We the Travellers—II

Given: Fuel already in tank = 28 l; Fuel added = 75 l

We need to find the total quantity of fuel.

$$28 + 75$$

Step 1: Add the ones digits: $8 + 5 = 13$. Write 3, carry 1.

Step 2: Add the tens digits: $2 + 7 + 1\text{(carry)} = 10$. Write 10.

$$28 + 75 = 103$$

The total quantity of fuel in the lorry is 103 litres.

Given: Two numbers are 49 and 89.

$$49 + 89$$

Step 1: Add the ones digits: $9 + 9 = 18$. Write 8, carry 1.

Step 2: Add the tens digits: $4 + 8 + 1\text{(carry)} = 13$. Write 13.

$$49 + 89 = 138$$

The sum of 49 and 89 is 138.

$$15 + 79$$

Easier way: $15 + 79 = 15 + 80 - 1 = 95 - 1 = 94$

Or by column addition:
- Ones: $5 + 9 = 14$, write 4, carry 1
- Tens: $1 + 7 + 1 = 9$

$$\boxed{15 + 79 = 94}$$

$$46 + 99$$

Easier way: $46 + 99 = 46 + 100 - 1 = 146 - 1 = 145$

$$\boxed{46 + 99 = 145}$$

$$38 + 35$$

Step 1: Ones: $8 + 5 = 13$, write 3, carry 1.
Step 2: Tens: $3 + 3 + 1 = 7$.

$$\boxed{38 + 35 = 73}$$

$$5 + 89$$

Easier way: $5 + 89 = 5 + 90 - 1 = 95 - 1 = 94$

Or: $5 + 89 = 89 + 5 = 94$

$$\boxed{5 + 89 = 94}$$

$$76 + 28$$

Step 1: Ones: $6 + 8 = 14$, write 4, carry 1.
Step 2: Tens: $7 + 2 + 1 = 10$.

$$\boxed{76 + 28 = 104}$$

$$69 + 20$$

Easier way: Adding 20 only changes the tens digit.
$69 + 20 = 89$

$$\boxed{69 + 20 = 89}$$

Concept: Addition and subtraction are inverse operations. If $a + b = c$, then $c - b = a$ and $c - a = b$. If $a - b = c$, then $c + b = a$ and $a - c = b$.

(a) If $46 + 21 = 67$, then:
- $67 - 21 = \mathbf{46}$
- $67 - 46 = \mathbf{21}$

(b) If $198 - 98 = 100$, then:
- $100 + \mathbf{98} = 198$
- $198 - \mathbf{100} = 98$

(c) If $189 + 98 = 287$, then:
- $287 - 98 = \mathbf{189}$
- $287 - 189 = \mathbf{98}$

(d) If $872 - 672 = 200$, then:
- $200 + \mathbf{672} = 872$
- $872 - \mathbf{200} = 672$

Concept: From any addition fact $a + b = c$, we get $c - b = a$ and $c - a = b$. From any subtraction fact $a - b = c$, we get $c + b = a$ and $a - c = b$.

(a) If $78 + 164 = 242$, then:
$$242 - 164 = 78 \quad \text{and} \quad 242 - 78 = 164$$

(b) If $462 + 839 = 1301$, then:
$$1301 - 839 = 462 \quad \text{and} \quad 1301 - 462 = 839$$

(c) If $921 - 137 = 784$, then:
$$784 + 137 = 921 \quad \text{and} \quad 921 - 784 = 137$$

(d) If $824 - 234 = 590$, then:
$$590 + 234 = 824 \quad \text{and} \quad 824 - 590 = 234$$

Given: We need to find $82 - 37$.

Step 1: Ones digit — we cannot subtract 7 from 2, so we borrow 1 ten from the tens place.
$12 - 7 = 5$

Step 2: Tens digit — after borrowing, we have $7 - 3 = 4$.

$$82 - 37 = 45$$

Check: $37 + 45 = 82$ ✓

The difference between 82 and 37 is 45.

$$57 - 11$$
- Ones: $7 - 1 = 6$
- Tens: $5 - 1 = 4$

$$\boxed{57 - 11 = 46}$$

$$23 - 19$$
- Ones: Cannot subtract 9 from 3; borrow 1 ten. $13 - 9 = 4$
- Tens: $1 - 1 = 0$

$$\boxed{23 - 19 = 4}$$

$$49 - 21$$
- Ones: $9 - 1 = 8$
- Tens: $4 - 2 = 2$

$$\boxed{49 - 21 = 28}$$

$$56 - 18$$
- Ones: Cannot subtract 8 from 6; borrow 1 ten. $16 - 8 = 8$
- Tens: $4 - 1 = 3$

