divelab/Sys2Bench huggingface dataset:
Question: I bought two books; for Rs.480. I sold one at a loss of 15% and other at a loss of 25% and then I found each book was sold at the same price. Find the cost of the book sold at a lower loss?
Options:
A)A)657
B)B)280
C)C)225
D)D)289
E)E)427
Rationale: x*(85/100)
= (480 - x)75/100
x = 225
Answer:C
Answer: C
Question: Which of the following has the most number of divisors?
Options:
A)344
B)377
C)176
D)888
E)271
Rationale: Explanation:
99 = 1 x 3 x 3 x 11
101 = 1 x 101
176 = 1 x 2 x 2 x 2 x 2 x 11
182 = 1 x 2 x 7 x 13
So, divisors of 99 are 1, 3, 9, 11, 33, .99
Divisors of 101 are 1 and 101
Divisors of 176 are 1, 2, 4, 8, 11, 16, 22, 44, 88 and 176
Divisors of 182 are 1, 2, 7, 13, 14, 26, 91 and 182.
Hence, 176 has the most number of divisors.
Answer: C) 176
Answer: C
Question: The cash difference between the selling prices of an article at a profit of 2% and 6% is Rs 3. The ratio of two selling prices is
Options:
A)51:53
B)52:53
C)53:54
D)54:55
E)None of these
Rationale: Explanation:
Let the Cost price of article is Rs. x
Required ratio = (102% of x) / (106% of x)
=102/106
=51/53 = 51:53.
Answer: A
Answer: A
Question: A jar contains 4 black and 3 white balls. If you pick two balls at the same time, what's the probability that one ball is black and one is white?
Options:
A)2/7
B)5/7
C)4/7
D)3/7
E)1/2
Rationale: P(1st black, 2nd white) = 4/7*3/6 = 4/14;
P(1st white, 2nd black) = 3/7*4/6 = 4/14.
P = 4/14 + 4/14 = 4/7.
Answer: C.
Answer: C
Question: If s and t are positive integers such that s/t = 64.14, which of the following could be the remainder when s is divided by t ?
Options:
A)9
B)16
C)18
D)21
E)32
Rationale: 0.14 = 7/50
The remainder must be divisible by 7.
The answer is D.
Answer: D
Question: A foreign language club at Washington Middle School consists of n students, 2/5 of whom are boys. All of the students in the club study exactly one foreign language. 1/3 of the girls in the club study Spanish and 4/5 of the remaining girls study French. If the rest of the girls in the club study German, how many girls in the club, in terms of n, study German?
Options:
A)2n/5
B)n/3
C)2n/25
D)2n/15
E)n/15
Rationale: 2/5 of the students are boys, thus 3/5 of the students are girls.
1/3 of the girls in the club study Spanish and 4/5 of the remaining girls study French.
Thus 1/5 of 2/3 = 2/15 of the girls study German.
Since girls comprise 3/5 of the students, then 3/5*2/15 = 2/25 are girls who study German.
The answer is C.
Answer: C
Question: Carol and Jordan draw rectangles of equal area. If Carol's rectangle measures 12 inches by 15 inches and Jordan's rectangle is 9 inches long, how wide is Jordan's rectangle, in inches?
Options:
A)17
B)18
C)19
D)20
E)21
Rationale: Area of first rectangle is 12*15=180 hence area of second would be 9x=180x
x=20
answer is D
Answer: D
Question: A is twice as good as workman as B and together they finish a piece of work in 18 days. In how many
days will B alone finish the work.
Options:
A)27 days
B)54 days
C)56 days
D)68 days
E)None of these
Rationale: Explanation:
As per question, A do twice the work as done by B.
So A:B = 2:1
Also (A+B) one day work = 1/18
To get days in which B will finish the work, lets calculate work done by B in 1 day =
=(118
∗
13)=154
[Please note we multiplied by 1/3 as per B share and total of raƟo is 1/3]
So B will finish the work in 54 days
Answer: B
Answer: B
Question: A, B and C completed a piece of work, A worked for 6 days, B for 9 days and C for 4 days. Their daily wages were in the ratio of 3:4:5. Find the daily wages of C, if their total earning was Rs.1554?
