 13, 21 and 34 Two of the slips are drawn at random from the box without replacement. What is the probability that the sum of the numbers on the two slips is equal to one of the numbers left in the box?
(A) 7/72
(B) 3/14
(C) 7/36
(D) 15/36
(E) 21/36
Rationale: Probability = No: of desired outcomes / Total No: of outcomes.
You are picking two slips out of 8 slips. So
Total No: of outcomes = 8C2 = 28
Desired outcome : sum of the numbers on the two slips is equal to one of the numbers left in the box. How many such outcomes are there?
If you look at the numbers closely, you will see that the following pair of numbers will give you the desired outcome.
(1,2) (2,3) (3,5) (5,8) (8,13) (13,21) . There are 6 such pairs. If the two numbers which I pick is from any of these 6 pairs, then I get my desired outcome.
So No: of desired outcomes = 6
Probability = 6/28 = 3/14
Answer :B
Answer: (B) 3/14
Question: Shan starts from Barabanki to Fatehpur, 1 hour after Ajay starts. Shahrukh meets Kajol 1.5 hours after Shahrukh starts. If the speed of Shan is at least 20km/hour fasterthan the speed of Kajol. What is the min speed of Shan to overtake Ajay, before he meets Kajol?
(A) 62.2 km/h
(B) 63 km/h
(C) 64.5 km/h
(D) 66.5 km/h
(E) 67.5 km/h
Rationale: Let t be the time after Kajol starts, when she meets Ajay, then
\inline t=\frac{300}{(x+y)}
This should be less than 2.5 or (x+y)>120
Since y= \inline \frac{3x}{2} \inline \Rightarrow y>72
This (y>72) is greater than 67.5 km/h and hence Shahrukh will always overtake Ajay before he meets Kajol.
E)
Answer: (E) 67.5 km/h
Question: The probability that a man will be alive for 10more yrs is 3/4 & the probability that his wife will alive for 10 more yrs is 1/5. The probability that none of them will be alive for 10 more yrs, is
(A) 1/2
(B) 1
(C) 2/3
(D) 1/5
(E) 2
Rationale: Sol. Required probability = Pg.) x P(B) = (1— D x(1 —I) =: x 1=1/5
Ans.(D)
Answer: (D) 1/5
Question: Grayson, a chemist, has 8 litres of a solution that contains 15% hydrochloric acid(HCI), plus unlimited 5% HCI solution. For a particular experiment, however, he needs to prepare some that contains 12% HCI. How much 5% HCI solution should be added to the 15% HCI solution to get the 12% HCI solution needed for the experiment?
(A) 7.6 litres
(B) 10 litres
(C) 12.8 litres
(D) 3.4 litres
(E) 8.2 litres
Rationale: 0.15(8)+0.05x = 0.12(8 + x)
Solve the equation.
0.15(8) + 0.05x = 0.12(8 + x)
1.2 + 0.05x = 0.96 + 0.12x
1.2 – 0.07x = 0.96
-0.07x = -0.24
x ≈ 3.4
Answer is D.
Answer: (D) 3.4 litres
Question: 3 men or 6 women can do a piece of work in 20 days. In how many days will 12 men and 8 women do the same work?
(A) 15/9 days
(B) 15/4 days
(C) 18/4 days
(D) 11/4 days
(E) 35/4 days
Rationale: 3M = 6W ---- 20 days
12M + 8W -----?
24W + 8 W = 32W ---?
6W ---- 20 32 -----?
6 * 20 = 32 * x => x = 15/4 days
Answer:B
Answer: (B) 15/4 days
Question: Find the C.I. on Rs. 15,625 for 9 months at 16% per annum compounded quarterly?
(A) Rs.1939
(B) Rs.1912
(C) Rs.1951
(D) Rs.1908
(E) Rs.1213
Rationale: P = Rs. 15625, n = 9 months = 3 quarters, R = 16% p.a. per quarter.
Amount = [15625 * (1 + 4/100)3]
= (15625 * 26/25 * 26/25 * 26/25)
= Rs. 17576 C.I.
= 17576 - 15625
= Rs.1951.