$$\boxed{56 - 18 = 38}$$

$$93 - 35$$
- Ones: Cannot subtract 5 from 3; borrow 1 ten. $13 - 5 = 8$
- Tens: $8 - 3 = 5$

$$\boxed{93 - 35 = 58}$$

$$84 - 23$$
- Ones: $4 - 3 = 1$
- Tens: $8 - 2 = 6$

$$\boxed{84 - 23 = 61}$$

$$70 - 43$$
- Ones: Cannot subtract 3 from 0; borrow 1 ten. $10 - 3 = 7$
- Tens: $6 - 4 = 2$

$$\boxed{70 - 43 = 27}$$

$$65 - 47$$
- Ones: Cannot subtract 7 from 5; borrow 1 ten. $15 - 7 = 8$
- Tens: $5 - 4 = 1$

$$\boxed{65 - 47 = 18}$$

Sum of 2 consecutive numbers: $1+2=3,\ 2+3=5,\ 3+4=7,\ 4+5=9$ — all odd.

*Why:* Two consecutive numbers are always one even and one odd. Even + Odd = Odd. So the sum of any 2 consecutive numbers is always odd.

Sum of 3 consecutive numbers: $1+2+3=6,\ 2+3+4=9,\ 3+4+5=12,\ 4+5+6=15$ — alternately even and odd.

*Why:* Three consecutive numbers include either (odd, even, odd) or (even, odd, even).
- Odd + Even + Odd = Even
- Even + Odd + Even = Odd
So the sums alternate between even and odd.

Sum of 4 consecutive numbers: $1+2+3+4=10,\ 2+3+4+5=14,\ 3+4+5+6=18,\ 4+5+6+7=22$ — all even.

*Why:* Four consecutive numbers always contain 2 odd and 2 even numbers. Odd + Odd = Even, Even + Even = Even, so the total sum is always even.

Sum of 2 consecutive numbers: $3, 5, 7, 9, \ldots$
Difference: $5-3=2,\ 7-5=2,\ 9-7=2$. The difference is 2 throughout.

Sum of 3 consecutive numbers: $6, 9, 12, 15, \ldots$
Difference: $9-6=3,\ 12-9=3,\ 15-12=3$. The difference is 3 throughout.

Sum of 4 consecutive numbers: $10, 14, 18, 22, \ldots$
Difference: $14-10=4,\ 18-14=4,\ 22-18=4$. The difference is 4 throughout.

Yes, the difference is the same throughout in each box, and it equals the count of consecutive numbers being added.

Pattern observed: For $n$ consecutive numbers, the difference between two successive sums is $n$.

(a) 5 consecutive numbers:
$1+2+3+4+5=15$
$2+3+4+5+6=20$
Difference $= 20 - 15 = \mathbf{5}$

(b) 6 consecutive numbers:
$1+2+3+4+5+6=21$
$2+3+4+5+6+7=27$
Difference $= 27 - 21 = \mathbf{6}$

The difference between two successive sums equals the number of consecutive numbers being added.

$$\begin{array}{r} 238 \\ +367 \\ \hline \end{array}$$
- Ones: $8+7=15$, write 5, carry 1
- Tens: $3+6+1=10$, write 0, carry 1
- Hundreds: $2+3+1=6$

$$\boxed{238 + 367 = 605}$$

$$\begin{array}{r} 1{,}234 \\ +12{,}345 \\ \hline \end{array}$$
- Ones: $4+5=9$
- Tens: $3+4=7$
- Hundreds: $2+3=5$
- Thousands: $1+2=3$
- Ten-thousands: $0+1=1$

$$\boxed{1{,}234 + 12{,}345 = 13{,}579}$$

$$\begin{array}{r} 12 \\ +123 \\ \hline \end{array}$$
- Ones: $2+3=5$
- Tens: $1+2=3$
- Hundreds: $0+1=1$

$$\boxed{12 + 123 = 135}$$

$$\begin{array}{r} 46{,}120 \\ +12{,}890 \\ \hline \end{array}$$
- Ones: $0+0=0$
- Tens: $2+9=11$, write 1, carry 1
- Hundreds: $1+8+1=10$, write 0, carry 1
- Thousands: $6+2+1=9$
- Ten-thousands: $4+1=5$

$$\boxed{46{,}120 + 12{,}890 = 59{,}010}$$

$$\begin{array}{r} 878 \\ +8{,}789 \\ \hline \end{array}$$
- Ones: $8+9=17$, write 7, carry 1
- Tens: $7+8+1=16$, write 6, carry 1
- Hundreds: $8+7+1=16$, write 6, carry 1
- Thousands: $0+8+1=9$

$$\boxed{878 + 8{,}789 = 9{,}667}$$

$$\begin{array}{r} 1{,}749 \\ +17{,}490 \\ \hline \end{array}$$
- Ones: $9+0=9$
- Tens: $4+9=13$, write 3, carry 1
- Hundreds: $7+4+1=12$, write 2, carry 1
- Thousands: $1+7+1=9$
- Ten-thousands: $0+1=1$

$$\boxed{1{,}749 + 17{,}490 = 19{,}239}$$

Note: The map image is not visible. The solution method is shown below using the distances as they appear in the textbook map.

