125876 + Interview Questions in General QA

1.	If the LCM of two numbers is 290 and their product is 7250, what will be its HCF?
Answer» Given LCM of two numbers = 290 Product of two numbers = 7250 HCF of two numbers = ? We know that, LCM × HCF = Product of two numbers 290 × HCF = 7250 HCF = \(\frac{7250}{290}\) = 25 ∴ HCF of two numbers = 25

1.

If the LCM of two numbers is 290 and their product is 7250, what will be its HCF?

Answer»

Given LCM of two numbers = 290

Product of two numbers = 7250

HCF of two numbers = ?

We know that, LCM × HCF = Product of two numbers

290 × HCF = 7250

HCF = \(\frac{7250}{290}\) = 25

∴ HCF of two numbers = 25

Discussion

2.	The LCM of 21, 35, 40 is ………………
Answer» Correct Answer is 210

Discussion

3.	Write the nearest number to 1234 which is divisible by 11.
Answer» If we divide 1234 by 11 we get remainder 2. So, 1234 – 2 = 1232 is divisible by 11 (Difference of sum of odd places digits and sum of even places digits of 1232 is ‘0’) Therefore, the nearest number to 1234 which is divisible by 11 is 1232.

3.

Write the nearest number to 1234 which is divisible by 11.

Answer»

If we divide 1234 by 11 we get remainder 2.

So, 1234 – 2 = 1232 is divisible by 11

(Difference of sum of odd places digits and sum of even places digits of 1232 is ‘0’)

Therefore, the nearest number to 1234

which is divisible by 11 is 1232.

Discussion

4.	Write the nearest number to 12345 which is divisible by 4.
Answer» Given number be 12345. A number to be divisible by 4, the number formed by the last two digits is divisible by 4. So, 12345 – 1 = 12344 is divisible by 4 12345 – 2 = 12343 is not divisible by 4 12345 – 3 = 12342 is not divisible by 4 12345 – 4 = 12341 is not divisible by 4 12345 – 5 = 12340 is divisible by 4 and 12345 + 1 = 12346 is not divisible by 4 12345 + 2 = 12347 is not divisible by 4 12345 + 3 = 12348 is divisible by 4 12340, 12344 and 12348 are divisible by 4. ∴ 12344 is the nearest number to 12345 which is divisible by 4.

4.

Write the nearest number to 12345 which is divisible by 4.

Answer»

Given number be 12345.

A number to be divisible by 4, the number formed by the last two digits is divisible by 4.

So, 12345 – 1 = 12344 is divisible by 4

12345 – 2 = 12343 is not divisible by 4

12345 – 3 = 12342 is not divisible by 4

12345 – 4 = 12341 is not divisible by 4

12345 – 5 = 12340 is divisible by 4

and

12345 + 1 = 12346 is not divisible by 4

12345 + 2 = 12347 is not divisible by 4

12345 + 3 = 12348 is divisible by 4

12340, 12344 and 12348 are divisible by 4.

∴ 12344 is the nearest number to 12345 which is divisible by 4.

Discussion

5.	Match the following :Divisibility RuleNumber1) Difference of the digits in the odd and even places is divisible by 11a) 102) Number formed by the last three digits(from the right) is divisible by 8b) 93) If the number has 0 in the units placec) 124) If the sum of the digits is divisible by 9d) 115) If the number is divisible by 3 and 4e) 8
Answer» Correct Answer is 1) – d) 11 2) – e) 8 3) – a) 10 4) – b) 9 5) – c) 12

5.

Match the following :Divisibility RuleNumber1) Difference of the digits in the odd and even places is divisible by 11a) 102) Number formed by the last three digits(from the right) is divisible by 8b) 93) If the number has 0 in the units placec) 124) If the sum of the digits is divisible by 9d) 115) If the number is divisible by 3 and 4e) 8

Answer»

Correct Answer is

1) – d) 11

2) – e) 8

3) – a) 10

4) – b) 9

5) – c) 12

Discussion

6.	Write all the possible numbers between 2000 and 2100, that are divisible by 11.
Answer» Numbers between 2000 and 2100 are 2001, 2002, 2003, ……… , 2097, 2098, 2099. If we divide 2000 by 11 we get remainder 9. By adding 2 to the 2000, then we get 2002. Check by 11 divisibility rule, 2002 is divisible by 11. (difference of sum of odd places digits and sum of even places digits is ‘0’.) Then 11 multiples after 2002 are 2013, 2024, 2035, 2046, 2057, 2068, 2079, 2090, 2101, …… 2112. Therefore, 2002, 2013, 2024, 2035, 2046, 2057, 2068, 2079 and 2090 are the numbers divisible by 11 in between 2000 and 2100.

