divelab/Sys2Bench huggingface dataset:
Question: The ratio, by volume, of soap to alcohol to water in a certain solution is 20:50:80. The solution will be altered so that the ratio of soap to alcohol is doubled while the ratio of soap to water is halved. If the altered solution will contain 100 cubic centimeters of alcohol, how many cubic centimeters of water will it contain?
Options:
A)520
B)640
C)700
D)750
E)800
Rationale: soap:alcohol
Initial ratio soap:alcohol: water --> 20:50:80
Initial soap:alcohol = 20/50 =20:50
after doubled soap:alcohol =2* 20/50 = 40:50
Initial soap:water = 20/80=20:80
after halved soap:water: 1/2 * 20/80 = 10/80 = 10:80
After soap: alcohol:water --> 40:50:320-->4:5:32
given alcohol 100 cubic centimeter.
ratio is 80:100:640(4:5:32)
For 100 cubic centimeter of Alcohol --- 640cubic cm water is required.
Answer-B
Answer: B
Question: A vessel is filled with liquid, 3 parts of which are water and 5 parts syrup. How much of the mixture must be drawn off and replaced with water so that the mixture may be half water and half syrup?
Options:
A)1/5
B)3/5
C)1/8
D)5/8
E)3
Rationale: Suppose the vessel initially contains 8 litres of liquid.
Let x litres of this liquid be replaced with water.
Quantity of water in new mixture = (3 -(3x/8)+ x)litres
Quantity of syrup in new mixture = (5 - 5x/8) litres
Therefore, (3 - (3x/8)	+ x)	= 5 -	5x/8)
= 5x + 24 = 40 - 5x
= 10x = 16
= x = 8/5
So, part of the mixture replaced = (8/5 x 1/8)= 1/5
Answer is A.
Answer: A
Question: All of the citizens of a country have a eight-character or nine-character national identification code that is created using the 26 letters of the alphabet and the 10 digits from 0 to 9. What is the maximum number of citizens who can be designated with these codes?
Options:
A)36^17
B)35(36^9)
C)37(36^9)
D)35(36^8)
E)37(36^8)
Rationale: The number of possible 8-digit codes is 36^8.
The number of possible 9-digit codes is 36^9.
The total sum is 36^8 + 36^9 = 36^8(1+36) = 37(36^8)
The answer is E.
Answer: E
Question: A and B started a partnership business investing some amount in the ratio of 3:5. C joined them after six months with an amount equal to that of B. In what proportion should the profit at the end of one year be distributed among A, B and C?
Options:
A)5 : 6 : 10
B)6 : 10 : 5
C)6 : 5 : 10
D)10 : 6 : 5
E)none
Rationale: Sol.
Let the initial investments of A and B be 3x amd 5x.
A : B : C = (3x * 12) : (5x * 12) : (5x * 6) = 36 : 60 : 30 = 6 : 10 : 5.
Answer B
Answer: B
Question: Find the compound ratio of (2:3), (6:11) and (11:5) is
Options:
A)3:2
B)2:1
C)1:2
D)4:5
E)2:3
Rationale: Required ratio = 2/3 * 6/11 * 11/5 = 2/1 = 4:5
Answer is D
Answer: D
Question: 1=1,2=2,3=9,4=16,5=25, then 9=?
Options:
A)1
B)255
C)3
D)445
E)235
Rationale: 1=1,2=2,3=9,4=16,5=25, then 9=?
9=3
check the third eqn.
ANSWER:C
Answer: C
Question: A miniature gumball machine contains 4 blue, 3 green, and 3 red gumballs, which are identical except for their colors. If the machine dispenses three gumballs at random, what is the probability that it dispenses one gumball of each color?
Options:
A)2/5
B)3/5
C)3/10
D)7/30
E)11/30
Rationale: The total number of ways to dispense 3 gumballs is 10C3=120.
The number of ways to choose one of each color is 4*3*3=36.
P(dispensing one of each color)=36/120=3/10.
The answer is C.
