divelab/Sys2Bench huggingface dataset:
Question: A can run 22.5 m while B runs 25 m. In a kilometre race B beats A by:
Options:
A)100
B)378
C)777
D)299
E)277
Rationale: Explanation:
When B runs 25 m, A runs \inline \fn_jvn {\color{Black} \frac{45}{2}} m
When B runs 1000 m, A runs \inline \fn_jvn {\color{Black} \left ( \frac{45}{2}\times \frac{1}{25}\times 1000 \right )}m = 900 m.
\inline \fn_jvn {\color{Black} \therefore }B beats A by 100 m.
Answer: A) 100 m
Answer: A
Question: A and B can do a work in 10 days and 15 days respectively. A starts the work and B joins him after 5 days. In how many days can they complete the remaining work?
Options:
A)6 days
B)2 days
C)8 days
D)4 days
E)3 days
Rationale: Work done by A in 5 days = 5/10 = 1/2
Remaining work = 1/2
Work done by both A and B in one day = 1/10 + 1/15 = 5/30 = 1/6
Remaining work = 1/2 * 6/1 = 3 days.
Answer: E
Answer: E
Question: The age of a father is 5/2 of his daughter. 5 years ago it was 3 times that of his daughter.what is daughter's present age?
Options:
A)20 YRS
B)25 YRS
C)15 YRS
D)10 YRS
E)18 YRS
Rationale: daughter's present age is x
father's present age is (5/2)*x
5 years ago father's age is (5x/2)-5=3(x-5)
Therefore,x=20 years
ANSWER:A
Answer: A
Question: In day before Yesterday it was Monday, What day will fall on day after tomorrow ?
Options:
A)Sunday
B)Tuesday
C)Saturday
D)Friday
E)Monday
Rationale: if day before yesterday was Monday then today is Wednesday
if today is wednesday then day after tommorrow will be Friday
ANSWER:D
Answer: D
Question: Pumping alone at their respective constant rates, one inlet pipe fills an empty tank to 1/2 of capacity in 9 hours and a second inlet pipe fills the same empty tank to 2/3 of capacity in 6 hours. How many hours will it take both pipes, pumping simultaneously at their respective constant rates, to fill the empty tank to capacity?
Options:
A)3.25
B)3.6
C)4.2
D)4.4
E)6
Rationale: Time taken by First tank Fill 1/2 the tank = 9 hours
i.e. Time taken by First tank to Fill 1 complete the tank = 18 hours
Time taken by Second tank Fill 2/3 the tank = 6 hours
i.e. Time taken by First tank to Fill 1 complete the tank = (3/2)*6 = 9 hours
in 1 Hour, Both tanks together Fill the tank = (1/18)+(1/9) =1/6 tank
i.e. Time taken by Both tank to Fill 1 complete the tank =6 hours
Answer: option E
Answer: E
Question: At 9 am a car (A) began a journey from a point, traveling at 40 mph. At 10 am another car (B) started traveling from the same point at 60 mph in the same direction as car (A). At what time will car B pass car A?
Options:
A)12p.m.
B)1p.m.
C)2p.m.
D)3p.m.
E)11p.m.
Rationale: If x hours have elapsed since 9AM:
A has traveled 40x
B has traveled 60(x - 1)
60(x - 1) = 40x
60x - 60 = 40x
60x - 40x = 60
20x = 60
x = 60/20
x = 3
ans: 9+3= 12p.m.
ANSWER:A
Answer: A
Question: If two dice are thrown simultaneously, then find the probability that the sum of numbers appeared on the dice is 2?
Options:
A)5/6
B)7/36
C)5/36
D)1/36
E)12/36
Rationale: The sum of numbers appeared is 2.
Therefore, the required probability = n(E)/n(S) = 1/36.
ANSWER:D
Answer: D
Question: Shekar scored 76, 65, 82, 67 and 75 marks in Mathematics, Science, Social studies, English and Biology respectively. What are his average marks?
