RikoteMaster/aqua-rat-mcqa huggingface dataset:
Question: 60 litres of diesel is required to travel 600 km using a 800 cc engine. If the volume of diesel required to cover a distance varies directly as the capacity of the engine, then how many litres of diesel is required to travel 800 km using 1200 cc engine?
A.80 litres
B.90 litres
C.120 litres
D.170 litres
Answer: C.120 litres
Explanation: Explanatory Answer
To cover a distance of 800 kms using a 800 cc engine, the amount of diesel required = 800/600*60 = 80 litres.
However, the vehicle uses a 1200 cc engine and the question states that the amount of diesel required varies directly as the engine capacity.
i.e., for instance, if the capacity of engine doubles, the diesel requirement will double too.
Therefore, with a 1200 cc engine, quantity of diesel required = 1200/800*80 = 120 litres.
Answer C
Question: If 2^z is a factor of 26!, which of the following could be the value of z?
I. 22
II. 23
III. 24
A.I ​ only
B.II only
C.I​ and II
D.I, II, and III
Answer: C.I​ and II
Explanation: 26! contains all numbers from 1 to 26. To find the factor with the highest power of 2, we count the number of multiples of 2, 4, 8 and 16.
Multiples of 2 = 13
Multiples of 4 = 6
Multiples of 8 = 3
Multiples of 16 = 1
Total count = 23
The answer is C.
Question: Find the missing number
5 : 24 :: 8 : ?
A.61
B.62
C.63
D.64
Answer: C.63
Explanation: 5^2 -1= 24
so 8^2 -1 =63
ANSWER:C
Question: What should be the least number to be added to the 3198 number to make it divisible by 8?
A.12
B.17
C.2
D.77
Answer: C.2
Explanation: Answer : 2
Option : C
Question: In the first 10 overs of a cricket game, the run rate was only 3.2. What should be the rate in the remaining 40 overs to reach the target of 292 runs?
A.6.5
B.6.22
C.6.29
D.6.39
Answer: A.6.5
Explanation: Required run rate
= [292 - (3.2 * 10)]/40
= 260/40
= 6.5
Answer: A
Question: In how many years will a sum of money doubles itself at 25% per annum on simple interest?
A.4%
B.8%
C.10%
D.12%
Answer: A.4%
Explanation: P = (P*25*R)/100
R = 4%
Answer:A
Question: Entry fee in an exhibition was Rs. 1. Later, this was reduced by 25% which increased the sale by 20%. The percentage increase in the number of visitors is :
A.33
B.37
C.60
D.28
Answer: C.60
Explanation: Explanation:
Let the total original sale be Rs. 100. Then, original number of visitors = 100.
New number of visitors = 120/0.75 = 160.
Increase % = 60 %.
Answer: C) 60 %
Question: A die is thrown twice and the sum of the numbers appearing is observed to be 6.
find the conditional probability that the number 4 has appeared at least once?
A.1/4
B.1/5
C.2/5
D.3/5
Answer: C.2/5
Explanation: Let E be the event that ‘number 4 appears at least once’ and F be
the event
that ‘the sum of the numbers appearing is 6’.
Then, E = {(4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (1,4), (2,4), (3,4),
(5,4), (6,4)}
and F = {(1,5), (2,4), (3,3), (4,2), (5,1)}
We have P(E) = 11/36
and P (F) = 5/36
Also EnF = {(2,4), (4,2)}
Therefore P(EnF) = 2/36
the required probability
P(E|F) = P(EnF)/P(F) = (2/36)/(5/36) = 2/5.
C)
Question: If a card is drawn from a well shuffled pack of cards, the probability of drawing a spade or a king is -
A.8/13
B.6/13
C.3/13
D.4/13
Answer: D.4/13
Explanation: Explanation:
P(Sá´œK) = P(S) + P(K) - P(Sâˆ©K), where S denotes spade and K denotes king.
P(Sá´œK) = 13/52 + 4/52 - 1/52 = 4/13
Answer: D
Question: Ben
ips a fair coin repeatedly. Given that Ben's first coin
ip is heads, compute the probability
Ben
ips two heads in a row before Ben
ips two tails in a row.