Answer: C
Question: A can do a work in 15 days and B in 20 days. If they work on it together for 8 days, then the fraction of the work that is left is
Options:
A)8/17
B)7/15
C)3/15
D)8/15
E)1/15
Rationale: Person ( A ) ( B ) ( A+B )
Time - ( 15 ) ( 20 ) (-)
Rate - ( 20 ) ( 15 ) ( 35 )
Work -(300) (300) (300)
therefore A+B requires (300/35) days to complete entire work
For 1st 4 days they work 35 * 8 = 280
Remaining work is 300 - 280 = 20
Remaining fraction of work is = 20/300 = 1/15
Answer E
Answer: E
Question: The ratio of the arithmetic mean of two numbers to one of the numbers is 3:5. What is the ratio of the smaller number to the larger?
Options:
A)1:5
B)1:4
C)1:3
D)1:2
E)2:3
Rationale: let the numbers be a,b
acc. to problem,
(a+b)/2∗a(a+b)/2∗a = 3/53/5
1/21/2 + b/2∗ab/2∗a = 3/53/5
b/ab/a = 1/51/5
Ans. A
Answer: A
Question: Find the cost of fencing around a circular field of diameter 34 m at the rate of Rs.2 a meter?
Options:
A)215
B)212
C)278
D)279
E)222
Rationale: 2 * 22/7 * 17 = 106
106 * 2
= Rs.212
Answer: B
Answer: B
Question: Entry fee in a zoo is $10. Later this was reduced by 50% which increased the sale by 40%. The percentage increase in the number of visitors is?
Options:
A)100%
B)50%
C)180%
D)200%
E)150%
Rationale: Let the total original sale be $1000
Then original number of visitors = 100
New number of visitors = 1400/5 = 280
Increase % = 180%
Answer is C
Answer: C
Question: If 20% of A is the same as 25% of B, then A : B is :
Options:
A)3:4
B)4:3
C)5:4
D)6:7
E)5:7
Rationale: Expl : 20% of A I= 25% of B =20A/100 = 25B/100 = 5/4 = 5:4
Answer:C
Answer: C
Question: There are three secretaries who work for four departments. If each of the four departments have one report to be typed out, and the reports are randomly assigned to a secretary, what is the probability R that all three secretary are assigned at least one report?
Options:
A)8/9
B)64/81
C)4/9
D)16/81
E)5/9
Rationale: I got the same answer: Here's my reasoning
First Report you have 3 choices
Second Report you have 2 choices
Third Report you have 1 choice
Fourth report 3 choices again
Then total number of ways is: 3*2*1*3=3^2*2
This is not correct. You have assumed that the 4th report must go to someone who already has a report. There is no such constraint. You can easily give the 1st and 2nd reports to secretary 1, 3rd report to secretary 2 and 4th report to secretary 3. But you have ignored all such cases.
The number of ways of ensuring at least one report goes to each secretary is 4C2 (select 2 reports out of 4 which go to the same person)*3C1 (select the person who must type 2 reports)*2! (since you have 2 reports left which you must distribute to the 2 remaining people such that each person gets one) = 36
Required probability R= 36/81.C
Answer: C
Question: In the "Big-Reds" parking lot there are 56 vehicles, 18 of them are buses and the rest are private cars. The color of 32 vehicles is red, from which 17 are buses. How many private cars can be found in the parking lot, which are not colored red?
Options:
A)1.
B)15
C)17
D)20
E)23
Rationale: Total Vehicles = 56
Buses = 18
Private Cars = 38
Total red vehicles = 32
Total red buses = 17
Total Red cars = 15
Total cars that are not red = 38 - 15 = 23
Correct Option: E
Answer: E
Question: In a certain sequence of numbers, a1, a2, a3, ..., an, the average (arithmetic mean) of the first m consecutive terms starting with a1 is m, for any positive integer m. If a1=1, what is a12?
