RikoteMaster/aqua-rat-mcqa huggingface dataset:
Question: The average age of a family of 6 members is 22 years.If the age of the youngest member is 7 years,what was the average age of the family at the birth of the youngest member?
A.15
B.18
C.21
D.12
Answer: A.15
Explanation: Present age of total members = 6 X 22 = 132
7 yrs back their ages were = 6 x 7 = 42
Ages at the birth of youngest member = 132 - 42 = 90
Therefore, avg age at the birth of youngest member = 90/6 = 15.
ANSWER:A
Question: If a man lost 4% by selling oranges at the rate of 12 a rupee at how many a rupee must he sell them to gain 44%?
A.8
B.7
C.5
D.2
Answer: A.8
Explanation: 96% ---- 12
144% ---- ?
96/144 * 12 = 8
ANSWER A
Question: If y > 0, (4y)/20 + (3y)/10 is what percent of y?
A.40%
B.50%
C.60%
D.70%
Answer: B.50%
Explanation: can be reduced to 2y/10+3y/10 =y/2=50%
Answer B
Question: If the sum of two numbers is 18 and the sum of their squares is 220, then the product of the numbers is
A.40
B.44
C.52
D.88
Answer: C.52
Explanation: Sol.
Let the numbers be x and y.
Then, (x+y) = 18 and x2 + y2 = 220.
Now, 2xy = (x+y)2 - (x2 + y2)
= (18)2 - 220= 324 - 220= 104
xy = 52.
Answer C
Question: Eight years ago, the ratio of ages of P and Q was 5:2. Six years from now, the ratio of their ages will be 4:3. What is P's age at present?
A.16
B.18
C.20
D.22
Answer: B.18
Explanation: 2(P-8)/5 = (Q-8). Then Q = 2(P-8)/5 + 8
3(P+6)/4 = Q+6
3(P+6)/4 = 2(P-8)/5 +14
15P+90 = 8P + 216
7P = 126
P = 18
The answer is B.
Question: A, B and C are partners. A receives 2/3 of profits, B and C dividing the remainder equally. A's income is increased by Rs.200 when the rate to profit rises from 5 to 7 percent. Find the Capital of B?
A.2888
B.2669
C.2500
D.2762
Answer: C.2500
Explanation: A:B:C = 2/3:1/6:1/6 = 4:1:1
x * 2/100 * 2/3 = 200
B capital = 15000*1/6 = 2500
Answer:C
Question: Which of the following is an integer?
I. 9! / 6!
II. 9! / 8!
III. 9! / 7!5!
A.I only
B.II only
C.III only
D.I and II only
Answer: D.I and II only
Explanation: 9!/6! = 9*8*7 which is an integer.
9!/8! = 9 which is an integer.
9!/7!5! = 9*8 / 5*4*3*2 = 3/5 which is not an integer.
The answer is D.
Question: John bought a shirt on sale for 25% off the original price and another 25 % off the discounted price. If the final price was $18, what was the price before the first discount?
A.$32
B.$34.31
C.$28.44
D.$67.54
Answer: A.$32
Explanation: let x be the price before the first discount. The price after the first discount is
x - 25%x (price after first discount)
A second discount of 25% of the discounted price after which the final price is 18
(x - 25%x) - 25%(x - 25%x) = 18
Solve for x
x = $32
correct answer A
Question: Two pipes A and B can fill a tank in 10 hours and 20 hours respectively. If both the pipes are opened simultaneously, how much time will be taken to fill the tank?
A.10/3
B.20/3
C.15/7
D.11/17
Answer: B.20/3
Explanation: Part filled by A in 1 hour = 1/10
Part filled by B in 1 hour = 1/20
Part filled by (A+B) in 1 hour = 1/10 + 1/20 = 3/20
Both the pipes together fill the tank in 20/3 hours
Answer is B
Question: At the wholesale store you can buy an 8-pack of hot dogs for $1.55, a 20-pack for $3.05, and a 450-pack for $22.95. What is the greatest number of hot dogs you can buy at this store with $200?
A.1,108
B.3,700
C.2,108
D.2,124
Answer: B.3,700
Explanation: We have $200 and we have to maximize the number of hot dogs that we can buy with this amount.
