WBBSE Solutions For Class 7 Maths Arithmetic Chapter 1 Problems On Fractions

Arithmetic Chapter 1 Revision Of Previous Lessons

Introduction

  1. The first four rules,
  2. Vulgar fractions,
  3. Decimal
  4. Recurring fractions,
  5. H.C.F. and L.C.M. of integers,
  6. H.C.F. and L.C.M. of fractions and decimals and
  7. Extraction of square roots of integers by factorisation and division.

First four rules

You know that addition, subtraction, multiplication and division are the first four rules. The entire structure of mathematics is based on these four rules. In every branch of mathematics namely arithmetic, algebra, geometry etc., we have to depend on these four rules. Let us discuss some problems on the first four rules.

Read and Learn More WBBSE Solutions For Class 7 Maths

Arithmetic Chapter 1 Revision Of Previous Lessons Exercise 1 Some Problems With The First Four Rules

Example 1. The sum of two numbers is 1040 and their difference is 304. Find the numbers.

Solution:

Given:

The sum of two numbers is 1040 and their difference is 304.

WBBSE Solutions For Class 7 Maths Arithmetic Chapter 1 Revision Of Previous Lessons Exercise 1 Problems On The First Four Rules Example 1

∴ The greater number = 1344 ÷ 2=672

Hence, the smaller number = 1040-672 = 368

∴ 672 and 368

The numbers are 672 and 368.

WBBSE Class 7 Fractions Solutions

Example 2. The sum of the two numbers is 1000. If the greater number is thrice the smaller number, find the numbers.

Solution:

Given:

The sum of the two numbers is 1000. If the greater number is thrice the smaller number,

The greater number+ the smaller number = 1000

WBBSE Solutions For Class 7 Maths Arithmetic Chapter 1 Problems On Fractions

∴ 3 x the smaller number + the smaller number = 1000

or, 4 x the smaller number = 1000 = 250

or, the smaller number = \(\frac{1000}{4}\) = 250

Hence, the greater number = 3 x 250 = 750.

∴ 750 and 250.

The numbers are 750 and 250.

WBBSE Class 7 Geography Notes WBBSE Solutions For Class 7 History
WBBSE Solutions For Class 7 Geography WBBSE Class 7 History Multiple Choice Questions
WBBSE Class 7 Geography Multiple Choice Questions WBBSE Solutions For Class 7 Maths

 

Example 3. The sum of two numbers is 645. If one of them is half of the other, then find the numbers.

Solution:

Given:

The sum of two numbers is 645. If one of them is half of the other,

Here, the smaller number is half of the greater number.

Hence, the greater number is twice the smaller number

Now, the greater number + the smaller number = 645

or, 2x the smaller number + the smaller number = 645

or, 3 x the smaller number = 645

or, the smaller number = \(\frac{645}{3}\) = 215

∴ The greater number = 215 x 2 = 430

∴ 430 and 215.

The numbers are 430 and 215.

Example 4. The sum of two numbers is 94792 and the greater number is 75368. Find the smaller number.

Solution:

Given:

The sum of two numbers is 94792 and the greater number is 75368.

The greater number + the smaller number = 94792

or, 75368+ the smaller number = 94792

or, the smaller number = 94792-75368 = 19424

∴  19424.

The smaller number is 19424.

Example 5. The sum of two numbers is 48925. If the smaller number is 12318, then find the greater number.

Solution:

Given:

The sum of two numbers is 48925. If the smaller number is 12318

The greater number + the smaller number = 48925

or, The greater number + 12318 = 48925

or, The greater number = 48925-12318 = 36607

∴  36607.

The greater number is 36607.

Example 6. The product of 36, 75 and another number is 218700. Find the number.

Solution:

Given:

The product of 36, 75 and another number is 218700.

Here, 36 x 75 x the required number=218700

or, the required number = \(\frac{218700}{36×75}\) = 81

∴ 81.

The number is 81.

Solved Problems for Class 7 Fractions

Example 7. In a multiplication the multiplier is 965, and 476005 being added to the product, the sum becomes 1 million. Find the multiplicand.

