divelab/Sys2Bench huggingface dataset:
Question: In a factory, an average of 65 TV's are produced per day for the fist 25days of the months. A few workers fell ill for the next 5days reducing the daily avg for the month to 60 sets/day . The average production per day for day last 5days is ?
Options:
A)20
B)35
C)48
D)50
E)59
Rationale: Production during these 5 days = Total production in a month - production in first 25 days.
= 30 x 60 - 25 x 65
= 175
∴ Average for last 5 days = 175 / 5
= 35
Aswer : B
Answer: B
Question: If a particular player is always chosen, in how many ways can a cricket-be chosen out of 15 players?
Options:
A)1236
B)1345
C)1358
D)1321
E)1365
Rationale: A particular player is always chosen, it means that 10 players are selected out of the remaining 14 players.
=. Required number of ways = 14C10 = 14C4
= 14!/4!x19! = 1365
E
Answer: E
Question: Aman started a business investing Rs. 70,000. Rakhi joined him after six months with an amount of Rs. 1,05,000 and Sagar joined them with Rs. 1.4 lakhs after another six months. The amount of profit earned should be distributed in what ratio among Aman, Rakhi and Sagar respectively, 3 years after Aman started the business?
Options:
A)7:6:10
B)12:15:16
C)42:45:56
D)Cannot be determined
E)None of these
Rationale: Aman : Rakhi : Sagar = (70000 * 36) : (105000 * 30) : (140000 * 24)
= 12:15:16
ANSWER:B
Answer: B
Question: A bookseller sells his books at a 20% markup in price. If he sells a book for $24.00, how much did he pay for it?
Options:
A)$14.40
B)$14.00
C)$10.00
D)$9.60
E)$20.00
Rationale: Let the cost price of book = x
Selling price of book = 24$
Markup % = 20
(120/100)x = 24
=> x = 20
Answer E
Answer: E
Question: What is the median from the below series 90, 92, 91, 88, 95, 88, 97, 87, and 98
Options:
A)80
B)85
C)91
D)97
E)100
Rationale: Ordering the data from least to greatest, we get:
87, 88, 88, 90, 91, 93, 95, 96, 98
The median quiz score was 91. (Four quiz scores were higher than 91 and four were lower.)
C
Answer: C
Question: 10% of the inhabitants of village having died of cholera.,a panic set in , during which 25% of the remaining inhabitants left the village. The population is then reduced to 4050. Find the number of original inhabitants.
Options:
A)6000
B)5000
C)3000
D)2000
E)1000
Rationale: Let the total number of orginal inhabitants be x.
((75/100))*(90/100)*x)=4050  (27/40)*x=4050
x=((4050*40)/27)=6000.
ANSWER A
Answer: A
Question: 100 × 10 – 100 + 2000 ÷ 100 = ?
Options:
A)29
B)780
C)920
D)979
E)None of these
Rationale: Solution
Given exp. = 100 × 10 - 100 + 20 = 1000 - 100 + 20 = 1020 - 100 = 920.
Answer C
Answer: C
Question: There are 5 sweets - Jammun, kaju, Peda, Ladu, Jilebi which can be consumed in 5 consecutive days. Monday to Friday. A person eats one sweet a day, based on the following constraints.
(i) Ladu not eaten on monday
(ii) If Jamun is eaten on Monday, Ladu should be eaten on friday.
(iii) Peda is eaten the day following the day of eating Jilebi
(iv) If Ladu eaten on tuesday, kaju should be eaten on monday
based on above, peda can be eaten on any day except
Options:
A)monday
B)tuesday
C)tuWesday
D)wednesday
E)friday
Rationale: From the (iii) clue, peda must be eaten after jilebi. so Peda should not be eaten on monday.
Answer:A
Answer: A
Question: A man takes 3 hours 45 minutes to row a boat 15 km downstream of a river and 2 hours 30 minutes to cover a distance of 5 km upstream. Find the speed of the current.
Options:
A)1 km/hr
B)2 km/hr
C)3 km/hr
D)4 km/hr
E)None of these
Rationale: Explanation:
First of all, we know that
speed of current = 1/2(speed downstream - speed upstream) [important]
So we need to calculate speed downstream and speed upstream first.
