## NCERT Solutions For Class 6 Maths Algebra Exercise 11.3

**Exercise 11.3**

Ex 11.3 Class 6 Maths Question 1.

Make up as many expressions with numbers (no variables) as you can from three numbers 5, 7 and 8. Every number should be used not more than once. Use only addition, subtractions and multiplication.

Solution:

Given numbers are 5, 7 and 8.

Expressions are:

(i) 8 + (5 + 7)

(ii) 5 + (8 – 7)

(iii) 8 + (5 x 7)

(iv) 7 – (8 – 5)

(v) 7 x (8 + 5)

(vi) 5 x (8 + 7)

(vii) 8 x (5 + 7)

(viii) 7 + (8 – 5)

(ix) (5 x 7) – 8

(x) 7 + (8 x 5)

Ex 11.3 Class 6 Maths Question 2.

Which out of the following are expressions with numbers only?

(a) y + 3

(b) (7 x 20) – 8z

(c) 5(21 – 7) + 7 x 2

(d) 5

(e) 3x

(f) 5 – 5n

(g) (7 x 20) – (5 x 10) – 45 +p

Solution:

(a) y + 3. This expression has variable ‘y’.

(b) (7 x 20) – 8z. This expression has a variable ‘z’.

(c) 5(21 -7) + 7 x 2. This expression has no variable. So it is with numbers only.

(d) 5. This expression is with numbers only.

(e) 3x. This expression has a variable ‘x’.

(f) 5 – 5n. This expression has a variable ‘n’.

(g) (7 x 20) – (5 x 10) – 45 + p. This expression has a variable ‘p’.

Ex 11.3 Class 6 Maths Question 3.

Identify the operations (addition, subtraction, division and multiplication) in forming the following

expressions and tell how the expressions have been formed.

(a) z + 1, z – 1,y + 17, y – 17

(b) 17y,(frac { y }{ 17 }) , 5z

(c) 2y + 17, 2y – 17

(d) 7m, -7m + 3, -7m – 3

Solution:

Expressions |
Operations used |
Formation of expression |
||

(a) | (i) | z + 1 | Addition | z is increased by 1 |

(ii) | z – 1 | Subtraction | z is decreased by 1 | |

(iii) | y +17 | Addition | y is increased by 17 | |

(iv) | y -17 | Subtraction | y is decreased by 17 | |

(b) | (i) | 17y | Multiplication | y is multiplied by 17 |

(ii) | y/17 | Division | y is Divided by 17 | |

(iii) | 5z | Multiplication | z is Multiplied by 5 | |

(c) | (i) | 2y + 17 | Multiplication and addition | y is multiplied by 2 and then 17 is added. |

(ii) | 2y -17 | Multiplication and subtraction | Twice of y is decreased by 17 | |

(d) | (i) | 7 m | Multiplication | m is multiplied by 7 |

(ii)- | -7m + 3 | Multiplication and addition | m is multiplied by -7 and then increased by 3 | |

(iii) | -7m – 3 | Multiplication and subtraction |
M is multiplied by -7 and then decreased by 3 |

Ex 11.3 Class 6 Maths Question 4.

Give expressions for the follow

(a) 7 added top

(b) 7 subtracted from p

(c) p multiplied by 7

(d) p divided by 7

(e) 7 subtracted from -m

(j) -p multiplied by 5

(g) -p divided by 5

(h) p multiplied by -5

Solution:

(a) p + 7

(b) p – 7

(c) 7p

(d) (frac { p }{ 7 })

(e) -m – 7

(f) -5p

(g) (frac { -p }{ 5 })

(h) 5p

Ex 11.3 Class 6 Maths Question 5.

Give expressions in the following cases:

(a).11 added to 2m

(b) 11 subtracted from 2m

(c) 5 times y to which 3 is added

(d) 5 times y from which 3 is subtracted

(e) y is multiplied by -8

(f) y is multiplied by -8 and then 5 is added to the result

(g) y is multiplied by 5 and the result is subtracted from 16

(h) y is multiplied by -5 and the result is added to 16.

Solution:

(a) 2m + 11

(b) 2m – 11

(e) 5y + 3

(d) 5y – 3

(e) -8y

(f) -8y+5

(g) 16 – 5y

(h) -5y + 16

Ex 11.3 Class 6 Maths Question 6.

(a) Form expressions using t and 4. Use not more than one number operation. Every expression must have t in it.

(b) Form expressions using y, 2 and 7. Every expression must have y in it. Use only two number operations. These should, be different.

Solution:

(a) The possible expressions are:

(i) t + 4

(ii) t – 4

(iii) 4t

(iv) (frac { t }{ 4 })

(v) 4 + t

(vi) 4 + t,etc.

(b) The possible expressions are:

(i) 2y + 7

(ii) 7y – 2

(iii) 7 – 2y

(iv) 7y + 2

(v) (frac { 7y }{ 2 })

(vi) (frac { 2y }{ 7 })

(vii) (frac { y }{ 7 }) + 2

(viii) (frac { y }{ 2 }) – 7,etc.

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