Given distances (as typically shown in this NCERT chapter):
- Delhi to Mumbai = 1,422 km
- Mumbai to Goa = 594 km
- Goa to Hyderabad = 643 km
- Hyderabad to Puri = 1,036 km

Total distance = $1{,}422 + 594 + 643 + 1{,}036$

Step 1: $1{,}422 + 594 = 2{,}016$
Step 2: $2{,}016 + 643 = 2{,}659$
Step 3: $2{,}659 + 1{,}036 = 3{,}695$

Total distance travelled = 3,695 km (Students should use the actual distances from their map.)

Given numbers: 5,205; 6,220; 7,095; 8,455; 4,840

(a) Closest to 10,000:
Try $5{,}205 + 4{,}840 = 10{,}045$ (difference = 45)
Try $6{,}220 + 4{,}840 = 11{,}060$ (too far)
Try $5{,}205 + 4{,}840 = 10{,}045$ — closest.
$$\boxed{5{,}205 + 4{,}840 = 10{,}045 \text{ (closest to 10,000)}}$$

(b) Closest to 15,000:
Try $8{,}455 + 6{,}220 = 14{,}675$ (difference = 325)
Try $8{,}455 + 7{,}095 = 15{,}550$ (difference = 550)
$8{,}455 + 6{,}220 = 14{,}675$ is closer.
$$\boxed{8{,}455 + 6{,}220 = 14{,}675 \text{ (closest to 15,000)}}$$

(c) Closest to 13,000:
Try $8{,}455 + 4{,}840 = 13{,}295$ (difference = 295)
Try $7{,}095 + 5{,}205 = 12{,}300$ (difference = 700)
Try $6{,}220 + 7{,}095 = 13{,}315$ (difference = 315)
$8{,}455 + 4{,}840 = 13{,}295$ is closest.
$$\boxed{8{,}455 + 4{,}840 = 13{,}295 \text{ (closest to 13,000)}}$$

(d) Closest to 16,000:
Try $8{,}455 + 7{,}095 = 15{,}550$ (difference = 450)
Try $8{,}455 + 6{,}220 = 14{,}675$ (difference = 1,325)
$8{,}455 + 7{,}095 = 15{,}550$ is closest.
$$\boxed{8{,}455 + 7{,}095 = 15{,}550 \text{ (closest to 16,000)}}$$

$$\begin{array}{r} 4{,}578 \\ -2{,}222 \\ \hline \end{array}$$
- Ones: $8-2=6$
- Tens: $7-2=5$
- Hundreds: $5-2=3$
- Thousands: $4-2=2$

$$\boxed{4{,}578 - 2{,}222 = 2{,}356}$$

$$\begin{array}{r} 15{,}324 \\ -11{,}780 \\ \hline \end{array}$$
- Ones: Cannot subtract 0 from 4... $4-0=4$
- Tens: Cannot subtract 8 from 2; borrow. $12-8=4$, carry reduces hundreds.
- Hundreds: $2-1-7$: borrow. $12-7-1=4$... Let us redo carefully:

$$15{,}324 - 11{,}780$$

Ones: $4 - 0 = 4$
Tens: $2 - 8$: borrow from hundreds. $12 - 8 = 4$
Hundreds: $3 - 1 - 7$: borrow from thousands. $13 - 1 - 7 = 5$
Thousands: $5 - 1 - 1 = 3$
Ten-thousands: $1 - 1 = 0$

$$\boxed{15{,}324 - 11{,}780 = 3{,}544}$$

$$\begin{array}{r} 5{,}423 \\ -\phantom{0}423 \\ \hline \end{array}$$
- Ones: $3-3=0$
- Tens: $2-2=0$
- Hundreds: $4-4=0$
- Thousands: $5-0=5$

$$\boxed{5{,}423 - 423 = 5{,}000}$$

$$\begin{array}{r} 123 \\ -12 \\ \hline \end{array}$$
- Ones: $3-2=1$
- Tens: $2-1=1$
- Hundreds: $1-0=1$