Speed = Distance / Time [important]
Speed upstream =(15/3 3/4)km/hr
=15×4/15=4km/hr
Speed Downstream = (5/2 1/2)km/hr
=5×2/5=2km/hr
So speed of current = 1/2(4−2)
=1km/hr
Option A
Answer: A
Question: A train 120 m long is running with a speed of 54 km per hour. In what time will it pass a telegraph post?
Options:
A)11 s
B)10 s
C)8 s
D)12 s
E)None of these
Rationale: We know from the formula Time = Distance/ Speed
Thus, Time = 120/ 54 x 5/18
or, Time = 8 sec.
ANSWER:C
Answer: C
Question: A candidate who gets 30% of the marks fails by 60 marks. But another candidate who gets 45% marks gets 30 marks more than necessary for passing. Find the number of marks for passing?
Options:
A)100 Marks
B)240 Marks
C)280 Marks
D)371 Marks
E)827 Marks
Rationale: 30% ------------ 60
45% ------------ 30
----------------------
15% ------------- 90
30% -------------- 180
180 + 60 = 240 Marks
Answer:B
Answer: B
Question: A can do a particular work in 6 days . B can do the same work in 8 days. A and B signed to do it for Rs. 3520. They completed the work in 3 days with the help of C. How much is to be paid to C?
Options:
A)s. 380
B)s. 600
C)s. 420
D)s. 440
E)s. 480
Rationale: Explanation :
Amount of work A can do in 1 day = 1/6
Amount of work B can do in 1 day = 1/8
Amount of work A + B can do in 1 day = 1/6 + 1/8 = 7/24
Amount of work A + B + C can do = 1/3
Amount of work C can do in 1 day = 1/3 - 7/24 = 1/24
work A can do in 1 day: work B can do in 1 day: work C can do in 1 day
= 1/6 : 1/8 : 1/24 = 4 : 3 : 1
Amount to be paid to C = 3520 × (1/8) = 440
Answer : Option D
Answer: D
Question: Running at the same constant rate, 6 identical machines can produce a total of 360 bottles per minute. At this rate, how many bottles could 10 such machines produce in 4 minutes?
Options:
A)648
B)1800
C)2400
D)10800
E)10900
Rationale: Let the required number of bottles be x.
More machines, More bottles (Direct Proportion)
More minutes, More bottles (Direct Proportion)
Machines	6	:	10	:: 360 : x
Time (in minutes)	1	:	4
6 x 1 x x = 10 x 4 x 360
x =	(10 x 4 x 360)/(6)
x = 2400.
ANSWER:C
Answer: C
Question: Farm A had an average output of 350 kgs of wheat per acre. Farm B had an average output of 300 kgs of wheat per acre. It is known that Farm B's area is 15 acres less than Farm A's. It is also known that Farm B in total harvested 2000 kgs of wheat more than Farm A. What is the area of Farm A?
Options:
A) 130 acres
B) 135 acres
C) 145 acres
D) 160 acres
E) 180 acres
Rationale: The area of Farm A = x acres;
The area of Farm B = x-15 acres.
Total output of Farm A = 350x kgs;
Total output of Farm B = 300(x-15) kgs.
Farm B in total harvested 2000 kgs of wheat more than Farm A: 350x+2000=300(x-15) --> x=130 acres.
Answer: a.
Answer: A
Question: Which letter will come exactly between the tenth letter from your left and the seventh letter from your right.
Without changing any order in the original from the alphabet?
Options:
A)U
B)Y
C)O
D)Y
E)R
Rationale: 10th letter from your left is J
7th letter from your right is T
The mid letter between J and T is O.
Answer: Option 'C'
Answer: C
Question: If f(x)=4x−1 and g(x)=2x+5 for all integers, which of the following is a possible value of g(f(x))?
Options:
A)-23
B)-17
C)0
D)8
E)19
Rationale: g(f(x)) = 2*f(x)+5 = 2*(4x-1)+5 = 8x+3
Of these five options, only 19 has the form 8x+3.
The answer is E.