Options:
A)65
B)69
C)75
D)85
E)73
Rationale: Explanation :
Average= (76+65+82+67+75)/5 = 365/5 =73
Hence average=73
Answer : E
Answer: E
Question: On Monday, a certain animal shelter housed 72 cats and dogs. By Friday, 1/6 of the cats and 1/5 of the dogs had been adopted; no new cats or dogs were brought to the shelter during this period. What is the greatest possible number of pets that could have been adopted from the animal shelter between Monday and Friday?
Options:
A)11
B)12
C)13
D)14
E)15
Rationale: To maximize the number of pets adopted, we should maximize the number of dogs adopted since 1/5 > 1/6.
The greatest multiple of 5 less than 72 is 60 (such that the remaining number is divisible by 6), so let's assume there were 60 dogs and 12 cats.
Then the maximum number of adopted pets is 1/5*(60) + 1/6*(12) = 12 + 2 = 14.
The answer is D.
Answer: D
Question: When a merchant imported a certain item, she paid a 7 percent import tax on the portion of the total value of the item in excess of $1,000. If the amount of the import tax that the merchant paid was $110.60, what was the total value of the item?
Options:
A)$2160
B)$2370
C)$2580
D)$2790
E)$2950
Rationale: Let x be the value of the item.
0.07*(x-1000) = 110.60
x = 2580
The answer is C.
Answer: C
Question: The roots of the equation ax2 + bx + c = 0 will be reciprocal if:
Options:
A)a = b,
B)a = bc,
C)c = a,
D)c = b
E)c = ab.
Rationale: Solution:
Let k be one of the root of the given equation.
According to the problem,
1/k will be the other root of the given equation.
We know that, product of the roots of the equation = c/a.
Therefore, k × 1/k = c/a.
or, 1 = c/a.
or, a = c [multiplying a on both sides].
The roots of the equation ax2 + bx + c = 0 will be reciprocal if a = c.
Therefore, a = c or c = a.
Answer: C
Answer: C
Question: When positive integer t is divided by 13, the remainder is 2. When n is divided by 8, the remainder is 5. How many such values are less than 180?
Options:
A) 0
B) 1
C) 2
D) 3
E) 4
Rationale: the equation that can be formed t is 13x+2=8y+5..
13x-3=8y...
as we can see x can take only odd values as the RHS will always be even..
Also x can take values till 13 as 13*14>180..
now we have to substitue x as 1,3,5,7,9,11,13...
once we find 7 fitting in , any other value need not be checked as every 4th value will give us answer so next value will be 15..
ans 1.. B
Answer: B
Question: Set S contains exactly 10 numbers and has an average (arithmetic mean) of 6.2. If one of the numbers in Set S is increased by 8, while all other numbers remain the same, what is the new average of Set S?
Options:
A)6.6
B)6.7
C)6.8
D)7.0
E)6.9
Rationale: Old set S - Total is avg * no of elements = 6.2*10 = 62
If one number is increased by 8 then total increased to 62+8=70
New avg - 70/10 = 7.0.
Hence answer is D.
Answer: D
Question: A small college reduced its faculty by approximately 20 percent to 195 professors. What was the original number of faculty members?
Options:
A)182
B)208
C)224
D)254
E)244
Rationale: If x is the original number of faculty members, then after 20% reduction in faculty members number is .80x
but we are given .80x = 195
x = 244
So the original number of faculty members is 244
Correct Answer - E
Answer: E
Question: Find the area of trapezium whose parallel sides are 20 cm and 18 cm long, and the distance between them is 5 cm.
Options:
A)187 cm2
B)185 cm2
C)85 cm2
D)85 cm2
E)95 cm2
Rationale: Explanation:
Area of a trapezium = 1/2 (sum of parallel sides) * (perpendicular distance between them) = 1/2 (20 + 18) * (5) = 95 cm2
Answer: Option E
Answer: E
Question: 0---:-|---:--|-:--:-|--:---|-:----1
On the number line above, the segment from 0 to 1 has been divided into fifths, as indicated by the large tick marks, and also into sevenths, as indicated by the small tick marks. What is the LEAST possible distance Q between any two of the tick marks?
