CAT 2017 DILR Questions — Slot 2

Set up the totals first. There are 800 pizzas in all. Party 3 gets 70%, i.e. $0.70 \times 800 = 560$. The remaining $800 - 560 = 240$ are split equally, so Party 1 and Party 2 get 120 each.

Fill the Thin Crust (T) column. Overall Thin Crust $= 0.375 \times 800 = 300$.

• Party 1: $0.6 \times 120 = 72$
• Party 2: $0.55 \times 120 = 66$
• Party 3 (by subtraction): $300 - 72 - 66 = 162$

Fill the Normal Cheese (NC) column. Overall Normal Cheese $= 0.52 \times 800 = 416$.

• Party 2: $0.3 \times 120 = 36$
• Party 3: $0.65 \times 560 = 364$
• Party 1 (by subtraction): $416 - 36 - 364 = 16$

Since Deep Dish $=$ Total $-$ Thin, and Extra Cheese $=$ Total $-$ Normal Cheese, the whole grid is now fixed:

The question asks for Thin Crust pizzas going to Party 3. Reading straight off the Thin column, Party 3's Thin Crust count is 162.

Answer: 162 (Choice B)

Set up the totals first. There are 800 pizzas in all. Party 3 gets 70%, i.e. $0.70 \times 800 = 560$. The remaining $800 - 560 = 240$ are split equally, so Party 1 and Party 2 get 120 each.

Fill the Thin Crust (T) column. Overall Thin Crust $= 0.375 \times 800 = 300$.

• Party 1: $0.6 \times 120 = 72$
• Party 2: $0.55 \times 120 = 66$
• Party 3 (by subtraction): $300 - 72 - 66 = 162$

Fill the Normal Cheese (NC) column. Overall Normal Cheese $= 0.52 \times 800 = 416$.

• Party 2: $0.3 \times 120 = 36$
• Party 3: $0.65 \times 560 = 364$
• Party 1 (by subtraction): $416 - 36 - 364 = 16$

Since Deep Dish $=$ Total $-$ Thin, and Extra Cheese $=$ Total $-$ Normal Cheese, the whole grid is now fixed:

The question asks for Normal Cheese pizzas going to Party 1. This was the residual we found: $416 - 36 - 364 = 16$.

Answer: 16 (Choice C)

Set up the totals first. There are 800 pizzas in all. Party 3 gets 70%, i.e. $0.70 \times 800 = 560$. The remaining $800 - 560 = 240$ are split equally, so Party 1 and Party 2 get 120 each.

Fill the Thin Crust (T) column. Overall Thin Crust $= 0.375 \times 800 = 300$.

• Party 1: $0.6 \times 120 = 72$
• Party 2: $0.55 \times 120 = 66$
• Party 3 (by subtraction): $300 - 72 - 66 = 162$

Fill the Normal Cheese (NC) column. Overall Normal Cheese $= 0.52 \times 800 = 416$.

• Party 2: $0.3 \times 120 = 36$
• Party 3: $0.65 \times 560 = 364$
• Party 1 (by subtraction): $416 - 36 - 364 = 16$

Since Deep Dish $=$ Total $-$ Thin, and Extra Cheese $=$ Total $-$ Normal Cheese, the whole grid is now fixed:

Now add the extra condition for Party 2: half of its Normal Cheese pizzas are Thin Crust.

Party 2 has $36$ Normal Cheese pizzas, so $\text{T-NC} = \text{D-NC} = \tfrac{1}{2}\times 36 = 18.$

Party 2's totals are Thin $=66$ and Deep Dish $=54$. The Extra Cheese counts are what's left after removing the Normal Cheese ones:

• T-EC $= 66 - 18 = 48$
• D-EC $= 54 - 18 = 36$

Difference $= 48 - 36 = \mathbf{12}$.

Answer: 12 (Choice B)

Set up the totals first. There are 800 pizzas in all. Party 3 gets 70%, i.e. $0.70 \times 800 = 560$. The remaining $800 - 560 = 240$ are split equally, so Party 1 and Party 2 get 120 each.

Fill the Thin Crust (T) column. Overall Thin Crust $= 0.375 \times 800 = 300$.

• Party 1: $0.6 \times 120 = 72$
• Party 2: $0.55 \times 120 = 66$
• Party 3 (by subtraction): $300 - 72 - 66 = 162$

Fill the Normal Cheese (NC) column. Overall Normal Cheese $= 0.52 \times 800 = 416$.

• Party 2: $0.3 \times 120 = 36$
• Party 3: $0.65 \times 560 = 364$
• Party 1 (by subtraction): $416 - 36 - 364 = 16$

Since Deep Dish $=$ Total $-$ Thin, and Extra Cheese $=$ Total $-$ Normal Cheese, the whole grid is now fixed:

First work out the four prices.

• T-EC $= 500$
• D-EC $= 500 + 50 = 550$
• T-NC and D-NC each cost $\tfrac{3}{5}$ of a D-EC pizza $= \tfrac{3}{5}\times 550 = 330$

Now split Party 1's pizzas under the given condition. Party 1 has $16$ Normal Cheese pizzas, of which $25\%$ are Deep Dish:

• D-NC $= 0.25 \times 16 = 4$, so T-NC $= 12$

Party 1's totals are Thin $=72$ and Deep Dish $=48$, so the Extra Cheese counts are:

• T-EC $= 72 - 12 = 60$
• D-EC $= 48 - 4 = 44$

Bill for Party 1:

Answer: Rs. 59,480 (Choice A)

Let $r_k$ be the rating of Cup $k$. All six ratings are different whole numbers from 1 to 10.

Step 1 — Which cups are even. Clue 3 says $r_3 = 2\,r_5$, so Cup 3 is even. Clue 5 says Cup 2 is even. Clue 4 says exactly two cups are even — so those two are Cup 2 and Cup 3, and Cups 1, 4, 5, 6 are all odd.

Step 2 — Fix Cups 2, 5, 3, 1. Cup 2 is the minimum and is even, so the smallest even value, $r_2 = 2$. Then $r_5$ is odd and $r_3 = 2\,r_5$:

• If $r_5 = 3$, then $r_3 = 6$. Cup 1 is odd, below 6 (clue 6: $r_3 > r_1$) and not 3, so $r_1 = 5$. ✓
• If $r_5 = 5$, then $r_3 = 10$ — the maximum. But the maximum belongs to Ooty (clue 2), and Ooty is not in Cup 3. ✗

So $r_1 = 5,\ r_2 = 2,\ r_3 = 6,\ r_5 = 3$.

