Chapter 4 of 15

Updated 6 August 2026

NCERT Solutions

We the Travellers—II — NCERT Solutions

CBSE · Class 5 · Mathematics

NCERT Solutions for We the Travellers—II, CBSE Class 5 Mathematics: 76 textbook questions solved step by step.

40 questions56 flashcards9 formulas & key relations5 concepts

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76 Questions Solved · 13 Sections

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Fuel Arithmetic

1A lorry has 28 litres of fuel in its tank. An additional 75 litres is filled. What is the total quantity of fuel in the lorry?Show solution

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.

2Find the sum of 49 and 89.Show solution

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.

Let Us Solve (Addition)

a15 + 79Show solution

$15 + 79$

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

Or by column addition:

Ones: $5 + 9 = 14$ , write 4, carry 1
Tens: $1 + 7 + 1 = 9$

$\boxed{15 + 79 = 94}$

b46 + 99Show solution

$46 + 99$

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

$\boxed{46 + 99 = 145}$

c38 + 35Show solution

$38 + 35$

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

$\boxed{38 + 35 = 73}$

d5 + 89Show solution

$5 + 89$

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

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

$\boxed{5 + 89 = 94}$

e76 + 28Show solution

$76 + 28$

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

$\boxed{76 + 28 = 104}$

f69 + 20Show solution

$69 + 20$

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

$\boxed{69 + 20 = 89}$

Relationship Between Addition and Subtraction

1Find the relationship between the numbers in the given statements and fill in the blanks appropriately.
(a) If 46 + 21 = 67, then 67 − 21 = ______ and 67 − 46 = ______.
(b) If 198 − 98 = 100, then 100 + ______ = 198 and 198 − ______ = 98.
(c) If 189 + 98 = 287, then 287 − 98 = ______ and 287 − 189 = ______.
(d) If 872 − 672 = 200, then 200 + ______ = 872 and 872 − ______ = 672.Show solution

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:

$200 + \mathbf{672} = 872$
$872 - \mathbf{200} = 672$

2In each of the following, write the subtraction and addition sentences that follow from the given sentence.
(a) If 78 + 164 = 242, then __________.
(b) If 462 + 839 = 1301, then __________.
(c) If 921 − 137 = 784, then __________.
(d) If 824 − 234 = 590, then __________.Show solution

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:
$590 + 234 = 824 \quad \text{and} \quad 824 - 590 = 234$

Let Us Solve (Subtraction)

1What is the difference between 82 and 37? Check your answer: Is 37 + ___ = 82?Show solution

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.

257 − 11 = ?Show solution

$57 - 11$

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

$\boxed{57 - 11 = 46}$

323 − 19 = ?Show solution

$23 - 19$

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

$\boxed{23 - 19 = 4}$

449 − 21 = ?Show solution

$49 - 21$

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

$\boxed{49 - 21 = 28}$

556 − 18 = ?Show solution

$56 - 18$

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

$\boxed{56 - 18 = 38}$

693 − 35 = ?Show solution

$93 - 35$

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

$\boxed{93 - 35 = 58}$

784 − 23 = ?Show solution

$84 - 23$

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

$\boxed{84 - 23 = 61}$

870 − 43 = ?Show solution

$70 - 43$

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

$\boxed{70 - 43 = 27}$

965 − 47 = ?Show solution

$65 - 47$

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

$\boxed{65 - 47 = 18}$

Sums of Consecutive Numbers

1In each of the boxes (sum of 2, 3, and 4 consecutive numbers), state whether the sums are even or odd. Explain why this is happening.Show solution

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.

2What is the difference between two successive sums in each box? Is it the same throughout?Show solution

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.

3What will be the difference between two successive sums for—
(a) 5 consecutive numbers
(b) 6 consecutive numbersShow solution

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.

Let Us Solve (Large Number Addition)

1a238 + 367Show solution

$\begin{array}{r} 238 \\ +367 \\ \hline \end{array}$

Ones: $8+7=15$ , write 5, carry 1
Tens: $3+6+1=10$ , write 0, carry 1
Hundreds: $2+3+1=6$

$\boxed{238 + 367 = 605}$

1b1,234 + 12,345Show solution

$\begin{array}{r} 1{,}234 \\ +12{,}345 \\ \hline \end{array}$

Ones: $4+5=9$
Tens: $3+4=7$
Hundreds: $2+3=5$
Thousands: $1+2=3$
Ten-thousands: $0+1=1$

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

1c12 + 123Show solution

$\begin{array}{r} 12 \\ +123 \\ \hline \end{array}$

Ones: $2+3=5$
Tens: $1+2=3$
Hundreds: $0+1=1$

$\boxed{12 + 123 = 135}$

1d46,120 + 12,890Show solution

$\begin{array}{r} 46{,}120 \\ +12{,}890 \\ \hline \end{array}$

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: $4+1=5$

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

1e878 + 8,789Show solution

$\begin{array}{r} 878 \\ +8{,}789 \\ \hline \end{array}$

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: $0+8+1=9$

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

1f1,749 + 17,490Show solution

$\begin{array}{r} 1{,}749 \\ +17{,}490 \\ \hline \end{array}$

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: $0+1=1$

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

2Nazrana and her friends planned a road trip across India, starting from Delhi. They first drove to Mumbai, then Goa, then Hyderabad, and finally Puri. Look at the distances marked on the map and help them find the total distance travelled. (Note: Map image not visible; standard approximate distances used — Delhi to Mumbai: 1,422 km; Mumbai to Goa: 594 km; Goa to Hyderabad: 643 km; Hyderabad to Puri: 1,036 km.)Show solution

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.)

