Number System

Find the difference in the place values of the two sevens in the number 8,72,574.

Answer

In the number 8,72,574 :

One 7 occurs at the ten-thousands place, so its place value = 7 × 10000 = 70000.

The other 7 occurs at the tens place, so its place value = 7 × 10 = 70.

Required difference = 70000 - 70 = 69930.

Hence, the difference in the place values of the two sevens = 69930.

Find the difference between the face value of 9 and the face value of 4 in the number 324798.

Answer

The face value of a digit is the digit itself, irrespective of its position in the number.

Face value of 9 = 9.

Face value of 4 = 4.

Required difference = 9 - 4 = 5.

Hence, the difference between the face values of 9 and 4 = 5.

By making a suitable chart, compare:

(i) 540276 and 369998

(ii) 6983245 and 6893254

Answer

(i) Writing the numbers in a place value chart :

Both numbers have an equal number of digits. Comparing the digits at the leftmost (lakhs) place, the first number is 5 and the second number is 3.

Since 5 > 3,

Hence, 540276 > 369998.

(ii) Writing the numbers in a place value chart :

Both numbers have an equal number of digits. At the leftmost (ten-lakhs) place, both have the same digit 6. At the next (lakhs) place, the first number is 9 and the second number is 8.

Since 9 > 8,

Hence, 6983245 > 6893254.

Compare the numbers written in the following table by writing them in ascending order:

Answer

The numbers represented by the rows of the chart are :

5432972, 23106293, 5223791, 23182634 and 54344782.

The numbers 5432972 and 5223791 have 7 digits each; comparing them, at the lakhs place 5223791 has 2 and 5432972 has 4, so 5223791 < 5432972.

The numbers 23106293 and 23182634 have 8 digits each; comparing them, the first differing digit (ten-thousands place) is 0 in 23106293 and 8 in 23182634, so 23106293 < 23182634.

The number 54344782 has the most digits (8 digits) with the greatest leftmost digit, so it is the greatest.

Arranging in ascending order :

Hence, 5223791 < 5432972 < 23106293 < 23182634 < 54344782.

Use a place value chart to compare the given numbers and write them in descending order: 543287 ; 5482900 ; 2732940 ; 43877 ; 78396 and 4999.

Answer

Writing the numbers in a place value chart :

The numbers 5482900 and 2732940 have 7 digits each (the most digits); comparing them at the ten-lakhs place, 5 > 2, so 5482900 is the greatest and 2732940 is the next greatest.

The number 543287 has 6 digits, so it is next.

The numbers 78396 and 43877 have 5 digits each; comparing them at the ten-thousands place, 7 > 4, so 78396 > 43877.

The number 4999 has the least number of digits (4 digits), so it is the smallest.

Arranging in descending order :

Hence, 54,82,900 > 27,32,940 > 5,43,287 > 78,396 > 43,877 > 4,999.

Find the smallest and the greatest numbers in each of the cases given below:

(i) 983, 5754, 84 and 5942

(ii) 32849, 53628, 5499 and 54909

Answer

(i) Among 983, 5754, 84 and 5942 :

84 has the least number of digits (2 digits), so it is the smallest.

5754 and 5942 have 4 digits each; comparing them at the hundreds place, 9 > 7, so 5942 is the greatest.

Hence, smallest number = 84 and greatest number = 5942.

(ii) Among 32849, 53628, 5499 and 54909 :

5499 has the least number of digits (4 digits), so it is the smallest.

The numbers 32849, 53628 and 54909 have 5 digits each. Comparing 53628 and 54909 at the thousands place, 4 > 3, so 54909 is the greatest.

Hence, smallest number = 5499 and greatest number = 54909.

Form the greatest and the smallest 4-digit numbers using the given digits, without repetition:

(i) 3, 7, 2 and 5

(ii) 6, 1, 4 and 9

(iii) 7, 0, 4 and 2

(iv) 1, 8, 5 and 3

(v) 9, 6, 0 and 7

Answer

To form the greatest number, write the digits in descending order. To form the smallest number, write the digits in ascending order (without placing 0 at the leftmost place).

(i) Digits : 3, 7, 2 and 5.

Greatest number = 7532 and smallest number = 2357.

Hence, greatest = 7532 and smallest = 2357.

(ii) Digits : 6, 1, 4 and 9.

Greatest number = 9641 and smallest number = 1469.

Hence, greatest = 9641 and smallest = 1469.

(iii) Digits : 7, 0, 4 and 2.

Greatest number = 7420 and smallest number = 2047 (0 is not placed at the leftmost place).

Hence, greatest = 7420 and smallest = 2047.

(iv) Digits : 1, 8, 5 and 3.

Greatest number = 8531 and smallest number = 1358.

Hence, greatest = 8531 and smallest = 1358.

(v) Digits : 9, 6, 0 and 7.

Greatest number = 9760 and smallest number = 6079 (0 is not placed at the leftmost place).

Hence, greatest = 9760 and smallest = 6079.

Form the greatest and the smallest 4-digit numbers using any four different digits, with the condition that digit 5 is always at the ten's place.

Answer

The digit 5 is fixed at the tens place, so the number is of the form _ _ 5 _.