Solution:

Given:

In a multiplication the multiplier is 965, and 476005 being added to the product, the sum becomes 1 million.

1 million = 1000000

When 476005 is added to the product it becomes 1000000.

Hence, the product = 1000000-476005 = 523995

Now, Multiplicand x Multiplier = 523995

or, multiplicand x 965 = 523995

or, multiplicand=523995 +965 = 543

∴  543.

The multiplicand is 543.

Example 8. What number divided by 372 gives the quotient 273 and the remainder 237?

Solution:

Given:

The number divided by 372 gives the quotient 273 and the remainder 237.

Here dividend = divisor x quotient + remainder

= 372 x 273 x 237 x = 101556 ÷ 237 = 101793

WBBSE Solutions For Class 7 Maths Arithmetic Chapter 1 Revision Of Previous Lessons Exercise 1 Problems On The First Four Rules Example 8

∴ 101793.

The number is 101793.

Example 9. Find the greatest number of 5 digits which is exactly divisible by 292.

Solution:

Given:

The greatest number of 5 digits which is exactly divisible by 292

The natural greatest number of 5 digits=99999

WBBSE Solutions For Class 7 Maths Arithmetic Chapter 1 Revision Of Previous Lessons Exercise 1 Problems On The First Four Rules Example 9

Thus, if the remainder 7 be subtracted from 99999, the result will be exactly divisible by 232.

∴ The required number= 99999-7= 99992

∴ 99992.

The greatest number is 99992.

Class 7 Maths Exercise on Fractions Solutions

Example 10. Find the least number of 6 digits which is exactly divisible by 215.

Solution:

Given:

The least number of 6 digits which is exactly divisible by 215

The natural least number of 6 digits = 100000

WBBSE Solutions For Class 7 Maths Arithmetic Chapter 1 Revision Of Previous Lessons Exercise 1 Problems On The First Four Rules Example 10

Now, 215-25 = 190

If 190 be added to 100000, the result will be divisible by 215.

∴ The required number = 100000+ 190 = 100190

∴ 100190.

The least number 100190.

Example 11. Five years ago, the father’s age was 8 times that of his son. Five years hence, the sum of their ages will be 65 years. Find their present ages.

Solution:

Given:

Five years ago, the father’s age was 8 times that of his son. Five years hence, the sum of their ages will be 65 years.

Since, after 5 years the sum of their ages will be 65 years

therefore, the sum of their present ages (65-25) years = (65-10) years 55 years.

Also, five years ago, the sum of their ages was (55-2x 5) years = 45 years.

At that time, the father’s age was 8 times that of son.

Hence, the sum of their ages was 9 times the then age of the son.

∴ At that time, the age of son was (459) or, 5 years and that of father was (5 x 8) years or 40 years.

∴ present age of father= (40+5) years = 45 years

and present age of son = (5+ 5) years = 10 years.

∴ Present age of son is 10 years and the present age of the father is 45 years.

Example 12. Find the quotient and the remainder, dividing by the factors of the divisor in the following case 7027 ÷ 105.

Solution:

WBBSE Solutions For Class 7 Maths Arithmetic Chapter 1 Revision Of Previous Lessons Exercise 1 Problems On The First Four Rules Example 12

Hence, the quotient = 66.

The actual remainder = the first remainder + the first divisor x the second remainder + the first divisor x the second divisor x the third remainder.

=1+3×2+3x5x6=1+6+90=97

∴ Quotient = 66, Remainder = 97.

Arithmetic Chapter 1 Revision Of Previous Lessons Exercise 1 Some Problems On Fractions

Example 1. Add together 339\(\frac{1}{10}\)+ 660\(\frac{9}{10}\) 

Solution: 339 \(\frac{1}{10}\)+ 660 \(\frac{9}{10}\)

= \(339+\frac{1}{10}+660+\frac{9}{10}\)

= \(339+660+\frac{1}{10}+\frac{9}{10}\)

= \(999+\frac{1-9}{10}\)

= \(999+\frac{10^1}{10}\)

= 999+1

= 1000

Example 2. Subtract 189 \(\frac{1}{11}\) – 99 \(\frac{1}{22}\)

