mathewhe/aqua_rat huggingface dataset:
Question: A can do a work in 6 days , B can do a work in 8 days and C can do it in 12 days. B left work after 3 days. For how many number of days should A and C should work together to complete the remaining work ?
(A) 0.1
(B) 0.2
(C) 0.3
(D) 0.4
(E) 0.5
Rationale: b work 1/8 *3=3/8
remaining work =1-3/8=5/8
a and c work together =1/6+1/12=3/12=1/4
remaining work is done in=8/5 *1/4=2/5=0.4
ANSWER:D
Answer: (D) 0.4
Question: Jane and Thomas are among the 8 people from which a committee of 4 people is to be selected. How many different possible committees of 4 people can be selected from these 8 people if at least one of either Jane or Thomas is to be selected?
(A) 28
(B) 46
(C) 55
(D) 63
(E) 70
Rationale: Solution:
J+T+6 = total 8 ,
we need to select at least one of the J and T or both= total 4 out of 8 - 4 out of 6
8C4 - 6C4
=55
Ans:C
Answer: (C) 55
Question: A reduction of 20% in the price of salt enables a lady to obtain 5 kgs more for Rs.100, find the original price per kg?
(A) 6
(B) 1
(C) 5
(D) 4
(E) 7
Rationale: 100*(20/100) = 20 --- 5
? --- 1 => Rs.4
100 --- 80
? ---4 => Rs.5
Answer: C
Answer: (C) 5
Question: If A is the product of all positive integers less than 31, than what is the greatest integer k for which A/18^k is an integer?
(A) 3
(B) 6
(C) 7
(D) 14
(E) 26
Rationale: 30! A must be divisible by (3*3*2)^k.
As I will find a lot of 2's among 30!'s factors I will focus on the 3's (that also occurs twice instead of only once as a factor in 18).
Every multiple of 3 will provide me with atleast one factor of 3. There aren't that many so I list them quickly.
3 6 9*
12 15 18*
21 24 27**
30
The stars are marked as they are multiples of 9* or 27** and thus provides two or three 3's respectively.
3=3 (occurs 10 times) +10
3*3=9 (occurs 3 times) +3(adding just one additional three each as they were not counted for in the multiples of 3)
3*3*3=27 (occurs 1 time) +1 (again, adding just one as they were previously not counted)
Total: 14.
For every k we add two 3's. So we must divide 14 by 2 to get the highest integer K before we receive a non-integer result.
Answer: 7 -> Answer choice: C.
Answer: (C) 7
Question: The car A has 1.5 times the price of car B which is Rs.250000 more. What is the price of car B?
(A) 100000
(B) 200000
(C) 500000
(D) 800000
(E) 1000000
Rationale: Let the price of car B be x, (x+250000/x)=3/2
Up on cross multiplication and solving for x, we get x=500000
Answer:C
Answer: (C) 500000
Question: A began a business with Rs. 85,000. He was joined afterwards by B with Ks. 40,000. For how much period does B join, if the profits at the end of the year are divided in the ratio of 3 : 1 ?
(A) 6.5 months
(B) 3.5 months
(C) 2.5 months
(D) 8.5 months
(E) 9.5 months
Rationale: Suppose B joined for x months . Then, ( 85000 * 12 )/(40000 * x) = 3. or x = (85000 * 12) / (40000 * 3) = 8.5
So, B joined for 8.5 months.
Answer: D
Answer: (D) 8.5 months
Question: In the xy-plane the point (-2 -3) is the centre of a circle, the point (-2, 1) lies inside the circle and the point (4, -3) lies outside the circle. If the radius r of the circle r is an integer then diameter ?
(A) 6
(B) 10
(C) 4
(D) 3
(E) 2
Rationale: Can be solved without much calculations.
You are given that (-2,-3) is the center of the circle. Point (4,-3) lies inside the circle ---> the radius is lesser than distance of (-2,-3) from (4,-3) ---> lesser than 6 units but the radius will also be greater than the distance of (-2,-3) from (-2,1) ----> greater than 4 units.
Thus the radius is >4 but <6 and as it is an integer, the only possible value of radius = 5 units.
Diameter =2r= 10
B is the correct answer.