Answer:C
Answer: (C) Rs.1951
Question: A 3-digit number 4a3 is added to another 3-digit number 984 to give a 4-digit number 13b7, which is divisible by 11. Then, (a + b) = ?
(A) 10
(B) 11
(C) 12
(D) 15
(E) 17
Rationale: Explanation:
4 a 3 |
9 8 4 } ==> a + 8 = b ==> b - a = 8
13 b 7 |
Also, 13 b7 is divisible by 11 (7 + 3) - (b + 1) = (9 - b)
(9 - b) = 0
b = 9
(b = 9 and a = 1) (a + b) = 10.
Answer: Option A
Answer: (A) 10
Question: If the annual increase in the population of a town is 10% and the present number of people is 11000, what will the population be in 2 years?
(A) 12100
(B) 15240
(C) 12456
(D) 11452
(E) 13310
Rationale: The required population is = 11000 (1 + 10/100)^2
= 11000 * 11/10 * 11/10
= 13310
Answer is E
Answer: (E) 13310
Question: In the following series how many times the sum of two consecutive digits is even?
1 2 3 4 6 3 4 2 5 9 3 6 7 4 1 2 3 6 7 6 5 4 3
(A) 5
(B) 2
(C) 3
(D) 4
(E) 6
Rationale: 4+6
4+2
5+9
9+3
so 4 times
ANSWER:D
Answer: (D) 4
Question: (4x - 7)(4x + 7) + (2x - 9)(2x + 9) =
(A) 20x2 - 2x - 100
(B) 20x2 - 130
(C) 12x2 - 32
(D) 12x2 + 32
(E) None
Rationale: Solution:
(4x - 7)(4x + 7) + (2x - 9)(2x + 9) =
[(4x)2 - 72] + [(2x)2 - 92] =
(16x2 - 49) + (4x2 - 81) =
16x2 - 49 + 4x2 - 81 =
20x2 - 130
Answer B
Answer: (B) 20x2 - 130
Question: Solve: 18800+470+20
(A) 2
(B) 5
(C) 4
(D) 0
(E) 1
Rationale: 18800+470+20=(18800/470)+20=40/20=2.
Answer is A.
Answer: (A) 2
Question: 45 men can complete a work in 18 days. Six days after they started working, 30 more men joined them. How many days will they now take to complete the remaining work?
(A) 6.6
(B) 10
(C) 15
(D) 7.2
(E) 8
Rationale: 45*18 men can complete the work in 1 day
1 man's 1 day work = 1/810
45 men 6 days work = 6*1/18 = 1/3
remaining work = 1-1/3 = 2/3
75 men's 1 day's work = 75/810 = 5/54
2/3 work is done by them in 54/5 * 2/3 = 7.2 days
Answer is D
Answer: (D) 7.2
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 16 hours and 81 hours respectively. The ratio of their speeds is?
(A) 4:9
(B) 9:4
(C) 4:5
(D) 4:1
(E) 4:2
Rationale: Let us name the trains A and B.
Then, (A's speed) : (B's speed)
= √b : √a = √81 : √16
= 9:4
Answer:B
Answer: (B) 9:4
Question: A bag contains 6 white and 4 black balls .2 balls are drawn at random. Find the probability that they are of same colour.
(A) 7/19
(B) 7/15
(C) 7/29
(D) 7/12
(E) 7/92
Rationale: Let S be the sample space
Then n(S) = no of ways of drawing 2 balls out of (6+4) = \inline {\color{Black}10C_{2}} = =45
Let E = event of getting both balls of same colour
Then,n(E) = no of ways (2 balls out of six) or (2 balls out of 4)
=\inline {\color{Black}6C_{2}+4C_{2}} = = 15+6 = 21
Therefore, P(E) = n(E)/n(S) = 21/45 = 7/15
Answer: B
Answer: (B) 7/15
Question: A train 100 m long crosses a platform 125 m long in 15 sec; find the speed of the train?
(A) 87 kmph
(B) 65 kmph
(C) 54 kmph
(D) 16 kmph
(E) 18 kmph
Rationale: D = 100 + 125 = 225
T = 15
S = 225/15 * 18/5 = 54 kmph
Answer:C
Answer: (C) 54 kmph
Question: Difference of two numbers is 1670. If 7.5% of the number is 12.5% of the other number , find the number ?