A.2/3
B.3/3
C.1/3
D.1/1
Answer: A.2/3
Explanation: If Ben successfully
ips two heads in a row, then Ben must have
ipped the coin an
even number of times. If Ben successfully
ips two tails in a row, then he must have
ipped
the coin an odd number of times. However, the coin must have been
ipped at least twice. The
desired probability is therefore
(1/3)/(1/2)=2/3
correct answer A
Question: Suppose 5 monkeys take 5 minutes to eat 5 bananas. How many monkeys would it take to eat 15 bananas in 15 minutes?
A.9
B.10
C.11
D.5
Answer: D.5
Explanation: One monkey takes 5 min to eat 1 banana , so in 15 mins 1 monkey will eat 3 bananas , so for 15 bananas in 15 min we need 15/3 =5 monkeys
ANSWER:D
Question: A cask is filled with alcohol and water in the ratio 5”3 Sixteen gallons of this are drawn off and the cask is filled with water and then the proportion of alcohol to water becomes 3:5 How many gallons does the cask hold ?
A.32
B.36
C.38
D.40
Answer: D.40
Explanation: ratio of alcohal and water is 25:15
Initially when Sixteen gallons of this are drawn off
ratio of alcohal and water is 15:9
and the cask is filled with water 15:25 which is 3:5
ANSWER:D
Question: The average age of husband, wife and their child 3 years ago was 27 years and that of wife and the child 5 years ago was 22 years. The present age of the husband is
A.40
B.36
C.45
D.55
Answer: B.36
Explanation: Explanation:
Sum of the present ages of husband, wife and child = (27 * 3 + 3 * 3) years = 90 years.
Sum of the present ages of wife and child = (22 * 2 + 5 * 2) years = 54 years.
Husband's present age = (90 - 54) years = 36 years
Option B
Question: Sand is poured into a box so that the box is being filled at the rate of 3 cubic feet per hour. If the empty rectangular box is 5 feet long, 4 feet wide, and 3 feet deep, approximately how many hours does it take to fill the box?
A.14
B.16
C.18
D.20
Answer: D.20
Explanation: The volume the box is: length*width*depth = 5*4*3= 60 cubic feet.
60 cubic feet / 3 cubic feet per hour = 20 hours.
It will take 20 hours to fill the box.
The answer is D.
Question: The present worth of a certain sum due sometime hence is Rs. 3400 and the true discount is Rs. 340. The banker's gain is:
A.Rs. 21
B.Rs. 17
C.Rs. 18
D.Rs. 34
Answer: D.Rs. 34
Explanation: Explanation :
BG = (TD)2/PW=(340)2/3400=340 × 340/3400 = 340/10 = Rs. 34 Answer : Option D
Question: A woman complete a journey in 5 hours. She travels first half of the journey at the rate of 21 km/hr and second half at the rate of 24 km/hr. Find the total journey in km.
A.334 km.
B.216 km.
C.112 km.
D.224 km.
Answer: C.112 km.
Explanation: 0.5x/21 + 0.5x/24 = 5
--> x/21 + x/24 = 10
--> x = 112 km.
C
Question: 10, 25, 45, 54, 60, 75, 80
A.10
B.45
C.49
D.54
Answer: C.49
Explanation: Each of the numbers except 54 is multiple of 5.
answer : C
Question: P and Q are the only two applicants qualified for a short-term research project that pays 480 dollars in total. Candidate P has more experience and, if hired, would be paid 50 percent more per hour than candidate Q would be paid. Candidate Q, if hired, would require 10 hours more than candidate P to do the job. Candidate P’s hourly wage is how many dollars greater than candidate Q’s hourly wage?
A.$5
B.$8
C.$11
D.$14
Answer: B.$8
Explanation: Let Q's hourly wage be x, then P's hourly wage is 1.5x
Let T be the number of hours that Q needs, then P needs T-10 hours to do the job.
Since they both are paid an equal total amount of $480:
x*T=1.5x*(T-10)
T = 30 hours and Q's hourly wage is 480/30 = $16
P's hourly wage is 480/(T-10) = $24 which is $8 per hour more.
