mathewhe/aqua_rat huggingface dataset:
Question: A can do a work in 9 days and B can do the same work in 18 days. If they work together, in how many days will they complete the work?
(A) 8
(B) 7
(C) 5
(D) 6
(E) 3
Rationale: One day's work of A and B
= 1/9 + 1/8
= (2+1)/18 = 1/6 So, the time taken to complete the work is 6 days.
Answer: C
Answer: (C) 5
Question: How many ways are there of placing 6 marbles in 4 bowls, if any number of them can be placed in each bowl?
(A) 6C4
(B) 6P4
(C) 4^6
(D) 6^4
(E) 6!
Rationale: Each marble has 4 options, so there are total of 4*4*4*4*4*4=4^6 ways.
Answer: C.
Answer: (C) 4^6
Question: What is the greatest number that divides 263, 935 and 1383 leaving a remainder of 7 in each case ?
(A) 30
(B) 31
(C) 32
(D) 35
(E) 37
Rationale: Answer
The required greatest number is the HCF of 263-7, 935-7, 1383-7
i.e. 256, 928 and 1376 HCF = 32
Correct Option: C
Answer: (C) 32
Question: The number which exceeds 16% of it by 126 is :
(A) 150
(B) 52
(C) 58
(D) 60
(E) 62
Rationale: Solution
Solution
Let the number be x.
x-16% of x =126
x-16/100x =126
x-4/25x=126
21/25x=126
x=(126x25/21)= 150
Answer A
Answer: (A) 150
Question: IF 2+7=57 ; 3+6=63 ; 5+9=206 Then 4+8=?
(A) 125
(B) 126
(C) 127
(D) 128
(E) 129
Rationale: 2^3 + 7^2= 57
3^3 + 6^2=63
5^3 + 9^2=206
and
4^3 + 8^2=128
ANSWER:D
Answer: (D) 128
Question: In a group of 80 children and 20 youngsters, each child got sweets that are 15% of the total number of children and each youngster got sweets that are 25% of the total number of children. How many sweets were there?
(A) 1140
(B) 1360
(C) 1180
(D) 1200
(E) 1210
Rationale: Number of sweets each child got = 15% of 80 = 15/100 * 80 = 12.
Number of sweets 80 children got = 80 * 12 = 960.
Number of sweets each youngster got = 25% of 80 = 25/100 * 80 = 20.
Number of sweets 20 youngsters got = 20 * 20 = 400.
Total number of sweets = 960 + 400 = 1360.
B)
Answer: (B) 1360
Question: The average monthly salary of 20 employees in an organisation is Rs. 1500. If the manager's salary is added, then the average salary increases by Rs. 100. What is the manager's monthly salary ?
(A) 3600
(B) 2988
(C) 2789
(D) 2887
(E) 1297
Rationale: Manager's monthly salary Rs. (1600 * 21 - 1500 * 20)
= Rs. 3600.
Answer: A
Answer: (A) 3600
Question: An alloy contains tin and copper in the ratio of 4:5 . If tin has 20% impurity and copper has 58% , what is the avg impurity percentage ?
(A) 20%
(B) 41.1%
(C) 35%
(D) 38%
(E) 43.1%
Rationale: Total of 9 kgs.
Tin = 4Kgs; copper = 5Kgs
Impurity in tin = 0.2 * 4 = 0.8
Impurity in copper = 0.58 * 5 = 2.9
Avg impurity = (0.8 + 2.9)/(4+5) = 3.7/9 = ~41%
ANSWER:B
Answer: (B) 41.1%
Question: Find the largest 4 digit number which isexactly divisible by 88?
(A) 8765
(B) 8907
(C) 9944
(D) 9954
(E) 9990
Rationale: Largest 4 digit number is 9999
After doing 9999 ÷ 88 we get remainder 55
Hence largest 4 digit number exactly divisible by 88 = 9999 - 55 = 9944
C
Answer: (C) 9944
Question: A train moves with a speed of 72 kmph. Its speed in metres per second is
(A) 10.8
(B) 18
(C) 30
(D) 20
(E) None
Rationale: Solution
Speed	= 72 Kmph
= (72 x 5/18)m/sec
= 20 m/sec.
