ICSE

ML Aggarwal Class 6 Solutions for ICSE Maths Chapter 1 Knowing Our Numbers Check Your Progress

Question 1.
Write the numeral for each of the following numbers and insert commas correctly:
(i) Six crore nine lakh forty seven.
(ii) One hundred four million seven hundred twenty two thousand three hundred ninety four.
Solution:
(i) 6,09,00,047
(ii) 104,722,394

Question 2.
Insert commas suitably and write the numebr 30189301 in words in Indian and International system of numeration.
Solution:
International system : 30,189,301
Three crore one lakh eighty nine thousand three hundred one Thirty million one hundred eighty nine thousand three hundred one

Question 3.
Find the difference between the place value and the face value of the digit 6 in the number 72601.
Solution:
Place value of 6

Question 4.
Write all possible two-digit number using the digits 4 and 0. repetition of digits is allowed.
Solution:
Possible digit numbers = 40, 44

Question 5.
Write all possible natural numbers using the digits 7, 0, 6. Repetition of digits is not allowed.
Solution:
The given digits are 7,0, 6 and repetition of digits is not allowed.
The one- digit numbers that can be formed are 7 and 6.
We are required to write 2-digit numbers.
Out of the given digits, the possible ways of choosing the two digits are
7, 0; 6, 0; 6, 7
Using the digits 7 and 0, the numbers are 70.
Similarily, Using the digits 6 and 0, the numbers are 60
Using the digits 6 and 7, the numbers are 67 and 76.
Hence, all possible 2-digit numbers are
60, 70, 67, 76

Now, We are required to write 3-digit numbers using the digits 7, 0, 6 and the repetition of the digits is not allowed. Keeping 0 at unit’s place, 3-digit number obtained are 670 and 760.

Keeping 6 at unit’s place, 3-digit number obtained are 706.
Keeping 7 at unit’s place, 3-digit number obtained are 607.
Hence, all possible 3-digit numbers are : 670, 760, 706 and 607.
All possible numbers using the digits 7, 0 and 6 are
6, 7, 76, 67, 70, 60, 706, 607, 760, 670.

Question 6.
Arrange the following numbers in ascending order:
3706, 58019, 3760, 59801, 560023
Solution:
3706, 3760, 58019, 59801, 560023

Question 7.
Write the greatest six-digit number using four different digits.
Solution:
Greatest six-digits number using four different digit is 999876.

Question 8.
Write the smallest eight-digit number using four different digits.
Solution:
Smallest eight-digit number = 10000023

Question 9.
Find the difference between the greatest and the smallest 4-digit numbers formed by the digits 0, 3, 6, 9.
Solution:
The greatest 4-digit number using 0, 3, 6, 9 = 9630
The smallest 4-digit number using 0, 3, 6, 9 = 3069
∴ Their Difference = 9630 – 3069 = 6561

Question 10.
Find the sum of the four-digit greatest number and the five-digit smallest number, each number having three different digits.
Solution:
Four digit greatest number with three different digits = 9987
Five digit smallest number with three different digit = 10002
∴ Their Sum = 9987 + 10002 = 19989

Question 11.
Write the greatest and the smallest four-digit numbers using four different digits with the conditions as given:
(i) Digit 3 always at hundred’s place.
(ii) Digit 0 always at ten’s place.
Solution:
(i) 9387; 1302
(ii)9807; 1203

Question 12.
A mobile number consists of ten digits. First four digits are 9, 9, 7 and 9. Make the smallest mobile number by using only one digit twice from the digits 8, 3, 5, 0, 6.
Solution:
The mobile number is 9979003568.

Question 13.
Two stitch a uniform, 1 m 75 cm cloth is needed. Out of 153 m cloth, how many uniforms can be stitched and how much cloth will remain?
Solution:
Total cloth 153 m = 15300 cm
To stich a uniform, cloth needed
= 1 m 75 cm = 175 cm
Total uniforms can be stiched = \(\frac{15300}{175}\)
\(=87 \frac{75}{175}\)
Hence, 87 uniforms can be stiched 75 cm cloth will remain extra.

