Class 7 Computational Thinking Chapter 1: Large Numbers Around Us – Solutions

Large-Numbers-Around-Us
CLASS 7 • COMPUTATIONAL THINKING

Chapter 1: Large Numbers Around Us

Complete solutions to all 10 questions with simple, step-by-step explanations.

▶ Watch the Complete Solution Video

In this post: We solve all 10 questions from Class 7 Computational Thinking Chapter 1,
Large Numbers Around Us, including number formation, place value, logical reasoning,
digit arrangements and large-number puzzles.

Questions & Step-by-Step Solutions

Q1

Find the highest possible value of Y

‘X’ is a 6-digit number formed using exactly three different digits. One digit appears once,
another twice, and the third three times. When 2 lakhs are added to X, the resulting number Y
is still a 6-digit number. What is the highest possible value of Y?
A) 977889
B) 999999
C) 999988
D) 999887

Solution

Since 2 lakhs = 200000, to keep Y a 6-digit number, X must be less than 800000.

The largest possible X satisfying the digit-frequency condition is:

X = 799988

Here, 7 appears once, 9 appears three times and 8 appears twice.

Y = 799988 + 200000 = 999988

Answer: C) 999988

Q2

Product of the numbers on a telephone dial-pad

What is the product of all the numbers on a telephone dial-pad?
A) 12345
B) 32451
C) 362880
D) None of these

Solution

The product of the digits 1 through 9 is:

1 × 2 × 3 × 4 × 5 × 6 × 7 × 8 × 9 = 362880

Answer: C) 362880

Q3

Find the difference between X and Y

Sam forms two different 6-digit numbers X and Y using digits from 0 to 9 without repetition. Exactly two digits are common. The digit 9 is in the same position in both numbers. Y ends in 0, has only one even digit, and its digits are in descending order. 6 is present in X, and the number of even digits on its left equals the number on its right. If X is the largest possible number, find X − Y.
A) 9310
B) 9210
C) 8420
D) 8310

Solution

Y ends in 0 and has only one even digit, so its other five digits must be the five odd digits
9, 7, 5, 3 and 1. Since they are in descending order:

Y = 975310

For X, 9 must be in the first position. To make X as large as possible while keeping exactly
two common digits and balancing the even digits around 6, we get:

X = 984620
X − Y = 984620 − 975310 = 9310

Answer: A) 9310

Q4

Find the value represented by the grey circle

If each of the rows follows the same theme, what will come in place of “?”?

Class 7 Computational Thinking Chapter 1 Question 4

Solution

Compare the first position where the two numbers differ in each row.

  • Row 1 gives grey digit ≤ 7.
  • Row 2 gives grey digit ≤ 3.
  • Row 3 gives grey digit ≤ 2.

For the last row, compare:

○○506○ < ○5○60○

The first difference is in the second position. Therefore the grey digit must be less than 5,
so its greatest possible value is 4.

Answer: B) ○ ≤ 4

Q5

Complete the number-word pattern

What will come in place of “?”?

Class 7 Computational Thinking Chapter 1 Question 5

Solution

The pattern subtracts the number of letters removed from the number of zeros in the corresponding
large number.

  • Crore has 7 zeros; “re” has 2 letters → 5 zeros.
  • Million has 6 zeros; “ion” has 3 letters → 3 zeros.
  • Thousand has 3 zeros; “d” has 1 letter → 2 zeros.
  • Billion has 9 zeros; “llion” has 5 letters → 4 zeros.

For lakh: 1 lakh has 5 zeros. Remove “kh” (2 letters):

5 − 2 = 3

Answer: C) 000

Q6

Arrange the six number tokens

The six tokens are 2, 5, 8, 1, 4, 7. Rearrange them to form a 6-digit number such that:
● The difference between the first and last digits is as small as possible
● The hundreds digit is double the thousands digit
● No consecutive digits appear next to each other
● All given tokens must be used exactly once
How many different 6-digit numbers between 2 lakhs and 8 lakhs can be formed under these conditions?
A) 1
B) 2
C) 3
D) More than 3

Solution

The hundreds digit must be double the thousands digit. After checking the arrangements between 2 lakhs and 8 lakhs and applying the remaining conditions, the smallest possible difference between the first and last digits is 1.

The two valid numbers are:

274851
752418

Both use all six tokens, have hundreds digit double the thousands digit, contain no consecutive adjacent digits, and have a first-to-last digit difference of 1.

Therefore, there are exactly two valid numbers.

Answer: B) 2

Q7

Find the maximum possible difference

Two five-digit numbers are formed using different digits from 0 to 9:

Number 1: _ 1 _ 9 _

Number 2: _ 4 _ 6 _

No two consecutive digits of the number series are present in the same number. What is the maximum possible difference?
A) 47525
B) 52663
C) 52429
D) 46733

Solution

The maximum difference is obtained with the valid pair:

Number 1 = 31597
Number 2 = 84260
84260 − 31597 = 52663

Answer: B) 52663

Q8

Determine E’s number card

A, B, C, D and E each have a different card among one hundred, one thousand, ten thousand, one lakh and ten lakhs. Use the clues to determine which card E has.

Solution

Number of zeros:

100 → 2 zeros
1,000 → 3 zeros
10,000 → 4 zeros
1,00,000 → 5 zeros
10,00,000 → 6 zeros

C must have an even number of zeros. If C had 6 zeros, D would need 8 zeros, which is impossible. Therefore C has 4 zeros.

  • C = 10,000
  • A = 100
  • D = 10,00,000
  • B = 1,000
  • Therefore E = 1,00,000

Answer: C) One lakh

Q9

Find Calvin’s profession and salary

Alex, Bob, Calvin and David have different professions—Doctor, Lawyer, Cricketer and Engineer— and different salaries—₹40,000, ₹50,000, ₹60,000 and ₹70,000 per year. Use the clues to determine Calvin’s profession and salary.

Solution

Bob is the lawyer and David is the engineer. The cricketer earns ₹70,000.
Alex earns more than Bob, while the doctor earns more than David.

Testing the possible salary and profession combinations consistently with all clues gives:

Calvin = Doctor
Calvin’s salary = ₹50,000

Answer: C) Doctor – ₹50,000

Q10

Find Nisha’s 6-digit password

The clues to the 6-digit password are:
35 tens, 31 thousands, 25 ones, 423 hundreds, 14 ten thousands.
What is Nisha’s password?
A) 213625
B) 203675
C) 212375
D) 213675

Solution

Convert each clue into its numerical value:

35 tens = 35 × 10 = 350
31 thousands = 31 × 1000 = 31,000
25 ones = 25
423 hundreds = 423 × 100 = 42,300
14 ten thousands = 14 × 10,000 = 140,000

Now add them:

140,000 + 42,300 + 31,000 + 350 + 25 = 213,675

Answer: D) 213675

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