RikoteMaster/aqua-rat-mcqa huggingface dataset:
Question: Timothy drove at an average speed of 10 mph for some time and then at an average speed of 50 mph for the rest of the journey. If he made no stops during the trip and his average speed for the entire journey was 40 mph, for what fraction of the total time did he drive at 50 mph?
A. 1/5
B. 1/3
C. 2/5
D. 3/4
Answer: D. 3/4
Explanation: We don't need to get into calculations for solving this question. We can use the concept of weighted averages.
We know that if the average speed for the entire journey was 40 mph, then he drove at 50 mph for a longer duration of time.
10 30
10--40------50
This shows that you can divide the entire journey into 4 equal parts. Thus, 3/4 part he drove at 50 mph and 1/4 part he drove at 10 mph.
Answer: D
Question: 1/2+[(2/3*3/8)+2]-9/16=
A.35/16
B.19/16
C.15/16
D.9/13
Answer: A.35/16
Explanation: 1/2+[(2/3*3/8)+2]-9/16=
1/2+[(1/4)+2]-9/16=
1/2+[9/4]-9/16=
8/16 + 36/16 - 9/16 = 35/16
A
Question: The mass of 1 cubic meter of a substance is 500 kilograms under certain conditions. What is the volume, in cubic centimeters, of 1 gram of this substance under these conditions? (1 kilogram = 1,000 grams and 1 cubic meter = 1,000,000 cubic centimeters)
A.0.8
B.1.25
C.2.0
D.2.5
Answer: C.2.0
Explanation: Density is mass divided by volume. So density of the given substance will be mass/volume=500kg/1m^3=800 kg/m^3 or 1g/1.5cm^3=0.5 g/cm^3.
Next, ask yourself if 500,000 g is equivalent to 1,000,000 cubic centimeters then 1 g is equivalent to how many cubic centimeters? --> 1 g - 1,000,000/500,000 = 10/5 = 2.0 cubic centimeters.
Answer is C
Question: In May, the grounds keeper at Spring Lake Golf Club built a circular green with an area of 25 square feet. In August, he doubled the distance from the center of the green to the edge of the green. What is the total area of the renovated green?
A.100PI
B.80
C.70
D.60
Answer: A.100PI
Explanation: AREA OF CIRCLE 25PI SQ FEET =PI R^2 THEREFORE R =5
NOW GREEN RADI DOUBLED I.E R =10 AREA =100 PI
A
Question: A man performs 1/2 of the total journey by rail, 1/3 by bus and the remaining 4km on foot. His total journey is
A.16km
B.10km
C.12km
D.24km
Answer: D.24km
Explanation: Explanation:
Let the journey be x km
Then, 1x/2 + 1x/3 +4 =x
5x +24 =6x
X = 24km
Answer: Option D
Question: The product of all the prime numbers less than 40 is closest to which of the following powers of 10 ?
A. 10^9
B. 10^8
C. 2*10^7
D. 10^6
Answer: C. 2*10^7
Explanation: 2*3*5*7*11*13*17*19
2*19*3*17*5*13*7*11
38*51*65*77
By approximation
40*50*70*70
2000*4900
By approximation
4000*5000
20,000,000
C
Question: If a and b are the roots of the equation x2 - 6x + 6 = 0, then the value of a2 + b2 is:
A.36
B.24
C.17
D.6
Answer: B.24
Explanation: Sol.(b) The sum of roots = a+ b = 6
Product of roots = ab = 6
Now, a2 + b2 = (a + b)2 - 2ab = 36 - 12 = 24
Answer B
Question: The value of x + x(xx) when x = 5
A.A) 120
B.B)130
C.C) 110
D.D) 316
Answer: B.B)130
Explanation: x + x(xx)
Put the value of x = 5 in the above expression we get,
5 + 5(55)
= 5 + 5(5 Ã— 5)
= 5 + 5(25)
= 5 + 125
= 130 The answer is (B)
Question: Find the value of a/b + b/a, if a and b are the roots of the quadratic equation x2 + 9x + 4 = 0?