The answer is B.
Question: A semicircular shaped window has diameter of 63cm. Its perimeter equals
A.162
B.766
C.298
D.277
Answer: A.162
Explanation: Explanation:
perimeter of window = {\color{Blue} \Pi }r+2r
= [{(22/7) * (63/2)} +63]
= 99+63
= 162 cm
Answer: A) 162
Question: The average age of the mother and her six children is 10 years which is reduced by 5 years if the age of the mother is excluded. How old is the mother?
A.40
B.41
C.42
D.43
Answer: A.40
Explanation: Total age of mother and 6 children = avg x n = 10 x (6+1)= 70
If mother is excluded new average is (10 - 5) = 5
so for 6 children = 6 x 5 =30
so mother's age is 70 - 30 =40
Answer: Option A
Question: The positive two-digit integers m and n have the same digits, but in reverse order. Which of the following must be a factor of m + n?
A.33
B.9
C.10
D.12
Answer: A.33
Explanation: Remember: When you take the difference between the two, it will always be 9. e.g 23-32=9, 89-98=9
and when you add both integers, the sum will always be a multiple of 11 e.g 23+32=55, 89+98= 187
33 is multiple of 11.
Answer : A
Question: If a train, travelling at a speed of 90 kmph, crosses a pole in 5 sec, then the length of train is?
A.281
B.125
C.288
D.266
Answer: B.125
Explanation: D = 90 * 5/18 * 5
= 125 m
Answer: B
Question: Aish started a business investing Rs. 25,000 in 2010, In 2011, he invested an additional amount of Rs. 10,000 and Roshan joined him with an amount of Rs. 35,000. In 2012, Aish invested another additional amount of Rs. 10,000 and Atul joined them with an amount of Rs. 35,000. What will be Roshan’s share in the profit of Rs. 1,50,000 earned at the end of 3 years from the start of the business in 2010?
A.Rs.45000
B.Rs.50000
C.Rs.60000
D.Rs.70000
Answer: B.Rs.50000
Explanation: Explanation :
Aish : Roshan : Atul
= (25000 x 12 + 35000 x 12 + 45000 x 12) : (35000 x 24) : (35000 x 12)
= 1260000 : 840000 : 420000 = 3 : 2 : 1
Roshan's share = Rs. 150000 x 2/6
= Rs. 50000.
Answer : B
Question: Find the sum of all odd numbers upto 100.
A.200
B.1000
C.2500
D.3140
Answer: C.2500
Explanation: Sol. The given numbers are 1, 3, 5, 7, ..., 99.
This is an A.P. with a = 1 and d = 2.
Let it contain n terms. Then,
1 + (n - 1) x 2 = 99 or n = 50.
Required sum = (n/2) (first term + last term)
= 50(1 + 99)/2 = 2500.
ANSWER C 2500
Question: How long does a train 110 m long traveling at 60 kmph takes to cross a bridge of 170 m in length?
A.16.6
B.16.0
C.16.3
D.16.8
Answer: D.16.8
Explanation: D = 110 + 170 = 280 m
S = 60 * 5/18 = 50/3
T = 280 * 3/50 = 16.8 sec
Answer: D
Question: The area of sector of a circle whose radius is 12 metro and whose angle at the center is 40° is?
A.53m2
B.51m2
C.50.2m2
D.45m2
Answer: C.50.2m2
Explanation: 40/360 * 22/7 * 12 * 12
= 50.2 m2
Answer:C
Question: If the average (arithmetic mean) of the four numbers K, 2K + 3, 4K – 4 and 5K + 1 is 63, what is the value of K?
A. 11
B. 15 3/4
C. 21
D. 23
Answer: C. 21
Explanation: K + 2K +3 + 4K - 4 + 5K +1 = 12K
(12K)/4 = 63
12K = 63 * 4 = 252
K = 252 / 12 = 21
Answer C.
Question: The average amount with a group of seven numbers is Rs. 30. If the newly joined member has Rs. 75 with him, what was the average amount with the group before his joining the group?