Options:
A)s.109
B)s.108
C)s.100
D)s.103
E)s.105
Rationale: 3x 4x 5x
6 9 4
18x + 36x + 20x = 1554
74x = 1554 => x = 21
5x = 105 Rs.
Answer:E
Answer: E
Question: Points A, B, C, and D, in that order, lie on a line. If AB = 2 cm, AC = 5 cm, and BD = 9 cm, what is CD, in centimeters?
Options:
A) 1
B) 2
C) 3
D) 6
E) 5
Rationale: Putting a value to each point, lets use the following:
A - 0
B - 2 (AB = 2)
C - 5 (AC = 5)
D - 11 (BD = 9)
CD is 11-5 = 6.
Ans D
Answer: D
Question: The sum of ages of 5 children born at the intervals of 3 years each is 60 years. What is the age of the youngest child?
Options:
A)3 years
B)4 years
C)6 years
D)7 years
E)8 years
Rationale: Let the ages of children be x, (x + 3), (x + 6), (x + 9) and (x + 12) years.
Then, x + (x + 3) + (x + 6) + (x + 9) + (x + 12) = 60
5x = 30
x = 6.
Age of the youngest child = x = 6 years.
C)
Answer: C
Question: If 20% of a number, then 120% of that number will be?
Options:
A)20
B)120
C)360
D)720
E)820
Rationale: Let the number x. Then,
20% of x = 120
x = (120 * 100)/20 = 600
120% of x = (120/100 * 600) = 720.
ANSWER:D
Answer: D
Question: Income and expenditure of a person are in the ratio 5 : 3. If the income of the person is Rs. 10000, then find his savings?
Options:
A)Rs.4000
B)Rs.3603
C)Rs.3639
D)Rs.3632
E)Rs.3602
Rationale: Let the income and the expenditure of the person be Rs. 5x and Rs. 3x respectively.
Income, 5x = 10000 => x = 2000
Savings = Income - expenditure = 5x - 3x =2x
So, savings = Rs.2*2000 = Rs. 4000
Answer:A
Answer: A
Question: Sanjay invested an amount of ` 16,000 for two years on compound interest and received an amount of ` 17,640 on maturity. What is the rate of interest per annum?
Options:
A)4%
B)5%
C)8%
D)9%
E)None of these
Rationale: The extra money for two years is 1,640. This can be broken up as 800 + (800 + 40). Since 40 = 5% of 800 and 800 = 5% of 16000
So, r = 5%.
Answer: B
Answer: B
Question: A shipping clerk has five boxes of different but unknown weights each weighing less than 100 kg. The clerk weighs the boxes in pairs. The weights obtained are 110, 112, 113, 114, 115, 116, 117, 118, 120 and 121 kg. What is the weight, in kg, of the heaviest box ?
Options:
A)60
B)61
C)64
D)Can't be determined
E)None
Rationale: Explanation :
Two boxes are selected at at time and weighed together. Thus, the selection order does not matter and selecting box 1 and then box 2 is the same as selecting box 2 and then box 1. So this is a combination problem.
2 boxes can be selected from 5 boxes in 5C2 ways, i.e. 10 ways. The given list of weighings is 10 in number. So we can assume that all the possible combinations of boxes have been weighed. Lets assume that all boxes are lined as per increasing weights and named as b1, b2, b3, b4, b5. From the given list (Arranges in increasing order of weights) ,
what we have is b1 + b2 = 110, as b1 is the lightest and b2 is the next lightest box heavier than b1. So the sum of these two should be the minimum value in the list.
The second lightest sum of weights, the second value in the list, will be the sum of the b1 + b3 = 112.
Similarly, the maximum value in the list should be the sum of weights of b4 and b5, giving b4 + b5 = 121.
The second last heaviest will be b3 + b5 = 120.
Adding the 4 equations, we get, 2 b1 + 2 b3 + 2 b5 + b2 + b4 = 463 - Equation 1.