Answer D
Answer: (D) 20
Question: Convert the 19/36 m/s into kilometers per hour?
(A) 1.7
(B) 1.5
(C) 1.3
(D) 1.1
(E) 1.9
Rationale: 19/36 m/s = 19/36 * 18/5 = 19/10
= 1.9 kmph.
Answer: E
Answer: (E) 1.9
Question: A group of five people are weighed with the results as follows (in pounds): 135, 140, 165, 178, 199. If the weight of the heaviest person is increased by 9 pounds, which of the following statements best describes the change in the mean and the median of the weights?
(A)  The mean and the median will remain unchanged.
(B)  The mean will remain unchanged but the median will increase.
(C)  The mean will increase but the median will remain unchanged.
(D)  The mean and the median will increase by the same amount.
(E)  The mean and the median will increase by different amounts.
Rationale: Old set = {135, 140, 165, 178, 199}.
New set = {135, 140, 165, 178, 208} --> the median (middle term) will remain the same and the mean will increase.
Answer: C.
Answer: (C)  The mean will increase but the median will remain unchanged.
Question: Jack can row at 9 kmph and still water. He takes 6 hours to row from A to B and back. What is the distance between A and B if the speed of the stream is 3 kmph?
(A) 30 km
(B) 25 km
(C) 24 km
(D) 28 km
(E) 20 km
Rationale: Let the distance between A and B be x km.
Total time = x/(9 + 3) + x/(9 - 3) = 6
=> x/12 + x/6 = 6 => (x + 2x)/12 = 6 => x = 24 km.
ANSWER:C
Answer: (C) 24 km
Question: On a certain date, Pat invested $10,000 at x percent annual interest, compounded annually. If the total value of the investment plus interest at the end of 12 years will be $40,000, in how many years, the total value of the investment plus interest will increase to $160,000?
(A) 15
(B) 16
(C) 18
(D) 20
(E) 24
Rationale: if i were to choose during the test, would go for 18 or 20. probably 18 cuz it wont take too long to get the value doubled....
I found a method: Rule of 72.
Given an x% return, it takes 10,000 to quadralope 12 years.
So according to the rule: 72/x is the no of years 10,000.00 took to double 20,000.00.
Again, 20,000.00 took to double 40,000.00 same (72/x) no of years.
72/x+ 72/x = 12
x = 12% (though rate here is not very much required).
Again, 40,000.00 takes the same (72/x) no of years to double 80,000.00.
72/x = 6 years.
So altogather:
10,000 - 20,000 = 6 years
20,000 - 40,000 = 6 years
40,000 - 80,000 = 6 years
80,000 - 160,000 = 6 years
total 24 years.
Answer E
Answer: (E) 24
Question: An article is bought for Rs.875 and sold for Rs.900, find the gain percent?
(A) 73 1/3%
(B) 63 1/3%
(C) 33 1/3%
(D) 2.8%
(E) 56.5%
Rationale: Explanation:
875 ---- 25
100 ---- ? => 2.8%
Answer:D
Answer: (D) 2.8%
Question: A company X spent 1/5 of its investment on machinery and 1/3 of the remaining on the promotion of its product. How much fraction of the investment is the company left with after purchasing machinery and advertising?
(A) 2/10
(B) 8/15
(C) 4/6
(D) 3/5
(E) 10/12
Rationale: A company spent 1/5 of its investment on machinery, 4/5 is remaining
1/3 of the remaining investment is spent on advertising, 1/3*4/5=4/15
Total amount spent by the company for machinery and advertising is 1/5+4/15
To make the denominator common we use 15, 3/15
3/15+4/15=7/15 was the total amount spent by the company X for machinery and advertising.
Remaining is 8/15 of the investment is left with the company
Answer is B
Answer: (B) 8/15
Question: For any 6 given numbers, which of the following is always equivalent to adding the 3 numbers together and then dividing the sum by 3?
I. Ordering the 3 numbers numerically, from highest to lowest, and then selecting the middle number.
II. Dividing each of the numbers by 3 and then adding the results together.
III. Multiplying each number by 6, adding the resulting products together, and then dividing the sum by 9.
(A) I only
(B) II only
(C) I and II only
(D) II and III only
(E) None of the above
Rationale: Let the 3 numbers be x, y and z.
