divelab/Sys2Bench huggingface dataset:
Question: If you divide 7^777 by 5, which remainder do you get?
Options:
A)0
B)1
C)2
D)3
E)4
Rationale: The units digit of 7^1 is 7.
The units digit of 7^2 is 9.
The units digit of 7^3 is 3.
The units digit of 7^4 is 1.
The units digit of 7^5 is 7.
etc...
The units digit of 7^n repeats in blocks of 4: {7, 9, 3, 1}
The remainder of 777/4 is 1.
So the units digit of 7^777 is 7.
The remainder of 7^777 / 5 is 2.
The answer is C.
Answer: C
Question: If y exceeds x by 10%, then x is less than y by?
Options:
A)16 2/9%
B)16 3/3%
C)16 2/3%
D)12 2/3%
E)9 1/11%
Rationale: X=100	y=110
110------10
100-------? => 9 1/11%
Answer:E
Answer: E
Question: A, B, C together started a business. A invested Rs.6000 for 5 months B invested Rs.3600 for 6 months and C Rs.7500 for 3 months. If they get a total profit of Rs.7410. Find the share of A?
Options:
A)2399
B)2889
C)2830
D)2878
E)3000
Rationale: Explanation:
60*5:36*6:75*3
100: 72: 75
100/247 * 7410 = 3000
Answer: E
Answer: E
Question: A gardener increased the length of a rectangle-shaped garden by 20 percent and decreased its width by 5 percent. The area of the new garden
Options:
A)has increased by 18 percent
B)has increased by 14 percent
C)has increased by 7 percent
D)is exactly the same as the old area
E)cannot be expressed in percentage terms without actual numbers
Rationale: Let L be the original length and let W be the original width.
The original area was L*W.
The new area is 0.95W*1.2L = (0.95)(1.2)L*W = 1.14*L*W
The answer is B.
Answer: B
Question: Walking across campus, a student interviewed a group of students. 25% of the students took a finance class last semester, 52% took a marketing class last semester, and 40% took neither a finance nor a marketing class last semester. What percent of the students in the group took both a finance and a marketing class?
Options:
A)60%
B)50%
C)25%
D)17%
E)10%
Rationale: {Total} = {finance} + {marketing} - {both} + {neither}
100 = 25 + 52 - {both} + 40
{both} = 17%.
Answer: D.
Answer: D
Question: In a business, A and C invested amounts in the ratio 2 : 1 , whereas the ratio between amounts invested by A and B was 3 : 2 . If Rs 157300 was their profit, how much amount did B receive?
Options:
A)48000
B)48200
C)48400
D)48600
E)None of these
Rationale: Explanation :
Assume that investment of C = x
Then, investment of A =2x
Investment of B = 4x/3
A:B:C = 2x : 4x/3 : x = 2 : 4/3 : 1 =6 : 4 : 3
B's share = 157300 * 4/(6+4+3) = 157300*4/13
= 12100*4 = 48400. Answer : Option C
Answer: C
Question: At what time between 2 and 3 o'clock will the hands of a clock be together?
Options:
A)120/28
B)120/22
C)120/19
D)120/11
E)120/112
Rationale: The hands of a clock would be together when the angle between The hour hand and minute hand is Zero. Now apply the formula: θ=∣∣∣30h−112m∣∣∣θ=|30h−112m|
Here θθ = 0
⇒11/2m – 30h = 0
⇒11/2m – 30×2 = 0
⇒ m = 120/11
Answer : D
Answer: D
Question: The purchase price of an article is $48. In order to include 15% of cost for overhead and to provide $12 of net profit, the markup should be
Options:
A)15%
B)25%
C)35%
D)40%
E)45%
Rationale: Cost price of article = 48$
% of overhead cost = 15
Net profit = 12 $
We need to calculate % markup
Net profit as % of cost price = (12/48)*100 = 25%
Total markup should be = 25 + 15 = 40%
Answer D
Answer: D
Question: (422+404)^2−(4×422×404)=?
