mathewhe/aqua_rat huggingface dataset:
Question: 7 men, 5 women and 8 children were given as assignment of distributing 2000 books to students in a school over a period of 3 days. All of them distributed books on the 1st day. One of the 2nd day 2 women and 3 children remained absent and on the 3rd day 3 men and 5 children remained absent. If the ratio of the no. of books distributed in a day by a man, a woman & a child was 5 : 4 : 2 respectively, a total of approximately how many books were distributed on the second day?
(A) 450
(B) 500
(C) 560
(D) 650
(E) 700
Rationale: Sol. Let the books distributed by man, a woman and a child be 5x, 4x and 2x respectively.
∴ No. of books distributed in 1st day
= 7×5x+5×4x+8×2x=71x
No. of books distributed in 1Ind day
=7×5x+3×4x+5×2x=57x
And no. of books distributed in IIIrd day
=4×5x+5×4x+3×2x=46x
71x + 57x + 46x = 2000, x = 2000/174
57x=2000/174×57=650
D
Answer: (D) 650
Question: What percent is 300 gm of 1 kg?
(A) 25%
(B) 40%
(C) 30%
(D) 8%
(E) 12%
Rationale: 1 kg = 1000 gm
300/1000 Ã— 100 = 30000/1000
=30 %
Answer is C
Answer: (C) 30%
Question: Pipe A can fill a tank in 5 hr, pipe B in 10hr and pipe C in 30hr. If all the pipes are open in how many hours will tank be filled?
(A) 1hr
(B) 2hr
(C) 3hr
(D) 4hr
(E) 5hr
Rationale: Part filled by (A+B+C) in 1hour = 1/5 + 1/10 + 1/30 = 1/3
All the 3 pipes together will fill the tank in 3hours
Answer is C
Answer: (C) 3hr
Question: The average weight of a class of 24 students is 35 kg. if the weight of the teacher be included, the average rises by 400 g. the weight of the teacher is
(A) 45 kg
(B) 50 kg
(C) 53 kg
(D) 55 kg
(E) None of these
Rationale: Solution
Weight of the teacher = (35.4 × 25 - 35 × 24) kg = 45 kg. Answer A
Answer: (A) 45 kg
Question: There are 6 boxes numbered 1, 2,3,4,5 and 6. Each box is to be filled up either with a red or a green ball in such a way that at least 1 box contains a green ball & the boxes containing green balls are consecutively numbered. The total no. of ways in which this can be done is?
(A) 18
(B) 20
(C) 21
(D) 23
(E) 26
Rationale: If only one of the boxes has a green ball, it can be any of the 6 boxes. So, this can be achieved in 6 ways.
If two of the boxes have green balls and then there are 5 consecutive sets of 2 boxes. 12, 23, 34, 45, 56.
Similarly, if 3 of the boxes have green balls, there will be 4 options.
If 4 boxes have green balls, there will be 3 options.
If 5 boxes have green balls, then there will be 2 options.
If all 6 boxes have green balls, then there will be just 1 options.
Total number of options = 6 + 5 + 4 + 3 + 2 + 1 = 21
C
Answer: (C) 21
Question: F is the set of the first n positive odd numbers, where n is a positive integer. Given that n > k, where k is also a positive integer, x is the maximum value of the sum of k distinct members of F, and y is the minimum value of the sum of k distinct members of F, what is x + y?
(A)  kn
(B)  kn + k^2
(C)  kn + 2k^2
(D)  2kn – k^2
(E)  2kn
Rationale: Probably the easiest way to solve this question would be to assume some values for n and k.
Say n=3, so F, the set of the first n positive odd numbers would be: F={1, 3, 5};
Say k=1, so X, the maximum value of the sum of K distinct members of F would simply be 5. Similarly, Y, the minimum value of the sum of K distinct members of F would simply be 1.
X+Y=5+1=6.
Now, substitute n=3 and k=1 in the options provided to see which one yields 6. Only asnwer choice E fits: 2kn=2*3*1=6.
Answer: E.