To form the greatest number, fill the remaining places (from the left) with the greatest possible different digits : 9, 8 and 7.

Greatest number = 9 8 5 7 = 9857.

To form the smallest number, fill the remaining places (from the left) with the smallest possible different digits, keeping 0 out of the leftmost place : 1, 0 and 2.

Smallest number = 1 0 5 2 = 1052.

Hence, greatest number = 9857 and smallest number = 1052.

Form the largest number with the digits 2, 3, 5, 9, 6 and 0, without repetition of any digit.

Answer

To form the largest number, write the given digits in descending order of their values.

Arranging 2, 3, 5, 9, 6 and 0 in descending order : 9, 6, 5, 3, 2, 0.

Largest number = 965320.

Hence, the largest number = 965320.

Write the smallest and the greatest 4-digit numbers, without repetition of any digit.

Answer

To form the smallest 4-digit number with different digits, use the smallest digits 1, 0, 2 and 3, keeping 0 out of the leftmost place :

Smallest number = 1023.

To form the greatest 4-digit number with different digits, use the greatest digits 9, 8, 7 and 6 in descending order :

Greatest number = 9876.

Hence, smallest number = 1023 and greatest number = 9876.

Find the sum of the largest and the smallest four-digit numbers.

Answer

The largest four-digit number = 9999.

The smallest four-digit number = 1000.

Required sum = 9999 + 1000 = 10999.

Hence, the sum of the largest and the smallest four-digit numbers = 10999.

Form the greatest and the smallest five-digit numbers with 8 in the hundred's place and with all the different digits .

Answer

The digit 8 is fixed at the hundreds place, so the number is of the form _ _ 8 _ _.

To form the greatest number, fill the remaining places (from the left) with the greatest different digits : 9, 7, 6 and 5.

Greatest number = 9 7 8 6 5.

To form the smallest number, fill the remaining places (from the left) with the smallest different digits, keeping 0 out of the leftmost place : 1, 0, 2 and 3.

Smallest number = 1 0 8 2 3.

Hence, greatest number = 97865 and smallest number = 10823.

(i) How many four-digit numbers are there between 999 and 3000?

(ii) How many four-digit numbers are there between 99 and 3000?

Answer

(i) The four-digit numbers between 999 and 3000 start from 1000 and end at 2999.

Number of such numbers = 2999 - 1000 + 1 = 2000.

Hence, there are 2000 four-digit numbers between 999 and 3000.

(ii) Every four-digit number is greater than 99, so the four-digit numbers between 99 and 3000 are the same as those from 1000 to 2999.

Number of such numbers = 2999 - 1000 + 1 = 2000.

Hence, there are 2000 four-digit numbers between 99 and 3000.

Form the greatest and the smallest 4-digit numbers using the digits 5, 4, 7 and 9 (without repeating the digits) and with the condition that:

(i) 7 is at the ones place.

(ii) 9 is at the tens place.

(iii) 4 is at the hundreds place.

Answer

(i) The digit 7 is fixed at the ones place, so the number is of the form _ _ _ 7.

Using the remaining digits 5, 4 and 9 :

Greatest number = 9 5 4 7 = 9547 (remaining digits in descending order).

Smallest number = 4 5 9 7 = 4597 (remaining digits in ascending order).

Hence, greatest number = 9547 and smallest number = 4597.

(ii) The digit 9 is fixed at the tens place, so the number is of the form _ _ 9 _.

Using the remaining digits 5, 4 and 7 :

Greatest number = 7 5 9 4 = 7594.

Smallest number = 4 5 9 7 = 4597.

Hence, greatest number = 7594 and smallest number = 4597.

(iii) The digit 4 is fixed at the hundreds place, so the number is of the form _ 4 _ _.

Using the remaining digits 5, 7 and 9 :

Greatest number = 9 4 7 5 = 9475.

Smallest number = 5 4 7 9 = 5479.

Hence, greatest number = 9475 and smallest number = 5479.

Population of a city was 3,54,976 in the year 2014. In the year 2015, it increased by 68,438. What was the population of the city at the end of the year 2015?

Answer

Population in the year 2014 = 3,54,976.

Increase in the year 2015 = 68,438.

Population at the end of 2015 = 3,54,976 + 68,438.

$\begin{array}{rr} & 3{,}54{,}976 \\ + & 68{,}438 \\ \hline & 4{,}23{,}414 \end{array}$

Hence, the population of the city at the end of the year 2015 = 4,23,414.

A = 7,43,000 and B = 8,00,100. Which is greater A or B? And by how much?

Answer

Both A and B have an equal number of digits. Comparing them at the lakhs place, B has 8 and A has 7.

Since 8 > 7, B is greater than A.

Difference = B - A = 8,00,100 - 7,43,000.

$\begin{array}{rr} & 8{,}00{,}100 \\ - & 7{,}43{,}000 \\ \hline & 57{,}100 \end{array}$

Hence, B is greater than A by 57,100.

A notebook has 56 pages. How many pages will 5326 such notebooks have?

Answer

Number of pages in one notebook = 56.

Number of pages in 5326 notebooks = 56 × 5326.