Answer: (B) 10
Question: Find the area of trapezium whose parallel sides are 20 cm and 18 cm long, and the distance between them is 10 cm.
(A) 190 cm2
(B) 885 cm2
(C) 285 cm2
(D) 825 cm2
(E) 725 cm2
Rationale: Area of a trapezium = 1/2 (sum of parallel sides) * (perpendicular distance between them)
= 1/2 (20 + 18) * (10)
= 190 cm2
Answer: A
Answer: (A) 190 cm2
Question: Two goods trains each 500 m long are running in opposite directions on parallel tracks. Their speeds are 30 km/hr and 30 km/hr respectively. Find the time taken by the slower train to pass the driver of the faster one?
(A) 22
(B) 88
(C) 48
(D) 99
(E) 60
Rationale: Relative speed = 30 + 30 = 60 km/hr.
60 * 5/18 = 50/3 m/sec.
Distance covered = 500 + 500 = 1000 m.
Required time = 1000 * 3/50 = 60 sec
Answer: E
Answer: (E) 60
Question: R campers fished for 6 hours. If m of the campers caught 2 fish apiece and the rest caught a combined total of n fish, how many fish did the R campers catch per hour?
(A) 2m + n(R-m)
(B) (2m + n(R-m))/3
(C) (2m + n(m-R))/3
(D) (2m + n)/3
(E) (2m + n)/6
Rationale: m of the campers caught 2 fish apiece in 6 hours --> total fish caught by m campers = 2m;
The rest caught a combined total of n fish.
Total fish caught in 6 hours = 2m+n --> per hour (2m+n)/6.
Answer: E.
Answer: (E) (2m + n)/6
Question: If 0 < x < 5 and -2 < y < 9, then the lowest possible value for xy is
(A) between 60 and 70
(B) between 0 and 60
(C) between - 50 and 50
(D) between 0 and -70
(E) less than -70
Rationale: If 0 < x < 5 and -2 < y < 9,
Say y = -1 (-2<y<9)
x = 4
So xy = -4 which is more than -70.
Answer is C.
Answer: (C) between - 50 and 50
Question: A boat can travel with a speed of 13 km/hr in still water. If the speed of the stream is 5 km/hr, find the time taken by the boat to go 70 km downstream.
(A) 4.25 hr
(B) 3.8 hr
(C) 8.25 hr
(D) 2.25 hr
(E) 2.50 hr
Rationale: speed of boat in still water = 13 km/hr
speed of the stream = 5 km/hr
Speed downstream = (13+5) = 18 km/hr
Time taken to travel 70 km downstream =70â„18 = 3.8 hours
Answer is B
Answer: (B) 3.8 hr
Question: There is a clock which shows false timing ie it shows 10 minutes less for each 1 hour....if the clock shows 12 noon now how many minutes will be lost when it shows 3 pm ?
(A) 16 minutes
(B) 26 minutes
(C) 36 minutes
(D) 46 minutes
(E) 56 minutes
Rationale: For 1 hour, 10minutes is slower, then for
06 miuntes - 1 minutes is slower
30 minutes - 5 minutes
Actual time false time
12 12
01 PM 12.50 PM (10 minutes slower)
02 PM 01.40 PM (12.50 + 60 min - 10 minutes slower)
03 PM 02.30 PM (01.40 + 60 min - 10 minutes slower)
03.36 PM 03.00 PM (02.30 + 36 min - 6 minutes slower )
So, 36 minutes will be lost when it is showing 3.00 PM. and the actual time is 3.36 PM
ANSWER:C
Answer: (C) 36 minutes
Question: A hat company ships its hats, individually wrapped, in 8-inch by 10-inch by 12-inch boxes. Each hat is valued at $7.50. If the company’s latest order required a truck with at least 336,000 cubic inches of storage space in which to ship the hats in their boxes, what was the minimum value of the order?
(A) $960
(B) $1,350
(C) $1,725
(D) $2,625
(E) $2,250
Rationale: Number of boxes= Total volume/volume of one box= 336,000/ (8*10*12)= 350
One box costs 7.50, so 350 box will cost= 350*7.5= 2625
D is the answer
Answer: (D) $2,625
Question: In a group of 8 semifinalists, all but 2 will advance to the final round. If in the final round only the top 3 will be awarded medals, then how many groups of medal winners are possible?