Step 3 — Cups 4 and 6. The remaining odd values are 7 and 9. Ooty has the highest rating and Cup 6 is Himachal (clue 1), so Ooty must be Cup 4: $r_4 = 9$ (Ooty) and $r_6 = 7$ (Himachal).

Solved ratings:

The four teas Munnar, Wayanad, Assam, Darjeeling fill Cups 1, 2, 3, 5, and by clue 7 must satisfy Assam > Wayanad > Munnar.

Second highest rating. The six ratings are 9, 7, 6, 5, 3, 2. In decreasing order, the highest is 9 and the second highest is 7.

Answer: 7

Let $r_k$ be the rating of Cup $k$. All six ratings are different whole numbers from 1 to 10.

Step 1 — Which cups are even. Clue 3 says $r_3 = 2\,r_5$, so Cup 3 is even. Clue 5 says Cup 2 is even. Clue 4 says exactly two cups are even — so those two are Cup 2 and Cup 3, and Cups 1, 4, 5, 6 are all odd.

Step 2 — Fix Cups 2, 5, 3, 1. Cup 2 is the minimum and is even, so the smallest even value, $r_2 = 2$. Then $r_5$ is odd and $r_3 = 2\,r_5$:

• If $r_5 = 3$, then $r_3 = 6$. Cup 1 is odd, below 6 (clue 6: $r_3 > r_1$) and not 3, so $r_1 = 5$. ✓
• If $r_5 = 5$, then $r_3 = 10$ — the maximum. But the maximum belongs to Ooty (clue 2), and Ooty is not in Cup 3. ✗

So $r_1 = 5,\ r_2 = 2,\ r_3 = 6,\ r_5 = 3$.

Step 3 — Cups 4 and 6. The remaining odd values are 7 and 9. Ooty has the highest rating and Cup 6 is Himachal (clue 1), so Ooty must be Cup 4: $r_4 = 9$ (Ooty) and $r_6 = 7$ (Himachal).

Solved ratings:

The four teas Munnar, Wayanad, Assam, Darjeeling fill Cups 1, 2, 3, 5, and by clue 7 must satisfy Assam > Wayanad > Munnar.

Which cup is Ooty? Ooty has the highest rating, 9, and from the solved table that is Cup 4.

Answer: 4

Let $r_k$ be the rating of Cup $k$. All six ratings are different whole numbers from 1 to 10.

Step 1 — Which cups are even. Clue 3 says $r_3 = 2\,r_5$, so Cup 3 is even. Clue 5 says Cup 2 is even. Clue 4 says exactly two cups are even — so those two are Cup 2 and Cup 3, and Cups 1, 4, 5, 6 are all odd.

Step 2 — Fix Cups 2, 5, 3, 1. Cup 2 is the minimum and is even, so the smallest even value, $r_2 = 2$. Then $r_5$ is odd and $r_3 = 2\,r_5$:

• If $r_5 = 3$, then $r_3 = 6$. Cup 1 is odd, below 6 (clue 6: $r_3 > r_1$) and not 3, so $r_1 = 5$. ✓
• If $r_5 = 5$, then $r_3 = 10$ — the maximum. But the maximum belongs to Ooty (clue 2), and Ooty is not in Cup 3. ✗

So $r_1 = 5,\ r_2 = 2,\ r_3 = 6,\ r_5 = 3$.

Step 3 — Cups 4 and 6. The remaining odd values are 7 and 9. Ooty has the highest rating and Cup 6 is Himachal (clue 1), so Ooty must be Cup 4: $r_4 = 9$ (Ooty) and $r_6 = 7$ (Himachal).

Solved ratings:

The four teas Munnar, Wayanad, Assam, Darjeeling fill Cups 1, 2, 3, 5, and by clue 7 must satisfy Assam > Wayanad > Munnar.

Extra condition: Munnar did not get the minimum rating. The minimum rating is 2 (Cup 2), so Munnar is not in Cup 2. Munnar must then sit in one of Cups 1, 3, 5 (ratings 5, 6, 3), and by clue 7 it is the smallest of Munnar, Wayanad, Assam.

• If Munnar = Cup 3 (rating 6): Wayanad and Assam would need ratings above 6, but the cups left (1, 2, 5) are 5, 2, 3 — none above 6. ✗
• If Munnar = Cup 1 (rating 5): two ratings above 5 are needed, but only 6 (among 2, 6, 3) is above 5. ✗
• If Munnar = Cup 5 (rating 3): ✓ Then Wayanad and Assam take Cups 1 and 3 (ratings 5 and 6). Since Assam > Wayanad, Assam = Cup 3 (6) and Wayanad = Cup 1 (5); Darjeeling takes Cup 2.

So Wayanad's rating is 5.

Answer: 5 (Choice B)

Let $r_k$ be the rating of Cup $k$. All six ratings are different whole numbers from 1 to 10.

Step 1 — Which cups are even. Clue 3 says $r_3 = 2\,r_5$, so Cup 3 is even. Clue 5 says Cup 2 is even. Clue 4 says exactly two cups are even — so those two are Cup 2 and Cup 3, and Cups 1, 4, 5, 6 are all odd.

Step 2 — Fix Cups 2, 5, 3, 1. Cup 2 is the minimum and is even, so the smallest even value, $r_2 = 2$. Then $r_5$ is odd and $r_3 = 2\,r_5$:

• If $r_5 = 3$, then $r_3 = 6$. Cup 1 is odd, below 6 (clue 6: $r_3 > r_1$) and not 3, so $r_1 = 5$. ✓
• If $r_5 = 5$, then $r_3 = 10$ — the maximum. But the maximum belongs to Ooty (clue 2), and Ooty is not in Cup 3. ✗

So $r_1 = 5,\ r_2 = 2,\ r_3 = 6,\ r_5 = 3$.

Step 3 — Cups 4 and 6. The remaining odd values are 7 and 9. Ooty has the highest rating and Cup 6 is Himachal (clue 1), so Ooty must be Cup 4: $r_4 = 9$ (Ooty) and $r_6 = 7$ (Himachal).

Solved ratings:

The four teas Munnar, Wayanad, Assam, Darjeeling fill Cups 1, 2, 3, 5, and by clue 7 must satisfy Assam > Wayanad > Munnar.

Extra condition: Wayanad's cup and Ooty's cup have consecutive numbers. Ooty is Cup 4, so Wayanad is in Cup 3 or Cup 5.

• If Wayanad = Cup 3 (rating 6): Assam must be above 6, but the remaining cups (1, 2, 5) are 5, 2, 3. ✗
• So Wayanad = Cup 5 (rating 3). Then Munnar must be below 3 → Munnar = Cup 2 (rating 2). Assam takes Cup 1 (5) or Cup 3 (6), and Darjeeling takes the other.

