Comprehension
The following table gives the tariff [in paise per kilo-watt-hour (kWh)] levied by the UPSEB in 1994-95, in four sectors and the regions within them. The table also gives the percentage change in the tariff as compared to 1991-92.
| Region 1 | Region 2 | Region 3 | Region 4 | Region 5 | ||||||
|---|---|---|---|---|---|---|---|---|---|---|
| P/kWh | % incr. | P/kWh | % incr. | P/kWh | % incr. | P/kWh | % incr. | P/kWh | % incr. | |
| Sector 1 | 425 | 15 | 472 | 5 | 420 | -4 | 415 | 8 | 440 | 10 |
| Sector 2 | 430 | 12 | 468 | 8 | 448 | 7 | 423 | -3 | 427 | 11 |
| Sector 3 | 428 | 8 | 478 | -4 | 432 | 6 | 441 | 10 | 439 | 8 |
| Sector 4 | 434 | -5 | 470 | 15 | 456 | 10 | 451 | 12 | 446 | -12 |
Question: 1
If the amount of power consumed by the various regions in sector 1 is the same, then as compared to 1991-92 the net tariff in 1994-95 was:
If the amount of power consumed by the various regions in sector 1 is the same, then as compared to 1991-92 the net tariff in 1994-95 was:
Show Hint
When all quantities are equal-weighted, percentage changes should be calculated using the average of absolute values, not the average of percentage changes.
Updated On: Jul 30, 2026
- increased by 6.5%
- decreased by 3.5%
- increased by 10.2%
- decreased by 7.3%
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The Correct Option is A
Solution and Explanation
We are given P/kWh values for 1994-95 and percentage increase compared to 1991-92. To find the net percentage change for the entire sector when all regions consume the same power:
For each region, the 1991-92 tariff = \(\frac{\text{Tariff in 1994-95}}{1 + \frac{% \text{ incr}}{100}}\)
\[ \begin{aligned} R_1 &: \frac{425}{1.15} = 369.565 \ \text{paise}
R_2 &: \frac{472}{1.05} = 449.524 \ \text{paise}
R_3 &: \frac{420}{0.96} = 437.5 \ \text{paise}
R_4 &: \frac{415}{1.08} = 384.259 \ \text{paise}
R_5 &: \frac{440}{1.10} = 400.0 \ \text{paise} \end{aligned} \] Average tariff in 1991-92 = \(\frac{369.565 + 449.524 + 437.5 + 384.259 + 400}{5} = 408.169 \ \text{paise}\)
Average tariff in 1994-95 = \(\frac{425 + 472 + 420 + 415 + 440}{5} = 434.4 \ \text{paise}\)
Percentage change = \(\frac{434.4 - 408.169}{408.169} \times 100 \approx 6.43%\)
Thus, approximately 6.5% increase.
For each region, the 1991-92 tariff = \(\frac{\text{Tariff in 1994-95}}{1 + \frac{% \text{ incr}}{100}}\)
\[ \begin{aligned} R_1 &: \frac{425}{1.15} = 369.565 \ \text{paise}
R_2 &: \frac{472}{1.05} = 449.524 \ \text{paise}
R_3 &: \frac{420}{0.96} = 437.5 \ \text{paise}
R_4 &: \frac{415}{1.08} = 384.259 \ \text{paise}
R_5 &: \frac{440}{1.10} = 400.0 \ \text{paise} \end{aligned} \] Average tariff in 1991-92 = \(\frac{369.565 + 449.524 + 437.5 + 384.259 + 400}{5} = 408.169 \ \text{paise}\)
Average tariff in 1994-95 = \(\frac{425 + 472 + 420 + 415 + 440}{5} = 434.4 \ \text{paise}\)
Percentage change = \(\frac{434.4 - 408.169}{408.169} \times 100 \approx 6.43%\)
Thus, approximately 6.5% increase.
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Question: 2
What was the approximate average tariff in region 3 in 1991-92?
What was the approximate average tariff in region 3 in 1991-92?
Show Hint
To find base-year values from a percentage increase, divide the current value by \(1 + \frac{\text{percentage increase}}{100}\).
Updated On: Jul 2, 2026
- 407
- 420
- 429
- None of these
Show Solution
Verified By Collegedunia
The Correct Option is B
Solution and Explanation
We need to compute the average tariff for region 3 across the four sectors for 1991-92.
Using \(\text{Tariff}_{91-92} = \frac{\text{Tariff}_{94-95}}{1 + \frac{% incr}{100}}\): \[ \begin{aligned} S_1 &: \frac{420}{0.96} = 437.5 \ \text{paise}
S_2 &: \frac{448}{1.07} \approx 418.692 \ \text{paise}
S_3 &: \frac{432}{1.06} \approx 407.547 \ \text{paise}
S_4 &: \frac{456}{1.10} \approx 414.545 \ \text{paise} \end{aligned} \] Average = \(\frac{437.5 + 418.692 + 407.547 + 414.545}{4} \approx 419.571 \ \text{paise}\)
Thus, approximately \(420\ \text{paise}\).
Using \(\text{Tariff}_{91-92} = \frac{\text{Tariff}_{94-95}}{1 + \frac{% incr}{100}}\): \[ \begin{aligned} S_1 &: \frac{420}{0.96} = 437.5 \ \text{paise}
S_2 &: \frac{448}{1.07} \approx 418.692 \ \text{paise}
S_3 &: \frac{432}{1.06} \approx 407.547 \ \text{paise}
S_4 &: \frac{456}{1.10} \approx 414.545 \ \text{paise} \end{aligned} \] Average = \(\frac{437.5 + 418.692 + 407.547 + 414.545}{4} \approx 419.571 \ \text{paise}\)
Thus, approximately \(420\ \text{paise}\).
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