Options:
A)1/70
B)1/35
C)2/35
D)1/12
E)1/7
Rationale: divisions are 0 1/7 2/7 3/7 4/7 5/7 6/7 1
1/5 2/5 3/5 4/5 5/5
expressing the same in terms of the lcm of 5,7 ,i.e 35
0 7/35 14/35 21/35 28/35 35/35
5/35 10/35 15/35 20/35 25/35 30/35
by comparing the divisions with each other we can see 1/35 is the shortest possible distance between any two selected divisions.
Q=1/35
Answer is B.
Answer: B
Question: AZ + CD = AAA, where AZ and CD are two-digit numbers and AAA is a three digit number; A, Z, C, and D are distinct positive integers. In the addition problem above, what is the value of C?
Options:
A) 1
B) 3
C) 7
D) 9
E) Cannot be determined
Rationale: Since AZ and CD are two-digit integers, their sum can give us only one three digit integer of a kind of AAA: 111.
So, A=1 and we have 1Z+CD=111
Now, C can not be less than 9, because no two-digit integer with first digit 1 (1Z<20) can be added to two-digit integer less than 90, so that to have the sum 111 (if CD<90, so if C<9, CD+1Z<111).
Hence C=9.
Answer: D.
Answer: D
Question: The weights of three boys are in the ratio 4 : 5 : 6. If the sum of the weights of the heaviest and the lightest boy is 48 kg more than the weight of the third boy, what is the weight of the lightest boy?
Options:
A)87 kg
B)38.4 kg
C)98.4kg
D)65.4 kg
E)87.4 kg
Rationale: Let the weights of the three boys be 4k, 5k and 6k respectively.
4k + 6k = 5k + 48
=> 5k = 48 => k = 9.6
Therefore the weight of the lightest boy
= 4k = 4(9.6) = 38.4 kg.
Answer:B
Answer: B
Question: A container holding 9 ounces of a solution that is 1 part alcohol to 2 parts water is added to a container holding 8 ounces of a solution that is 1 part alcohol to 3 parts water. What is the ratio of alcohol to water in the resulting solution?
Options:
A)2:5
B)3:7
C)3: 5
D)5:12
E)7:3
Rationale: Container 1 has 9 ounces in the ratio 1:2
or,
x+2x=9 gives x(alcohol)=3 and remaining water =6
container 2 has 8 ounces in the ratio 1:3
or,
x+3x=8 gives x(alcohol)=2 and remaining water =6
mixing both we have alcohol=3+2 and water =6+6
ratio thus alcohol/water =5/12
Answer D
Answer: D
Question: A fruit seller sells mangoes at the rate of Rs.12 per kg and thereby loses 15%. At what price per kg, he should have sold them to make a profit of 5%
Options:
A)Rs 8.82
B)Rs 9.82
C)Rs 10.82
D)Rs 14.82
E)None of these
Rationale: Explanation:
85 : 12 = 105 : x
x= (12×105/85)
= Rs 14.82
Option D
Answer: D
Question: Difference between the length &breadth of a rectangle is 23 m. If its perimeter is 206 m, then its area is?? We have: (l - b) = 23 and 2(l + b) = 206 or (l + b) = 103?
Options:
A)2500 m^2
B)25
C)2520 metre sq
D)2540 m^2
E)2590 m^2
Rationale: Solving the two equations, we get: l = 63 and b = 40.
Area = (l x b) = (63 x 40) m2 = 2520 m ^2
C
Answer: C
Question: How many numbers from 15 to 65 are exactly divisible by 5?
Options:
A)5
B)7
C)9
D)10
E)12
Rationale: Option 'D'
15/5 = 3 and 65/5 = 13 ==> 13-3 = 10 Numbers
Answer: D
Question: Find Unit digit of(7^71*6^59*3^65)
Options:
A)18
B)55
C)66
D)4
E)22
Rationale: Explanation:
Unit digit of 7^71 is 3
Unit digit of 6^59 is 6
Unit digit of 3^65 is 3
Unit digit of(7^71*6^59*3^65)=(3*6*3)=4.
D
Answer: D
Question: 3 buffaloes eat as much as 4 cows or 2 oxen. At a farm there are 15 buffaloes, 8 oxen and 24 cows. The fodder for these cattle is available for 24days. If 8 more cows & 9 more buffaloes are brought in, how long will the fodder last?