(A) 20
(B) 56
(C) 120
(D) 560
(E) 720
Rationale: 1) Is it important here that how many go into finals... NO....
what is important is HOW many get medals..3
finally these 3 can be any of the 8 - 8C3=8!5!3!=8∗7∗63∗2=568C3=8!5!3!=8∗7∗63∗2=56
2) Is order important...NO
we are looking for groups only..
ans 56 ANSWER:B
Answer: (B) 56
Question: In how much time would the simple interest on a certain sum be 0.125 times the principal at 10% per annum?
(A) 1 1/4
(B) 1 1/8
(C) 1 1/2
(D) 1 1/3
(E) 1 1/5
Rationale: Let sum = x. Then, S.I. = 0.125x = 1/8 x, R = 10%
Time = (100 * x) / (x * 8 * 10) = 5/4 = 1 1/4 years.
Answer: A
Answer: (A) 1 1/4
Question: Which of the following inequalities is equivalent to −8 < x < 6?
(A) |x - 1| < 7
(B) |x + 2| < 6
(C) |x + 3| < 5
(D) |x - 2| < 6
(E) None of the above
Rationale: We know that |x| < a means -a < x < a, where Sum of lower limit of x (i.e -a) and the upper limit of x (i.e a), is 0
Given is, -8 < x < 6, let's say by adding y to this inequality we will get into the above format
-8+y < x+y < 6+y
Now, to move this into the mod format, we need to have (-8+y) + (6+y) = 0 => y = 1
Thus, -7 < x+1 < 7 => |x+1| < 7.
Hence, answer is E
Answer: (E) None of the above
Question: Rich has 3 green, 2 red and 3 blue balls in a bag. He randomly picks 5 from the bag WITH replacement. What is the probability that of the 5 drawn balls, Rich has picked 1 red, 2 green, and 2 blue balls?
(A) (2/8)∗(3/8)^2∗(3/8)^2∗5!/(2!∗2!)
(B) (2/8)∗(3/8)^2∗(3/8)^2∗5!
(C) (2/8)∗(3/8)^2∗(3/8)2
(D) (28)∗(38)∗(38)^2(28)∗(38)∗(38)^2
(E) (38)2∗(38
Rationale: CASE I.
Rich has 3 green, 2 red and 3 blue balls in a bag. He randomly picks 5 from the bag WITH replacement. What is the probability that of the 5 drawn balls, Rich has picked 1 red, 2 green, and 2 blue balls?
The probability would be: 2/8*3/8*3/8*3/8*3/8*5!/2!2!=14.8%
CORRECT ANSWER:A
Answer: (A) (2/8)∗(3/8)^2∗(3/8)^2∗5!/(2!∗2!)
Question: A satellite is composed of 30 modular units, each of which is equipped with a set of sensors, some of which have been upgraded. Each unit contains the same number of non-upgraded sensors. If the number of non-upgraded sensors on one unit is 1/6 the total number of upgraded sensors on the entire satellite, what fraction of the sensors on the satellite have been upgraded?
(A) 5/6
(B) 1/5
(C) 1/6
(D) 1/7
(E) 1/24
Rationale: Let x be the number of upgraded sensors on the satellite.
The number of non-upgraded sensors per unit is x/6.
The number of non-upgraded sensors on the whole satellite is 30(x/6) = 5x.
The fraction of sensors which have been upgraded is x / (x+5x) = x/6x = 1/6
The answer is C.
Answer: (C) 1/6
Question: Sakshi can do a piece of work in 30 days.Tanya is 25% more efficient than Sakshi. The number of days taken by Tanya to do the same piece of work :
(A) 15
(B) 16
(C) 14
(D) 25
(E) 10
Rationale: Solution
Ratio of times taken by Sakshi and Tanya = 125 : 100 = 5 : 4 .
Suppose Tanya taken x days to do the work.
5 : 4 : : 30 : x ⇒ x = (30x4/5) ⇒ x = 24 days.
Hence,Tanya takes 16 days is complete the work.
Answer C
Answer: (C) 14
Question: If n = 8^8 – 8, what is the units digit of n?