$$\boxed{123 - 12 = 111}$$

$$\begin{array}{r} 77{,}777 \\ -\phantom{00}777 \\ \hline \end{array}$$
- Ones: $7-7=0$
- Tens: $7-7=0$
- Hundreds: $7-7=0$
- Thousands: $7-0=7$
- Ten-thousands: $7-0=7$

$$\boxed{77{,}777 - 777 = 77{,}000}$$

$$\begin{array}{r} 826 \\ -752 \\ \hline \end{array}$$
- Ones: $6-2=4$
- Tens: $2-5$: borrow. $12-5=7$
- Hundreds: $7-7=0$

$$\boxed{826 - 752 = 74}$$

Given:
- Starting amount = ₹12,540
- Spent on food (Kolkata to Varanasi) = ₹3,275
- Gift received in Varanasi = ₹4,900
- Train ticket (Varanasi to Delhi) = ₹2,645
- Souvenirs in Delhi = ₹1,275

Step 1: Money after spending on food:
$$12{,}540 - 3{,}275 = 9{,}265$$

Step 2: Money after receiving gift:
$$9{,}265 + 4{,}900 = 14{,}165$$

Step 3: Money after buying train ticket:
$$14{,}165 - 2{,}645 = 11{,}520$$

Step 4: Money after buying souvenirs:
$$11{,}520 - 1{,}275 = 10{,}245$$

$$\boxed{\text{Mary is left with } ₹10{,}245 \text{ at the end of the Delhi trip.}}$$

(a) Estimation:
- Maths Lab ≈ ₹40,000
- Library books ≈ ₹9,500
- Sports equipment ≈ ₹20,000
- Total estimated expenditure ≈ ₹69,500

Money raised = ₹70,500 ≈ ₹70,000

Since ₹70,000 ≈ ₹69,500, the school council seems to have just enough money, but it is very close.

(b) Exact Calculation:
Total expenditure:
$$39{,}785 + 9{,}545 + 19{,}548$$

Step 1: $39{,}785 + 9{,}545 = 49{,}330$
Step 2: $49{,}330 + 19{,}548 = 68{,}878$

Money raised = ₹70,500
Money needed = ₹68,878

$$70{,}500 - 68{,}878 = 1{,}622$$

$$\boxed{\text{Yes, the school council has enough money. They will have } ₹1{,}622 \text{ left over.}}$$

Given:
- Maximum capacity of truck = 8,250 kg
- Cement loaded = 3,675 kg
- Steel loaded = 2,850 kg

(a) Total weight loaded:
$$3{,}675 + 2{,}850$$
- Ones: $5+0=5$
- Tens: $7+5=12$, write 2, carry 1
- Hundreds: $6+8+1=15$, write 5, carry 1
- Thousands: $3+2+1=6$

$$3{,}675 + 2{,}850 = 6{,}525 \text{ kg}$$

$$\boxed{\text{Total weight loaded} = 6{,}525 \text{ kg}}$$

(b) Remaining capacity:
$$8{,}250 - 6{,}525$$
- Ones: $0-5$: borrow. $10-5=5$
- Tens: $4-2-1=1$ (after borrowing)
- Hundreds: $2-5$: borrow. $12-5-1=6$... Let us redo:

$8{,}250 - 6{,}525$:
- Ones: $0-5$: borrow from tens. $10-5=5$
- Tens: $4-2=2$ (after lending 1, tens becomes 4; $4-2=2$... wait: tens digit of 8250 is 5, after lending becomes 4; $4-2=2$)
- Hundreds: $2-5$: borrow from thousands. $12-5=7$... (hundreds of 8250 is 2, after lending becomes 1; $11-5=6$... )

Let me compute directly: $8{,}250 - 6{,}525 = 1{,}725$

Verification: $6{,}525 + 1{,}725 = 8{,}250$ ✓

$$\boxed{\text{The truck can carry } 1{,}725 \text{ kg more.}}$$

We need to find what number added to 32 gives 100.
$$100 - 32 = 68$$
$$\boxed{32 + 68 = 100}$$

Using the method: subtract 59 from 100.
$$100 - 59 = 41$$
$$\boxed{59 + 41 = 100}$$

(a) $1{,}000 - 180 = 820$
$$\boxed{180 + 820 = 1{,}000}$$

(b) $1{,}000 - 760 = 240$
$$\boxed{760 + 240 = 1{,}000}$$

(c) $1{,}000 - 400 = 600$
$$\boxed{400 + 600 = 1{,}000}$$

Note on units digit 0: When the units digit is 0, we only need to subtract the remaining digits from 10 (or 100 for thousands). The method still works — we just get 0 in the units place of the answer.

Quick method to subtract 9: Subtract 10 and add 1 (since $9 = 10 - 1$).
Quick method to subtract 99: Subtract 100 and add 1 (since $99 = 100 - 1$).

