S.chand publication New Learning Composite mathematics solution of class 7 Chapter 6 Linear Equations in One Variable Exercise 6C

 Exercise 6C


Q1 | Ex-6C |Class 7 |S.Chand | New Learning Composite maths |Linear Equations in One Variable |Ch-6 |myhelper

Question 1

The number decreased by 7 is 25, find the number.

Sol :

Let , the number be x

x-7=25

x=25+7

x=32

Ans: The number is 32



Q2 | Ex-6C |Class 7 |S.Chand | New Learning Composite maths |Linear Equations in One Variable |Ch-6 |myhelper

Question 2

The number increased by 10 is 16, find the number.

Sol :

Let, the number be x

x+10=16

x=16-10

x=6



Q3 | Ex-6C |Class 7 |S.Chand | New Learning Composite maths |Linear Equations in One Variable |Ch-6 |myhelper

Question 3

The number is multiplied by 2, and 8 is added to the product, the result is 50, find the number.

Sol :

Let, the number be x

(2x)+8=50

$x=\frac{42}{2}=21$



Q4 | Ex-6C |Class 7 |S.Chand | New Learning Composite maths |Linear Equations in One Variable |Ch-6 |myhelper

Question 4

The number is subtracted from 23, the result is 10 less than twice the number, find the number.

Sol :

Let , the number be x

23-x=2x-10

-x-2x=-10-23

-3x=-33

$x=\frac{33}{3}=11$



Q5 | Ex-6C |Class 7 |S.Chand | New Learning Composite maths |Linear Equations in One Variable |Ch-6 |myhelper

Question 5

Two-thirds of the number is increased by 9, the sum will be 19, find the number.

Sol :

Let, the number be x

$\frac{2}{3}x+9=19$

$\frac{2x}{3}=19-9$

$\frac{2x}{3}=10$

2x=30

$x=\frac{30}{2}$

x=15

Ans: The number is 15



Q6 | Ex-6C |Class 7 |S.Chand | New Learning Composite maths |Linear Equations in One Variable |Ch-6 |myhelper

Question 6

The sum of the one eight, one twelveth and one-twentieth parts is 31, find the number.

Sol :

Let the number be x

$\frac{x}{8}+\frac{x}{12}+\frac{x}{20}=31$

$\frac{15x+10x+6x}{20}=31$

31x=31×120

$x=\frac{31\times 120}{31}$

x=120

The number is 120



Q7 | Ex-6C |Class 7 |S.Chand | New Learning Composite maths |Linear Equations in One Variable |Ch-6 |myhelper

Question 7

Seema’s little sister is now 5 years old. If her age is 2 years more than 1/4 of Seema’s age, how old is Seema.

Sol :

Let, seema's age be x

∴According to statement

$\frac{x}{4}+2=5$

$\frac{x}{4}=5-2$

$\frac{x}{4}=3$

x=3×4=12

Seema's age is 12 year old



Q8 | Ex-6C |Class 7 |S.Chand | New Learning Composite maths |Linear Equations in One Variable |Ch-6 |myhelper

Question 8

One number is 9 more than the other. If their sum is 63, what are the numbers?

Sol :

Let, one number be x

∴Other number=(x+9)

∴x+(x+9)=63

2x=63-9

2x=54

$x=\frac{54}{2}$

∴x=27

∴Other number=27+9=36

The number are 27 and 36



Q9 | Ex-6C |Class 7 |S.Chand | New Learning Composite maths |Linear Equations in One Variable |Ch-6 |myhelper

Question 9

The sum of two numbers is 64, and the second number is 16 less than the first. Find the numbers.

Sol :

Let , the first number be x

∴Second number be (x-16)

x+(x-16)=64

2x=64+16

2x=80

$x=\frac{80}{2}=40$

x=40


∴Second number be (x-16)=(4+-16)

=24

Numbers are 40 and 24



Q10 | Ex-6C |Class 7 |S.Chand | New Learning Composite maths |Linear Equations in One Variable |Ch-6 |myhelper

Question 10

One number is 7 times another number. If 14 is added to the sum of the numbers, the result is 38. Find the number.

Sol :

Let, the first number be x

Second number be 7x

x+7x+14=38

8x=38-14

$x=\frac{24}{8}=3$


Second number be =(7×3)=21

The numbers are 3 and 21



Q11 | Ex-6C |Class 7 |S.Chand | New Learning Composite maths |Linear Equations in One Variable |Ch-6 |myhelper

Question 11

John is 7 years older than his sister, and the sum of their ages is 21 years. Find the ages.

Sol :

Let, Sister's age be x

∴John's age (x+7)

ATQ,

x+(x+7)=21

2x=21-7

2x=14

$x=\frac{14}{2}=7$

x=7


John's sister's age 7 years

John's age=(7+7) years=14 years



Q12 | Ex-6C |Class 7 |S.Chand | New Learning Composite maths |Linear Equations in One Variable |Ch-6 |myhelper

Question 12

Find three consecutive numbers whose sum is 108. (Hint. Let the consecutive numbers be x, x + 1, x+ 2)

Sol :

Let, the consecutive number be x, (x+1), (x+2)

x+(x+1)+(x+2)=108

3x+3=108

3x=108-3

$x=\frac{105}{3}=35$

The consecutive numbers are 35, (35+1), (35+2)

=35, 36, 37 



Q13 | Ex-6C |Class 7 |S.Chand | New Learning Composite maths |Linear Equations in One Variable |Ch-6 |myhelper

Question 13

Find two consecutive odd numbers whose sum is 56. (Hint. Let the consecutive odd numbers be 2x + 1, 2x + 3)

Sol :

Let, the consecutive odd number be (2x+1),(2x+3)

(2x+1)+(2x+3)=56

2x+1+2x+3=56

4x+4=56

4x=52

$x=\frac{52}{4}=13$

The consecutive odd numbers are {(2×13)+1}, {(2×13)43}

=27, 29



Q14 | Ex-6C |Class 7 |S.Chand | New Learning Composite maths |Linear Equations in One Variable |Ch-6 |myhelper

Question 14

Find two consecutive even numbers whose sum is 86. (Hint. Let the consecutive even numbers be 2x, 2(x + 1)?