Options:
A)324
B)342
C)343
D)348
E)322
Rationale: Explanation :
(a+b)^2−4ab=(a^2+2ab+b^2)−4ab=(a^2−2ab+b^2)=(a−b)^2
Given Equation is in the form (a+b)^2−4ab where a = 422 and b = 404
Hence answer = (a+b)^2−4ab=(a−b)62=(422−404)^2=18^2=324
Answer is A
Answer: A
Question: Find the area of trapezium whose parallel sides are 20 cm and 18 cm long, and the distance between them is 15 cm.
Options:
A)200
B)234
C)285
D)345
E)244
Rationale: Area of a trapezium = 1/2 (sum of parallel sides) * (perpendicular distance between them) = 1/2 (20 + 18) * (15) = 285 cm2
Answer: Option C
Answer: C
Question: If a and b are the roots of the equation x2 - 9x + 20 = 0, find the value of a2 + b2 + ab?
Options:
A)57
B)59
C)61
D)63
E)65
Rationale: a2 + b2 + ab = a2 + b2 + 2ab - ab
i.e., (a + b)2 - ab
from x2 - 9x + 20 = 0, we have
a + b = 9 and ab = 20. Hence the value of required expression (9)2 - 20 = 61.
Option C
Answer: C
Question: there are 10 girls and 20 boys in a classroom. what is the ratio of girls to boys?
Options:
A)1/2
B)1/3
C)1/5
D)10/30
E)2/5
Rationale: If girls is 10 and boys is 20, then 10/20.
so ratio of girls to boys is = 10/20 = 1/2
ANSWER:A
Answer: A
Question: Rahul is 30 years older than his daughter Mary. In 20 years Rahul will be twice as old as Mary. What is Mary current age.
Options:
A)12
B)14
C)10
D)16
E)18
Rationale: Now: Mary = x, Rahul =x+30
In 20 years Mary = x+20, Rahul= x+30+20 or 2(x+20)
x+30+20= 2(x+20)
x+50=2x+40
50-40=2x-x
x= 10
Mary is 10 years old
Answer : C
Answer: C
Question: In how many T ways can 4 white and 3 black chess pieces be arranged in a row such that they occupy alternate places? Assume that the pieces are distinct.
Options:
A)T=288
B)T=144
C)T=12
D)48
E)96
Rationale: 4 white chess pieces can be arranged in 4! ways and 4 black pieces can be arranged in 3 ! ways.
W_W_W_W
Number of ways = 4!*3! = 24 * 6 = 144
Answer B
Answer: B
Question: The first term in a sequence is 1 and the second term is 21. From the third term on each term is the average (arithmetic mean) of all preceding terms. What is the 38th term in the sequence?
Options:
A)8
B)11
C)21
D)38
E)49
Rationale: The third term is 11 since this is the average of 1 and 21.
Then the next terms will all be 11 since the average stays at 11.
The answer is B.
Answer: B
Question: A reduction of 20% in the price of salt enables a lady to obtain 10kgs more for Rs.100, find the original price per kg?
Options:
A)3.7
B)2.8
C)2.5
D)4.7
E)2.7
Rationale: C
100*(20/100) = 20 --- 10
? --- 1 => Rs.2
100 --- 80
? --- 2 => Rs.2.5
Answer: C
Question: During 2001, a stock lost 40 percent of its value. During the following year, the stock's value increased by 200 percent. Which of the following is the percent change in the stock's value during those two years?
Options:
A)60 percent decrease
B)40 percent decrease
C)20 percent increase
D)80 percent increase
E)140 percent increase
Rationale: Let the value of the stock in 2000 be 100, then in 2001 its value would be 100*(1-40/100)=100*0.6=60 and in 2002 its value would be 60*(1+200/100)=180. Hence during those two years the value of the stock increased from 100 to 180, so by 80%.
Answer: D.
Answer: D
Question: 2, 6, 14, 26, 42, ?
Options:
A)62
B)56
C)74
D)57
E)32
Rationale: The pattern is + 4, + 8, + 12, + 16, +20
Answer : A.
Answer: A
Question: The average of five integers is 63, and none of these integers is greater than 100. If the average of three of the integers is 65, what is the least possible value of one of the other two integers?