Question 14.
Medicine is packed in boxes, each weighing 4 kg 500 g. How many such boxes can be loaded in a van which cannot carry beyond 800 kg?
Solution:
Weight of 1 box = 4 kg 500 gm
= 4 × 1000 + 500 = 4500 gm
Van can carry upto 800 kg
= 800 × 1000 = 800000 gm
Number of boxes that can be loaded in van
= \(\frac{800000}{4500}\) = 177.8 i.e. 177 boxes
A van can lc d 177 boxes of medicines.

Question 15.
Estimate : 6554 – 677 by estimating the numbers to their nearest
(i) thousands
(ii) hundreds
(iii) greatest places
Also point out the most reasonable estimate.
Solution:
(i) Thousands (6554 – 677)
⇒ 7000- 1000 = 6000
(ii) Hundreds (6554 – 677)
⇒ 6600 – 700 = 5900
(iii) Greatest places (6554 – 677)
⇒ 7000 – 700 = 6300
Estimation to their nearest hundreds is most reasonable.

ML Aggarwal Class 6 Solutions for ICSE Maths

ML Aggarwal Class 6 Solutions for ICSE Maths Chapter 2 Whole Numbers Ex 2.2

Question 1.
Fill in the blanks to make each of the following a true statement:
(i) 378 + 1024 = 1024 + …….
(ii) 337 + (528 + 1164) = (337 + ……..) + 1164
(iii) (21 + 18) + ……….. = (21 + 13) + 18
(iv) 3056 + 0 = ……….. = 0 + 3056
Solution:
(i) 378 + 1024= 1024 + 378 (Commutative property of addition)
(ii) 337 + (528 + 1164) = (337 + 528) + 1164 (Associative law of addition)
(iii) (21 + 18) + 13 = (21 + 13) + 18 (Associative law of addition)
(iv) 3056 + 0 = 3056 = 0 + 3056

Question 2.
Add the following numbers and check by reversing the order of addends :
(i) 3189 + 53885
(ii) 33789 + 50311.
Solution:
(i) 3189 + 53885 = 57074
Check 53885 + 3189 = 57074
∴57074

(ii) 33789 + 50311 = 84100
Check 50311 + 33789 = 84100
∴ 84100

Question 3.
By suitable arrangements, find the sum of:
(i) 311,528,289
(ii) 723, 834, 66, 277
(iii) 78, 203, 435, 7197, 422.
Solution:
(i) 311, 528, 289
Sum (311 +289)+ 528
= 600+ 528= 1128

(ii) 723 + 834 + 66 + 277
= (723 + 277) + (834 + 66)
= 1000 + 900 = 1900

(iii) 78, 203, 435, 7197, 422
Sum = (78 + 422) + (203 + 7197) + 435
= 500 + 7400 + 435
= 7900 + 435 = 8335

Question 4.
Fill in the blanks to make each of the following a true statement:
(i) 375 × 57 = 57 × ……….
(ii) (33 × 16) × 25 = 33 × (…….. × 25)
(iii) 37 × 24 = 37 × 18 + 37 × …………
(iv) 7205 × 1 = …………. = 1 × 7205
(v) 366 × 0 =
(vi) …………… × 579 = 0
(vii) 473 × 108 = 473 × 100 + 473 × ………….
(viii) 684 × 97 = 684 × 100 – …………… × 3
(ix) 0 ÷= 5 =
(x) (14 – 14) ÷ 7 = ………….
Solution:
(i) 375 × 57 = 57 × 375 (Commutative property of multiplication)
(ii) 33 × 16) × 25 = 33 × (16 × 25) (Associative law of multiplication)
(iii) 37 × 24 = 37 × 18 + 37 × 6 (Distributive law of multiplication)
(iv) 7205 × 1 = 7205 = 1 × 7205
(v) 366 × 0 = 0
(vi) 0 × 579 = 0
(vii) 473 × 108 = 473 × 100 + 473 × 8
(viii) 684 × 97 = 684 × 100 – 684 × 3
(ix) 0 ÷ 5 = 0
(x) (14 – 14) ÷ 7 = 0