A.s. 25.6
B.s. 45/2
C.s. 16.6
D.s. 26
Answer: B.s. 45/2
Explanation: Total members in the group = 7
Average amount = Rs. 30
Total amount with them = 7 * 30 = Rs. 210
One number has Rs. 50. So, the amount with remaining 6 people = 210 - 75 = Rs. 135
The average amount with them = 135/6 = Rs. 45/2
ANSWER:B
Question: With a uniform speed a car covers the distance in 8 hours. Had the speed been increased by 3 km/hr, the same distance could have been covered in 7 1/2 hours. What is the distance covered?
A.360 km
B.480 km
C.278 km
D.297 km
Answer: A.360 km
Explanation: Let the distance be x km. Then,
x/(7 1/2) - x/8 = 3
2x/15 - x/8 = 3 => x = 360 km.
Answer:A
Question: The sum of the first 50 positive even integers is 2,550. What is the sum of the odd integers from 101 to 200, inclusive?
A.5,050
B.7,500
C.10,500
D.15,000
Answer: B.7,500
Explanation: 101+103+.......199
If we remove 100 from each of these it will be sum of 1st 100 odd numbers.
so 101+103+.......199 = 50 * 100 + (1+3+5+7+......)
sum of 1st 100 natural numbers = (100 * 101) / 2 = 5050
Sum of 1st 50 positive even integers = 2550
sum of 1st 100 odd numbers = 5050 - 2550 = 2500
so 101+103+.......199 = 50 * 100 + (1+3+5+7+......) = 5000 + 2500 = 7500
B is the answer.
Question: The measures of angles A, B and C of a triangle are in the ratio 3:4:5. What is the measure, in degrees, of the largest angle?
A.35°
B.65°
C.85°
D.75°
Answer: D.75°
Explanation: Let the sizes of angles A, B and C be written as follows
A = 3 x, B = 4 x and C = 5 x , where x is a common factor.
The sum of all three angles is equal to 180°. hence
3 x + 4 x + 5 x = 180°
Solve the above equation for x
x = 15
The size of the largest angle is equal to
5 x = 5 * 15 = 75°
correct answer D
Question: X does a work in 30 days. Y does the same work in 45 days. In how many days they together will do the same work?
A.18
B.12
C.20
D.30
Answer: A.18
Explanation: X's 1day's work= 1/30
Y's 1 day's work=1/45
(X+Y)'s 1day's work= (1/30+1/45)=1/18
both together will finish the work in 18 days.
correct option is A
Question: The average of five consecutive odd numbers is 61. What is the difference between the highest and lowest numbers
A.7
B.8
C.6
D.4
Answer: B.8
Explanation: Explanation:
Let the numbers be x, x + 2, x + 4, x + 6 and x + 8.
Then [x + (x + 2) + (x + 4) + (x + 6) + (x + 8) ] / 5 = 61.
or 5x + 20 = 305 or x = 57.
So, required difference = (57 + 8) - 57 = 8
Answer: B
Question: The amount of time that three people worked on a special project was in the ratio of 1 to 2 to 3. If the project took 120 hours, how many more hours did the hardest working person work than the person who worked the least?
A.35 hours
B.40 hours
C.36 hours
D.38 hours
Answer: B.40 hours
Explanation: Let the persons be A, B, C.
Hours worked:
A = 1*120/6 = 20 hours
B = 2*120/6 = 40 hours
C = 3*120/6 = 60 hours
C is the hardest worker and A worked for the least number of hours. So the difference is 60-20 = 40 hours.
Answer : B
Question: The simple interest in 14 months on a certain sum at the rate of 6 per cent per annum is 250 more than the interest on the same sum at the rate of 8 per cent in 8 months. How much amount was borrowed?
A.15000
B.25000
C.7500
D.14500
Answer: A.15000
Explanation: Let the amount be x.
From the question, x×14×6/1200−x×8×8/1200=250
∴ x = 15000
Answer A
Question: The surface area of a sphere is 4πR2, where R is the radius of the sphere. If the area of the base of a hemisphere is 3, what is the surface area W of that hemisphere?
