RikoteMaster/aqua-rat-mcqa huggingface dataset:
Question: If the side length of Square B is three times that of Square A, the area of Square B is how many times the area of Square A?
A.9
B.8
C.6
D.3
Answer: A.9
Explanation: Let x be the side length of Square A.
Then the area of Square A is x^2.
The area of Square B is (3x)^2=9x^2.
The answer is A.
Question: 5 cats are about to jump on 5 different mice. What is the probability that exactly one cat will jump on a single mouse?
A.5*5
B.1/5!
C.5!/5^5
D.5!/5^2
Answer: C.5!/5^5
Explanation: Each cat out of 5 has 5 options, hence total # of outcomes is 5^5;
Favorable outcomes will be 5!, which is # of ways to assign 5 different mice to 5 cats:
1-2-3-4-5 (mice)
A-B-C-D-E (cats)
B-A-C-D-E (cats)
B-C-A-D-E (cats)
...
So basically # of arrangements of 5 distinct objects: 5!.
P=favorable/total=5!/5^5
Answer: C.
Question: A dishonest dealer professes to sell goods at the cost price but uses a weight of 800 grams per kg, what is his percent?
A.29%
B.25%
C.95%
D.26%
Answer: B.25%
Explanation: 800 --- 200
100 --- ? => 25%
Answer: B
Question: What is the minimum number of square marbles required to tile a floor of length 5 metres 78 cm and width 3 metres 06 cm?
A.153
B.187
C.540
D.748
Answer: A.153
Explanation: The marbles used to tile the floor are square marbles.
Therefore, the length of the marble=width of the marble.the length of the marble=width of the marble.
As we have to use whole number of marbles, the side of the square should a factor of both 5 m 78 cm and 3m 06. And it should be the highest factor of 5 m 78 cm and 3m 06.
5 m 78 cm = 578 cm and 3 m 06 cm = 306 cm.
The HCF of 578 and 306 = 34.
Hence, the side of the square is 34.
The number of such square marbles required,
=578×306/34×34
=153 marbles
Option(A)
Question: A man rows downstream 30 km and upstream 18 km, taking 5 hours each time. What is the velocity of the stream (current)?
A.1.2 km/hr
B.1.5 km/hr
C.2.5 km/hr
D.1.8 km/hr
Answer: A.1.2 km/hr
Explanation: 30/(B+5) = 18/(B-5) = 5
? B+5 = 6 km/hr
B - 5 = 3.6 km/hr
? 5 = 1.2 km/hr
Answer: A.
Question: Boy purchased two flats for Rs.5,00,000. On one he gains 15 % while on the other he losses 15%. How much does he gain or lose?
A.2%
B.2.12%
C.2.25%
D.2.30%
Answer: C.2.25%
Explanation: Generally in such cases, there is always loss.
So always remember, when two materials are sold and if one material gets profit and the other gets a loss, then use the trick shown below to calculate the loss.
Loss% = Common loss and gain% 2 = x 2
10 10
Therefore, here common loss and gain % = 15%
Hence,
Loss% = 15 2 = 2.25%
C
Question: If a is a positive integer, and if the units' digit of a^2 is 4 and the units' digit of (a + 1)^2 is 1, what is the units' digit of (a + 2)^2 ?
A.	0
B.	2
C.	4
D.	6
Answer: A.	0
Explanation: A for me.
_1² = _1
_2² = _4
_3² = _9
_4² = _6
_5² = _5
_6² = _6
_7² = _9
_8² = _4
_9² = _1
_0² = _0
On the list, the only number that squared has the units digit = 4 preceding a number that squared has the units digit = 1 is the _8.
So, a = _8, that has square 4. The next square is 1 andnext nextsquare is 0.
Question: On my sister's 15th birthday, she was 159 cm in height, having grown 6% since the year before. How tall was she the previous year?
A.150 cm
B.140 cm
C.142 cm
D.154 cm
Answer: A.150 cm
Explanation: Explanation :
Given that height on 15th birthday = 159 cm and growth = 6%
Let the previous year height = x
Then height on 15th birthday = x ×100+6/100=x×106/100
⇒159 = x × (106/100)
⇒�x=159×100/106=1.5×100=150;cm
Answer : Option A
Question: Find the ratio in which rice at Rs. 7.20 a kg be mixed with rice at Rs. 5.70 a kg to produce a mixture worth Rs. 6.30 a kg.