Answer: E
Question: A vendor sells 30 percent of the apples he had and throws away 20 percent of the remainder. The next day, the vendor sells 50 percent of the remaining apples and throws away the rest. In total, what percent of his apples does the vendor throw away?
Options:
A)40
B)42
C)44
D)46
E)48
Rationale: Let x be the original number of apples.
On day one, the vendor throws away (0.2)(0.7)x=0.14x.
The remaining apples are (0.8)(0.7)x = 0.56x.
On day two, the vendor throws away (0.5)(0.56)x=0.28x.
The vendor throws away a total of 0.14x + 0.28x = 0.42x.
The vendor throws away 42 percent of the apples.
The answer is B.
Answer: B
Question: A car traveled 340 miles per tankful of gasoline on the highway and 336 miles per tankful of gasoline in the city. If the car traveled 6 fewer miles per gallon in the city than on the highway, how many miles per gallon did the car travel in the city?
Options:
A)504
B)416
C)321
D)220
E)170
Rationale: Let the speed in highway be h mpg and in city be c mpg.
h = c+6
h miles are covered in 1 gallon
340 miles will be covered in 340/h.
Similarly c miles are covered in 1 gallon
336 miles will be covered in 336/c.
Both should be same (as car's fuel capacity does not change with speed)
=> 336/c = 340/h
=> 336/c = 340/(c+6)
=> 336c+336*6=340c
=>c=336*6/4=504
Answer A.
Answer: A
Question: A can do a piece of work in 14 days and B in 20 days. They began the work together but 5 days before the completion of the work, A leaves. The work was completed in?
Options:
A)15 3/17 days
B)11 3/17 days
C)12 3/17 days
D)19 3/17 days
E)13 3/17 days
Rationale: B
(x â€“ 5)/14 + x/20 = 1
x = 11 3/17 days
Answer: B
Question: Two cars cover the same distance at the speed of 60 and 64 kmps respectively. Find the distance traveled by them if the slower car takes 1 hour more than the faster car?
Options:
A)228
B)288
C)960
D)882
E)271
Rationale: 60(x + 1) = 64x
X = 15
60 * 16 = 960 km
Answer: C
Answer: C
Question: The ratio between the number of sheep and the number of horses at the Stewar farm is 2 to 7. If each of horse is fed 230 ounces of horse food per day and the farm needs a total 12880 ounces of horse food per day. what is number sheep in the form??
Options:
A)18
B)16
C)32
D)56
E)58
Rationale: et no of sheep and horses are 2k and 7k
no of horses= 12880/230=56
now 7K=56 and k=8
no of sheep=(2*8) =16
ANSWER:B
Answer: B
Question: A box contains eight apples out of which 4 are rotten. If four apples are chosen at random, find the probability that all the four apples are rotten?
Options:
A)1/113
B)1/90
C)1/80
D)1/70
E)None of these
Rationale: Out of nine apples , five are good and four are rotten.
Required probability = 4C4/8C4 = 1/70
D
Answer: D
Question: Which of the following equations has a root in common with x^2−7x+10=0 ?
Options:
A)x^2+5=0
B)x^2-x-1=0
C)x^2-6x+5=0
D)x^2-2x-1=0
E)x^2-1=0
Rationale: If we carefully look at the given equation we can arrange it in the following manner:
(x-5)(x-2) = 0
so the two roots are 5 and 2.
Now put 5 and 2 in given equations. The equation in which one of them gives value 0, that will be our answer.
(A) putting 5: we get 30; Putting 2 we get 9. Reject this option [by the way x^2+1 cannot have real roots]
(B) putting 5: we get 19; Putting 2 we get 1. Reject this option too [by the way x^2-x-1 cannot have real roots]
(C) putting 5: we get 0. This is the equation.
We are lucky. No need to check other options.
C is the answer.
Answer: C
Question: The difference between simple and compound interest on Rs. 1600 for one year at 10% per annum reckoned half-yearly is?
Options:
A)8
B)3
C)9
D)4
E)2
Rationale: S.I. = (1600 * 10 * 1)/100 = Rs. 160
C.I. = [1600 * (1 + 5/100)2 - 1600]
= Rs. 164 Difference = (164 - 160)
= Rs.4.