Sum of all the weights in the given list is 1160. This sum is the sum of each of the cases b1+b2, b1+b3, b1+b4, b1+b5, b2+b3, b2+b4, b2+b5, b3+b4, b3+b5, b4+b5. Thus sum of all this will be 4(b1 + b2 + b3 + b4 + b5) = 1160. Thus, b1 + b2 + b3 + b4 + b5 = 290 - Equation 2.
Subtracting equation 2 from equation 1, we get, b1 + b3 + b5 = 173.
We know that b1 + b3 = 112, So b5 = 173 - 112 = 61 = Desired answer.
Answer : B
Answer: B
Question: X completes a work in 60 days and Y complete the same work in 30 days. If both of them work together, then the number of days required to complete the work will be?
Options:
A)30 days.
B)50 days.
C)25 days.
D)10 days.
E)20 days.
Rationale: If X can complete a work in x days and Y can complete the same work in y days, then, both
of them together can complete the work in x y/ x+ y days
Therefore, here, the required number of days = 60 Ã— 30/90 = 20 days.
E)
Answer: E
Question: Tina’s cat weighs 3.2 kilograms. Her dog weighs 4 times as much as her cat. How much does her dog weigh in kilograms?
Options:
A)A)10.4kg
B)B)38kg
C)C)12.8kg
D)D)48kg
E)E)20kg
Rationale: 3.2*4=12.8kg.Answer is C.
Answer: C
Question: Anintegratedteam consists of 2 members from the engineering team and 2 members from the design team. If the engineering team has 6 members and the design team has 3 members, how manyintegratedteam formations are possible?
Options:
A) 36
B) 45
C) 76
D) 90
E) 112
Rationale: 2 out of 6 times 2 out of 3 => 6*5/2*1 * 3*2/2*1 = 45
B
Answer: B
Question: In a school of 850 boys, 44% of Muslims, 14% Hindus, 10% Sikhs and the remaining of other communities. How many belonged to the other communities?
Options:
A)A)125
B)B)627
C)C)153
D)D)721
E)E)272
Rationale: 44 + 14 + 10 = 68%
100 – 68 = 32%
850 * 32/100 = 272
Answer:E
Answer: E
Question: Which of the following values of x satisfies the inequality 2x2 – 3x + 1 > 0?
Options:
A)-1 < x < -1⁄2
B)1⁄2 < x < 1
C)x > 1 or x < 1⁄2
D)-1⁄2 < x < 1
E)None of these
Rationale: 2x2 – 3x + 1 = 0.
or (2x – 1)(x – 1) = 0
Hence, α = 1⁄2 and β = 1
Now, the given inequality is 2x2 – 3x + 1 > 0,
Hence, sign scheme will be as follows:
--------|---------------------|----------------
+ve 1⁄2 –ve 1 +ve
∴ x > 1 or x < 1⁄2
[Note:If the inequality were 2x2 – 3x + 1 < 0; the answer will be 1⁄2 < x < 1]
Answer C
Answer: C
Question: Sripad has scored average of 68 marks in three objects. In no subjects has he secured less than 58 marks. He has secured more marks in Maths than other two subjects. What could be his maximum score in Maths ?
Options:
A)79
B)28
C)38
D)27
E)88
Rationale: Assuming Sripad has scored the least marks in subject other than science,
Then the marks he could secure in other two are 58 each.
Since the average mark of all the 3 subject is 68.
i.e (58+58+x)/3 = 68
116 + x = 204
x = 88 marks.
Therefore, the maximum marks he can score in maths is 88.
Answer: E
Answer: E
Question: There are 4 more women than men on Centerville's board of education. If there are 8 members on the board, how many are women?
Options:
A) 3
B) 4
C) 6
D) 7
E) 8
Rationale: Given: M=W-4 so, W+(W-4)=8 --> W=6.
Answer: C.
Answer: C
Question: 96 is divided into two parts in such a way that seventh part of first and ninth part of second are equal. Find the smallest part?
Options:
A)66
B)77
C)42
D)88
E)12
Rationale: x/7 = y/9 => x:y = 7:9
7/16 * 96 = 42.Answer: C
Answer: C
Question: If a is a positive integer and 10^a – 74 in decimal notation has digits whose sum is 440, what is the value of a?