A.1 : 3.
B.2 : 3.
C.2 : 4
D.1 : 4
Answer: B.2 : 3.
Explanation: Cost of 1 kg of 1st kind=720 p
Cost of 1 kg of 2nd kind=570 p
Mean Price=630 p
Cost of 1 kg of 1st kind - Mean Price=720-630=90
Cost of 1 kg of 2nd kind- Mean Price=570-630=60
Required ratio = 60 : 90 = 2 : 3.
Answer:B
Question: The last day of a century cannot be
A.Monday
B.Wednesday
C.Tuesday
D.sunday
Answer: C.Tuesday
Explanation: 100 years contain 5 odd days.
Last day of 1st century is Friday.
200 years contain (5 x 2) 3 odd days.
Last day of 2nd century is Wednesday.
300 years contain (5 x 3) = 15 1 odd day.
Last day of 3rd century is Monday.
400 years contain 0 odd day.
Last day of 4th century is Sunday.
This cycle is repeated.
Last day of a century cannot be Tuesday or Thursday or Saturday.
Answer: Option C
Question: A train starts full of passengers at the first station it drops 1/3 of the passengers and takes 280 more at the second station it drops one half the new total and takes twelve more. On arriving at the third station it is found to have 246 passengers. Find the no of passengers in the beginning?
A.292
B.180
C.192
D.282
Answer: D.282
Explanation: Let no of passengers in the beginning be X
After first station no passengers=(X-X/3)+280=2X/3 +280
After second station no passengers =1/2(2X/3+280)+12
1/2(2X/3+280)+12=246
2X/3+280=2*234 = 468 - 280 = 188
2X/3=188
X=282. So the answer is option D) 282.
Question: A man goes downstream at 12 kmph, and upstream8 kmph. The speed of the stream is
A.2kmph
B.4kmph
C.16kmph
D.2.5kmph
Answer: A.2kmph
Explanation: Speed of the stream = 1/2(12-8)kmph
=2kmph.
correct option A
Question: Siddharth wants to borrow Rs.9000 at rate of interest 6% p.a. at S.I and lend the same amount at C.I at same rate of interest for two years. What would be his income in the above transaction?
A.s.21.60
B.s.32.40
C.s.21.18
D.s.21.98
Answer: B.s.32.40
Explanation: Amount of money Siddharth borrowed at S.I at 6% p.a. for two years = Rs.9,000
He lend the same amount for C.I at 6% p.a. for two years.
=> Siddharth's income = C.I - S.I
= p[1 + r/ 100]n - p - pnr/100
= p{ [1 + r/ 100]2 - 1 - nr/100
= 9,000{ [1 + 6/100]2 - 1 - 12/100}
= 9,000 {(1.06)2- 1 - 0.12} = 9,000(1.1236 - 1 - 0.12)
= 9,000 (0.0036) = 9 * 3.6
= Rs.32.40
Answer: B
Question: How many such pairs of letters are there in the word EXPRESSION which have as many letters between them in the word as in the alphabet ?
A.4
B.5
C.3
D.2
Answer: A.4
Explanation: (E,N)
(P,S)
(O.N)
(X,S)
ANSWER:A
Question: What is the value of 100p2 ?
A.9801
B.12000
C.5600
D.9900
Answer: D.9900
Explanation: Explanation :
100p2 = 100 x 99 = 9900. Answer : Option D
Question: What is the greatest value of n such that 18^n is a factor of 17! ?
A.1
B.2
C.3
D.4
Answer: C.3
Explanation: Another way to solve this question :
We know that 18 = 2 *3^2.
Hence, we need to find total number of 3s in 18! ( As out of 3 and 2 , we will have count of 3 least)
No. of 3s = 18/3 =6/3 =2.
Add the numbers in Bold, you will get total number of 3s = 8.
Hence, total number of 3^2 = 8/2 = 4.