Options:
A)100
B)55
C)23
D)19
E)1
Rationale: As stated above, (a1+a2+a3......+a12)/12=12
therefore a1+a2+a3.......a12=144 (1)
using the same logic, we got a1+a2+a3..........+a11=121 (2)
(2)-(1) we got a12=23
C
Answer: C
Question: The number of female students in a college in the previous year was 350. If in the current year the number of male students in the college doubled over that of the previous year but the proportion of male students halved, then what is the number of female students in the college in the current year?
Options:
A)1000
B)1200
C)1400
D)1600
E)1800
Rationale: Let x be the number of male students last year. The number of male students this year is 2x.
Let y be the number of female students this year.
x/350 = 2(2x/y)
y = 1400
The answer is C.
Answer: C
Question: The average weight of a group of boys is 30 kg. After a boy of weight 40 kg joins the group, the average weight of the group goes up by 1 kg. Find the number of boys in the group originally ?
Options:
A)14
B)5
C)6
D)7
E)9
Rationale: Let the number off boys in the group originally be x.
Total weight of the boys = 30x
After the boy weighing 40 kg joins the group, total weight of boys = 30x + 40
So 30x + 40 = 31(x + 1) = > x = 9.
Answer: E
Answer: E
Question: Tough and Tricky questions: Word Problems.
It takes Sarah 12 minutes to do one oil change, and it takes Alice 8 minutes. If Sarah and Alice both begin doing oil changes at exactly 9:00 a.m. without pausing, when is the first time at which they will finish changing the oil of a car simultaneously?
Options:
A)09:24 a.m.
B)10:42 a.m.
C)12:00 a.m.
D)1:00 p.m.
E)2:00 p.m.
Rationale: LCM of 812
8 = 2 * 2 * 2
12 = 2 * 2 * 3
LCM = 2 * 2 * 2 * 3 = 24 minutes
924AM
A
Answer: A
Question: The toll for crossing a certain bridge is $0.75 each crossing. Drivers who frequently use the bridge may instead purchase a sticker each month for $13.00 and then pay only $0.10 each crossing during that month. If a particular driver will cross the bridge twice on each of x days next month and will not cross the bridge on any other day, what is the least value of x for which this driver can save money by using the sticker?
Options:
A)12
B)15
C)16
D)28
E)29
Rationale: Option #1: $0.75/crossing....Cross twice a day = $1.5/day
Option #2: $0.30/crossing....Cross twice a day = $0.6/day + $13 one time charge.
If we go down the list of possible answers, you can quickly see that 14 days will not be worth purchasing the sticker. 1.5x14 (21) is cheaper than 0.6x14 +13 (21.4)...it's pretty close so let's see if one more day will make it worth it... If we raise the number of days to 15, the sticker option looks like a better deal...1.5x15 (22.5) vs 0.6x15 + 13 (22). Answer: A
Answer: A
Question: if sum of two no. is 12 and subtraction to that is 4.so how much min. should added to smaller no. to make it even?
Options:
A)1
B)2
C)0.5
D)1.5
E)0
Rationale: let smaller no. be x.
bigger no. =12-x.
12-x-x=4.
12-2x=4
2x=8,x=4.
so 4 is itself even no.so,zero should be added.
answer E
Answer: E
Question: Find the value for x from below equation: x/3=-2?
Options:
A)-6
B)1
C)-2
D)-3
E)4
Rationale: 1. Multiply both sides by 3:
x*3/3= -2/3
2. Simplify both sides:
x = -6
A
Answer: A
Question: 40% of a number is more than 20% of 650 by 190. Find the number?
Options:
A)299
B)277
C)800
D)289
E)271
Rationale: (40/100) * X – (20/100) * 650 = 190
2/5 X = 320
X = 800
Answer:C
Answer: C
Question: The units digit of (35)^(87) + (93)^(49) is:
Options:
A)2
B)4
C)6
D)8
E)0
Rationale: The units digit of powers of 3, cycles in a group of 4: {3, 9, 7, 1}
49 has the form 4k+1, so the units digit of 93^49 is 3.
The units digit of powers of 5 is always 5.
3+5 = 8, so the units digit is 8.
The answer is D.
Answer: D
Question: There are 9 fictions and 6 non-fictions. How many cases are there such that 2 fictions and 2 non-fictions are selected from them?