Question: Which of the following is equivalent to (x + y + z)/3
I. Ordering the 3 numbers numerically, from highest to lowest, and then selecting the middle number. --> Not always equivalent Eg: 111 and 141
II. Dividing each of the numbers by 3 and then adding the results together. --> x/3 + y/3 + z/3 = (x + y + z)/3 --> Equivalent
III. Multiplying each number by 6, adding the resulting products together, and then dividing the sum by 9. --> (6x + 6y + 6z)/9 = 2(x + y + z)/3 --> Not equivalent
Answer: D
Answer: (D) II and III only
Question: A bag contains ten marbles of the same size: 3 are identical green marbles, 2 are identical red marbles, and the other 5 are five distinct colors. If 5 marbles are selected at random, how many distinct combinations E of five marbles could be drawn?
(A)  41
(B)  51
(C)  62
(D)  72
(E)  82
Rationale: there will be different cases, which we will take one by one..
1)case 1 when all five colours are different-7c5=21..
2)case 2 when three different colours and 2 of same type=6c3*2=30..
3)case 3 when two different colours and 3 of same type=6c2=15..
4)case 4 when one different colours and 2 of two different colors=5c1*4!/2!2!=15...
5)case 4 when 3 of one type and 2 of other colour=1...
total E-82 ans E
Answer: (E)  82
Question: In how many different ways can 3 identical orange hats and 3 identical purple hats be distributed among 6 children such that each child receives a hat?
(A) 215
(B) 20
(C) 997952
(D) 6169
(E) 241
Rationale: 1st Child: 6 has options
2nd Child: 5 has options…
Therefore, for all kids: 6 x 5 x 4 x 3 x 2 = 720 arrangements.
Since the orange are identical, we divide by 3!; Since the purple are identical, we divide by another 3!
So: in all, 720/[ 3! X 3! ] = 20 ways.
(e.g. Answer: B)
Answer: (B) 20
Question: If the volume of two cubes are in the ratio 27:8, the ratio of their edges is:
(A) 3:1
(B) 3:2
(C) 3:5
(D) 3:7
(E) None of these
Rationale: Explanation:
Let the edges be a and b of two cubes, then
a3/b3=27/8
=>(a/b)3=(3/2)3
a/b=3/2
=>a:b=3:2
Option B
Answer: (B) 3:2
Question: A cistern of capacity 8000 litres measures externally 3.3 m by 2.6 m by 1.5 m and its walls are 5 cm thick. The thickness of the bottom is:
(A) 90 cm
(B) 5 dm
(C) 1 m
(D) 1.1 cm
(E) None of these
Rationale: Explanation:
Let the thickness of the bottom be x cm.
Then , [(330 - 10) × (260 - 10) × (150 - x)] = 8000 × 1000
=> 320 × 250 × (150 - x) = 8000 × 1000
=> (150 - x) = 8000×1000/320=
100
=> x = 50 cm = 5 dm.
Answer: B
Answer: (B) 5 dm
Question: If a = -0.6, which of the following is true?
(A) a < a^2 < a^3
(B) a < a^3 < a^2
(C) a^2 < a < a^3
(D) a^3 < a < a^2
(E) a^2 < a^3 < a
Rationale: a=-0.6;
a^2=0.36;
a^3=-0.216;
Answer : B
Answer: (B) a < a^3 < a^2
Question: Robin is traveling from one end of a forest to the other. In order to find his way back, he is leaving morsels of bread in the following pattern: 2 morsels of waffle, 3 morsels of rye, and 1 morsel of white. The pattern repeats after he leaves the morsel of white. If Robin drops 2,000 morsels of bread, what are the last 3 morsels of bread that he drops?
(A) white − waffle − waffle
(B) waffle − waffle − rye
(C) rye − white − waffle
(D) rye − rye − rye
(E) rye − rye − white
Rationale: In a single round Robin drops 6 morsels.
Remainder (2000/6) = 2
Final action = 2 drops of waffle --> Options B, C, D and E can be eliminated
Answer: A
Answer: (A) white − waffle − waffle
Question: The average of 35 numbers is 25. If each number is multiplied by 5, find the new average?