A.17
B.88
C.14
D.65
Answer: B.88
Explanation: a/b + b/a = (a2 + b2)/ab = (a2 + b2 + a + b)/ab
= [(a + b)2 - 2ab]/ab
a + b = -9/1 = -9
ab = 4/1 = 4
Hence a/b + b/a = [(-9)2 - 2(4)]/4 = 73/4 = 18.25.
B)
Question: The average of five numbers is 27. If one number is excluded, the average becomes 25. The excluded number is?
A.8
B.6
C.58
D.35
Answer: D.35
Explanation: Excluded number
= (27 * 5) - (25 * 4)
= 135 - 100 = 35.
Answer:D
Question: A train crosses a post in 15 seconds and a platform 100 m long in 25 seconds. Its length is
A.150 m
B.300 m
C.400 m
D.180 m
Answer: A.150 m
Explanation: Explanation:
Assume x is the length of the train and v is the speed
x/v = 15 => v = x/15
(x+100)/v = 25 => v = (x+100)/25
Ie, x/15 = (x+100)/25
=> 5x = 3x+ 300
=> x = 300/2 = 150
Answer: Option A
Question: A reduction of 20% in the price of salt enables a lady to obtain 10kgs more for Rs.100, find the original price per kg?
A.Rs.3.2
B.Rs.2.9
C.Rs.2.5
D.Rs.5.2
Answer: C.Rs.2.5
Explanation: 100*(20/100) = 20 --- 10
? --- 1 => Rs.2
100 --- 80
? --- 2 => Rs.2.5
Answer:C
Question: In a kilometer race, A beats B by 50 meters or 10 seconds. What time does A take to complete the race?
A.238
B.190
C.127
D.279
Answer: B.190
Explanation: Explanation:
Time taken by B run 1000 meters = (1000 * 10)/50 = 200 sec.
Time taken by A = 200 - 10 = 190 sec.
Answer: B
Question: The sum of two numbers is 45. Their difference is 5. The bigger number is
A.20
B.25
C.30
D.35
Answer: B.25
Explanation: X + Y = 45
X - Y = 5
Adding 2X = 50 ==> X = 25 and Y = 45 - 25 = 20
Bigger number is X = 25
Answer B.
Question: The population of a town is 8000. It decreases annually at the rate of 10% p.a. What will be its population after 1 years?
A.5100
B.5120
C.7200
D.5400
Answer: C.7200
Explanation: Formula :
( After =100 denominator
Ago = 100 numerator)
8000 Ã— 90/100= 7200
Answer : C
Question: A soft drink company had 6000 small and 14000 big bottles in storage. If 20% of small 23% of big bottles have been sold, then the total bottles remaining in storage is
A.15360
B.15720
C.15060
D.14930
Answer: B.15720
Explanation: 6000 + 14000 - (0.2*6000 + 0.22*14000) = 15720.
Answer: B.
Question: Find the odd man out
4, 10, 22, 40, 84, 94
A.22
B.40
C.94
D.84
Answer: D.84
Explanation: Differnce is multiple of 6
4 and 10 diff is 6
10 and 22 diff is 12
22 and 40 diff is 18
40+24 = 64 but in series it is mentioned as 84
64+30 = 94
ANSWER:D
Question: Three times the first of three consecutive odd integers is 3 more than twice the third. The third integer is :
A.9
B.11
C.13
D.15
Answer: D.15
Explanation: let the three integers be x, x+2, x+4
3x = 2(x+4) + 3 ==> 3x = 2x + 8 +3 ==> x = 11
so the third integer is x+4 = 15.
so the correct option is D.
Question: A train running at the speed of 60 km/hr crosses a pole in 12 seconds. What is the length of the train?
A.120 metres
B.180 metres
C.324 metres
D.200 metres
Answer: D.200 metres
Explanation: Speed=(60 x 5/18)/m/sec	=	50/3	m/sec.
Length of the train = (Speed x Time).
Length of the train =	50 x 12/3 m = 200 m.
ANSWER : OPTION D
Question: A person covered one-fourth of the total distance at 26 kmph and remaining distance at 24 kmph. What is the average speed for the total distance?
A.21 (1/3)kmph
B.22 (1/3)kmph
C.24 (1/3)kmph
D.26 (1/3)kmph
Answer: A.21 (1/3)kmph
Explanation: A
21 (1/3)kmph
Let the total distance be x km
total time taken = (x/4)/16 + (3x/4)/24 = x/64 + x/32 = 3x/64
Average speed = x/(3x/64) = 64/3kmph = 21 (1/3)kmph.