Answer:D
Answer: D
Question: How many numbers greater than 1000 can be formed with the digits 3,4,6,8,9 if a digit cannot occur more than once in a number?
Options:
A)240
B)120
C)360
D)60
E)600
Rationale: This is choosing 4 from 5 (any 4 digit number chosen from 3,4,6,8,9 will be>1000) plus 5 from 5 (any 5 digit number will be >1000), where order is important.
So the number of ways we can arrange the given digits so that our resulting number is greater than 1000 such that no digit occurs more than once, is:
5P4 + 5P5 = 240
A
Answer: A
Question: A fair coin is tossed 9 times. What is the probability of getting more heads than tails in 9 tosses?
Options:
A)1/2
B)63/128
C)4/7
D)61/256
E)63/64
Rationale: On each toss, the probability of getting a head is 1/2 and the probability of getting a tail is 1/2.
There is no way to get the same number of heads and tails on an odd number of tosses.
There will either be more heads or more tails.
Then there must be more heads on half of the possible outcomes and more tails on half of the possible outcomes.
P(more heads) = 1/2
The answer is A.
Answer: A
Question: A certain fraction is equivalent to 2/5. If the numerator of the fraction is increased by 3 and the denominator is doubled, the new fraction is equivalent to 1/3. What is the sum of the numerator and denominator of the original fraction?
Options:
A) 21/4
B) 35
C) 28
D) 26
E) 21
Rationale: x/y = 2/5 -> 1
(x+3)/2y = 1/3 -> 2
Divide 1 by 2 :
=> 2x/(x+3) = 6/5
=> 10x = 6x + 18
=> x = 3/2
=> y = 5/2 * 3/2= 15/4
so x + y = 21/4
The answer is A
Answer: A
Question: Dawson’s salary is reduced by 20% and then reduced salary is increased by 20%. Find ,how many percentage his present salary is less as compared to his previous salary?
Options:
A)0%
B)1%
C)2%
D)3%
E)4%
Rationale: 20*20
-------- = 4%
100
Answer: E
Answer: E
Question: Vlaudia can choose any two of four different candles and any 8 of 9 different flowers for a centerpiece arrangement. Given these choices, how many candle + flower groupings can she select?
Options:
A)54
B)72
C)96
D)144
E)432
Rationale: Vlaudia can choose any two of four different candles and any 8 of 9 different flowers for a centerpiece arrangement. 2 candles from 4 candles: 4C2: 6
8 flowers from 9 flowers: 9C8 : 9
So total grouping
9*6= 54.
Answer is A.
Answer: A
Question: The proportion of water to alcohol in Solution A is 4:1 and the proportion of water to alcohol in Solution B is 3:1. If an equal amount of each solution is mixed together, what is the concentration of alcohol in the new solution?
Options:
A)16.5%
B)18.5%
C)20.5%
D)22.5%
E)24.5%
Rationale: Let V be the total volume of the new solution.
Then a volume of V/2 was added from each solution A and B.
The amount of alcohol added to the new solution was:
(1/5)(V/2)+(1/4)(V/2)=V/10+V/8=9V/40.
The concentration of alcohol is 9/40=22.5%
The answer is D.
Answer: D
Question: A trader bought a car at 20% discount on its original price. He sold it at a 40% increase on the price he bought it. What percent of profit did he make on the original price?
Options:
A)62%
B)42%
C)12%
D)22%
E)82%
Rationale: Explanation:
Original price = 100
CP = 80
S = 80*(140/100) = 112
100 - 112 = 12%
Answer: C
Answer: C
Question: In a partnership, A invests 1⁄6 of the capital for 1⁄6 of the time, B invests 1⁄3 of the capital for 1⁄3 of the time and C, the rest of the capital for whole time. Find A’s share of the total profit of 2,300.
