Ncert solutions

Ncert solutions

Ncert solutions

Grade 7

Q1) Set up equations and solve them to find the unknown numbers in the following cases:

(a) Add 4 to eight times a number; you get 60.

(b) One-fifth of a number minus 4 gives 3.

(c) If I take three-fourths of a number and add 3 to it, I get 21.

(d) When I subtracted 11 from twice a number, the result was 15.

(e) Munna subtracts thrice the number of notebooks he has from 50, he finds the

result to be 8.

(f) Ibenhal thinks of a number. If she adds 19 to it and divides the sum by 5, she

will get 8.

(g) Anwar thinks of a number. If he takes away 7 from \frac{5}{2} of the number, the result is 23.

Solution 1:

(a) Let the no be x

8x+4 = 60

8x= 60-4

8x = 54

x = 7

(b) Let the no be x

\frac{1}{5}x\ -4\ =\ 3

\frac{x}{5}=\ 3+4\

\frac{x}{5}=\ 7

x= 7\times5

x = 35

(c) Let the no be x

\frac{3x}{4}+3\ =\ 21

\frac{3x}{4}=\ 21-3

\frac{3x}{4}=18\

x\ =\ \frac{18\times4}{3}

x= 24

(d) Let the no be x

2x -11 = 15

2x= 15+11

2x = 26

x=\ \frac{26}{2\ }

x= 13

(e) Let the no of notebook be x

50-(3x) = 8

-3x = 8-50

3x= 42

x\ =\ \frac{42}{3}

x= 14

(f) Let the no be x

\frac{x+19}{5}=\ 8\

x+19\ =\ 8\times5

x+19 = 40

x= 40 -19

x= 21

(g) Let the no be x

\frac{5}{2}x-7\ =\ 23\

\frac{5}{2}x\ =\ 23\ +7\

\frac{5}{2}x=\ 30

5x\ =\ 30\times2

x\ =\ \frac{60}{5}

x= 12

Still have questions? Our expert teachers can help you out

Book a free class nowQ1) Set up equations and solve them to find the unknown numbers in the following cases:

(a) Add 4 to eight times a number; you get 60.

(b) One-fifth of a number minus 4 gives 3.

(c) If I take three-fourths of a number and add 3 to it, I get 21.

(d) When I subtracted 11 from twice a number, the result was 15.

(e) Munna subtracts thrice the number of notebooks he has from 50, he finds the

result to be 8.

(f) Ibenhal thinks of a number. If she adds 19 to it and divides the sum by 5, she

will get 8.

(g) Anwar thinks of a number. If he takes away 7 from \frac{5}{2} of the number, the result is 23.

Solution 1:

(a) Let the no be x

8x+4 = 60

8x= 60-4

8x = 54

x = 7

(b) Let the no be x

\frac{1}{5}x\ -4\ =\ 3

\frac{x}{5}=\ 3+4\

\frac{x}{5}=\ 7

x= 7\times5

x = 35

(c) Let the no be x

\frac{3x}{4}+3\ =\ 21

\frac{3x}{4}=\ 21-3

\frac{3x}{4}=18\

x\ =\ \frac{18\times4}{3}

x= 24

(d) Let the no be x

2x -11 = 15

2x= 15+11

2x = 26

x=\ \frac{26}{2\ }

x= 13

(e) Let the no of notebook be x

50-(3x) = 8

-3x = 8-50

3x= 42

x\ =\ \frac{42}{3}

x= 14

(f) Let the no be x

\frac{x+19}{5}=\ 8\

x+19\ =\ 8\times5

x+19 = 40

x= 40 -19

x= 21

(g) Let the no be x

\frac{5}{2}x-7\ =\ 23\

\frac{5}{2}x\ =\ 23\ +7\

\frac{5}{2}x=\ 30

5x\ =\ 30\times2

x\ =\ \frac{60}{5}

x= 12

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