Question: The ratio of incomes of two person P1 and P2 is 5 : 4 and the ratio of their expenditures is 3 : 2. If at the end of the year, each saves Rs.1200, then what is the income of P1?
A.s.800
B.s.2400
C.s.3000
D.s.3200
Answer: C.s.3000
Explanation: Let the income of P1 and P2 be Rs. 5x and Rs.4x respectively and let their
expenditures be Rs.3y and 2y respectively.
Then, 5x – 3y = 1200 …(i) and 4x – 2y = 1200 ……..(ii)
On multiplying (i) by 2, (ii) by 3 and subtracting, we get : 2x = 1200 -> x = 600
P1’s income = Rs 5*600 = Rs.3000
ANSWER:C
Question: If w/x = 1/3 and w/y = 2/3, then (x + y)/y =
A.2
B.4
C.5
D.3
Answer: D.3
Explanation: Ratio 1:
3w = x
Ratio 2:
3w = 2y
x = 2y
(x+y)/y = (2y+y)/y = y(2+1)/y = 3
Answer is D
Question: Mike, Jim and Bob are all professional fisherman. Mike can catch 15 fish in one hour, Jim can catch twice as much and Bob can catch 50% more than Jim. If the three started to fish together and after 40 minutes Mike and Bob left, how many fish did the three fishermen catch in one hour?
A.64
B.70
C.85
D.98.
Answer: B.70
Explanation: All of them catch fishes in relation to number 15....
2/3*15+2*15+2*1.5*15*2/3=70
Answer is B
Question: If a*b*c=130, b*c*d = 65, c*d*e=1000 and d*e*f=150 the (a*f)/(c*d) = ?
A.1/2
B.1/4
C.3/4
D.1/3
Answer: D.1/3
Explanation: Explanation :
aâˆ—bâˆ—c/bâˆ—câˆ—d= 130/65 => a/d = 2
dâˆ—eâˆ—f/câˆ—dâˆ—e= 150/1000 => f/c = 1/4
a/d* f/c = 2 * 15/10 = 1/3
Answer : D
Question: Two unbiased coin are tossed .what is the probability of getting atmost one head?
A.1/3
B.3/4
C.5/7
D.8/9
Answer: B.3/4
Explanation: Here S={HH,HT,TH,TT}
Let Ee=event of getting one head
E={TT,HT,TH}
P(E)=n(E)/n(S)=3/4
Ans: B
Question: Anne bought doughnuts for a class breakfast party. She bought 9 chocolate doughnuts, 7 coconut doughnuts, and 9 jam-filled doughnuts. How many doughnuts did Anne buy in all?
A.25
B.36
C.39
D.21
Answer: A.25
Explanation: Add the numbers of doughnuts.
9 + 7 + 9 =25.
Answer is A.
Question: Four people are to be seated on a bench. How many different sitting arrangements are possible if Erik must sit next to Joe?
A.4
B.6
C.8
D.12
Answer: C.8
Explanation: Consider E and J one. But, in this arrangement they can sit in two different combinations.
E first and J second or J first and E second.
Total we have (E and J) and the two other persons= 4 arrangements to do.
Total arrangements= 4*2= 8
C is the answer
Question: If the sides of a rectangle are increased by 25%, what is the percentage increase in the area?
A.54.25%
B.56.25$
C.53.25%
D.55.25%
Answer: B.56.25$
Explanation: If sides are a and b, after increase sides would be 1.25a and 1.25 b.
Percentage increase in area = (1.25a*1.25b - ab) *100/ab = 56.25%
Answer : B
Question: The set S consists of 5 numbers: {1,2,3,4,5}. If all possible subsets including the null set are created and one subset is chosen at random, what is the probability that the subset has 3 as its largest number?
A.1/2
B.1/4
C.1/8
D.1/16
Answer: C.1/8
Explanation: The set S has 2^5 = 32 subsets.
The number 5 is in half of these subsets.
Of the remaining 16 subsets, 4 is an element in 8 of them.