Options:
A)5
B)15
C)20
D)21
E)30
Rationale: The sum of two other integers = 63*5-65*3 = 120
the least possible value of one integer when the other integer has the highest value of 100 =120 - 100 = 20
ANSWER:C
Answer: C
Question: According to the direction on a can of frozen orange juice concentrate is to be mixed with 3 cans of water to make orange juice . How many 6 - ounce cans of the concentrate are required to prepare 200 6-ounce servings of orange juice?
Options:
A)25
B)34
C)50
D)67
E)50
Rationale: Orange juice concentrate: water::1:3
total quantity of orange juice = 200*6 = 1200 oz
so Orange juice concentrate: water :: 300 oz:900 oz
no. of 6 oz can = 300 oz/6 oz = 50
Answer E, 50 cans
Answer: E
Question: There are n members in a certain department, including Michael. Two representatives are to be selected to attend a company conference. If there are 36 possible combinations in which Michael is not selected, what is the value of n?
Options:
A)8
B)9
C)10
D)11
E)12
Rationale: The combinations of two people, in which Michael was not selected = 36
The number of ways two people can be selected from m people = m*(m-1)/2
Let m be the number of people excluding Michael.
Then m*(m-1) = 72 = 9*8
Thus, n = m + 1(Michael) = 10
The answer is C.
Answer: C
Question: A rectangular lawn of dimensions 80 m * 60 m has two roads each 10 m wide running in the middle of the lawn, one parallel to the length and the other parallel to the breadth. What is the cost of traveling the two roads at Rs.3 per sq m?
Options:
A)Rs.3600
B)Rs.3700
C)Rs.3800
D)Rs.3900
E)Rs.4000
Rationale: Area = (l + b – d) d
(80 + 60 – 10)10 => 1300 m2
1300 * 3 = Rs.3900
ANSWER:D
Answer: D
Question: A book was sold for Rs 27.50 with a profit of 10%. if it were sold for Rs25.75, then what would be % of profit or loss?
Options:
A)Profit 5%
B)Loss 5%
C)Profit 3%
D)Loss 3%
E)Profit 4%
Rationale: SP=Rs 27.50: profit =10%.
CP=Rs {(100/110)*27.50}=Rs 25.
When sp =Rs 25.75 ,profit =Rs(25.75-25)=Rs 0.75
Profit%={(0.75/25)*100}%=25/6%=3%
Answer is C.
Answer: C
Question: A thief steels a car at 6a.m. and drives it at 50kmph. The theft is discovered at 7a.m. and the owner sets off in another car at 70 kmph. When will he overtake the thief?
Options:
A)9a.m.
B)9.30a.m.
C)10a.m.
D)11.45a.m.
E)10.15a.m.
Rationale: Suppose the thief overtakes x hours after 6a.m.
Distance covered by the thief in x hrs = distance covered by the owner in (x-1 hours)
50x = 70(x-1)
20x = 70
x = 7/2 hrs
Thief is overtaken at 9.30a.m.
Answer is B
Answer: B
Question: Six persons A,B,C,D,E &F went to soldier cinema. There are six consecutive seats. A sits in the first seat followed by B, followed by C and so on. If A taken on of the six seats, then B should sit adjacent to A. C should sit adjacent to A or B. D should sit adjacent to A, B or C and so on. How many possibilities are there ?
Options:
A)30
B)32
C)34
D)36
E)38
Rationale: First we have to arrange A, there is 1 method for it, for second person B, there are 2 methods one left and one right, for third 2 methods are there same as for B similarly for others, so total number of methods = 1*2*2*2*2*2 = 32
ANSWER:B
Answer: B
Question: a man owns 2/3 of market reserch beauro buzness,and sells 3/4 of his shares for 30000rs,what is the value of buzness?
Options:
A)75000
B)160000
C)170000
D)60000
E)70000
Rationale: If value of business=x
Total sell (2x/3)(3/4)=30000
->x=60000
ANSWER:D
Answer: D
Question: In what time will a train 100 m long cross an electric pole, it its speed be 162 km/hr?
Options:
A)2.5
B)2.9
C)2.4
D)2.8
E)2.22
Rationale: Speed = 162 * 5/18 = 45 m/sec
Time taken = 100/45
= 2.22 sec.