$\begin{array}{rr} & 5{,}326 \\ \times & 56 \\ \hline & 31{,}956 \\ & 2{,}66{,}300 \\ \hline & 2{,}98{,}256 \end{array}$

Hence, 5326 notebooks will have 2,98,256 pages.

A person has 75,000 sheets of paper. If each sheet makes 8 pages of a notebook, how many notebooks of 200 pages can be made using the above sheets?

Answer

Number of sheets = 75,000.

Total number of pages obtained = 8 × 75,000.

$\begin{array}{rr} & 75{,}000 \\ \times & 8 \\ \hline & 6{,}00{,}000 \end{array}$

Number of pages in one notebook = 200.

Number of notebooks made = 6,00,000 ÷ 200.

$\begin{array}{l} 200\overline{\smash{\big)}\phantom{.}600000\smash{\big(}}3000 \\ \phantom{+()}\underline{-600} \\ \phantom{++))}\phantom{000}000 \end{array}$

Hence, 3000 notebooks of 200 pages can be made.

Add 1,76,209 ; 4,50,923 and 44,83,947.

Answer

Required sum = 1,76,209 + 4,50,923 + 44,83,947.

$\begin{array}{rr} & 1{,}76{,}209 \\ & 4{,}50{,}923 \\ + & 44{,}83{,}947 \\ \hline & 51{,}11{,}079 \end{array}$

Hence, the required sum = 51,11,079.

A cricket player has scored 7,849 runs so far in test matches. He wishes to complete 10,000 runs; how many more runs does he need?

Answer

Runs to be completed = 10,000.

Runs already scored = 7,849.

Runs needed = 10,000 - 7,849.

$\begin{array}{rr} & 10{,}000 \\ - & 7{,}849 \\ \hline & 2{,}151 \end{array}$

Hence, the player needs 2151 more runs.

In an election two candidates A and B are the only contestants. If candidate A scored 9,32,567 votes and candidate B scored 9,00,235 votes, by how much margin did A win or loose the election?

Answer

Votes scored by A = 9,32,567.

Votes scored by B = 9,00,235.

Since 9,32,567 > 9,00,235, candidate A won the election.

Margin = 9,32,567 - 9,00,235.

$\begin{array}{rr} & 9{,}32{,}567 \\ - & 9{,}00{,}235 \\ \hline & 32{,}332 \end{array}$

Hence, A won the election by 32332 votes.

Find the difference between the largest and the smallest five digit numbers that can be formed using the digits 5, 1, 6, 3 and 2 without repeating any digit.

Answer

Largest five-digit number (digits in descending order) = 65321.

Smallest five-digit number (digits in ascending order) = 12356.

Required difference = 65321 - 12356.

$\begin{array}{rr} & 65{,}321 \\ - & 12{,}356 \\ \hline & 52{,}965 \end{array}$

Hence, the required difference = 52965.

A man has ₹ 1,57,184 with him. He placed an order for purchasing 80 articles at ₹ 125 each. How much money will remain with him after the purchase?

Answer

Money that man has = ₹ 1,57,184.

Cost of 80 articles = 80 × ₹ 125.

$\begin{array}{rr} & 125 \\ \times & 80 \\ \hline & 10{,}000 \end{array}$

Remaining Money = ₹ 1,57,184 - ₹ 10,000.

$\begin{array}{rr} & 1{,}57{,}184 \\ - & 10{,}000 \\ \hline & 1{,}47{,}184 \end{array}$

Hence, ₹ 1,47,184 will remain with him after the purchase.

A student multiplied 8,035 by 87 instead of multiplying by 78. By how much was his answer greater than or less than the correct answer?

Answer

The student's answer = 8,035 × 87.

The correct answer = 8,035 × 78.

Difference = 8,035 × 87 - 8,035 × 78

= 8,035 × (87 - 78)

= 8,035 × 9.

$\begin{array}{rr} & 8{,}035 \\ \times & 9 \\ \hline & 72{,}315 \end{array}$

Since 87 > 78, the student's answer was greater than the correct answer.

Hence, his answer was greater than the correct answer by 72,315.

Mohani has 30 m cloth and she wants to make some shirts for her son. If each shirt requires 2 m 30 cm cloth, how many shirts, in all, can be made and how much length of the cloth will be left?

Answer

Total length of cloth = 30 m = 3000 cm.

Cloth required for one shirt = 2 m 30 cm = 230 cm.

Number of shirts = 3000 ÷ 230.

$\begin{array}{l} 230\overline{\smash{\big)}\phantom{.}3000\smash{\big(}}13 \\ \phantom{+()}\underline{-230} \\ \phantom{++))}\phantom{0}700 \\ \phantom{+()}\phantom{0}\underline{-690} \\ \phantom{230,\big),}\phantom{00}10 \end{array}$

So, 13 shirts can be made and 10 cm of cloth is left as remainder.

Cloth used for 13 shirts = 13 × 230 cm.

$\begin{array}{rr} & 230 \\ \times & 13 \\ \hline & 690 \\ & 2{,}300 \\ \hline & 2{,}990 \end{array}$

Cloth left = 3000 cm - 2990 cm.