Hence, answer is 3. C
Question: A rectangular box of volume V has width, depth and height in the ratio of 2:2:1 (in that order). What is the width as a function of V?
A.(3V/2)^(1/4)
B.(7V/6)^(1/4)
C.(9V/2)^(1/4)
D.(16V/4)^(1/4)
Answer: D.(16V/4)^(1/4)
Explanation: We are given width = 2, depth = 2, height = 1.
Then Volume V = Width x Depth x Height = 2*2*1 = 4
i.e. V = 4
The correct option should result in 2 (the width) on substituting the value of V=4.
Checking options with V=4:
A) (3V/2)^(1/4) = (6/2)^(1/4) = 3^(1/4) INCORRECT ANSWER
B) (7V/6)^(1/4) = (28/6)^(1/4) INCORRECT ANSWER
C) (9V/2)^(1/4) = (36/2)^(1/4) = 18^(1/4) INCORRECT ANSWER
D) (16V/4)^(1/4) = (64/4)^(1/4) = 16^(1/4) = 2 CORRECT ANSWER
E) (4V/2)^(1/4) = (16/2)^(1/4) = 8^(1/4) INCORRECT ANSWER
Answer: Option D
Question: A man covers a distance of 1200 km in 70 days resting 9 hours a day, if he rests 10 hours a day and walks with speed 1Â½ times of the previous in how many days will he cover 840 km ?
A.25 days
B.75 days
C.35 days
D.15 days
Answer: C.35 days
Explanation: Explanation:
Distance d = 1200km
let S be the speed
he walks 15 hours a day(i.e 24 - 9)
so totally he walks for 70 x 15 = 1050hrs.
S = 1200/1050 => 120/105 = 24/21 => 8/7kmph
given 1 1/2 of previous speed
so 3/2 * 8/7= 24/14 = 12/7
New speed = 12/7kmph
Now he rests 10 hrs a day that means he walks 14 hrs a day.
time = 840 x 7 /12 => 490 hrs
=> 490/14 = 35 days
So he will take 35 days to cover 840 km.
Answer: C
Question: A boy gets some rupees from his mother and spends them on 5 stores. He spends one rupee more than half of the money. How much money he had at the time of entering the shop.
A.60
B.61
C.62
D.63
Answer: C.62
Explanation: (x+2)/(2^n)=2
where n=no. of shops
so we get ,(x+2)/(2^5)=2
x=62
ANSWER:C
Question: If apples cost x dollars per m dozen, how many dollars will it cost to buy n apples?
A.xn/(12m)
B.xn/m
C.m/(xn)
D.12m/(xn)
Answer: A.xn/(12m)
Explanation: let it 2 dollars for 2 dozens(24) and we are asked to find the value n apples...let n be 4...
In order to buy 24 apples it took 2 dollars...then for 4 apples it'll take 4*2/12*2 = 1/3.. i.e. n*x/12*m.
Answer: option A is correct answer...
Question: Interest Rate: 4% monthly Starting Balance: 124 Time Passed: 4 months How much interest has accrued if calculated as compound interest? What is the new total balance?
A.21 & 145
B.22 & 146
C.23 & 145
D.20 & 140
Answer: A.21 & 145
Explanation: Compound Interest: Total Balance = P(1 + R)T P = principle = starting balance = $124 R = interest rate = 4% T = time = 4 years Total balance = principle × (1 + interest rate)time = 124 × (1 + (4 / 100))4 = 145 Interest accrued = total balance - starting balance = 145 - 124 = 21
Answer A
Question: Find the perimeter and area of a square of side 19 cm.
A.324
B.361
C.400
D.441
Answer: B.361
Explanation: We know that the perimeter of square = 4 Ã— side
Side= 19 cm
Therefore, perimeter = 4 Ã— 19 cm = 76 cm
Now, area of the square = (side Ã— side) sq. units
= 19 Ã— 19 cmÂ²
= 361 cmÂ²
ANSWER : B
Question: The value of a machine depreciates at 20% per annum. If its present value is Rs. 1,50,000, at what price should it be sold after two years such that a profit of Rs. 28,000 is made?