Answer:E
Answer: E
Question: A basket has 5 oranges and 4 oranges. Three fruits are picked at random. The probability that at least 2 oranges are picked is
Options:
A)23/41
B)1/23
C)25/42
D)34/57
E)3/41
Rationale: Total fruits = 9
Since there must be at least two oranges,
(⁵C₂ * ⁴C₁)/⁹C₃ + ⁵C₃/⁹C₃ = 25/42
C
Answer: C
Question: The volumes of two cones are in the ratio 1 : 40 and the radii of the cones are in the ratio of 1 : 2. What is the length of the wire?
Options:
A)8:5
B)8:9
C)2:2
D)8:2
E)2:8
Rationale: The volume of the cone = (1/3)πr2h
Only radius (r) and height (h) are varying.
Hence, (1/3)π may be ignored.
V1/V2 = r1^2 . h1/r2^2. h2 => 1/40
= (1)^2h1/(2)^2h2
=> h1/h2 = 8/5
i.e. h1 : h2 = 8:5
Answer: A
Answer: A
Question: 1, 2, 3, 10, ?, 9802
Options:
A)99
B)8
C)73
D)28
E)27
Rationale: Answer: A) 99
Answer: A
Question: A shopkeeper sells 300 metres of cloth for Rs. 9000 at a loss of Rs.6 per metre. Find his cost price for one metre of cloth?
Options:
A)s.59
B)s.58
C)s.36
D)s.46
E)s.13
Rationale: SP per metre = 9000/300
= Rs. 30 Loss per metre
= Rs. 6 CP per metre = 30 + 6
= Rs.36
Answer: C
Answer: C
Question: Andrew travelling to 7 cities. Gasoline prices varied from city to city.
$1.75, $1.61, $1.79, $2.11, $1.96, $2.09, $1.83. What is the median gasoline price?
Options:
A)$1
B)$1.83
C)$1.92
D)$2.13
E)$2.15
Rationale: Ordering the data from least to greatest, we get:
$1.61, $1.75, $1.79, $1.83, $1.96, $2.09, $2.11
The median gasoline price is $1.83. (There were 3 states with higher gasoline prices and 3 with lower prices.)
B
Answer: B
Question: Anu can do a work in 6 days and Binu alone in 9 days. Anu and Binu undertook to do it for Rs.4800. With help of Minu, they completed the work in 3 days. How much is to be paid to Minu and Anu?
Options:
A)s.750, Rs.2250
B)s.2000, Rs.750
C)s.750, Rs.2000
D)s.800, Rs.2400
E)s.900, Rs.1250
Rationale: Minuâ€™s one dayâ€™s work = 1/3 - 1/6 + 1/9= 1/3 - 5/18 = 1/18
Anuâ€™s wages: Binuâ€™s wages: Minuâ€™s wages = 1/6 : 1/9 : 1/18 = 6: 4: 2
Minuâ€™s share = Rs.4800* 2/12 = Rs.800
Anuâ€™s share = Rs.4800* 6 /12 = Rs.2400
ANSWER:D
Answer: D
Question: A bowl was filled with 10 ounces of water, and 0.007 ounce of the water evaporated each day during a 50-day period. What percent of the original amount of water evaporated during this period?
Options:
A)0.004%
B)0.04%
C)0.40%
D)4%
E)3.5%
Rationale: Total amount of water evaporated each day during a 50-day period = .007 * 50
=.007 * 100/2
= .7/2
= .35
percent of the original amount of water evaporated during this period = (.35/10) * 100%
= 3.5 %
Answer E
Answer: E
Question: xa=yb=zc and y2=xz. Find b.
Options:
A)2ac/a+c
B)7ac/a+c
C)1ac/a+c
D)6ac/a+c
E)3ac/a+c
Rationale: ANSWER : A
Answer: A
Question: If (y - 1)(z + 1) + 4(z - 3) = 0, then which of the following must be true?
Options:
A) y = -4 and z = -3
B) y = 4 or z = -3
C) y = -4 or z = 3
D) y = 1 and z = 3
E) y = 1 or z = - 3
Rationale: yz + y - 2z - 1 + 4z - 12 = 0
yz + y + 3z - 13 = 0
yz + y + 3z = 13
So by checking options we get ,
D.