$\begin{array}{rr} & 3{,}000 \\ - & 2{,}990 \\ \hline & 10 \end{array}$

Hence, 13 shirts can be made and 10 cm of cloth will be left.

The distance between two places A and B is 3 km 760 m. A boy travels from A to B and then B to A every day. How much distance does he travel in 8 days?

Answer

Distance between A and B = 3 km 760 m = 3760 m.

Distance travelled in one day (A to B and B to A) = 2 × 3760 m.

$\begin{array}{rr} & 3{,}760 \\ \times & 2 \\ \hline & 7{,}520 \end{array}$

Distance travelled in 8 days = 8 × 7520 m.

$\begin{array}{rr} & 7{,}520 \\ \times & 8 \\ \hline & 60{,}160 \end{array}$

60160 m = $\dfrac{60160}{1000}$ = 60.160 km.

Hence, the boy travels 60.160 km in 8 days.

An oil-tin contains 6 litre 60 ml oil. How many identical bottles can this oil fill, if the capacity of each bottle is 30 ml?

Answer

Quantity of oil = 6 litre 60 ml = 6060 ml (since 1 litre = 1000 ml).

Capacity of each bottle = 30 ml.

Number of bottles = 6060 ÷ 30.

$\begin{array}{l} 30\overline{\smash{\big)}\phantom{0}6060\smash{\big(}}202 \\ \phantom{())}\underline{-60} \\ \phantom{++}\phantom{00}60 \\ \phantom{+)}\phantom{00}\underline{-60} \\ \phantom{30,\big),}\phantom{000}0 \end{array}$

Hence, 202 identical bottles can be filled.

The sale receipt of a company in a certain year was ₹ 83,73,540. In the following year, it decreased by ₹ 7,84,670.

(i) What was the sale receipt of the company during the following year?

(ii) What was the total sale receipt of the company during these two years?

Answer

(i) Sale receipt in the first year = ₹ 83,73,540.

Decrease in the following year = ₹ 7,84,670.

Sale receipt in the following year = ₹ 83,73,540 - ₹ 7,84,670.

$\begin{array}{rr} & 83{,}73{,}540 \\ - & 7{,}84{,}670 \\ \hline & 75{,}88{,}870 \end{array}$

Hence, the sale receipt during the following year = ₹ 75,88,870.

(ii) Total sale receipt during the two years = ₹ 83,73,540 + ₹ 75,88,870.

$\begin{array}{rr} & 83{,}73{,}540 \\ + & 75{,}88{,}870 \\ \hline & 1{,}59{,}62{,}410 \end{array}$

Hence, the total sale receipt during these two years = ₹ 1,59,62,410.

A number exceeds 8,59,470 by 3,00,999. What is the number?

Answer

The required number = 8,59,470 + 3,00,999.

$\begin{array}{rr} & 8{,}59{,}470 \\ + & 3{,}00{,}999 \\ \hline & 11{,}60{,}469 \end{array}$

Hence, the required number = 11,60,469.

Round off (approximate) each of the following to the nearest ten:

(i) 62

(ii) 265

(iii) 543

(iv) 6296

Answer

To round off to the nearest ten, look at the ones digit. If it is less than 5, replace it by 0; if it is 5 or more, increase the tens digit by 1 and replace the ones digit by 0.

(i) 62 → ones digit is 2 (less than 5) → 60.

(ii) 265 → ones digit is 5 → 270.

(iii) 543 → ones digit is 3 (less than 5) → 540.

(iv) 6296 → ones digit is 6 (more than 5) → 6300.

Round off (approximate) each of the following to the nearest hundred:

(i) 748

(ii) 784

(iii) 6998

(iv) 59997

Answer

To round off to the nearest hundred, look at the tens digit. If it is less than 5, replace the tens and ones digits by 0; if it is 5 or more, increase the hundreds digit by 1 and replace the tens and ones digits by 0.

(i) 748 → tens digit is 4 (less than 5) → 700.

(ii) 784 → tens digit is 8 (more than 5) → 800.

(iii) 6998 → tens digit is 9 (more than 5) → 7000.

(iv) 59997 → tens digit is 9 (more than 5) → 60000.

Round off (approximate) each of the following to the nearest thousand:

(i) 6475

(ii) 6732

(iii) 9948

(iv) 89994

Answer

To round off to the nearest thousand, look at the hundreds digit. If it is less than 5, replace the hundreds, tens and ones digits by 0; if it is 5 or more, increase the thousands digit by 1 and replace the digits on its right by 0.

(i) 6475 → hundreds digit is 4 (less than 5) → 6000.

(ii) 6732 → hundreds digit is 7 (more than 5) → 7000.

(iii) 9948 → hundreds digit is 9 (more than 5) → 10000.

(iv) 89994 → hundreds digit is 9 (more than 5) → 90000.

Round off (approximate):

(i) 578 to the nearest ten.

(ii) 578 to the nearest hundred.

(iii) 4327 to the nearest thousand.

(iv) 32974 to the nearest ten thousand.

(v) 27487 to the nearest ten thousand.

Answer

(i) To round off 578 to the nearest ten, look at the ones digit. It is 8 (more than 5), so increase the tens digit by 1 and replace the ones digit by 0.