Answer: (E)  2kn
Question: Three candidates contested an election and received 1136, 7636 and 16628 votes respectively. What percentage of the total votes did the winning candidate got?
(A) 65.4%
(B) 56%
(C) 57%
(D) 58%
(E) 59%
Rationale: Total number of votes polled = (1136 + 7636 + 16628) = 25400
So, Required percentage = 16628/25400 * 100 = 65.4%
ANSWER : A
Answer: (A) 65.4%
Question: The average weight of 3 packets A, B and C is 84 kg. Another packet D added into the group and the average now becomes 80 kg. If another packet E, whose weight is 3 kg more than that of D, replaces A then the average weight of B, C, D and E becomes 79 kg. The weight of A is:
(A) 70 kg
(B) 72 kg
(C) 75 kg
(D) 80 kg
(E) None of these
Rationale: Explanation :
A+B+C = (84x3) = 252kg, A+B+C+D=(80x4) = 320kg.
D = (320 - 252) = 68 , E = (68+3) =7l.
B+C+D+E = (79x4)=316.
(A+B+C+D) -(B+C+D+E)=(320 - 316)kg = 4kg.
A - E = 4 = A = (4+E)=75kg
Answer : C
Answer: (C) 75 kg
Question: The price of a T.V. set worth 20,000 is to paid in 20 instalments of 1000 each. If the rate of interest be 6% per annum, and the first instalment be paid at the time of purchase, then the value of the last instalment covering the interest as well will be :
(A) 1050
(B) 2050
(C) 3000
(D) None of these
(E) 2020
Rationale: Money paid in cash = 1000.
Balance payment = (20000 – 1000) = 19000.
Answer D
Answer: (D) None of these
Question: if the ratio of two number is 6:7 and LCM of the number is 450 then what is the number.
(A) 11
(B) 20
(C) 25
(D) 30
(E) 35
Rationale: product of two no = lcm *hcf
6x*7x=450*x
x=11
ANSWER:A
Answer: (A) 11
Question: A reduction of 20% in the price of salt enables a lady to obtain 10kgs more for Rs.500, find the original price per kg?
(A) s.12.7
(B) s.12.4
(C) s.12.5
(D) s.12.1
(E) s.12.9
Rationale: 100*(20/100) = 20 --- 10
? --- 1 => Rs.2
500 --- 80
? --- 2 => Rs.12.5
Answer:C
Answer: (C) s.12.5
Question: Mr X, a businessman, had income in the year 1995 such that he earned a profit of 20% on his investment in the business. In the year 1996 his investment was less by 3000 but still had the same income (Income = Investment + Profit) as that in 1995. Thus the per cent profit earned in 1996 increased by 6%. What was his investment in 1995?
(A) 100000
(B) 100500
(C) 63000
(D) Data inadequate
(E) None of these
Rationale: Let the investment of X in 1995 be x.
∴ Profit = x⁄5
∴ Income = (x + x⁄5) = 6⁄5x
Investment of company X in 1996 would be (x – 3000) From the question,
(x - 3000) × 126⁄100 = 6⁄5x ⇒ x = 63000
Answer C
Answer: (C) 63000
Question: At a local gym there were 55 people working out at 4 p.m., of those people 29 did cardio-vascular exercises and 20 did weight training. If 13 did neither cardio-vascular exercises nor weight training, then how many people did both cardio-vascular exercises and weight training?
(A) 62
(B) 49
(C) 13
(D) 12
(E) 7
Rationale: The total number =55
Cardio=29
Weight=20
Neither =13
N(CUW)= N(c) +N(w)-N(C intersection W)+N(neither C or W)
55=29+20-(Intersection)+13=7
Answer:E
Answer: (E) 7
Question: A group of students decided to collect as many paise from each member of group as is the number of members. If the total collection amounts to Rs. 32.49, the number of the member is the group is:
(A) 57
(B) 67
(C) 77
(D) 87
(E) 97
Rationale: Explanation:
Money collected = (32.49 x 100) paise = 3249 paise.