Of the remaining 8 subsets, 3 is an element in 4 of them.
Thus 3 is the largest number in 4 subsets of S.
The probability that 3 is the largest number is 4/32 = 1/8.
The answer is C.
Question: The age of father 10 years ago was thirce the age of his son.Ten years hence,father’s age will be twice that of his son.The ration of their present ages is :
A.5:2
B.7:3
C.9:2
D.13:4
Answer: B.7:3
Explanation: Solution
Let the ages of father and son 10 year ago be 3x and x years respectively.
Then,(3x + 10)+10 = 2[(x +10)+10 ⇔ 3x + 20 = 2x + 40 ⇔ x =20.
∴ Required ratio =(3x +10) : (x + 10) = 70 : 30 : 7 : 3. Answer B
Question: Mary’s annual income is $15,000 and John’s annual income is $16,000. By how much must Mary’s annual income increase so that it constitutes 50% of Mary and John’s combined income?
A. $1,000
B. $4,000
C. $7,000
D. $11,000
Answer: A. $1,000
Explanation: Let Mary's income increase by x, then the equation will be
15,000 + x= (50/100)* (15,000 + x + 16000 )
15,000 + x = (1/2) * (31,000 + x)
30,000 + 2x = x + 31,000
x =1,000
So, Answer will be A
Question: The area of the largest circle that can be drawn inside a square of side 28 Cm in length is:
A.600
B.646
C.616
D.636
Answer: C.616
Explanation: Radius of required circle = 14 Cm
its area = 22/7 × 14 × 14
44 × 14 = 616 Cm(power)2
Answer is C.
Question: Find the curved surface area, if the radius of a cone is 7 m and slant height is 14 m?
A.308
B.369
C.434
D.476
Answer: A.308
Explanation: Cone curved surface area = Ï€rl
22/7 Ã— 7 Ã— 14 = 22 Ã— 14 = 308 m(power2)
Answer is A.
Question: Two brothers took the GMAT exam, the higher score is U and the lower one is V. If the difference between the two scores is equal to their average, what is the value of V/U ?
A.3
B.2
C.1/2
D.1/3
Answer: D.1/3
Explanation: Answer is D : 1/3
U - V = (U + V)/2
Solving for V/U = 1/3
D
Question: The sale price bags listed for Rs.2000 after successive discount is 25% and 15% is?
A.415
B.425
C.435
D.445
Answer: B.425
Explanation: 2000*(75/100)*(85/100)
= 425
Answer: B
Question: The perimeters of two squares are 36 cm and 32 cm. Find the perimeter of a third square whose area is equal to the difference of the areas of the two squares?
A.16.4 cm
B.24 cm
C.88 cm
D.55 cm
Answer: A.16.4 cm
Explanation: We know perimeter of square = 4(side)
So Side of first square = 40/4 = 10 cm
Side of second square = 36/4 = 9 cm
Area of third Square = 9*9 - 8*8
= 17 cm
So side of third square = 4.1 [because area of square = side*side]
Perimeter = 4*Side = 4*4.1 = 16.4 cm
Answer:A
Question: In a class of 28 students in an examination in maths 4 students scored 95 marks each, 3 get zero each and the average of the rest was 45. What is the average of the whole class?
A.A)47.3
B.B)45.6
C.C)44
D.D)48
Answer: A.A)47.3
Explanation: Explanation:
Total marks obtained by a class of 25 students
= 4*95 + 3*0 + 21 * 45 =1325
Average marks of whole class = 1325/28 =47.3
Answer: Option A
Question: Find the average marks of all the students in classes A and B, if :
classes A = 43 students (average marks of students is 65)
classes B = 54 students (average marks of students is 72)
A.41.9
B.68.9
C.65.9
D.61.9
Answer: B.68.9
Explanation: Sum of the marks for the class of 43 students = 43 * 65 = 2795
Sum of the marks for the class of 54 students = 54 * 72 = 3888
Sum of the marks for the class of 97 students =
2795 + 3888 = 6683
Average marks of all the students = 6683/97
= 68.9
Answer:B
Question: Amar takes as much time in running 18 meters as a car takes in covering 48 meters. What will be the distance covered by Amar during the time the car covers 1.2 km?