A. 6/π
B. 9/π
C. 6
D. 9
Answer: D. 9
Explanation: Given Area of the base of a hemisphere is 3 = PI * R^2
Thus R = Sqrt ( 3 / PI ) .
Surface area of whole sphere = 4*PI*R^2 .
= 4 * PI * 3 / PI
= 12 .
Since the hemisphere is half of a sphere the Surface area of the hemisphere = 12 / 2
= 6 ( curved part , not including the flat rounded base ) .
But the total surface area = 6 + Area of the base of a hemisphere .
= 6 + 3
= 9.
Answer is D !!
Question: If P is a prime number greater than 3, find the remainder when P^2 + 12 is divided by 12.
A.6
B.1
C.0
D.8
Answer: B.1
Explanation: Every prime number greater than 3 can be written 6N+1 or 6N-1.
If P = 6N+1, then P^2 + 13 = 36N^2 + 12N + 1 + 12 = 36N^2 + 12N + 12 + 1
If P = 6N-1, then P^2 + 13 = 36N^2 - 12N + 1 + 12 = 36N^2 - 12N + 12 + 1
When divided by 12, it must leave a remainder of 1.
The answer is B.
Question: A person has to cover a distance of 6 km in 45 minutes. If he covers one-half of the distance in two-thirds of the total time; to cover the remaining distance in the remaining time, his speed (in km/hr) must be :
A.7 km/hr
B.8 km/hr
C.11 km/hr
D.12 km/hr
Answer: D.12 km/hr
Explanation: Sol.
Remaing distance = 3 km and
Remaining time = [1/3 * 45] min = 15 min = 1/4 hour.
∴ Required speed = (3*4) km/hr = 12 km/hr.
Answer D
Question: What is the value of x if (64^2)-(36^2)=20*x?
A.120
B.100
C.140
D.130
Answer: C.140
Explanation: 20*x=(64+36)(64-36)
20*x=100*28
x=(100*28)/20
x=5*28
x=140
Option C
Question: A present value of a machine is $128000. Its value depletiation rate is 25% per annum then find the machine value after 3 years
A.A)$48000
B.B)$54000
C.C)$36520
D.D)$32450
Answer: B.B)$54000
Explanation: P= $128000
R = 25%
T = 3 years
Machine value after 3years = P/(1-R/100)^T
=128000*3/4 * 3/4 * 3/4
=$54000
Answer is B
Question: There are 18 balls in a box: 10 balls are green, 6 are blue and 2 are white. Then 1 green and 1 blue balls are taken from the box and put away. What is the probability that a blue ball is selected at random from the box?
A.5/16
B.4/13
C.3/13
D.5/13
Answer: A.5/16
Explanation: If 1 green and 1 blue ball are taken from the box, then there are 9 green, 5 blue and 2 white balls, a total of 16 balls. If one ball is selected at random, the probability that it is blue is given by 5 / 16
Answer : A
Question: A, B and C enter into a partnership. They invest Rs. 40,000, Rs. 80,000 and Rs. 1,20,000 respectively. At the end of the first year, B withdraws Rs. 40,000, while at the end of the second year, C withdraws Rs. 80,000. IN what ratio will the profit be shared at the end of 3 years?
A.3:4:8
B.3:4:7
C.3:4:5
D.3:4:1
Answer: B.3:4:7
Explanation: A:B:C = (40000 * 36) : (80000 * 12 + 40000 * 24) : (120000 * 24 + 40000 * 12)
= 144:192:336 = 3:4:7
Answer: B
Question: In how many ways can three boys can be seated on five chairs?
A.72
B.28
C.60
D.58
Answer: C.60
Explanation: There are three boys.
The first boy can sit in any of the five chairs (5 ways).
5
Now there are 4 chairs remaining. The second boy can sit in any of the four chairs (4 ways).
5	4
Now there are 3 chairs remaining. The third boy can sit in any of the three chairs (3 ways).
5	4	3
Hence, total number of ways in which 3 boys can be seated on 5 chairs
=5×4×3=60
Ans: C
Question: How many paying stones, each measuring 2 1/2 m * 2 m are required to pave a rectangular court yard 30 m long and 16 1/2 m board?
