CLASS 7 · GANITA PRAKASH PART I · CHAPTER 1
Read, compare, estimate and calculate with lakhs, crores and other large numbers.
59 worked answers · short steps first, explanations when you need them
Textbook pages 2–5
Question 1
Add one after the largest number with a given number of digits.
999+1=1,000; 9,999+1=10,000; 99,999+1=1,00,000.
Question 2
Divide 1,00,000 by the number tasted each day, then divide by 365 for years.
One per day: 1,00,000 days ≈274 years. Two per day: 50,000 days ≈137 years. Three per day: about 33,334 days ≈91.3 years.
Question 3
Subtract the smaller amount from the larger amount in each comparison.
25,000 less; 6,000 more; increase = 31,000.
Question 4
Each floor is about 4 m, so multiply 10×4.
Subtract the building height from each larger height.
Building ≈40 m; the statue is 140 m taller; the waterfall is 410 m taller. The waterfall is about 113 such 4 m floors high.
Question 5
Read the groups as lakh, thousand, then the final three digits.
Three lakh six hundred; five lakh four thousand eighty-five; twenty-seven lakh thirty thousand; seventy lakh fifty-three thousand one hundred thirty-eight.
Question 6
Place each named amount in its correct place-value group; include zeroes for empty places.
1,23,456; 4,07,704; 50,05,050; 10,00,235.
Textbook pages 5–7
Question 1
Divide each target by 1,000.
10; 53; 90; 100; 153 presses.
Question 2
Divide each target by 10.
50; 78; 100; 370; 1,000; 10,000; 435 presses.
Question 3
Divide each target by 100.
4; 37; 100; 530; 900; 976; 1,000; 582 presses.
Question 4
The ten-button outputs are multiples of 10, and thousand-button outputs are multiples of 1,000.
The hundred-button outputs are also multiples of 100, so they are multiples of 10.
No. Every multiple of 100 is also a multiple of 10, so the tens calculator can make it; not every multiple of 100 is a multiple of 1,000.
Question 5
Break each total into sums of allowed button values; regroup to get another expression.
5,072=5×1,000+7×10+2; also 4×1,000+10×100+7×10+2. 8,300=8×1,000+3×100; also 5×1,000+33×100.
Question 6
Use the largest place-value buttons first.
The digit in each place gives the required press count.
5,072: 14 presses; 8,300: 11; 40,629: 21; 56,354: 23; 66,666: 30; 3,67,813: 28.
Question 7
For the largest, balance +100, +10 and +1 presses so 30 clicks stay below 1,000.
For the smallest, use the fewest +10 presses needed to reach 100, then fill the remaining clicks with +1.
For 997, regroup the place-value total to change the number of clicks.
Largest: 8×100+19×10+3×1=993 (30 clicks). Smallest: 8×10+22×1=102 (30 clicks). Also, 997=8×100+19×10+7×1 uses 34 clicks, different from 9×100+9×10+7×1, which uses 25.
Question 321-clicks
Add the button contributions in each plan.
Compare each total with 321.
32×10+1=321. Also, 2×100+12×10+1=321. Both plans make 321.
Question figure-it-out-two-ways
Write each target as a sum of available button values.
Regroup one place-value amount to make a second plan.
Examples: 8,300=8×1,000+3×100=5×1,000+33×100. 40,629=40×1,000+6×100+2×10+9=30×1,000+106×100+29. 56,354=56×1,000+3×100+5×10+4=46×1,000+103×100+54. 66,666=66×1,000+6×100+66=6×10,000+6×1,000+6×100+6×10+6. 3,67,813=3×1,00,000+6×10,000+7×1,000+8×100+10+3=36×10,000+78×1,000+13.
Textbook pages 8–9
Question 1
Use 1 lakh=1,00,000; 1 crore=1,00,00,000; 1 billion=1,000,000,000.
One thousand lakh=10,00,00,000 (8 zeroes); one lakh=5; one hundred thousand=5; one crore=7; one billion=9.
Question 2
Group digits from the right: 3-2-2 in Indian notation, groups of 3 internationally.