Hence, 578 to the nearest ten = 580.

(ii) To round off 578 to the nearest hundred, look at the tens digit. It is 7 (more than 5), so increase the hundreds digit by 1 and replace the tens and ones digits by 0.

Hence, 578 to the nearest hundred = 600.

(iii) To round off 4327 to the nearest thousand, look at the hundreds digit. It is 3 (less than 5), so replace the hundreds, tens and ones digits by 0 and keep the other digit as it is.

Hence, 4327 to the nearest thousand = 4000.

(iv) To round off 32974 to the nearest ten thousand, look at the thousands digit. It is 2 (less than 5), so replace the thousands, hundreds, tens and ones digits by 0 and keep the other digit as it is.

Hence, 32974 to the nearest ten thousand = 30000.

(v) To round off 27487 to the nearest ten thousand, look at the thousands digit. It is 7 (more than 5), so increase the ten thousands digit by 1 and replace the digits on its right by 0.

Hence, 27487 to the nearest ten thousand = 30000.

Round off (approximate) each of the following to the nearest ten, nearest hundred and nearest thousand.

(i) 864

(ii) 1249

(iii) 54,547

(iv) 68,076

(v) 56,293

(vi) 7,293

(vii) 89,24,379

Answer

For the nearest ten, look at the ones digit; for the nearest hundred, look at the tens digit; for the nearest thousand, look at the hundreds digit. If that digit is less than 5, replace it (and the digits on its right) by 0; if it is 5 or more, increase the digit on its left by 1 and replace the digits on its right by 0.

(i) 864 :

Nearest ten → ones digit is 4 (less than 5) → 860. Nearest hundred → tens digit is 6 (more than 5) → 900. Nearest thousand → hundreds digit is 8 (more than 5) → 1000.

∴ 860, 900 and 1000.

(ii) 1249 :

Nearest ten → ones digit is 9 (more than 5) → 1250. Nearest hundred → tens digit is 4 (less than 5) → 1200. Nearest thousand → hundreds digit is 2 (less than 5) → 1000.

∴ 1250, 1200 and 1000.

(iii) 54,547 :

Nearest ten → ones digit is 7 (more than 5) → 54550. Nearest hundred → tens digit is 4 (less than 5) → 54500. Nearest thousand → hundreds digit is 5 (equal to 5) → 55000.

∴ 54550, 54500 and 55000.

(iv) 68,076 :

Nearest ten → ones digit is 6 (more than 5) → 68080. Nearest hundred → tens digit is 7 (more than 5) → 68100. Nearest thousand → hundreds digit is 0 (less than 5) → 68000.

∴ 68080, 68100 and 68000.

(v) 56,293 :

Nearest ten → ones digit is 3 (less than 5) → 56290. Nearest hundred → tens digit is 9 (more than 5) → 56300. Nearest thousand → hundreds digit is 2 (less than 5) → 56000.

∴ 56290, 56300 and 56000.

(vi) 7,293 :

Nearest ten → ones digit is 3 (less than 5) → 7290. Nearest hundred → tens digit is 9 (more than 5) → 7300. Nearest thousand → hundreds digit is 2 (less than 5) → 7000.

∴ 7290, 7300 and 7000.

(vii) 89,24,379 :

Nearest ten → ones digit is 9 (more than 5) → 8924380. Nearest hundred → tens digit is 7 (more than 5) → 8924400. Nearest thousand → hundreds digit is 3 (less than 5) → 8924000.

∴ 8924380, 8924400 and 8924000.

Approximate each of the following to the nearest ten:

(i) ₹ 562

(ii) 837 m

(iii) 545 cm

(iv) ₹ 27

Answer

To round off to the nearest ten, look at the ones digit. If it is less than 5, replace it by 0; if it is 5 or more, increase the tens digit by 1 and replace the ones digit by 0.

(i) ₹ 562 → ones digit is 2 (less than 5) → ₹ 560.

(ii) 837 m → ones digit is 7 (more than 5) → 840 m.

(iii) 545 cm → ones digit is 5 (equal to 5) → 550 cm.

(iv) ₹ 27 → ones digit is 7 (more than 5) → ₹ 30.

List all the natural numbers which can be rounded off to 30.

Answer

A number rounds off to 30 (to the nearest ten) when its ones digit decides whether it moves to 30 or away from it.

The numbers from 25 to 29 have ones digit 5 or more, so they round up to 30.

The numbers from 30 to 34 have ones digit less than 5, so they round down to 30.

Hence, the required numbers are 25, 26, 27, 28, 29, 30, 31, 32, 33 and 34.

List all the whole numbers which are approximate to 50.

Answer

A number rounds off to 50 (to the nearest ten) depending on its ones digit.

The numbers from 45 to 49 have ones digit 5 or more, so they round up to 50.

The numbers from 50 to 54 have ones digit less than 5, so they round down to 50.

Hence, the required numbers are 45, 46, 47, 48, 49, 50, 51, 52, 53 and 54.

Write the smallest and the largest numbers which are rounded off to 90.

Answer

A number rounds off to 90 (to the nearest ten) depending on its ones digit.