Solution: 189 \(\frac{1}{11}\) – 99 \(\frac{1}{22}\)

= \(189+\frac{1}{11}-\left(99+\frac{1}{22}\right)\)

= \(189+\frac{1}{11}-99-\frac{1}{22}\)

= \((189-99)+\left(\frac{1}{11}-\frac{1}{22}\right)\)

= \(90+\frac{2-1}{22}\)

= \(90+\frac{1}{22}\)

∴  90\(\frac{1}{22}\)

West Bengal Board Class 7 Arithmetic Assistance

Example 3. Find the product 99990 \(\frac{494}{495}\) x 99

Solution: 999 \(\frac{494}{495}\)  × 99

= \(999+\frac{494}{495}\)

= \(999+1-\frac{1}{495} \times 99\)

= \(\left(1000-\frac{1}{495} \times 99\right)\)

= \(99000-\frac{1}{5}\)

= \(99000-1+1-\frac{1}{5}\)

= \(98999+\frac{4}{5}\)

∴ 98999\(\frac{4}{5}\)

Example 4.  Divide 29 \(\frac{8}{23}\) ÷ \(\frac{15}{17}\) 

Solution: 29 \(\frac{8}{23}\) ÷ \(\frac{15}{17}\)

= \(\frac{675}{23} \div \frac{15}{17}\)

= \(\frac{675}{23} \times \frac{17}{15}\)

= \(\frac{765}{23}\)

∴ 33 \(\frac{6}{23}\)

Example 5. Simplify: 

WBBSE Solutions For Class 7 Maths Arithmetic Chapter 1 Revision Of Previous Lessons Exercise 1 Problems On Fractions Example 5

Solution:

WBBSE Solutions For Class 7 Maths Arithmetic Chapter 1 Revision Of Previous Lessons Exercise 1 Problems On Fractions Example5

 

∴ \(\frac{2}{3}\)

Example 6. Add together 7.23 + 15.735 + 235.7

Solution:

Given

7.23 + 15.735 + 235.7

WBBSE Solutions For Class 7 Maths Arithmetic Chapter 1 Revision Of Previous Lessons Exercise 1 Problems On Fractions Example 6

 

Example 7. Subtract 52.917 from 218.2

Solution:

Given:

52.917 And 218.2

WBBSE Solutions For Class 7 Maths Arithmetic Chapter 1 Revision Of Previous Lessons Exercise 1 Problems On Fractions Example 7

 

Example 8. Multiply 33.123 by 1.25

Solution:

Given:

33.123 And 1.25

WBBSE Solutions For Class 7 Maths Arithmetic Chapter 1 Revision Of Previous Lessons Exercise 1 Problems On Fractions Example 8

 

Example 9. Divide 67.072 by 16.

Solution:

Given:

67.072 And16

WBBSE Solutions For Class 7 Maths Arithmetic Chapter 1 Revision Of Previous Lessons Exercise 1 Problems On Fractions Example 9

 

Example 10. Find the value of 15.8794:

  1. To two places of decimal.
  2. Correct to two places of decimal.

Solution:

  1. The value of 15.8794 to two places of decimal = 15.87.
  2. The value of 15.8794 correct to two places of decimal = 15.88.

Example 11. Convert into recurring decimal: 3 \(\frac{7}{22}\)

Class Vii Math Solution Wbbse

Solution: 3 \(\frac{7}{22}\) = \(\frac{73}{22}\)

WBBSE Solutions For Class 7 Maths Arithmetic Chapter 1 Revision Of Previous Lessons Exercise 1 Problems On Fractions Example 11

 

Example 12. Convert into vulgar fraction 7.028

Solution: 7.028

WBBSE Solutions For Class 7 Maths Arithmetic Chapter 1 Revision Of Previous Lessons Exercise 1 Problems On Fractions Example 12

∴ 7 \(\frac{14}{495}\)

Example 13. A man gave \(\frac{1}{3}\) of his property to his son and \(\frac{1}{5}\) of his property to his daughter. If 1267 was still left with him, what was his property?

Solution: The man gave (\(\frac{1}{3}\)+\(\frac{1}{5}\))

or, \(\frac{8}{15}\) part of his property to his son and daughter.

