What is the solution to the linear equation? p = p p = 1 p = 2 p = 8 p = 10
Exercise 4.one
Question 1:
Complete the final column of the table:
Answer:
(i) 10 + iii = 0
LHS = ten + 3
Put x = 3, we get
LHS = 3 + 3 = 6 ≠ RHS
And so, the equation is non satisfied.
(2) x + 3 = 0
LHS = x + 3
Put ten = 0, we get
LHS = 0 + 3 = 3 ≠ RHS
So, the equation is not satisfied.
(three) x + 3 = 0
LHS = 10 + three
Put x = -3, we get
LHS = -3 + 3 = 0 = RHS
So, the equation is satisfied.
(4) x - vii = 1
LHS = ten - seven
Put x = 7, we get
LHS = 7 - seven = 0 ≠ RHS
And so, the equation is not satisfied.
(v) x - 7 = ane
LHS = x - 7
Put x = 8, we become
LHS = viii - 7 = i = RHS
And so, the equation is satisfied.
(6) 5x = 25
LHS = 5x
Put x = 0, we get
LHS = 5 * 0 = 0 ≠ RHS
So, the equation is not satisfied.
(7) 5x = 25
LHS = 5x
Put ten = 5, we go
LHS = five * five = 25 = RHS
And so, the equation is satisfied.
(viii) 5x = 25
LHS = 5x
Put x = -5, we go
LHS = 5 * (-5) = -25 ≠ RHS
So, the equation is not satisfied.
(ix) m/iii = 2
LHS = thousand/3
Put m = -6, we get
LHS = -6/3 = -2 ≠ RHS
And then, the equation is not satisfied.
(x) grand/iii = 2
LHS = k/three
Put m = 0, we get
LHS = 0/3 = 0 ≠ RHS
And so, the equation is non satisfied.
(eleven) m/3 = 2
LHS = m/3
Put m = 6, we get
LHS = 6/3 = 2 = RHS
So, the equation is satisfied.
Question 2:
Check whether the value given in the brackets is a solution to the given equation or not:
(a) n + 5 = 19 (northward = i) (b) 7n + 5 = xix (n = -2) (c) 7n + 5 = 10 (north = ii)
(d) 4p – 3 = 13 (p = 1) (e) 4p – iii = xiii (p = -four) (f) 4p – 3 = 13 (p = 0)
Answer:
(a) n + 5 = 19 (n = 1)
LHS = northward + v
Put n = 1 in LHS, we get
LHS = 1 + 5 = half dozen ≠ RHS
And then, north = ane is non the solution of given equation.
(b)7n + 5 = 19 (n = -2)
LHS = 7n + v
Put northward = -two in LHS, we get
LHS = 7 * (-2) + v = -14 + 5 = -9 ≠ RHS
Then, northward = -2 is non the solution of given equation.
(a) 7n + five = 19 (n = 2)
LHS = 7n + 5
Put n = 2 in LHS, we get
LHS = 7 * 2 + v = fourteen + 5 = 19 = RHS
So, n = 2 is the solution of given equation.
(d) 4p – 3 = 13 (p = ane)
LHS = 4p - 13
Put p = one in LHS, we go
LHS = iv * 1 - three = 4 - three = i ≠ RHS
And then, p = ane is not the solution of given equation.
(eastward) 4p – 3 = xiii (p = -4)
LHS = 4p - 13
Put p = -4 in LHS, we get
LHS = 4 * (-iv) - 3 = -16 - three = -nineteen ≠ RHS
So, p = -iv is not the solution of given equation.
(f) 4p – 3 = 13 (p = 0)
LHS = 4p - 13
Put p = 0 in LHS, nosotros get
LHS = iv * 0 - 3 = 0 - three = -3 ≠ RHS
So, p = 0 is not the solution of given equation.