The numbers from 85 to 89 have ones digit 5 or more, so they round up to 90.

The numbers from 90 to 94 have ones digit less than 5, so they round down to 90.

So, the numbers that round off to 90 lie from 85 up to 94.

Hence, the smallest number is 85 and the largest number is 94.

Write the smallest and the largest numbers which are approximate to 130.

Answer

A number rounds off to 130 (to the nearest ten) depending on its ones digit.

The numbers from 125 to 129 have ones digit 5 or more, so they round up to 130.

The numbers from 130 to 134 have ones digit less than 5, so they round down to 130.

So, the numbers that round off to 130 lie from 125 up to 134.

Hence, the smallest number is 125 and the largest number is 134.

Estimate the sum after rounding off each pair of numbers to the nearest ten:

(i) 34 and 87

(ii) 78 and 18

(iii) 96 and 55

(iv) 76 and 62

(v) 457 and 175

(vi) 527 and 267

Answer

(i) 34 → 30 and 87 → 90.

Required sum = 30 + 90 = 120.

(ii) 78 → 80 and 18 → 20.

Required sum = 80 + 20 = 100.

(iii) 96 → 100 and 55 → 60.

Required sum = 100 + 60 = 160.

(iv) 76 → 80 and 62 → 60.

Required sum = 80 + 60 = 140.

(v) 457 → 460 and 175 → 180.

Required sum = 460 + 180 = 640.

(vi) 527 → 530 and 267 → 270.

Required sum = 530 + 270 = 800.

Estimate the sum after rounding off each pair of numbers to the nearest hundred:

(i) 336 and 782

(ii) 270 and 495

(iii) 4280 and 5295

(iv) 30047 and 39287

Answer

(i) 336 → 300 and 782 → 800.

Required sum = 300 + 800 = 1100.

(ii) 270 → 300 and 495 → 500.

Required sum = 300 + 500 = 800.

(iii) 4280 → 4300 and 5295 → 5300.

Required sum = 4300 + 5300 = 9600.

(iv) 30047 → 30000 and 39287 → 39300.

Required sum = 30000 + 39300 = 69300.

Estimate the sum after rounding off the following pairs of numbers to the nearest thousand:

(i) 53826 and 36455

(ii) 56802 and 22475

Answer

(i) 53826 → 54000 and 36455 → 36000.

Required sum = 54000 + 36000 = 90,000.

(ii) 56802 → 57000 and 22475 → 22000.

Required sum = 57000 + 22000 = 79,000.

Estimate the following differences after rounding off each number to the nearest ten:

(i) 82 - 27

(ii) 96 - 36

(iii) 508 - 248

Answer

(i) 82 → 80 and 27 → 30.

Required difference = 80 - 30 = 50.

(ii) 96 → 100 and 36 → 40.

Required difference = 100 - 40 = 60.

(iii) 508 → 510 and 248 → 250.

Required difference = 510 - 250 = 260.

Estimate each difference after rounding off each given number to the nearest hundred:

(i) 769 - 314

(ii) 856 - 687

(iii) 6352 - 2086

Answer

(i) 769 → 800 and 314 → 300.

Required difference = 800 - 300 = 500.

(ii) 856 → 900 and 687 → 700.

Required difference = 900 - 700 = 200.

(iii) 6352 → 6400 and 2086 → 2100.

Required difference = 6400 - 2100 = 4300.

Estimate each difference after rounding off each given number to the nearest thousand:

(i) 45974 - 38766

(ii) 76003 - 48399

Answer

(i) 45974 → 46000 and 38766 → 39000.

Required difference = 46000 - 39000 = 7000.

(ii) 76003 → 76000 and 48399 → 48000.

Required difference = 76000 - 48000 = 28000.

Estimate each of the following products by rounding off each number to the nearest ten:

(i) 49 × 52

(ii) 63 × 38

(iii) 27 × 54

(iv) 53 × 85

(v) 74 × 67

(vi) 25 × 33

Answer

(i) 49 → 50 and 52 → 50.

Required product = 50 × 50 = 2500.

(ii) 63 → 60 and 38 → 40.

Required product = 60 × 40 = 2400.

(iii) 27 → 30 and 54 → 50.

Required product = 30 × 50 = 1500.

(iv) 53 → 50 and 85 → 90.

Required product = 50 × 90 = 4500.

(v) 74 → 70 and 67 → 70.

Required product = 70 × 70 = 4900.

(vi) 25 → 30 and 33 → 30.

Required product = 30 × 30 = 900.

Estimate each of the following products by rounding off each number to the nearest hundred:

(i) 477 × 213

(ii) 624 × 236

(iii) 333 × 247

(iv) 537 × 283

(v) 382 × 127

(vi) 472 × 328

Answer

(i) 477 → 500 and 213 → 200.

Required product = 500 × 200 = 1,00,000.

(ii) 624 → 600 and 236 → 200.

Required product = 600 × 200 = 1,20,000.

(iii) 333 → 300 and 247 → 200.

Required product = 300 × 200 = 60,000.

(iv) 537 → 500 and 283 → 300.

Required product = 500 × 300 = 1,50,000.