6.

Write all the possible numbers between 2000 and 2100, that are divisible by 11.

Answer»

Numbers between 2000 and 2100 are 2001, 2002, 2003, ……… , 2097, 2098, 2099.

If we divide 2000 by 11 we get remainder 9.

By adding 2 to the 2000, then we get 2002.

Check by 11 divisibility rule, 2002 is divisible by 11. (difference of sum of odd places digits and sum of even places digits is ‘0’.)

Then 11 multiples after 2002 are 2013, 2024, 2035, 2046, 2057, 2068, 2079, 2090, 2101, …… 2112.

Therefore, 2002, 2013, 2024, 2035, 2046, 2057, 2068, 2079 and 2090 are the numbers divisible by 11 in between 2000 and 2100.

Discussion

7.	Using divisibility rules, determine which of the following numbers are divisible by 11. i) 6446 ii) 10934 iii) 7138965 iv) 726352
Answer» i) 6446 If the difference between the sum of the digits at odd places and the sum of the digits at even places of a number is either 0 or a multiple of 11. Then the number is divisible by 11. Sum of the digits at odd places = 6 + 4 = 10 Sum of the digits at even places = 4 + 6 = 10 Difference 10 – 10 = 0 So, 6446 is divisible by 11. ii) 10934 Sum of the digits at odd places = 4 + 9 + 1 = 14 Sum of the digits at even places = 3 + 0 = 3 Difference = 14 – 3 = 11 is a 11 multiple by divisibility rule for 11. So, 10934 is divisible by 11. iii) 7138965 Sum of the digits at odd places = 5 + 9 + 3 + 7 = 24 Sum of the digits at even places = 6 + 8 + 1 = 15 Difference = 24 – 15 = 9 is not a multiple by divisibility rule for 11 So, 7138965 is not divisible by 11. iv)726352 Sum of the digits at odd places = 2 + 3 + 2 = 7 Sum of the digits at even places = 5 + 6 + 7 = 18 Difference = 18 – 7 = 11 is a 11 multiple by divisibility rule for 11. So, 726352 is divisible by 11.

7.

Using divisibility rules, determine which of the following numbers are divisible by 11. i) 6446 ii) 10934 iii) 7138965 iv) 726352

Answer»

i) 6446

If the difference between the sum of the digits at odd places and the sum of the digits at even places of a number is either 0 or a multiple of 11.

Then the number is divisible by 11.

Sum of the digits at odd places = 6 + 4 = 10

Sum of the digits at even places = 4 + 6 = 10

Difference 10 – 10 = 0

So, 6446 is divisible by 11.

ii) 10934

Sum of the digits at odd places = 4 + 9 + 1 = 14

Sum of the digits at even places = 3 + 0 = 3

Difference = 14 – 3 = 11

is a 11 multiple by divisibility rule for 11.

So, 10934 is divisible by 11.

iii) 7138965

Sum of the digits at odd places = 5 + 9 + 3 + 7 = 24

Sum of the digits at even places = 6 + 8 + 1 = 15

Difference = 24 – 15 = 9

is not a multiple by divisibility rule for 11

So, 7138965 is not divisible by 11.

iv)726352

Sum of the digits at odd places = 2 + 3 + 2 = 7

Sum of the digits at even places = 5 + 6 + 7 = 18

Difference = 18 – 7 = 11

is a 11 multiple by divisibility rule for 11.

So, 726352 is divisible by 11.