∴ Number of members = √(3249) = 57.
Answer: A
Answer: (A) 57
Question: A can do a piece of work in 10 days and B can do it in 15 days and C can do it 20 days. They started the work together and A leaves after 2 days and B leaves after 4 days from the beginning. How long will work lost?
(A) 1 days
(B) 12 days
(C) 52 days
(D) 60 days
(E) 10 2/3 days
Rationale: Explanation:
2/10 + 4/15 + x/20 = 1
x = 32/3 = 10 2/3 days
Answer: E
Answer: (E) 10 2/3 days
Question: A certain company employs 9 senior officers and 4 junior officers. If a committee is to be created, that is made up of 3 senior officers and 1 junior officer, how many different committee are possible?
(A) 8
(B) 24
(C) 58
(D) 336
(E) 210
Rationale: Number of ways to select 3 senior officers from 9 of them = 9C3 = 84 ways
Next we can determine the number of ways to select 1 junior officer.
number of ways to select 1 junior officer from 4 of them = 4C1 = 4 ways
Thus the number of ways to select 3 senior officers and 1 junior officer is 84 x 4 = 336 ways.
Answer: D
Answer: (D) 336
Question: A sum of money at simple interest amounts to Rs. 815 in 3 years and to Rs. 854 in 4 years. The sum is?
(A) 287
(B) 278
(C) 698
(D) 279
(E) 275
Rationale: S.I. for 1 year = (854 - 815)
= Rs. 39
S.I. for 3 years = 39 * 3
= Rs. 117
Principal = (815 - 117)
=Rs.698.
Answer: C
Answer: (C) 698
Question: A three digit number divisible by 7 becomes divisible by 3 when 16 is subtracted from it. The largest such number is :
(A) 994
(B) 981
(C) 963
(D) 951
(E) 942
Rationale: Out of all the 5 options, only 994 is NOT divisible by 3. All others are divisible
So Answer = A (No further calculation required)
Addition of any two non-divisible numbers by 3 gives the resultant divisible by 3
16 is non-divisible by 3; we are subtracting a number from that so that the resultant becomes divisible by 3
Applying the above rule,It means that the number which we are going to add should be non-divisible by 3
So comes the answer = 994
Answer : A
Answer: (A) 994
Question: John has 6 pairs of shoes; each pair is a different color. If Kim randomly selects 2 shoes without replacement from the 12 shoes, what is the probability that she will select 2 shoes of the same color?
(A) 2/5
(B) 1/5
(C) 1/9
(D) 1/8
(E) 1/11
Rationale: Can be tackled in this way as well: Probability of selecting any 1 out of 12 shoes = 12/12 = 1
Probability of selecting the next shoe (out of 12 available) having the same color = 1/9 (as after selecting the 1st one, there is only 1 another shoe left with the same color).
Thus the total probability = 1*1/11=1/11.
E is the correct answer.
Answer: (E) 1/11
Question: The average weight of 9 persons increases by 1.5 kg. If a person weighing 65 kg is replaced by a new person, what could be the weight of the new person?
(A) 76 kg
(B) 77 kg
(C) 78.5 kg
(D) Data inadequate
(E) None of these
Rationale: total weight increases = 9 × 1.5 = 13.5 kg
so the weight of new person = 65 + 13.5 = 78.5 kg
Answer C
Answer: (C) 78.5 kg
Question: Three competing juice makers conducted a blind taste test with mall shoppers. The shoppers could choose to taste any number of the three brands of juices but had to select at least one juice that they liked. If 100 shoppers liked brand J, 200 shoppers liked brand K, 700 shoppers liked brand L, 350 shoppers liked exactly 2 juices, and 100 shoppers liked all three juices, how many shoppers took the taste test?