A.Rs. 1,10,000
B.Rs. 1,20,000
C.Rs. 1,24,000
D.Rs. 1,21,000
Answer: C.Rs. 1,24,000
Explanation: The value of the machine after two years = 0.8 * 0.8 * 1,50,000 = Rs. 96,000
SP such that a profit of Rs. 24,000 is made = 96,000 + 28,000 = Rs. 1,24,000
ANSWER:C
Question: A family consists of two grandparents, two parents and three grandchildren. The average age of the grandparents is 65 years, that of the parents is 32 years and that of the grandchildren is 5 years. The average age of the family is
A.29 1/7 years
B.29 6/7 years
C.39 6/7 years
D.19 6/7 years
Answer: B.29 6/7 years
Explanation: Total age of the grandparents = 65 × 2=130
Total age of the parents = 32 × 2=64
Total age of the grandchildren = 5 × 3=15
Average age of the family = (130+64+15)/7
=209/7
=29 6/7 years
ANSWER:B
Question: A rectangular lawn of length 200m by 120m has two roads running along its center, one along the length and the other along the width. If the width of the roads is 5m what is the area W covered by the two roads?
A.400
B.1550
C.1575
D.1600
Answer: C.1575
Explanation: Area Covered by Road Along the Length = 5*200 = 1000 Square Meter
Area Covered by Road Along the Width = 5*120 = 600 Square Meter
Common Area in both Roads (where the roads intersect) = Square with Side 5 meter = 5*5 = 25
Total Area of the Roads W = 1000+600-25 = 1575
Answer: option C
Question: A man can row his boat with the stream at 20 km/h and against the stream in 8 km/h. The man's rate is?
A.1 kmph
B.7 kmph
C.6 kmph
D.5 kmph
Answer: C.6 kmph
Explanation: DS = 20
US = 8
S = ?
S = (20 - 8)/2 = 6 kmph
Answer: C
Question: The owner of a furniture shop charges his customer 32% more than the cost price. If a customer paid Rs. 5400 for a computer table, then what was the cost price of the computer table?
A.Rs.4090
B.Rs.4067
C.Rs.6290
D.Rs.6725
Answer: A.Rs.4090
Explanation: CP = SP * (100/(100 + profit%))
= 5400(100/132) = Rs.4090.
Answer: A
Question: The difference between compound interest and simple interest on a certain amount of money at 5% per annum for 2 years is 15. Find the sum :
A.4500
B.7500
C.5000
D.6000
Answer: D.6000
Explanation: Sol.(d) Let the sum be 100.
Therefore, SI = 100×5×2100=10100×5×2100=10
and CI = 100(1+5100)2−100100(1+5100)2−100
∴ =100×21×2120×20−100=414=100×21×2120×20−100=414
Difference of CI and SI = 41⁄4 - 10 = 1⁄4
If the difference is 1⁄4 , the sum = 100
=> If the difference is 15, the sum
= 400 × 15 = 6000 Answer D
Question: C and D work at a hospital with 4 other workers. For an internal review, 2 of the 6 workers will be randomly chosen to be interviewed. What is the probability that C and D will both be chosen?
A.A)1/3
B.B)1/4
C.C)1/15
D.D)3/8
Answer: C.C)1/15
Explanation: Total number of people = 6
Probability of selecting C first and then D is 1/6∗1/5=1/30
Probability of selecting D first and then C is 1/6∗1/5=1/30
Therefore probability of selecting Cand D for the review is 1/30+1/30=1/15
Answer is C
Question: How many ways are there to award a gold, silver and bronze medal to 15 contending teams?
A.15 × 14 × 13
B.10!/(3!7!)
C.10!/3!
D.360
Answer: A.15 × 14 × 13
Explanation: we clearly know that there can be only one winning team which deserves the gold medal. we can do the selection in 15 ways
if gold medal is given to 1 of the teams then only possible 14 teams can be considered for silver medal. we can do selection in 14 ways
similarly if gold and silver medals are awarded then only remaining 13 teams can be considered for a bronze medal. we can do the selection in 13 ways
Total number of ways to select the 3 possible medal winners = 15 * 14 * 13
Correct answer - A
Question: The tax on a commodity is diminished by 16% and its consumption increased by 15%. The effect on revenue is?