Therefore, (1 – \(\frac{8}{15}\))

or, \(\frac{7}{15}\) part of his property was left with him.

∴ \(\frac{7}{15}\) part of his property = ₹ 1267

∴ The whole property
= ₹1267 x \(\frac{15}{7}\) = ₹ 181 × 15 = ₹ 2715

∴ ₹ 2715.

Example 14. A post has \(\frac{1}{4}\)  of its length in mud and \(\frac{1}{5}\) of it in water and 11 metres above water. Find the whole length of the post.

Solution: (\(\frac{1}{4}\) + \(\frac{1}{5}\)) or (\(\frac{5+4}{20}\))

or, \(\frac{9}{20}\) of the post is in mud and water.

∴ (1- \(\frac{9}{20}\) or, \(\frac{11}{20}\)) of the post is above water.

∴ \(\frac{11}{20}\) of the whole post = 11 metres

∴ Lenght of the whole post = 11 x \(\frac{20}{11}\) meters.

∴ 20 meters.

Wbbse Class 7 Maths Solutions

Example 15. A man does 0.24 of a piece of work on the first day, 0.18 of it on the second day and 0.08 of it on the third day. How much of the work remains undone?

Solution:

Given:

A man does 0.24 of a piece of work on the first day, 0.18 of it on the second day and 0.08 of it on the third day.

In the first three days the man does (0.24 +0.18 +0.08) or 0.5 of the work.

Hence, (10.5) or 0.5 of the work remains undone.

∴ 0.5 of the work.

Example 16. 0.25 of the soldiers of an army died in combat, 0.15 was injured, 0.3 was captivated and the remaining 1200 soldiers escaped. How many soldiers were there in the army?

Solution:

Given:

0.25 of the soldiers of an army died in combat, 0.15 was injured, 0.3 was captivated and the remaining 1200 soldiers escaped.

(0.25 +0.15 + 0.3) or, 0.7 of the soldiers died, was injured and captivated.

Hence, (1-0.7) or, 0.3 of the soldiers escaped.

∴ 0.3 of the soldiers = 1200

∴ Total soldiers in the army = 1200 x \(\frac{10}{3}\) = 4000
∴4000.

Arithmetic Chapter 1 Revision Of Previous Lessons Exercise 1 Some Problems On H.C.F. and L.C.M.

Example 1. Find the L.C.M. of \(\frac{3}{4}\), \(\frac{5}{8}\), \(\frac{15}{16}\)

Solution:

WBBSE Solutions For Class 7 Maths Arithmetic Chapter 1 Revision Of Previous Lessons Exercise 1 Problems On Least Common Multiplication Example 1

∴ 3 \(\frac{3}{4}\)

Example 2. Find the H.C.F. of \(\frac{3}{5}\), 2 \(\frac{1}{10}\), \(\frac{6}{25}\)

Solution:

WBBSE Solutions For Class 7 Maths Arithmetic Chapter 1 Revision Of Previous Lessons Exercise 1 Problems On Highest Common Fraction Example 2

 

∴ \(\frac{3}{50}\)

Example 3. Find the L.C.M of 0.015, 0.42, and 1.5

Solution: 0.015 = 0.015
0.45 = 0.045
1.5 = 1.500

Now, let us find the L.C.M of 15, 450 and 1500

WBBSE Solutions For Class 7 Maths Arithmetic Chapter 1 Revision Of Previous Lessons Exercise 1 Problems On Least Common Multiplication Example 3

Now, 2 x 3 5 x 3 x 10 = 4500

Thus, the L.C.M of 15, 450 and 1500 is 4500

Hence the required L.C.M = 45

Example 4. Find the H.C.f of 0.4, 0.08, 0.016

Solution: 0.4 = 0.400
0.08 = 0.080
0.016 = 0.016

Let us now find the H.C.F of 400,80, 16

WBBSE Solutions For Class 7 Maths Arithmetic Chapter 1 Revision Of Previous Lessons Exercise 1 Problems On Highest Common Fraction Example 4

Thus, the H.C.F. of 400, 80, and 16 is 16

Hence, the required H.C.F. = 0.016

∴ 0.016.

