Algebra word Problems(LEVEL - 3 )
Q1–10 of 45A school has to buy at least 15 chairs within a budgetary ceiling of ₹2000. A chair with arms costs ₹160 and one without arms costs ₹100. What is the maximum number of chairs with arms that the school can buy?
Goal: Maximize chairs with arms ($x$).
Equations/Inequalities:
Total chairs: $x + y \ge 15 \implies y \ge 15 - x$
Total cost: $160x + 100y \le 2000$
Calculation:
Since $x$ must be an integer, $x_{\max} = 8$.
Correct Answer: (b) 8
On Monday, a certain animal shelter housed 55 cats and dogs. By Friday, exactly 1/5 of the cats and 1/4 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?
Given: Total pets $C + D = 55$.
Condition: Adopted $= \frac{C}{5} + \frac{D}{4}$.
Calculation:
To maximize adoption while ensuring $C$ is a multiple of 5 and $D$ is a multiple of 4:
Since $D$ must be a multiple of 4, $55 - C$ must be divisible by 4.
If $C = 15 \implies D = 40 \implies \text{Adopted} = \frac{15}{5} + \frac{40}{4} = 3 + 10 = 13$
If $C = 35 \implies D = 20 \implies \text{Adopted} = \frac{35}{5} + \frac{20}{4} = 7 + 5 = 12$
If $C = 55 \implies D = 0 \implies \text{Adopted} = 11$
Maximum adopted = $13$.
Out of two-thirds of the total number of basketball matches, a team has won 17 matches and lost 3 of them. What is the maximum number of matches that the team can lose and still win more than three-fourth of the total number of matches if it is true that no match can end in a tie?
Given:
Played so far $= \frac{2}{3}N$
Won $= 17$, Lost $= 3 \implies \frac{2}{3}N = 20 \implies N = 30$ matches.
Condition: Total wins must be $> \frac{3}{4}(30) = 22.5 \implies$ At least $23$ wins.
Calculation:
Minimum total wins needed $= 23$
Wins remaining needed $= 23 - 17 = 6$
Matches remaining $= 30 - 20 = 10$
Maximum additional losses $= 10 - 6 = 4$
Manick visited his cousin Aniket during the summer vacation. In the mornings, they both would go for swimming. In the evenings, they would play tennis. They would engage in at most one activity per day i.e., either they went swimming or played tennis each day. There were days when they took rest and stayed home all day long. There were 32 mornings when they did nothing, 18 evenings when they stayed at home, and a total of 28 days when they swam or played tennis. What duration of the summer vacation did Manick stay with Aniket?
Given:
Total days $= N$
Swam in morning ($S$), Tennis in evening ($T$), Rest ($R$)
32 mornings did nothing $\implies$ No swim $\implies T + R = 32$
18 evenings stayed home $\implies$ No tennis $\implies S + R = 18$
$S + T = 28$
Calculation:
Adding all three:
Total duration $= 39$ days.
Correct Answer: (b) 39 days
10 cows can graze in a field for 15 days and 20 cows can graze in the same field for 10 days. For how many days can 30 cows graze in the field?
Model: Work/Grass growth formula: $G + m \cdot t = c \cdot t$ (where $G$ is initial grass, $m$ is growth rate per day, $c$ is number of cows).
$G + 15m = 10 \times 15 = 150$
$G + 10m = 20 \times 10 = 200$
Solving for $G$ and $m$:
Subtract (2) from (1): $5m = -50 \implies m = -10$ (Grass is decaying/drying up at rate $10$).
$G = 150 - 15(-10) = 300$.
For 30 cows for $d$ days:
Grass in a lawn grows equally thick and at a uniform rate. It takes 24 days for 70 cows and 60 days for 30 cows to eat the whole of the grass. How many cows are needed to eat the grass in 96 days?
Formula: $G + m \cdot t = c \cdot t$
$G + 24m = 70 \times 24 = 1680$
$G + 60m = 30 \times 60 = 1800$
Calculation:
Subtracting gives: $36m = 120 \implies m = \frac{10}{3}$
$G = 1680 - 24\left(\frac{10}{3}\right) = 1680 - 80 = 1600$
For $C$ cows in 96 days:
Directions (Questions 7–8):These questions are based on the following information:
In a holy city, there are ten shrines and a certain number of holy lakes. A group of pilgrims stayed in the city for a few days and visited the shrines and lakes during their stay. At the end of the stay it turned out that all each shrine was visited exactly by 4 pilgrims and each lake was visited exactly by 6 pilgrims. Each pilgrim visited exactly 5 shrines and 3 lakes.
Q. The number of lakes in the holy city is
Let:
$S = 10$ (Shrines)
$L = \text{Lakes}$
$P = \text{Pilgrims}$
Total shrine visits:
Total lake visits:
Answers: (a) 4
Directions (Questions 7–8):These questions are based on the following information:
In a holy city, there are ten shrines and a certain number of holy lakes. A group of pilgrims stayed in the city for a few days and visited the shrines and lakes during their stay. At the end of the stay it turned out that all each shrine was visited exactly by 4 pilgrims and each lake was visited exactly by 6 pilgrims. Each pilgrim visited exactly 5 shrines and 3 lakes.
Q. The number of pilgrims in the group is
Let:
$S = 10$ (Shrines)
$L = \text{Lakes}$
$P = \text{Pilgrims}$
Total shrine visits:
Total lake visits:
Answers:(c) 8
From a number of apples, a man sells half the number of existing apples plus 1 to the first customer, sells 1/3rd of the remaining apples plus 1 to the second customer and 1/5th of the remaining apples plus 1 to the third customer. He then finds that he has 3 apples left. How many apples did he have originally?
Before 3rd customer:
$$\left(x_3 - \left(\frac{1}{5}x_3 + 1\right)\right) = 3 \implies \frac{4}{5}x_3 - 1 = 3 \implies \frac{4}{5}x_3 = 4 \implies x_3 = 5$$Before 2nd customer:
$$\left(x_2 - \left(\frac{1}{3}x_2 + 1\right)\right) = 5 \implies \frac{2}{3}x_2 - 1 = 5 \implies \frac{2}{3}x_2 = 6 \implies x_2 = 9$$Before 1st customer (Original):
$$\left(x_1 - \left(\frac{1}{2}x_1 + 1\right)\right) = 9 \implies \frac{1}{2}x_1 - 1 = 9 \implies \frac{1}{2}x_1 = 10 \implies x_1 = 20$$
Correct Answer: (c) 20
Ravi has two examinations on Wednesday – Engineering Mathematics in the morning and Engineering Drawing in the afternoon. He has a fixed amount of time to read the textbooks of both these subjects on Tuesday. During this time he can read 80 pages of Engineering Mathematics and 100 pages of Engineering Drawing. Alternatively, he can also read 50 pages of Engineering Mathematics and 250 pages of Engineering Drawing. Assume that the amount of time it takes to read one page of the textbook of either subject is constant. Ravi is confident about Engineering Drawing and wants to devote full time to reading Engineering Mathematics. The number of Engineering Mathematics textbook pages he can read on Tuesday is
Let $M$ be time per page of Math and $D$ be time per page of Drawing.
Given Total Time $T$:
$80M + 100D = T$
$50M + 250D = T$
Equating:
Total time in terms of Math pages:
So, he can read 100 pages of Math in total.
Correct Answer: (b) 100