(A) 1,300
(B) 1,000
(C) 900
(D) 700
(E) 450
Rationale: Given: Atleast 1 juice was liked by the participants. --> Neither = 0
Assume, I -->No overlap between the sets
II --> Overlap between 2 sets
III --> Overlap between 3 sets
I + 2*(II) + 3*(III) = 100 + 200 + 700
I + 2*(350) + 3*(100) = 1000
I = 0
Total number of shoppers who took the taste test = I + II + III = 0 + 350 +100 = 450
Answer: E
Answer: (E) 450
Question: The length of a train and that of a platform are equal. If with a speed of 90 k/hr, the train crosses the platform in one minute, then the length of the train (in meters) is?
(A) 757
(B) 758
(C) 718
(D) 750
(E) 738
Rationale: Speed = [90 * 5/18] m/sec = 25 m/sec; Time = 1 min. = 60 sec.
Let the length of the train and that of the platform be x meters.
Then, 2x/60 = 25 è x = 25 * 60 / 2
=750
Answer:D
Answer: (D) 750
Question: Two trains are traveling on a collision course. If train A is traveling at a speed of 350 mph and train B is traveling 28% slower, how much time will it take the trains to collide if the initial distance between the two is 1655.5 miles?
(A) Two hours and 30 minutes.
(B) One hour and 10 minutes.
(C) Two hours and 45 minutes.
(D) Three hours and 15 minutes.
(E) Four hours and 20 minutes.
Rationale: Answer is C. Two hours and 45 minutes.
Since they are racing towards each other, the speed is accumulative, ie. 350 mph + 350*0.72 = 602 (0.72 because the second train is 28% slower)
So time before collision is total distance / total speed = 1655.5/602 = 2 hours and 45 mins
Answer: (C) Two hours and 45 minutes.
Question: The regular price per can of a certain brand of soda is $0.55. If the regular price per can is discounted 25 percent when the soda is purchased in 24-can cases, what is the price of 70 cans of this brand of soda purchased in 24-can cases?
(A) $16.32
(B) $18.00
(C) $21.60
(D) $24.48
(E) $28.87
Rationale: The discounted price of one can of soda is (0.75)($0.55), or $0.4125
Therefore, the price of 70 cans of soda at the discounted price would be (70)($0.4125) = 28.875
Answer: E
Answer: (E) $28.87
Question: A certain article of clothing was discounted during a special sale to 3/4 of its original retail price. When the clothing didn't sell, it was discounted even further to 2/5 of its original retail price during a second sale. By what percent did the price of this article of clothing decrease from the first sale to the second sale?
(A) 50%
(B) 33.33%
(C) 25%
(D) 46.66%
(E) 12.5%
Rationale: Say the original retail price of the item was $200.
The price after the first sale = 3/4 * $200 = $150.
The price after the second sale = 2/5 * $200 = $80.
The percent change from the first sale to the second = (150 -80) / 150 = 1/3 = 46.66%.
Answer: D.
Answer: (D) 46.66%
Question: Mike invested Rs. 3750 in a scheme for 3 years at compound interest rate 6.5% p.a. How much amount will Mike get on maturity of the fixed deposit?
(A) s.4581.29
(B) s.4450
(C) s.4500
(D) s.4500.25
(E) s.4529.81
Rationale: Amount = [3750 * (1 + 6.5/100)3]
= 3750 *106.5/100 * 106.5/100 *106.5/100
= Rs.4529.81
Answer:E
Answer: (E) s.4529.81
Question: A part-time employee’s hourly wage was increased by 20%. She decided to decrease the number of hours worked per week so that her total income did not change. By approximately what percent should the number of hours worked be decreased?
(A) 9%
(B) 17%
(C) 25%
(D) 50%
(E) 100%
Rationale: Let's plug in somenicenumbers and see what's needed.
Let's say the employee used to make $1/hour and worked 100 hours/week
So, the TOTAL weekly income was $100/week
After the 20% wage increase, the employee makes $1.20/hour
We want the employee's income to remain at $100/week.
So, we want ($1.20/hour)(new # of hours) = $100
Divide both sides by 1.20 to get:
new # of hours = 100/1.20
≈ 83 hours
So, the number of hours decreases from 100 hours to (approximately) 83 hours.
This represents a 17% decrease (approximately) .
