Two thirds of people in a party are lawyers, whose average I.Q. is 120. The rest are engineers whose I.Q. is 180. What is the average I.Q. of all persons in the room?
Answer:
140
150
130
160
Use a weighted average. Let the total number of people be 3 parts: 2 parts lawyers and 1 part engineers. $\bar I=\frac{2(120)+1(180)}{3}$ $\boxed{140}$
A rectangular soccer field is twice as long as it is wide. If the perimeter of the soccer field is 300 yards, what are the field’s dimensions?
Answer:
50 yd by 100 yd
40 yd by 80 yd
45 yd by 90 yd
60 yd by 120 yd
Let the width be $w$ and the length be $2w$. The perimeter is: $2(w+2w)=300$ $6w=300$, so $w=50$ yd and length $=100$ yd. $\boxed{50\text{ yd by }100\text{ yd}}$
Dana and her husband recently installed an inground rectangular swimming pool measuring 40 ft by 30 ft. They want to add a brick border of uniform width around all sides of the pool. How wide can they make the brick border if they purchased enough brick to cover an area of 296 ft2?
Answer:
2.0 ft
1.5 ft
2.5 ft
3.0 ft
Let the uniform border width be $x$. The outside dimensions are $(40+2x)$ by $(30+2x)$. Border area is: $(40+2x)(30+2x)-40(30)=296$ $4x^2+140x=296$ $x^2+35x-74=0$ $x=2$ ft $\boxed{2.0\text{ ft}}$
In a chemical reaction, the amount of starting material C cm3 left after t minutes is given by C=400e-0.006t. Determine the time taken for the concentration to decrease by half.
Answer:
116 minutes
98 minutes
89 minutes
107 minutes
Half the starting amount means $C=200$: $200=400e^{-0.006t}$ $0.5=e^{-0.006t}$ $\ln(0.5)=-0.006t$ $t=\frac{0.6931}{0.006}=115.5$ min $\boxed{116\text{ minutes}}$
A certain power plant consumes 1500 tons of coal in 4.0 weeks. There is a stockpile of 10,000 tons of coal available when the plant starts operating. After 3.0 weeks in operation, an additional boiler, capable of using 2300 tons in 3.0 weeks, is put on line with the first boiler. In how many more weeks will the stockpile of coal be consumed?
Answer:
7.8 weeks
5.5 weeks
10.8 weeks
12.3 weeks
The first boiler uses $1500/4=375$ tons/week. In 3 weeks it consumes: $375(3)=1125$ tons Coal left: $10000-1125=8875$ tons The added boiler uses $2300/3=766.67$ tons/week, so combined use is: $375+766.67=1141.67$ tons/week Remaining time: $8875/1141.67=7.77$ weeks $\boxed{7.8\text{ weeks}}$
Cars A and B leave a town at the same time and travel in opposite directions. Car A travels at the speed of 40kph and car B at 60kph. How long will the two cars be 350km apart?
Answer:
3.5 hrs
2.5 hrs
5.5 hrs
4.5 hrs
The cars separate at the combined speed because they travel in opposite directions: $40+60=100$ kph Time to be 350 km apart: $t=350/100$ $\boxed{3.5\text{ hrs}}$
After M students took the test, there was a total of 64% correct answers. If the test contains 50 questions, what is the least number of questions that the next student has to get right to bring the total of correct answers to 70%?
3M + 35
3M + 20
4M + 15
4M + 20
After $M$ students, total answers are $50M$ and correct answers are $0.64(50M)=32M$. Let the next student get $x$ correct. To bring the total to 70%: $32M+x=0.70(50M+50)$ $32M+x=35M+35$ $\boxed{x=3M+35}$
If the farmer sells 75 of his chickens, his stock of feed will last 20 more days than planned.
If he buys 100 more chickens, he will run out of feed 15 days earlier than planned.
If no chickens were sold or bought, the farmer will be exactly on schedule.
How many chickens does he have?
300
60
240
275
Let $N$ be chickens and $D$ planned days. Feed amount is proportional to chicken-days. $(N-75)(D+20)=ND$ $(N+100)(D-15)=ND$ From the first, $20N-75D=1500$. From the second, $100D-15N=1500$. Solving gives: $\boxed{N=300}$
If x is even and x! is odd, what is the value of x?
0
1
2
4
For $x!$ to be odd, none of the factors from 1 to $x$ can include 2. Since $x$ is even, the only even nonnegative integer satisfying this is $x=0$, and $0!=1$ is odd. $\boxed{0}$
Two cities A and B are 30 km apart. From both of these cities, simultaneously, two hikers start their journeys toward each other. They are walking at a constant speed of 5 km/h each.
A fly leaves city A. It flies at 10 km/h and passes the hiker from its city. When it reaches the B hiker, it turns around and flies back to the A hiker. It keeps doing so until the fly lands on the shoulder of the A hiker as he continues his journey to B.
How many kilometers has the fly flown?
30 km
25 km
37.5 km
45 km
The two hikers approach each other at $5+5=10$ km/hr, so they meet after: $30/10=3$ hr The fly travels continuously at 10 km/hr for 3 hr: $10(3)=30$ km $\boxed{30\text{ km}}$
Suppose you receive x dollars in January. Each month thereafter you receive $100 more than you received the month before.
Write a factored polynomial that describes the total dollar amount received from January through April.
4x + 600
x + 300
4x + 300
x + 600
The monthly amounts from January through April are: $x$, $x+100$, $x+200$, and $x+300$ Total: $x+(x+100)+(x+200)+(x+300)=4x+600$ $\boxed{4x+600}$
Larry, Jerwin, Elvira, and Yuriy are doing business together. They agreed their money will be kept in a safe, and in order to open it the majority of them must be present.