A.2%
B.3.8%
C.3.4%
D.3.6%
Answer: C.3.4%
Explanation: 100 * 100 = 10000
84 * 115 = 9660
-----------
10000-----------340
100-----------? => 3.4% decrease
Answer: C
Question: If t > 0, which of the following could be true?
I. t^3 > t^2
II. t^2 = t
III. t^2 > t^3
A.I only
B.III
C.IIIII
D.All of the above
Answer: D.All of the above
Explanation: 1. t^3 > t^2 Well this is true for all t > +1 (but not for values 0 < t < 1)
2. t^2 = t This is true for only one positive number, 1
3. t^2 > t^3 Again this is true only for values of t such that 0 < t < 1
So the answer is D, since there exists at least one value of t which satisfies the equation/ in equation(s)
Question: Marts income is 60 percent more than Tims income and Tims income is 50 percent less than Juans income. What percentage of Juans income is Marts income
A.124%
B.b) 120 %
C.c) 96 %
D.d) 80 %
Answer: D.d) 80 %
Explanation: M = (160/100)T
T = (50/100)J
=> M = (80/100)J
Answer D.
Question: 5358 x 53 = ?
A.283974
B.283758
C.273298
D.273258
Answer: A.283974
Explanation: A
5358 x 53 = 5358 x (50 + 3)
= 5358 x 50 + 5358 x 3
= 267900 + 16074
= 283974.
Question: A dishonest milkman wants to make a profit on the selling of milk. He would like to mix water (costing nothing) with milk costing Rs.33 per litre so as to make a profit of 20% on cost when he sells the resulting milk and water mixture for Rs.36 In what ratio should he mix the water and milk?
A.1:20
B.1:10
C.1:8
D.1:4
Answer: B.1:10
Explanation: cost needed to net a 20% profit:
(36-x)/x = .2
x=30
actual cost:
33
solution (x=liters of water needed to be added to the 1liter of milk):
33/(1+x)=30
x = 1/10
so to get the cost down to 30
Milk:Water
1:1/10
or (10/10):(1/10)
ANSWER:B
Question: What is the unit digit in 7105 ?
A.1
B.5
C.7
D.9
Answer: C.7
Explanation: Unit digit in 7105 = unit digit in [(74)26 * 7]
But, unit digit in (74)26 = 1
unit digit in 7105 = (1 * 7) = 7
ANSWER:C
Question: The manager of a theater noted that for every 10 admissions tickets sold, the theater sells 3 bags of popcorn at 2.40 $ each, 4 sodas at 1.50 $ each and 2 candy bars at 1.00$ each. To the nearest cent, what is the average (arithmetic mean) amount of these snacks sales per ticket sold?
A.1.48$
B.1.52$
C.1.60$
D.1.64$
Answer: B.1.52$
Explanation: For every 10 tickets amount of snacks sold is 3*2.40+4*1.5+2*1=$15.2, hence amount of the snacks sales per ticket is $15.2/10=~$1.52.
Answer: B.
Question: There were two candidates in an election. Winner candidate received 52% of votes and won the election by 288 votes. Find the number of votes casted to the winning candidate?
A.776
B.3744
C.299
D.257
Answer: B.3744
Explanation: W = 52% L = 48%
52% - 48% = 4%
4% -------- 288
52% -------- ? => 3744
Answer:B
Question: Three containers A,B and C are having mixtures of milk and water in the ratio of 1:5 and 3:5 and 5:7 respectively. If the capacities of the containers are in the ratio of all the three containers are in the ratio 5:4:5, find the ratio of milk to water, if the mixtures of all the three containers are mixed together.
A.53:115
B.55:115
C.115:55
D.5:115
Answer: A.53:115
Explanation: Assume that there are 500,400 and 500 liters respectively in the 3 containers.
Then ,we have, 83.33, 150 and 208.33 liters of milk in each of the three containers.
Thus, the total milk is 441.66 liters.
Hence, the amount of water in the mixture is 1400-441.66=958.33liters.