Answer : B
Answer: (B) 17%
Question: 220, 218, 214, 208, 200, 190, ?
(A) 166
(B) 178
(C) 182
(D) 190
(E) 186
Rationale: Explanation :
220
220 - 2 = 218
218 - 4 = 214
214 - 6 = 208
208 - 8 = 200
200 - 10 = 190
190 - 12 = 178
Answer : Option B
Answer: (B) 178
Question: The average of 13 numbers is 60. Average of the first 7 of them is 55 and that of the last 7 is 61. Find the 8th number?
(A) 32
(B) 83
(C) 45
(D) 53
(E) 64
Rationale: Sum of all the 13 numbers = 13 * 60 = 780
Sum of the first 7 of them = 7 * 55 = 385
Sum of the last 7 of them = 7 * 61 = 427
So, the 8th number = 427 + 385 - 780 = 32.
Answer:A
Answer: (A) 32
Question: If 60% of of a number is 36, then the number is
(A) 223
(B) 378
(C) 100
(D) 298
(E) 2778
Rationale: Explanation:
Let the number be x.Then
60% of {\color{Black} \frac{3}{5}}of x=36.
{\color{Black} \Rightarrow }{\color{Black} \frac{60}{100}\times \frac{3}{5}\times x=36}
{\color{Black} \Rightarrow x=100}
Required number is 100.
Answer: C) 100
Answer: (C) 100
Question: We invested a total of $1,000. We invested one part of the money at 3% and the rest of the money at 4%. The total investment with interest at the end of the year was $1,036.70. How much money did we invest at 3%?
(A) $300
(B) $330
(C) $360
(D) $390
(E) $420
Rationale: Let x be the money invested at 3%.
1.03x+1.04(1000-x)=1036.70.
0.01x=1040-1036.70.
0.01x=3.30.
x=330.
The answer is B.
Answer: (B) $330
Question: In a partnership between A, B and C. A's capital is Rs.5000. If his share of a profit of Rs.800 is Rs.200 and C's share is Rs.150, what is B's capital?
(A) 19878
(B) 12798
(C) 26908
(D) 11250
(E) 22073
Rationale: 200 + 150 = 350
800 - 350 = 450
200 ---- 5000
450 ---- ? => 11250
Answer: D
Answer: (D) 11250
Question: P is three times as fast as Q and working together, they can complete a work in 12 days. In how many days can Q alone complete the work?
(A) 16
(B) 77
(C) 88
(D) 55
(E) 778
Rationale: P = 3Q
P + Q = 3Q + Q = 4Q
These 4Q people can do the work in 12 days, which means Q can do the work in 48 days.
Hence, P can do the work in 16 days.
Answer:A
Answer: (A) 16
Question: A manufacturer produces a certain men's athletic shoe in integer sizes from 8 to 17. For this particular shoe, each unit increase in size corresponds to a 1/5-inch increase in the length of the shoe. If the largest size of this shoe is 40% longer than the smallest size, how long, in inches, is the shoe in size 15?
(A) 5.4
(B) 5.9
(C) 6.3
(D) 6.7
(E) 7.5
Rationale: Let x be the length of the size 8 shoe.
Then 0.4x = 9/5
x = 4.5 inches
The size 15 shoe has a length of 4.5 + 7/5 = 5.9 inches
The answer is B.
Answer: (B) 5.9
Question: A train 100 m long is running at a speed of 68 kmph. How long does it take to pass a man who is running at 8 kmph in the same direction as the train?
(A) 6 sec
(B) 9 sec
(C) 12 sec
(D) 15 sec
(E) 18 sec
Rationale: Speed of the train relative to man = (68 - 8) kmph
= (60* 5/18) m/sec = (50/3)m/sec
Time taken by the train to cross the man
= Time taken by It to cover 100 m at 50/3 m / sec
= 100 *3/ 50 sec = 6sec
Answer : A
Answer: (A) 6 sec
Question: What is the probability of getting sum 10 with the help of three dice?