The safe is provided with several locks and each person is given keys to certain locks.
Find the number of locks required and the number of keys each must have.
6 locks, 3 keys
3 locks, 1 key
4 locks, 2 keys
5 locks, 4 keys
A majority of 4 people means any 3 can open the safe, but no pair can. Use one lock for each pair of people excluded from opening it: $\binom{4}{2}=6$ locks Each person is missing the locks assigned to pairs containing them, which is 3 locks, so each person has the other 3 keys. $\boxed{6\text{ locks, }3\text{ keys}}$
A group of engineering examinees plans to have an excursion. They hired a mini-bus for ₱56,000, but for unknown reasons two of them were not able to join the trip, and the share of each remaining member increased by ₱3,200.
How many were there in the original group?
7
8
9
10
Let the original number be $n$. Original share is $56000/n$; after two miss the trip, share is $56000/(n-2)$ and increases by 3200: $\frac{56000}{n-2}-\frac{56000}{n}=3200$ $56000\frac{2}{n(n-2)}=3200$ $n(n-2)=35$ $\boxed{n=7}$
Labor laws in a certain country require factory owners to give every worker a holiday whenever one of them has a birthday and to hire without discrimination on grounds of sex.
Except for these holidays, they work 365 days per year.
The owner wants to maximize the expected total number of man-days worked per year in the factory.
How many workers should the factory have?
Zoologists have determined empirically that the total number of endangered species y may be approximated by the formula:
y = 1883 / (1 + 7.892(1.097)−t)
where t is the number of years past 1980.
How many more species does the formula estimate will be added to the endangered list for 2020 than the actual number given for 2007?
A farmer selling eggs at ₱50 per dozen gains 20%.
If he sells the eggs at the same price after the cost rises by 12.5%, calculate his new gain.
6.65%
6.58%
6.89%
6.12%
Selling at P50 per dozen gives 20% gain, so original cost is: $C=50/1.20=41.6667$ New cost after 12.5% rise: $C_n=41.6667(1.125)=46.875$ New gain percent: $\frac{50-46.875}{46.875}(100)=6.67\%$ $\boxed{6.65\%}$
The letters in the word TRIANGLE are to be arranged.
How many ways can these letters be arranged in a line?
50240
362880
40320
5040
In part 1, how many of those do the letters E and G appear in adjacent positions?
10080
12030
11540
9080
How many ways can these letters be arranged in a circle?
4020
5230
6210
5040
Part 1.
TRIANGLE has 8 distinct letters. Number of linear arrangements: $8!=40320$. $\boxed{40320}$
Part 2.
Treat E and G as one block. Then there are 7 objects to arrange: the EG block plus the other 6 letters. The block can be arranged as EG or GE, so multiply by 2: $7!\times2=10080$ $\boxed{10080}$
Part 3.
For circular arrangements of 8 distinct letters, fix one letter and arrange the remaining 7. Number of arrangements: $(8-1)!=7!=5040$. $\boxed{5040}$
Question Bank: t2089
MSTE - Algebra / Worded Problems / Besavilla CE Pre-Board Math & Surveying
Three people P, Q and R contribute to a fund. P provides 3/5 of the total, Q provides 2/3 of the remainder and R provides P8. Determine the total fund.
P45
P53
P60
P57
P65
Let the total fund be $T$. P contributes $\frac{3}{5}T$, leaving $\frac{2}{5}T$. Q contributes $\frac{2}{3}$ of the remainder: $Q=\frac{2}{3}\left(\frac{2}{5}T\right)=\frac{4}{15}T$ R gets the remaining fraction: $T-\frac{3}{5}T-\frac{4}{15}T=\frac{2}{15}T$ $\frac{2}{15}T=8 \Rightarrow T=60$ $\boxed{\text{P}60}$
Question Bank: w2
MSTE - Algebra / Worded Problems / MSTE May 2019
To stimulate his son in the pursuit of calculus, a math professor offered to pay him P400 for every problem correctly solved and to fine him P250 for every incorrect solution. At the end of 26 problems, neither owed any money to the other. How many did the boy solve correctly?
Ian wants to calculate how many baristas are required in his coffee shop. He operates 313 days a year for 11 hours per day. Ian wants a capacity cushion of 12% to deal with unexpected fluctuation in demand. This year Ian is forecasting 500,000 customers with an average time of 3 minutes.
Many printing presses are designed with large plates that print a fixed number of pages as a unit. Each unit is called a signature. A particular press prints a signature of 16 pages each. Suppose $C(p)$ is the cost of printing a book of $p$ pages, assuming each signature printed costs P1,400. What is the cost of printing a book of 128 pages?
P179,200
P11,200
P24,300
P5,200
Number of signatures $= \frac{128}{16} = 8$. Cost $= 8 \times 1400 = \boxed{P\,11{,}200}$
Question Bank: w14
MSTE - Algebra / Worded Problems / MSTE May 2019
The Wollomombi Falls in Australia have a height of 1100 ft. A pebble is thrown upward from the top of the falls with an initial velocity of 20 ft/sec. The height of the pebble $h$ after $t$ sec is given by the equation $h = -16t^2 + 20t + 1100$. How long after the pebble is thrown will it hit the ground?
10.58
8.94
12.37
13.69
The pebble hits the ground when $h = 0$: $-16t^2 + 20t + 1100 = 0$ $t = \frac{-20 - \sqrt{20^2 - 4(-16)(1100)}}{2(-16)}$ $\boxed{t = 8.94\text{ s}}$