Options:
A)90
B)120
C)540
D)180
E)200
Rationale: Number of ways of selecting 2 fiction books = 9C2
Number of ways of selecting 2 non fiction books = 6C2
9C2 * 6C2 = 36 * 15 = 540
Answer: C
Answer: C
Question: A man gets a simple interest of Rs.500 on a certain principal at the rate of 5% p.a in two years. Find the compound interest the man will get on twice the principal in two years at the same rate.
Options:
A)276.25
B)256.25
C)256.29
D)236.25
E)256.22
Rationale: Let the principal be Rs.P
S.I at 5% p.a in 8 years on Rs.P = Rs.500
(P)(8)(5)/100 = 500
P = 1250
C.I on Rs.2P i.e., Rs.2500 at 5% p.a in two years
=2500{ [1 + 5/100]2 - 1} = 2500{ 212 - 202 /202}
= 2500/400(441 - 400)
= 25/4(41) = 1025/4 = Rs.256.25.Answer:B
Answer: B
Question: In a function they are distributing noble prize. In how many ways can 3 prizes be distributed among 7 boys when No boy gets more than one prize?
Options:
A)112
B)114
C)117
D)210
E)217
Rationale: Sol. In this case, repetitions are not allowed. So, the first prize can be given in 7 ways. The second in 6 ways and the third in 5 ways. But fundamental principle, (7 x 6 x 5) ways = 210 ways
D
Answer: D
Question: There are three guys who have equal number of bullets. All of them shoot 6 bullets. The remaining bullets are equal to the total number of bullets remaining is equal to the bullets each had after division. What was the original number of bullets each one had?
Options:
A)18
B)19
C)20
D)21
E)27
Rationale: let initially each was having x bullets
they shot 6 bullets
so 3x-18=x
x=9
so each was having 9 bullets.
total was 27
ANSWER:E
Answer: E
Question: An estate pay would be divided among five sons with the remainder donated to charity. One stipulation in the estate is that no one among the sons can receive an amount within 15% of another's amount. If one of the sons received $10,000, what is the smallest possible range between the highest and lowest amounts (rounded to the nearest dollar) among the five sons?
Options:
A)$4096
B)$4804
C)$7892
D)$6736
E)$5736
Rationale: 1st - 10,000.
2nd - 0.85*10,000 = 8500
3rd - 0.85*8,500 = 7,225
4th - 0.85*7,225 = ~6,141.25
5th - 0.85*6,141.25 = ~5,220.0625
Range = 10,000 - 5,220 = 4,780
Answer: B.
Answer: B
Question: Ms. Feldman invested $10,000 for 1 year, part at 7% and part at 6.25%.
If she earned a total interest of $656.50, how much was invested at each rate?
Options:
A)4200 & 5800
B)4600 & 5400
C)9000 & 1000
D)7000 & 3000
E)6000 & 4000
Rationale: Interest rate of 4200/- =294
Interest rate of 5800/- =362.50
Answer is A
Answer: A
Question: A boat can travel with a speed of 24 km/hr in still water. If the speed of the stream is 4 km/hr, find the time taken by the boat to go 196 km downstream
Options:
A)3 hours
B)4 hours
C)5 hours
D)6 hours
E)7 hours
Rationale: Explanation:
Speed of the boat in still water = 24 km/hr
speed of the stream = 4 km/hr
Speed downstream = (22+5) = 28 km/hr
Distance travelled downstream = 196 km
Time taken = distance/speed = 196/28 = 7 hours
Answer: Option E
Answer: E
Question: If x > y^4 > z^7, which of the following statements could be true? I.x>y>z II.z=x III.x<z
Options:
A)I,III only
B)I only
C)II only
D)III only
E)I,II only
Rationale: Answer : B
Answer: B
Question: A focus group is currently made up of x red balls and y green balls. If 5 red balls and 4 green balls are added to the group, and if one ball is selected at random from the larger focus group, then what is the probability that a red ball is selected?