Hence,
the ratio of milk to water is 441.66:958.33 => 53:115(using division by .3333)
The calculation thought process should be
(441*2+2):(958*3+1)=1325:2875
Dividing by 25 => 53:115.
ANSWER A 53:115
Question: For 600 km journey it will take 8hrs, if 120 km is done by bus and the rest by car. It takes 20 minutes more, if 200 km is done by busand the rest by car. The ratio of the speed of the train to that of the cars is?
A.1 : 3
B.3 : 4
C.2 : 3
D.2 : 5
Answer: B.3 : 4
Explanation: Let the speed of the train be x km/hr and that of the car be y km/hr.
Then,	120	+	480	= 8 1	+	4	=	1	....(i)
x	y	x	y	15
And,	200	+	400	=	25 1	+	2	=	1	....(ii)
x	y	3	x	y	24
Solving (i) and (ii), we get: x = 60 and y = 80.
Ratio of speeds = 60 : 80 = 3 : 4.
B
Question: Robin is traveling from one end of a forest to the other. In order to find her way back, she is leaving morsels of bread in the following pattern: 2 morsels of GRAIN, 3 morsels of white, and 1 morsel of rye. The pattern repeats after she leaves the morsel of rye. If Robin drops 2,000 morsels of bread, what are the last 3 morsels of bread that she drops?
A.rye − wheat − wheat
B.wheat − wheat − white
C.white − rye − wheat
D.RYE-GRAIN-GRAIN
Answer: D.RYE-GRAIN-GRAIN
Explanation: Q is nothing BUT asking about remainder..
total morsels thrown before the pattern is repeated = 2+3+1 = 6..
so REMAINDER when 2000 is div by 6..
1998 is EVEN and also div by 3, so remainder is 2..
Last 3 of morsels are 1998-1999-2000
so he uses the last of morsel of pattern in 1998 two of the first morsels of the pattern in 1999 and 2000, and the pattern is G, G, W, W, W, R..
ans R-G-G
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: Solution
Let the three integers be x, x +2 and x + 4. Then, 3x = 2 (x + 4) + 3 ⇔ x = 11.
∴ Third integer = x + 4 = 15.
Answer D
Question: A tap can fill a tank in 6 hours. After half the tank is filled, three more similar taps are opened. What is the total time taken to fill the tank completely?
A.3 hrs 45 min
B.4 hrs
C.2 hrs
D.3 hrs 15 min
Answer: A.3 hrs 45 min
Explanation: time taken by one tap to fill half = 3 hrs
part filled by four taps in 1 hr (4*1/6)=2/3
remain =(1-1/2)=1/2
=>2/3:1/2=>1:x
tot time = 3 hrs 45 min
ANSWER A
Question: Total 30 cows 10 cow gives each 2 liter milk 10 cow gives each 3/4 liter milk 10 cow gives each 1/4 liter milk this is split into 3 son per each 10 cows & 10 liter milk how?
A.10
B.12
C.15
D.16
Answer: A.10
Explanation: 10 cow 2 liter each =20 liter
10cow 3/4 liter each= 3/4=0.758*10=7.5
10 cow 1/4 liter each = 1/4=0.25*10=2.5
add 20+7.5+2.5=30
milk split into 3 son each 10 liter then 30/3=10
ANSWER:A
Question: In an office, at various times during the day the boss gives the secretary a letter to type, each time putting the letter on top of the pile in the secretary's inbox. Secretary takes the top letter and types it. Boss delivers in the order 1, 2, 3, 4, 5 which cannot be the order in which secretary types?
A.4, 5, 2, 3, 9
B.4, 5, 2, 3, 1
C.4, 5, 2, 3, 0
D.4, 5, 2, 3, 2
Answer: B.4, 5, 2, 3, 1
Explanation: 4, 5, 2, 3, 1
Answer:	B
Question: An accurate clock shows 8 o'clock in the morning. Through how may degrees will the hour hand rotate when the clock shows 2 o'clock in the afternoon?