Now test the options under this arrangement:

• Cup 5 contains Assam — Cup 5 is Wayanad. False.
• Cup 1 contains Darjeeling — happens when Assam = Cup 3. This can be true. ✓
• Wayanad has a rating of 6 — Wayanad's rating is 3. False.
• Darjeeling got the minimum rating — Darjeeling is in Cup 1 or 3 (rating 5 or 6); the minimum, 2, is Cup 2. False.

Answer: Cup 1 contains tea from Darjeeling (Choice B)

Compare a candidate sequence with the original position-by-position. A position is out of place if the finger there differs from the original finger in that position. At most two positions may be out of place.

• 0 out of place: the original sequence itself — 1 way.
• 1 out of place: impossible. If only one position had a different finger, that finger would have to be a duplicate, but all five fingers are distinct — you can never change exactly one position of an arrangement. — 0 ways.
• 2 out of place: pick the two positions that differ and swap their fingers. The number of ways to choose 2 positions out of 5 is $\binom{5}{2} = 10.$

Total allowed = 1 + 0 + 10 = 11.

💡 Teacher tip: In any rearrangement, you can never have exactly one thing out of place — the misfit always needs a partner to swap with.

Answer: 11

Now the rule is different: each finger may end up at most one position away from its original spot. If a finger moves one step, the finger it pushes into must move the opposite step. So every allowed sequence is built from non-overlapping swaps of neighbours — each finger either stays put or trades places with an immediate neighbour.

Let $a_n$ be the count for $n$ fingers. Look at the last finger: either it stays (leaving $a_{n-1}$ ways for the first $n-1$), or it swaps with its neighbour (leaving $a_{n-2}$ ways for the rest). So $a_n = a_{n-1} + a_{n-2}, \quad a_1 = 1,\ a_2 = 2.$

This gives $a_3 = 3,\ a_4 = 5,\ a_5 = 8$.

Listing them for five fingers confirms 8: the original; the four single neighbour-swaps (positions 1-2, 2-3, 3-4, 4-5); and the three double swaps that don't overlap (1-2 & 3-4, 1-2 & 4-5, 2-3 & 4-5).

Answer: 8 (Choice C)

Now there are six scans: one finger appears twice and the other four once. As before, compare position-by-position with the original and allow at most two positions out of place.

• 0 out of place: the original — 1 way.
• 1 out of place: impossible, same reason as before — 0 ways.
• 2 out of place: choose two positions and swap their contents. There are $\binom{6}{2} = 15$ pairs of positions. But one of those pairs holds the two identical copies of the doubled finger — swapping them changes nothing, so it isn't a new sequence. Remove it: $15 - 1 = 14$ genuine swaps.

Total allowed = 1 + 0 + 14 = 15.

Answer: 15

Original sequence: L R L T I M — the finger L appears twice (positions 1 and 3). The rule is the "each item moves at most one position" rule from before, now on six slots, and the two L's count as the same finger.

As established, sequences where every item stays within one place of its origin are exactly the ones built from non-overlapping neighbour swaps. For six slots that count is the next Fibonacci number: $a_6 = a_5 + a_4 = 8 + 5 = 13.$

Breaking the 13 down: the original (1), five single neighbour-swaps, six pairs of disjoint swaps, and one triple of disjoint swaps.

Do the two L's cause any duplicates? The two L's start at positions 1 and 3. Because a neighbour swap only moves an item to an adjacent slot, an L from position 1 can reach slots 1–2, while an L from position 3 can reach slots 2–4 — they can never both land on the same target in two different ways. Writing the 13 sequences out confirms they are all distinct:

LRLTIM, RLLTIM, LLRTIM, LRTLIM, LRLITM, LRLTMI, RLTLIM, RLLITM, RLLTMI, LLRITM, LLRTMI, LRTLMI, RLTLMI.

So the number of allowed sequences is 13.

Answer: 13 (Choice C)

Value up the estate. Bank deposits $=70$, house $=50$, three flats $=3\times 30 = 90$, and $n$ gold coins $=n$ (each worth 1 lakh). Total value $= 70 + 50 + 90 + n = (210 + n)$ lakh.

This scenario: the total value is shared equally, and the gold coins are shared equally.

Each child then gets value $\dfrac{210+n}{3}$ and coins $\dfrac{n}{3}$. So each child's non-coin value (house + flats + bank) is $\frac{210+n}{3} - \frac{n}{3} = \frac{210}{3} = 70 \text{ lakh}.$

Each child must therefore build exactly 70 lakh out of {house 50, flats 30 each, bank 70}. The only way to place the one house and all three flats is:

• House + 20 bank $= 70$
• 2 flats + 10 bank $= 70$
• 1 flat + 40 bank $= 70$

So the bank amounts are 10, 20, 40. Neeta got the least and Geeta the most:

The question asks for Seeta's bank deposits. Seeta is the child who took the house plus 20 lakh in bank deposits.

Answer: Rs. 20 lakh (Choice C)

Value up the estate. Bank deposits $=70$, house $=50$, three flats $=3\times 30 = 90$, and $n$ gold coins $=n$ (each worth 1 lakh). Total value $= 70 + 50 + 90 + n = (210 + n)$ lakh.

This scenario: the total value is shared equally, and the gold coins are shared equally.

Each child then gets value $\dfrac{210+n}{3}$ and coins $\dfrac{n}{3}$. So each child's non-coin value (house + flats + bank) is $\frac{210+n}{3} - \frac{n}{3} = \frac{210}{3} = 70 \text{ lakh}.$

Each child must therefore build exactly 70 lakh out of {house 50, flats 30 each, bank 70}. The only way to place the one house and all three flats is:

• House + 20 bank $= 70$
• 2 flats + 10 bank $= 70$
• 1 flat + 40 bank $= 70$

So the bank amounts are 10, 20, 40. Neeta got the least and Geeta the most:

The question asks how many flats Neeta received. Neeta is the child who took 2 flats (plus 10 lakh in bank).

Answer: 2

Value up the estate. Bank deposits $=70$, house $=50$, three flats $=3\times 30 = 90$, and $n$ gold coins $=n$ (each worth 1 lakh). Total value $= 70 + 50 + 90 + n = (210 + n)$ lakh.

This scenario: value is shared in the ratio $1:2:3$ (Neeta: Seeta: Geeta) and coins in the ratio $2:3:4$.