(A) 0
(B) 1
(C) 8
(D) 2
(E) 4
Rationale: 8 ^ 8 - 8 = 8 (8 ^ 7 - 1)
==> 8 (2 ^ 21 - 1)
Last digit of 2 ^ 21 is 2 based on what explanation LiveStronger is saying. 2 ^ 24 - 1 yields 2 - 1 = 1 as the unit digit. Now on multiply this with 8, we get unit digit as 8
Answer : C
Answer: (C) 8
Question: By mixing two qualities of pulses in the ratio 2: 3 and selling the mixture at the rate of Rs 22 per kilogram, a shopkeeper makes a profit of 10 %. If the cost of the smaller quantity be Rs 14 per kg, the cost per kg of the larger quantity is:
(A) 33
(B) 88
(C) 24
(D) 77
(E) 21
Rationale: Explanation:
Cost Price of 5 kg = Rs.(14*2 + x*3) = (28 + 3x).
Sell price of 5 kg = Rs. (22x5) = Rs. 110.
[{110 - (28 + 3x)}/(28 + 3x) ]* 100 =10
[82-3x/28 + 3x]= 1 / 10
820 - 30x = 28 +3x
33x = 792
x = 24
Answer: C) Rs 24
Answer: (C) 24
Question: In a stream running at 2kmph,a motor boat goes 6km upstream and back again to the starting point in 33 minutes.find the speed of the motorboat in still water.
(A) 22
(B) 24
(C) 26
(D) 28
(E) None of them
Rationale: let the speed of the motarboat in still water be x kmph.then,
6/x+2 +6/x-2=33/60
11x2-240x-44=0
11x2-242x+2x-44=0
(x-22)(11x+2)=0
x=22.
Answer is A
Answer: (A) 22
Question: Four years ago,the father’s age was three times the age of his son.The total of the ages of the father and the son after four years,will be 64 years.What is the father’s age at present ?
(A) 32 years
(B) 36 years
(C) 44 years
(D) None of these
(E) 49 years
Rationale: Solution
Let the ages of father and son 4 years ago be 3x and x years respectively.
Then,[(3x + 4) + 4] + [(x + 4)+ 4]= 64 ⇔ 4x = 48 ⇔ x =12.
∴ Father's present age = 3x = 36 years. Answer D
Answer: (D) None of these
Question: In a certain game, you pick a card from a standard deck of 52 cards. If the card is a spade, you win. If the card is not a spade, the person replaces the card to the deck, reshuffles, and draws again. The person keeps repeating that process until he picks a spade, and the point is to measure how many draws did it take before the person picked a spade and won. What is the probability that there will be at least four draws involved in a win, i.e. someone picking her first spade on the fourth draw or later?
(A)  1/2
(B)  9/16
(C)  27/64
(D)  13/16
(E)  15/16
Rationale: Probability of picking a spade on any draw = 1/4
Probability of NOT picking a heart on the first draw AND on the second draw And 3rd draw = [1-(1/4)] X [1-(1/4)] X [1-(1/4)]= 3/4 X 3/4 X 3/4 = 27/64
Answer : C
Answer: (C)  27/64
Question: The population of a town is 15000. It decreases annually at the rate of 20% p.a. What will be its population after 2 years?
(A) 4300
(B) 4500
(C) 5120
(D) 9600
(E) 5366
Rationale: 15000 × 80/100 × 80/100 = 9600
Answer : D
Answer: (D) 9600
Question: Virat buys an old car for Rs. 50000 and spends Rs. 6000 on its repairs. If he sells the scooter for Rs. 70000, his gain percent is:
(A) 25%
(B) 11 (1/9) %
(C) 12 (2/3) %
(D) 12%
(E) None of these
Rationale: Explanation:
C.P. = Rs. (50000 + 6000) = Rs. 56000
S.P. = Rs. 70000.
Gain% = ( (14000 / 56000) * 100 ) % = 25%
ANSWER A
Answer: (A) 25%
Question: A pipe can fill a cistern in 88 minutes. Find the time in which 1/11 part of the cistern will be filled?
