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17 Jun 2021, 04:15
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A
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
65%
(hard)
Question Stats:
63% (02:53) correct
37%
(03:03)
wrong
based on 120
sessions
History
Date
Time
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Not Attempted Yet
Mukesh, Manish, Lalu and Jaggi bought a MOKIA mobile for £60. Mukesh paid one–half of the sum of the amounts paid by the other persons. Manish paid one–third of the sum of the amounts paid by the other persons. Lalu paid one–fourth of the sum of the amounts paid by the other persons. How much did Jaggi have to pay?
(A) £ 13
(B) £ 15
(C) £ 17
(D) £ 23
(E) None of these
(A) £ 13
(B) £ 15
(C) £ 17
(D) £ 23
(E) None of these
17 Jun 2021, 09:07
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Bunuel
Solution :
We are given,
- • Mukesh, Manish, Lalu and Jaggi bought a MOKIA mobile.
• Total amount paid for the mobile = £60
When Mukesh paid one–half of the sum of the amounts paid by the other persons
Let the sum of amounts paid by others = \(x\)
Then, amount paid by Mukesh = \(\frac{1}{2} * x\)
We know,
The total amount paid by all = 60
Mukesh + others = 60
⟹ \(\frac{1}{2} * x\) + \(x\) = 60
⟹ \(\frac{x}{2}\) + \(x\) = 60
⟹ \(\frac{3x}{2}\) = 60
⟹ x = \(\frac{60 * 2}{3}\)
⟹ x = 40
Therefore, amount paid by Mukesh = \(\frac{1}{2} * 40\) = £20
When Manish paid one–third of the sum of the amounts paid by the other persons
Let the sum of amounts paid by others = \(y\)
Then, amount paid by Manish = \(\frac{1}{3} * y\)
We know,
The total amount paid by all = 60
Manish + others = 60
⟹ \(\frac{1}{3} * y\) + \(y\) = 60
⟹ \(\frac{y}{3}\) + \(y\) = 60
⟹ \(\frac{4y}{3}\) = 60
⟹ y = \(\frac{60 * 3}{4}\)
⟹ y = 45
Therefore, amount paid by Manish = \(\frac{1}{3} * 45\) = £15
When Lalu paid one–fourth of the sum of the amounts paid by the other persons
Let the sum of amounts paid by others = \(z\)
Then, amount paid by Lalu = \(\frac{1}{4} * z\)
We know,
The total amount paid by all = 60
Lalu + others = 60
⟹ \(\frac{1}{4} * z\) + \(z\) = 60
⟹ \(\frac{z}{4}\) + \(z\) = 60
⟹ \(\frac{5z}{4}\) = 60
⟹ z = \(\frac{60 * 4}{5}\)
⟹ z = 48
Therefore, amount paid by Manish = \(\frac{1}{4} * 48\) = £12
Thus, the amount paid by Jaggi = 60 - 20 - 15 - 12 = 13
Hence, the correct answer is Option A.
18 Jun 2021, 00:11
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Mukesh(Mu), Manish(Ma), Lalu(L), and Jaggi(J)
=> Mu + Ma + L + J = £60
Mukesh paid one–half of the sum of the amounts paid by the other persons: \(Mu =\frac{ Ma + L + J }{ 2}\)
=> 2(Mu) = Ma + L+ J
=> Mu + 2(Mu) = £60
=> 3Mu = £60
=> Mu = £20
=> Ma + L + J = £60 - £20 = £40 or L + J = £40 - Ma
Manish paid one–third of the sum of the amounts paid by the other persons: 3(Ma) = 20 + L + J
=> 3Ma = £20 + £40 - Ma
=> 4Ma = £60
=> Ma = £15
=> L + J = £60 - £20 - £15 = £25
Lalu paid one–fourth of the sum of the amounts paid by the other persons: 4(L) = Mu + Ma + J
=> 4L - J = 35
Adding both equations, we get 5L = 60 or L = 12
Therefore, J = 25 - 12 = 13
Answer A
=> Mu + Ma + L + J = £60
Mukesh paid one–half of the sum of the amounts paid by the other persons: \(Mu =\frac{ Ma + L + J }{ 2}\)
=> 2(Mu) = Ma + L+ J
=> Mu + 2(Mu) = £60
=> 3Mu = £60
=> Mu = £20
=> Ma + L + J = £60 - £20 = £40 or L + J = £40 - Ma
Manish paid one–third of the sum of the amounts paid by the other persons: 3(Ma) = 20 + L + J
=> 3Ma = £20 + £40 - Ma
=> 4Ma = £60
=> Ma = £15
=> L + J = £60 - £20 - £15 = £25
Lalu paid one–fourth of the sum of the amounts paid by the other persons: 4(L) = Mu + Ma + J
=> 4L - J = 35
Adding both equations, we get 5L = 60 or L = 12
Therefore, J = 25 - 12 = 13
Answer A
