- NCERT SOLUTIONS FOR CLASS 10 MATHS EXERCISE 3.3 CHAPTER 3 QUESTION NUMBER 3 ALL SOLUTIONS
- MATHS NCERT EX 3.3 CLASS 10 SOLUTIONS
प्रश्न - ( 1 ) निम्न समस्याओं में रेखिक समीकरण युग्म बनाइये और उनके हल प्रतिस्थापन विधि द्वारा ज्ञात कीजिये ?
( i ) दो संख्याओं का अंतर 26 हे , और एक संख्या दूसरी की तीन गुनी हे। उन्हें ज्ञात कीजिये ?
हल : माना की एक संख्या x हे।
तब दूसरी संख्या y = 3x ----> ( 1 ) { जहाँ y > x क्यूंकि यह तीन गुनी हे }
इनका अंतर , y - x = 26 -------> ( 2 )
y का मान समीकरण ( 1 ) से समीकरण ( 2 ) में रखने पर।
=> 3x - x = 26
=> 2x = 26
=> x = 26 / 2
=> x = 13
=> y = 3x
=> y = 3 * 13
y = 39
उत्तर : 13 , 39
उत्तर की जाँच : 39 - 13 = 26
(ii ) दो सम्पूरक कोणों में बड़ा कोण , छोटे कोण से 18 अधिक हे। उन्हें ज्ञात कीजिये ?
हल : हम जानते हे की सम्पूरक कोणों का योग 180 अंश होता हे।
माना छोटा कोण x हे। तब बड़ा कोण y = x + 18 -----> ( 1 ) होगा।
प्रश्न अनुसार ,
छोटा कोण + बड़ा कोण = 180
x + y = 180 --------> ( 2 )
y का मान समीकरण ( 1 ) से समीकरण ( 2 ) में रखने ( प्रतिस्थापित ) पर ,
x + x + 18 = 180
=> 2 x = 180 - 18
=> 2 x = 162
=> x = 162 / 2
=> x = 81 = छोटा कोण
बड़ा कोण = x + 18
= 81 + 18
= 99
अभीष्ट कोण 81 और 99 हे। 🙏धन्यवाद।
( iii ) एक क्रिकेट टीम के कोच ने 7 बल्ले तथा 6 गेंदे 3800 रुपये में खरीदी। बाद में उसने 3 बल्ले तथा 5
गेंदे 1750 रुपये खरीदी। प्रत्येक बल्ले एवं प्रत्येक गेंद का मूल्य ज्ञात कीजिये।
हल : माना की एक बल्ले का मूल्य x रुपये हे , एवं एक गेंद का मूल्य y रुपये हे ,
तब प्रश्न्नानुसार , 7x + 6y = 3800 ------- > ( 1 )
3x + 5y = 1750 ------- > ( 2 )
समीकरण ( 2 ) से , x = (1750 - 5y ) / 3 ------- > (3 )
x का यह मान समीकरण ( 1 ) में रखने पर ,
=> 7x + 6y = 3800
=> 7 ( 1750 - 5y ) / 3 + 6y = 3800
=> 12250 - 35y + 18y = 11400
=> -17y = 11400 - 12250
=> - 17y = - 850
=> y = 850 / 17
y = 50 रुपये = गेंद की कीमत
y का यह मान समीकरण ( 3 ) में रखने पर ,
=> x = (1750 - 5y ) / 3
=> x = ( 1750 - 5 * 50 ) / 3
=> x = (1750 - 250 ) / 3
=> x = 1500 / 3
=> x = 500 रुपये = बल्ले की कीमत
( iv ) एक नगर में टेक्सी के भाड़े में एक नियत भाड़े के अतिरिक्त चली गयी दुरी पर भाड़ा सम्मिलित किया जाता हे। 10 km की दूरी के लिए भाड़ा 105 रुपये हे , तथा 15 km दुरी के लिए भाड़ा 155 रुपये हे। नियत भाड़ा तथा प्रति km भाड़ा क्या हे ? एक व्यक्ति को 25 km यात्रा करने के लिए कितना भाड़ा देना होगा ?