So the value shares are $\dfrac{210+n}{6},\ \dfrac{210+n}{3},\ \dfrac{210+n}{2}$, and the coin values are $\dfrac{2n}{9},\ \dfrac{3n}{9},\ \dfrac{4n}{9}$.

Each child's non-coin value (house + flats + bank) is the value share minus the coin value:

• Neeta: $\dfrac{210+n}{6} - \dfrac{2n}{9} = 35 - \dfrac{n}{18}$
• Seeta: $\dfrac{210+n}{3} - \dfrac{n}{3} = 70$
• Geeta: $\dfrac{210+n}{2} - \dfrac{4n}{9} = 105 + \dfrac{n}{18}$

Who holds the three flats (worth 90)? Only Geeta's non-coin value ($105 + n/18$) is large enough to contain 90, so Geeta took all three flats — and, as told, not the house.

Place the house. Seeta's non-coin value is exactly 70, which must be the house (50) plus 20 in bank. So Seeta = house + 20 bank, and Neeta is left with bank only, $35 - \dfrac{n}{18}$.

Use the last clue — a child other than Geeta got 30 lakh in bank. Seeta has 20, so that child is Neeta: $35 - \frac{n}{18} = 30 \ \Rightarrow\ \frac{n}{18} = 5 \ \Rightarrow\ n = 90.$

With $n = 90$, Geeta's bank $= 105 + \dfrac{90}{18} - 90 = 105 + 5 - 90 = 20$.

Final distribution ($n = 90$ coins):

(Check: values $50:100:150 = 1:2:3$; coins $20:30:40 = 2:3:4$. Both ratios hold.)

The question asks for the number of gold coins. From the clue on Neeta's bank deposit, $n = \mathbf{90}$.

Answer: 90 (Choice B)

Value up the estate. Bank deposits $=70$, house $=50$, three flats $=3\times 30 = 90$, and $n$ gold coins $=n$ (each worth 1 lakh). Total value $= 70 + 50 + 90 + n = (210 + n)$ lakh.

This scenario: value is shared in the ratio $1:2:3$ (Neeta: Seeta: Geeta) and coins in the ratio $2:3:4$.

So the value shares are $\dfrac{210+n}{6},\ \dfrac{210+n}{3},\ \dfrac{210+n}{2}$, and the coin values are $\dfrac{2n}{9},\ \dfrac{3n}{9},\ \dfrac{4n}{9}$.

Each child's non-coin value (house + flats + bank) is the value share minus the coin value:

• Neeta: $\dfrac{210+n}{6} - \dfrac{2n}{9} = 35 - \dfrac{n}{18}$
• Seeta: $\dfrac{210+n}{3} - \dfrac{n}{3} = 70$
• Geeta: $\dfrac{210+n}{2} - \dfrac{4n}{9} = 105 + \dfrac{n}{18}$

Who holds the three flats (worth 90)? Only Geeta's non-coin value ($105 + n/18$) is large enough to contain 90, so Geeta took all three flats — and, as told, not the house.

Place the house. Seeta's non-coin value is exactly 70, which must be the house (50) plus 20 in bank. So Seeta = house + 20 bank, and Neeta is left with bank only, $35 - \dfrac{n}{18}$.

Use the last clue — a child other than Geeta got 30 lakh in bank. Seeta has 20, so that child is Neeta: $35 - \frac{n}{18} = 30 \ \Rightarrow\ \frac{n}{18} = 5 \ \Rightarrow\ n = 90.$

With $n = 90$, Geeta's bank $= 105 + \dfrac{90}{18} - 90 = 105 + 5 - 90 = 20$.

Final distribution ($n = 90$ coins):

(Check: values $50:100:150 = 1:2:3$; coins $20:30:40 = 2:3:4$. Both ratios hold.)

The question asks for Geeta's bank deposits. From the final table, Geeta received 20 lakh in bank deposits.

Answer: 20

The extra charge on a single seat depends on its row-zone and seat-type:

Step 1 — Jayanta, Ajit, Byomkesh. They sit in the same seat-letter, three consecutive rows (in that order), and all pay different amounts. For three consecutive rows to give three different prices, they must straddle three different price-zones — the only such run is rows 10, 11, 12. So Jayanta = row 10, Ajit = row 11, Byomkesh = row 12.

Step 2 — Which letter? A middle seat costs 0 in both rows 10 and 11, so a middle letter can't give three different prices. So their shared letter is an Aisle seat:

• Jayanta (row 10, aisle) = 500
• Ajit (row 11, aisle) = 400
• Byomkesh (row 12, aisle) = 1000

Step 3 — Manik sits next to Jayanta, i.e. also in row 10, in the other aisle seat: 500.

Step 4 — Total check drives the rest. Seven friends pay Rs. 4600 in total and one pays nothing. So far J + A + B + Manik = 500 + 400 + 1000 + 500 = 2400, leaving 2200 for the other four (Gargi, Kikira, Prodosh, Tapesh), one of whom pays 0.

• Gargi sits next to Kikira (a Window + Middle adjacent pair). The only way an adjacent Window–Middle pair is expensive enough is a premium row (1, 12 or 13), where every seat is 1000. So Gargi = 1000 and Kikira = 1000 (2000 together).
• Prodosh and Tapesh share a letter in consecutive rows with different prices; that leaves 200 + 0, i.e. a Window seat in row 20 (200) and row 21 (0). So Prodosh = 200, Tapesh = 0.

Now 2000 + 200 + 0 = 2200 ✓, and the seat-type tally is 3 Window, 4 Aisle, 1 Middle ✓.

Solved seating:

This question — Manik's row. Manik sits next to Jayanta, who is in row 10, so Manik is in row 10.

Answer: Row 10 (Choice A)

The extra charge on a single seat depends on its row-zone and seat-type:

Step 1 — Jayanta, Ajit, Byomkesh. They sit in the same seat-letter, three consecutive rows (in that order), and all pay different amounts. For three consecutive rows to give three different prices, they must straddle three different price-zones — the only such run is rows 10, 11, 12. So Jayanta = row 10, Ajit = row 11, Byomkesh = row 12.

Step 2 — Which letter? A middle seat costs 0 in both rows 10 and 11, so a middle letter can't give three different prices. So their shared letter is an Aisle seat:

• Jayanta (row 10, aisle) = 500
• Ajit (row 11, aisle) = 400
• Byomkesh (row 12, aisle) = 1000

Step 3 — Manik sits next to Jayanta, i.e. also in row 10, in the other aisle seat: 500.