Indian: 40,50,678, forty lakh fifty thousand six hundred seventy-eight. International: 4,050,678, four million fifty thousand six hundred seventy-eight.
Question 3
Re-group from the right using 3-2-2 for Indian notation and 3-3-3 for international notation.
4,81,21,620; 2,00,22,002; 24,68,13,579; 34,50,00,543; 1,02,03,04,050. International: 48,121,620; 20,022,002; 246,813,579; 345,000,543; 1,020,304,050.
Question 4
Translate each named group into its numerical value, then add the groups.
1,01,01,010; 1,00,10,01,001; 10,20,30,040; 9,08,07,00,600.
Question 5
Convert both quantities in each pair to the same unit or numeral.
30 thousand < 3 lakh; 500 lakh > 5 million; 800 thousand < 8 million; 640 crore < 60 billion.
Textbook pages 10–13
Question 1
Match the rounding choice to the consequence of being slightly above or below.
Examples: round up when ordering food for guests; round down when estimating an affordable purchase; either direction for approximate travel distance; use exact values for a bank balance.
Question 2
Look at the digit immediately to the right of the place being rounded. Round up if it is 5 or more; otherwise round down.
3,87,69,957 → 3,87,70,000; 3,87,70,000; 3,88,00,000; 3,90,00,000; 4,00,00,000. 29,05,32,481 → 29,05,32,000; 29,05,30,000; 29,05,00,000; 29,10,00,000; 29,00,00,000.
Question 3
Add the numbers by place value.
Compare the exact sum with each benchmark.
Sum=8,82,810: above 8,00,000 and 8,50,000, but below 8,83,128. The 9,00,000 estimate is closer than 8,00,000? Distances: 17,190 vs 82,810, so 9,00,000 is closer.
Question 4
Subtract by place value.
Compare the result with the given benchmarks.
Difference=9,73,108: it is above all three benchmarks. It is closer to 10,00,000 (26,892 away) than to 9,00,000 (73,108 away).
Question 5
Subtract the 2001 population from the 2011 population for each city.
Compare ratios for doubling and divide Mumbai’s population by Patna’s.
Most listed cities grew. Pune: 31,15,431 in 2011, about 6 lakh higher than 2001. Bengaluru increased most (41,24,644). Bengaluru, Hyderabad, Surat and Vadodara almost doubled. Mumbai/Patna ≈7.39.
Question all-five-neighbours
The nearest-thousand interval around 5,00,00,000 runs from 4,99,99,500 to 5,00,00,499.
This interval is narrower than the corresponding intervals for the larger rounding units, so it is the overlap.
For example, 5,00,00,001 works. The overlap contains every integer from 4,99,99,500 through 5,00,00,499, inclusive: 5,00,00,499−4,99,99,500+1=1,000 numbers.
Textbook pages 14–18
Question 1
Regroup the factors: 2×50=100; 125=1,000÷8; 125×8=1,000 and 40×25=1,000.
176,800; 9,000; 1,000,000.
Question 2
Use 25=100÷4 and 250=1,000÷4.
300; 6,000; 30,000; 30,000; one pair is 1,200×1,00,000.
Question 3
Use smallest and largest possible factors to bound the product.
Two 2-digit numbers give 3 or 4 digits. Two 3-digit numbers cannot give a 4-digit product (smallest is 100×100=10,000). A 4-digit number times a 2-digit number can give 5 digits, e.g. 1,000×10=10,000.
Question 4
Use 365 days per year and convert units.
Compare each result with the target.
Using the chapter’s estimate 2,100×70,000=147,000,000 km for the Sun’s distance, 147,000,000÷1,000=147,000 days, or about 403 years—longer than a typical lifetime. Paper: 500,000 g=500 kg. Births/day: 250×60×24=360,000, below one million. Coins/day: 86,400, also below one million.
Question 5
Multiply each pair carefully and compare the repeated digit patterns.
Continue by adding one repeated digit to each factor as shown.