Let's try to find out what is the maximum number of hot dogs that we can buy for a lesser amount of money, which in this case is 450 for $22.95.
For the sake of calculation, let's take $23. 23x8 gives 184 ,i.e. a total of 450x8 = 3600 hot dogs . We are left with ~$16. Similarly, let's use $3 for calculation. We can buy 5 20-pack hot dogs (3x5), a total of 20x5 = 100 hot dogs. So we have 3700 hot dogs.
2108 looks far-fetched (since we are not likely to be left with > $1.55). Hence, (b) 3700 (ANSWER B)
Question: 4, 6, 12, 14, 28, 30, ?
A.20
B.40
C.60
D.80
Answer: C.60
Explanation: C
60
The given sequence is a combination of two series 4, 12, 28, .... and 6, 14, 30, .... The pattern is +8, +16, +32. So, the missing number = (28 + 32) = 60
Question: In a sample of books, 10 percent are 100-200 pages long, and 60 percent have more than 100 pages. What percentage of the books are more than 200 pages?
A.40
B.50
C.60
D.70
Answer: B.50
Explanation: 60 percent are more than 100 pages --> This means 60 percent of books are either 100-200 pages or more than 200 pages
Already, we know % of 100-200 pages= 10 %
So, % of 200+ page books = 60 % - 10 % = 50 % .. => (B).. Is it correct ?
Question: A certain sum of money is divided among A, B and C so that for each Rs. A has, B has 65 paisa and C 40 paisa. If C's share is Rs.40, find the sum of money?
A.Rs.208
B.Rs.279
C.Rs.228
D.Rs.205
Answer: D.Rs.205
Explanation: A:B:C = 100:65:40
= 20:13:8
8 ---- 40
41 ---- ? => Rs.205
Answer:D
Question: The speed of a car increases by 2 kms after every one hour. If the distance travelling in the first one hour was 35 kms. what was the total distance travelled in 12 hours?
A.456 kms
B.482 kms
C.552 kms
D.556 kms
Answer: C.552 kms
Explanation: Explanation:
Total distance travelled in 12 hours =(35 + 37 + 39 + .....upto 12 terms)
This is an A.P with first term, a=35, number of terms,
n = 12 , d = 2.
Required distance = 12/2[2 × 35 + {12 - 1) × 2] =6(70 + 23)
= 552 kms.
Answer: C
Question: A tank can be filled by a tap in 20 minutes and by another tap in 60 minutes. Both the taps are kept open for 10 minutes and then the first tap is shut off. After this, the tank will be completely filled in what time ?
A.10 mins
B.15 mins
C.20 mins
D.25 mins
Answer: C.20 mins
Explanation: Explanation:
How we can solve this question ?
First we will calculate the work done for 10 mins, then we will get the remaining work, then we will find answer with one tap work, As
Part filled by Tap A in 1 min = 1/20
Part filled by Tap B in 1 min = 1/60
(A+B)'s 10 mins work =
10∗(1/20+1/60)
=10∗4/60=2/3
Remaining work = 1−2/3=1/3
METHOD 1
=>1/60:1/3=1:X
=>X=20
METHOD 2
1/60 part filled by B in=1min1/3 part will be filled in=1/3/1/60=60/3=20
Option C
Question: If x and y are integers and xy = 330, x or y must be divisible by which of the following?
A.3
B.4
C.6
D.20
Answer: A.3
Explanation: xy = 330 = 2 * 3 * 5 * 11
One of x or y must be divisible by 3.
The answer is A.
Question: What will be remainder when (67^67 + 67) is divided by 68 ?
A.65
B.66
C.67
D.71
Answer: B.66
Explanation: (xn + 1) will be divisible by (x + 1) only when n is odd.
(67^67 + 1) will be divisible by (67 + 1)
(67^67 + 1) + 66, when divided by 68 will give 66 as remainder.
