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An elevator descends into a mine shaft at the rate of 6 m/min. If the descent starts from 10 m above the ground level, how long will it take to reach minus 350 m.

7. An elevator descends into a mine shaft at the rate of 6 m/min. If the descent starts from 10 m above the ground level, how long will it take to reach – 350 m.

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HARSH KANKARIA

6. In a class test (+ 3) marks are given for every correct answer and (–2) marks are given for every incorrect answer and no marks for not attempting any question. (i) Radhika scored 20 marks. If she has got 12 correct answers, how many questions has she attempted incorrectly? (ii) Mohini scores –5 marks in this test, though she has got 7 correct answers. How many questions has she attempted incorrectly?

Given,

Marks for every correct answer = +3

Marks for every wrong answer = - 2

(i) According to question,

Marks obtained by Radhika = 20

Number of correct answers = 12

\therefore Marks obtained for correct answers = 12\times3 = 36

\therefore Marks obtained for incorrect answers = Total marks – Marks obtained for correct answers

\\ = 20 - 36 \\= - 16

\therefore Number of incorrect answers = (-16)\div (-2)

\\ = \frac{16}{2} \\ = 8

Therefore, Radhika attempted 8 questions incorrectly.

(ii) 

According to question,

Marks obtained by Mohini = -5

Number of correct answers = 7

\therefore Marks obtained for correct answers = 7\times3 = 21

\therefore Marks obtained for incorrect answers = Total marks – Marks obtained for correct answers

\\ = (-5) - 21 \\= - 26

\therefore Number of incorrect answers = (-26)\div (-2)

\\ = \frac{26}{2} \\ = 13

Therefore, Mohini attempted 13 questions incorrectly.

 

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HARSH KANKARIA

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5. The temperature at 12 noon was 10°C above zero. If it decreases at the rate of 2°C per hour until midnight, at what time would the temperature be 8°C below zero? What would be the temperature at mid-night?

Given,

Temperature at 12 noon = 10^{\circ}C

Final temperature = -8^{\circ}C

\therefore Decrease in temperature = 10 - (-8) = 18^{\circ}C

Time taken for the temperature  to decrease by 2^{\circ}C1\ hour

\therefore Time taken for the temperature  to decrease by 18^{\circ}C

=\frac{1}{2}\times18\ hour

= 9\ hours

Now,

Time until midnight = 12\ hours

\therefore Decrease in temperature in 12\ hours = 2\times12 = 24^{\circ}C

Therefore, temperature at midnight = (10 - 24)^{\circ}C = -14^{\circ}C

Therefore, the temperature at mid-night will be 14^{\circ}C below zero.

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HARSH KANKARIA

4. Write five pairs of integers (a, b) such that a ÷ b = –3. One such pair is (6, –2) because 6 ÷ (–2) = (–3).

Five pairs of integers (a, b) such that a \div b = -3 are:

(-6,2); (9,-3); (-9,3) ; (3,-1) ; (-3,1)

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HARSH KANKARIA

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3.  Fill in the blanks:
                (a) 369 ÷ _____ = 369            (b) (–75) ÷ _____ = –1
                (c) (–206) ÷ _____ = 1             (d) – 87 ÷ _____ = 87
                (e) _____ ÷ 1 = – 87                (f) _____ ÷ 48 = –1
                (g) 20 ÷ _____ = –2                 (h) _____ ÷ (4) = –3

(a) Given, 369 \div \underline{\ \ \ \ \ \ } = 369

A number divided by 1 gives the number itself.

369 \div \underline{1} = 369          

 

(b) (-75) \div \underline{\ \ \ \ \ } = -1

The product is negative, therefore there must be odd number of negative integers. 

(-75) \div \underline{75} = -1
                

(c) (-206) \div \underline{\ \ \ \ }= 1

A number divided by itself gives 1

(-206) \div \underline{(-206)}= 1           

 

(d) (-87) \div \underline{\ \ \ \ }= 87

The product is positive, therefore there must be even number of negative integers. 

(-87) \div \underline{(-1) }= 87
                

(e)\underline{\ \ \ \ } \div 1 = - 87

A number divided by 1 gives the number itself.                