A.267
B.180
C.287
D.177
Answer: B.180
Explanation: Angle traced by the hour hand in 6 hours
=(360/12)*6
Answer: B
Question: A factory produces x widgets per day. The factory's fixed costs are $7000 per day. The price per widget is $80 and the variable costs are $20 per widget. How many widgets need to be produced for profits of $6200 a day?
A.42.33
B.90.33
C.168
D.220
Answer: D.220
Explanation: profits=6200=60x-8000 --> x=220
Answer D.
Question: Mani buys Rs. 20 shares paying 9% dividend. The man wants to have an interest of 12% on his money. The market value of each share must be:
A.10
B.13
C.15
D.20
Answer: C.15
Explanation: shares = 20%
==> 9%
interest of 12%==>15
ANSWER C
Question: The sum of the first 100 numbers, 1 to 100 is always divisible by
A.2
B.2 and 4
C.2, 4 and 8
D.2, 4 and 9
Answer: A.2
Explanation: Explanation:
The sum of the first 100 natural numbers = (100 x 101)/2
= (50 x 101)
As 101 is an odd number and 50 is divisible by 2, so the sum is always divisible by 2.
Answer: Option A
Question: A train when moves at an average speed of 40 kmph, reaches its destination on time. When its average speed becomes 35 kmph, then it reaches its destination 15 minutes late. Find the length of journey.
A.70 km
B.80 km
C.40 km
D.20 km
Answer: A.70 km
Explanation: Sol.
Difference between timings = 15 min = 1/4 hr.
Let the length of journey be x km.
Then, x/35 - x/40 = 1/4
⇔ 8x - 7x = 70
⇔ x = 70 km.
Answer A
Question: Two positive integers differ by 4, and sum of their reciprocals is 6. Then one of the numbers is
A.a) 3
B.b) 1
C.c) 5
D.4
Answer: D.4
Explanation: Algebraic approach:
Let n be the smaller integer => 1/n + 1/(n+4) = 6
or ((n+4)+n)/n(n+4) =6 or (n^2+4n)*6 =2n+4 or n=2 as n cannot be -negative
Solve for n => n=4. Hence,
D
Question: The average of 25 results is 24. The average of first 12 of those is 14 and the average of last 12 is 17. What is the 13th result?
A.740
B.228
C.690
D.780
Answer: B.228
Explanation: Solution: Sum of 1st 12 results = 12*14
Sum of last 12 results = 12*17
13th result = x (let)
Now,
12*14+12*17+x = 25*24
Or, x = 228.
Answer: Option B
Question: In a consumer survey, 65% of those surveyed liked at least one of three products: 1, 2, and 3. 50% of those asked liked product 1, 30% liked product 2, and 20% liked product 3. If 5% of the people in the survey liked all three of the products, what percentage of the survey participants liked more than one of the three products?
A.30
B.10
C.15
D.20
Answer: A.30
Explanation: pick 100 and x= all set with exactly two items
65=30+50+20 -(X)-10
X=25
so the answer is 25+5/100= 30%
Answer A
Question: At the moment there are 54,210 tagged birds in a certain wildlife refuge. If exactly 30 percent of all birds in the refuge are tagged, what percent of the untagged birds must be tagged so that half of all birds in the refuge are tagged?
A.66.67
B.66
C.67
D.65.99
Answer: A.66.67
Explanation: All birds = 54210
Currently Tagged = 54210 * 30/100 = 16263
So Untagged birds = 54210 - 16263 = 37947
Half of all birds = 27105
The number of birds to be tagged to make half of all birds tagged = 27105 - 16263 = 10842
So now the question remains - 10842 is how much percentage of untagged birds (16263)
= 10842*100/16263
= 66.67
Answer (A)
Question: Renuka got married 8 years ago. Today her age is 4/3 times of marriage. Her daughter age is 1/8 times her age. her daughter age is
A.3 years
B.4 years
C.6 years
D.8 years
Answer: B.4 years
Explanation: Renuka got married 8 years ago. Today her age is 4/3 times of marriage.
If her age at time of marriage was x, then
x*4/3 = x+8x/3=8
x=24 yrs
Her present age = 24+8=32 yrs
Daughter's age = 32/8= 4 years
ANSWER:B