हल : माना की नियत भाड़ा x रुपये हे एवं प्रति km भाड़े की दर y रुपये हे ,
तब प्रश्नानुसार , x + 10 y = 105 ------> ( 1 )
एवं x + 15 y = 155 -------> ( 2 )
अब समीकरण ( 1 ) से , x + 10 y = 105 ------> ( 1 )
x = 105 - 10y -------> (3 )
x का यह मान समीकरण ( 2 ) में रखने पर ,
x + 15 y = 155
105 - 10 y + 15y = 155
105 + 5y = 155
5y = 155 -105
5y = 50
y = 50 / 5
y = 10 रुपये = प्रति km भाड़े की दर
y का यह मान समीकरण ( 3 ) में रखने पर ,
x = 105 - 10y -------> (3 )
x = 105 - 10 * 10
x = 105 - 100
x = 5 रुपये = नियत भाड़ा
x = 5 , y = 10 उत्तर
( v) यदि किसी भिन्न के अंश व हर दोनों में 2 जोड़ दिया जाये , तो वह 9 / 11 हो जाती हे। यदि अंश और हर दोनों में 3 जोड़ दिया जाये तो वह 5 / 6 हो जाती हे। वह भिन्न ज्ञात कीजिये ?
हल :माना की भिन्न का अंश x तथा हर y हे तब भिन्न = x / y होगी।
तब प्रश्नानुसार , (x + 2 ) / ( y + 2 ) = 9 / 11
11 ( x + 2 ) = 9 ( y + 2 )
11x + 22 = 9y + 18
11 x - 9y = 18 - 22
11x - 9y = -4 -------- > ( 1 )
एवं
( x + 3 ) / (y + 3 ) = 5 / 6
6 ( x + 3 ) = 5 ( y + 3 )
6x + 18 = 5y + 15
6x - 5y = 15 - 18
6x - 5y = -3 ------- > (2 )
समीकरण ( 2 ) से ,
6x = -3 + 5y
x = (5y -3) / 6 -------- > ( 3 )
x का यह मान समीकरण ( 1 ) में रखने पर ,
11x - 9y = -4 -------- > ( 1 )
11 * (5y -3 ) / 6 - 9y = -4
(55y - 33) / 6 - 9y = -4
55y -33 - 54 y = -24
y = -24 + 33
y = 9
y का यह मान समीकरण ( 3 ) में रखने पर ,
x = (5y -3) / 6 -------- > ( 3 )
x = (5 * 9 - 3 ) / 6
x = 45 - 3 / 6
x = 42 / 6
x = 7
तब , भिन्न = x / y = 7 / 9 होगी।
( vi ) पांच वर्ष बाद जेकब की आयु उसके पुत्र की आयु से तीन गुनी हो जाएगी। पांच वर्ष पूर्व जेकब की आयु उसके पुत्र की आयु से सात गुनी थी। उनकी वर्तमान आयु क्या हे।
हल : माना की जेकब की वर्तमान आयु x वर्ष एवं उसके पुत्र की वर्तमान y वर्ष हे।
पांच वर्ष बाद ,
जेकब की आयु = x + 5 वर्ष
उसके पुत्र की आयु = y + 5 वर्ष
तब प्रश्नानुसार ,
x + 5 = 3 ( y + 5 )
=> x + 5 = 3 y + 15
=> x - 3y = 15 - 5
=> x - 3y = 10 -------- > ( 1 )
एवं , पांच वर्ष पूर्व ,
जेकब कि आयु = x - 5 वर्ष
उसके पुत्र की आयु = y - 5 वर्ष
पुनः प्रश्नानुसार ,
x - 5 = 7 ( y - 5 )
x - 5 = 7y - 35
=> x - 7y = - 35 + 5
=> x - 7y = -30 ---------> ( 2 )
=> x = 7y -30 ------ > ( 3 )
समीकरण (3 ) से x का यह मान समीकरण ( 1 ) में रखने पर ,
x - 3y = 10 -------- > ( 1 )
7y - 30 -3y = 10
=> 4y = 10 +30
=> 4y = 40
=> y = 40 / 4
=> y = 10 = पुत्र की वर्तमान आयु
y का यह मान समीकरण (3 ) में रखने पर ,
x - 3y = 10
=> x - 3 * 10 = 10
=> x - 30 = 10
=> x = 10 + 30
=> x = 40 = जेकब की वर्तमान आयु
ENGLISH TRANSLATION :
Question - ( 1 ) Form a pair of linear equations in the following problems and find their solution by substitution method?