Question iii:
Solve the following equations by trial and error method:
(i) 5p + 2 = 17 (ii) 3m – fourteen = 4
Answer:
(i) 5p + ii = 17
Put p = 1 in LHS
v * 1 + ii = v + two = 7 ≠ RHS
Put p = two in LHS
5 * 2 + 2 = 10 + 2 = 12 ≠ RHS
Put p = three in LHS
5 * 3 + 2 = 15 + 2 = 17 = RHS
Hence, p = 3 is the solution of the given method.
(ii) 3m – 14 = 4
Put thousand = four in LHS
3 * 4 – fourteen = 12 – 14 = -2 ≠ RHS
Put thousand = 5 in LHS
3 * five – 14 = 15 – 14 = 1 ≠ RHS
Put k = 6 in LHS
3 * vi – 14 = xviii – 14 = four = RHS
Hence, g = half dozen is the solution of the given method.
Question 4:
Write equations for the following statements:
(i) The sum of numbers ten and 4 is 9.
(ii) 2 subtracted from y is 8.
(iii) Ten times a is 70.
(iv) The number b divided by five gives 6.
(v) 3-4th of t is xv.
(vi) Seven times thou plus 7 gets you 77.
(vii) 1-fourth of a number 10 minus 4 gives 4.
(eight) If you take away 6 from 6 times y, you get 60.
(ix) If y'all add 3 to one-third of z, you get 30.
Respond:
(i) The sum of numbers 10 and iv is 9.
=> x + four = 9
(ii) 2 subtracted from y is 8.
=> y – ii = viii
(iii) Ten times a is seventy.
=> 10a = 70
(iv) The number b divided by 5 gives 6.
=> b/5 = vi
(v) Iii-4th of t is 15.
=> 3t/4 = 15
(half-dozen) Seven times m plus seven gets you 77.
=> 7m + 7 = 77
(seven) One-fourth of a number x minus 4 gives 4.
=> x/iv – iv = 4
(viii) If you lot take away six from vi times y, you lot get 60.
=> 6y – half-dozen = threescore
(ix) If you add 3 to i-third of z, you become 30.
=> z/three + 3 = 30
Question 5:
Write the following equations in statement grade:
(i) p + 4 = xv (ii) grand – 7 = 3 (3) 2m = seven (four) m/five = iii
(v) 3m/5 = 6 (six) 3p + 4 = 25 (vii) 4p – 2 = 18 (viii) p/2 + 2 = 8
Answer:
(i) p + 4 = 15
The sum of numbers p and 4 is 15.
(ii) m – 7 = 3
seven subtracted from m is 3.
(iii) 2m = seven
Two times grand is 7.
(iv) m/5 = iii
The number k is divided past 5 gives 3.
(v) 3m/v = vi
Three-fifth of the number m is vi.
(half-dozen) 3p + 4 = 24
Three times p plus four gets 25.
(vii) 4p – two = 18
If you take away ii from iv times p, you lot go 18.
(viii) p/ii + two = 8
If you added 2 to one-half is p, you lot get 8.
Question 6:
Ready upwards an equation in the post-obit cases:
(i) Irfan says that he has 7 marbles more than than five times the marbles Parmit has. Irfan has 37 marbles.
(Tale m to be the number of Parmit'southward marbles.)
(two) Laxmi'south father is 49 years old. He is iv years older than three times Laxmi's historic period. (Take axmi'due south historic period to exist y years.)
(iii) The instructor tells the class that the highest marks obtained past a student in her class are twice the everyman marks plus 7.
The highest score is 87. (Take the lowest score to be 50. )
(iv) In an isosceles triangle, the vertex angle is twice either base angle. (Permit the base angle be B in degrees.
Remember that the sum of angles of a triangle is 1800.)
Answer:
(i) Allow k exist the number of Parmit's marbles.
=> 5m + 7 = 27
(ii) Let the age of Laxmi be y years.
=> 3y + 4 = 49
(iii) Let the lowest score be l.
