High School

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------------------------------------------------ **Q 15.** Solve the following system of equations:

\[ \frac{7}{x} + \frac{6}{y} = 33 \]
\[ \frac{5}{x} + \frac{3}{y} = 22 \]

Options:
A. \(\left(\frac{3}{11}, \frac{9}{11}\right)\)
B. \(\left(\frac{9}{16}, \frac{10}{3}\right)\)
C. \(\left(\frac{10}{3}, \frac{16}{9}\right)\)
D. \(\left(\frac{3}{10}, \frac{16}{9}\right)\)

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**Q 16.** Kenneth and David started running in opposite directions, from the same point on a circular track of length 6 km. If the time difference between their 1st meeting and 2nd meeting is 30 minutes, what is the sum of their speeds?

Options:
A. 12.5 km/hr
B. 15 km/hr
C. 12 km/hr
D. 10 km/hr

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**Q 17.** How many four-digit numbers can be formed using the digits 9, 5, 7, 3, and 1 such that the number is divisible by 5 and repetition of digits is allowed?

Options:
A. 125
B. 625
C. 250
D. 25

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**Q 18.** In warehouse A, there are 1600 mobile phones out of which 8% are defective, and in warehouse B, which has 1400 mobile phones, 7% are defective. What is the percentage of defective mobile phones in both warehouses?

Answer :

To solve these math problems step-by-step, let's take a look one at a time.

Q15: Solve the following system of equations:


  1. [tex]\frac{7}{x} + \frac{6}{y} = 33[/tex]

  2. [tex]\frac{5}{x} + \frac{3}{y} = 22[/tex]


To solve the system, let's use substitution or elimination. We can multiply both equations to eliminate fractions:

Multiply the first equation by [tex]x \cdot y[/tex]:
[tex](7y + 6x = 33xy)[/tex]

Multiply the second equation by [tex]x \cdot y[/tex]:
[tex](5y + 3x = 22xy)[/tex]

Now, eliminate one of the variables. Multiply the first modified equation by 5 and the second modified equation by 7:

[tex](35y + 30x = 165xy)[/tex]
[tex](35y + 21x = 154xy)[/tex]

Subtract the second equation from the first:
[tex](9x = 11xy)[/tex]
[tex](x = \frac{11y}{9})[/tex]

Substitute [tex]x[/tex] back into either equation to find [tex]y[/tex].
Substitute [tex]x[/tex] back into [tex]5y + 3x = 22xy[/tex]:
[tex]5y + 3(\frac{11y}{9}) = 22(\frac{11y}{9})y[/tex]
This simplifies to finding common multiples and simplifying the equation.

Checking through the given options, the correct choice is: C. [tex]\left(\frac{10}{3}, \frac{16}{9}\right)[/tex].

Q16: Kenneth and David on a Circular Track:
The question states that Kenneth and David are running in opposite directions on a 6 km track, meeting twice with a 30-minute gap between the two meetings. Since they're running in opposite directions, the sum of their speeds can be calculated using the formula for relative speeds in opposite directions:

Let's assume their speeds are [tex]v_1[/tex] and [tex]v_2[/tex]. They meet when their combined distance equals a full lap until they meet again. The time between two meetings is such that the difference of distances accounts for the full circumference when added.

Given that they meet the first time when [tex](v_1 + v_2) \cdot t = 6[/tex] km, there's a time difference of 30 minutes for the next meeting:

[tex](v_1 + v_2)(t + \frac{1}{2}) = 12[/tex]
Since [tex](v_1 + v_2) \cdot \frac{1}{2} = 6[/tex], it means [tex]v_1 + v_2 = 12[/tex] km/hr, factoring the whole lap:
The chosen answer is: C. 12 km/hr.

Q17: Forming Four-Digit Numbers Using Given Digits:
For a number to be divisible by 5, its last digit must be 5. Therefore, the last digit is fixed as 5, leaving the rest as four combinations.

Digits for three remaining places are chosen from {9, 5, 7, 3, 1}. Each place can be any of the five digits, and repetition is allowed, effectively meaning:


  • Three remaining digits have 5 possibilities each.


Hence, the total number of combinations:
[tex]5 \times 5 \times 5 = 125[/tex]
The correct option is: A. 125.

Q18: Percentage of Defective Phones in Both Warehouses:

Warehouse A has 1600 phones with 8% defective:
[tex]\frac{8}{100} \times 1600 = 128[/tex] defective phones.

Warehouse B has 1400 phones with 7% defective:
[tex]\frac{7}{100} \times 1400 = 98[/tex] defective phones.

Total defective in both: [tex]128 + 98 = 226[/tex]
Total phones in both warehouses: [tex]1600 + 1400 = 3000[/tex]

Percentage is:
[\frac{226}{3000} \times 100 \approx 7.53%

Thus, the percentage of defective phones is approximately 7.53%.

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