(A) 125
(B) 189
(C) 297
(D) 267
(E) 298
Rationale: Sum of the 35 numbers = 35 * 25 = 875
If each number is multiplied by 5, the sum also gets multiplied by 5 and the average also gets multiplied by 5.
Thus, the new average = 25 * 5
= 125.
Answer:A
Answer: (A) 125
Question: What is the product of all positive factors of 15?
(A) 225
(B) 324
(C) 1521
(D) 3042
(E) 5832
Rationale: Positive factors of 15 are 1, 3, 5, 15.
So the product is: 15*5*3 = 225.
Answer: A
Answer: (A) 225
Question: Excluding stoppages, the average speed of a bus is 60 km/hr and including stoppages, the average speed of the bus is 20 km/hr. For how many minutes does the bus stop per hour?
(A) 22
(B) 88
(C) 77
(D) 40
(E) 99
Rationale: In 1hr, the bus covers 60 km without stoppages and 20 km with stoppages.
Stoppage time = time take to travel (60 - 20) km i.e 40 km at 60 km/hr.
stoppage time = 40/60 hrs = 40 min.
Answer: D
Answer: (D) 40
Question: 10 play kabadi, 20 play kho kho only, 5 play both gmaes. Then how many in total?
(A) 30
(B) 35
(C) 38
(D) 40
(E) 45
Rationale: 10 play kabadi =>n(A)=10,
5 play both gmaes.=> n(AnB)=5
20 play kho kho only,=>n(B)=n(B only)+n(AnB)=20+5=25
Total => n(AUB)= n(A)+n(B)-n(AnB)
= 10+25-5
= 30
ANSWER:A
Answer: (A) 30
Question: four concentric circles r given .the radius of 1st circle is x.next is 2x,then 3x and 4x. given that area b/w 2nd and 1st is A and 4th and 3rd circles being B.find ratio of A to B
(A) 2:7
(B) 1:7
(C) 3:5
(D) 3:7
(E) 3:4
Rationale: area of first circle = pi*(x)^2 = pi*x^2
area of second circle = pi*(2x)^2 = 4*pi*x^2
area of third circle = pi*(3x)^2 = 9*pi*x^2
area of fourth circle = pi*(4x)^2 = 16*pi*x^2
area b/w 1st and 2nd circles = 3*pi*x^2
area b/w 3rd and 4th circles = 7*pi*x^2
so ratio = 3:7
ANSWER:D
Answer: (D) 3:7
Question: A train is 385 meter long is running at a speed of 45 km/hour. In what time will it pass a bridge of 140 meter length
(A) 42 seconds
(B) 30 seconds
(C) 40 seconds
(D) 50 seconds
(E) None of these
Rationale: Explanation:
Speed = 45 Km/hr = 45*(5/18) m/sec
= 25/2 m/sec
Total distance = 385+140 = 525 meter
Time = Distance/speed
=525∗2/25=42seconds
Option A
Answer: (A) 42 seconds
Question: For all integers a and b, a%b = (a + b)(a - b). If 5%x = 21, then which of the following could be a value of x?
(A) -4
(B) -3
(C) 2
(D) 3
(E) 6
Rationale: 5%x = (5+x)(5-x)
25-x^2 = 21
x^2 = 4
x=-2,2
Ans: C
Answer: (C) 2
Question: A and B invests Rs.3000 and Rs.4000 respectively in a business. If A doubles his capital after 6 months. In what ratio should A and B divide that year's profit?
(A) 9:6
(B) 9:8
(C) 9:0
(D) 9:2
(E) 9:1
Rationale: (3*6 + 6*6): (4*12)
54:48 => 9:8
Answer: B
Answer: (B) 9:8
Question: The speed of a boat in upstream is 40 kmph and the speed of the boat downstream is 60 kmph. Find the speed of the boat in still water and the speed of the stream?
(A) 50,10 kmph
(B) 78,10 kmph
(C) 70,70 kmph
(D) 78,10 kmph
(E) 70,12 kmph
Rationale: Speed of the boat in still water
= (40+60)/2
= 50 kmph. Speed of the stream
= (60-40)/2
= 10 kmph.
Answer:A
Answer: (A) 50,10 kmph
Question: The ratio of the ages of three people is 4:6:7. Six years ago, the sum of their ages was 67. Find their present ages.