Options:
A)x/(x + 4)
B)y/(x + 5)
C)(x + 5)/(x + y + 9)
D)(y + 4)/(x + y + 9)
E)(x + 4)/(x + y + 9)
Rationale: x red balls
y green balls
x+4+y+5 is the total
=x+y+9
x+5 is the number of red balls
so (x + 5)/(x + y + 9)
the answer is C
Answer: C
Question: A train 275 m long is running with a speed of 60 km/hr. In what time will it pass a man who is running at 6 km/hr in the direction opposite to that in which the train is going?
Options:
A)8
B)6
C)5
D)15
E)1
Rationale: Speed of train relative to man = 60 + 6 = 66 km/hr.
= 66 * 5/18 = 55/3 m/sec.
Time taken to pass the men = 275 * 3/55 = 15 sec.
Answer: D
Answer: D
Question: Two pipes A and B can fill a cistern in 37.5 minutes and 45 minutes respectively. This cistern will be filled in half an hour if both the pipes are opened together initially and pipe B is then turned off after X minutes. What is X?
Options:
A)10 minutes
B)15 minutes
C)12 minutes
D)9 minutes
E)8 minutes
Rationale: Explanation:
Let B be turned off after X minutes.
Part of water will be filled by A + B in x minutes + After tap B is closed rest filled by A in (30-x) minutes = 1
=> x/(time taken to fill the container by each separately) + 30 - x/(time taken to fill the container by A alone) = 1
=> x(2/75 + 1/45) + (30-x)2/75 = 1
=> 11x + 180 – 6x = 225
Therefore, x = 9 minutes
ANSWER: D
Answer: D
Question: Mira is making telescopes, each consisting of 2 lenses, 1 tube, and 1 eyepiece. Lenses can be purchased only in packs of 50, tubes only in packs of 10, and eyepieces only in packs of 30. However, half of the lenses in each pack are not usable for telescopes. If all parts are used only for the telescopes, what is the minimum number of lenses Mira must purchase to make a set of telescopes with no leftover components other than the unusable lenses?
Options:
A)75
B)150
C)300
D)1200
E)7,500
Rationale: I'd use backsolving for this question. Check option C first:
300 lenses = 150 usable lenses = 75 tubes = 75 eyepieces. We cannot buy 75 tubes or 75 eyepieces, so that not to have leftovers.
A and B are automatically out. Check D:
1200 lenses = 600 usable lenses = 300 tubes = 300 eyepieces. We CAN buy 300 tubes and 300 eyepieces, so that not to have leftovers.
Answer: D.
Answer: D
Question: Two trains, each 100 m long, moving in opposite directions, cross other in 8 sec. If one is moving twice as fast the other, then the speed of the faster train is?
Options:
A)26 km/hr
B)17 km/hr
C)60 km/hr
D)77 km/hr
E)46 km/hr
Rationale: Let the speed of the slower train be x m/sec.
Then, speed of the train = 2x m/sec.
Relative speed = ( x + 2x) = 3x m/sec.
(100 + 100)/8 = 3x => x = 25/3.
So, speed of the faster train = 50/3 = 50/3 * 18/5
= 60 km/hr.
Answer:C
Answer: C
Question: J is 25% less than P and 20% less than T. T is x% less than P. What is the value of x?
Options:
A)93.5
B)90
C)6.75
D)6.25
E)2
Rationale: Let, P = 400
then J = (75/100)*400 = 300
also J = (80/100)*T
i.e. T = 300*100/80 = 375
and T = [1-(x/100)]*P
i.e. 100-x = 100*T/P = 100*375/400 = 93.75
i.e. x = 6.25
Answer: Option D
Answer: D
Question: The monthly rent of a shop of dimension 10feet × 8feet is Rs.2400. What is the annual rent per square foot of the shop?
Options:
A)43
B)56
C)68
D)360
E)92
Rationale: Sol. Monthly rent per square feet = 2400/(10*8) = 30
& Annual rent per square feet
=12*30=360
Answer : D
Answer: D
Question: Tomy's age 18 years hence will be thrice his age four years ago. Find Tomy's present age?