Sol :

Let , the consecutive even numbers are 2x, 2(x+1)

2x+2(x+1)=86

2x+2x+2=86

4x=86-2

4x=84

$x=\frac{84}{4}=21$

The consecutive even numbers are (2×21), 2(21+1)

=42,44



Q15 | Ex-6C |Class 7 |S.Chand | New Learning Composite maths |Linear Equations in One Variable |Ch-6 |myhelper

Question 15

Rakesh is now 3 times as old as his younger brother Pinku, but in 3 more years he will be only twice as old as Pinku will be then, How old are they now?

Sol :

Let, Pinku's age be x

∴Rakesh's age=3x


After 3 years

Age of pinku=x+3

Age of Rakesh=3x+3


ATQ,

3x+3=2(x+3)

3x+3=2x+6

3x-2x=6-3

x=3


Now ,Pinku's age =3 years

Rakesh's age =(3×3)=9 years



Q16 | Ex-6C |Class 7 |S.Chand | New Learning Composite maths |Linear Equations in One Variable |Ch-6 |myhelper

Question 16

In a basketball, a field goal counts 2 points and a foul shot one point. Shreyas scored 14 points in a game. He made 3 times as many field goals as foul shots. On how many shots of each kind did he score?

Sol :

Let us say number of foul shots made by Shreyas be x

∴Number of goals be 3x

ATQ,

14=2(3x)+x

14=6x+x

7x=14

x=2


Ans: Number of foul shots made=2

Number of field goals=6



Q17 | Ex-6C |Class 7 |S.Chand | New Learning Composite maths |Linear Equations in One Variable |Ch-6 |myhelper

Question 17

The length of a rectangle is 6 m less than twice its breadth. If the perimeter of the rectangle is 54 m. Find its length and breadth.

Sol :

Let, the breadth of rectangle be x

Length=2x-6


2{x+(2x+6)}=54

2x+4x-12=54

6x=54+12

$x=\frac{66}{6}=11$

∴The breadth of rectangle be 11m

Length of rectangle={(2×11)-6}m

=22-6=16m



Q18 | Ex-6C |Class 7 |S.Chand | New Learning Composite maths |Linear Equations in One Variable |Ch-6 |myhelper

Question 18

I have some 5-rupee coins and some 2-rupee coins is 4 times that of 5 rupee coins. If I have Rs. 117 in all, find the number of coins of each denominations.

Sol :

Let, the number of 5 rupee coin be x

Number of 2 rupee coin be 4x

5x+(2×4x)=117

13x=117

$x=\frac{117}{13}=9$

Number of 5 rupee coin be 9

Number of 2 rupee coin be 36



Q19 | Ex-6C |Class 7 |S.Chand | New Learning Composite maths |Linear Equations in One Variable |Ch-6 |myhelper

Question 19

A number is divided into two parts such that one part is 10 more than the other. If the two parts are in the ratio 5:3. Find the numbers.

Sol :

Let, one number be x

Then the other number is (x+10)


ATQ

$\frac{x+10}{x}=\frac{5}{3}$

3(x+10)=5x

3x+30=5x

-2x=-30

$x=\frac{30}{2}=15$

∴One number is 15

Other number is 25



Q20 | Ex-6C |Class 7 |S.Chand | New Learning Composite maths |Linear Equations in One Variable |Ch-6 |myhelper

Question 20

Pratik left one-third of his property for his son, one fourth for his daughter and the remainder for his wife. If wife’s share in the property was worth Rs. 10 lakh, find the total worth of Pratik’s property.

Sol :

Let, Pratik total property be x

∴Pratik's son's property$=\frac{x}{3}$

Pratik's daughter property$=\frac{x}{4}$


∴$x-\left(\frac{x}{3}+\frac{x}{4}\right)=10$

$x-\left(\frac{4x+3x}{12}\right)=10$

$x-\frac{7x}{12}=10$

$\frac{12x-7x}{12}=10$

$\frac{5x}{12}=10$

$x=\frac{120}{5}=24$

Ans : Pratik's total property=24 lakh



Q21 | Ex-6C |Class 7 |S.Chand | New Learning Composite maths |Linear Equations in One Variable |Ch-6 |myhelper

Question 21

One-fourth of a herd of deer have gone to the forest. One-third of the total number are grazing in a field and remaining 15 are drinking water on the bank of a river. Find the total number of deer.

Sol :

Let, total of  of deer=x

ATQ,

$x-\left(\frac{x}{4}+\frac{x}{3}\right)=15$

$x-\left(\frac{3x+4x}{12}\right)=15$

$x-\frac{7x}{12}=15$

$\frac{12x-7x}{12}=15$

5x=15×12

$x=\frac{15\times 12}{5}=36$

Ans Total of of deer =36

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