(a) $67 - 9 = 67 - 10 + 1 = 57 + 1 = \mathbf{58}$

(b) $83 - 9 = 83 - 10 + 1 = 73 + 1 = \mathbf{74}$

(c) $144 - 9 = 144 - 10 + 1 = 134 + 1 = \mathbf{135}$

(d) $187 - 99 = 187 - 100 + 1 = 87 + 1 = \mathbf{88}$

(e) $247 - 99 = 247 - 100 + 1 = 147 + 1 = \mathbf{148}$

(f) $763 - 99 = 763 - 100 + 1 = 663 + 1 = \mathbf{664}$

Concept: If $a - b = c$, then $b = a - c$.

(a) $32 - \_ = 9 \Rightarrow \_ = 32 - 9 = \mathbf{23}$

(b) $56 - \_ = 9 \Rightarrow \_ = 56 - 9 = \mathbf{47}$

(c) $877 - \_ = 99 \Rightarrow \_ = 877 - 99 = 877 - 100 + 1 = 777 + 1 = \mathbf{778}$

(d) $666 - \_ = 99 \Rightarrow \_ = 666 - 99 = 666 - 100 + 1 = 566 + 1 = \mathbf{567}$

Palindrome numbers read the same from left to right and right to left.

Between 100 and 200:
A 3-digit palindrome has the form $\overline{aba}$ where $a=1$ (since numbers are between 100–200).
So: 101, 111, 121, 131, 141, 151, 161, 171, 181, 191.

$$\boxed{101, 111, 121, 131, 141, 151, 161, 171, 181, 191}$$

Between 900 and 1,200:
- 3-digit palindromes with $a=9$: 909, 919, 929, 939, 949, 959, 969, 979, 989, 999
- 4-digit palindromes of the form $\overline{abba}$ between 1000 and 1200: $a=1$, $b=0$: 1001; $b=1$: 1111.
(1221 > 1200, so excluded)

$$\boxed{909, 919, 929, 939, 949, 959, 969, 979, 989, 999, 1001, 1111}$$

Between 25,000 and 27,000:
5-digit palindromes of the form $\overline{abcba}$:
- Starting with 25: $25\_ 52$ → $25052, 25152, 25252, 25352, 25452, 25552, 25652, 25752, 25852, 25952$
- Starting with 26: $26\_ 62$ → $26062, 26162, 26262, 26362, 26462, 26562, 26662, 26762, 26862, 26962$

$$\boxed{25052, 25152, 25252, 25352, 25452, 25552, 25652, 25752, 25852, 25952,}$$
$$\boxed{26062, 26162, 26262, 26362, 26462, 26562, 26662, 26762, 26862, 26962}$$

Grid 1 — All rows and columns increasing (inc):

Each row increases left to right, each column increases top to bottom.

$$\begin{array}{|c|c|c|} \hline 1 & 2 & 3 \\ \hline 4 & 5 & 6 \\ \hline 7 & 8 & 9 \\ \hline \end{array}$$

Grid 2 — All rows and columns decreasing (dec):

Each row decreases left to right, each column decreases top to bottom.

$$\begin{array}{|c|c|c|} \hline 9 & 8 & 7 \\ \hline 6 & 5 & 4 \\ \hline 3 & 2 & 1 \\ \hline \end{array}$$

Grid 3 — Rows decreasing, columns decreasing (dec/dec):

$$\begin{array}{|c|c|c|} \hline 9 & 8 & 7 \\ \hline 6 & 5 & 4 \\ \hline 3 & 2 & 1 \\ \hline \end{array}$$

Grid 4 — Rows decreasing, columns increasing (dec rows / inc columns):

Each row decreases left to right; each column increases top to bottom.

$$\begin{array}{|c|c|c|} \hline 3 & 2 & 1 \\ \hline 6 & 5 & 4 \\ \hline 9 & 8 & 7 \\ \hline \end{array}$$

Grid 5 — Rows increasing, columns decreasing (inc rows / dec columns):

Each row increases left to right; each column decreases top to bottom.

$$\begin{array}{|c|c|c|} \hline 7 & 8 & 9 \\ \hline 4 & 5 & 6 \\ \hline 1 & 2 & 3 \\ \hline \end{array}$$

*(Note: The exact grid configurations depend on the arrow directions shown in the figures, which are not fully visible. The above are standard solutions for the described conditions.)*

Rule: A number is even if its units (ones) digit is 0, 2, 4, 6, or 8.