B)
Question: If A : B= 3 : 5 and B : C = 7 : 9, then A : B : C is:
A.2:4:8
B.21:35:45
C.5:10:15
D.12:24:36
Answer: B.21:35:45
Explanation: A:B=3:5, B:C=7:9, A:B:C=3Ã—7:7Ã—5:5Ã—9
â‡’ 21:35:45
Answer: B.
Question: The expenses of organizing a garden party increased from Rs.9000 to Rs.12000 when the number of registered candidates increased from 25 to 40. Find the total cost of organizing the party if there are 50 final registrations.
A.Rs.11000
B.Rs.12000
C.Rs.13000
D.Rs.14000
Answer: D.Rs.14000
Explanation: We know that the expenses of party increases not directly proportional but directly relational.
Assume the fixed component of expense is K rupees and Variable component is M rupees.
Then the total cost is given by
K + 25 x M = 9000 ............(1)
When there are 40 registrations the total cost is
K + 40 x M = 12000 .........(2)
By substracting (1) from (2)
⇒15M=3000⇒M=200⇒15M=3000⇒M=200
So the variable cost per head is Rs.200
To find the fixed cost we can substitute Rs.200 in either (1) or (2), then K = 4000
If there are 50 registrations then the total cost = 4000 + 50 x 200 = Rs.14000
Answer D
Question: As part of a promotion designed to attract visitors, an amusement park gives out 24,600 free tickets per year. Usually, the number of tickets given away each month is constant. Due to an employee error, only 7,000 tickets were given out in the first four months of the year. How many tickets should the amusement park give out in the fifth month so that it will begin the sixth month back on its regular giveaway schedule?
A.2,050
B.2,079
C.2,200
D.3,250
Answer: D.3,250
Explanation: 24600/12=2050
2050*5=10250
10250-7000=3250
Answer D
Question: The mass of 1 cubic meter of a substance is 300 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.1.33
B.1.25
C.8.0
D.12.5
Answer: A.1.33
Explanation: Density is mass divided by volume. So density of the given substance will be mass/volume=800kg/1m^3=300 kg/m^3 or 1g/1.33cm^3=0.3 g/cm^3.
Next, ask yourself if 300,000 g is equivalent to 1,000,000 cubic centimeters then 1 g is equivalent to how many cubic centimeters? --> 1 g - 1,000,000/300,000 = 10/3 = 1.33 cubic centimeters.
Answer is A
Question: What will be the vulgar fraction of 0.4
A.3/5
B.2/5
C.3/2
D.3/7
Answer: B.2/5
Explanation: Explanation:
0.4 = 40/100 = 2/5
Option B
Question: In how many different number of ways 5 boys and 3 girls can sit on a bench such that girls always sit together.
A.A)720
B.B)730
C.C)4230
D.D)4320
Answer: D.D)4320
Explanation: 4320
Option 'D'
Question: A man runs 1 mile at a speed of r mph. If he runs at a speed of r+20 miles per hour, he would run the same distance in 4 minutes less. What is the value of r?
A.20 mph
B.8 mph
C.7 mph
D.10 mph
Answer: D.10 mph
Explanation: Convert minutes to hours: 4 min = 4/60 min = 1/15 hours.
Equation: 1/r = 1/(r+20) + 1/15.
Plug options into the equation. D fits: 1/10 = 1/(10+20) + 1/15.
Answer: D
Question: The London Racetrack needs to submit its 3 fastest horses to the Kentucky Derby out of 25 horses. However, all of their information was lost and they don't know any of the horse's times. Similarly, they all look identical so they can't remember who's fastest.
They can only race 5 horses at once, so what is the fewest number of races they can conduct to find the 3 fastest horses?
A.6
B.7
C.5
D.4
Answer: B.7
Explanation: 7 races are required.
First you divide the 25 horses into 5 groups of 5. You conduct the 5 races and take all of the fastest horses in those races and have a race with them, giving you the fastest horse. Then you take the remaining 24 horses (excluding the fastest) and remove the 4th and 5th horses in the first set of 5 races (since they definitely have 3 horses faster than them), leaving you with 14 horses. Next you can remove all of the horses that were beat in the preliminary race by the horses that got 4th and 5th in the championship race, leaving you with 8 horses. Finally, you can remove the horses that remain that lost to the 3rd place horse in the final race in the preliminary race and the horse that got 3rd in the preliminary to the horse that got 2nd in the championship race, leaving you with 5 horses.