Options:
A)12 years
B)13 years
C)15 years
D)17 years
E)18 years
Rationale: Let MTomy's present age be 'x' years.
x+ 18 = 3(x- 4)
=> 2x = 30 => x = 15 years.
C
Answer: C
Question: Sum of 42 odd numbers is?
Options:
A)1764
B)1774
C)1784
D)1794
E)1754
Rationale: sum of 1st n odd no.s = 1+3+5+7+... = n^2
so, Sum of 1st 42 odd numbers = 42^2 = 1764
ANSWER:A
Answer: A
Question: A and B put in Rs.300 and Rs.400 respectively into a business. A reinvests into the business his share of the first year's profit of Rs.196 where as B does not. In what ratio should they divide the second year's profit?
Options:
A)39:40
B)39:49
C)24:25
D)39:47
E)39:41
Rationale: Explanation:
3: 4
A = 3/7*196 = 84
384: 400
24:25
Answer:C
Answer: C
Question: A certain no. when divided by 35leaves a remainder 25, what is the remainder if the same no.be divided by 15?
Options:
A)2
B)4
C)7
D)8
E)9
Rationale: Explanation:
35 + 25 = 60/15 = 4 (Remainder)
B
Answer: B
Question: The probability that event A occurs is 0.4, and the probability that events A and B both occur is 0.25. If the probability that either event A or event B occurs is 0.6, what is the probability that event B will occur?
Options:
A)0.05
B)0.15
C)0.45
D)0.5
E)0.55
Rationale: P(A or B) = P (A) + P(B) - p(a n b)
0.6= 0.4 + P(B) - 0.25
P(B) = 0.45
ANSWER:C
Answer: C
Question: A right triangle ARC has to be constructed in the xy-plane so that the right angle is at A and AR is parallel to x axis. The coordinates of A, R and C are integers and satisfy the inequalities -1 ≤ x ≤ 7 and 1 ≤ y ≤ 7. How many different triangles can be constructed with these properties?
Options:
A)63
B)336
C)567
D)3024
E)5040
Rationale: All the cordinates are integer hence,
possible AR values are
{-1,0}, {-1,1} ....{-1,7}: 8 ways
....
{7,-1}, {7,0} ......{7,6}: 8 ways
9*8 ways = 72
for AC values can be
{1,2}, {1,3} .... {1,7} : 6ways
......
{7,1}, {7,2} ..... {7,6} : 6 ways
7*6 = 42
Total = 72 * 42 = 3024 hence D
Answer: D
Question: In how many different number of ways a Committee of 3 person of can be selected from 5 men and 3 women such that atleast 1 women is included in the committee
Options:
A)36
B)46
C)50
D)52
E)56
Rationale: 1W 2M 2W 1M 3W
= 3C1 × 5C2 + 3C2 × 5C1 + 3C3s
= 3 × (5 × 4)/2 + 3 × 5 + 1
= 30 + 15 + 1 = 46
Total 5M 3W
8C3 = 56
5C3 = 10
At least one women = total - with out women
At least one women = 56 - 10 = 46
B)
Answer: B
Question: An error 2% in excess is made while measuring the side ofa square. The % of error in the calculated area of the square is?
Options:
A)4%
B)4.04%
C)4.23%
D)4.56%
E)5%
Rationale: 100 cm is read as 102 cm.
A1 = (100 x 100) cm2 and A2 (102 x 102) cm2.
(A2 - A1) = [(102)2 - (100)2]
= (102 + 100) x (102 - 100)
= 404 cm2.
Percentage error = 404 x 100 % = 4.04%
100 x 100
B
Answer: B
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?
Options:
A)240
B)377
C)2908
D)287
E)201
Rationale: Speed = (54 * 5/18) m/sec = 15 m/sec. Length of the train = (15 x 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: A
Answer: A
Question: he speed of a train is 90 kmph. What is the distance covered by it in 10 minutes?