(A) 12,18,21
(B) 16,24,28
(C) 20,30,35
(D) 24,36,42
(E) 28,42,49
Rationale: Let the present ages be 4x, 6x, 7x.
(4x-6) + (6x-6) + (7x-6) = 67
17x = 85
x = 5
Their present ages are 20, 30, and 35.
The answer is C.
Answer: (C) 20,30,35
Question: A sells his goods 50% cheaper than B but 50% dearer than C. The cheapest is?
(A) 33.8
(B) 33.7
(C) 33.3
(D) 33.2
(E) 33.1
Rationale: Let B = 100
A = 50
C * (150/100) = 50
3C = 100
C = 33.3 then 'C' Cheapest
Answer:C
Answer: (C) 33.3
Question: P,Q and R together earn Rs.1980 in 9 days. P and R can earn Rs.600 in 5 days. Q and R in 7 days can earn Rs.910. How much amount does R can earn per day?
(A) s.30
(B) s.70
(C) s.90
(D) s.100
(E) s.120
Rationale: Explanation :
Amount Earned by P,Q and R in 1 day = 1980/9 = 220 ---(1)
Amount Earned by P and R in 1 day = 600/5 = 120	---(2)
Amount Earned by Q and R in 1 day = 910/7 = 130	---(3)
(2)+(3)-(1) => Amount Earned by P , Q and 2R in 1 day
- Amount Earned by P,Q and R in 1 day = 120+130-220 = 30
=>Amount Earned by R in 1 day = 30
Answer : Option A
Answer: (A) s.30
Question: In John's School, there are 80boys of age 14 each, 70boys of age 15 each & 50boys of age 13 each and another 60boys of age 12each. The average age of the boys of the school is
(A) 13
(B) 13.6
(C) 13.8
(D) 14.6
(E) 15.7
Rationale: Average age of boys = Sum of the ages of all boys/ Total number of boys
= (80 x14)+(70x15)+(50x13)+(60x12)/(80+70+50+60)
= (1120+1050+650+720)/260
=3540/260 =13.6
B
Answer: (B) 13.6
Question: The ratio of the arithmetic mean of two numbers to one of the numbers is 4:7. What is the ratio of the smaller number to the larger number?
(A) 1:8
(B) 1:7
(C) 1:6
(D) 1:5
(E) 1:4
Rationale: For two numbers, the arithmetic mean is the middle of the two numbers.
The ratio of the mean to the larger number is 4:7, thus the smaller number must have a ratio of 1.
The ratio of the smaller number to the larger number is 1:7.
The answer is B.
Answer: (B) 1:7
Question: A cistern can be filled by a tap in 3 hours while it can be emptied by another tap in 9 hours. If both the taps are opened simultaneously, then after how much time will the cistern get filled ?
(A) 4.5 hrs
(B) 5 hrs
(C) 6.5 hrs
(D) 7.2 hrs
(E) 9.2 hrs
Rationale: Net part filled in 1 hour= 1/3 -1/9= 2/9
Therefore the cistern will be filled in 9/2 hours or 4.5 hours.
ANSWER:A
Answer: (A) 4.5 hrs
Question: A sum of Rs.21000 is divided into three parts such that the simple interests accrued on them for three, four and five years respectively may be equal. Find the amount deposited for 5 years.
(A) 13000
(B) 11000
(C) 16000
(D) 14000
(E) 12000
Rationale: Let the amounts be x, y, z in ascending order of value. As the interest rate and interest accrued are
same for 3 years 4 years and 5 years i.e. 3x = 4y = 5z = k.
L.C.M. of 3,4,5 = 60 So x:y:z: = 20000 : 15000 :12000
The amount deposited for 5 years = 12000
ANSWER:E
Answer: (E) 12000
Question: Kate and Danny each have $10. Together, they flip a fair coin 5 times. Every time the coin lands on heads, Kate gives Danny $1. Every time the coin lands on tails, Danny gives Kate $1. After the five coin flips, what is the probability Q that Kate has more than $10 but less than $15?
(A) 5/16
(B) 1/2
(C) Q=12/30
(D) Q=15/32
(E) 3/8
Rationale: For 21 heads Kate will end up with < $10 and we want her to win. Therefore, only possibilities are 3 or 4 heads.