(A) 5min
(B) 2min
(C) 8min
(D) 1min
(E) 10min
Rationale: Full cistern filled in = 88 minutes
1/11 part filled in = 88*1/11 = 8minutes
Answer is C
Answer: (C) 8min
Question: A grocer has a sale of Rs. 5420, Rs. 5660, Rs. 6200, Rs. 6350 and Rs. 6500 for 5 consecutive months. Find the sale he should have in the sixth month, so that he gets an average sale of Rs. 6000?
(A) Rs. 5879
(B) Rs. 5872
(C) Rs. 5871
(D) Rs. 5870
(E) Rs. 5876
Rationale: Explanation:
Total sale for 5 months = Rs. (5420 + 5660 + 6200 + 6350 + 6500) = Rs. 30,130
Therefore, required sale = Rs. [(6000 * 6) – 30,130] = Rs. (36000 – 30,130) = Rs. 5870
ANSWER: D
Answer: (D) Rs. 5870
Question: John has on his shelf four books of poetry, four novels, and two reference works. Suppose from these ten books, we were to pick two books at random. What is the probability that we pick one novel and one reference work?
(A) 1/2
(B) 2/5
(C) 3/10
(D) 7/20
(E) 8/45
Rationale: For the denominator, we are going to pick two books from among ten total: a combination of two from ten. Again, we will use the formula:
nC2 = n(n-1)/2
which, for profound mathematical reasons we need not address here, is also the formula for the sum of the first (n – 1) positive integers. Here
10C2=10(9)/2=5*9=45
That’s the total number of pairs of books we could pick from the ten on the shelf. That’s our denominator.
Now, the numerator. We want one novel and one reference work. Well, there are four novels and two reference works, so by the FCP, the number of ways we can pick this is 4 x 2 = 8. That’s the total possible number of pairs involving exactly one of these four novels and exactly one of these two reference works. That’s our numerator.
probability= 8/45
Answer = E
Answer: (E) 8/45
Question: How many seconds will a 800 meter long train moving with a speed of 63 km/hr take to cross a man walking with a speed of 3 km/hr in the direction of the train ?
(A) 48
(B) 52
(C) 38
(D) 42
(E) 45
Rationale: Explanation:
Here distance d = 800 mts
speed s = 63 - 3 = 60 kmph = 60 x 5/18 m/s
time t = 800x 18 / 60x 5= 48 sec.
ANSWER IS A
Answer: (A) 48
Question: City A's population is 68000, decreasing at a rate of 80 people per year. City B having population 42000 is increasing at a rate of 120 people per year. In how many years both the cities will have same population?
(A) 130years
(B) 140years
(C) 150years
(D) 160years
(E) 170years
Rationale: city A population = 68000
city B population = 42000
68000 decreasing at a rate 0f 80p = 42000 increasing at a rate of 120p
68000-80p=42000+120p
200p=26000
p=130years
ANSWER:A
Answer: (A) 130years
Question: Alex deposited x dollars into a new account that earned 7 percent annual interest, compounded annually. One year later Alex deposited an additional x dollars into the account. If there were no other transactions and if the account contained w dollars at the end of two years, which of the following expresses x in terms of w ?
(A) w/(1+1.08)
(B) w/(1.08+1.16)
(C) w/(1.16+1.24)
(D) w/(1.07+1.07^2)
(E) w/(1.08^2+1.08^2)
Rationale: Account at the end of the first year would be 1.07x dollars. At this time x dollars was deposited, hence the account at the beginning of the second year would be (1.07x+x) dollars. Account at the end of the second year would be (1.07x+x)*1.07=w --> x(1.07^2+1.07)=w --> x=w/(1.07+1.07^2).
Answer: D.
Answer: (D) w/(1.07+1.07^2)
Question: For each 6-month period during a light bulb's life span, the odds of it not burning out from over-use are half what they were in the previous 6-month period. If the odds of a light bulb burning out during the first 6-month period following its purchase are 1/5, what are the odds of it burning out during the period from 6months to 1 year following its purchase?