Options:
A)25 kmph
B)15 kmph
C)17 kmph
D)95 kmph
E)14 kmph
Rationale: 90 * 10/60 = 15 kmph
Answer:B
Answer: B
Question: In 2010, Arman’s birthday was on a Thursday. What day of the week was it 66 days after his birthday?
Options:
A)Thursday
B)Friday
C)Saturday
D)Sunday
E)Monday
Rationale: Day of week after 66 days after Arman's birthday = 7*9 + 3
Saturday as the period will have 9 weeks and 3 days .
The 3 days in the end will be Friday - Saturday and Sunday
Answer D
Answer: D
Question: What is the 9th term in this sequence?
1,3,4,11,19,30,53,87
Options:
A)120
B)129
C)140
D)180
E)169
Rationale: 1+3 =4 +0 =4
4+3 =7+4 =11
11+4 =15+4 =19
19+11=30+0 =30
30+19=49+4 =53
53+30=83+4 =87
87+53=140+0=140
so 9th term is 140
ANSWER:C
Answer: C
Question: Find the fraction which has the same ratio to 1/2 that 2/5 has to 3/7
Options:
A)1/2
B)2/12
C)14/30
D)3/8
E)7/9
Rationale: A : 1/2 = 2/5 : 3/7
As the product of the means is equal to the product of the extremes.
A*3/7 = 1/2 * 2/5
A*3/7 = 2/10
A = 14/30 => A = 14/30
ANSWER:C
Answer: C
Question: The ratio of the ages of Maala and Kala is 3 : 5. The total of their ages is 3.2 decades. The proportion of their ages after 0.8 decades will be
[1 Decade = 10 years]
Options:
A)6:5
B)5:7
C)4:5
D)7:9
E)3:6
Rationale: Let, Maala’s age = 3A and Kala’s age = 5A
Then 3A + 5A = 32
A = 4
Maala’s age = 12 years
and Kala’s age = 20 years
Proportion of their ages after 8 is = (12 + 8) : (20 + 8)
= 20 : 28
= 5 : 7
Answer : B
Answer: B
Question: The smallest 3 digit prime number is
Options:
A)A)103
B)B)101
C)C)109
D)D)113
E)E)None
Rationale: The smallest 3-digit number is 100, which is divisible by 2.
100 is not a prime number.
101 < 11 and 101 is not divisible by any of the prime numbers 2, 3, 5, 7, 11.
101 is a prime number.
Hence 101 is the smallest 3-digit prime number.
B)
Answer: B
Question: John can complete a given task in 20 days. Jane will take only 12 days to complete the same task. John and Jane set out to complete the task by beginning to work together. However, Jane was indisposed 5 days before the work got over. In how many days did the work get over from the time John and Jane started to work on it together?
Options:
A)95/8
B)93/8
C)90/8
D)12
E)10
Rationale: In such questions, you need to start from the end. Last 4 days John works alone and completes 5 * (1/20) = 1/4 of the work.
So 3/4 of the work should have been completed by the two of them together before Jane left.
Their combined rate of work is 1/20 + 1/12 = 8/60
Time taken to complete 3/4 of the work = (3/4)/(8/60) = 45/8 days.
So total number of days taken to complete the work = 45/8 + 5 = 95/8 days.
A
Answer: A
Question: What is the next number of the following sequence
4, 16, 12, 48, 44, 176 (...)
Options:
A)142
B)146
C)154
D)168
E)172
Rationale: The series is
4*4=16
16-4=12
12*4=48
48-4=44
44*4=176
176-4=172
ANSWER: E
Answer: E
Question: A man saves 20% of his monthly salary. If an account of dearness of things he is to increase his monthly expenses by 20%, he is only able to save Rs. 200 per month. What is his monthly salary?
Options:
A)Rs.5000
B)Rs.25000
C)Rs.7500
D)Rs.8500
E)Rs.9500
Rationale: Income = Rs. 100
Expenditure = Rs. 80
Savings = Rs. 20
Present Expenditure 80*(20/100) = Rs. 96
Present Savings = 100 – 96 =Rs. 4
100 ------ 4
? --------- 200 => 5000
ANSWER:A
Answer: A