(v) 382 → 400 and 127 → 100.

Required product = 400 × 100 = 40,000.

(vi) 472 → 500 and 328 → 300.

Required product = 500 × 300 = 1,50,000.

Estimate each of the following products by rounding off the first number correct to the nearest ten and the other number correct to the nearest hundred:

(i) 28 × 287

(ii) 432 × 128

(iii) 48 × 165

(iv) 72 × 258

(v) 83 × 664

(vi) 44 × 250

Answer

(i) 28 → 30 (nearest ten) and 287 → 300 (nearest hundred).

Required product = 30 × 300 = 9000.

(ii) 432 → 430 (nearest ten) and 128 → 100 (nearest hundred).

Required product = 430 × 100 = 43000.

(iii) 48 → 50 (nearest ten) and 165 → 200 (nearest hundred).

Required product = 50 × 200 = 10000.

(iv) 72 → 70 (nearest ten) and 258 → 300 (nearest hundred).

Required product = 70 × 300 = 21000.

(v) 83 → 80 (nearest ten) and 664 → 700 (nearest hundred).

Required product = 80 × 700 = 56000.

(vi) 44 → 40 (nearest ten) and 250 → 300 (nearest hundred).

Required product = 40 × 300 = 12000.

Estimate each of the following quotients by rounding off each number to the nearest ten:

(i) 87 ÷ 28

(ii) 84 ÷ 23

(iii) 77 ÷ 22

(iv) 198 ÷ 24

(v) 355 ÷ 26

(vi) 444 ÷ 42

(vii) 843 ÷ 33

Answer

(i) 87 → 90 and 28 → 30.

Required quotient = $\dfrac{90}{30}$ = 3.

(ii) 84 → 80 and 23 → 20.

Required quotient = $\dfrac{80}{20}$ = 4.

(iii) 77 → 80 and 22 → 20.

Required quotient = $\dfrac{80}{20}$ = 4.

(iv) 198 → 200 and 24 → 20.

Required quotient = $\dfrac{200}{20}$ = 10.

(v) 355 → 360 and 26 → 30.

Required quotient = $\dfrac{360}{30}$ = 12.

(vi) 444 → 440 and 42 → 40.

Required quotient = $\dfrac{440}{40}$ = 11.

(vii) 843 → 840 and 33 → 30.

Required quotient = $\dfrac{840}{30}$ = 28.

There are 768 houses in a town and about 7 people live in each house. Round off these two numbers to the nearest ten and estimate the total number of people living in the town.

Answer

768 after rounding off to the nearest ten = 770

7 after rounding off to the nearest ten = 10

Estimated total number of people = 770 × 10 = 7,700

Hence, the estimated total number of people living in the town is 7,700.

There are 514 mango trees in an orchard. The owner plans to plant 178 more mango trees. Estimate how many mango trees will be there in the orchard after rounding off to the nearest hundred?

Answer

514 after rounding off to the nearest hundred = 500

178 after rounding off to the nearest hundred = 200

Estimated total number of mango trees = 500 + 200 = 700

Hence, there will be approximately 700 mango trees in the orchard.

A machine manufactures 5,782 screws every day. If we round off the number of screws produced in a day to the nearest thousand, approximately how many screws will the machine manufacture in the month of April?

Answer

5,782 after rounding off to the nearest thousand = 6,000

Number of days in the month of April = 30

Estimated number of screws manufactured = 6,000 × 30 = 1,80,000

Hence, the machine will manufacture approximately 1,80,000 screws in the month of April.

The weight of a box is 4 kg 859 g. What will be the approximate weight of 150 such boxes if we round off the weight of each box to 5 kg?

Answer

Weight of each box after rounding off = 5 kg

Estimated weight of 150 such boxes = 150 × 5 = 750 kg

Hence, the approximate weight of 150 boxes is 750 kg.

The sum of place value of two 3s in the number 73423 is:

1. 76423

2. 73426

3. 3003

4. none of these

Answer

In the number 73423 :

The place value of 3 in the thousands place = 3000

The place value of 3 in the ones place = 3

Sum of the place values = 3000 + 3 = 3003

Hence, option 3 is the correct option.

53729 is greater than 53972:

1. true

2. false

3. none of these

Answer

Comparing 53729 and 53972, the first two digits (5 and 3) are the same.

Comparing the next digit, 7 < 9.

So, 53729 < 53972.

Therefore, the given statement is false.

Hence, option 2 is the correct option.

The difference between the largest and the smallest 3-digit numbers is:

1. 10,000

2. 899

3. 1,099

4. none of these

Answer

The largest 3-digit number = 999

The smallest 3-digit number = 100

Required difference = 999 - 100 = 899

Hence, option 2 is the correct option.

The number of two digit numbers from 40 to 70 is:

1. 30

2. 31

3. 29

4. none of these

Answer

The two digit numbers from 40 to 70 (both included) are 40, 41, 42, ..., 70.

Required count = 70 - 40 + 1 = 31

Hence, option 2 is the correct option.

If A = 83, B = 12 and C = 17, the value of A - B + C is:

1. 54

2. 112

3. 88

4. none of these

Answer

A - B + C

= 83 - 12 + 17

= 71 + 17

= 88

Hence, option 3 is the correct option.