You can then run a final race where the 1st and 2nd place horses are the 2nd and 3rd fastest. Then you know the 3 fastest horses.
Question: How many positive integer solutions does the equation 8x+2y = 100 have?
A.50
B.84
C.16
D.35
Answer: B.84
Explanation: formula: (constant)/(lcm of two nos)
=100/(8*2)=84
ANSWER:B
Question: At the end of year X, automobile installment credit accounted for 36% of all outstanding consumer installment credit. At that time automobile finance companies extended $75 billion of credit, or 1/2 of the automobile installment credit. How many billion dollars of consumer installment credit was outstanding at that time?
A.412.16
B.150
C.400
D.416.67
Answer: D.416.67
Explanation: System of Equations
a = (36/100) c
(1/2) a = 75 --> a = 150
Substitution
150 = (36/100) c
c = (100/36) 150
Calculate 150/36 * 100
The Correct Answer is D
- The correct answer will be slightly smaller than 513. Therefore, the correct answer is 475.
Question: A group of students decided to collect as many paise from each member of group as is the number of members. If the total collection amounts to Rs. 4356, the number of the member is the group is:
A.57
B.66
C.77
D.87
Answer: B.66
Explanation: Money collected = (43.56 x 100) paise = 4356 paise.
Number of members =Square root of 4356 = 66.
Answer: Option B
Question: If 64(2^x) = 1 then x=
A.–2
B.–1
C.0
D.-6
Answer: D.-6
Explanation: 2^x = 1/64
2^x = 1/2^6
2^x= 2^-6
x = -6
D
Question: Marla starts running around a circular track at the same time Nick starts walking around the same circular track. Marla completes 30 laps around the track per hour and Nick completes 15 laps around the track per hour. How many minutes after Marla and Nick begin moving will Marla have completed 4 more laps around the track than Nick?
A.5
B.16
C.12
D.15
Answer: B.16
Explanation: Maria's rate - 30 laps per hour --> 30/60 laps/min
Nick's rate - 15 laps per hour --> 15/60 laps/min
lets set equations:
30/60*t=4 (since Maria had to run 4 laps before Nick would start)
15/60*t=0 (Hick has just started and hasn't run any lap yet)
(30/60-15/60)*t=4-0 (since Nick was chasing Maria)
t=16 min needed Maria to run 4 laps
Answer : B
Question: What is the equation of a circle of radius 7 units centered at (3, 2)?
A.x^2 + y^2 - 6x – 4y = 36
B.x^2 + y^2 - 6x + 4y = 23
C.x^2 + y^2 + 6x + 4y = 23
D.x^2 + y^2 - 6x – 4y = - 23
Answer: A.x^2 + y^2 - 6x – 4y = 36
Explanation: the equation of a circle with center at (a,b) and radius R is (X-a)^2+(Y-b)^2=R^2
=> answer = (X-3)^2+(Y-2)^2=7^2
solving,,we get the equation in answer choice A
Question: On a windy day, out of every 3 sand dunes formed, only 1 remains. Out of 5 blown-out
sand-dunes, 1 has treasure and only 2 out of 7 sand dunes formed have lucky coupons. Find
the probability that the blown-out sand dune will contain both the things.
A.2/25
B.4/75
C.7/75
D.3/79
Answer: D.3/79
Explanation: Probability of sand dune blown out = 2/3
Probability that it has treasure = 1/5
Probability that it has lucky coupon = 2/7
total probability = 2/3 * 1/5 * 2/7 = 3/79
ANSWER:D
Question: A motorcyclist goes from Bombay to Pune, a distance of 192 kms at an average of 32 kmph speed. Another man starts from Bombay by car 2 ½ hours after the first, and reaches Pune ½ hour earlier. What is the ratio of the speed of the motorcycle and the car?
A.2:8
B.1:2
C.2:5
D.6:9
Answer: B.1:2
Explanation: B
1:2
T = 192/32 = 6 h
T = 6 - 3 = 3
Time Ratio = 6:3 = 2:1
Speed Ratio = 1:2
Question: The side of a rhombus is 26 m and length of one of its diagonals is 20 m. The area of the rhombus is?