A.600 m
B.200 m
C.300 m
D.400 m
Answer: C.300 m
Explanation: C
300 m
Distance covered by Amar = 18/4.8 (1.6km) = 3/8(1200) = 300 m
Answer is C
Question: Danny and Steve are running towards each other, each one from his own house. Danny can reach Steve's house in 35 minutes of running, which is half the time it takes Steve to reach Danny's house. If the two started to run at the same time, how much time longer will it take Steve to reach the halfway point between their houses than Danny to reach the halfway point between their houses?
A.17.5 minutes.
B.35 minutes.
C.52.5 minutes.
D.70 minutes.
Answer: A.17.5 minutes.
Explanation: Danny's time is 35 minutes. The time to reach halfway is 17.5 minutes.
Steve's time is 70 minutes. The time to reach halfway is 35 minutes.
The time difference is 35 - 17.5 = 17.5 minutes
The answer is A.
Question: Pipes A and B can fill a tank in 5 and 6 hours respectively. Pipe C can empty it in 12 hours. If all the three pipes are opened together, then the tank will be filled in?
A.3
B.3 9/11
C.3 9/17
D.4
Answer: C.3 9/17
Explanation: Net part filled in 1 hour 1	+	1	-	1 =	17	.
5	6	12	60
The tank will be full in	60/17	hours i.e., 3 9/17	hours.
C)
Question: The telephone company wants to add an area code composed of 2 letters to every phone number. In order to do so, the company chose a special sign language containing 24 different signs. If the company used 22 of the signs fully and two remained unused, how many additional area codes can be created if the company uses all 24 signs?
A.246
B.248
C.92
D.15,128
Answer: C.92
Explanation: # of 2-letter codes possible from 24 different signs = 24*24.
# of 2-letter codes possible from 22 different signs = 22*22.
The difference = 24^2 - 22^2 = (24 - 22)(24 + 22) = 92.
Answer: C.
Question: Company T produces two kinds of stereos: basic and deluxe. Of the stereos produced by Company T last month, 2/3 were basic and the rest were deluxe. If it takes 7/5 as many hours to produce a deluxe stereo as it does to produce a basic stereo, then the number of hours it took to produce the deluxe stereos last month was what fraction of the total number of hours it took to produce all the stereos?
A.5/17
B.7/17
C.4/17
D.3/17
Answer: B.7/17
Explanation: # of basic stereos was 2/3 of total and # of deluxe stereos was 1/3 of total, let's assume total=15, then Basic=10 and Deluxe=5.
Now, if time needed to produce one Deluxe stereo is 1 unit than time needed to produce one Basic stereo would be 7/5 units. Total time for Basic would be 10*1=10 and total time for Deluxe would be 5*7/5=7 --> total time for both of them would be 10+7=17 --> deluxe/total=7/17.
B
Question: Andrew purchased 10kg of grapes at the rate of 82 per kg and 6 kg of mangoes at the rate of 62 per kg. How much amount did he pay to the shopkeeper?
A.1000
B.1055
C.1378
D.1075
Answer: C.1378
Explanation: Cost of 10 kg grapes = 82 × 10 = 820.
Cost of 6 kg of mangoes = 62 × 6 = 558.
Total cost he has to pay = 820 + 558 = 1378
C
Question: If log10 2 = 0.3010, then log2 10 is equal to:
A.1000/301
B.0.301
C.600/301
D.0.6992
Answer: A.1000/301
Explanation: log2 10 = 1/log10 2 = 1/0.3010 = 10000/3010 = 1000/301
ANSWER A
Question: Lionel left his house and walked towards Walt's house, 48 miles away. Two hours later, Walt left his house and ran towards Lionel's house. If Lionel's speed was 2 miles per hour and Walt's 6 miles per hour, how many miles had Lionel walked when he met Walt?
A.12
B.15
C.20
D.24
Answer: B.15
Explanation: In the first 2 hours Lionel at the rate of 2 miles per hour covered distance=rate*time=2*2=4 miles. So, the distance between him and Walt was 48-4=44 miles when Walt left his house. Now, their combined rate to cover this distance was 2+6=8 miles per hour, hence they will meet (they will cover that distance) in time=distance/rate=44/8=5.5 hours. Total time that Lionel was walking is 2+5.5=7.5 hours, which means that he covered in that time interval distance=rate*time=2*7.5=15 miles.