I made an educated guess and it worked fine.
Ans- 'D'
MGMAT's anagram helped here as well.
HHHHT = 5!/4!*1! = 5
HHHTT = 5!/3!*2! = 10
Total acceptable cases = 15
Total cases = 32
P = 15/32.D
Answer: (D) Q=15/32
Question: Three unbiased coins are tossed. What is the probability of getting at least 2 heads?
(A) 1/4
(B) 1/2
(C) 1/3
(D) 1/8
(E) 1/6
Rationale: Here S={TTT, TTH, THT, HTT, THH, HTH, HHT, HHH}
Let E=event of getting at least two heads = {THH, HTH, HHT, HHH}
P(E) = n(E)/n(S) = 4/8 = 1/2
Answer B.
Answer: (B) 1/2
Question: On a school’s Annual day sweets were to be equally distributed amongst 112 children. But on that particular day, 32 children were absent. Thus the remaining children got 6 extra sweets. How many sweets was each child originally supposed to get ?
(A) 17
(B) 18
(C) 29
(D) 15
(E) 19
Rationale: Let 'K' be the total number of sweets.
Given total number of students = 112
If sweets are distributed among 112 children,
Let number of sweets each student gets = 'L'
=> K/112 = L ....(1)
But on that day students absent = 32 => remaining = 112 - 32 = 80
Then, each student gets '6' sweets extra.
=> K/80 = L + 6 ....(2)
from (1) K = 112L substitute in (2), we get
112L = 80L + 480
32L = 480
L = 15
Therefore, 15 sweets were each student originally supposed to get.
Answer: D
Answer: (D) 15
Question: If one person completes a journey in 10 hrs. He travels 1st half of the journey at the rate of 21 km/hr and 2nd half at therate of 24 km/hr. Find the total journey in km.
(A) 200 km
(B) 212 km
(C) 224 km
(D) 230 km
(E) 256 km
Rationale: distance = speed x time
Let time taken to travel the first half = x hr
then time taken to travel the second half = (10 - x) hr
Distance covered in the the first half = 21x
Distance covered in the the second half = 24(10 - x)
But distance covered in the the first half = Distance covered in the the second half
=> 21x = 24(10 - x)
=> 21x = 240 - 24x
=> 45x = 240
=> 9x = 48
=> 3x = 16
⇒x=163Hence Distance covered in the the first half = 21x=21×163=7×16=112 kmTotal distance = 2×112=224 km
C
Answer: (C) 224 km
Question: The circumferences of two circles are 264 meters and 352 meters. Find the difference between the areas of the larger and the smaller circles.
(A) 388
(B) 277
(C) 4312
(D) 2887
(E) 2992
Rationale: Let the radii of the smaller and the larger circles be s m and l m respectively.
2∏s = 264 and 2∏l = 352
s = 264/2∏ and l = 352/2∏
Difference between the areas = ∏l2 - ∏s2
= ∏{1762/∏2 - 1322/∏2}
= 1762/∏ - 1322/∏
= (176 - 132)(176 + 132)/∏
= (44)(308)/(22/7) = (2)(308)(7) = 4312 sq m.Answer: C
Answer: (C) 4312
Question: Four staff members at a certain company worked on a project. The amounts of time that the four staff members worked on the project were in the ratio 2 to 3 to 5 to 6. If one of the four staff members worked on the project for 150 hours, which of the following CANNOT be the total number of hours that the four staff members worked on the project?
(A) 800
(B) 1200
(C) 480
(D) 192
(E) 400
Rationale: Four members worked in ration 2:3:5:6, hence as everyone mentioned, individual work could be taken as 2x, 3x,5x, and 6x. Also this gives us total work as 16x.
But we are told that one of these individual works is 30hrs. hence, possible scenarios, if
(1)2x =150 => 16x = 1200 (2) 3x =150 => 16x = 800 (3) 5x =150 => 16x = 480 (4) 6x =150 => 16x = 400
Hence Answer is D 192 which can not be any of these.
Answer: (D) 192
Question: 145 liters of a mixture of milk and water contains in the ratio 3:2. How much water should now be added so that the ratio of milk and water becomes 3:4?