Options:
A)100
B)200
C)300
D)400
E)None of these
Rationale: Remaining capital = 1 - (1⁄6 + 1⁄3) = 1⁄2
Ratio of their profit
= 1⁄6 × [1⁄6 × 12] : 1⁄3 × [1⁄3 × 12] : 1⁄2 × 12
= 1⁄3 : 4⁄3 : 6 = 1 : 4 : 18
∴ A's share = 1 /1+4+18×2300 = 100
Answer A
Answer: A
Question: A and B can do a piece of work in 10 days and 10 days respectively. They work together for 10 days and B leaves. In how many days the whole work is completed?
Options:
A)30days
B)40days
C)50days
D)60days
E)70days
Rationale: Explanation:
(A+B)’s 10 days work = 10[1/10+1/10]=10[1+1/10]= 2
A complete remaining work in 2*10 = 20
Total work = 10+20 = 30days
Answer: Option A
Answer: A
Question: A furniture store owner decided to drop the price of her recliners by 20% to spur business. By the end of the week she had sold 40% more recliners. What is the percentage increase of the gross?
Options:
A)10%
B)15%
C)11.2%
D)7%
E)13%
Rationale: Say a recliner is actually worth $100
If she sells 100 recliners then she earns $10000
After the discount of 20%, she will earn $80 per recliner and she sells 40% more ie.,140 recliners
Hence her sales tields 140*80=$11200
Increase in sales=11200-10000=$1120
So% increase = 1120*100/10000 =11.20%
C is the answer
Answer: C
Question: In a class, the number of girls is 20% more than that of the boys. The strength of the class is 66. If 4 more girls are admitted to the class, the ratio of the number of boys to that of the girls is
Options:
A)1:2
B)3:4
C)1:4
D)3:5
E)None
Rationale: Solution: Girls:boys = 6:5;
Hence, girls = 6*66/11 = 36;
Boys = 30;
New ratio, 30:(36+4) = 3:4.
Answer: Option B
Answer: B
Question: A batsman scored 150 runs which included 3 boundaries and 8 sixes. What percent of his total score did he make by running between the wickets?
Options:
A)50%
B)40%
C)60%
D)70%
E)80%
Rationale: Number of runs made by running = 150 - (3 x 4 + 8 x 6)
= 150 - (60)
= 90
Now, we need to calculate 90 is what percent of 150.
=> 90/150 x 100 = 60 %
ANSWER:C
Answer: C
Question: (1.00003)(0.99997) - (1.00002)(0.99998) =
Options:
A)0
B)-5(10^-10)
C)3(10^-10)
D)10^-5
E)3(10^-5)
Rationale: (1.00003*0.99997)-(1.00002*0.99998)
= (1+0.00003)(1-0.00003) - (1+0.00002)(1-0.00002)
= 1 - (0.00003)^2 - 1 + (0.00002)^2
= -(0.0000000009) + (0.0000000004)
= - 0.0000000005= -5(10^-10)
Answer B
Answer: B
Question: The student manager of the university theater estimated that for every 6 movie tickets sold on Valentine's Weekend, the theater will sell 3 packs of grain crackers at $2.25 each, 4 bottles of a beverage at $1.50 each, and 4 chocolate bars at $1.00 each. Find the approximate average (arithmetic mean) amount of estimated snack sales per movie ticket sold.
Options:
A)$1.40
B)$1.58
C)$2.00
D)$2.64
E)$2.79
Rationale: 3*2.25 + 4 *1.50+ 4 * 1 = 6.75+6+4 = 16.75.
This is the sale for 6 tickets
Avg sale per ticket: 16.75/6 = $2. 79
Answer : E
Answer: E
Question: A deck of 9 cards contains 2 red cards, 3 blue cards, and 4 green cards. 3 cards are randomly drawn with replacement from the deck. What is the probability that all three cards are the same color?
Options:
A)11/81
B)64/729
C)5/7
D)27/729
E)2/5
Rationale: Note that cards are drawn with replacement. So every time we draw from 9 cards and it's possible to draw 3 red cards, 3 blue cards or 3 green cards:
P=(2/9)3+(3/9)3+(4/9)3=11/81
Answer: A.
Answer: A
Question: The H.C.F. of two numbers is 11 and their L.C.M. is 7700. If one of the numbers is 275, then the other is:
Options:
A)278
B)288
C)298
D)308
E)318
Rationale: Other numberOther number =(11x7700)/275 =308
answer :D
Answer: D
Question: The compound ratio of 1:3, 3:2 and 4:5?