Discussion

8.	If the HCF and LCM of two numbers are 1 and 91, then the product of two numbers is ………………
Answer» Correct Answer is 91

Discussion

9.	Match the following :Divisibility RuleNumber1) If sum of the digits of a number divisible by 3a) 62) Number has 0, 2, 4, 6, 8 in units placeb) 43) Number formed by the last two digits (from the right) is divisible by 4c) 54) If the number is divisible by 2 and 3d) 35) If the number Has 0 or 5 in units placee) 2
Answer» Correct Answer is 1) – d) 3 2) – e) 2 3) – b) 4 4) – a) 6 5) – c) 5

9.

Match the following :Divisibility RuleNumber1) If sum of the digits of a number divisible by 3a) 62) Number has 0, 2, 4, 6, 8 in units placeb) 43) Number formed by the last two digits (from the right) is divisible by 4c) 54) If the number is divisible by 2 and 3d) 35) If the number Has 0 or 5 in units placee) 2

Answer»

Correct Answer is

1) – d) 3

2) – e) 2

3) – b) 4

4) – a) 6

5) – c) 5

Discussion

10.	Think about the situation where we work with the rounded off number and when we need the exact number.
Answer» In distance, speed, time etc., we can estimate the number. But in rent, bill, salary etc. we need exact number.

Discussion

11.	The nearest number to 2020 which is divisible by 11 is ……………..
Answer» Correct Answer is 2024

Discussion

12.	Two milk cans have 60 and 165 liters of milk. Find a can of maximum capacity which can exactly measure the milk in two cans.
Answer» Capacity of first can = 60 l Capacity of second can = 165 l Capacity of required can = HCF (60,165) HCF of 60 and 165 is 15. Therefore maximum capacity of the required can is 15 liters.

12.

Two milk cans have 60 and 165 liters of milk. Find a can of maximum capacity which can exactly measure the milk in two cans.

Answer»

Capacity of first can = 60 l

Capacity of second can = 165 l

Capacity of required can = HCF (60,165)

HCF of 60 and 165 is 15.

Therefore maximum capacity of the required can is 15 liters.

Discussion

13.	The greatest 3-digited number which is divisible by 26, 14, 91 leaves the remainder 7 is ……………..
Answer» Correct Answer is 182

Discussion

14.	Make 3 different 3 digit numbers using 2, 3, 4 where each digit can be used only once, Check which of these numbers is divisible by 9.
Answer» 3 different 3-digit numbers using 2, 3, 4 are 234, 342, 243 Given digits 2, 3 & 4; their sum = 2 + 3 + 4 = 9, Any number formed by these digits is always divisible by 9. a) Sum of the digits of 234 = 2 + 3 + 4 = 9 is divisible by 9. If the sum of the digits of the number is divisible by 9. So, 234 is divisible by 9. b) Sum of the digits of 342 = 3 + 4 + 2 = 9 is divisible by 9. So, 342 is divisible by 9. c) Sum of the digits of 243 = 2 + 4 + 3 = 9 is divisible by 9. So, 243 is divisible by 9.

14.

Make 3 different 3 digit numbers using 2, 3, 4 where each digit can be used only once, Check which of these numbers is divisible by 9.

Answer»

3 different 3-digit numbers using 2, 3, 4 are 234, 342, 243

Given digits 2, 3 & 4; their sum = 2 + 3 + 4 = 9,

Any number formed by these digits is always divisible by 9.

a) Sum of the digits of 234

= 2 + 3 + 4 = 9 is divisible by 9.

If the sum of the digits of the number is divisible by 9.

So, 234 is divisible by 9.

b) Sum of the digits of 342 = 3 + 4 + 2 =

9 is divisible by 9.

So, 342 is divisible by 9.

c) Sum of the digits of 243 = 2 + 4 + 3 =

9 is divisible by 9.

So, 243 is divisible by 9.

Discussion

15.	Rounding numbers to the nearest tens.(i) Which flag is closer to 10 ? (ii) Which flag is closer to 20 ?
Answer» (i) flag 13 is closer to 10. (ii) Flag 17 is closer to 20.

15.

Rounding numbers to the nearest tens.(i) Which flag is closer to 10 ? (ii) Which flag is closer to 20 ?

Answer»

(i) flag 13 is closer to 10.

(ii) Flag 17 is closer to 20.