=> 2l + 7 = 87
(iv) Let the base angle of the isosceles triangle be b,
So, vertex angle = 2b
At present, 2b + b + b = 1800
=> 4b = 1800 [Bending sum property of triangle]
Exercise 4.two
Question 1:
Give first the step you will utilise to separate the variable then solve the equations:
(a) x – 1 = 0 (b) x + 1 = 0 (c) x - 1 = 5 (d) ten + 6 = ii
(e) y – 4 = -7 (f) y – iv = 4 (1000) y + four = 4 (h) y + four = -4
Answer:
(a) x – i = 0
=> x – 1 + ane = 0 + 1 [Calculation i on both sides]
=> x = 1
(b) x + one = 0
=> x + 1 – i = 0 – one [Subtracting 1 on both sides]
=> 10 = -i
(c) 10 - one = 5
=> 10 – 1 + ane = v + 1 [Adding 1 on both sides]
=> x = 6
(d) x + 6 = 2
=> ten + 6 – 6 = two – six [Subtracting 6 on both sides]
=> ten = -4
(due east) y – 4 = -vii
=> y – 4 + 4 = -seven + 4 [Adding 4 on both sides]
=> y = -3
(f) y – four = four
=> y – 4 + iv = 4 + 4 [Calculation iv on both sides]
=> y = eight
(chiliad) y + iv = 4
=> y + 4 – 4 = 4 – 4 [Subtracting 4 on both sides]
=> y = 0
(h) y + 4 = -4
=> y + 4 – 4 = -4 – 4 [Subtracting 4 on both sides]
=> y = -eight
Question ii:
Give first the step you lot will use to separate the variable and then solve the equations:
(a) 3l = 42 (b) b/2 = 6 (c) p/7 = 4 (d) 4x = 25
(due east) 8y = 36 (f) z/3 = v/4 (thousand) a/5 = seven/15 (h) 20t = -x
Reply:
(a) 3l = 42
=> 3l/3 = 42/iii [Dividing both sides past 3]
=> l = 12
(b) b/ii = 6
=> (b/2) * 2 = 6 * two [Multiplying both sides by ii]
=> b = 12
(c) p/seven = 4
=> (p/7) * vii = four * seven [Multiplying both sides by 7]
=> p = 28
(d) 4x = 25
=> 4x/four = 25/iv [Dividing both sides by 4]
=> x = 25/4
(e) 8y = 36
=> 8y/8 = 36/viii [Dividing both sides past 8]
=> y = nine/ii
(f) z/3 = 5/4
=> (z/iii) * 3 = (v/4) * 3 [Multiplying both sides past 3]
=> z = 15/four
(g) a/5 = 7/15
=> (a/5) * 5 = (vii/xv) * 5 [Multiplying both sides past 5]
=> a = 35/fifteen [35 and fifteen are dividing by 5]
=> a = vii/3
(h) 20t = -x
=> 20t/20 = -10/20 [Dividing both sides by xx]
=> t = -1/ii
Question 3:
Give first the step you will use to split up the variable and and then solve the equations:
(a) 3n – 2 = 46 (b) 5m + seven = 17 (c) 20p/iii = twoscore (d) 3p/10 = 6
Answer:
(a) 3n – two = 46
=> 3n – 2 + ii = 46 + 2 [Adding 2 on both sides]
=> 3n = 48
=> 3n/3 = 48/3 [Dividing both sides past iii]
=> due north = xvi
(b) 5m + seven = 17
=> 5m + 7 – 7 = 17 – vii [Subtracting vii on both sides]
=> 5m = 10
=> 5m/five = 10/5 [Dividing both sides past v]
=> one thousand = two
(c) 20p/3 = 40
=> (20p/iii) * 3 = 40 * iii [Multiplying both sides past three]
=> 20p = 120
