(solved)Question 10 SSC-CGL 2018 June 4 Shift 1 | 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?

Parveen, Feb 07, 2020

Question 10
SSC-CGL 2018 June 4 Shift 1

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?

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

Solution in Short

Basically, we have to first determine $x$.

By shear hit and trial: 120 reduced by 1/2 is 60, whereas 40 increased by 1/2 is also 60. Hence x = 1/2 or 50%.

Next "x% of 210" is 105, and "(x + 20)% of 180" is 126. The percentage difference between 126 and 105 can be found as 16-2/3% answer!

Solution in Detail

we can set an equation to determine $ x $

[1] 120 less x% is: $\displaystyle 120 - \bigg(\frac{x}{100} \times 120\bigg) $

[2] 40 plus x% is: $\displaystyle 40 + \bigg(\frac{x}{100} \times 40\bigg)$

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

Next, $\displaystyle x\% \text{ of } 210$ means $\displaystyle 50\% \times 210$

$\displaystyle \text{. . .}$ which comes to 105.

And, $\displaystyle (x + 20)\% \text{ of } 180 = 126$

Finally: "What % is 105 less than 126"?

We have $\displaystyle \frac{126 - 105}{105} \times 100 = 16 \frac23\%$

Solution by Alternate Method

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.