Answer: B.
Question: If by selling 110 apples, the cost price of 120 apples is realised. The gain % is?
A.9%
B.10%
C.11%
D.12%
Answer: C.11%
Explanation: 1 apple=1 rup
cp of 110 app=110 rs
sp of 110 app=120 rs
gain=120-110=10
gain%=10/100*110=11%
ANSWER:C
Question: If a:b=3:1 and b:c=1:4 find a:b:c?
A.1:6:2
B.3:1:4
C.2:3:4
D.2:3:3
Answer: B.3:1:4
Explanation: a:b = 3:1, b:c = 1:4
3:1
1:4
(a = a × b, b = b × b and c = b × c)
a:b:c = 3:1:4
B
Question: Anne earned $3 an hour baby-sitting, and $4 an hour working in the garden. Last week she did
baby-sitting for 5 hours and garden work for 3 hours. How much more money does she need to
buy a game that costs $35?
A.$8
B.$12
C.$6
D.$21
Answer: A.$8
Explanation: 5 x $3 = $15 for baby-sitting
3 x $4 = $12 for garden work
$15 + $12 = $27 she has
$35 - $27 = $8 more needed to
buy the game
correct answer A
Question: What is the CP of Rs 100 stock at 6 discount, with 1/5% brokerage?
A.96.9
B.96.3
C.94.2
D.96.7
Answer: C.94.2
Explanation: Explanation:
Use the formula,
CP= 100 â€“ discount + brokerage%
CP= 100-6+1/5
94.2
Thus the CP is Rs 94.2.
ANSWER: C
Question: Jack, Jill, and Sandy each have one try to make a basket from half court. If their individual probabilities of making the basket are 1/6, 1/7, and 1/8 respectively, what is the probability that all three will make a basket?
A.1/336
B.5/336
C.11/336
D.1/168
Answer: A.1/336
Explanation: The probability that all three will make a basket is 1/6*1/7*1/8=1/336.
The answer is A.
Question: A bag contains 12 white and 18 black balls. 2balls are drawn in succession. What is the probability that first is white and second is black?
A.21/112
B.36/145
C.3/115
D.2/115
Answer: B.36/145
Explanation: The probability that first ball is white:
=12C130C1=12C130C1
=1230=1230
=25=25
Since, the ball is not replaced; hence the number of balls left in bag is 29.
Hence, the probability the second ball is black:
=18C129C1=18C129C1
=1829=1829
Required probability,
=(25)×(1829)=(25)×(1829)
=36/145
B
Question: If a + xy = a and x is not equal to 0, which of the following must be true?
A.y=0
B.x>y
C.y<x
D.x-y=0
Answer: A.y=0
Explanation: a + xy = a --> xy=0. Since x is not equal to 0, thus y = 0.
Answer: A
Question: On Saturday morning, John will begin a camping vacation and he will return home at the end of the first day on which it rains. If on the first three days of the vacation the probability of rain on each day is 0.4, what is the probability that Malachi will return home at the end of the day on the following Monday?
A.0.008
B.0.125
C.0.144
D.0.512
Answer: C.0.144
Explanation: Re-phrasing the question:
What is the probability of: Saturday: No rain, Sunday: No rain, Monday: Rain
Probability of rain = 0.4
Therefore, probability of no rain = 0.6
0.6*0.6*0.4 = 0.144
Hence C
Question: If four cards are drawn at random from a well shuffled pack of cards, what is the probability that each card is an ace?
A.6/52C4
B.4/52C4
C.1/52C4
D.3/52C4
Answer: C.1/52C4
Explanation: Four cards can be drawn from a pack in 52C4 ways.
Let E be the event of each card being an ace.
This can be done in 4C4 i.e, 1 way.
So P(E) = 1/52C4
ANSWER:C
Question: A football team lost 5 yards and then gained 9. What is the team's progress?
A.1
B.3
C.2
D.4
Answer: D.4
Explanation: For lost, use negative. For gain, use positive.
Progress = -5 + 9 = 4 yards
correct answer is D)4
Question: Three candidates contested an election and received 1136, 7636 and 11628 votes respectively. What percentage of the total votes did the winning candidate go?