Discussion

16.	What is the least number which when divided by the numbers 3, 5, 6, 8, 10 and 12 leaves in each case a remainder 2, but when divided by 13 leaves no remainder.(a) 312 (b) 962 (c) 1562 (d) 1586
Answer» Answer is: (b) 962

Discussion

17.	Find the smallest number that must be added to 128, so that it becomes exactly divisible by 5.
Answer» Given number be 128. A number to be divisible by 5, its unit digit must be ‘0’ or ‘5’. So, 128 + 2 = 130 is divisible by 5 128 + 7 = 135 is divisible by 5 In these two numbers, 2 is the smallest number. The smallest number to be added is 2.

17.

Find the smallest number that must be added to 128, so that it becomes exactly divisible by 5.

Answer»

Given number be 128.

A number to be divisible by 5, its unit digit must be ‘0’ or ‘5’.

So, 128 + 2 = 130 is divisible by 5

128 + 7 = 135 is divisible by 5

In these two numbers, 2 is the smallest number.

The smallest number to be added is 2.

Discussion

18.	The smallest number which and when 7 is subtracted from it is exactly divisible by 7, 8, 9 is ……………
Answer» Correct Answer is 511

Discussion

19.	A trader paid? 18,57,750 for buying 150 television sets. Find out the cost of one television set.
Answer» Cost of 150 television set = ₹ 18,57,750 Thus, cost of 1 television set = 1857750/150 = ₹ 12,385

19.

A trader paid? 18,57,750 for buying 150 television sets. Find out the cost of one television set.

Answer»

Cost of 150 television set = ₹ 18,57,750

Thus, cost of 1 television set = 1857750/150 = ₹ 12,385

Discussion

20.	8596 cows and 7015 buffaloes are there in a village. Find out which cattle is more than the other and how much? (Rounding the number to 100).
Answer» Number of cows = 8596 Round off 8596 to nearest hundred = 8600 Number of buffaloes = 7015 Round off 7015 to nearest hundred = 7000 Difference = 8600 – 7000 = 1600 Thus, cows are more than others, approximately 1600

20.

8596 cows and 7015 buffaloes are there in a village. Find out which cattle is more than the other and how much? (Rounding the number to 100).

Answer»

Number of cows = 8596

Round off 8596 to nearest hundred = 8600

Number of buffaloes = 7015

Round off 7015 to nearest hundred = 7000

Difference = 8600 – 7000 = 1600

Thus, cows are more than others, approximately 1600

Discussion

21.	Write the following numbers ascending order.(i) 8435, 4835, 13584, 5348, 25843 (ii) 1100, 1001, 1011, 1010 (iii) 50500, 50050, 55555, 50505 (iv) 58695376, 58685376,58695306, 58685378
Answer» Writing the numbers in ascending order : (i) 4835 < 5348 < 8435 < 13584 < 25843 (ii) 1001 < 1010 < 1011 < 1100 (iii) 50050 < 50500 < 50505 < 55555 (iv) 58685376 < 58685378 < 58695306 < 58695376

21.

Write the following numbers ascending order.(i) 8435, 4835, 13584, 5348, 25843 (ii) 1100, 1001, 1011, 1010 (iii) 50500, 50050, 55555, 50505 (iv) 58695376, 58685376,58695306, 58685378

Answer»

Writing the numbers in ascending order :

(i) 4835 < 5348 < 8435 < 13584 < 25843

(ii) 1001 < 1010 < 1011 < 1100

(iii) 50050 < 50500 < 50505 < 55555

(iv) 58685376 < 58685378 < 58695306 < 58695376

Discussion

22.	An airplane travels 685 kilometers in an hour. How much distance would it cover in 36 hours?
Answer» In 1 hour, distance travelled by airplane = 685 km Thus, in 36 hours distance travelled by airplane. = 685 × 36 km = 24,660 km

22.

An airplane travels 685 kilometers in an hour. How much distance would it cover in 36 hours?

Answer»

In 1 hour, distance travelled by airplane = 685 km

Thus, in 36 hours distance travelled by airplane. = 685 × 36 km = 24,660 km

Discussion

23.	Write 876259, 876295, 876952, 879652, 879256 in ascending order.
Answer» Writing the numbers in ascending order : 876259 < 876295 < 876952 < 879256 < 879652

23.

Write 876259, 876295, 876952, 879652, 879256 in ascending order.