Answer: D
Question: A trained covered x km at 40 kmph and another 2x km at 20 kmph. Find the average speed of the train in covering the entire 6x km.
Options:
A)16
B)48
C)24
D)19
E)12
Rationale: Total time taken = x/40 + 2x/20 hours
= 5x/40 = x/8 hours
Average speed = 6x/(x/8) = 48 kmph
Answer: B
Answer: B
Question: The sum of a number and the number preceding it is 33. By how much is two less than six times the
number?
Options:
A)196
B)94
C)90
D)100
E)120
Rationale: two numbers must be 16 and 17
16+17=33
required number is 17
six times of this number= 6*17=102
two less than 102= 102-2= 100
ANSWER:D
Answer: D
Question: A seller has 1000 kg of rice part of which he sells at 8% profit and the rest at 18% profit. He gains 14%. The quantity sold at 18% profit is?
Options:
A)300 kg
B)350 kg
C)550 kg
D)600 kg
E)700 kg
Rationale: By rule of alligation,
Profit% by selling 1st part Profit% by selling 2nd part
8 18
Net % profit
14
18-14=4 14-8=6
=> Quantity of part1 : Quantity of part2 = 4 : 6 = 2 : 3
Total quantity is given as 1000 kg. So quantity of part2 (quantity sold at 18% profit)
=1000×35
= 600 kg
D
Answer: D
Question: An equity analyst needs to select 3 stocks for the upcoming year and rank these securities in terms of their investment potential. If the analyst has narrowed down the list of potential stocks to 7, in how many ways can she choose and rank her top 3 picks?
Options:
A)21
B)35
C)210
D)420
E)840
Rationale: Number of ways of choosing 3 stocks = 7C3
Number of ways of ranking the chosen stocks = 3!
Hence total ways of choosing and ranking = 7C3 * 3! = 7!/4! = 7*6*5 = 210
Correct Option: C
Answer: C
Question: How many positive integers less than 300 can be formed using the numbers 1, 2, 3 and 5 for the digits?
Options:
A) 48
B) 52
C) 66
D) 68
E) 84
Rationale: Notice that we can find the number of 2 and 3 digit numbers by just assuming the first digit can also be zero:
0 1 1
1 2 2
2 3 3
2 5 5
Number of possibilities = 3*4*4 = 48.
Then, just add up the number of 1 digits numbers = 4, so total is 48+4 = 52.
Answer: B
Answer: B
Question: 70, 75, 80, 85, 90, 105, 105, 130, 130, 130
The list shown consists of the times, in seconds, that it took each of 10 schoolchildren to run a distance of 400 on of meters. If the standard devastation of the 10 running times is 25.4 seconds, rounded to the nearest tenth of a second, how many of the 10 running times are more than 1 standard deviation below the mean of the 10 running times?
Options:
A)one
B)two
C)three
D)four
E)five
Rationale: The most time consuming part in this question is to define the mean. Under exam pressure and time pressure it is very easy to make mistake.
it is easier to group numbers: 130*3=390; 105*2=210; 75+85=160; 70+80=150; 90;
Next stage combine results, again using more convenient ways to calculate: 390+210=600; 160+150=310; 90. 600+310+90=1000. Since there are 10 numbers the mean is 100. Questions asks to find the quantity of numbers one SD BELOW the mean, which is 100-25,4=74,6. There are only two numbers below 74,6. The answer is B
Answer: B
Question: A certain company assigns employees to offices in such a way that some of the offices can be empty and more than one employee can be assigned to an office. In how many ways can the company assign 2 employees to 3 different offices?