WBBSE Class 7 Chapter 1 Fractions Guide

Example 5. Three bells toll together and then at intervals of 3 \(\frac{1}{3}\), 2 \(\frac{3}{50}\) and 1 \(\frac{2}{3}\) seconds respectively. When will they toll together again?

Solution: The required time will be the L.C.M. of \(\frac{1}{3}\), 2 \(\frac{3}{50}\) and 1 \(\frac{2}{3}\) seconds i.e., the L.C.M. of \(\frac{10}{3}\), \(\frac{5}{2}\) and \(\frac{5}{3}\) seconds.

Now, the L.C.M. of \(\frac{10}{3}\), \(\frac{5}{2}\) and \(\frac{5}{3}\) seconds.

= \(\frac{L.C.M of 10,5,5}{H.C.F of 3,2,3}\) = \(\frac{10}{1}\) = 10

∴ They will toll together again after 10 seconds.

Example 6. What is the smallest number which is exactly divisible by 1 \(\frac{2}{7}\), 1 \(\frac{1}{35}\), \(\frac{45}{49}\)?

Solution: Here the numbers are

\(\frac{2}{7}\), 1 \(\frac{1}{35}\), \(\frac{45}{49}\)

i.e.., \(\frac{9}{7}\), \(\frac{36}{35}\), \(\frac{45}{49}\)

Now, the L.C.M of \(\frac{9}{7}\), \(\frac{36}{35}\), \(\frac{45}{49}\)

= \(\frac{L.C.M of 9,36,45}{H.C.F of 7,35,49}\)

= \(\frac{180}{7}\) = 25 \(\frac{5}{7}\)

∴ 25 \(\frac{5}{7}\)

Example 7. By what greatest fraction are \(\frac{14}{17}\), 3 \(\frac{1}{17}\), \(\frac{28}{34}\) exactly divisible giving integers as quotients?

Solution: Here the numbers are \(\frac{14}{17}\), 3 \(\frac{1}{17}\), \(\frac{28}{34}\)

i.e.., \(\frac{14}{17}\), \(\frac{52}{17}\), \(\frac{14}{17}\)

Now, the H.C.F. of \(\frac{14}{17}\), \(\frac{52}{17}\), \(\frac{14}{17}\)

WBBSE Solutions For Class 7 Maths Arithmetic Chapter 1 Revision Of Previous Lessons Exercise 1 Problems On H.C.F And L.C.M Example 7

 

= \(\frac{H.C.F of 14,52,14}{L.C.M of 17,17,17}\)

= \(\frac{2}{17}\)

∴ \(\frac{2}{17}\)

Wbbse Class 7 Maths Solutions

Example 8. By what greatest quantity must \(\frac{7}{8}\) kg, \(\frac{14}{15}\) kg and 1 \(\frac{1}{20}\) kg be divided so that the quotients are integers?

Solution: Let us first find the H.C.F. of \(\frac{7}{8}\) kg, \(\frac{14}{15}\) kg and 1 \(\frac{1}{20}\)

i.e.., of \(\frac{7}{8}\) kg, \(\frac{14}{15}\) kg and 1 \(\frac{1}{20}\)

Now, H.C.f of \(\frac{7}{8}\) kg, \(\frac{14}{15}\) kg and 1 \(\frac{1}{20}\)

WBBSE Solutions For Class 7 Maths Arithmetic Chapter 1 Revision Of Previous Lessons Exercise 1 Problems On H.C.F And L.C.M Example 8

= \(\frac{H.C.F of 7,14,21}{L.C.M of 8,15,20}\)

= \(\frac{7}{120}\)

∴ 2 x 2 x 5 x 2 x 3 = 120

Hence, the required quantity = \(\frac{7}{120}\) kg

∴ \(\frac{7}{120}\) kg

Example 9. By what greatest number must 2.1, 2.8, and 3.5 be divided to get integral quotients?

Solution: Here, the required number will be the H.C.F. of 2.1, 2.8, and 3.5.
Let us first find the H.C.F. of 21, 28, and 35.