A.99
B.88
C.77
D.665
Answer: A.99
Explanation: 30 * 33/2 = 5/2 * 2 * x => x = 99
Answer:A
Question: What is the sum of the integers from -150 to 155, inclusive?
A.865
B.955
C.1050
D.765
Answer: D.765
Explanation: In an arithmetic progression,
the nth term is given by tn=a+(n-1)d
here tn=155, a=-150, d=1
hence, 155=-150+(n-1)
or n=306
Sum of n terms can be calculated by sn=n/2(a+l)
a=first term, l=last term, n=no. of terms
sn=306*(-150+155)/2
sn=306*5/2 = 765
Answer : D
Question: The difference between compound and simple interest on a certain sum of money for 3 years at 6 2/3% p.a is Rs.184. Find the sum?
A.13518
B.13519
C.13527
D.13500
Answer: D.13500
Explanation: P = (184*106) / [6 2/3 * 6 2/3 *(300*6 2/3)]
P = 13500
Answer: D
Question: . A question paper consists of five problems, each problem having three internal choices. In how many ways can a candidate attempt one or more problems?
A.338
B.2767
C.1023
D.1877
Answer: C.1023
Explanation: Explanation:
Given that, the question paper consists of five problems. For each problem, one or two or three or none of the choices can be attempted.
Hence, the required number of ways = 45 - 1.
= 210 - 1 = 1024 - 1 = 1023
Answer:C
Question: Tough and Tricky questions: Word Problems.
It takes Sarah 44 minutes to do one oil change, and it takes Alice 20 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?
A.10:00 a.m.
B.10:42 a.m.
C.12:00 a.m.
D.12:40 p.m.
Answer: D.12:40 p.m.
Explanation: LCM of 2044
20 = 2 * 2 * 5
44 = 2 * 2 * 11
LCM = 2 * 2 * 5 * 11 = 220 = 3 hour 40 minutes
1240PM
D
Question: A lady grows Broccoli in her garden that is in the shape of a square. Each Broccoli takes 1 square foot of area in her garden. This year, she has increased her output by 79 Broccoli when compared to last year. The shape of the area used for growing the Broccoli has remained a square in both these years. How many Broccoli did she produce this year?
A.1600
B.1500
C.1650
D.1700
Answer: A.1600
Explanation: Explanatory Answer
The shape of the area used for growing Broccoli has remained a square in both the years.
Let the side of the square area used for growing Broccoli this year be X ft.
Therefore, the area of the ground used for cultivation this year = X2 sq.ft.
Let the side of the square area used for growing Broccoli last year be Y ft.
Therefore, the area of the ground used for cultivation last year = Y2 sq.ft.
As the number of Broccoli grown has increased by 79, the area would have increased by 79 sq ft because each Broccoli takes 1 sq ft space.
Hence, X2 - Y2 = 79
(X + Y)(X - Y) = 79.
79 is a prime number and hence it will have only two factors. i.e., 79 and 1.
Therefore, 79 can be expressed as product of 2 numbers in only way = 79 * 1
i.e., (X + Y)(X - Y) = 79 * 1
So, (X + Y) should be 79 and (X - Y) should be 1.
Solving the two equations we get X = 40 and Y = 39.
Therefore, number of Broccoli produced this year = X2 = 402 = 1600.
Alternative Approach : Use answer choices
The area in both the years are squares of two numbers.
That rules out choice B,C and D. as 1500,1650 and 1700 are not the square of any number.
Check Choice A: If this year's produce is 1600, last year's produce would have been 1600 - 79 = 1521
1521 is the square of 39.
So, 1600 is the answer.
Choice A
Question: In the xy-plane, line h passes through point (3,-2) and intersect perpendicularly with line n that passes through points (0,3) and (3,5). What is the equation of line h?
A.3x+2y-5=0
B.3x+2y+5=0
C.2x-3y-12=0
D.2x+3y=0
Answer: A.3x+2y-5=0
Explanation: If two lines intersect perpendicularly, we get -1 when we multiply two slopes together. The slope of a line n is (5-3)/(3-0)=2/3. Hence, the slope of a line h should be -3/2. Since it crosses (3,-2), if we calculate we get 3x+2y=5. Hence, the correct answer is A.