Options:
A)8
B)10
C)12
D)18
E)20
Rationale: Sol. 2 oxen = 3 buffaloes = 4 cows 8 oxen = 16 cows and 15 buffaloes = 20 cows 16 cows + 20 cows + 24 cows = 60 cows Since fodder is available for 60 cows for 24 days
So fodder will be available for (60 + 8 + 12) = 80 cows c 24x60 .0 .
for-^—=18 days.
D
Answer: D
Question: A train 500m long takes 10 sec to cross a man walking at 6 kmph in a direction opposite to that of the train. Find the speed of the train?
Options:
A)A)175kmph
B)B)150kmph
C)C)162kmph
D)D)176kmph
E)E)100kmph
Rationale: Let the speed of the train be x kmph
Speed of the train relative to man = x+6 = (x+6)*5/18 m/sec
500/[(x+6)*5/18] = 10
10(x+6) = 1800
x = 176kmph
Answer is D
Answer: D
Question: For a light that has an intensity of 60 candles at its source, the intensity in candles, S, of the light at a point d d feet from the source is given by the formula S=60k/d^2 60, where k is a constant. If the intensity of the light is 40 candles at a distance of 3 feet from the source, what is the intensity of the light at a distance of 20 feet from the source?
Options:
A)3/10 candles
B)9/10 candles
C)1 1/3 candles
D)2 candles
E)3 candles
Rationale: S= 60k/d^2
At 3 feet
40 = 60k/9 so k = 6
At 20 feet
S = 60 *6/20^2 = 9/10
B
Answer: B
Question: A boat goes 100 km downstream in 4 hours, and 75 Km upstream in 15 hours. The speed of the stream is?
Options:
A)10
B)20
C)25
D)35
E)15
Rationale: 100 --- 10 DS = 25
? ---- 1
75 ---- 15 US = 5
? ----- 1 S = (25 - 5)/2
= 10 kmph.
Answer:A
Answer: A
Question: P and Q started a business investing Rs. 85,000 and Rs. 35,000 respectively. In what ratio the profit earned after 2 years be divided between P and Q respectively?
Options:
A)17:6
B)17:0
C)17:7
D)17:2
E)17:3
Rationale: P:Q = 85000 : 35000
= 17:7.
Answer: C
Answer: C
Question: The average of 20 numbers is zero. Of them, at the most, how many may be greater than zero?
Options:
A)0
B)1
C)9
D)10
E)19
Rationale: Average of 20 numbers = 0.
Sum of 20 numbers (0 x 20) = 0.
It is quite possible that 19 of these numbers may be positive and if their sum is a then 20th number is (-a).
answer: option E
Answer: E
Question: A dice has one of the first 6 prime number on each its six sides ,with no two sides having the same number .the dice is rolled 10 times and the results added.the addition is most likely to be closet to
Options:
A)41
B)48
C)60
D)68
E)70
Rationale: if die is rolled then avg score = (2+3+5+7+11+13 )/6 = 41/6
so, most likely sum for 10 times = 41/6 * 10 = 68
ANSWER:D
Answer: D
Question: How much time will it take for an amount of 1800 to yield 81 as interest at 4.5% per annum of simple interest?
Options:
A)3 years
B)4 years
C)2 years
D)1 years
E)7 years
Rationale: Time = (100 x 81)/(1800 x 4.5) years = 1 years.
answer D
Answer: D
Question: The ratio between the number of sheep and the number of horses at the Stewart farm is 1 to 7, If each horse is fed 230 ounces of horse food per day and the farm needs a total 12,880 ounces of horse food per day, what is the number of sheep in the farm?
Options:
A)18
B)28
C)32
D)56
E)8
Rationale: Let the number of sheeps and horses be 1x and 7x.
Now total number of horses = total consumption of horse food/consumption per horse = 12880/230 = 56, which is equal to 7x. => x = 8
Sheeps = 1x = 1*8 = 8.
Hence E
Answer: E
Question: In how many ways can the letters of the word 'LEADER' be arranged?
Options:
A)360
B)450
C)455
D)375
E)480
Rationale: The word 'LEADER' contains 6 letters, namely 1L, 2E, 1A, 1D and 1R.