11²=121, 111²=12,321, 1111²=1,234,321. 66×61=4,026; 666×661=440,226; 6666×6661=44,402,226. 3×5=15; 33×35=1,155; 333×335=111,555. 101²=10,201; 102²=10,404; 103²=10,609.
Question 6
Regroup factors to make powers of ten when possible.
For division, cancel common factors or check by multiplication.
1,250×380=475,000; 2,100×70,000=147,000,000; 6,400×62,500=400,000,000. 1,395,000÷150=9,300; 105,000,000÷700=150,000; 52,000,000,000÷130=400,000,000.
Question why-multiply-by-five
Since 5=10÷2, replace 5 with 10÷2.
Calculate 116×10=1,160, then halve 1,160.
n×5=n×(10÷2)=(n×10)÷2. For example, 116×5=1,160÷2=580.
Textbook pages 19–24
Question 1
A multiple of 5 must end in 0 or 5. Put the larger remaining digits first for the maximum.
For the minimum even number, use the smallest nonzero first and keep the final digit even.
Largest multiple of 5: 9,876,543,210. Smallest even: 1,023,456,798.
Question 2
Repeat a number word with many spoken letters in each place.
77,77,777 is one maximum: “seventy-seven lakh seventy-seven thousand seven hundred seventy-seven” has 60 letters when spaces and hyphens are not counted. Each group uses a number word with a largest letter count: seventy-seven (12), seven hundred seventy-seven (24), plus lakh (4) and thousand (8).
Question 3
Place digits in descending order except for a final pair that is out of order.
Example: 987,654,312. Swapping the final 1 and 2 gives 987,654,321, which is larger.
Question 4
At each position, choose the largest digit available while leaving enough digits for the remaining places.
5,534,512,345.
Question 6(a)
Digits through 99 use 9+90×2=189 places.
The next 811 digits cover 270 complete three-digit numbers plus one digit.
Digits through 99,999 total 488,889; the remaining 511,111 digits enter the six-digit numbers.
1,000th digit is the first digit of 370 (because 100+270=370). The millionth digit is the first digit of 185,185: the first 85,185 six-digit numbers use 511,110 digits, so digit 1,000,000 is the first digit of the next number.
Question 6(c)
Count each occurrence of 5 as you write the integers in order.
The count reaches 5,000 while writing 13,495.
The 5,000th occurrence is in 13,495.
Question 6
Use the 10,000 button for the ten-thousands part, then 100 for the remainder.
2×10,000+8×100; 9×10,000+21×100; 12×10,000+5×100; 653×10,000; 702×10,000+57×100.
Question 7
One billion is 1,000,000,000 and one lakh is 100,000.
Divide 1,000,000,000 by 100,000.
10,000 lakhs.
Question 8
Convert 180 m to 180,000 mm.
Divide distance by the daily rate or elapsed time.
180,000 coins; about 12–13.3 days (roughly 12–14 days); about 1,233 km/day and 51.25 km/hour.
Question 10
Count the segments for each digit in the printed pattern.
For a rearrangement puzzle, move only the allowed number of segments and verify every new digit.
These image-based challenges need their printed matchstick patterns for exact counts; many larger results can be valid.
Question 9
For the largest sum, give the largest digits to the highest place values in both addends.
For the smallest difference, make the two numbers as close as possible from left to right.
Largest sum: 99,88,776+65,544=1,00,54,320. Smallest difference: 11,22,334−99,887=10,22,447. Each row uses only the number cards available in the two sets; add or subtract the resulting numerals to check.
Question 10-card-expressions
Try products first to reach the correct scale, then add or subtract remaining cards.
Compare each result with its target to see the difference.
Examples: (a) 4,000×(20+5)+13,000=1,13,000, which is 3,000 above. (b) 1,50,000+70,000−(4,000×5)=2,00,000 exactly. (c) 300×(13,000−((70,000+1,50,000)÷20))−(4,000×5)=5,80,000 exactly; every card is used once. (d) (70,000×20)−1,50,000−4,000−(300×5)=12,44,500, which is 500 below. (e) 70,000+(13,000÷20)+4,000×((1,50,000÷300)+5)=20,90,650, which is 150 below the target.