(A) 5/27
(B) 2/9
(C) 1/2
(D) 4/9
(E) 2/3
Rationale: P(of not burning out in a six mnth period)=1/2 of P(of not burning out in prev 6 mnth period)
P(of burning out in 1st 6 mnth)= 1/5
--->P( of not burning out in 1st 6 mnth)=1-1/5=4/5
---->P(of not burning out in a six mnth period)=1/2 *4/5=4/10=2/5
--->P(of burning out in a six mnth period)=1-2/5=3/5
now
P( of burning out in 2nd six mnth period)=P( of not burning out in 1st six mnth)*P(of burning out in a six mnth)
=4/5 * 3/5=1/2
Ans C
Answer: (C) 1/2
Question: Arrange the following in a logical order:
1. Birth
2. College
3. Marriage
4. School
5. Employment
(A) 1, 3, 5, 4, 2
(B) 1, 4, 2, 5, 3
(C) 2, 5, 3, 1, 4
(D) 2, 4, 1, 5, 3
(E) None of these
Rationale: Explanation:
The given words when arranged in the order of various events as they occur in man's life, the sequence is : Birth, School, College, Employment, Marriage.
Sequence is : 1, 4, 2, 5, 3.
ANSWER B
Answer: (B) 1, 4, 2, 5, 3
Question: A basket has 5 apples and 4 oranges. Three fruits are picked at random. The probability that at least 2 apples are picked is
(A) 25/42
(B) 25/49
(C) 25/46
(D) 25/40
(E) 25/41
Rationale: Total fruits = 9
Since there must be at least two apples,
(5C2 * 4C1)/9C3 + 5C2/3C3 = 25/42.
Answer: A
Answer: (A) 25/42
Question: The average of seven numbers is 55. If one number is excluded, the average becomes 52. The excluded number is:
(A) 55
(B) 57
(C) 70
(D) 73
(E) 74
Rationale: Explanation :
Excluded number = (55 x 7) (52 x 6) = 385 - 312 = 73
Answer : D
Answer: (D) 73
Question: Two cylinders are of the same height. Their radii are in the ratio 1: 3. If the volume of the first cylinder is 40 cc. Find the volume of the second cylinder?
(A) 360
(B) 982
(C) 279
(D) 270
(E) 289
Rationale: r1 = x r2 = 3x
Π * x2 * h = 40
Π9 x2h = 40 * 9 = 360
Answer:A
Answer: (A) 360
Question: a, b, c are integers. |a| ≠ |b| ≠ |c| and -10 ≤ a, b, c ≤ 10. What will be the maximum possible value of [abc – (a+b+c)]?
(A) 524
(B) 693
(C) 731is answer
(D) 970
(E) None of the above
Rationale: Is the answer bank correct? Shouldn't option C be 713? C
Answer: (C) 731is answer
Question: For how many values of Q, is |||Q - 5| -10| -5| = 2?
(Those ls are mods)
(A)  0
(B)  2
(C)  4
(D)  8
(E)  More than 8
Rationale: I think its 8
|||Q-5|-10|-5| =2
let |Q-5| = a which makes above
||a-10|-5| =2
let |a-10| = b which makes
|b-5| = 2
now for the above b can take 3, 7
for every b =3 a can have 13, 7
and for b = 7 a can have 17 and 3
so 'a' has four solutions 13, 7, 17 and 3
for a = 13; Q has 18 or -8 thus has 2 for every combination hence 4x2 = 8
answer D
Answer: (D)  8
Question: It takes Avery 3 hours to build a brick wall while Tom can do it in 3 hours. If the two start working together and after an hour Avery leaves, how much time will it take Tom to complete the wall on his own?
(A) 20
(B) 60
(C) 30
(D) 40
(E) 50
Rationale: Avery takes 3 hours
Tom takes 3 hours
Efficiency of Avery is 1/3 units/hr
Efficiency of Tom is 1/3 units/hr
Combined efficiency of Tom and Avery is 1/3 +1/3 = 2/3 units/hr
Since they worked for 1 hour they completed 2/3 units of work and 1/3 units of work is left which is to be completed by Tom ( Since Avery left )
So Time taken by Tom to complete the remaining work will be 1/3/1/3 hours => 1*60 = 60 minutes...
Answer will be (B)
Answer: (B) 60
Question: If n + 11 is a positive odd integer, which of the following must be true about the sum of the next four integers?
(A) It is a multiple of 3
(B) It is a multiple of 4
(C) It is a prime number
(D) It is a multiple of 2
(E) It is a multiple of 5
Rationale: n+11 is a positive odd integer. This means n will be an odd integer
The next 4 integers would be: n + 11, n + 12, n + 13, n + 14
Sum = 4n + 50
This sum will always be even.