Question: For which of the following functions does f(−z)=−f(z) for all real number values of x?
A.z^8−z^4
B.z^5/(z^2+1)
C.z^5/z^7
D.z^9/(z^5+1)
Answer: B.z^5/(z^2+1)
Explanation: all even root cannot be matched eliminated
put z=2
f(−z)=(−2)^5/(−2^)2+1=−32/5, and
−f(z)=−2^5/2^2+1=−32/5, for a match.
ANSWER: B
Question: 111 = O,
222 = E,
333 = T,
444 = S,
Reading and analyzing the above, can you find out the following ?
555 = ?
A.0
B.1
C.3
D.2
Answer: A.0
Explanation: Solution:
0
Explanation:
1 * 1 * 1 = One
2 * 2 * 2 = Eight
3 * 3 * 3 = Twenty seven
4 * 4 * 4 = Sixty Four
For all the above, you can see that the first letter of the sum is the code.
Thus 5 * 5 * 5 = One twenty five
Thus 555 = O.
Answer A
Question: Ten years ago, the age of Anand was one-third the age of Bala at that time. The present age of Bala is 8 years more than the present age of Anand. Find the present age of Anand?
A.12
B.14
C.15
D.16
Answer: B.14
Explanation: Explanation:
Let the present ages of Anand and Bala be 'a' and 'b' respectively.
a - 10 = 1/3 (b - 10) --- (1)
b = a + 8
Substituting b = a + 8 in first equation,
a - 10 = 1/3 (a -2) => 3a - 30 = a - 2
=> 2a = 28 => a = 14.
Answer: B
Question: If 2994 ÷ 14.5 = 172, then 29.94 ÷ 1.45 = ?
A.17.2
B.18.2
C.19.2
D.15.2
Answer: A.17.2
Explanation: 29.94/ 1.45 = 299.4/14.5
= (2994/14.5) x 1/10) [ Here, Substitute 172 in the place of 2994/14.5 ]
= 172/10
= 17.2
Answer is A.
Question: A company has two types of machines, type R and type S. Operating at a constant rate, a machine of type R does a certain job in 18 hrs and a machine of type S does the same job in 9 hours. If the company used the same number of each type of machine to do the job in 2 hours, how many machines of type R were used?
A.A)3
B.B)4
C.C)6
D.D)9
Answer: A.A)3
Explanation: Rate of machine R =1/18
Rate of machine S =1/9
since same no of machines used for R and S to do the same work in 2 hrs
So collective rate needed to finish the work in 2 hrs= 1/2
Let the no of machine be x
So, x/18 +x/9 =1/2
3x/18=1/2
x=18/6=3
So no of machine R is 3
Answer A
Question: A vessel of capacity 2 litre has 30% of alcohol and another vessel of capacity 6 litre had 40% alcohol. The total liquid of 8 litre was poured out in a vessel of capacity 10 litre and thus the rest part of the vessel was filled with the water. What is the new concentration of Mixture?
A.30%.
B.71%.
C.49%.
D.29%.
Answer: A.30%.
Explanation: 30% of 2 litres = 0.6 litres
40% of 6 litres = 2.4 litres
Therefore, total quantity of alcohol is 3.0 litres.
This mixture is in a 10 litre vessel. Hence, the concentration of alcohol in this 10 litre vessel is 30%
ANSWER:A
Question: The sum of the fourth and twelfth term of an arithmetic progression is 20. What is the sum of the first 15 terms of the arithmetic progression?
A.3
B.1/3
C.2
D.9
Answer: C.2
Explanation: The sum of the first n terms of a G.P. is given by , where 'a' is the first term of the G.P., 'r' is the common ratio and 'n' is the number of terms in the G.P.
Therefore, the sum of the first 6 terms of the G.P will be equal to
And sum of the first 3 terms of the G.P. will be equal to
Use the ratio between these two sums to find 'r'
The ratio of the sum of the first 6 terms : sum of first 3 terms = 9 : 1
i.e.
Or r3 + 1 = 9
r3 = 8
r = 2
The answer is C