Req = 6!/1!2!1!1!1! = 360
ANSWER A
Answer: A
Question: Two pipes A and B can fill a cistern in 12 and 15 minutes respectively. Both are opened together but after 1 minutes A is turned off. After how much more time will the cistern be filled?
Options:
A)8 1/7
B)8 1/1
C)8 1/4
D)8 1/2
E)12 3/4
Rationale: 1/12 + (1 + x)/15 = 1
x = 12 3/4
Answer: E
Answer: E
Question: A shopkeeper sells 20% of his stock at 10% profit ans sells the remaining at a loss of 5%. He incurred an overall loss of Rs. 350. Find the total worth of the stock?
Options:
A)20029
B)20000
C)20289
D)17500
E)20026
Rationale: Let the total worth of the stock be Rs. x.
The SP of 20% of the stock = 1/5 * x * 1.1 = 11x/50
The SP of 80% of the stock = 4/5 * x * 0.95 = 19x/25 = 38x/50
Total SP = 11x/50 + 38x/50 = 49x/50
Overall loss = x - 49x/50 = x/50
x/50 = 350 => x = 17500
Answer: D
Answer: D
Question: If n = 2×3×5×7×11×13×19, then which of the following statements must be true?
I. n^2 is divisible by 600
II. n + 19 is divisible by 19
III. (n + 4)/2 is even
Options:
A)I only
B)II only
C)III only
D)I and III
E)None of the above
Rationale: Statement I: 600 = (2^3)*(3)*(5^2) this is NOT true.
Statement 2 Since n is divisible by 19 so n+19 is also divisible by 19. So this is true.
Statement 3: (n+4)/2 = n/2 + 2
So, ODD + 2 cannot be even.
So this is NOT true.
Hence option (B).
Answer: B
Question: From the set of numbers {p, t, u, v, w, x, y, z}, how many different combinations can we have without the t, the x, or the u in them? Ex:. (x,y), (x), (w,z,y,x), etc... and note that (x,y)=(y,x)
Options:
A)8
B)15
C)16
D)31
E)32
Rationale: Any letter (p, v, w, y, z) can be included or not.
Thus there are 2^5 combinations - 1 empty combination = 31 combinations
The answer is D.
Answer: D
Question: An air-conditioning unit costs $470. On December there was a discount for Christmas of 16%. Six months later, the holiday season was over so the company raised the price of the air-conditioning by 10%. How much will an air-conditioning unit cost in November?
Options:
A)$434
B)$470
C)$472
D)$484
E)$491
Rationale: if its previous November(before discount) then price is $470.
but if its November of next year then
16% discount on $470=470(1-16/100)=$394.8
again a corrected raised price of 10% over $394.8=394.8(1+10/100)=434.28~$434
Ans A
Answer: A
Question: A basket contains Rs 216 in the form of one rupee, 50 paise and 25 paise coins in the ratio of 3 : 4 : 4. The number of 50 paise coins is ?
Options:
A)96
B)128
C)144
D)100
E)90
Rationale: Let number of one rupee coins = 3K
Number of 50 paise coins = 4K
Number of 25 paise coins = 4K
Therefore 3K + 0.50 x 4K + 0.25 × 4K = 216
=> 3K + 2K + 1K = 216
=> K = 36
Therefore number of 50 paise coins = 4K = 144
ANSWER:C
Answer: C
Question: Jerry’s average (arithmetic mean) score on the first 3 of 4 tests is 90. If Jerry wants to raise his average by 2 points, what score must he earn on the fourth test?
Options:
A)87
B)89
C)90
D)98
E)95
Rationale: Total score on 3 tests = 90*3 = 270
Jerry wants the average to be = 92
Hence total score on 4 tests should be = 92*4 = 368
Score required on the fourth test = 368 - 270 = 98
Option D
Answer: D
Question: The greatest number that divides 60 and 190 leaving remainder 6 and 10 respectively is:
Options:
A)15
B)18
C)19
D)22
E)23
Rationale: explanation:
60-6= 54 ,190-10= 180
Highest number that divides 54 and 180 is HCF of numbers
HCF of 54 nad 180= 18
Answer:B
Answer: B
Question: 3 buffaloes eat as much as 4 cows or 2 oxen. At a farm, there are 15 buffaloes, 8 oxen, and 24 cows. The fodder for these cattle is available for 48 days. If 60 more cows and 30 more buffaloes are brought in, how many days will the fodder last?