(A) 2660, 1000
(B) 3660, 2000
(C) 3000, 4160
(D) 2490, 4150
(E) 2505, 4175
Rationale: Let the numbers be x and y. Then, 7.5 % of x =12.5% of y
X = 125*y/75 = 5*y/3.
Now, x-y =1670
5*y/3 –y =1670
2*y/3= 1670
y =[ (1670*3)/2] =2505.
One number = 2505, Second number =5*y/3 =4175.
Answer E.
Answer: (E) 2505, 4175
Question: Their are two boxes containing a total number of 60 balls, each box has an equal number of red and blue balls. What is the probability of getting a red ball from one box and a blue ball from the other
(A) 1/2
(B) 1/4
(C) 1/16
(D) 1/8
(E) 1
Rationale: The number of possible selections is 60.
Let E be the event of getting an even number on one die and an odd number on the other. Since there exist an equal number of blue and red balls, probability of picking a blue or red ball is = 30/60 = 1/2
P(E) = 1 - 1/2
= 1/2.
Answer:A
Answer: (A) 1/2
Question: Find the simple interest on Rs.500 for 9 months at 6 paisa per month?
(A) 337
(B) 270
(C) 287
(D) 266
(E) 281
Rationale: I = (500*9*6)/100
= 270
Answer: B
Answer: (B) 270
Question: Peter invested an amount of Rs. 12,000 at the rate of 10% p.a simple interest and another amount at the rate of 20% p.a. simple interest. The total interest earned at the end of one year on the total amount invested became 14% p.a. Find the total amount invested?
(A) 20000
(B) 3888
(C) 2677
(D) 29812
(E) 2322
Rationale: Let the second amount be Rs. x. Then,
(12000 * 10 * 1)/100 + (x * 20 * 1)/100 = [(12000 + x) * 14 * 1] / 100
120000 + 20x = 16800 + 14x
x = 8000
Total investment = 12000 + 8000 = Rs. 20,000.
Answer: A
Answer: (A) 20000
Question: Jim and Renee will play one game of Rock, Paper, Scissors. In this game, each will select and show a hand sign for one of the three items. Rock beats Scissors, Scissors beat Paper, and Paper beats Rock. Assuming that both Jim and Renee have an equal chance of choosing any one of the hand signs, what is the probability that Renee will win?
(A) 5/6
(B) 2/3
(C) 1/2
(D) 5/12
(E) 3/9
Rationale: There are 3*3=9 combinations (cases) possible. In 3 cases there will be a tie (rock-rock, scissors/scissors, paper/paper). Now, out of 6 cases left Jim and Renee have equal chances of winning, so in 3 cases Jim will win and in other 3 Renee will win. So the probability Renee winning the game is (favorable outcomes) / (total # of outcomes) = 3/9
Answer: E.
Answer: (E) 3/9
Question: A statue is being carved by a sculptor. The original piece of marble weighed 250 kg. In the first week 30 percent is cut away. In the second week 20 percent of the remainder is cut away. In the third week the statue is completed when 25 percent of the remainder is cut away. What is the weight of the final statue?
(A) 105 kg
(B) 103 kg
(C) 108 kg
(D) 125 kg
(E) 117 kg
Rationale: A
105 kg
250 × 0.7 × 0.8 × 0.75 = 105 kg.
Answer: (A) 105 kg
Question: A chair is bought for Rs.600/- and sold at a loss of 30% find its selling price
(A) s.420/-
(B) s.550/-
(C) s.560/-
(D) s.590/-
(E) s.580/-
Rationale: 100 % ------> 600 (100 * 6 = 600)
70 % ------> 540 (70 * 6 = 420)
Selling price = Rs.420/-
A)
Answer: (A) s.420/-
Question: 30 people attend a party. 4 men are single and the rest are there with their wives. There are no children in the party. In all 22 women are present. Then the number of married men at the party is ?