Step 4 — Total check drives the rest. Seven friends pay Rs. 4600 in total and one pays nothing. So far J + A + B + Manik = 500 + 400 + 1000 + 500 = 2400, leaving 2200 for the other four (Gargi, Kikira, Prodosh, Tapesh), one of whom pays 0.

• Gargi sits next to Kikira (a Window + Middle adjacent pair). The only way an adjacent Window–Middle pair is expensive enough is a premium row (1, 12 or 13), where every seat is 1000. So Gargi = 1000 and Kikira = 1000 (2000 together).
• Prodosh and Tapesh share a letter in consecutive rows with different prices; that leaves 200 + 0, i.e. a Window seat in row 20 (200) and row 21 (0). So Prodosh = 200, Tapesh = 0.

Now 2000 + 200 + 0 = 2200 ✓, and the seat-type tally is 3 Window, 4 Aisle, 1 Middle ✓.

Solved seating:

This question — Jayanta's extra. Jayanta is in an aisle seat in row 10, where an aisle seat costs Rs. 500.

Answer: Rs. 500 (Choice C)

The extra charge on a single seat depends on its row-zone and seat-type:

Step 1 — Jayanta, Ajit, Byomkesh. They sit in the same seat-letter, three consecutive rows (in that order), and all pay different amounts. For three consecutive rows to give three different prices, they must straddle three different price-zones — the only such run is rows 10, 11, 12. So Jayanta = row 10, Ajit = row 11, Byomkesh = row 12.

Step 2 — Which letter? A middle seat costs 0 in both rows 10 and 11, so a middle letter can't give three different prices. So their shared letter is an Aisle seat:

• Jayanta (row 10, aisle) = 500
• Ajit (row 11, aisle) = 400
• Byomkesh (row 12, aisle) = 1000

Step 3 — Manik sits next to Jayanta, i.e. also in row 10, in the other aisle seat: 500.

Step 4 — Total check drives the rest. Seven friends pay Rs. 4600 in total and one pays nothing. So far J + A + B + Manik = 500 + 400 + 1000 + 500 = 2400, leaving 2200 for the other four (Gargi, Kikira, Prodosh, Tapesh), one of whom pays 0.

• Gargi sits next to Kikira (a Window + Middle adjacent pair). The only way an adjacent Window–Middle pair is expensive enough is a premium row (1, 12 or 13), where every seat is 1000. So Gargi = 1000 and Kikira = 1000 (2000 together).
• Prodosh and Tapesh share a letter in consecutive rows with different prices; that leaves 200 + 0, i.e. a Window seat in row 20 (200) and row 21 (0). So Prodosh = 200, Tapesh = 0.

Now 2000 + 200 + 0 = 2200 ✓, and the seat-type tally is 3 Window, 4 Aisle, 1 Middle ✓.

Solved seating:

This question — Gargi's extra. Gargi's Window–Middle pair with Kikira sits in a premium row (1, 12 or 13), where every seat costs Rs. 1000, so Gargi paid Rs. 1000 whichever of the two seats she took.

Answer: Rs. 1000 (Choice D)

The extra charge on a single seat depends on its row-zone and seat-type:

Step 1 — Jayanta, Ajit, Byomkesh. They sit in the same seat-letter, three consecutive rows (in that order), and all pay different amounts. For three consecutive rows to give three different prices, they must straddle three different price-zones — the only such run is rows 10, 11, 12. So Jayanta = row 10, Ajit = row 11, Byomkesh = row 12.

Step 2 — Which letter? A middle seat costs 0 in both rows 10 and 11, so a middle letter can't give three different prices. So their shared letter is an Aisle seat:

• Jayanta (row 10, aisle) = 500
• Ajit (row 11, aisle) = 400
• Byomkesh (row 12, aisle) = 1000

Step 3 — Manik sits next to Jayanta, i.e. also in row 10, in the other aisle seat: 500.

Step 4 — Total check drives the rest. Seven friends pay Rs. 4600 in total and one pays nothing. So far J + A + B + Manik = 500 + 400 + 1000 + 500 = 2400, leaving 2200 for the other four (Gargi, Kikira, Prodosh, Tapesh), one of whom pays 0.

• Gargi sits next to Kikira (a Window + Middle adjacent pair). The only way an adjacent Window–Middle pair is expensive enough is a premium row (1, 12 or 13), where every seat is 1000. So Gargi = 1000 and Kikira = 1000 (2000 together).
• Prodosh and Tapesh share a letter in consecutive rows with different prices; that leaves 200 + 0, i.e. a Window seat in row 20 (200) and row 21 (0). So Prodosh = 200, Tapesh = 0.

Now 2000 + 200 + 0 = 2200 ✓, and the seat-type tally is 3 Window, 4 Aisle, 1 Middle ✓.

Solved seating:

This question — who paid nothing. From the table, Tapesh is in a window seat in row 21, where no extra charge applies, so Tapesh paid 0.

Answer: Tapesh (Choice D)

The queen at c5 attacks along its column (c), its row (5) and its four diagonals, stopping at the first piece it meets in each direction. Here no piece blocks another, so check each one:

• c2 — same column c, path clear. Attacked.
• g1 — on the down-right diagonal c5 → d4 → e3 → f2 → g1, path clear. Attacked.
• g3 — from c5, the column difference is 4 and the row difference is 2, so it is not on the same row, column, or a diagonal. Not attacked.
• g5 — same row 5, path clear. Attacked.
• a3 — on the down-left diagonal c5 → b4 → a3, path clear. Attacked.

Pieces under attack: c2, g1, g5, a3 — that is 4.

💡 Teacher tip: A square is on a diagonal of the queen exactly when the horizontal and vertical distances are equal. That single check settles g1, a3 (equal) versus g3 (unequal) instantly.

Answer: 4 (Choice C)

The six pieces are at a1, a3, b4, d7, h7, h8.

A queen attacks a piece if they share a row, a column, or a diagonal — and if two pieces lie on the same line, only the nearer one is hit (the far one is blocked). Reading columns as numbers a = 1, b = 2, …, h = 8, two squares share a diagonal when their column + row sums are equal, or their column − row differences are equal.

Count how many pieces the queen attacks from each option (respecting blocking):

• f8 — row 8 reaches h8; the down-left diagonal f8 → e7 → d6 → c5 → b4 stops there (a3 is behind b4). = 2.
• a7 — column a reaches a3 (a1 is behind a3); row 7 reaches d7 (h7 is behind d7). = 2.
• c1 — row 1 reaches a1; the up-left diagonal c1 → b2 → a3. = 2.
• d3 — column d reaches d7; row 3 reaches a3; the up-right diagonal d3 → e4 → f5 → g6 → h7. = 3.