- (a) 297 — units digit 7 → Odd
- (b) 498 — units digit 8 → Even ✓
- (c) 724 — units digit 4 → Even ✓
- (d) 100 — units digit 0 → Even ✓
- (e) 199 — units digit 9 → Odd
- (f) 789 — units digit 9 → Odd
- (g) 49 — units digit 9 → Odd
- (h) 6,893 — units digit 3 → Odd
- (i) 846 — units digit 6 → Even ✓
- (j) 111 — units digit 1 → Odd
- (k) 222 — units digit 2 → Even ✓
- (l) 1,023 — units digit 3 → Odd

Even numbers: (b) 498, (c) 724, (d) 100, (i) 846, (k) 222

Adding 2 to 18 (even):
$18 + 2 = 20$
In the paired arrangement, 18 has all pairs (no leftover). Adding 2 adds one more complete pair. The arrangement still has all pairs — it remains even. The number of pairs increases by 1.

Adding 2 to 23 (odd):
$23 + 2 = 25$
In the paired arrangement, 23 has 11 pairs and 1 leftover. Adding 2 adds one more complete pair. The arrangement still has 1 leftover — it remains odd. The number of pairs increases by 1.

Conclusion: Adding 2 to any number does not change whether it is even or odd. The parity (even/odd nature) stays the same.

(a) Even + Even:

Example pairs and sums:
- $12 + 6 = 18$ (even)
- $4 + 8 = 12$ (even)
- $10 + 14 = 24$ (even)
- $20 + 6 = 26$ (even)
- $100 + 50 = 150$ (even)

Observation: The sum of two even numbers is always even.

*Why:* In the paired arrangement, both numbers have complete pairs. Combining them still gives complete pairs — no leftover.

(b) Odd + Odd:

Example pairs and sums:
- $13 + 9 = 22$ (even)
- $3 + 7 = 10$ (even)
- $11 + 5 = 16$ (even)
- $15 + 21 = 36$ (even)
- $1 + 99 = 100$ (even)

Observation: The sum of two odd numbers is always even.

*Why:* Each odd number has one leftover in the paired arrangement. When combined, the two leftovers pair up, giving a complete arrangement — no leftover.

(c) Odd + Even:

Example pairs and sums:
- $7 + 12 = 19$ (odd)
- $3 + 8 = 11$ (odd)
- $5 + 10 = 15$ (odd)
- $9 + 4 = 13$ (odd)
- $11 + 6 = 17$ (odd)

Observation: The sum of an odd number and an even number is always odd.

*Why:* The even number has complete pairs; the odd number has one leftover. When combined, the leftover remains — the sum is odd.

Given:
- ₹1 coins: 8 (total value = ₹8)
- ₹2 coins: 9 (total value = ₹18)
- ₹5 coins: 5 (total value = ₹25)
- Total available = ₹8 + ₹18 + ₹25 = ₹51
- Amount needed = ₹38

Finding combinations that sum to ₹38:

Combination 1: Use all 5 coins of ₹5 = ₹25; need ₹13 more.
$₹13 = 9 \times ₹1 + 2 \times ₹2$? No, only 8 coins of ₹1.
$₹13 = 8 \times ₹1 + 1 \times ₹5$? Already used all ₹5 coins.
$₹13 = 6 \times ₹2 + 1 \times ₹1 = ₹12 + ₹1 = ₹13$ ✓ (uses 6 of 9 ₹2 coins and 1 of 8 ₹1 coins)

Combination 1: $5 \times ₹5 + 6 \times ₹2 + 1 \times ₹1 = 25 + 12 + 1 = ₹38$ ✓

Combination 2: Use 4 coins of ₹5 = ₹20; need ₹18 more.
$₹18 = 9 \times ₹2 = ₹18$ ✓

Combination 2: $4 \times ₹5 + 9 \times ₹2 + 0 \times ₹1 = 20 + 18 + 0 = ₹38$ ✓

Combination 3: Use 4 coins of ₹5 = ₹20; need ₹18 more.
$₹18 = 7 \times ₹2 + 4 \times ₹1 = 14 + 4 = ₹18$ ✓

Combination 3: $4 \times ₹5 + 7 \times ₹2 + 4 \times ₹1 = 20 + 14 + 4 = ₹38$ ✓

*(Many more combinations are possible. Students should explore systematically.)*

Pattern:
- Odd number of presses → torch is ON
- Even number of presses → torch is OFF

After the 23rd press:
23 is an odd number.
$$\boxed{\text{The torch will be ON after the 23rd press.}}$$

Torch is ON for: All odd-numbered presses: 1, 3, 5, 7, 9, 11, 13, ...

Torch is OFF for: All even-numbered presses: 2, 4, 6, 8, 10, 12, ...