(A) 5
(B) 6
(C) 4
(D) 8
(E) 9
Rationale: Total people = number of men + number of women
30 = 4 + number of married men + 22
number of married men = 30- 22-4 = 4 men
ANSWER: C
Answer: (C) 4
Question: The average age of a committee of 8 members is 40 years. A member aged 55 years
retired and his place was taken by another member aged 39 years . The average age of
present committee is;
(A) 39 years
(B) 38 years
(C) 36 years
(D) 35 years
(E) 34 years
Rationale: Exp. Total age of the committee = 40*8 = 320,
Total age when a member is retired
and a new one was joined = 320-55 +39, =304
Average age of present committee = 304/8 = 38.
Answer: B
Answer: (B) 38 years
Question: Three positive integers a, b, and c are such that their average is 22 and a ≤ b ≤ c. If the median is (a + 11), what is the least possible value of c?
(A) 21
(B) 23
(C) 25
(D) 27
(E) 29
Rationale: Solution:
We know that the average value is 22 therefore (a+b+c)/3 = 22
so a+b+c = 66
and b = a + 11
therefore a + (a + 11) + c = 66
The least value of c is when c = b
so take c = a + 11
hence, a + (a + 11) + (a + 11) = 66
i.e., a = 44/3 a non integer therefore c is not equal to b
so take c = b+1 therefore c = a+12
a + (a + 11) + (a + 12) = 66
i.e., a= 43/3 again a non integer therefore c is not equal to b+1
next take c= b+2 therefore c = a+13
a + (a + 11) + (a + 13) = 66
i.e., a=42/3 = 14
hence, a = 14, b = 25, and c = 27
Answer: D
Answer: (D) 27
Question: John was thrice as old as Tom 6 years ago. John will be 15/13 times as old as Tom in 6 years. How old is Tom today?
(A) 5
(B) 7
(C) 9
(D) 11
(E) 13
Rationale: J-6 = 3(T-6), so J = 3T-12
J+6 = 15/13 * (T+6)
13J + 78 = 15T + 90
13(3T-12) + 78 = 15T + 90
24T = 168
T = 7
The answer is B.
Answer: (B) 7
Question: A computer is programmed to multiply consecutive even integers 2*4*6*8*…*n until the product is divisible by 2211, What is the value of n?
(A) 22
(B) 134
(C) 62
(D) 122
(E) 672
Rationale: Factorise 2211.. 3*11*67..
So n has to be a multiple of largest prime number,61..
So n=2*61=134..
ans :B
Answer: (B) 134
Question: The calendar for the year 2007 will be the same for the year
(A) 2017
(B) 2018
(C) 2014
(D) 2016
(E) 2015
Rationale: Explanation:
For a year to have the same calendar with 2007 ,the total odd days from 2007 should be 0.
Year:	2007	2008	2009	2010	2011	2012	2013	2014	2015	2016	2017
Odd Days: 1 2 1 1 1 2 1 1 1 2 1
Take the year 2014 given in the choice.
Total odd days in the period 2007-2013 = 5 normal years + 2 leap year
= 5 x 1 + 2 x 2 = 9 odd days
= 2 odd day (As we can reduce multiples of 7 from odd days which will not change anything)
Take the year 2016 given in the choice.
Number of odd days in the period 2007-2015 = 7 normal years + 2 leap year
= 7 x 1 + 2 x 2 = 11 odd days
= 4 odd days
(Even if the odd days were 0, calendar of 2007 will not be same as the calendar of 2016 because 2007 is not a leap year whereas 2016 is a leap year. In fact, you can straight away ignore this choice due to this fact without even bothering to check the odd days)
Take the year 2017 given in the choice.
Number of odd days in the period 2007-2016 = 7 normal years + 3 leap year
= 7 x 1 + 3 x 2 = 13 odd days
= 6 odd days
Take the year 2018 given in the choice.
Number of odd days in the period 2007-2017 = 8 normal years + 3 leap year
= 8 x 1 + 3 x 2 = 14 odd days
= 0 odd day (As we can reduce multiples of 7 from odd days which will not change anything)
Also, both 2007 and 2018 are not leap years.
Since total odd days in the period 2007-2017 = 0 and both 2007 and 2018 are of same type, 2018 will have the same calendar as that of 2007
Answer: Option B
Answer: (B) 2018