Question 5.
Determine the following products by suitable arrangement:
(i) 4 × 528 × 25
(ii) 625 × 239 × 16
(iii) 125 × 40 × 8 × 25
Solution:
(i) 4 × 528 × 25 = 4 × 25 × 528
= 100 × 528 = 52800

(ii) 625 × 239 × 16 = 625 × 16 × 239
= 10000 × 239 = 2390000

(iii) 125 × 40 × 8 × 25 = 125 × 8 × 40 × 25
= 1000 × 1000 = 1000000

Question 6.
Find the value of the following:
(i) 54279 × 92 + 54279 × 8
(ii) 60678 × 262 – 60678 × 162
Solution:
(i) 54279 × 92 + 54279 × 8
= 54279 (92 + 8)
= 54279 × 100 = 5427900

(ii) 60678 × 262 – 60678 × 162
= 60678 (262 – 162)
= 60678 × 100 = 6067800

Question 7.
Find the following products by using suitable properties:
(i) 739 × 102
(ii) 1938 × 99
(iii) 1005 × 188
Solution:
(i) 739 × 102
= 739 × (100 + 2)
= 739 × 100 + 739 × 2
= 73900 + 1478 = 75378

(ii) 1938 × 99
= 1938 × (100- 1)
= 1938 × 100 – 1938 × 1
= 193800 – 1938 = 191862

(iii) 1005 × 188
= (1000 + 5) × (100 + 88)
= 1000 × 100 + 1000 × 88 + 5 × 100 + 88 × 5
= 100000 + 88000 + 500 + 440 = 188940

Question 8.
Divide 7750 by 17 and check the result by division algorithm.
Solution:
7750 ÷ 17

On dividing 7750 by 17, we get
Quotient = 455 and Remainder = 15
Check by division algorithm:
Divident = Divisior × Quotient + Remainder
= 17 × 455 + 15 = 7750

Question 9.
Find the number which when divided by 38 gives the quotient 23 and remainder 17.
Solution:
Divisor = 38,Quotient = 23
Remainder = 17
Dividend = divisor × quotient + remainder
= 38 × 23 + 17 = 874 + 17 = 891

Question 10.
Which least number should be subtracted from 1000 so that the difference is exactly divisible by 35.
Solution:
On dividing 1000 by 35
we get quotient = 28 and remainder 20

So, 20 should be subtracted from 1000.

Question 11.
Which least number should be added to 1000 so that 53 divides the sum exactly.
Solution:

On dividing 1000 by 53, we get quotient = 18 and remainder = 46. To get the remainder 0, we should add 53 – 46 = 7 to 1000.
∴ 7

Question 12.
Find the largest three-digit number which is exactly divisible by 47.
Solution:
Largest three digit no. = 999

On dividing 999 by 47, we get
Quotient = 21 and Remainder =12
So on subtracting 12 from 999, we get
999 – 12 = 987

Question 13.
Find the smallest five-digit number which is exactly divisible by 254.
Solution:
Smallest 5 digit number = 10000

On dividing 10000 by 254, we get
Remainder = 94
So 254 – 94 = 160 should be added to 10000 to get the smallest 5 digit number divisible by 254.
∴ 10000 + 160 = 10160

Question 14.
A vendor supplies 72 litres of milk to a student’s hostel in the morning and 28 litres of milk in the evening every day. If the milk costs?39 per litre, how much money is due to the vendor per day?
Solution:
Supply of milk in morning = 72 litres
Supply of milk in evening = 28 litres
Cost of per litre milk = ₹ 39
Money of per day = ₹ 39 (72 l + 28 l)
= ₹ 39 × 100 = ₹ 3900

Question 15.
State whether the following statements are true (T) or false (F):
(i) If the product of two whole numbers is zero, then atleast one of them will be zero.
(ii) If the product of two whole numbers is 1, then each of them must be equal to 1.
(iii) If a and b are whole numbers such that a ≠ 0 and b ≠ 0, then ab may be zero.
Solution:
(i) True
(ii) True
(iii)False

Question 16.
Replace each * by the correct digit in each of the following:

Solution:

ML Aggarwal Class 6 Solutions for ICSE Maths

ML Aggarwal Class 6 Solutions for ICSE Maths Chapter 2 Whole Numbers Ex 2.3

Question 1.
Using shorter method, find
(i) 3246 + 9999
(ii) 7501 + 99999
(iii) 5377 – 999
(iv) 25718 – 9999
(v) 123 × 999
(vi) 203 × 9999
Solution:
(i) 3246 + 9999
= (3246 – 1) + (9999 + 1) (Adding and subtracting 1)
= 3245 + 10000 = 13245

(ii) 7501 + 99999
= (7501 – 1) + (99999 + 1) (Adding and subtracting 1)
= 7500+ 100000 = 107500

(iii) 5377 – 999
= 5377 – (1000- 1)
= 5377 – 1000 + 1 = 4377 + 1 = 4378

(iv) 25718 – 9999
= 25718 – (10,000 – 1)
= 15718 + 1 = 15719

(v) 123 × 999
= 123 × (1000 – 1) (By subtracting 1)
= 123 × 1000 – 1 × 123 = 123000 – 123 = 122877

(vi) 203 × 9999
= 203 × (10,000 – 1) (By subtracting 1)
= 203 × 10,000 – 203 × 1 = 2030000 – 203 = 2029797

Question 2.
Without using a diagram, find
(i) 9th square number
(ii) 7th triangular number
Solution:
(i) 9th square number = ?
The first square number is 1 × 1 = 1
The second square number is 2 × 2 = 4
The third square number is 3 × 3 = 9
Similarly 9th square number is 9 × 9 = 81

(ii) 7th triangular number = ?
First triangular number = 1
Second triangular number = 1 + 2 = 3
Third triangular number = 1 + 2 + 3 = 6
Fourth triangular number = 1 + 2 + 3 + 4 = 10
Similarly 7th triangular number = 1 + 2 + 3 + 4 + 5 + 6 + 7 = 28

Question 3.
(i) Can a rectangular number be a square number?
(ii) Can a triangular number be a square number?
Solution:
(i) Yes, 9 is a square as well as rectangular number.
(ii) Yes, 8th triangular number = 36, which is a square number.

Question 4.
Observe the following pattern and fill in the blanks:
1 × 9 + 1 = 10
12 × 9 + 2= 110
123 × 9 + 3 = 1110
1234 × 9 + 4 = ……….
12345 × 9 + 5 = …………..
Solution:
1 × 9 + 1 = 10
12 × 9 + 2= 110
123 × 9 + 3 = 1110
1234 × 9 + 4 = 11110
12345 × 9 + 5 = 111110

Question 5.
Observe the following pattern and fill in the blanks:
9 × 9 + 7 = 88
98 × 9 + 6 = 888
987 × 9 + 5 = 8888
9876 x 9 + 4 = …………
98765 × 9 + 3 = ……….
Solution:
9 × 9 + 7 = 88
98 × 9 + 6 = 888
987 × 9 + 5 = 8888
9876 × 9 + 4 = 88888
98765 × 9 + 3 = 888888

ML Aggarwal Class 6 Solutions for ICSE Maths

ML Aggarwal Class 7 Solutions for ICSE Maths Chapter 1 Integers Ex 1.2

Question 1.
Write a pair of integers whose:
(i) sum is -3
(ii) difference is -5
(iii) difference is 4
Solution:
Pair of integers whose,
(i) sum is -3 = -6, 3 (-6 + 3 = -3)
(ii) difference is -5 = -2, 3 (-2 – 3 = -5)
(iii) difference is 4 = -3, -7 [-3 – (-7) = -3 + 7 = 4]

Question 2.
(i) Write a pair of negative integers whose difference is 5.
(ii) Write a negative integer and a positive integer whose sum is -8.
(iii) Write a negative integer and a positive integer whose difference is -3.
Solution:
(i) Let two negative integers whose difference is 5 be -7, -12.
Now the difference is -7 – (-12)
= -7 + 12 = 5
(ii) Let two integers one negative and other positive but sum is -8 are -13, 5
= -13 + 5 = -8
(iii) Let two integer, one negative and other positive whose difference is -3 be -1, 2
Difference = -1 – (2) = -1 – 2 = -3