The maximum, 3 pieces, comes from d3.

Answer: d3 (Choice D)

The six pieces are at a1, a3, b4, d7, h7, h8.

A queen attacks a piece if they share a row, a column, or a diagonal — and if two pieces lie on the same line, only the nearer one is hit (the far one is blocked). Reading columns as numbers a = 1, b = 2, …, h = 8, two squares share a diagonal when their column + row sums are equal, or their column − row differences are equal.

A square is safe (the queen there attacks nothing) only if it shares no row, no column, and no diagonal with any of the six pieces. Note blocking never rescues a square — the nearest piece on a shared line is always hit.

Rule out rows and columns first.

• Rows used by pieces: 1, 3, 4, 7, 8 → safe rows are 2, 5, 6.
• Columns used by pieces: a, b, d, h → safe columns are c, e, f, g.

That leaves 3 × 4 = 12 candidate squares. Now remove any that share a diagonal with a piece. The pieces' diagonal sums (col + row) are {2, 4, 6, 11, 15, 16} and their differences (col − row) are {0, −2, −3, 1}.

Exactly 4 squares — e2, f2, g2, g5 — attack no piece.

Answer: 4 (Choice C)

The queen is alone on the board at d5. A square is attacked if it is in row 5, in column d, or on either diagonal through d5. Using column numbers (a = 1 … h = 8), d5 has column + row = 9 and column − row = −1, which name its two diagonals.

Count the attacked squares (not counting d5 itself):

• Row 5: a5, b5, c5, e5, f5, g5, h5 → 7.
• Column d: d1, d2, d3, d4, d6, d7, d8 → 7.
• Diagonal with sum 9: a8, b7, c6, e4, f3, g2, h1 → 7.
• Diagonal with difference −1: a2, b3, c4, e6, f7, g8 → 6.

These four lines meet only at d5 itself, so there is no double-counting. Attacked squares = 7 + 7 + 7 + 6 = 27.

Safe squares = 64 total − 1 (the queen's own square) − 27 (attacked) = 36.

Answer: 36 (Choice C)

A bar chart gives the number of dorms at each repair cost:

Repairs are grouped as Light (1 or 2), Moderate (3 or 4) and Extensive (5 or 6).

What each dorm is allowed (from clue 1).

• Odd dorms (1, 3, 5, 7, 9) cannot be light → they are Moderate or Extensive.
• Even dorms (2, 4, 6, 8, 10) cannot be moderate → they are Light or Extensive.
• Dorms divisible by 3 (3, 6, 9) cannot be extensive → they are Light or Moderate.

Combining these:

• Dorm 3 and Dorm 9 → Moderate only (3 or 4).
• Dorm 6 → Light only (1 or 2).
• Dorms 1, 5, 7 → Moderate or Extensive.
• Dorms 2, 4, 8, 10 → Light or Extensive.

Using clue 2 (dorms 4–9 all get different costs, dorm 7 the maximum, dorm 8 the minimum):

1. Dorm 7 = 6 (max), Dorm 8 = 1 (min).
2. Dorm 6 is light and cost 1 is taken, so Dorm 6 = 2.
3. Dorm 4 is even (so 2 or 5) and 2 is taken, so Dorm 4 = 5.
4. Dorms 5 and 9 are moderate, different, so they take 3 and 4 in some order.

The costs still to be placed among dorms 1, 2, 3, 10 are one more 1, two more 3s and one more 6, i.e. the set {1, 3, 3, 6}.

1. Dorm 3 is moderate → 3.
2. Dorm 1 is odd (can't be 1) → 3.
3. Dorms 2 and 10 are the leftover {1, 6} in some order.

Solved table (only the two pairs stay flexible):

Within each pair the two dorms take opposite values: {Dorm 5, Dorm 9} = {3, 4}, and {Dorm 2, Dorm 10} = {1, 6}.

Now test each statement against the solved table.

• Dorm 1 needs a moderate repair — Dorm 1 = 3, which is moderate. Always true.
• Dorm 5 will cost no more than Rs. 4 Cr — Dorm 5 is 3 or 4, both ≤ 4. Always true.
• Dorm 7 needs an extensive repair — Dorm 7 = 6, which is extensive. Always true.
• Dorm 10 will cost no more than Rs. 4 Cr — Dorm 10 is 1 or 6. If it is 6, it exceeds 4. So this is not guaranteed.

The statement that need not be true is about Dorm 10.

Answer: Dorm 10 repair will cost no more than Rs. 4 Crores (Choice D)

A bar chart gives the number of dorms at each repair cost:

Repairs are grouped as Light (1 or 2), Moderate (3 or 4) and Extensive (5 or 6).

What each dorm is allowed (from clue 1).

• Odd dorms (1, 3, 5, 7, 9) cannot be light → they are Moderate or Extensive.
• Even dorms (2, 4, 6, 8, 10) cannot be moderate → they are Light or Extensive.
• Dorms divisible by 3 (3, 6, 9) cannot be extensive → they are Light or Moderate.

Combining these:

• Dorm 3 and Dorm 9 → Moderate only (3 or 4).
• Dorm 6 → Light only (1 or 2).
• Dorms 1, 5, 7 → Moderate or Extensive.
• Dorms 2, 4, 8, 10 → Light or Extensive.

Using clue 2 (dorms 4–9 all get different costs, dorm 7 the maximum, dorm 8 the minimum):

1. Dorm 7 = 6 (max), Dorm 8 = 1 (min).
2. Dorm 6 is light and cost 1 is taken, so Dorm 6 = 2.
3. Dorm 4 is even (so 2 or 5) and 2 is taken, so Dorm 4 = 5.
4. Dorms 5 and 9 are moderate, different, so they take 3 and 4 in some order.

The costs still to be placed among dorms 1, 2, 3, 10 are one more 1, two more 3s and one more 6, i.e. the set {1, 3, 3, 6}.

1. Dorm 3 is moderate → 3.
2. Dorm 1 is odd (can't be 1) → 3.
3. Dorms 2 and 10 are the leftover {1, 6} in some order.

Solved table (only the two pairs stay flexible):

Within each pair the two dorms take opposite values: {Dorm 5, Dorm 9} = {3, 4}, and {Dorm 2, Dorm 10} = {1, 6}.

Add up the odd-numbered dorms (1, 3, 5, 7, 9):

• Dorm 1 = 3, Dorm 3 = 3, Dorm 7 = 6 — these are fixed.
• Dorms 5 and 9 together are 3 and 4 (in some order), so their sum is 7 no matter which way round.

Total = 3 + 3 + 7 + 6 = 19.