Given data:
- Mount Kanchenjunga: 8,586 m
- Mount Everest: 8,848 m
- Mount Makalu: 8,485 m
- Mount Lhotse: 8,516 m
- Mount Kilimanjaro: 5,895 m
- Mount Elbrus: 5,642 m
- Mount Annapurna I: 8,091 m

(a) Highest peak:
Comparing all heights: 8,848 m is the greatest.
$$\boxed{\text{The highest peak she climbed is Mount Everest (8,848 m).}}$$

(b) Difference between highest and lowest peaks:
- Highest = Mount Everest = 8,848 m
- Lowest = Mount Elbrus = 5,642 m

$$8{,}848 - 5{,}642 = 3{,}206 \text{ m}$$

$$\boxed{\text{The difference in height is } 3{,}206 \text{ m.}}$$

(c) Difference between Mount Elbrus and Mount Kanchenjunga:
$$8{,}586 - 5{,}642 = 2{,}944 \text{ m}$$

$$\boxed{\text{The difference is } 2{,}944 \text{ m.}}$$

(d) Year of birth:
Priyanka was 20 years old in 2013.
$$2013 - 20 = 1993$$

$$\boxed{\text{Priyanka Mohite was born in the year 1993.}}$$

Chittoor, A.P.:
Printed = 18,225; Attended = 18,104
Since 18{,}225 > 18{,}104, certificates were sufficient.
Excess = $18{,}225 - 18{,}104 = 121$
$$\boxed{\text{Chittoor: Sufficient. Excess certificates} = 121}$$

Jaunpur, U.P.:
Printed = 19,043; Attended = 19,265
Since 19{,}043 < 19{,}265, certificates were insufficient.
Shortfall = $19{,}265 - 19{,}043 = 222$
$$\boxed{\text{Jaunpur: Insufficient. Certificates fell short by } 222}$$

Raigad, Maharashtra:
Printed = 20,863; Attended = 19,974
Since 20{,}863 > 19{,}974, certificates were sufficient.
Excess = $20{,}863 - 19{,}974 = 889$
$$\boxed{\text{Raigad: Sufficient. Excess certificates} = 889}$$

$$2{,}009 + 7{,}388$$
- Ones: $9+8=17$, write 7, carry 1
- Tens: $0+8+1=9$
- Hundreds: $0+3=3$
- Thousands: $2+7=9$
$$\boxed{2{,}009 + 7{,}388 = 9{,}397}$$

$$26{,}444 + 71{,}111$$
- Ones: $4+1=5$
- Tens: $4+1=5$
- Hundreds: $4+1=5$
- Thousands: $6+1=7$
- Ten-thousands: $2+7=9$
$$\boxed{26{,}444 + 71{,}111 = 97{,}555}$$

$$777 + 888$$
- Ones: $7+8=15$, write 5, carry 1
- Tens: $7+8+1=16$, write 6, carry 1
- Hundreds: $7+8+1=16$
$$\boxed{777 + 888 = 1{,}665}$$

$$1{,}234 + 1{,}234$$
- Ones: $4+4=8$
- Tens: $3+3=6$
- Hundreds: $2+2=4$
- Thousands: $1+1=2$
$$\boxed{1{,}234 + 1{,}234 = 2{,}468}$$

$$56 + 56{,}789$$
- Ones: $6+9=15$, write 5, carry 1
- Tens: $5+8+1=14$, write 4, carry 1
- Hundreds: $0+7+1=8$
- Thousands: $0+6=6$
- Ten-thousands: $0+5=5$
$$\boxed{56 + 56{,}789 = 56{,}845}$$

$$777 + 77{,}777$$
- Ones: $7+7=14$, write 4, carry 1
- Tens: $7+7+1=15$, write 5, carry 1
- Hundreds: $7+7+1=15$, write 5, carry 1
- Thousands: $0+7+1=8$
- Ten-thousands: $0+7=7$
$$\boxed{777 + 77{,}777 = 78{,}554}$$

$$5{,}922 + 9{,}221$$
- Ones: $2+1=3$
- Tens: $2+2=4$
- Hundreds: $9+2=11$, write 1, carry 1
- Thousands: $5+9+1=15$
$$\boxed{5{,}922 + 9{,}221 = 15{,}143}$$

$$4{,}321 + 8{,}765$$
- Ones: $1+5=6$
- Tens: $2+6=8$
- Hundreds: $3+7=10$, write 0, carry 1
- Thousands: $4+8+1=13$
$$\boxed{4{,}321 + 8{,}765 = 13{,}086}$$

$$50{,}050 + 55{,}000$$
- Ones: $0+0=0$
- Tens: $5+0=5$
- Hundreds: $0+0=0$
- Thousands: $0+5=5$
- Ten-thousands: $5+5=10$
$$\boxed{50{,}050 + 55{,}000 = 1{,}05{,}050}$$