Options:
A)5
B)6
C)7
D)8
E)9
Rationale: Ans. Each employee can go into any of the two offices. Thus we have
=> 3 * 3 = 9
Answer : E
Answer: E
Question: Simplify the following 515*515*515-221*221*221/(515*515+515*221+221*221)
Options:
A)294
B)394
C)284
D)274
E)264
Rationale: 515*515*515-221*221*221/(515*515+515*221*221+221)
=[(515)^3 - (221)^3 ] / [(515)^2 + 515*221 + (221)^2]
if x = 515 & y = 221 then exp is
(x^3 - y^3)/(x^2 + xy + y^2) which is equal to (x-y)
(x^3 - y^3)/(x^2 + xy + y^2)=(x-y)
[(515)^3 - (221)^3 ] / [(515)^2 + 515*221 + (221)^2] = (515-221)= 294
ANSWER:A
Answer: A
Question: f1 (x) = x 0 ≤ x ≤ 1
= 1 x ≥ 1
= 0 otherwise
f2 (x) = f1(–x) for all x
f3 (x) = –f2(x) for all x
f4 (x) = f3(–x) for all x
How many of the following products are necessarily zero for every x:
f1(x) f2 (x), f2 (x) f3 (x), f2(x) f4 (x)
Options:
A)0
B)1
C)2
D)3
E)4
Rationale: Explanation :
We have to consider positive and negative numbers for all the cases.
f1(x) is positive for positive numbers, and 0 for negative numbers. (0 for x = 0)
f2(x) is 0 for positive numbers, and positive for negative numbers. (0 for x = 0)
f3(x) is 0 for positive numbers, and negative for negative numbers. (0 for x = 0)
f4(x) is negative for positive numbers, and 0 for negative numbers. (0 for x = 0)
So, we see that, out of the 3 products in the question, f1(x)*f2(x) and f2(x)*f4(x) are always zero, for any x.
Answer : C
Answer: C
Question: If the simple interest on a certain sum of money for 6 years is one–fifth of the sum, then the rate of interest per annum is
Options:
A)3.3%
B)7.2%
C)6.1%
D)5.0%
E)3.1%
Rationale: Explanation:
Let the Principal(P) be x
Then, Simple Interest(SI) = x/5
Time(T) = 6 years
Rate of interest per annum(R) = (100×SI)/PT
= (100 × (x/5)/(x × 6) = 20/6 = 3.3%
Answer: Option A
Answer: A
Question: The number of people at Ovations Bar in the hour from 12 p.m. to 1 p.m. was 50% greater than the number of people there from 11 a.m. to 12 p.m. The number of the people at the bar from 11 a.m. to 12 p.m. was 25% less than the number there from 10 a.m. to 11 a.m. If 75 people were at the bar from 11 a.m. to 1 p.m., which of the following is the number of people in the bar from 10 a.m. to 11 a.m.?
Options:
A)10
B)20
C)30
D)40
E)50
Rationale: No of people from 10 - 11 is x
No of people from 11 - 12 is 3x/4
No of people from 12 to 1 is (3/2)(3x/4)
Given that 3x/4 + (3/2)(3x/4) = (3x/4)(5/2) = 75
Then x = 40
D
Answer: D
Question: The speed at which a man can row a boat in still water is 22 kmph. If he rows downstream, where the speed of current is 3 kmph, what time will he take to cover 80 metres?
Options:
A)12.59
B)11.53
C)15.69
D)14.56
E)18.98
Rationale: Speed of the boat downstream = 22 + 3 = 25 kmph= 25 * 5/18 = 6.94 m/sHence time taken to cover 80 m = 80/6.94= 11.53 seconds.Answer:B
Answer: B
Question: In measuring the sides of a rectangle, one side is taken 20% in excess and the other 10% in deficit. Find the error per cent in area calculated from the measurement?
Options:
A)12% deficit
B)10% excess
C)8% excess
D)9% excess
E)none of these
Rationale: Since Side1 x Side2 = Area
Therefore, error% in area
= ( x + y + xy/100) %
= [20 - 10 + ( -10 x 20)/100]% or 8%
i.e., 8% excess
ANSWER:C
Answer: C
Question: Insert the missing number
2, 4, 8, ..., 32, 64, 128, 256
Options:
A)15
B)14
C)16
D)13
E)18
Rationale: series is 2* Previous Number.
ANSWER:C
Answer: C
Question: Population of a city in 20004 was 1000000. If in 2005 there isan increment of 15%, in 2006 there is a decrements of 35 % and in 2007 there is an increment of 45%, then find the population of city atthe end of the year 2007
Options:
A)3436521
B)3546577
C)3546546
D)7565643
E)1083875
Rationale: Required population = P (1 + R1/100)(1 - R2/100)(1 + R3/100)
= P (1 + 15/100)(1 - 35/100)(1 + 45/100)
= 1083875
E
Answer: E
Question: Mark and Ann together were allocated n boxes of cookies to sell for a club project. Mark sold 7 boxes less than n and Ann sold 2 boxes less than n. If Mark and Ann have each sold at least one box of cookies, but together they have sold less than n boxes, what is the value of n?