Options:
A)1:9
B)1:7
C)1:2
D)2:5
E)1:4
Rationale: 1/3 * 3/2 * 4/5
= 2/5
=2:5
Answer: D
Answer: D
Question: How many numbers are divisible by both 2 and 3 up to 300?
Options:
A)50
B)100
C)150
D)200
E)250
Rationale: Divide 300 by 6, as the numbers which are divisible by both 2 and 3 are also divisible by 6.
The quotient obtained after dividing 300 by 6 is 50, hence answer is 50.
Option A
Answer: A
Question: Find the least number must be subtracted from 427398 so that remaining no.is divisible by 15?
Options:
A)3
B)5
C)6
D)9
E)8
Rationale: On dividing 427398 by 15 we get the remainder 3, so 3 should be subtracted
A
Answer: A
Question: what is the product of first three whole no.
Options:
A)0
B)1
C)6
D)2
E)3
Rationale: first three whole no. are 0,1,2.
remember whole no. start from 0 and natural no. from 1
so product will 0 only.
answer A
Answer: A
Question: The result when a number subtracted from 150 is the same as the number added to 68. what is the number?
Options:
A)110
B)100
C)109
D)218
E)210
Rationale: Answer
Let the number be x.
Then, x - 68= 150- x
â‡’ 2x = 150+ 68= 218
â‡’ 2x = 218
âˆ´ x = 109
Correct Option: C
Answer: C
Question: The ratio of the incomes of Chetan and Dinesh is 3:4. The ratio of their expenditures is 5:7. If each of them saves Rs.200, find the incomes of both?
Options:
A)1600
B)2888
C)2662
D)9222
E)7122
Rationale: The savings of Chetan and Dinesh are 3x - 5y and 4x - 7y respectively.
3x - 5y = 200 --- (1)
4x - 7y = 200 --- (2)
Multiplying (1) by 7 and (2) by 5 and subtracting the resultant equation (2) from resultant equation (1), we get x = 400.
The incomes of Chetan and Dinesh are 3x = Rs.1200 and 4x = Rs.1600 respectively.
Answer: A
Answer: A
Question: A sum of Rs.2200 has been divided among A, B and C such that A gets 1/4 of what B gets and B gets 1/5 of what C gets. What is B's share?
Options:
A)Rs. 341
B)Rs. 364
C)Rs. 372
D)Rs. 352
E)Rs. 354
Rationale: Let C's share =x
Then B's share =x/5
A's share =(x/5)×1/4=x/20
x+(x/5)+(x/20)=2200
⇒x(1+(1/5)+(1/20))=2200
⇒x(25/20)=2200
⇒x(5/4)=2200
⇒x/4=440
⇒x=1760
B's share=x/5=1760/5=352
Answer: D
Answer: D
Question: P can do the work in 10 days and Q can do the same work in 15 days. If they work together for 4 days, what is the fraction of work that is left?
Options:
A)1/5
B)1/4
C)1/3
D)2/5
E)2/3
Rationale: 4/10 + 4/15 = 20/30 = 2/3 (completed work)
The work that is left is 1 - 2/3 = 1/3
The answer is C.
Answer: C
Question: On his first 3 tests, Rajeev received an average score of X points. If on his fourth test, he exceeds his previous average score by 24 points, what is his average score for his first 4 tests?
Options:
A)X
B)X+6
C)X+4
D)X+3
E)X+2
Rationale: Total of 1st 3 tests= 3X
4th test score = X + 24
Total of all 4 tests = 4X + 24
Average = X + 6
Answer : B
Answer: B
Question: Find the value of (1/(2*3))+(1/(3*4))+(1/(4*5))+(1/(5*6))+…..+ ((1/(9*10)).
Options:
A)1/5
B)2/5
C)4/5
D)3/5
E)None of them
Rationale: Given expression=((1/2)-(1/3))+((1/3)-(1/4))+((1/4)-(1/5))+
((1/5)-(1/6))+….+ ((1/9)-(1/10))
=((1/2)-(1/10))=4/10 = 2/5.