Answer»

Writing the numbers in ascending order :

876259 < 876295 < 876952 < 879256 < 879652

Discussion

24.	A cricketer made 15030 runs in Test Cricket and 18999 runs in One day Cricket. How many runs were made in both types of games?
Answer» Rim scored in Test Cricket = 15,030 Run scored in One day Cricket = 18,999 Thus total runs scored = 15,030 + 18,999 = 34,029

24.

A cricketer made 15030 runs in Test Cricket and 18999 runs in One day Cricket. How many runs were made in both types of games?

Answer»

Rim scored in Test Cricket = 15,030

Run scored in One day Cricket = 18,999

Thus total runs scored = 15,030 + 18,999 = 34,029

Discussion

25.	Members of a self-employment group makes 1,385 Papads daily. How many Papads would they make in August?
Answer» Papads made in 1 day = 1,385 Number of days in August month = 31 Thus, total papads made in August month = 1,385 × 31 = 42,935

25.

Members of a self-employment group makes 1,385 Papads daily. How many Papads would they make in August?

Answer»

Papads made in 1 day = 1,385 Number of days in August month = 31

Thus, total papads made in August month = 1,385 × 31 = 42,935

Discussion

26.	Learn these 9 + 1 = 10 99 + 1 = 100 999 + 1 = …………….. 9,999 + 1 = …………….. 99,999 + 1 = …………….. 9,99,999 + 1 = …………….. 99,99,999 + 1 = 1,00,00,000 10 × 10 = 100 100 × 10 = 1,000 1,000 × 10 = 10,000 10,000 × 10 = 1,00,000 1,00,000 × 10 = 10,00,000 10,00,000 × 10 = 1,00,00,000
Answer» 9 + 1 = 10 99 + 1 = 100 999 + 1 = 1,000 9,999 + 1 10,000 99,999 + 1 = 1,00,000 9,99,999 + 1 = 10,00,000 99,99,999 + 1 = 1,00,00,000

26.

Learn these 9 + 1 = 10 99 + 1 = 100 999 + 1 = …………….. 9,999 + 1 = …………….. 99,999 + 1 = …………….. 9,99,999 + 1 = …………….. 99,99,999 + 1 = 1,00,00,000 10 × 10 = 100 100 × 10 = 1,000 1,000 × 10 = 10,000 10,000 × 10 = 1,00,000 1,00,000 × 10 = 10,00,000 10,00,000 × 10 = 1,00,00,000

Answer»

9 + 1 = 10

99 + 1 = 100

999 + 1 = 1,000

9,999 + 1 10,000

99,999 + 1 = 1,00,000

9,99,999 + 1 = 10,00,000

99,99,999 + 1 = 1,00,00,000

Discussion

27.	Write 619407, 704916, 691740, 619704 in descending order.
Answer» Writting the numbers in descending order : 704916 > 691740 > 619704 > 619407

27.

Write 619407, 704916, 691740, 619704 in descending order.

Answer»

Writting the numbers in descending order :

704916 > 691740 > 619704 > 619407

Discussion

28.	Take six digits of your choice and form different numbers and compare them.
Answer» Taking 6 digits, 7, 4, 8, 3, 5, 9 and forming the number – 345789,754398,987543,839547,495378,589374 On comparing – 987543 > 839547 > 754398 > 589374 > 495378 > 345789

Discussion

29.	Find the difference between the biggest and smallest numbers obtained by using all the digits 5, 3, 9, 7, 4 once.
Answer» Largest number = 97,543 and smallest number = 34,579 Thus, difference between two numbers = 97,543 – 34,579 = 62,964

29.

Find the difference between the biggest and smallest numbers obtained by using all the digits 5, 3, 9, 7, 4 once.

Answer»

Largest number = 97,543 and smallest number = 34,579

Thus, difference between two numbers = 97,543 – 34,579 = 62,964

Discussion

30.	Given the numbers 5, 7, 0, 6, 1, 3 and 4. Form five numbers of seven digits using these.
Answer» Using digit 5, 7, 0, 6, 1, 3 and 4 seven digit, five numbers are as follows : 1630457, 4753160, 7105364, 5047316, 3507641

Discussion

31.	Using digits 6, 1, 4 make largest and smallest number then find their difference.
Answer» Largest number = 641 Smallest number – 146 Thus, difference = 641 – 146 = 495

31.