Options:
A)4
B)5
C)6
D)7
E)8
Rationale: if n =8
mark sold 1 box
and Ann sold 6 boxes
total 7 < 8
Answer:E
Answer: E
Question: The sum of the present ages of father and his son is 60 years. Six years ago, father's age was five times the age of the son. After 6 years, son's age will be?
Options:
A)11 years
B)17 years
C)18 years
D)20 years
E)58 years
Rationale: Let the present ages of son and father be x and (60 - x) years respectively.
Then, (60 - x) - 6 = 5(x - 6)
6x = 84 => x = 14
Son's age after 6 years = (x + 6) = 20 years.
Answer:D
Answer: D
Question: Angelo and Isabella are both salespersons. In any given week, Angelo makes $550 in base salary plus 8 percent of the portion of his sales above $2,000 for that week. Isabella makes 10 percent of her total sales for any given week. For what amount of weekly sales would Angelo and Isabella earn the same amount of money?
Options:
A) 19,500
B) 24,500
C) 25,500
D) 26,500
E) 27,500
Rationale: OFFICIAL SOLUTION:
The problem asks for the amount of weekly sales it takes for Angelo and Isabella to earn the same amount of money. You can write an equation that sets Angelo’s and Isabella’s weekly earnings equal to each other, with x representing weekly sales. Weekly earnings for each salesperson equal base salary plus commission. So Angelo’s earnings are 550 + (0.08)(x – 2,000), and Isabella’s are 0.10x. Set up the equation and solve:
550 + (0.08)(x – 2,000) = 0.10x
Distribute the 0.08:
550 + 0.08x – 160 = 0.10x
Combine terms and subtract 0.08x from both sides:
390 = 0.02x
Divide both sides by 0.02:
19,500 = x
Your answer is A.
Answer: A
Question: What is the value of (12 + 22 + 32 + 42 + ----- + 102)
Options:
A)295
B)324
C)385
D)391
E)399
Rationale: Explanation:
(12 + 22 + ….. + n2) = (1/6) n(n + 1) (2n + 1)
Here, n = 10
Therefore,
(12 + 22 + ….. + 102) = (1/6) 10 (10 + 1) ( 2 x 10 + 1)
= (1/6) x 10 x 11 x 21
= 385
ANSWER: C
Answer: C
Question: 6 men and 8 women can complete a work in 10 days. 26 men and 48 women can finish the same work in 2 days. 15 men and 20 women can do the same work in - days.
Options:
A)4 days
B)6 days
C)2 days
D)8 days
E)10 DAYS
Rationale: Explanation :
Let work done by 1 man in 1 day = m and work done by 1 woman in 1 day = b
Work done by 6 men and 8 women in 1 day = 1/10
=> 6m + 8b = 1/10
=> 60m + 80b = 1 --- (1)
Work done by 26 men and 48 women in 1 day = 1/2
=> 26m + 48b = ½
=> 52m + 96b = 1--- (2)
Solving equation 1 and equation 2. We get m = 1/100 and b = 1/200
Work done by 15 men and 20 women in 1 day
= 15/100 + 20/200 =1/4
=> Time taken by 15 men and 20 women in doing the work = 4 days
Answer : Option A
Answer: A
Question: C and D are tying water balloons for a party. 5 out of every 8 balloons that C ties does not pop, and 2 out of every 3 balloons that D ties does not pop. If they each tie 120 balloons, what fraction of the 240-pack will pop?
Options:
A)85/240
B)70/240
C)155/240
D)140/240
E)60/240
Rationale: Ratio of A's balloons that do not pop: 5/8
Ratio of B's balloons that do not pop: 2/3
Ratio of A and B's balloons that do not pop: 5/8+2/3=31/24
Total number of A's and B's balloons that do not pop: 120x(31/24)=155/240
Fraction of balloons that pop: 1-155/240=85/240
Answer is A
Answer: A