(i) The difference of two numbers is 26, and one number is three times the other.
Know them?
Solution : Let x be a number.
Then the second number y = 3x ----> ( 1 ) {where y > x because it is tripled }
Their difference, y - x = 26 -------> ( 2 )
Substituting the value of y in Equation ( 1 ) to Equation (2) .
=> 3x - x = 26
=> 2x = 26
=> x = 26 / 2
=> x = 13
=> y = 3x
=> y = 3 * 13
=> y = 39
Answer: 13, 39
Check Answer : 39 - 13 = 26
(ii) The larger of two supplementary angles is 18 more than the smaller angle. Know them?
Solution : We know that the sum of supplementary angles is 180 degrees.
Let the smaller angle be x. Then the greater angle will be
y = x + 18 -----> ( 1 ).
According to the question,
Small angle + big angle = 180
x + y = 180 --------> ( 2 )
Substituting the value of y from Equation ( 1 ) to Equation (2) ,
=> x + x + 18 = 180
=> 2 x = 180 - 18
=> 2 x = 162
=> x = 162 / 2
=> x = 81 = smaller angle
Greater angle = x + 18
= 81 + 18
= 99
Required angles are 81 and 99. Thank you
(iii) The coach of a cricket team bought 7 bats and 6 balls for Rupee 3800. Later he took 3 bats and 5 Bought lilies for Rupee 1750. Find the cost of each bat and each ball.
Solution: Let the cost of a bat be Rupee x, and that of a ball is rupee y,
Then according to the question, 7x + 6y = 3800 ------- > ( 1 )
3x + 5y = 1750 ------- > ( 2 )
From Eqn. ( 2 ) , x = (1750 - 5y ) / 3 ------- > (3 )
Substituting this value of x in Equation ( 1 ) ,
=> 7x + 6y = 3800
=> 7 ( 1750 - 5y ) / 3 + 6y = 3800
=> 12250 - 35y + 18y = 11400
=> -17y = 11400 - 12250
=> - 17y = - 850
=> y = 850 / 17
y = Rupee 50 = cost of the ball
Substituting this value of y in equation ( 3 ) ,
=> x = (1750 - 5y ) / 3
=> x = ( 1750 - 5 * 50 ) / 3
=> x = (1750 - 250 ) / 3
=> x = 1500 / 3
=> x = Rupee 500 = cost of the bat
(iii) The coach of a cricket team bought 7 bats and 6 balls for rupee 3800. Later he took 3 bats and 5
Bought lilies for rupee 1750. Find the cost of each bat and each ball.
Solution: Let the cost of a bat be rupee x, and that of a ball is Rupee y,
Then according to question, 7x + 6y = 3800 ------- > ( 1 )
3x + 5y = 1750 ------- > ( 2 )
From Eq. ( 2 ) , x = (1750 - 5y ) / 3 ------- > (3 )
Substituting this value of x in Equation ( 1 ) ,
=> 7x + 6y = 3800
=> 7 ( 1750 - 5y ) / 3 + 6y = 3800
=> 12250 - 35y + 18y = 11400
=> -17y = 11400 - 12250
=> - 17y = - 850
=> y = 850 / 17
y = Rupee 50 = cost of the ball
Substituting this value of y in equation ( 3 ) ,
=> x = (1750 - 5y ) / 3
=> x = ( 1750 - 5 * 50 ) / 3
=> x = (1750 - 250 ) / 3
=> x = 1500 / 3
=> x = Rupee 500 = cost of the bat
(iv) Taxi fare in a city includes the fare for the distance travel in addition to a fixed fare. The fare for a distance of 10 km is Rupee 105, and for a distance of 15 km is Rupee 155. What is the fixed fare and the fare per km? How much will a person have to pay for traveling 25 km?