Options:
A)40
B)44
C)45
D)46
E)50
Rationale: Answer E = 50
We have (10^a)-74, where the sum of the digits is 440.
100-74 = 26, so we already have 2+6 = 8 from the 440.
Because we have a power of 10, we will have numbers like 100, 1000, 10000 and so on. This minus 74 rests 26 and a lot of 9s. E.g: 1000-74 = 9926.
So dividing the 432/9 = 48, that`s the number of 9s. and we have 2 other numbers (2 and 6) wich were 00 before the subtraction.
So we have 48 + 2 as an a = 50
Answer: E
Question: Two men start together to walk a certain distance, one at 4 kmph and another at 3 kmph.The former arrives half an hour before the latter. Find the distance.
Options:
A)6 km
B)7 km
C)8 km
D)9 km
E)None of these
Rationale: Let the distance be x km. Then,
x/3 - x/4 =1/2
(4x - 3x)/12 = 1/2
x = 6 km
ANSWER:A
Answer: A
Question: Find compound interest on Rs. 8000 at 15% per annum for 2 years 4 months, compounded annually
Options:
A)3109
B)3115
C)3250
D)3500
E)4000
Rationale: Time = 2 yrs 4 mos = 2(4/12) yrs = 2(1/3) yrs.
Amount = Rs. [8000 X (1+(15/100))^2 X (1+((1/3)*15)/100)]
= Rs. 11109.
:. C.I. = Rs. (11109 - 8000) = Rs. 3109.
ANSWER A
Answer: A
Question: Given that there are 11 cricket players per team, how many ways can you select 10 cricket players from 2 teams if no more than 1 players can be selected from each team?
Options:
A)150
B)200
C)121
D)140
E)150
Rationale: Now from each group we have to select 5 persons only: 11C1 * 11C1= 11 * 11 = 121
Total ways=313444 ways.
Answer C
Answer: C
Question: The average amount with a group of seven numbers is Rs. 30. If the newly joined member has Rs. 50 with him, what was the average amount with the group before his joining the group?
Options:
A)Rs. 25.6
B)Rs. 26.6
C)Rs. 16.6
D)Rs. 26
E)Rs. 25
Rationale: Total members in the group = 7
Average amount = Rs. 30
Total amount with them = 7 * 30 = Rs. 210
One number has Rs. 50. So, the amount with remaining 6 people = 210 - 50 = Rs. 160
The average amount with them = 160/6 = Rs. 26.6
ANSWER:B
Answer: B
Question: Which of the following is closest to 1/5+1/55+1/555
Options:
A)1/4.54
B)1/3.5
C)1/2.5
D)1/3.6
E)1/3.7
Rationale: 1/5+1/55+1/555=0.2+0.018+0.0018=0.2198=1/4.54
Answer is A
Answer: A
Question: Find the ones digit of 73^330
Options:
A)3
B)5
C)6
D)7
E)9
Rationale: Cyclicity of 3 is 3,9,7,1after 4 multiplication again the cycle repeats.
So divide 330 by 4 and we get 87 as quotient and 2 as remainder. So cycle will run for 87 times and then 2 times more. So pick up the 2nd item from the cycle.
Hence Answer B.
Answer: B
Question: I came across this problem in the GMATprep software and got it totally wrong!
A company consists of 6 senior and 4 junior staff officers. If a committee is created with 4 senior and 1 junior staff officers, how many possible committees are there?
Options:
A)12
B)44
C)56
D)60
E)200
Rationale: 6C4*4C1
=60=D
Answer: D
Question: A worker earns $20 on the first day and spends $15 on the second day. The worker earns $20 on the third day and spends $15 on the fourth day. If this pattern continues, on which day will the worker first reach a net total of $80?
Options:
A)24
B)25
C)27
D)30
E)32
Rationale: Every two days, the net total is $5.
After 24 days, the worker will have $60.
On day 25, the worker will receive $20 for a net total of $80.
The answer is B.