3,294 + 1,320 after rounding off to the nearest 10 is:

1. 4,430

2. 4,420

3. 4,426

4. none of these

Answer

3,294 after rounding off to the nearest ten = 3,290

1,320 after rounding off to the nearest ten = 1,320

Required sum = 3,290 + 1,320 = 4,610

Since 4,610 is not among the given options,

Hence, option 4 is the correct option.

The difference between 652 and 4,038 after rounding off each given number to the nearest hundred is:

1. 3,300

2. 3,400

3. 3,390

4. 3,350

Answer

652 after rounding off to the nearest hundred = 700

4,038 after rounding off to the nearest hundred = 4,000

Required difference = 4,000 - 700 = 3,300

Hence, option 1 is the correct option.

2,228 rounded off to the nearest hundred minus 228 to the nearest ten is:

1. 1,790

2. 1,970

3. 2,170

4. 2,220

Answer

2,228 after rounding off to the nearest hundred = 2,200

228 after rounding off to the nearest ten = 230

Required difference = 2,200 - 230 = 1,970

Hence, option 2 is the correct option.

Sum of digits 3 and 8 used in the numbers 5,03,489 is:

1. 3,800

2. 3,030

3. 11

4. none of these

Answer

The digits referred to are 3 and 8.

Sum of the digits = 3 + 8 = 11

Hence, option 3 is the correct option.

80 is multiplied with 23 instead of 32; the result of multiplication differs by:

1. 80 × (23 - 32)

2. 720

3. 80 × 32 - 23

4. none of these

Answer

Correct product = 80 × 32 = 2,560

Product obtained = 80 × 23 = 1,840

Difference = 2,560 - 1,840 = 720

Hence, option 2 is the correct option.

Statement 1: The smallest number of 6-digits using only the first three digits is 100022.

Statement 2: To get the largest number using a given set of digits, the largest digit is written at the right most and then the remaining digits are written in descending order of their values to the right.

Which of the following options is correct?

1. Both the statements are true.

2. Both the statements are false.

3. Statement 1 is true, and statement 2 is false.

4. Statement 1 is false, and statement 2 is true.

Answer

Statement 1 : The first three digits are 0, 1 and 2. To form the smallest 6-digit number using these digits, the smallest non-zero digit (1) is placed at the left most place, followed by 0s, and 2 is placed at the ones place. So, the smallest 6-digit number formed is 100002, and not 100022. Hence, Statement 1 is false.

Statement 2 : To get the largest number using a given set of digits, the largest digit is written at the left most place (and not the right most place), and then the remaining digits are written in descending order of their values to the right. Hence, Statement 2 is false.

Hence, option 2 is the correct option.

25 - 42 + 63 = ?

Statement 1: 25 - 42 + 63 before rounding off the given numbers = 88 - 42 = 46 = 50

Statement 2: After rounding off the given numbers to the nearest 10;
25 - 42 + 63 = 30 - 40 + 60 = 50

1. Both the statements are true.

2. Both the statements are false.

3. Statement 1 is true, and statement 2 is false.

4. Statement 1 is false, and statement 2 is true.

Answer

Statement 1 : Without rounding off, 25 - 42 + 63 = 88 - 42 = 46. The statement then writes 46 = 50, which is incorrect since 46 ≠ 50. Hence, Statement 1 is false.

Statement 2 : Rounding off each number to the nearest 10, we get 25 → 30, 42 → 40 and 63 → 60. So, 30 - 40 + 60 = 90 - 40 = 50. Hence, Statement 2 is true.

So, Statement 1 is false and Statement 2 is true.

Hence, option 4 is the correct option.

Number 89623

Assertion (A): The sum of place values of digits 6 and 2 is 620.

Reason (R): The required sum = 600 + 2 = 602.

1. A is true, R is false.

2. A is false, R is true.

3. Both A and R are true.

4. Both A and R are false.

Answer

Assertion (A) : In 89623, the digit 6 is at the hundreds place, so its place value = 600, and the digit 2 is at the tens place, so its place value = 20. Their sum = 600 + 20 = 620. So, A is true.

Reason (R) : The reason takes the place value of 2 as 2 and writes the sum as 600 + 2 = 602. But the place value of 2 is 20, not 2. So, R is false.

So, A is true and R is false.

Hence, option 1 is the correct option.

Assertion (A): In the number 709, the sum of place value and face value of 0 is 0.

Reason (R): The place value of a digit depends upon the position it occupies in the number, however the face value of a digit is always the digit itself regardless of the place it occupies in the given number.

1. A is true, R is false.

2. A is false, R is true.

3. Both A and R are true.

4. Both A and R are false.

Answer

Assertion (A) : In 709, the digit 0 is at the tens place, so its place value = 0 × 10 = 0, and its face value = 0. Sum = 0 + 0 = 0. So, A is true.

Reason (R) : The place value of a digit depends on its position in the number, whereas the face value of a digit is always the digit itself, irrespective of its position. This is a correct statement, so R is true.

So, both A and R are true.

Hence, option 3 is the correct option.