💡 Teacher tip: When two unknowns are just a fixed pair in some order, you don't need to resolve the order — their sum is already locked.

Answer: 19

A bar chart gives the number of dorms at each repair cost:

Repairs are grouped as Light (1 or 2), Moderate (3 or 4) and Extensive (5 or 6).

What each dorm is allowed (from clue 1).

• Odd dorms (1, 3, 5, 7, 9) cannot be light → they are Moderate or Extensive.
• Even dorms (2, 4, 6, 8, 10) cannot be moderate → they are Light or Extensive.
• Dorms divisible by 3 (3, 6, 9) cannot be extensive → they are Light or Moderate.

Combining these:

• Dorm 3 and Dorm 9 → Moderate only (3 or 4).
• Dorm 6 → Light only (1 or 2).
• Dorms 1, 5, 7 → Moderate or Extensive.
• Dorms 2, 4, 8, 10 → Light or Extensive.

Using clue 2 (dorms 4–9 all get different costs, dorm 7 the maximum, dorm 8 the minimum):

1. Dorm 7 = 6 (max), Dorm 8 = 1 (min).
2. Dorm 6 is light and cost 1 is taken, so Dorm 6 = 2.
3. Dorm 4 is even (so 2 or 5) and 2 is taken, so Dorm 4 = 5.
4. Dorms 5 and 9 are moderate, different, so they take 3 and 4 in some order.

The costs still to be placed among dorms 1, 2, 3, 10 are one more 1, two more 3s and one more 6, i.e. the set {1, 3, 3, 6}.

1. Dorm 3 is moderate → 3.
2. Dorm 1 is odd (can't be 1) → 3.
3. Dorms 2 and 10 are the leftover {1, 6} in some order.

Solved table (only the two pairs stay flexible):

Within each pair the two dorms take opposite values: {Dorm 5, Dorm 9} = {3, 4}, and {Dorm 2, Dorm 10} = {1, 6}.

Bring in the new conditions: 4 dorms are women's dorms costing Rs. 20 Cr in total, and only one of Dorms 1–5 is a women's dorm.

First, the total cost of all 10 dorms is $1(2) + 2(1) + 3(3) + 4(1) + 5(1) + 6(2) = 34 \text{ Cr}.$

So the four women's dorms cost 20 Cr and the six men's dorms cost 14 Cr.

We need one dorm from {1, 2, 3, 4, 5} and three dorms from {6, 7, 8, 9, 10} whose costs add to 20. Test the flexible pairs:

• Dorms 6, 7, 8 are fixed at 2, 6, 1. Their three-dorm combinations with dorm 9 or 10 must be checked.
• The only assignment that reaches 20 is Dorm 5 = 4, Dorm 9 = 3, Dorm 2 = 1, Dorm 10 = 6. Then the women's dorms are Dorm 4 (5), Dorm 7 (6), Dorm 9 (3), Dorm 10 (6), totalling 5 + 6 + 3 + 6 = 20, with exactly one of them (Dorm 4) from 1–5. ✓

Every other split of the pairs fails to make three dorms from {6–10} sum correctly.

Hence Dorm 9 = 3.

Answer: 3

A bar chart gives the number of dorms at each repair cost:

Repairs are grouped as Light (1 or 2), Moderate (3 or 4) and Extensive (5 or 6).

What each dorm is allowed (from clue 1).

• Odd dorms (1, 3, 5, 7, 9) cannot be light → they are Moderate or Extensive.
• Even dorms (2, 4, 6, 8, 10) cannot be moderate → they are Light or Extensive.
• Dorms divisible by 3 (3, 6, 9) cannot be extensive → they are Light or Moderate.

Combining these:

• Dorm 3 and Dorm 9 → Moderate only (3 or 4).
• Dorm 6 → Light only (1 or 2).
• Dorms 1, 5, 7 → Moderate or Extensive.
• Dorms 2, 4, 8, 10 → Light or Extensive.

Using clue 2 (dorms 4–9 all get different costs, dorm 7 the maximum, dorm 8 the minimum):

1. Dorm 7 = 6 (max), Dorm 8 = 1 (min).
2. Dorm 6 is light and cost 1 is taken, so Dorm 6 = 2.
3. Dorm 4 is even (so 2 or 5) and 2 is taken, so Dorm 4 = 5.
4. Dorms 5 and 9 are moderate, different, so they take 3 and 4 in some order.

The costs still to be placed among dorms 1, 2, 3, 10 are one more 1, two more 3s and one more 6, i.e. the set {1, 3, 3, 6}.

1. Dorm 3 is moderate → 3.
2. Dorm 1 is odd (can't be 1) → 3.
3. Dorms 2 and 10 are the leftover {1, 6} in some order.

Solved table (only the two pairs stay flexible):

Within each pair the two dorms take opposite values: {Dorm 5, Dorm 9} = {3, 4}, and {Dorm 2, Dorm 10} = {1, 6}.

Bring in the new conditions: 4 women's dorms cost Rs. 20 Cr in total, and only one of Dorms 1–5 is a women's dorm.

The total cost of all 10 dorms is $1(2) + 2(1) + 3(3) + 4(1) + 5(1) + 6(2) = 34 \text{ Cr},$ so the four women's dorms must total 20 and the six men's dorms 14.

Picking one dorm from 1–5 and three from 6–10 to reach 20, the only workable choice is Dorm 5 = 4, Dorm 9 = 3, Dorm 2 = 1, Dorm 10 = 6, which makes the women's dorms {4, 7, 9, 10} with costs 5 + 6 + 3 + 6 = 20.

Check the options against this set:

• Dorm 2 — not in the set.
• Dorm 5 — not in the set.
• Dorm 8 — not in the set.
• Dorm 10 — in the set. ✓

Answer: Dorm 10 (Choice D)

How to read the table. Each row is an old elective and each column is a new one, so a cell counts students who moved from the row's elective to the column's elective. That means:

• a row total = students in that elective before the change,
• a column total = students in that elective after the change.

Every blank is either 0 or 1. We use the four given clues to pin them down.