Options:
A)15 days
B)18 days
C)21 days
D)24 days
E)27 days
Rationale: 2 oxen = 3 buffaloes = 4 cows
Also: 15 buffaloes + 8 oxen + 24 cows
= 10 oxen + 8 oxen + 12 oxen = 30 oxen
There is enough fodder for 1 oxen for 30*48 days.
60 cows + 30 buffaloes = 30 oxen + 20 oxen = 50 oxen
The new total is equal to 80 oxen instead of 30 oxen.
30*48 / 80 oxen = 18 days
The answer is B.
Answer: B
Question: The unit digit in the product (722 * 774 * 889 * 223) is:
Options:
A)2
B)7
C)6
D)8
E)1
Rationale: Explanation:
Unit digit in the given product = Unit Digit in (2*4*9*3) = 6
ANSWER: C
Answer: C
Question: After playing 40 games, the Lesley basketball team had won 50% of the games. The additional games played gave the team a 20% total winning average. How many additional games must the team have won to obtain this average?
Options:
A)2
B)15
C)7
D)10
E)9
Rationale: Explanation: Identify the parts, set up equation, and solve.
Number of games initially won by the team:
50% * 40 = 20
Additional games won by the team leaving a 20% average. (Let x represent the unknown number of additional games won.) Therefore, the total number of games played is 40+x, and the total number of games won is 20+x. This gives the equation as:
20%*(40+x)=20+x
Solve: Let's get rid of the percent sign
20/100*(40+x)=20+x
1/5 *(40+x)=20+x
Get rid of the fraction:
5[1/5 * (40+x)]=5(20+x)
1 * (40+x) =100+5x
40+x =100+5x
(40-100)+x = (100-100)+5x
-60+x= 5x
-60+(x-x)+5x-x
-60=4x
15 = x
Answer: B
Answer: B
Question: To asphalt 1km road, 30 men spent 12 days working 8 hours per day. How many days, 20 men will spend to asphalt a road of 2km working 15 hours a day?
Options:
A)23
B)22
C)21
D)20
E)19.2
Rationale: Man-hours required to asphalt 1 km road=30*12*8=2880
Man-hours required to asphalt 2 km road=2880*2=5760
Man-hours available per day=20*15=300
Therefore number of days=5760/300=19.2 days
Ans=E
Answer: E
Question: A batsman makes a score of 50 runs in the 6th inning and thus increases his average by 2. Find his average after 6th inning.
Options:
A)40
B)50
C)60
D)70
E)80
Rationale: Let the average after 6th inning = x
Then, average after 5th inning = x-2
5(x-2)+50 = 6x
x = 10-50 = 40
Answer is A
Answer: A
Question: Two men A and B start from place X walking at 4 ½ kmph and 5 ¾ kmph respectively. How many km apart they are at the end of 6 ½ hours if they are walking in the same direction?
Options:
A)7 km
B)6 km
C)8 1/8 km
D)9 km
E)5 km
Rationale: RS = 5 ¾ - 4 ½ = 1 ¼
T = 6 ½ h.
D = 5/4 * 13/2 = 65/8
= 8 1/8 km
Answer:C
Answer: C
Question: Jacob brought a scooter for a certain sum of money. He spent 10% of the cost on repairs and sold the scooter for a profit of Rs. 1100. How much did he spend on repairs if he made a profit of 20%?
Options:
A)332
B)787
C)500
D)277
E)181
Rationale: Let the C.P. be Rs. x. Then, 20% of x = 1100
20/100 * x = 1100 => x = 5500
C.P. = Rs. 5500, expenditure on repairs = 10%
Actual price = Rs. (100 * 5500)/110 = Rs. 5000
Expenditures on repairs = (5500 - 5000) = Rs. 500.
Answer:C
Answer: C
Question: 1397 x 1397 = ?
Options:
A)1951607
B)1951609
C)1951610
D)1951611
E)1951613
Rationale: 1397 x 1397	= (1397)2
= (1400 - 3)2
= (1400)2 + (3)2 - (2 x 1400 x 3)
= 1960000 + 9 - 8400
= 1960009 - 8400
= 1951609.