Answer: B
Question: Village X has a population of 68000, which is decreasing at the rate of 1200 per year. Village Y has a population of 42000, which is increasing at the rate of 800 per year .in how many years will the population of the two villages be equal?
Options:
A)15
B)19
C)11
D)18
E)13
Rationale: Let the population of two villages be equal after p years
Then, 68000-1200p=42000+800p
2000p=26000
p=13
Answer is E.
Answer: E
Question: Three pipes, A, B, & C are attached to a tank. A & B can fill it in 20 & 30 minutes respectively while C can empty it in 20 minutes. If A, B & C are kept open successively for 1 minute each, how soon will the tank be filled?
Options:
A)2 hours
B)4 hours
C)1 hour 30 min
D)5 hours
E)6 hours
Rationale: in three minute 1/20+1/30-1/20=2/60 = 1/30 part is filled
3 min--------1/30 parts
x min--------- 1 part(full)
x=90 min =1 hour 30 min
ANSWER:C
Answer: C
Question: In a fort, there are 1200 soldiers. If each soldier consumes 3 kg per day, the provisions available in the fort will last for 30 days. If some more soldiers join, the provisions available will last for 25 days given each soldier consumes 2.5 kg per day. Find the number of soldiers joining the fort in that case?
Options:
A)273
B)528
C)539
D)538
E)723
Rationale: Assume X soldiers join the fort. 1200 soldiers have provision for 1200(days for which provisions last them)(rate of consumption of each soldier)
= 1200(30)(3) kg
Also provisions available for (1200 + x) soldiers is (1200 + x)(25)(2.5) kg
As the same provisions are available
=> 1200(30)(3) = (1200 + x)(25)(2.5)
x = [1200(30)(3)]/[(25)(2.5)] - 1200
x = 528
Answer: Option B
Answer: B
Question: Total of ages of A,B and C at present is 90 years. If ratio of this ages is 2 : 3 : 4 what will be the age of C after 2 years?
Options:
A)45 years
B)42 years
C)40 years
D)37 years
E)None of the above
Rationale: 2+3+4=9
4*(90/9)=40yrs
after 2yrs=42
ANSWER:B
Answer: B
Question: If $30,000 interest is invested in 10 percent simple annual interest for 2 years, which of the following represents the total amount of interest, in dollars, that will be earned by this investment in the 2 years?
Options:
A)$6000
B)$4000
C)$5000
D)$3000
E)$2000
Rationale: 30000*10/100*2 = $6000
Answer : A
Answer: A
Question: Teas worth Rs. 126 per kg and Rs. 135 per kg are mixed with a third variety in the ratio 1 : 1 : 2. If the mixture is worth Rs 156 per Kg , the price of the third variety per Kg will be?
Options:
A)Rs. 147.50
B)Rs. 785.50
C)Rs. 178.50
D)Rs. 258.50
E)None of these
Rationale: Explanation :
Since first and second varieties are mixed in equal proportions.
So, their average price =Rs. (126+135)/2.
=>Rs. 130.50.
So, the mixture is formed by mixing two varieties, one at Rs. 130.50 per kg and the other at say, Rs. x per kg in the ratio 2 : 2, i.e., 1 : 1.
We have to find x.
By the rule of alligation, we have:
Cost of 1 kg Cost of 1 kg
of 1st kind of 2nd kind
(Rs. 130.50) (Rs. x)
\ /
Mean Price
(Rs. 156)
/ \
xâˆ’156 22.50
=>xâˆ’(156/22.50)=1.
=>xâˆ’156=22.50.
=>x= 178.50 Rs.
Answer : C
Answer: C
Question: What is the % change in the area of a rectangle when its length increases by 35% and its width decreases by 35%?
Options:
A)0%
B)20% increase
C)20% decrease
D)12% decrease
E)Insufficient data
Rationale: (135/10)*(65/10) = 8775/100~88 of original area
0.88 is a 12% decrease from 100/100 ->D
Answer: D
Question: A jar contains 7 marbles consisting of an equal number of red, green, and blue marbles. Three marbles are removed from the jar and discarded. What is the probability that only two colors will remain in the jar after the Three marbles have been removed?