Question 3.
Write two integers which are smaller than -5 but their difference is greater than -5.
Solution:
Let two integer which are smaller than -5 but their difference is greater than -5
= -6, -8 as -6 – (-8)
= -6 + 8
= 2 > -5

Question 4.
In a quiz, team A scored -30, 20, 0 and team B scored 20, 0, -30 in three successive rounds. Which team socred more? Can we
say that we can add integers in any order?
Solution:
In a quiz,
Team A scored -30, 20, 0 and
Team B scored 20, 0, -30 in three rounds
Sum of scores of A team = -30 + 20 + 0 = -10
Sum of scores of B team = 20 + 0 – 30 = -10
The scores of both the team are equal i.e. -10
Yes, by adding the scores in any order, the result will be the same.

Question 5.
Find the sum of integers -72, 237, 84, 72, -184, -37.
Solution:
Sum of integers -72, 237, 84, 72, -184, -37
= -72 + 237 + 84 + 72 + -184 + -37
= 237 + 84 + 72 + (-72 – 184 – 37)
= (393) + (-293)
= 393 – 293
= 100

ML Aggarwal Class 7 Solutions for ICSE Maths

ML Aggarwal Class 7 Solutions for ICSE Maths Chapter 4 Exponents and Powers Ex 4.1

Question 1.
Fill in the blanks:
(i) In the expression 3⁷, base = …….. and exponent = ……..
(ii) In the expression (-7)⁵, base = ………. and exponent = …….
(iii) In the expression \(\left( \frac { 2 }{ 5 } \right) ^{ 11 }\), base = …… and exponent = …….
(iv) If base is 6 and exponent is 8, then exponential form = ……..
Solution:
(i) In the expression 3⁷, base = 3 and exponent = 7.
(ii) In the expression (-7)⁵, base = -7 and exponent = 5.
(iii) In the expression \(\left( \frac { 2 }{ 5 } \right) ^{ 11 }\), base = -7 and exponent = 11.
(iv) If base is 6 and exponent is 8, then exponential form = 6⁸.

Question 2.
Find the value of the following:
(i) 2⁶
(ii) 5⁵
(v) (-6)⁴
(vi) \(\left( \frac { 2 }{ 3 } \right) ^{ 4 }\)
(v) \(\left( \frac { -2 }{ 3 } \right) ^{ 5 }\)
(vi) (-2)⁹
Solution:
(i) 2⁶ = 2 × 2 × 2 × 2 × 2 × 2 = 64
(ii) 5⁵ = 5 × 5 × 5 × 5 × 5 = 3125
(iii) (-6)⁴ = (-6) × (-6) × (-6) × (-6) = 1296

(vi) (-2)⁹ = (-2) × (-2) × (-2) × (-2) × (-2) × (-2) × (-2) × (-2) × (-2) = -512

Question 3.
Express the following in the exponential form:
(i) 6 × 6 × 6 × 6 × 6
(ii) t × t × t
(iii) 2 × 2 × a × a × a × a
(iv) a × a × a × c × c × c × c × d
Solution:
(i) 6 × 6 × 6 × 6 × 6 = 6⁵
(ii) t × t × t = t³
(iii) 2 × 2 × a × a × a × a = 2² × a⁴ = 2²a⁴
(iv) a × a × a × c × c × c × c × d = a³ × c⁴ × d = a³c⁴d

Question 4.
Simplify the following:
(i) 7 × 10³
(ii) ²⁵ × 9
(iii) 3³ × 10⁴
Solution:
(i) 7 x 10³ = 7 × 10 × 10 × 10 = 7 × 1000 = 7000
(ii) 2⁵ × 9 = 2 × 2 × 2 × 2 × 2 × 9 = 32 × 9 = 288
(iii) 3³ × 10⁴ = 3 × 3 × 3 × 10 × 10 × 10 × 10 = 27 × 10000 = 270000