Solution : Let the fixed fare be Rupee x and the rate of fare per km be Rupee y,
Then according to the question, x + 10 y = 105 ------> ( 1 )
and x + 15 y = 155 -------> ( 2 )
Now from equation ( 1 ) , x + 10 y = 105 ------> ( 1 )
x = 105 - 10y -------> (3 )
Substituting this value of x in Equation (2),
x + 15 y = 155
105 - 10 y + 15y = 155
105 + 5y = 155
5y = 155-105
5y = 50
y = 50 / 5
y = Rupee 10 = rate of fare per km
Substituting this value of y in equation ( 3 ) ,
x = 105 - 10y -------> (3 )
x = 105 - 10 * 10
x = 105 - 100
x = Rs.5 = fixed fare
x = 5 , y = 10 Answer
(v) If 2 is added to both the numerator and denominator of a fraction, it becomes 9/11. If 3 is added to both the numerator and the denominator, it becomes 5/6. Find that fraction?
Solution : Let the numerator of the fraction be x and the denominator of the fraction is y, then the fraction = x / y.
Then according to the question, (x + 2 ) / ( y + 2 ) = 9 / 11
11 ( x + 2 ) = 9 ( y + 2 )
11x + 22 = 9y + 18
11 x - 9y = 18 - 22
11x - 9y = -4 -------- > ( 1 )
And
( x + 3 ) / (y + 3 ) = 5 / 6
6 ( x + 3 ) = 5 ( y + 3 )
6x + 18 = 5y + 15
6x - 5y = 15 - 18
6x - 5y = -3 ------- > (2 )
From equation (2),
6x = -3 + 5y
x = (5y -3) / 6 -------- > ( 3 )
Substituting this value of x in Equation ( 1 ) ,
11x - 9y = -4 -------- > ( 1 )
11 * (5y -3 ) / 6 - 9y = -4
(55y - 33) / 6 - 9y = -4
55y -33 - 54 y = -24
y = -24 + 33
y = 9
Substituting this value of y in equation ( 3 ) ,
x = (5y -3) / 6 -------- > ( 3 )
x = (5 * 9 - 3 ) / 6
x = 45 - 3 / 6
x = 42 / 6
x = 7
Then, fraction = x / y = 7 / 9.
(vi) After five years Jacob's age will be three times that of his son. Five years ago Jacob's age was seven times that of his son. What is his present age?
Solution : Let the present age of Jacob be x years and that of his son be y years.
five years later,
Jacob's age = x + 5 years
Age of his son = y + 5 years
Then according to the question,
x + 5 = 3 ( y + 5 )
=> x + 5 = 3 y + 15
=> x - 3y = 15 - 5
=> x - 3y = 10 -------- > ( 1 )
and, five years ago,
Jacob's age = x - 5 years
Age of his son = y - 5 years
As per question again,
x - 5 = 7 ( y - 5 )
x - 5 = 7y - 35
=> x - 7y = - 35 + 5
=> x - 7y = -30 ---------> ( 2 )
=> x = 7y -30 ------ > ( 3 )
Substituting this value of x from Equation (3) in Equation (1),
x - 3y = 10 -------- > ( 1 )
7y - 30 -3y = 10
=> 4y = 10 +30
=> 4y = 40
=> y = 40 / 4
=> y = 10 = present age of son
Substituting this value of y in Equation (3),
x - 3y = 10
=> x - 3 * 10 = 10
=> x - 30 = 10
=> x = 10 + 30
=> x = 40 = Jacob's present age