Using digits 6, 1, 4 make largest and smallest number then find their difference.

Answer»

Largest number = 641

Smallest number – 146

Thus, difference = 641 – 146 = 495

Discussion

32.	Fill in the blanks (i) 1 million = …………….. lakh (ii) 1 crore = …………….. million(iii) 1 km = …………….. m (iv) 1 gm = …………….. mg (v) 1 litre = …………….. ml
Answer» (i) 10 (ii) 10 (iii) 1000 (iv) 1000 (v) 1000

32.

Fill in the blanks (i) 1 million = …………….. lakh (ii) 1 crore = …………….. million(iii) 1 km = …………….. m (iv) 1 gm = …………….. mg (v) 1 litre = …………….. ml

Answer»

(i) 10

(ii) 10

(iii) 1000

(iv) 1000

(v) 1000

Discussion

33.	Smallest four digit number will be : (i) 9,900 (ii) 1,099 (iii) 9,999 (iv) 1000
Answer» (iv) The smallest number is 1000

Discussion

34.	Keeping the place value of number six at the same place and jumbling the digits of number 6350947; the smallest number shall be: (i) 6975430 (ii) 6043579 (iii) 6034579 (iv) 6034759
Answer» Required number will be 6034579. Thus, option (iii) is correct.

Discussion

35.	The largest five digit number using digits 7, 8 and 9 is : (i) 98978 (ii) 99897 (iii) 99987 (iv) 98799
Answer» Required number will be 99987. Thus, option (iii) is correct.

Discussion

36.	What number will we get when we add 1 to the largest number of six digit?
Answer» We know that largest 6-digit number is 9,99,999. So we add 1 to 9,99,999 and we get 9,99,999 + 1 = 10,00,000

36.

What number will we get when we add 1 to the largest number of six digit?

Answer»

We know that largest 6-digit number is 9,99,999.

So we add 1 to 9,99,999 and we get 9,99,999 + 1 = 10,00,000

Discussion

37.	Population of a town in first year is 2,90,770 and in second year is 3,07,640. Find total population of both years. (Round off nearest 100).
Answer» Population in 1st year= 2,90,770 Round off to nearest hundred = 2,90,800 Population in 2nd year = 3,07,640 Round off to nearest hundred = 3,07,600 Thus, total population = 2,90,800 + 3,07,600 = 5,98,400 Total population of two years = 5,98,400 (approx.)

37.

Population of a town in first year is 2,90,770 and in second year is 3,07,640. Find total population of both years. (Round off nearest 100).

Answer»

Population in 1st year= 2,90,770

Round off to nearest hundred = 2,90,800

Population in 2nd year = 3,07,640

Round off to nearest hundred = 3,07,600

Thus, total population = 2,90,800 + 3,07,600 = 5,98,400

Total population of two years = 5,98,400 (approx.)

Discussion

38.	Using 7, 9, 2, 8 largest number formed will be : (i) 7,928 (ii) 2,879 (iii) 9,872 (iv) 9,782
Answer» (iii) The largest number 9872

Discussion

39.	75,000 sheets of paper are available for making Practice books. From every sheet, 8 pages of Practice book are made. Every Practice book has 200 pages. How many Practice books can be made from the available sheets of paper?
Answer» Number of sheets = 75,000 Number of pages formed by 1 sheet = 8 ∴ Number of pages formed by 75,000 sheets = 75,000 × 8 = 6,00,000 No. of pages each Practice book has = 200 Thus, number of books made = 600000/200 = 3,000

39.

75,000 sheets of paper are available for making Practice books. From every sheet, 8 pages of Practice book are made. Every Practice book has 200 pages. How many Practice books can be made from the available sheets of paper?