A.488
B.429
C.418
D.480
Answer: D.480
Explanation: 262 – 102 = 242
d1 = 20 d2
= 48
1/2 * 20 * 48
= 480
Answer: D
Question: How many litres of pure acid are there in 12 litres of a 40% solution
A.3.6
B.4.8
C.3.5
D.2.3
Answer: B.4.8
Explanation: Explanation:Question of this type looks a bit typical, but it is too simple, as below... It will be 12 * 40/100 = 4.8 Answer: Option B
Question: In 2010, Arman’s birthday was on a Thursday. What day of the week was it 69 days after his birthday?
A.Thursday
B.Wednesday
C.Saturday
D.Sunday
Answer: B.Wednesday
Explanation: Day of week after 69 days after Arman's birthday = 7*9 + 6
Saturday as the period will have 9 weeks and 6 days .
The 6 days in the end will be Friday - Saturday - Sunday- Monday-Tuesday and Wednesday
Answer B
Question: The volume of a sphere with radius r is (4/3)*pi*r^3 and the surface area is 4*pi*r^3. If a sperical balloon has a volume of 2304 pi cubic centimeters, what is hte surface area of the balloon in square centimeters?
A.a. 40
B.b. 100
C.c. 400
D.d. 1,800
Answer: D.d. 1,800
Explanation: The surface area is 4.pi.r^2 (its area remember not volume)
as 4/3.pi.r^3=2304pi
r=12
so area = 4.pi.r^2= 576.pi= 576 x 3.14 = 1800 (approx)
D
Question: A bullet train 140 m long is running at 60 kmph. In how much time will it pass a platform 160 m long?
A.34 Seconds
B.18 Seconds
C.41 Seconds
D.24 Seconds
Answer: B.18 Seconds
Explanation: Distance travelled = 140 + 160m = 300m
Speed = 60 * 5/8 = 50/3m
Time = 300 * 3/50 = 18 Seconds
ANSWER B
Question: The length of the bridge, which a train 130 metres long and travelling at 45 km/hr can cross in 30 seconds, is:
A.200 meters
B.225 meters
C.245 meters
D.250 meters
Answer: C.245 meters
Explanation: Given, Length of the train = 130 meters
Speed of the train = 45 km/hr = 45 * 518 m/sec = 252 m/sec
Time = 30 secs
Let the length of the bridge be x meters .
The distance travelled by the train while crossing the bridge will be sum of the length of train and Length of the bridge = (x+130) meters
so , 252 = x+13030
= 30*25 = 2x + 260
= 750 - 260 = 2x
= 490 = 2x
=x = 245 meters. ANSWER C
Question: Consider a sequence of numbers given by the expression 7 + (s - 1) * 5, where n runs from 1 to 80.What is the sum of this series ?
A.409
B.1636
C.16360
D.16000
Answer: C.16360
Explanation: terms in this sequence are 7,12,17----402
now since this is an a.p. with a common difference of 5. therefore its sum can be given as
s(a+l)/2----------------1)
s= total no. of terms =80
a= first term=7
l= last term=402
subsituting values in the expression 1 we have
80(7+402)/2
= 40(409)
=16360=C
Question: A father tells his son, "I was of your present age when you were born". If the father is 36 now, how old was the boy five years back ?
A.13
B.15
C.17
D.20
Answer: A.13
Explanation: Explanation:
Let the father's age be x and the son's age be y.
Then, x - y = y or x = 2y,
Now, x = 36. So, 2y = 36 or y = 18.
Therefore Son's present age = 18 years.
So, son's age 5 years ago = 13 years.
Answer: Option A
Question: The ratio of buses to cars on River Road is 1 to 10. If there are 90 fewer buses than cars on River Road, how many cars are on River Road?