WBBSE Solutions For Class 7 Maths Arithmetic Chapter 1 Revision Of Previous Lessons Exercise 1 Problems On H.C.F And L.C.M Example 9

Thus, H.C.F of 21, 28, and 35 is 7

Hence, H.C.F of 2.1, 2.8, and 3.5 is 0.74

∴ the required number is 0.7

Class 7 Maths Exercise 1 Solved Examples

Example 10. What least number when divided by 1.9, 0.95, and 0.076, gives integers as quotients?

Solution: The required number is the L.C.M. of 1.9, 0.95, and 0.076.
Now, 1.9 = 1.900
0.95 = 0.950
0.076 = 0.076

Now, let us find the L.C.M. of 1900, 950 and 76

WBBSE Solutions For Class 7 Maths Arithmetic Chapter 1 Revision Of Previous Lessons Exercise 1 Problems On H.C.F And L.C.M Example 10

∴ L.C.M. of 1900, 950 and 76
=2×2×5×5×19 = 1900

∴ L.C.M. of 1.9, 0.950, 0.076 is 1.9

∴ The required number is 1.9.

Example 11. The L.C.M. and H.C.F. of two fractions are \(\frac{4}{5}\) and \(\frac{2}{15}\) respectively. If one of them be \(\frac{2}{5}\) find the other.

Solution: We know that,

Product of two numbers = Their L.C.M.XH.C.F.

Hence, \(\frac{2}{5}\) x the other number = \(\frac{4}{5}\) x \(\frac{2}{15}\)

or, the other number= \(\frac{4}{5}\) x \(\frac{2}{15}\) x \(\frac{5}{2}\) = \(\frac{4}{15}\)

∴ \(\frac{4}{15}\)

Example 12. The product of two numbers is 0.84. If their L.C.M. be 4.2, find their H.C.F.

Solution:

L.C.M. H.C.F. = Product of the numbers

or, 4.2 x H.C.F. = 0.84

or, H.C.F. = \(\frac{0.84}{4.2}\) = 0.2

Example 13. From what least number must 5 be subtracted so that the remainder may be exactly divisible by 48, 64, 90, 120?

Solution:

WBBSE Solutions For Class 7 Maths Arithmetic Chapter 1 Revision Of Previous Lessons Exercise 1 Problems On H.C.F And L.C.M Example 13

∴ L.C.M. of 48, 64, 90 and 120
=2×2×2×2×3 x 5 x 4 x 3 = 2880

Hence, the required number = 2880 +5=2885

∴ The number is 2885.

Example 14. What number nearest to 100000 is exactly divisible by 2, 3, 4, 5, 6, and 7?

Solution:

WBBSE Solutions For Class 7 Maths Arithmetic Chapter 1 Revision Of Previous Lessons Exercise 1 Problems On H.C.F And L.C.M Example 14

Now, 100000-40=99960 and 100000+ (420-40)= 100000+ 380 = 100380 are the two numbers divisible by 420.

Between these two numbers 99960 is nearer to 100000. Hence, 99960 is the required number.

∴ 99960

Example 15. Find the greatest number that will divide 8718, 16299, and 25396 leaving the remainder 1, 2, and 3 respectively.

Solution: Since, 8718, 16299, and 25396 when divided by the required number leaves the remainder 1 , 2 and 3 respectively therefore (8718- 1), i.e., 8717, (16299- 2)

i.e., 16297 and (25396- 3)

i.e., 25393 are exactly divisible by the required number.

Hence, H.C.F. of 8717, 16297 and 25393 is the required number.

WBBSE Solutions For Class 7 Maths Arithmetic Chapter 1 Revision Of Previous Lessons Exercise 1 Problems On H.C.F And L.C.M Example 15

∴ The required number is 379.

Example 16. 175 mangoes and 105 oranges can be equally divided among certain number of boys and girls. Find the number.

Solution:

WBBSE Solutions For Class 7 Maths Arithmetic Chapter 1 Revision Of Previous Lessons Exercise 1 Problems On H.C.F And L.C.M Example 16

Hence, the number of boys and girls may be 35 or any factor of 35 i.e., 5, 7.
∴ 5, 7, 35.