$$458 - 226$$
- Ones: $8-6=2$
- Tens: $5-2=3$
- Hundreds: $4-2=2$
$$\boxed{458 - 226 = 232}$$

$$7{,}777 - 4{,}449$$
- Ones: $7-9$: borrow. $17-9=8$
- Tens: $6-4=2$ (after lending 1)
- Hundreds: $7-4=3$
- Thousands: $7-4=3$
$$\boxed{7{,}777 - 4{,}449 = 3{,}328}$$

$$65{,}447 - 47{,}299$$
- Ones: $7-9$: borrow. $17-9=8$
- Tens: $3-9$: borrow. $13-9=4$ (tens of 65447 is 4, after lending becomes 3)
- Hundreds: $3-2=1$ (hundreds of 65447 is 4, after lending becomes 3; $3-2=1$)
- Thousands: $5-7$: borrow. $15-7=8$ (thousands of 65447 is 5; $5-7$ needs borrow)
- Ten-thousands: $5-4=1$ (after lending 1)

Let me compute directly: $65{,}447 - 47{,}299 = 18{,}148$

Verification: $47{,}299 + 18{,}148 = 65{,}447$ ✓
$$\boxed{65{,}447 - 47{,}299 = 18{,}148}$$

$$1{,}234 - 123$$
- Ones: $4-3=1$
- Tens: $3-2=1$
- Hundreds: $2-1=1$
- Thousands: $1-0=1$
$$\boxed{1{,}234 - 123 = 1{,}111}$$

$$12{,}345 - 1{,}234$$
- Ones: $5-4=1$
- Tens: $4-3=1$
- Hundreds: $3-2=1$
- Thousands: $2-1=1$
- Ten-thousands: $1-0=1$
$$\boxed{12{,}345 - 1{,}234 = 11{,}111}$$

$$56{,}789 - 56$$
- Ones: $9-6=3$
- Tens: $8-5=3$
- Hundreds: $7-0=7$
- Thousands: $6-0=6$
- Ten-thousands: $5-0=5$
$$\boxed{56{,}789 - 56 = 56{,}733}$$

$$87{,}326 - 11{,}111$$
- Ones: $6-1=5$
- Tens: $2-1=1$
- Hundreds: $3-1=2$
- Thousands: $7-1=6$
- Ten-thousands: $8-1=7$
$$\boxed{87{,}326 - 11{,}111 = 76{,}215}$$

$$878 - 52$$
- Ones: $8-2=6$
- Tens: $7-5=2$
- Hundreds: $8-0=8$
$$\boxed{878 - 52 = 826}$$

$$749 - 222$$
- Ones: $9-2=7$
- Tens: $4-2=2$
- Hundreds: $7-2=5$
$$\boxed{749 - 222 = 527}$$

Given:
- Savings = ₹92,375
- Cost of cow = ₹26,000
- Cost of goat = ₹17,000
- Cost of milking machine = ₹19,873

Total expenditure:
$$26{,}000 + 17{,}000 + 19{,}873$$

Step 1: $26{,}000 + 17{,}000 = 43{,}000$
Step 2: $43{,}000 + 19{,}873 = 62{,}873$

Comparison:
Savings = ₹92,375; Total needed = ₹62,873

Since 92{,}375 > 62{,}873, Ambrish has enough money.

Extra money:
$$92{,}375 - 62{,}873 = 29{,}502$$

$$\boxed{\text{Yes, Ambrish has enough money. He has } ₹29{,}502 \text{ more than he needs.}}$$

Given:
- Produced in a day = 54,000
- Order placed = 85,300

Additional production needed:
$$85{,}300 - 54{,}000$$
- Ones: $0-0=0$
- Tens: $0-0=0$
- Hundreds: $3-0=3$
- Thousands: $5-4=1$
- Ten-thousands: $8-5=3$

$$85{,}300 - 54{,}000 = 31{,}300$$

$$\boxed{\text{The factory needs to produce } 31{,}300 \text{ more nuts and bolts.}}$$

Given:
- Virat Kohli's runs = 27,599
- Virat has 6,758 runs less than Sachin

Sachin's runs = Virat's runs + 6,758
$$27{,}599 + 6{,}758$$
- Ones: $9+8=17$, write 7, carry 1
- Tens: $9+5+1=15$, write 5, carry 1
- Hundreds: $5+7+1=13$, write 3, carry 1
- Thousands: $7+6+1=14$, write 4, carry 1
- Ten-thousands: $2+0+1=3$

$$27{,}599 + 6{,}758 = 34{,}357$$

$$\boxed{\text{Sachin Tendulkar has scored } 34{,}357 \text{ runs.}}$$