=> 20p/20 = 120/20 [Dividing both sides by twenty]
=> p = 6
(d) 3p/10 = 6
=> (3p/ten) * 10 = 6 * 10 [Multiplying both sides by ten]
=> 3p = 60
=> 3p/iii = 60/vi [Dividing both sides by 6]
=> p = 10
Question 4:
Solve the post-obit equation:
(a) 10p = 100 (b) 10p + 10 = 100 (c) p/4 = 5 (d) –p/three = v (e) 3p/four = six (f) 3s = -ix
(grand) 3s + 12 = 0 (h) 3s = 0 (i) 2q = 6 (j) 2q – 6 = 0 (k) 2q + half dozen = 0 (l) 2q + 6 = 12
Reply:
(a) 10p = 100
=> 10p/ten = 100/10 [Dividing both sides by 10]
=> p = 10
(b) 10p + x = 100
=> 10p + 10 – x = 100 – 10 [Subtracting 10 on both sides]
=> 10p = 90
=> 10p/10 = 90/ten [Dividing both sides by 10]
=> p = 9
(c) p/4 = v
=> (p/4) * 4 = 5 * 4 [Multiplying both sides by 4]
=> p = 20
(d) –p/iii = 5
=> (–p/3) * (-3) = 5 * (-iii) [Multiplying both sides by (-three)]
=> p = -15
(e) 3p/4 = 6
=> (3p/4) * iv = vi * 4 [Multiplying both sides by iv]
=> 3p = 24
=> 3p/3 = 24/3 [Dividing both sides past 3]
=> p = 8
(f) 3s = -9
=> 3s/iii = -ix/3 [Dividing both sides past 3]
=> south = -3
(k) 3s + 12 = 0
=> 3s + 12 – 12 = 0 – 12 [Subtracting 12 on both sides]
=> 3s = -12
=> 3s/3 = -12/three [Dividing both sides past 3]
=> s = -4
(h) 3s = 0
=> 3s/iii = 0/3 [Dividing both sides by iii]
=> s = 0
(i) 2q = half dozen
2q/2 = vi/2 [Dividing both sides past 2]
=> q = 3
(j) 2q – six = 0
=> 2q – half dozen + half dozen = 0 + 6 [Adding half dozen on both sides]
=> 2q = half-dozen
=> 2q/2 = 6/2 [Dividing both sides by 2]
=> q = iii
(k) 2q + 6 = 0
=> 2q + 6 - 6 = 0 – half dozen [Subtracting six on both sides]
=> 2q = -six
=> 2q/2 = -half dozen/2 [Dividing both sides past ii]
=> q = -iii
(50) 2q + 6 = 12
=> 2q + 6 - 6 = 12 – 6 [Subtracting 6 on both sides]
=> 2q = 6
=> 2q/two = half-dozen/2 [Dividing both sides past 2]
=> q = iii
Exercise iv.three
Question ane:
Solve the post-obit equations:
(a) 2y + v/2 = 37/two (b) 5t + 28 = ten (c) a/v + 3 = 2 (d) q/4 + vii = 5 (e) 5x/2 = 10
(f) 5x/ii = 25/4 (g) 7m + nineteen/2 = 13 (h) 6z + ten = -2 (i) 3l/2 = ii/3 (j) 2b/3 – five = three
Answer:
(a) 2y + 5/2 = 37/2
=> 2y = 37/ii – five/2
=> 2y = (37 - 5)/two
=> 2y = 32/2
=> 2y = sixteen
=> y = 16/ii
=> y = 8
(b) 5t + 28 = 10
=> 5t = 10 – 28
=> 5t = -18
=> t = -eighteen/5
(c) a/5 + three = 2
=> a/5 = 2 – 3
=> a/5 = -i
=> a = -one * 5
=> a = -5
(d) q/iv + 7 = five
=> q/4 = five – 7
=> q/4 = -2
=> q = -2 * 4
=> q = -8
(east) 5x/2 = 10
=> 5x = ten * 2
=> 5x = xx
=> 10 = 20/v
=> x = four
(f) 5x/ii = 25/4
=> 5x = (25/4) * 2
=> 5x = 50/4
=> 5x = 25/2
=> 10 = 25/(5 * two)
=> x = 25/10
=> x = 5/2 [25 and x are divided by 5]