A.50
B.40
C.100
D.30
Answer: C.100
Explanation: B/C=1/10
C-B=90.........> B=C-90
(C-90)/C=1/10
Testing answers. Clearly Eliminate ABDE
Put C=100.........> (100-90)/100=10/100=1/10
Answer: C
Question: A man borrows Rs. 20,000 at 10% compound interest. At the end of every year he pays Rs. 2000 as part repayment. How much does he still owe after three such installments?
A.Rs.24000
B.Rs.20000
C.Rs.15000
D.Rs.10000
Answer: B.Rs.20000
Explanation: Explanation:
Amount after 3 years on Rs.20000 at 10% compound interest
= 20000(1 + 10/100)3 = 20000(11/10)3 = 26620--- (1)
He paid Rs.2000 after 1st year.
Hence Rs.2000 and its compound interest for 2 years (i.e., amount on 2000 after 2 year) need to be reduced from (1)
Similarly, he paid Rs.2000 after 2nd year.
Hence Rs.2000 and and its compound interest for 1 year
(i.e., amount on 2000 after 1 year) need to be reduced from (1)
Similarly, he paid Rs.2000 after 3rd year.
Hence this Rs.2000 also need to be reduced from (1)
Hence, remaining amount
= 26620 − 2000 (1+10/100)2 − 2000(1+10/100)1 − 2000
= 26620 − 2000(11/10)2 − 2000(11/10) − 2000
= 26620 −2420 − 2200 − 2000 = 20000
i.e, he still owes Rs.20000 even after three installments
Answer: Option B
Question: If A lends Rs.3500 to B at 10% per annum and B lends the same sum to C at 11.5% per annum then the gain of B in a period of 3 years is?
A.157.20
B.157.29
C.157.50
D.157.30
Answer: C.157.50
Explanation: (3500*1.5*3)/100
=> 157.50
Answer: C
Question: A and B are two partially filled buckets of water. If 8 liters are transferred from A to B, then A would contain one-third of the amount of water in B. Alternatively, if 8 liters are transferred from B to A, B would contain one-half of the amount of water in A. Bucket A contains how many liters of water?
A.88/5
B.13
C.17
D.21
Answer: A.88/5
Explanation: Let Bucket A be A and Bucket B be B
Scenario 1
A-8 = 1/3 (B+8) ----> 3A-24 =B+8
Scenario 2
B-8 = 1/2 (A+8) -----> 2B-16 = A+8
From Scenario 1, B = 3A-32
Substitute B with this information in Stmt 2
2 (3A-32) -16 = A+8 ------> 6A -64-16 = A+8 ------> 6A- A = 80+8 ---> 5A= 88
A= 88/5, Answer choice A
Question: On 8th Feb, 1995 it was Wednesday. The day of the week on 8th Feb, 1994 was
A.wednesday
B.Tuesday
C.Thursday
D.Friday
Answer: B.Tuesday
Explanation: 1994 being an ordinary year, it has 1 odd day. So, the day on 8th Feb, 1995 is one day beyond the day on 8th Feb, 1994. But, 8th Feb, 1995 was Wednesday. 8th Feb, 1994 was Tuesday.
Answer B
Question: In how many different ways can the letters of the word 'DETAIL' be arranged such that the consonants must occupy only the odd positions?
A.None of these
B.64
C.120
D.36
Answer: D.36
Explanation: OEOEOE
3 consonants DTL can be arranged on odd places in 3!=6 ways. Similarly 3 vowels can be arranged on even places in 3!=6 ways.
Total = 6*6 = 36.
Answer: D.
Question: A, B, C and D enter into partnership. A subscribes 1/3 of the capital B 1/4, C 1/5 and D the rest. How much share did A get in a profit of Rs.2490 ?
A.Rs. 832
B.Rs. 830
C.Rs. 822
D.Rs. 812
Answer: B.Rs. 830
Explanation: Let the total amount in the partnership be 'x'.
Then A's share = x/3
B's share = x/4
C's share = x/5
D's share = x - (x/3 + x/4 +x/5) = 13x/60
A : B : C : D = x/3 : x/4 : x/5 : 13x/60 = 20 : 15 : 12 : 13
A's share in the profit of Rs. 2490 = 20 (2490/60) = Rs. 830.
Answer: B
Question: If y/(x - z) = (y + x)/z = x/y then find x : y : z ?