1. Row E1 has no blanks, so E1 before $= 9+5+10+1+4+2 = 31$. Clue 1 says E1 had 6 more than E4 before, so E4 before $=25$. Row E4 already shows $3+2+14+4=23$, so its two blanks must add to 2 — both are 1.
2. Clue 3 says E4 had 2 more than E6 before, so E6 before $=23$. Row E6 shows $7+3+2+9=21$, so its two blanks are both 1.
3. Column E2 has no blanks, so E2 after $= 5+34+6+3+5+7+16 = 76$. Clue 2 says E2 rose by 30, so E2 before $=46$. Row E2 already totals 46, so its two blanks are both 0.
4. Clue 3 also says E2 was 10 more than E3 before, so E3 before $=36$. Row E3 shows $2+6+25+2=35$, so exactly one of its blanks is 1.
5. Using Clue 1's "after" part (E4 after $=$ E1 after $+3$) to balance the columns fixes the last blanks: in row E3 the move to E5 is 1 (and to E4 is 0), and row E5's blanks come out as E5→E1 $=1$, E5→E3 $=1$, E5→E6 $=1$ (E5 before $=38$).

Completed movement table:

Before vs After enrollment:

The question asks how many electives lost students. Compare each column's before and after:

• E1: $31 \to 18$ — down
• E2: $46 \to 76$ — up
• E3: $36 \to 79$ — up
• E4: $25 \to 21$ — down
• E5: $38 \to 45$ — up
• E6: $23 \to 61$ — up

Only E1 and E4 fell, so 2 electives had a decrease.

Answer: 2 (Choice C)

How to read the table. Each row is an old elective and each column is a new one, so a cell counts students who moved from the row's elective to the column's elective. That means:

• a row total = students in that elective before the change,
• a column total = students in that elective after the change.

Every blank is either 0 or 1. We use the four given clues to pin them down.

1. Row E1 has no blanks, so E1 before $= 9+5+10+1+4+2 = 31$. Clue 1 says E1 had 6 more than E4 before, so E4 before $=25$. Row E4 already shows $3+2+14+4=23$, so its two blanks must add to 2 — both are 1.
2. Clue 3 says E4 had 2 more than E6 before, so E6 before $=23$. Row E6 shows $7+3+2+9=21$, so its two blanks are both 1.
3. Column E2 has no blanks, so E2 after $= 5+34+6+3+5+7+16 = 76$. Clue 2 says E2 rose by 30, so E2 before $=46$. Row E2 already totals 46, so its two blanks are both 0.
4. Clue 3 also says E2 was 10 more than E3 before, so E3 before $=36$. Row E3 shows $2+6+25+2=35$, so exactly one of its blanks is 1.
5. Using Clue 1's "after" part (E4 after $=$ E1 after $+3$) to balance the columns fixes the last blanks: in row E3 the move to E5 is 1 (and to E4 is 0), and row E5's blanks come out as E5→E1 $=1$, E5→E3 $=1$, E5→E6 $=1$ (E5 before $=38$).

Completed movement table:

Before vs After enrollment:

The question asks for the after-change enrollments of E1 to E6. These are exactly the column totals of the completed table: 18, 76, 79, 21, 45, 61.

Answer: 18, 76, 79, 21, 45, 61 (Choice D)

How to read the table. Each row is an old elective and each column is a new one, so a cell counts students who moved from the row's elective to the column's elective. That means:

• a row total = students in that elective before the change,
• a column total = students in that elective after the change.

Every blank is either 0 or 1. We use the four given clues to pin them down.

1. Row E1 has no blanks, so E1 before $= 9+5+10+1+4+2 = 31$. Clue 1 says E1 had 6 more than E4 before, so E4 before $=25$. Row E4 already shows $3+2+14+4=23$, so its two blanks must add to 2 — both are 1.
2. Clue 3 says E4 had 2 more than E6 before, so E6 before $=23$. Row E6 shows $7+3+2+9=21$, so its two blanks are both 1.
3. Column E2 has no blanks, so E2 after $= 5+34+6+3+5+7+16 = 76$. Clue 2 says E2 rose by 30, so E2 before $=46$. Row E2 already totals 46, so its two blanks are both 0.
4. Clue 3 also says E2 was 10 more than E3 before, so E3 before $=36$. Row E3 shows $2+6+25+2=35$, so exactly one of its blanks is 1.
5. Using Clue 1's "after" part (E4 after $=$ E1 after $+3$) to balance the columns fixes the last blanks: in row E3 the move to E5 is 1 (and to E4 is 0), and row E5's blanks come out as E5→E1 $=1$, E5→E3 $=1$, E5→E6 $=1$ (E5 before $=38$).

Completed movement table:

Before vs After enrollment:

The question asks which elective changed the most, as a percentage of its original size. Compute $\dfrac{\text{After} - \text{Before}}{\text{Before}}$ for each:

The biggest swing is E6 at about $+165\%$.

Answer: E6 (Choice D)

How to read the table. Each row is an old elective and each column is a new one, so a cell counts students who moved from the row's elective to the column's elective. That means:

• a row total = students in that elective before the change,
• a column total = students in that elective after the change.

Every blank is either 0 or 1. We use the four given clues to pin them down.

1. Row E1 has no blanks, so E1 before $= 9+5+10+1+4+2 = 31$. Clue 1 says E1 had 6 more than E4 before, so E4 before $=25$. Row E4 already shows $3+2+14+4=23$, so its two blanks must add to 2 — both are 1.
2. Clue 3 says E4 had 2 more than E6 before, so E6 before $=23$. Row E6 shows $7+3+2+9=21$, so its two blanks are both 1.
3. Column E2 has no blanks, so E2 after $= 5+34+6+3+5+7+16 = 76$. Clue 2 says E2 rose by 30, so E2 before $=46$. Row E2 already totals 46, so its two blanks are both 0.
4. Clue 3 also says E2 was 10 more than E3 before, so E3 before $=36$. Row E3 shows $2+6+25+2=35$, so exactly one of its blanks is 1.
5. Using Clue 1's "after" part (E4 after $=$ E1 after $+3$) to balance the columns fixes the last blanks: in row E3 the move to E5 is 1 (and to E4 is 0), and row E5's blanks come out as E5→E1 $=1$, E5→E3 $=1$, E5→E6 $=1$ (E5 before $=38$).

Completed movement table:

Before vs After enrollment:

Apply the re-enrolment rule. Any elective (except E7) that ended below 20 students forces everyone who left it to come back. Looking at the after-change row, only E1 ended below 20 (at 18).

E1 started with 31 students; 9 stayed (the E1→E1 cell), so $31 - 9 = 22$ had left. They now return to E1, leaving whichever elective they had switched to. From row E1, those 22 went: 5 to E2, 10 to E3, 1 to E4, 4 to E5, 2 to E6.

Subtract them back out:

(E4 lands at exactly 20, so it does not trigger another round.)

In decreasing order: E2 (71) > E3 (69) > E6 (59) > E5 (41) > E1 (40) > E4 (20).

Answer: E2, E3, E6, E5, E1, E4 (Choice A)