# Ncert Solutions For Class 7 Maths Ex 1.4

## Ncert Solutions For Class 7 Maths Chapter 1 Ex 1.4

Question 1:

Evaluate:

i) (-20)¸10

(-20) ¸ 10 = (20)×110=20×110=2$(-20) \times \frac{1}{10}=-\frac{20\times 1}{10}=-2$

ii) 60¸(-6)

60 ¸ (-6) = 60×(16)=60×16=10$60\times (-\frac{1}{6})=-\frac{60\times 1}{6}=-10$

iii) (-45) ¸ (-9)

(-45) ¸ (-9) = (45)×(19)=45×19=5$(-45)\times (-\frac{1}{9})=\frac{45\times 1}{9}=5$

iv) 14¸[(-2) + 1]

14 ¸ [(-2) + 1] = 14 ¸ (-1) = 13×(11)=13$13\times (\frac{-1}{1})=-13$

v) [(-48)¸12] ¸ 2

[(-48) ¸ 12] ¸ 2 =

[(48)×112]×12=(3612)×12=(4)×12=2$[(-48)\times \frac{1}{12}]\times \frac{1}{2}=(\frac{-36}{12})\times \frac{1}{2}=(-4)\times \frac{1}{2}=-2$

Question 2:

Verify that x ¸ (y + z)  (x ¸ y) + (x ¸ z) for the following values of x, y and z.

i) x = 18, y = 6, z = 3

ii) x = (-6), y = 1, z = 1

i) Given: x ¸ (y + z) ≠ (x ¸ y) + (x ¸ z) and x = 18, y = 6, z = 3

Putting the given values in L.H.S. = 18 ¸ (6 + 3) = 18×19=18×19=189=2$18\times \frac{1}{9}=\frac{18\times 1}{9}=\frac{18}{9}=2$

Putting the given values in R.H.S. = (18 ¸ 6) + (18 ¸ 3) = (18×16)+(18×13)=3+6=9$(18\times \frac{1}{6})+(18\times \frac{1}{3})=3+6=9$

Since, L.H.S. ≠ R.H.S.

Hence, verified.

ii) Given: x ¸ (y + z) ≠ (x ¸ y) + (x ¸ z) and x = (-6), y = 1, z = 1

Putting the given values in L.H.S. = (-6) ¸ (1 + 1) = -3

Putting the given values in R.H.S. = [(-6) ¸ 1] + [(-6) ¸ 3] = -8

Since, L.H.S. ≠ R.H.S.

Hence, verified.

Question 3:

Fill up the blanks:

i) 460¸________ = 460

ii) (-69)¸_______ = (-1)

iii) (-123) ¸ ________ = 1

iv) (-34)¸________ = 34

v) _______¸1 = -45

vi) _______¸34 = -1

vii) 40 ¸ ________ = -4

viii) ________ ¸ (3) = -3

i) 460 ¸1= 460

ii) (-69) ¸69= (-1)

iii) (-123) ¸ (-123) = 1

iv) (-34) ¸(-1)= 34

v)(-45)¸ 1 = -45

vi)(-34)¸ 34 = -1

vii) 40 ¸ (-10) = -4

viii) (-9) ¸ (3) = -3

Question 4:

Write four pairs of integers (x,y) such that x ¸ y = -2.

i) (2,-1)

ii) (-2,1)

iii) (4,-2)

iv) (-4,2)

Question 5:

The temperature at noon was 10oC. Until midnight, it decreases at the rate of 2oC per hour. At what time the temperature will be -8oC?

The temperature is represented in the following number line:

The temperature decreases 2oC = 1 hour

The temperature decreases 1oC = ½ hour

The temperature decreases 18oC = ½ x 18 = 9 hours

Total time = 12 noon + 9 hr = 21 hours = 9 pm

Thus, at 9pm the temperature would be -8oC.

Question 6:

A class test is taken in which (+3) marks are given for every correct answer, (-2) marks are given for every incorrect answer and no marks for not attempting any question.

i) Rohini scored 20 marks. If she got 12 correct answers, how many questions has she attempted incorrectly?

ii) Radhika scored (-5) marks. She got 7 correct answers. Calculate the number of questions she attempted incorrectly.

Given:

Marks obtained for one correct answer = 3

Marks obtained for one incorrect answer = -2

i) Marks obtained for 12 correct answers = 3 x 12 = 36

Rohini scored 20 marks.

Therefore, marks obtained for incorrect answers = 20 – 36 = -16

Number of incorrect answers = (-16) ¸ (-2) = 8

Thus, Rohini has attempted 8 incorrect questions.

ii) Marks given for seven correct answers = 3 x 7 = 21

Marks obtained for incorrect answers = -5 -21 = -26

Number of incorrect answers = (-26) ¸ (-2) = 13

Thus, Radhika attempted 13 incorrect questions.

Question 7:

The rate of descending of an elevator into a mine shaft is 6 m/min. If the descent starts from 10 m above the ground level, how long will it take to reach -350 m?

Total distance covered by mine shaft = 10 m – (-350 m) = 10 + 350 = 360 m

According to the question, time taken to cover a distance of 6m = 1 min

So, time taken to cover a distance of 1 m = 1/6 min

Therefore, time taken to cover a distance of 360 m = 16×360$\frac{1}{6}\times 360$

= 60 minutes

= 1 hour

Thus, the elevator will reach -350 m, from 10 m, in one hour.