Answer is B.
Answer: B
Question: If 314,314 football shoes need to be placed in boxes and each box can contain at most 6 pair of football shoes. How many pair of football shoes will be in the last unfilled box?
Options:
Options:
A)1football shoes
B)3 football shoes
C)4 football shoes
D)7 football shoes
E)9 football shoes
Rationale: The number of football shoes that can be placed in boxes of 6 football shoes is the first number that is divisible by 6 that occurs before 314,314.
In order to divide the sum in 6 parts, the amount must be divisible by 6
Divisibility rule of 6: The sum of the digits must be divisible by 6
Sum of digits of 314,314 = 16 and 6 is divisible by 6.
Hence, we need to remove 4 to this number for it to be divisible by 6
Correct Option: c
Answer: C
Question: A man bought a horse and a cart. If he sold the horse at 10 % loss and the cart at 20 % gain, he would not lose anything; but if he sold the horse at 5% loss and the cart at 5% gain, he would lose Rs. 10 in the bargain. The amount paid by him was Rs._______ for the horse and Rs.________ for the cart.
Options:
A)Cost price of horse = Rs. 230 & the cost price of cart = 900.
B)Cost price of horse = Rs. 900 & the cost price of cart = 700.
C)Cost price of horse = Rs. 400 & the cost price of cart = 200.
D)Cost price of horse = Rs. 300 & the cost price of cart = 800.
E)Cost price of horse = Rs. 800 & the cost price of cart = 800.
Rationale: Let x be the cost price of the horse and y be the cost price of the cart.
In the first sale there is no loss or profit. (i.e.) The loss obtained is equal to the gain.
Therefore (10/100) * x = (20/100) * y
X = 2 * y -----------------(1)
In the second sale, he lost Rs. 10. (i.e.) The loss is greater than the profit by Rs. 10.
Therefore (5 / 100) * x = (5 / 100) * y + 10 -------(2)
Substituting (1) in (2) we get
(10 / 100) * y = (5 / 100) * y + 10
(5 / 100) * y = 10
y = 200
From (1) 2 * 200 = x = 400
Answer:C
Answer: C
Question: The average age of the district level hockey team of eleven is 22 years.The average age gets increased by 3 years when the coach age is also included.what is the age of the coach ?
Options:
A)45 yrs.
B)53 yrs.
C)58 yrs.
D)48 yrs.
E)55 yrs.
Rationale: Explanation:
Total age of players in team = 22 x 11 = 242
When coach is included, total members = 12, Avg age decreses by one becomes 25
Total age when coach is included = 25x12 = 300
Age of coach = 300-242 = 58 yrs.
Answer: C
Answer: C
Question: When anunfaircoin is tossed twice, the probability of getting one tails and one heads is 1/2. What is the probability of getting two heads and two tails if the coin is tossed 4 times?
Options:
A)2/3
B)1/3
C)1/4
D)1/6
E)1/9
Rationale: Let the probability of heads be x and the probability of tails be 1-x.
So, we are given that P(one tails and one heads) = 2*x*(1-x) = 1/2.
P(two heads and two tails) = 4!/(2!2!)*x^2*(1-x)^2 = 6*(x*(1-x))^2. We are multiplying by 4!/(2!2!) because HHTT scenario can occur in several ways: HHTT, HTHT, THHT, ... (permutation of 4 letters HHTT where 2 T's and 2 H's are the same).
Since from 2*x*(1-x) = 1/2 it follows that x*(1-x) = 1/4, then 4*(x*(1-x))^2 = 4*(1/6)^2 = 1/9.
Answer: E.
Answer: E
Question: If 1000 microns = 1 decimeter, and 1,000,000angstroms = 1 decimeter, how many angstroms equal 1 micron?
Options:
A)10
B)1000
C)100
D)0.01
E)0.0001
Rationale: 1000 microns = 1 decimeter, and 1,000,000angstroms = 1 decimeter
1000 microns = 1,000,000angstroms
1 micron = 1,000,000/1,000 = 1,000
Answer : B
Answer: B