3Find 2 numbers among 5,205; 6,220; 7,095; 8,455; and 4,840 whose sum is closest to:
(a) 10,000
(b) 15,000
(c) 13,000
(d) 16,000Show solution

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.
$\boxed{8{,}455 + 7{,}095 = 15{,}550 \text{ (closest to 16,000)}}$

Let Us Solve (Large Number Subtraction)

1a4,578 – 2,222Show solution

$\begin{array}{r} 4{,}578 \\ -2{,}222 \\ \hline \end{array}$

Ones: $8-2=6$
Tens: $7-2=5$
Hundreds: $5-2=3$
Thousands: $4-2=2$

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

1b15,324 – 11,780Show solution

$\begin{array}{r} 15{,}324 \\ -11{,}780 \\ \hline \end{array}$

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: $1 - 1 = 0$

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

1c5,423 – 423Show solution

$\begin{array}{r} 5{,}423 \\ -\phantom{0}423 \\ \hline \end{array}$

Ones: $3-3=0$
Tens: $2-2=0$
Hundreds: $4-4=0$
Thousands: $5-0=5$

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

1d123 – 12Show solution

$\begin{array}{r} 123 \\ -12 \\ \hline \end{array}$

Ones: $3-2=1$
Tens: $2-1=1$
Hundreds: $1-0=1$

$\boxed{123 - 12 = 111}$

1e77,777 – 777Show solution

$\begin{array}{r} 77{,}777 \\ -\phantom{00}777 \\ \hline \end{array}$

Ones: $7-7=0$
Tens: $7-7=0$
Hundreds: $7-7=0$
Thousands: $7-0=7$
Ten-thousands: $7-0=7$

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

1f826 – 752Show solution

$\begin{array}{r} 826 \\ -752 \\ \hline \end{array}$

Ones: $6-2=4$
Tens: $2-5$ : borrow. $12-5=7$
Hundreds: $7-7=0$

$\boxed{826 - 752 = 74}$

2Mary's train journey to Delhi.
Mary starts from Kolkata with ₹12,540. She spends ₹3,275 on food during her trip to Varanasi. In Varanasi, her uncle gives her ₹4,900. She then spends ₹2,645 on the train ticket to Delhi. She spends ₹1,275 on souvenirs in Delhi. How much money is Mary left with at the end of the Delhi trip?Show solution

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:
$11{,}520 - 1{,}275 = 10{,}245$

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

3Members of a school council have raised ₹70,500. They plan to set up a Maths Lab worth ₹39,785, buy library books worth ₹9,545, and purchase sports equipment worth ₹19,548.
(a) Estimate whether the school council has raised enough money.
(b) Check your estimate with calculations.Show solution

(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

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

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

4A truck can carry 8,250 kg of goods. A factory loads 3,675 kg of cement and 2,850 kg of steel on it.
(a) What is the total weight loaded onto the truck?
(b) How much more weight can the truck carry before reaching its maximum capacity?

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: $6{,}525 + 1{,}725 = 8{,}250$ ✓

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

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Quick Sums and Differences

1Help Sukanta fill in the blanks: 32 + ______ = 100

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2Try Piku's method for 59: 59 + ___ = 100

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3877 + ___ = 1,000 and 666 + ___ = 1,000

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44,103 + ___ = 10,000 and 5,555 + ___ = 10,000

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5(a) 180 + ___ = 1,000
(b) 760 + ___ = 1,000
(c) 400 + ___ = 1,000

(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.

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6Subtract 9 or 99 quickly:
(a) 67 − 9
(b) 83 − 9
(c) 144 − 9
(d) 187 − 99
(e) 247 − 99
(f) 763 − 99

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

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7Find the missing number to get 9 or 99 as the answer:
(a) 32 − ___ = 9
(b) 56 − ___ = 9
(c) 877 − ___ = 99
(d) 666 − ___ = 99

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

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Let Us Think and Solve

1List all palindrome numbers between 100 and 200.
List all palindrome numbers between 900 and 1,200.
List all palindrome numbers between 25,000 and 27,000.