A.1 : 2 : 3
B.3 : 2 : 1
C.4 : 2 : 3
D.2 : 4 : 7
Answer: C.4 : 2 : 3
Explanation: Given, y/(x - z) = (y + x)/z
yz = xy + x2- yz - xz .. (1)
Also (x/y) = (y/(x - z)) => x2 - xz = y2 .. (2)
Using (1) and (2), we get yz = xy - yz + y2
2yz = xy + y2
2z = x + y
Only option (c) satisfies the equation
ANSWER:C
Question: Determine the value of (27/31*31/27)*3
A.0
B.1
C.2
D.3
Answer: D.3
Explanation: Solution:
Both fractions should be reduced before performing arithmetic operations. We get (27/31*31/27)3=1*3=3
Answer D
Question: What is the area of square field whose side of length 15 m?
A.225 sq m
B.186 sq m
C.586 sq m
D.287 sq m
Answer: A.225 sq m
Explanation: 15 * 15
= 225 sq m
Answer: A
Question: Rs.8000 become Rs.9261 in a certain interval of time at the rate of 5% per annum of C.I. Find the time?
A.2
B.9
C.8
D.3
Answer: D.3
Explanation: n:
9261 = 8000(21/20)N
(21/20)3 = (21/20)N => N = 3
Answer:D
Question: Raju age after 14 years will be 3 years more than 5 times his age 5 years back, What is the present age of Raju
A.9 years
B.8 years
C.5 years
D.6 years
Answer: A.9 years
Explanation: Explanation:
Clearly,
x+14 = 5(x-5)+3
<=> 4x = 36 => x = 9
Option A
Question: Jane covered a distance of 340 miles between city A and city taking a total of 5 hours. If part of the distance was covered at 60 miles per hour speed and the balance at 80 miles per hour speed, how many hours did she travel at 60 miles per hour?
A.2 Hours
B.3 Hours
C.4 Hours
D.5 Hours
Answer: B.3 Hours
Explanation: X is the hours which Jane is travelled with 60 miles per hour. then 5-X is the hours which Jane travelled with 80 miles per hour. Total Distance becomes
(X*60)+(5-X)*80 = 340. Solving this X will be 3 Hours
ANSWER:B
Question: If the length and breadth of a room are increased by y feet each, the perimeter increases by 16 feet. Find y
A.8
B.2
C.6
D.4
Answer: D.4
Explanation: 2(l+b)=x;
2(l+y+b+y)=x+16;
2(l+b)+4y=x+16;
x+4y=x+16;
y=4
ANSWER:D
Question: The difference between a number and its three fifth is 50. What is the number?
A.75
B.100
C.125
D.180
Answer: C.125
Explanation: Solution
Let the number be x.
Then, x - 3 / 5x = 50
‹=› 2 / 5 x = 50
‹=›x = (50 x 5 / 2)
‹=›125.
Answer C
Question: Out of 4 consonants and 2 vowels, how many words of 3 consonants and 1 vowel can be formed?
A.192
B.950
C.252
D.2150
Answer: A.192
Explanation: Solution
Number of ways of selecting (3 consonants out of 4) and (1 vowel out of 2)
4c3 x 2c1= (4/1)x(2/1) =8
Number of group,each having 3 consonants and 1 vowel = 8 .
Each group contains 4 letter.
Number of ways of arranging 4 letters among themselves
= 4 ! = ( 4 × 3 × 2 × 1) = 24 .
So, Required number of words = (24 × 8) = 192 .Answer A
Question: In a certain store, the profit is 320% of the cost. If the cost increases by 25% but the selling price remains constant, approximately what percentage of the selling price is the profit?
A.30%
B.70%
C.100%
D.250%
Answer: B.70%
Explanation: Let C.P.= Rs. 100. Then, Profit = Rs. 320, S.P. = Rs. 420.
New C.P. = 125% of Rs. 100 = Rs. 125
New S.P. = Rs. 420.
Profit = Rs. (420 - 125) = Rs. 295
Required percentage = (295/420 * 100)% = 1475/21 % = 70% (approximately)
Answer = B