(A) 12 liters
(B) 32 liters
(C) 41 liters
(D) 50 liters
(E) 58 liters
Rationale: Milk = 3/5 * 145 = 87 liters
Water = 58 liters
87 : (58+P) = 3:4
174 + 3P = 348 => P = 58
58 liters of water are to be added for the ratio become 3:4.
Answer:E
Answer: (E) 58 liters
Question: In a company 200 persons, working 6 hours per day complete 1/3 rd of the work in 7 days. Employee number of additional persons, working 4 hrs per day, required to complete the remaining work in 15 days is :
(A) 60
(B) 80
(C) 90
(D) 70
(E) 50
Rationale: Explanation :
Solution: Let the number of persons completing the work in 15 days be x.
Work done = 1/3, remaining work = (1-1/3) = 2/3.
less hours per day, more men required (IP)
more work, more men required (DP)
more days, less men required (IP)
hours per day 4 : 6
work 1/3 : 2/3 } : : 200 : x
days 15 : 7
.'. 4*1/3 * 15*x = 6 * 2/3 * 7 * 200
=> 60x = 16800
=> x = 16800/60 = 280.
Additional men required = (280-200) = 80.
Answer : B
Answer: (B) 80
Question: Which oif the following is a prime number?
(A) 45
(B) 39
(C) 27
(D) 33
(E) 31
Rationale: Clearly, 31 is a prime number. So choice E is correct.
Answer: (E) 31
Question: Find the odd man out. 24,36,52,72,96
(A) 52
(B) 72
(C) 36
(D) 24
(E) 96
Rationale: All except 52 are multiples of 6
Answer is A
Answer: (A) 52
Question: A village bank offers 5% compound interest calculated on half-yearly basis. A regular customer deposits Rs. 1600 each on 1st January and 1st July of a year. At the end of the year, the amount the customer would have gained by the way of interest is:
(A) 121
(B) 130
(C) 125
(D) 135
(E) 140
Rationale: after 1 yr Rs 1600 at 5% at half yearly
P(1+(R/2)/100)^2T = 1600(1+(5/2)/100)^2*1 = 1600(1+1/40)^2--1
amt after 1/2 yr on Rs 1600 at 5%
==> 1600(1+(5/2)/100)^2*1/2 = 1600(1+1/40)--2
total amt after 1 yr from 1 and 2
1600(41/40)(81/40) = 41* 81 = 3321
Comp int = 3321- 3200 = Rs 121
ANSWER A
Answer: (A) 121
Question: The cost price of 12 articles is equal to the selling price of 8 articles. What is the profit percent?
(A) 40%
(B) 45%
(C) 50%
(D) 55%
(E) 60%
Rationale: 8*SP = 12*CP
SP = 1.5*CP
The profit percent is 50%.
The answer is C.
Answer: (C) 50%
Question: A box contains 12 pairs of shoes (24 shoes in total). If two shoes are selected at random, what it is the probability that they are matching shoes?
(A) 1/190
(B) 1/23
(C) 1/19
(D) 1/10
(E) 1/9
Rationale: The problem with your solution is that we don't choose 1 shoe from 24, but rather choose the needed one AFTER we just took one and need the second to be the pair of it. So, the probability would simply be: 1/1*1/23(as after taking one at random there are 23 shoes left and only one is the pair of the first one)=1/23
Answer: B
Answer: (B) 1/23
Question: In a group of ducks and cows, the total number of legs are 20 more than twice the no. of heads. Find the total no.of buffaloes.
(A) 10
(B) 12
(C) 13
(D) 15
(E) 16
Rationale: Let the number of buffaloes be x and the number of ducks be y
=> 4x + 2y = 2 (x + y) + 20
=> 2x = 20 => x = 10
A
Answer: (A) 10
Question: Two spherical balls lie on the ground touching. If one of the balls has a radius of 5 cm, and the point of contact is 7 cm above the ground, what is the radius of the other ball?
(A) 10 cm
(B) 25/2 cm
(C) 35/3 cm
(D) 15 cm
(E) none of the these
Rationale: similar triangle properties..
2/r+5 = 5/r-5
giving r = 35/3.
Answer: C
Answer: (C) 35/3 cm