Options:
A)1/45
B)3/35
C)1/81
D)1/35
E)1/25
Rationale: Jar contains {4-red, 4-green, 4-blue} marbles. We are asked to find the probability that after removing 3 marbles only 2 colours remain in the jar, i.e., the probability of removing EITHER red, OR green, OR blue.
Sample space = 7C3 = 35.
Probability of choosing 3 red = 3C3/7C3 = 1/35.
Probability of choosing 3 green = 3C3/7C3 = 1/35.
Probability of choosing 3 blue = 3C3/7C3 = 1/35.
Required probability = 1+1+1/495 = 3/35 Ans (B).
Answer: B
Question: A canoe has two oars, left and right. Each oar either works or breaks. The failure or non-failure of each oar is independent of the failure or non-failure of the other. You can still row the canoe with one oar. The probability that the left oar works is 3/5. The probability that the right oar works is also 3/5. What is the probability R that you can still row the canoe?
Options:
A)9/25
B)10/25
C)6/10
D)2/3
E)21/25
Rationale: Simply look at the question from the other side. What is the probability that you can’t row the canoe? This would be 2/5 x 2/5 = 4/25.
Using the idea that the probability of something happening is 1 – the probability that it doesn’t happen, you can use the following equation to reach the right answer R: 1 – 4/25 = 21/25.Answer choice E.
Answer: E
Question: In your bucket you have 6.35 $, and you want to travel to the next state by train and its ticket costs 1.35 $. How many times can you use this train between the two stations?
Suppose that the cost of ticket is the same in both ways!
Options:
A)4
B)5
C)6
D)7
E)8
Rationale: As the ticket costs 1.35 $ for one way, so the two-way trip costs 2.7$, and the two tow-way trips cost 5.4$.
So if you have 6.35$, you are able to go and return just two times according to your money in your bucket.
So the correct choice is A
Answer: A
Question: a wall clock rings after every 1 hr and a wrist watch beep after every 25 min.they voice together at 12 a.m,time taken to voice again simultaneously?
Options:
A)4 a.m
B)5 a.m
C)6 a.m
D)7 a.m
E)8 a.m
Rationale: 1 hr = 60 minutes
So, L.C.M. of 60 and 25 is = 300
300 minutes = 5 hr
So after 5 hr that is at 5 a.m both will voice again simultaneously.
ANSWER:B
Answer: B
Question: For integers U and V, when U is divided by V, the remainder is even. Which of the following must be true?
Options:
A)U is odd
B)At least one of U and V is odd
C)The sum U + V is odd
D)U and V share no common factors other than 1
E)UV is odd.
Rationale: At least one of U and V is odd
B
Answer: B
Question: Lisa's grade was 75th percentile of the 40 grades for her class. Of the 50 grades from another class, 8 were higher than Lisa's, and the rest were lower. If no other grade was the same as Lisa's grade, then Lisa's grade was what percentile of the grades of the two classes of two classes combined?
Options:
A)62nd
B)74th
C)80th
D)87th
E)92nd
Rationale: 62nd/74th/80th/87th/92nd
Correct option: C
Answer: C
Question: 16 is divided into two parts in such a way that seventh part of first and ninth part of second are equal. Find the smallest part?
Options:
A)14
B)12
C)11
D)8
E)7
Rationale: x/7 = y/9 => x:y = 7:9
7/16 * 16 = 7
ANSWER:E
Answer: E
Question: The sum of the squares of three numbers is 225, while the sum of their products taken two at a time is 200. Their sum is:
Options:
A)10
B)30
C)40
D)25
E)None of these
Rationale: x^+y^2+z^2=225
xy+yz+zx=200
as we know.. (x+y+z)^2=x^2+y^2+z^2+2(xy+yz+zx)
so (x+y+z)^2=225+(2*200)
(x+y+z)^2=625
so x+y+z=25
ANSWER:D
Answer: D
Question: In a shop, the marked price of an article is worked out in such a way that it generates a profit of 33133313. What should be the discount percent allowed on the marked price during a sale, so that the final profit made is 20%?
