Quantitative Aptitude Test - 3

The first pipe and third pipe fill it together in \(20\) minutes.

The first pipe fills it in \(\frac{1}{\frac{1}{20}-\frac{1}{60}}=30\) minutes

The second pipe and third pipe fill it in \(15\) minutes.

The second pipe fills it in \(\frac{1}{\frac{1}{15}-\frac{1}{60}}=20\) minutes

So, the time required for the operation \(=\frac{1}{\frac{1}{30}+\frac{1}{20}-\frac{1}{60}}=15\) minutes

As log₆ (x² + 22x + 153) = 2,
--> x² + 22x + 153 = 6²
--> x² + 22x + 117 = 0
--> (x + 13) (x + 9) = 0
x = -13 or -9.

If b = 1: 10 = 2 × 5, possible values of 'a' are from 9 to 18.
If b = 2: 10² = 2² × 5² .
So, the value of 'a' should be such that a! has at least two multiples of 5.
So, the value of 'a' should be greater than 9.
Possible values of 'a' are from 10 to 17.
If b = 3: 10³ = 2³ × 5³ .
So, the value of 'a' should be such that a! has at least three multiples of 5.
So, the value of 'a' should be greater than 14.
Possible values of 'a' are 15 and 16.
So, total number of pairs, (a, b) = 10 + 8 + 2 = 20.

Since, one angle is 65°, one of the other angles has to be 115o, since, a pair of opposite sides are parallel. Let the other two angles be x° and (180 – x)°.
If x < 65, then (180 – x)° > 115º and if x > 65, then (180 – x)° < 115º.
Since, we are looking for the largest possible angle of the trapezium, xº, 65º, 115º, (180 – xº) are in A.P.
⇒ 65° – x = 115 – 65º, ⇒ x = 15º
⇒ Largest possible angle = 180º – 15º = 165º.

Cost of a tubelight = Rs. t and cost of a bulb = Rs. b
t + b = 52 ...(i)
Cost of bulb drops by 20% and cost of tubelight increases by 50%
⇒ 1.5t + 0.8b = 50 ...(ii)
Solving equations in (i) and (ii) t = Rs.12, b = Rs.40