 

Arithmetic Chapter 1 Revision Of Previous Lessons Exercise 1 Some Problems On Square Root

Example 1. Find the square root of 1225 by factorisation.

Solution:

Given:

1225

WBBSE Solutions For Class 7 Maths Arithmetic Chapter 1 Revision Of Previous Lessons Exercise 1 Problems On Square Root Example 1

Now, 1225 = 5 x 5 x 7 x 7=52 x 72

Hence, √1225 = 5 x 7=35

∴ 35.

The square root of 1225 = 35.

Example 2. Find the square root of 15625 by the division method.

Solution:

Given:

15625

WBBSE Solutions For Class 7 Maths Arithmetic Chapter 1 Revision Of Previous Lessons Exercise 1 Problems On Square Root Example 2

∴ 125

The square root of 15625 = 125

Example 3. By what least number must 450 be multiplied to give a perfect square?

Solution:

Given:

450

WBBSE Solutions For Class 7 Maths Arithmetic Chapter 1 Revision Of Previous Lessons Exercise 1 Problems On Square Root Example 3

Now, 450 = 2 x 3 x3x5x5=2x 32 x 52

So, if we multiply 450 by 2 then it will be a perfect square.

Understanding Fractions for Class 7 Students

Example 4. By what least number must 1323 be divided so that the quotient will be a perfect square?

Solution:

Given:

1323

WBBSE Solutions For Class 7 Maths Arithmetic Chapter 1 Revision Of Previous Lessons Exercise 1 Problems On Square Root Example 4

Now, 1323 = 3 x 32 x 72

So, if we divide 1323 by 3 then it will be a perfect square.

Math Solution Of Class 7 Wbbse

Example 5. The product of two numbers is 1575 and their quotinet is \(\frac{9}{7}\); find the numbers.

Solution:

Given:

The product of two numbers is 1575 and their quotinet is \(\frac{9}{7}\)

Let the smaller number be A then the greater number is \(\frac{9}{7}\) A.

Now, A x \(\frac{9}{7}\) A = 1575

or, A2 = \(\frac{1575 X 7}{9}\) =175×7=1225

or, A = √1225=35

∴ The smaller number = 35 and the greater number = \(\frac{9}{7}\) × 35 = 9 x 5 = 45
∴ 35 and 45.

Example 6.  77847 soldiers were arranged in a square. After being arranged 6 soldiers were in excess. What was the number of soldiers in the front row?

Solution:

Given:

77847 soldiers were arranged in a square. After being arranged 6 soldiers were in excess.

Since there was an excess of 6 soldiers after they were arranged in the form of a square therefore,(77847 – 6) or, 77841 soldiers made the perfect square. Then the number of soldiers in the front row = √77841=279

WBBSE Solutions For Class 7 Maths Arithmetic Chapter 1 Revision Of Previous Lessons Exercise 1 Problems On Square Root Example 6

∴ The number of soldiers in the front row = 279.

Step-by-Step Solutions for Class 7 Fraction Problems

Example 7. Each member of a club contributed as many 25 paise as the total number of members and the total contribution was 400. Find the number of members of the club.

Solution:

Given:

Each member of a club contributed as many 25 paise as the total number of members and the total contribution was 400.

₹400 = 40000 paise
Now, number of 25 paise in 40000 paise = \(\frac{40000}{25}\) = 1600

∴ Number of members of the club = √1600 = 40 perfect square.

∴ 40.

Number of members of the club = 40.

Example 8. The length of the sides of two squares are 15 cm and 20 cm respectively. A new square is formed whose area is equal to the sum of the areas of these two squares. Find the length of the side of the new square.

Solution:

Given:

The length of the sides of two squares are 15 cm and 20 cm respectively. A new square is formed whose area is equal to the sum of the areas of these two squares.

Area of the first square= (15)2 sq cm = 225 sq cm.

Area of the second square = (20)2 sq cm = 400 sq cm.

Area of the new square (225+ 400) sq cm 625 sq cm.

∴The length of the side of the new square = √625 cm = 25 cm

∴ 25 cm.

The length of the side of the new square is 25 cm.

 

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