Question 10 SSC CGL 2018 June 4 Shift 1

Posted by Parveen(Hoven),
Aptitude Trainer and Software Developer

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Question 10 SSC-CGL 2018 June 4
[NOTE: only math questions solved]

Question: If 120 is reduced by x%, the same result will be obtained if 40 is increased by x%. Then x% of 210 will be what percentage less than (x + 20) % of 180?

1. 18
2. 16-2/3
3. 20
4. 33-1/3

Method 1

Basically, we have to first determine $x$.

We can observe by hit and trial that 120 - 1/2 of 120 = 40 + 1/2 of 40. So x = 50.

Next, x% of 210 = 50% of 210 = 105

And, (x + 20)% of 180 = (50 + 20)% of 180 = 126

Hence, answer is $\frac{126 - 105}{126} \times 100$ = 16-2/3% Ans.

Method 2

we can set an equation to determine $ x $

$\displaystyle 120 - \bigg(\frac{x}{100} \times 120\bigg) = 40 + \bigg(\frac{x}{100} \times 40\bigg)$

Solving, we can get $ x = \frac12$ or 50%

Then we can proceed as above to obtain the answer.

Method 3

If you have a deeper understanding of the concept of speed.

This question has an analogy with the classic question of two trains travelling towards each other, and we have to find the time to meet. Why is it so? Take a train travelling at $\displaystyle v$ kph. For each km travelled, it takes $ (\frac{1}{v})^{\text{th}}$ [fraction] of the journey time. A bit of reflection will help you understand the whole idea.

So, our trains are (120 - 40) = 80 km apart, and they are approaching at a relative speed of 120 + 40 = 160 km/h. Time to meet is $ \frac{80 \text{km}}{160 \text{km/h}}$ = 1/2 or 50% of an hour.

Then we can proceed as in method 1 to obtain the answer.