Options:
A)13 1/3
B)12 1/2
C)10%
D)8 1/2
E)6 2/3
Rationale: soln:-
lets consider the cost price of an article as $60 (take a number whose 33.33% can be calculated easily)
so marked price = 1.3333*60 = 80
profit to be earned = 20%
final selling price = 1.2*60 = 72
therefore discount to be given = 8/80 = 10%
answer - C
Answer: C
Question: If x^2 +2x = 15, then one possible value of x – 4 =
Options:
A)-8
B)-5
C)-3
D)-1
E)5
Rationale: x^2 +2x = 15
i.e. x^2 +2x - 15 = 0
i.e. x^2 +5x -3x - 15 = 0
i.e. (x+5)(x-3) = 0
i.e. x =-5 or 3
i.e. x-4 = -5-4 or 3-4
i.e. x-4 = - 9 or -1
Answer: Option D
Answer: D
Question: A sum of Rs.1460 is divided into 14 students so that each boy gets Rs.120 while each girl gets Rs.100. Find the number of boys and number of girls
Options:
A)8B, 5G
B)5B, 8G
C)3B, 9G
D)9B, 4G
E)9B, 5G
Rationale: 120B+100G=1460-----i
B+G=12-------ii
from i
100B+20B+100G=1460
100(B+G)+20B=1460
100(14)+20B=1460
By simplifing we get
B=3;G=9
ANSWER:C
Answer: C
Question: The probability of Sam passing the exam is 1/4. The probability of Sam passing the exam and Michael passing the driving test is 1/6. What is the probability of Michael passing his driving test?
Options:
A)1/24
B)1/2
C)1/3
D)2/3
E)2/5
Rationale: let p(s) and p(m) denote probability of sam passing the exam and michael passing the driving test respectively
as per the question
p(s)=1/4 and
p(s) * p(m)=1/6
therefore,
p(m)=(1/6)*4=2/3
Answer:D
Answer: D
Question: The average age of a husband and his wife was 23 years at the time of their marriage. After five years they have a one year old child. The average age of the family now is
Options:
A)19 years
B)23 years
C)28.5 years
D)29.3 years
E)None
Rationale: Solution
Sum of the present ages on husband, wife and child	=(23x2+5x2)+1
= 57 years
Required average
= (57/3)
= 19 years.
Answer A
Answer: A
Question: If 6 men can reap 60 acres of land in 10 days, how many acres of land can 12 men reap in 20 days?
Options:
A)120
B)360
C)240
D)182
E)1000
Rationale: 6 men 60 acres 10 days
12 men ? 20 days
60 * 12/6 * 20/10
60 * 2* 2
60 * 4 = 240
Answer:C
Answer: C
Question: The true discount on Rs. 1760 due after a certain time at 12% per annum is Rs. 160. The time after which it is due is
Options:
A)11
B)66
C)77
D)10
E)81
Rationale: P.W. = Rs. (1760 -160) = Rs. 1600.
S.I. on Rs. 1600 at 12% is Rs. 160.
Time = 100 x 160 =	5	years = 5	x 12	months = 10 months.
1600 x 12	6	6
Answer: D
Answer: D
Question: A 1200 m long train crosses a tree in 120 sec, how much time will I take to pass a platform 700 m long?
Options:
A)180 sec
B)190 sec
C)170 sec
D)175 sec
E)185 sec
Rationale: L = S*T
S= 1200/120
S= 10 m/Sec.
Total length (D)= 1900 m
T = D/S
T = 1900/10
T = 190 Sec
ANSWER:B
Answer: B
Question: If one positive integer is greater than another positive integer by 4, and the difference of their cubes is 208, what is their sum?
Options:
A)8
B)10
C)12
D)14
E)16
Rationale: 1^3 = 1
2^3 = 8
3^3 = 27
4^3 = 64
5^3 = 125
6^3 = 216
The two numbers are 2 and 6.
The answer is A.
Answer: A
Question: There are 5women and 6 men in an office. In how many ways can a person can be selected?
Options:
A)10
B)11
C)12
D)15
E)20
Rationale: Number of ways in which a person can be selected = 5+6 = 11 ways
Answer is B
Answer: B