A.57%
B.60%
C.65%
D.90%
Answer: A.57%
Explanation: Solution
Total no of votes polled
= ( 1136 + 7636 + 11628)
= 20400.
Required percentage	= [11628 / 20400 x 100]%
= 57%.
Answer A
Question: Machine P can print one lakh books in 8 hours. Machine Q can print the same number of books in 10 hours while machine R can print the same in 12 hours. All the machines started printing at 9 A.M. Machine P is stopped at 11 A.M. and the remaining two machines complete work. Approximately at what time will the printing of one lakh books be completed?
A.3 pm
B.2 pm
C.1:00 pm
D.11 am
Answer: C.1:00 pm
Explanation: Explanation :
Work done by P in 1 hour = 1/8
Work done by Q in 1 hour = 1/10
Work done by R in 1 hour = 1/12
Work done by P,Q and R in 1 hour = 1/8 + 1/10 + 1/12 = 37/120
Work done by Q and R in 1 hour = 1/10 + 1/12 = 22/120 = 11/60
From 9 am to 11 am, all the machines were operating.
Ie, they all operated for 2 hours and work completed = 2 × (37/120) = 37/60
Pending work = 1- 37/60 = 23/60
Hours taken by Q an R to complete the pending work = (23/60) / (11/60) = 23/11
which is approximately equal to 2
Hence the work will be completed approximately 2 hours after 11 am ; ie around 1 pm
Answer : Option C
Question: The average weight of a group of persons increased from 48 kg to 51 kg, when two persons weighing 78 kg and 93 kg join the group. Find the initial number of members in the group?
A.87
B.67
C.23
D.75
Answer: C.23
Explanation: Let the initial number of members in the group be n.
Initial total weight of all the members in the group = n(48)
From the data,
48n + 78 + 93 = 51(n + 2) => 51n - 48n = 69 => n = 23
Therefore there were 23 members in the group initially.
Answer: C
Question: The value for A=(0.889 × 55)/9.97 to the nearest tenth is
A.0.5
B.4.63
C.4.9
D.7.7
Answer: C.4.9
Explanation: if we read the Q, we can easily home on to the answer,,
the numerator is clearly between 40 and 50..
denomiator is close to 10..
so A=(0.889 × 55)/9.97 is between 4 and 5..
It may tempt us to solve it since ther are two values between 4 and 5..
but the catch is innearest tenth
4.63 can be called nearest to hundreth and no tenth, so can be eliminated..
4.9 is our answer...
C
Question: A man took a loan at rate of 12% per annum simple interest. After 3 years he had to pay 9000 interest. The principal amount borrowed by him was.
A.Rs 14000
B.Rs 25000
C.Rs 16000
D.Rs 17000
Answer: B.Rs 25000
Explanation: Explanation:
S.I.=Pâˆ—Râˆ—T/100
=>P=S.I.âˆ—100/Râˆ—T
=>P=9000âˆ—100/12âˆ—3=Rs25000
Option B
Question: A motor pool has 500 vehicles of which 30 percent are trucks. 20 percent of all the vehicles in the motor pool are diesel, including 15 trucks. What percent of the motor pool is composed of vehicles that are neither trucks nor diesel?
A.165%
B.90%
C.53%
D.55%
Answer: C.53%
Explanation: Trucks = 30% of 500 = 150
Other Vehicles (Excluding Trucks) = 500-150 = 350
Diesel Vehicle = 20% of 500 = 100 (Including 15 Trucks)
Other Diesel Vehicles (Excluding Trucks) = 100-15 = 85
Trucks that are NOT diesel = 150 - 15 = 135
Other Vehicles that are NOT diesel Vehicles = 350-85 = 265
Required % = (265/500)*100 = 53%
Answer: option C
Question: Find the average of all prime numbers between 30 and 50.
A.30
B.33.6
C.39.8
D.44.9
Answer: C.39.8
Explanation: Sol.
There are five prime numbers between 30 and 50.
They are 31, 37, 41, 43 and 47
∴ Required average = [31 + 37 + 41 + 43 + 47 / 5] = 199/ 5 = 39.8
Answer C