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24 Aug 2024, 14:54
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C
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In a toy factory, there are 3 different types of machines: P, Q and R that operate at their individual constant rates to paint identical toys. One machine of type P and one machine. of type Q operating together paint 1,000 toys in 2 hours. However, one machine of type P and one machine of type R operating together take double the time taken by one machine of type P and one machine of type Q operating together to paint 1,000 toys. If one machine of type Q and two machines of type R operating together paint 1,000 toys in 3 hours, then how much time will one machine of type P, one machine of type Q and one machine of type R take operating together to paint 1,000 toys?
(A) 19/36
(B) 17/19
(C) 36/19
(D) 2
(E) 36/17
(A) 19/36
(B) 17/19
(C) 36/19
(D) 2
(E) 36/17
24 Aug 2024, 20:04
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\(P+Q=\frac{1}{2}\)
\(P+R=\frac{1}{4}\)
\(Q+2R=\frac{1}{3}\)
\(Q=\frac{1}{3}-2R\), substituting this in first equation will give, \(P+\frac{1}{3}-2R=\frac{1}{2}\)
\(P-2R=\frac{1}{6}\), from this we can get P & R, and in turn Q
\(P=\frac{8}{36}\), \(Q=\frac{10}{36}\), \(R=\frac{1}{36}\)
\(P+Q+R=\frac{8}{36}+\frac{10}{36}+\frac{1}{36}=\frac{19}{36}\)
Time taken is \(\frac{36}{19}\)
Answer is C
\(P+R=\frac{1}{4}\)
\(Q+2R=\frac{1}{3}\)
\(Q=\frac{1}{3}-2R\), substituting this in first equation will give, \(P+\frac{1}{3}-2R=\frac{1}{2}\)
\(P-2R=\frac{1}{6}\), from this we can get P & R, and in turn Q
\(P=\frac{8}{36}\), \(Q=\frac{10}{36}\), \(R=\frac{1}{36}\)
\(P+Q+R=\frac{8}{36}+\frac{10}{36}+\frac{1}{36}=\frac{19}{36}\)
Time taken is \(\frac{36}{19}\)
Answer is C
27 Aug 2024, 12:16
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To solve the problem, we need to determine the rates of machines P, Q, and R based on the information provided.
1. Machine P and Q together:
- They paint 1,000 toys in 2 hours.
- Their combined rate is:
$$\text{Rate}_{P+Q} = \frac{1000 \text{ toys}}{2 \text{ hours}} = 500 \text{ toys/hour}$$
2. Machine P and R together:
- They take double the time of P and Q to paint 1,000 toys.
- Since P and Q take 2 hours, P and R take:
$$2 \times 2 = 4 \text{ hours}$$
- Their combined rate is:
$$\text{Rate}_{P+R} = \frac{1000 \text{ toys}}{4 \text{ hours}} = 250 \text{ toys/hour}$$
3. Machine Q and two machines of R together:
- They paint 1,000 toys in 3 hours.
- Their combined rate is:
$$\text{Rate}_{Q+2R} = \frac{1000 \text{ toys}}{3 \text{ hours}} = \frac{1000}{3} \text{ toys/hour}$$
Now, we can set up equations based on the rates:
Let:
- Rate of machine P = $ p $ toys/hour
- Rate of machine Q = $ q $ toys/hour
- Rate of machine R = $ r $ toys/hour
From the first equation:
$$p + q = 500 \quad \text{(1)}$$
From the second equation:
$$p + r = 250 \quad \text{(2)}$$
From the third equation:
$$q + 2r = \frac{1000}{3} \quad \text{(3)}$$
Now, we can solve these equations step by step.
From equation (1):
$$q = 500 - p \quad \text{(4)}$$
Substituting (4) into (3):
$$(500 - p) + 2r = \frac{1000}{3}$$
$$500 - p + 2r = \frac{1000}{3}$$
To eliminate the fraction, multiply the entire equation by 3:
$$1500 - 3p + 6r = 1000$$
Rearranging gives:
$$6r = 3p - 500$$
$$r = \frac{3p - 500}{6} \quad \text{(5)}$$
Now substitute (5) into (2):
$$p + \frac{3p - 500}{6} = 250$$
Multiply through by 6 to eliminate the fraction:
$$6p + 3p - 500 = 1500$$
$$9p - 500 = 1500$$
$$9p = 2000$$
$$p = \frac{2000}{9} \text{ toys/hour}$$
Now substitute $ p $ back into (4) to find $ q $:
$$q = 500 - \frac{2000}{9}$$
Convert 500 to a fraction:
$$q = \frac{4500}{9} - \frac{2000}{9} = \frac{2500}{9} \text{ toys/hour}$$
Substituting $ p $ into (5) to find $ r $:
$$r = \frac{3\left(\frac{2000}{9}\right) - 500}{6}$$
$$= \frac{\frac{6000}{9} - 500}{6}$$
Convert 500 to a fraction:
$$= \frac{\frac{6000}{9} - \frac{4500}{9}}{6} = \frac{\frac{1500}{9}}{6} = \frac{1500}{54} = \frac{250}{9} \text{ toys/hour}$$
Now, we can find the combined rate of P, Q, and R:
$$\text{Rate}_{P+Q+R} = p + q + r$$
$$= \frac{2000}{9} + \frac{2500}{9} + \frac{250}{9} = \frac{2000 + 2500 + 250}{9} = \frac{4750}{9} \text{ toys/hour}$$
Finally, to find the time taken to paint 1,000 toys:
$$\text{Time} = \frac{1000 \text{ toys}}{\text{Rate}_{P+Q+R}} = \frac{1000}{\frac{4750}{9}} = 1000 \times \frac{9}{4750} = \frac{9000}{4750} = \frac{1800}{950} = \frac{360}{190} = \frac{72}{38} = \frac{36}{19} \text{ hours}$$
Thus, one machine of type P, one machine of type Q, and one machine of type R together will take $$\frac{36}{19}$$ hours to paint 1,000 toys.