B)
Answer: B
Question: Average of five consecutive even numbers is 35. Find the greatest number in these five numbers?
Options:
A)32
B)33
C)39
D)40
E)41
Rationale: 39
Option 'C'
Answer: C
Question: A, B and C can do a work in 90, 30 and 45 days respectively. If they work together, in how many days will they complete the work?
Options:
A)12
B)10
C)15
D)8
E)13
Rationale: Option C
Explanation:
One days's work of A, B and C = 1/90 + 1/30 + 1/45
= (1 + 3 + 2)/90 = 1/15
A, B and C together can do the work in 15 days.
Answer: C
Question: What is the remainder when (71)(73) is divided by 8?
Options:
A)1
B)2
C)4
D)5
E)7
Rationale: (71)(73) = (72-1)(72+1) = 72^2 - 1 which is 1 less than a multiple of 8.
Then the remainder will be 7.
The answer is E.
Answer: E
Question: A shopkeeper buys two articles for Rs.1000 each and then sells them, making 30% profit on the first article and 30% loss on second article. Find the net profit or loss percent?
Options:
A)200
B)278
C)282
D)202
E)300
Rationale: Profit on first article = 30% of 1000 = 300.
This is equal to the loss he makes on the second article. That, is he makes neither profit nor loss.
Answer: E
Answer: E
Question: Lucia is a wonderful grandmother.Her age is b/w 50 to 70.Each of her sons have as many sons as they have brothers.Their combined number gives Lucia's present age?what is her age?
Options:
A)85
B)55
C)84
D)64
E)62
Rationale: If she has x sons
each son has x-1 brothers.
Total no of grandchildren=x(x-1)
Total No OF Sons+Grandchildren= x+x(x-1)=x^2
=>Lucia's age is a perfect square between 50 and 70. The only number that satisfies this condition is 64.
ANSWER:D
Answer: D
Question: Three boys are ages 4, 6 and 7 respectively. Three girls are ages 5, 8 and 9, respectively. If two of the boys and two of the girls are randomly selected and the sum of the selected children's ages is t, what is the difference between the probability that t is even and the probability that t is odd?
Options:
A) 1/9
B) 1/6
C) 2/9
D) 1/4
E) 1/2
Rationale: Age of Boys t:4, 6, 7
Sum of ages taken 2 at a time: 10,13,11
Ages of Girls:5, 8, 9
Sum of ages taken 2 at a time: 13,17,14
9 Combinations of sum between sets(10,12,11)(13,17,14)
=23,27,24- 16,30,17- 24,28,25
Prob(Even)= 5/9
Prob(Odd) =4/9
Answer=5/9 - 4/9 = 1/9
Answer: A
Question: A picnic attracts 240 persons. There are 80 more men than women, and 80 more adults than children. How many men are at this picnic?
Options:
A)240
B)75
C)110
D)120
E)200
Rationale: Adult + Children = 240
Let, Children = y
then, Adult = y+80
i.e. y+(y+80) = 240
i.e. y =80
i.e. Adult = 80+80 = 160
Adults include only Men and Women
i.e. Men + Women = 160
Let women, W = x
then Men, M = x+80
i.e. x+(x+20) = 2x+80 = 160
i.e. x = 40
i.e. Men, M = 40+80 = 120
Answer: Option D
Answer: D
Question: Find four consecutive even integers such that the sum of the second and fourth integers is one and a half times the sum of the first and third integers.
Options:
A)n, 2n, 4n, 6n
B)n, 2, 4, 6
C)n, 2n, 3n, 4n
D)n, n+2, n+4, n+6
E)n, n^2, n^4, n^6
Rationale: Sum of Second and Fourth Integers: (n + 2) + (n + 6) = 2n + 8
Sum of First and Third Integers: n + (n + 4) = 2n + 4
Since sum of the second and fourth integers is twice than the sum of the first and third integers, it follows that:
2n + 8 = 1.5(2n + 4)
2n + 8 = 3n + 6
3n − 2n = 8 − 6
n = 2
Answer: D
Answer: D