Question 11
Convert metres to millimetres: 1 m=1,000 mm.
Divide the total height by the thickness of one coin.
180 m=180,000 mm. At 1 mm per coin, 180,000 coins are needed.
Question 12
At 1,000 km/day: 12,000÷1,000=12 days.
At 900 km/day: 12,000÷900≈13.3 days.
The journey would take roughly 12–14 days, depending on its daily speed.
Question 13
Divide 13,560 km by 11 days.
Divide the daily distance by 24 hours.
13,560÷11≈1,233 km per day. Then 1,233÷24≈51 km per hour.
Question 14
Divide each height by 40 m.
For a range, divide both endpoints.
Eagles: 4,500÷40=112.5 to 6,000÷40=150 building-heights. Everest: 8,850÷40≈221. Aeroplanes: 10,000÷40=250 to 12,800÷40=320 building-heights.
Question 5
Check the one-digit names: one/two share o; two/three share t; three/four share r; and so on.
Notice that at place-value changes, a word such as hundred or thousand is often shared; at a change like 999 to 1,000, nine and one share n and e.
There is no such pair in the usual English number names, so no finite counting point will find one. For example, 999 is “nine hundred ninety-nine” and 1,000 is “one thousand”; both names contain n and e. The repeated place-value words and carry patterns keep a letter in common.
Question size-of-a-lakh
Compare a lakh with a person’s lifetime, a city crowd, and quantities such as hairs or fish eggs.
A quantity can seem large in one situation and small in another.
A lakh is large for counting one item each day: 1,00,000 days is about 274 years. It is small beside a stadium that holds over a lakh people or the many hairs on a head. So whether it feels large depends on what we compare it with.
Question population-and-moon-estimates
Buses: 1,00,000×50=50,00,000 seats, less than Mumbai’s 1,24,42,373 people.
Ships: 5,000×2,500=1,25,00,000 seats, slightly more than Mumbai’s population.
In 10 years at 365 days per year: 100×365×10=3,65,000 km.
No for the buses: 50 lakh seats are not enough for 1,24,42,373 people. Yes for the ships, if each has 2,500 available seats: 1,25,00,000 seats are enough. No for the Moon trip: 3,65,000 km is 19,400 km short of 3,84,400 km.
Textbook pages 25
Question 1
Count each digit: 4 uses 4, 2 uses 5, 0 uses 6, 1 uses 2 and 9 uses 6 sticks.
Add the five counts.
4+5+6+2+6=23 sticks.
Question example-5108
Count the sticks in each digit: 5 uses 5, 1 uses 2, 0 uses 6 and 8 uses 7.
Add: 5+2+6+7.
5108 uses 20 sticks.
Question 2
In the seven-segment pattern, add one stick to change 0 into 8.
Add one stick to change 1 into 7.
One answer is 42,879: change the 0 to 8 and the 1 to 7. Exactly two sticks are added, and 42,879 is greater than 42,019.
Question 3
Keep the original digit order. Compare each possible insertion from left to right.
Placing the extra 1 after the 2 gives 421,019, which is larger than the other placements.
Insert the extra 1 after the 2: 421,019. This is largest because it keeps 4 and 2 in the two highest places, then makes the third digit 1 instead of 0.
Question 4
Keep the original digits in order.
Try each gap, including before the first digit and after the last.
The distinct results are 142,019; 412,019; 420,119; and 420,191. (Inserting beside the existing 1 gives the same numeral either way.)
Question 5
Use the chapter’s matchstick shapes for each digit.
Move four segments while checking that the total number of sticks is unchanged.
One possible result is 66,006. For example, change 3 to 6 (remove 1 and add 2 sticks), 8 to 0 (remove 1), 9 to 0 (remove 1 and add 1), and 0 to 6 (remove 1 and add 1). That is four removed and four added sticks. The book also gives 88,078.
Question 6
The digit 7 needs three sticks.
Use eight 7s and add their stick counts.
77,777,777 uses 8×3=24 sticks.