(k) 7m + 19/ii = 13
=> 7m = 13 – nineteen/2
=> 7m = (13 * ii - nineteen)/2
=> 7m = (26 - 19)/ii
=> 7m = 7/2
=> k = 7/(2 * 7)
=> grand = seven/14
=> m = 1/2 [7 and 14 are divided by vii]
(h) 6z + 10 = -2
=> 6z = -2 – 10
=> 6z = -12
=> z = -12/6
=> z = -2
(i) 3l/ii = 2/iii
=> 3l = (ii/three) * two
=> 3l = 4/three
=> l = 4/(3 * 3)
=> l = iv/nine
(j) 2b/3 – 5 = 3
=> 2b/iii = 3 + 5
=> 2b/3 = viii
=> 2b = 8 * 3
=> 2b = 24
=> b = 24/2
=> b = 12
Question 2:
Solve the following equations:
(a) two(ten + 4) = 12 (b) 3(n - 5) = 21 (c) 3(n - 5) = -21 (d) 3 - 2(2 - y) = 7
(e) -4(2 - x) = 9 (f) four(2 - 10) = 9 (k) 4 + v(p - 1) = 34 (h) 34 - 5(p - 1) = 4
Respond:
(a) 2(x + 4) = 12
=> 2x + 2*4 = 12
=> 2x + 8 = 12
=> 2x = 12 – 8
=> 2x = iv
=> x = 4/2
=> x = 2
(b) iii(n - 5) = 21
=> 3n – 3 * v = 21
=> 3n – 15 = 21
=> 3n = 21 + xv
=> 3n = 36
=> n = 36/3
=> n = 12
(c) 3(n - 5) = -21
=> 3n – 3 * 5 = -21
=> 3n – 15 = -21
=> 3n = -21 + fifteen
=> 3n = -six
=> n = -half dozen/3
=> n = -2
(d) three - ii(2 - y) = 7
=> -2(2 - y) = 7 – 3
=> -two(two - y) = 4
=> -2 * 2 + 2 y = 4
=> -4 + 2y = 4
=> 2y = four + iv
=> 2y = 8
=> y = 8/ii
=> y = 4
(east) -4(two - x) = 9
=> -4 * 2 + 4x = 9
=> -viii + 4x = 9
=> 4x = nine + 8
=> 4x = 17
=> x = 17/4
(f) 4(two - x) = 9
=> 4 * 2 – 4x = 9
=> 8 – 4x = 9
=> -4x = 9 – 8
=> -4x = 1
=> 4x = -1
=> x = -1/4
(m) 4 + 5(p - 1) = 34
=> 4 + 5p – 5 * 1 = 34
=> 4 + 5p – 5 = 34
=> 5p – ane = 34
=> 5p = 34 + i
=> 5p = 35
=> p = 35/five
=> p = 7
(h) 34 - 5(p - 1) = 4
=> 34 – 5p + v * ane = 4
=> 34 – 5p + 5 = 4
=> 39 – 5p = 4
=> -5p = 4 – 39
=> -5p = -35
=> 5p = 35
=> p = 35/5
=> p = 7
Question iii:
Solve the following equations:
(a) 4 = 5(p - ii) (b) -4 = 5(p - 2) (c) -16 = -five(2 - p) (d)10 = 4 + 3(t + 2)
(e) 28 = four + iii(t + 5) (f) 0 = xvi + 4(m - 6)
Respond:
(a) 4 = 5(p - 2)
=> 4 = 5p – 5 * 2
=> 4 = 5p – 10
=> 5p = 4 + 10
=> 5p = xiv
=> p = 14/5
(b) -4 = v(p - two)
=> -iv = 5p – five * 2
=> -iv = 5p – ten
=> 5p = ten – 4
=> 5p = 6
=> p = vi/5
(c) -sixteen = -v(two - p)
=> -xvi = -v * 2 + 5p
=> -sixteen = -10 + 5p
=> 5p = -16 + 10
=> 5p = -6
=> p = -6/v
(d) 10 = 4 + iii(t + 2)
=> 10 = 4 + 3t + 3 * 2
=> 10 = 4 + 3t + half-dozen
=> 10 = 10 + 3t
=> 3t = 10 – 10
=> 3t = 0
=> t = 0/three
=> t = 0
(e) 28 = iv + 3(t + five)
=> 28 = four + 3t + 3 * v
=> 28 = 4 + 3t + xv
=> 28 = 3t + 19
=> 28 - 19 = 3t
=> 3t = 9
=> t = 9/3
=> t = 3
(f) 0 = 16 + iv(thousand - 6)
=> 0 = 16 + 4m – 4 * half-dozen
=> 0 = 16 + 4m - 24
=> 0 = 4m – 8
=> 4m = 8
=> yard = 8/4
=> m = ii
Question 4:
(a) Construct three equations starting with x = ii.