\underline{(-87)} \div 1 = - 87

 

(f) \underline{\ \ \ \ \ } \div 48 = -1

The product is negative, therefore there must be odd number of negative integers. 

\underline{(-48)} \div 48 = -1
                

(g) 20\div \underline{\ \ \ \ \ } = -2

The product is negative, therefore there must be odd number of negative integers. 

20\div \underline{(-10)} = -2               

 

(h) \underline{\ \ \ \ \ } \div (4) = -3

The product is negative, therefore there must be odd number of negative integers. 

\underline{\left (\frac{-4}{3} \right ) } \div (4) = -3

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HARSH KANKARIA

2. Verify that a \div (b + c) \neq (a\div b)+ (a\div c) for each of the following values of a, b and c.
     (a) a = 12, b = – 4, c = 2            (b) a = (–10), b = 1, c = 1

a \div (b + c) \neq (a\div b)+ (a\div c)

(a) a = 12, b = – 4, c = 2   

L.H.S = \\ a \div (b + c)

\\ = 12 \div [(-4)+2] \\ = 12 \div (-2) \\ = - \left (\frac{12}{2} \right ) \\ = -6

R.H.S = (a\div b)+ (a\div c)

\\ = [12\div (-4)]+ (12\div 2) \\ = \left [-\left (\frac{12}{4} \right ) \right ] + \left (\frac{12}{2} \right ) \\ = (-3)+6 \\ = 3

Therefore. L.H.S \neq R.H.S

Hence verified.

 

(b) a = (–10), b = 1, c = 1

L.H.S = \\ a \div (b + c)

\\ = (-10) \div [1+1] \\ = (-10) \div 2 \\ = - \left (\frac{10}{2} \right ) \\ = -5

R.H.S = (a\div b)+ (a\div c)

\\ = [(-10)\div 1]+ ((-10)\div 1) \\ = \left [-\left (\frac{10}{1} \right ) \right ] + \left [-\left (\frac{10}{1} \right ) \right ] \\ = (-10)+(-10) \\ = -20

Therefore. L.H.S \neq R.H.S

Hence verified.

 

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HARSH KANKARIA

NEET 2024 Most scoring concepts

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1.   Evaluate each of the following:
                (a) (–30) ÷ 10            (b) 50 ÷ (–5)             (c) (–36) ÷ (–9)
                (d) (– 49) ÷ (49)        (e) 13 ÷ [(–2) + 1]     (f ) 0 ÷ (–12)
                (g) (–31) ÷ [(–30) + (–1)]
                (h) [(–36) ÷ 12] ÷ 3    (i) [(– 6) + 5)] ÷ [(–2) + 1]

(_24)divide5

 

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Manzoor ah

Q. Is   (i) 1 ÷ a = 1?
     (ii) a ÷ (–1) = – a? for any integer a.
Take different values of a and check.

(i) 1 \div a = 1 is true only for a =1.

(ii) a \div (-1) = - a

Taking, a =1; 1\div (-1) = -1

a =2; 2\div (-1) = -2

a =-1; (-1)\div (-1) = 1

Hence, this is true for every integer .

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Find:    (a) (–36) ÷ (– 4)        (b) (–201) ÷ (–3)        (c) (–325) ÷ (–13)

When we divide a negative integer by a negative integer, we first divide them as whole numbers and then put a positive sign (+).

(a)

\\ (-36) \div (- 4) \\ = +\left (\frac{36}{4} \right ) \\ = 9       

 

(b)

\\ (-201) \div (- 3) \\ = +\left (\frac{201}{3} \right ) \\ = 67      

 

(c)

\\ (-325) \div (- 13) \\ = +\left (\frac{325}{13} \right ) \\ = 25

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HARSH KANKARIA

Find:     (a) 125 ÷ (–25)        (b) 80 ÷ (–5)        (c) 64 ÷ (–16)

When we divide a positive integer by a negative integer, we first divide them as whole numbers and then put a minus sign (–) before the quotient.

(a) 125 \div (-25) = -\left (\frac{125}{25} \right ) = -5       

(b) 80\div (-5) = -\left (\frac{80}{5} \right ) = -16        

(c) 64\div (-16) = -\left (\frac{64}{16} \right ) = -4

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