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: $26\_ 62$ → $26062, 26162, 26262, 26362, 26462, 26562, 26662, 26762, 26862, 26962$

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

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2In a 3×3 grid, arrange the numbers 1 to 9 such that each row and each column has numbers in increasing (inc) order. Then fill the grid such that each row and column has numbers in decreasing (dec) order. Also fill grids with mixed inc/dec conditions as indicated.

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.)

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Even and Odd Numbers

1Circle the numbers that are even:
(a) 297 (b) 498 (c) 724 (d) 100 (e) 199 (f) 789 (g) 49 (h) 6,893 (i) 846 (j) 111 (k) 222 (l) 1,023

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

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2Observe the paired arrangement for 18 and 23.
Add 2 to 18. What changes or does not change in the arrangement?
Add 2 to 23. What changes or does not change in the arrangement?

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.

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3What do you notice about the sums in each of the following cases?
(a) 12 and 6 are a pair of even numbers. Choose 5 such pairs of even numbers. Add the numbers in each of the pairs.
(b) 13 and 9 are a pair of odd numbers. Choose 5 such pairs of odd numbers. Add the numbers in each of the pairs.
(c) 7 and 12 are a pair of odd and even numbers. Choose 5 such pairs of odd and even numbers. Add the numbers in each of the pairs.

(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.

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Let Us Think

1Jincy opened her piggy bank. She found 8 coins of ₹1, 9 coins of ₹2 and 5 coins of ₹5. She wants to buy stickers worth ₹38. What possible combination of coins can she use to pay the exact amount?

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.)

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2Raghu presses a torch switch. Press 1: ON. Press 2: OFF. Press 3: ON... Will the torch be ON or OFF after the 23rd press? For what number of presses will the torch be ON? For what number of presses will it be OFF?

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3Mountain climbing — Priyanka Mohite's peaks:
(a) Which is the highest peak she climbed?
(b) What is the difference in height between the highest and lowest peaks she has climbed?
(c) What is the difference between heights of Mount Elbrus and Mount Kanchenjunga?
(d) If Priyanka was 20 years old when she summited Mount Everest in 2013, in which year was she born?

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.
$2013 - 20 = 1993$

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

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Math Metric Mela

1For each district, find out if the number of certificates were sufficient. If insufficient, calculate how many certificates fell short. If extra, calculate how many certificates were in excess.
- Chittoor, A.P.: Printed 18,225; Attended 18,104
- Jaunpur, U.P.: Printed 19,043; Attended 19,265
- Raigad, Maharashtra: Printed 20,863; Attended 19,974

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 = $20{,}863 - 19{,}974 = 889$
$\boxed{\text{Raigad: Sufficient. Excess certificates} = 889}$

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Let Us Do

1aAdd: 2,009 + 7,388

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1bAdd: 26,444 + 71,111

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1cAdd: 777 + 888

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1dAdd: 1,234 + 1,234

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1eAdd: 56 + 56,789

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1f_firstAdd: 777 + 77,777

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1f_secondAdd: 5,922 + 9,221

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1gAdd: 4,321 + 8,765

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1hAdd: 50,050 + 55,000

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2aSubtract: 458 − 226

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2bSubtract: 7,777 − 4,449

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2cSubtract: 65,447 − 47,299

$65{,}447 - 47{,}299$

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

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2dSubtract: 1,234 − 123

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2eSubtract: 12,345 − 1,234

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2f_firstSubtract: 56,789 − 56

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2f_secondSubtract: 87,326 − 11,111

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2gSubtract: 878 − 52

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2hSubtract: 749 − 222

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3Ambrish saved ₹92,375 over a year to buy cows and goats. He buys a cow for ₹26,000 and a goat for ₹17,000. He also buys a milking machine for ₹19,873. Does he have enough money to buy these? How much more or less does he have than he needs?

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

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

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4A factory produces 54,000 nuts and bolts in a day. An order is placed for 85,300 nuts and bolts. How many more nuts and bolts does the factory need to produce to complete the order?

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: $8-5=3$

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

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

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5Virat Kohli has scored 27,599 runs. He has 6,758 runs less than Sachin Tendulkar. How many runs has Sachin Tendulkar scored?

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: $2+0+1=3$

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

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

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Frequently Asked Questions

What are the important topics in We the Travellers—II for CBSE Class 5 Mathematics?

Key topics in We the Travellers—II include Making sums equal, Fuel arithmetic and place value addition, Relationship between addition and subtraction, Consecutive numbers and their sums. Study these first, then practise questions on each for Class 5 exams.

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We the Travellers—II — NCERT Solutions

Studying We the Travellers—II? Get the full interactive chapter.

Fuel Arithmetic

Let Us Solve (Addition)

Relationship Between Addition and Subtraction

Let Us Solve (Subtraction)

Sums of Consecutive Numbers

Let Us Solve (Large Number Addition)

Let Us Solve (Large Number Subtraction)

Quick Sums and Differences

Let Us Think and Solve

Even and Odd Numbers

Let Us Think

Math Metric Mela

Let Us Do

Frequently Asked Questions

Sources & Official References

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Practice Quiz

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