1. Machine P and Q together:
- They paint 1,000 toys in 2 hours.
- Their combined rate is:
$$\text{Rate}_{P+Q} = \frac{1000 \text{ toys}}{2 \text{ hours}} = 500 \text{ toys/hour}$$
2. Machine P and R together:
- They take double the time of P and Q to paint 1,000 toys.
- Since P and Q take 2 hours, P and R take:
$$2 \times 2 = 4 \text{ hours}$$
- Their combined rate is:
$$\text{Rate}_{P+R} = \frac{1000 \text{ toys}}{4 \text{ hours}} = 250 \text{ toys/hour}$$
3. Machine Q and two machines of R together:
- They paint 1,000 toys in 3 hours.
- Their combined rate is:
$$\text{Rate}_{Q+2R} = \frac{1000 \text{ toys}}{3 \text{ hours}} = \frac{1000}{3} \text{ toys/hour}$$
Now, we can set up equations based on the rates:
Let:
- Rate of machine P = $ p $ toys/hour
- Rate of machine Q = $ q $ toys/hour
- Rate of machine R = $ r $ toys/hour
From the first equation:
$$p + q = 500 \quad \text{(1)}$$
From the second equation:
$$p + r = 250 \quad \text{(2)}$$
From the third equation:
$$q + 2r = \frac{1000}{3} \quad \text{(3)}$$
Now, we can solve these equations step by step.
From equation (1):
$$q = 500 - p \quad \text{(4)}$$
Substituting (4) into (3):
$$(500 - p) + 2r = \frac{1000}{3}$$
$$500 - p + 2r = \frac{1000}{3}$$
To eliminate the fraction, multiply the entire equation by 3:
$$1500 - 3p + 6r = 1000$$
Rearranging gives:
$$6r = 3p - 500$$
$$r = \frac{3p - 500}{6} \quad \text{(5)}$$
Now substitute (5) into (2):
$$p + \frac{3p - 500}{6} = 250$$
Multiply through by 6 to eliminate the fraction:
$$6p + 3p - 500 = 1500$$
$$9p - 500 = 1500$$
$$9p = 2000$$
$$p = \frac{2000}{9} \text{ toys/hour}$$
Now substitute $ p $ back into (4) to find $ q $:
$$q = 500 - \frac{2000}{9}$$
Convert 500 to a fraction:
$$q = \frac{4500}{9} - \frac{2000}{9} = \frac{2500}{9} \text{ toys/hour}$$
Substituting $ p $ into (5) to find $ r $:
$$r = \frac{3\left(\frac{2000}{9}\right) - 500}{6}$$
$$= \frac{\frac{6000}{9} - 500}{6}$$
Convert 500 to a fraction:
$$= \frac{\frac{6000}{9} - \frac{4500}{9}}{6} = \frac{\frac{1500}{9}}{6} = \frac{1500}{54} = \frac{250}{9} \text{ toys/hour}$$
Now, we can find the combined rate of P, Q, and R:
$$\text{Rate}_{P+Q+R} = p + q + r$$
$$= \frac{2000}{9} + \frac{2500}{9} + \frac{250}{9} = \frac{2000 + 2500 + 250}{9} = \frac{4750}{9} \text{ toys/hour}$$
Finally, to find the time taken to paint 1,000 toys:
$$\text{Time} = \frac{1000 \text{ toys}}{\text{Rate}_{P+Q+R}} = \frac{1000}{\frac{4750}{9}} = 1000 \times \frac{9}{4750} = \frac{9000}{4750} = \frac{1800}{950} = \frac{360}{190} = \frac{72}{38} = \frac{36}{19} \text{ hours}$$
Thus, one machine of type P, one machine of type Q, and one machine of type R together will take $$\frac{36}{19}$$ hours to paint 1,000 toys.