Option D
Answer: (D) It is a multiple of 2
Question: A sum of money becomes double itself in 8 years at simple interest. How many times will it become 10 years at the same rate?
(A) 2 1/4
(B) 2 7/4
(C) 2 1/8
(D) 2 9/4
(E) 2 1/4
Rationale: P ---- 2P ---- 8 years
2 1/4 P ---- 10 years
Answer: A
Answer: (A) 2 1/4
Question: If 213 × 16 = 3408, then 1.6 × 21.3 is equal to :
(A) 0.3408
(B) 3.408
(C) 34.08
(D) 340.8
(E) None of these
Rationale: Solution
1.6 × 21.3 = (16/10x 213/10)= (16x 213/100) = 3408/100 = 34.08. Answer C
Answer: (C) 34.08
Question: Mary is about as many days as John in weeks, and Mary is as many months as Robert in years. Mary, John and Robert together are 144 years. Can you tell me Robert’s age in years?
(A) 60 years
(B) 86 years
(C) 108 years
(D) 96 years
(E) 84 years
Rationale: Let r be Robert’s age in years. If j is John’s age in years, then John is 52j weeks old. If m is Mary’s age in years, them Mary is 365m days old. Thus,365m = 52jSince Mary is 12m months old,12m = rSince Mary, John and Robert together are 144 years,m + j + r = 144,The above system of 3 equations in 3 unknowns (m, j and r) can be solved as follows:r/12 + 365r/(52 x 12) + r = 144 or52r + 365r + 624r = 624 x 144 orr = 624 x 144/1041 = 86 yearsAnswer: B
Answer: (B) 86 years
Question: A, B and C invest in the ratio of 3 : 4: 5. The percentage of return on their investments are in the ratio of 6 : 5 : 4. Find the total earnings, If B earns Rs. 100 more than A :
(A) 2900
(B) 7250
(C) 2767
(D) 1998
(E) 2771
Rationale: Explanation:
A B C
investment 3x 4x 5x
Rate of return 6y% 5y% 4y%
Return \inline \frac{18xy}{100} \inline \frac{20xy}{100} \inline \frac{20xy}{100}
Total = (18+20+20) = \inline \frac{58xy}{100}
B's earnings - A's earnings = \inline \frac{2xy}{100} = 100
Total earning = \inline \frac{58xy}{100} = 2900
Answer: A) Rs.2900
Answer: (A) 2900
Question: A box contains 4 red, 3 white and 2 blue balls. Three balls are drawn at random. Find out the number of ways of selecting the balls of different colours?
(A) 62
(B) 48
(C) 12
(D) 24
(E) None of these
Rationale: Explanation :
1 red ball can be selected in 4C1 ways
1 white ball can be selected in 3C1 ways
1 blue ball can be selected in 2C1 ways
Total number of ways
= 4C1 x 3C1 x 2C1
=4 x 3 x 2
= 24. Answer : Option D
Answer: (D) 24
Question: For what value of x will the function f(x) = -3(x - 10)(x - 4) have a maximum value? Find the maximum value.
(A) 17
(B) 37
(C) 57
(D) 27
(E) 47
Rationale: f(x) = -3(x - 10)(x - 4) = -3x2*square* + 42x - 120 : expand and obtain a quadratic function
h = -b/2a = -42/-6 = 7 : h is the value of x for which f has a maximum value
f(h) = f(7) = 27 : maximum value of f.
correct answer D
Answer: (D) 27
Question: The odds against ram solving the problem are 4 to 3 and the odds in favour of krishna solving the same are 6 to 5. What is the chance that it will be solved when they both try?
(A) 1/4
(B) 5/21
(C) 16/21
(D) 23/28
(E) 57/77
Rationale: Odds against Ram are 4:3 so probability that he will solve the problem is 3/7.
Odds in favor of Krishna are 6:5 so probability that he will solve it is 6/11
It will be solved means that either Ram or Krishna or both will be able to solve it.
P(A or B) = P(A) + P(B) - P(A and B) = 3/7 + 6/11 - (3/7)*(6/11) = 57/77
E
Answer: (E) 57/77