(A) 1/4
(B) 1/8
(C) 1/16
(D) 1/24
(E) 1/32
Rationale: (1,3,6) 3!=6 (2,2,6) =3!/2=3
(1,4,5) 3!=6 (2,4,4)=3
(2,3,5) 3!=6 (3,3,4)=3
prob. of getting sum 10 =(6+6+6+3+3+3)/216 =1/8 Ans
ANSWER:B
Answer: (B) 1/8
Question: On a certain transatlantic crossing, 15 percent of a ship's passengers held round-trip tickets and also took their cars abroad the ship. If 60 percent of the passengers with round-trip tickets did not take their cars abroad the ship, what percent of the ship's passengers held round-trip tickets?
(A) 37.5%
(B) 40%
(C) 50%
(D) 60%
(E) 66.6%
Rationale: 0.15P = RT + C
0.6(RT) = no C
=> 0.40(RT) had C
0.15P = 0.40(RT)
RT/P = 37.5%
Answer - A
Answer: (A) 37.5%
Question: A bus started its journey from Mumbai and reached Pune in 54min with its average speed of 55km/hr. If the average speed of the bus is increased by 6km/hr, how much time will it take to cover the same distance?
(A) 45 min. 30sec
(B) 43min.
(C) 45min.
(D) 44 min. 30sec
(E) 46min.
Rationale: Sol. Distance between Mumbai and Pune=(55*54)/60=99/2
Average speed of the bus is increased by 6 km/hr then the
speed of the bus = 60 km/hr
Required time = 99/2*60/60 = 44 min. 30sec
D
Answer: (D) 44 min. 30sec
Question: A number is selected at random from the first 30 natural numbers. What is the probability that the number is a multiple of either 3 or 17?
(A) 17/30
(B) 2/5
(C) 7/15
(D) 4/15
(E) 11/30
Rationale: Number of Multiples of 3 from 1 through 30 = 30/3 = 10
Number of Multiples of 17 from 1 through 30 = 30/17 = 1
Number of Multiples of 3 and 17 both from 1 through 30 = Number of Multiples of 17*3(=51) = 0
Total favourable cases = 10 + 1 - 0 = 11
Probability = 11 / 30
Answer: option E
Answer: (E) 11/30
Question: Average between two sets of numbers is closer to the set withmore numbers?
(A) 75
(B) 80
(C) 85
(D) 90
(E) 96
Rationale: If on a test three people answered 90% of the questions correctly and two people answered 80% correctly, then the average for the group is not 85% but rather 3×90+2×805=4305=86.3×90+2×805=4305=86.
Here, 90 has a weight of 3 => it occurs 3 times.
Whereas 80 has a weight of 2 => it occurs 2 times.
So the average is closer to 90 than to 80 as we have just calculated.
B
Answer: (B) 80
Question: In how many ways can a group of 12 friends be seated round 2 tables if one of the tables can seat 8 and the other can seat 4 people?
(A) 15C5 * 9! *4!
(B) 15C3 * 8! *3!
(C) 15C4 * 9! *3!
(D) 15C2 * 10! *3!
(E) 12C4 * 7! * 3!
Rationale: 8 people can be selected from 12 people in 12C8 ways.
Remaining 4 people can be selected from 4 people in 4C4 ways.
Now, arranging 8 people on a round table = (8-1) ! = 7!
Arranging 4 people on a round table = (4-1) ! = 3!
Hence, total number of ways = 12C8 * 7! * 4C4 * 3!
= 12C4 * 7! * 3!
E
Answer: (E) 12C4 * 7! * 3!
Question: What are the last two digits of 63*35*37*82*71*41*53?
(A)  10
(B)  30
(C)  40
(D)  70
(E)  80
Rationale: We know that to find the last two digits, we need to find the remainder we get when we divide the product by 100.
Remainder of (63*35*37*82*71*41)/ 100
Note that we can simplify this expression by canceling out the 5 and 2 in the numerator and denominator. But before we do that, here is an important note:
Note: We cannot just cancel off the common terms in the numerator and denominator to get the remainder. But, if we want to cancel off to simplify the question, we can do it, provided we remember to multiply it back again.