Question 5.
Simplify the following:
(i) (-3) × (-2)³
(ii) (-3)² × (-5)²
(iii) (-2)³ × (-10)⁴
(iv) (-1)⁹
(v) 25² × (-1)³¹
(vi) 4² × 3³ × (-1)¹²²
Solution:
(i) (-3) × (-2)³ = (-3) × (-2) × (-2) × (-2) = 24
(ii) (-3)² × (-5)² = (-3) × (-3) × (-5) × (-5) = 9 × 25 = 225
(iii) (-2)³ × (-10)⁴ = (-2) × (-2) × (-2) × 10 × 10 × 10 × 10 = -8 × 10000 = -80000
(iv) (-1)⁹ = -1 (∵ 9 is odd)
(v) 25² × (-1)³¹ = 25 × 25 × (-1) (∵ 31 is odd)
= 625 × (-1) = -625
(vi) 4² × 3³ × (-1)¹²²
= 4 × 4 × 3 × 3 × (1) = 144 × 1 = 144 (∵ 122 is even)

Question 6.
Identify the greater number in each of the following:
(i) 4³ or 3⁴
(ii) 7³ or 3⁷
(iii) 4⁵ or 5⁴
(iv) 2¹⁰ or 10²
Solution:
(i) 4³ or 3⁴
4³ = 4 × 4 × 4 = 64
3⁴ = 3 × 3 × 3 × 3 = 81
3⁴ is greater
(ii) 7³ or 3³
7³ = 7 × 7 × 7 = 343
3⁷ = 3 × 3 × 3 × 3 × 3 × 3 × 3 = 2187
3⁷ is greater
(iii) 4⁵ or 5⁴
4⁵ = 4 × 4 × 4 × 4 × 4 = 1024
5⁴ = 5 × 5 × 5 × 5 = 625
45 is greater
(iv) 2¹⁰ or 10²
2¹⁰ = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 1024
10² = 10 × 10 = 100
2¹⁰ is greater

Question 7.
Write the following numbers as powers of 2:
(i) 8
(ii) 128
(iii) 1024
Solution:
(i) 8 = 2 × 2 × 2 = 2³
(ii) 128 = 2 × 2 × 2 × 2 × 2 × 2 × 2 = 2⁷

(iii) 1024 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 2¹⁰

Question 8.
To what power (-2) should be raised to get 16?
Solution:
16 = (-2) × (-2) × (-2) × (-2) = (-2)⁴
Power = 4

Question 9.
Write the following numbers as powers of (-3):
(i) 9
(ii) -27
(iii) 81
Solution:
(i) 9 = (-3) × (-3) = (-3)²
(ii) -27 = (-3) × (-3) × (-3) = (-3)³
(iii) 81 = (-3) × (-3) × (-3) × (-3) = (-3)⁴

Question 10.
Find the value of x in each of the following:
(i) 7^x = 343
(ii) 3^x = 729
(in) (-8)^x = -512
(iv) (-4)^x = -1024
(v) \(\left( \frac { 2 }{ 5 } \right) ^{ x }\) = \(\frac { 32 }{ 3125 }\)
(vi) \(\left( \frac { -3 }{ 4 } \right) ^{ x }\) = \(\frac { -243 }{ 1024 }\)
Solution:

Question 11.
Write the prime factorization of the following numbers in the exponential form:
(i) 72
(ii) 360
(iii) 405
(iv) 540
(v) 2280
(vi) 3600
(vii) 4725
(viii) 8400
Solution:
(i) 72 = 2 × 2 × 2 × 3 × 3 = 2³ × 3²

(ii) 360 = 2 × 2 × 2 × 3 × 3 × 5 = 2³ × 3² × 5

(iii) 405 = 3 × 3 × 3 × 3 × 5 = 3⁴ × 5¹

(iv) 540 = 2 × 2 × 3 × 3 × 3 × 5 = 2² × 3³ × 5¹

(v) 2280 = 2 × 2 × 2 × 3 × 5 × 19 = 2³ × 3¹ × 5¹ × 19¹

(vi) 3600 = 2 × 2 × 2 × 2 × 3 × 3 × 5 × 5 = 2⁴ × 3² × 5²

(vii) 4725 = 3 × 3 × 3 × 5 × 5 × 7 = 3³ × 5² × 7

(viii) 8400 = 2 × 2 × 2 × 2 × 3 × 5 × 5 × 7 = 2⁴ × 3¹ × 5² × 7¹

ML Aggarwal Class 7 Solutions for ICSE Maths