Answer»

Number of sheets = 75,000

Number of pages formed by 1 sheet = 8

∴ Number of pages formed by 75,000 sheets = 75,000 × 8 = 6,00,000

No. of pages each Practice book has = 200

Thus, number of books made = 600000/200 = 3,000

Discussion

40.	Using 2, 4, 1 six digit smallest number will be : (i) 124112 (ii) 111124 (iii) 112141 (iv) 424211
Answer» (ii) Six digit smallest number will be 111124

Discussion

41.	Round off 78 to nearest ten will be :(i) 70 (ii) 75 (iii) 80 (iv) 78
Answer» (iii) The nearest ten will be 80

Discussion

42.	Round off 946 to nearest hundred will be :(i) 940 (ii) 945 (iii) 950 (iv) 900
Answer» (iv) The nearest hundred will be 900

Discussion

43.	Write the following numbers in words. (i) 5,782 (ii) 75,879 (iii) 3,89,087 (iv) 21,32,452 (v) 7,68,92,479 (vi) 50,60,798
Answer» In words, numbers are written as : (i) Five thousand seven hundred eighty two (ii) Seventy five thousand eight hundred seventy nine (iii) Three lakh eighty nine thousand eighty seven (iv) Twenty one lakh thirty two thousand four hundred fifty two (v) Seven crore sixty eight lakh ninety two thousand four hundred seventy nine (vi) Fifty lakh sixty thousand seven hundred ninety eight

43.

Write the following numbers in words. (i) 5,782 (ii) 75,879 (iii) 3,89,087 (iv) 21,32,452 (v) 7,68,92,479 (vi) 50,60,798

Answer»

In words, numbers are written as :

(i) Five thousand seven hundred eighty two

(ii) Seventy five thousand eight hundred seventy nine

(iii) Three lakh eighty nine thousand eighty seven

(iv) Twenty one lakh thirty two thousand four hundred fifty two

(v) Seven crore sixty eight lakh ninety two thousand four hundred seventy nine

(vi) Fifty lakh sixty thousand seven hundred ninety eight

Discussion

44.	A river 3 m deep and 40 m wide is flowing at the rate of 2 km per hour. How much water will fall into the sea in a minute?
Answer» Radius of water flow = 2 km per hour = (2000/60) m/min = (100/3) m/min Depth (h) of river = 3m Width (b) of river = 40m Volume of water followed in 1 min = 100/3 x 40 x 3 m³ Thus, 1 minute 4000 m³ = 4000000 liters of water will fall in sea.

44.

A river 3 m deep and 40 m wide is flowing at the rate of 2 km per hour. How much water will fall into the sea in a minute?

Answer»

Radius of water flow = 2 km per hour = (2000/60) m/min

= (100/3) m/min

Depth (h) of river = 3m

Width (b) of river = 40m

Volume of water followed in 1 min = 100/3 x 40 x 3 m³

Thus, 1 minute 4000 m³ = 4000000 liters of water will fall in sea.

Discussion

45.	Water in a canal 30 dm wide and 12 dm deep, is flowing with a velocity of 100 km per hour. How much area will it irrigate in 30 minutes if 8 cm of standing water is desired?
Answer» Given, Width of canal = 30 dm = 3m Depth of canal = 12 dm = 1.2 m Velocity of flow = 100 km/h Length of cuboid thus formed = distance travelled by water in 30 minute at speed 100km/h \(=100\times\cfrac{30}{60}\) = 50 km = 50000m Volume of water = 50000 × 3 × 1.2 m³ Water accumulated in field forms a cuboid of base area equals to area of field and height = \(\cfrac{8}{100}m\) = hence, area of field = \(\cfrac{5000\times3\times1.2\times100}8\) = 2250000m²

45.

Water in a canal 30 dm wide and 12 dm deep, is flowing with a velocity of 100 km per hour. How much area will it irrigate in 30 minutes if 8 cm of standing water is desired?

Answer»

Given,

Width of canal = 30 dm = 3m

Depth of canal = 12 dm = 1.2 m

Velocity of flow = 100 km/h

Length of cuboid thus formed = distance travelled by water in 30 minute at speed 100km/h

\(=100\times\cfrac{30}{60}\) = 50 km = 50000m

Volume of water = 50000 × 3 × 1.2 m³

Water accumulated in field forms a cuboid of base area equals to area of field and height = \(\cfrac{8}{100}m\)

= hence, area of field = \(\cfrac{5000\times3\times1.2\times100}8\) = 2250000m²

Discussion

46.	A river 3 m deep and 40 m wide is flowing at the rate of 2 km per hour. How much water will fall into the sea in a minute ?
Answer» Depth of river, h = 3m breadth, b = 40 m 1 Km. = 1000 m ∴ 2 Km. = 2000 m. Speed of the river per hour is 2 Km. For 60 minutes, 2000m For 1 minute …?… \(= \frac{2000}{60}\) = \(\frac{100}{3}\)m/minute ∴ Depth of river in 1 minute, l = \(\frac{100}{3}\)m. Volume of water flows in 1 minute, v = l × b × h = \(\frac{100}{3}\) × 40 × 3 = 4000 m³ ∴ In One minute, 4000 m³ water reaches the sea.