So say, we want to find the remainder when 14 is divided by 10 i.e. 14/10 (remainder 4). But we cancel off the common 2 to get 7/5. The remainder here will be 2 which is not the same as the remainder obtained by dividing 14 by 10. But if we multiply 2 back by 2 (the number we canceled off), the remainder will become 2*2 = 4 which is correct.
Take another example to reinforce this – what is the remainder when 85 is divided by 20? It is 5.
We might rephrase it as – what is the remainder when 17 is divided by 4 (cancel off 5 from the numerator and the denominator). The remainder in this case is 1. We multiply the 5 back to 1 to get the remainder as 5 which is correct.
So keeping this very important point in mind, let’s go ahead and cancel the common 5 and 2.
We need the
Remainder of (63*7*37*41*71*41*5*2)/10*5*2
Remainder of (63*7*37*41*71*41)/10
Now using concept 2, let’s write the numbers in form of multiples of 10
Remainder of (60+3)*7*(30+7)*(40+1)*(70+1)*(40+1)/10
Remainder of 3*7*7*1*1*1/10
Remainder of 147/10 = 7
Now remember, we had canceled off 10 so to get the actual remainder so we need to multiply by 10: 7*10 = 70.
When 63*35*37*82*71*41 is divided by 100, the remainder is 70. So the last two digits of 63*35*37*82*71*41 must be 80.
Answer (E)
Answer: (E)  80
Question: What is the are of an equilateral triangle of side 4 cm?
(A) 64√5
(B) 4√3
(C) 64√4
(D) 64√3
(E) 64√2
Rationale: Area of an equilateral triangle = √3/4 S2
If S = 4, Area of triangle = √3/4 * 4 * 4 = 4√3 cm2;
Answer: B
Answer: (B) 4√3
Question: If "PLANT" is given a code of 63, then what is the sum of "SOIL" and "STEM"?
(A) 112
(B) 100
(C) -2
(D) 2
(E) 120
Rationale: Case – Codes : A - 1, B - 2, C - 3, D - 4, E - 5, F - 6, G - 7, H - 8, I - 9, J - 10, K - 11, L - 12, M - 13, N - 14, O - 15, P - 16, Q - 17, R - 18, S - 19, T - 20, U - 21, V - 22, W - 23, X - 24, Y - 25, Z - 26
SOIL-55
STEM-57
Answer: A
Answer: (A) 112
Question: Set J consists of the terms {a, b, c, d, e}, where e > d > c > b > a >0.1. Which of the following operations would decrease the standard deviation of Set J?
(A) Multiply each term by e/d
(B) Divide each term by e/c
(C) Multiply each term by −1
(D) Divide each term by d/e
(E) Multiply each term by 1
Rationale: CONCEPT: Standard Deviation is Defined as Average Deviation of Terms in the set from the Mean value of the set. i.e.
1) It depends on the separation between the successive terms of the set
2) If a Constant Value is Added/Subtracted in every terms of set then the Separation between successive terms does NOT change Hence S.D. remains Constant
3) If a Constant Value is Multiplied in every terms then the Separation between succesive terms gets multiplied by the constant Hence S.D. remains gets multiplied by same Number
B
Answer: (B) Divide each term by e/c
Question: A can contains a mixture of liquids A and B is the ratio 7:5. When 6 litres of mixture are drawn off and the can is filled with B, the ratio of A and B becomes 7:9. How many liter of liquid A was contained by the can initially?
(A) 12
(B) 18
(C) 14
(D) 20
(E) 22
Rationale: Ci*Vi = Cf*Vf
(7/12)*(V1 - 6) = (7/16)*V1
(V1 - 6)/V1 = 3/4
6 accounts for the difference of 1 on ratio scale so Initial volume = V1 = 4*6 = 24 litres.
7/12 of the initial mixture was liquid A so liquid A was (7/12)*24 = 14 litres.
Answer : C
Answer: (C) 14