46.

A river 3 m deep and 40 m wide is flowing at the rate of 2 km per hour. How much water will fall into the sea in a minute ?

Answer»

Depth of river, h = 3m

breadth, b = 40 m

1 Km. = 1000 m

∴ 2 Km. = 2000 m.

Speed of the river per hour is 2 Km.

For 60 minutes, 2000m

For 1 minute …?…

\(= \frac{2000}{60}\)

= \(\frac{100}{3}\)m/minute

∴ Depth of river in 1 minute, l = \(\frac{100}{3}\)m.

Volume of water flows in 1 minute,

v = l × b × h

= \(\frac{100}{3}\) × 40 × 3 = 4000 m³

∴ In One minute, 4000 m³ water reaches the sea.

Discussion

47.	A river 3 m deep and 40 m wide is flowing at the rate of 2 km per hour. How much water will fall into the sea in a minute?
Answer» Water flow of a river = 2 km per hour = \((\frac {2000}{60}) \)m/min or \((\frac {100}{3}) \) m/min [we know: 1 km = 1000 m and 1 hour = 60 mins] Depth of the river (h) = 3 m Width of the river (b) = 40 m Volume of the water flowing in 1 min = \(\frac {100}{3}\) x 40 x 3 = 4000 m³ Or 4000 m³ = 4000000 litres Therefore, in 1 minute 4000000 litres of water will fall in the sea.

47.

A river 3 m deep and 40 m wide is flowing at the rate of 2 km per hour. How much water will fall into the sea in a minute?

Answer»

Water flow of a river = 2 km per hour = \((\frac {2000}{60}) \)m/min or \((\frac {100}{3}) \) m/min

[we know: 1 km = 1000 m and 1 hour = 60 mins]

Depth of the river (h) = 3 m

Width of the river (b) = 40 m

Volume of the water flowing in 1 min = \(\frac {100}{3}\) x 40 x 3 = 4000 m³

Or 4000 m³ = 4000000 litres

Therefore, in 1 minute 4000000 litres of water will fall in the sea.

Discussion

48.	Check whether 33 – 32 is divisible by 3. Explain
Answer» 33 – 32 = 27 – 9 = 18 → 1 + 8 = 9 ⇒ 9/3 (R = 0) ∴ It is divisible by ‘3’.

48.

Check whether 33 – 32 is divisible by 3. Explain

Answer»

33 – 32 = 27 – 9 = 18 → 1 + 8 = 9 ⇒ 9/3 (R = 0)

∴ It is divisible by ‘3’.

Discussion

49.	2A8 is a number divisible by 2, what might be the value of A’?
Answer» If the units digit of a number be 0, 2, 4, 6, 8 then it is divisible by 2. ∴ 2A8 is divisible by 2 for any value of A. ∴ A = (0, 1, 2 ………………….9)

49.

2A8 is a number divisible by 2, what might be the value of A’?

Answer»

If the units digit of a number be 0, 2, 4, 6, 8 then it is divisible by 2.

∴ 2A8 is divisible by 2 for any value of A.

∴ A = (0, 1, 2 ………………….9)

Discussion

50.

Check whether 22 + 23+24 is divisible by 2 or 4 or by both 2 and 4?

Answer»

Number	Divisible by 2	Y/N	Divisible by 4	Y/N
2² + 2³ + 2⁴	= 4 + 8 + 16 = 28 =28 \(\rightarrow\) 8/2(R = 0)	Yes	=28 \(\rightarrow\) 28/4(R = 0)	Yes

∴ 2² + 2³ + 2³ is divisible by both 2 and 4.

Discussion

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