(b) Construct 3 equations starting with 10 = -2.
Answer:
(a) Construct three equations starting with x = 2.
(i) x = 2
Multiplying both sides by 10, we get
10x = 10 * 2
=> 10x = 20
Calculation 2 on both sides, we go
10x + 2 = 20 + 2
=> 10x + ii = 22
(two) x = 2
Multiplying both sides by 5, we get
5x = v * 2
=> 5x = 10
Subtracting ii on both sides, we get
5x - 3 = ten - 3
=> 5x - 3 = 7
(iii) x = ii
Dividing both sides by 5, we get
ten/5 = 2/5
(b) 3 equations starting with x = -2.
(i) ten = -two
Multiplying both sides past three, we get
3x = -2 * iii
=> 3x = -6
(ii) x = -2
Multiplying both sides by iii, we get
3x = -two * 3
=> 3x = -6
Adding 7 on both sides, we become
3x + 7 = -six + 7
=> 3x + 7 = one
(iii) x = -2
Multiplying both sides by three, we go
3x = -two * 3
=> 3x = -six
Adding ten on both sides, we get
3x + 10 = -6 + 10
=> 3x + ten = iv
Practice 4.4
Question one:
Fix equations and solve them to find the unknown numbers in the following cases:
(a) Add iv to eight times a number; y'all go sixty.
(b) I-fifth of a number minus 4 gives 3.
(c) If I have three-fourth of a number and add 3 to it, I get 21.
(d) When I subtracted xi from twice a number, the result was 15.
(east) Munna subtracts thrice the number of notebooks he has from l, he finds the result to be 8.
(f) Ibenhal thinks of a number. If she adds 19 to it divides the sum by 5, she volition get 8.
(g) Respond thinks of a number. If he takes away 7 from v/two of the number, the result is 11/2.
Answer:
(a) Allow the number be x.
According to question,
8x + 4 = lx
=> 8x = 60 – iv
=> 8x = 56
=> x = 56/viii
=> 10 = 7
And then, the number is 7
(b) Permit the number by y.
According to question,
y/5 – 4 = iii
=> y/5 = 3 + 4
=> y/five = 7
=> y = 7 * five
=> y = 35
So, the number is 35
(c) Allow the number be z.
Co-ordinate to question,
3z/four + 3 = 21
=> 3z/4 = 21 – three
=> 3z/four = 18
=> 3z = 18 * 4
=> 3z = 72
=> z = 72/3
=> z = 24
So, the number is 24
(d) Let the number be x.
According to question,
2x – 11 = 15
=> 2x = 15 + eleven
=> 2x = 26
=> 10 = 26/2
=> x = 13
And so, the number is 13
(e) Le the number be m.
Co-ordinate to question,
50 – 3m = 8
=> -3m = 8 – l
=> -3m = -42
=> 3m = 42
=> m = 42/3
=> m = 14
So, the number is 14
(f) Let the number exist n.
Co-ordinate to question,
(north + 19)/5 = 8
=> n + 19 = eight * five
=> n + xix = 40
=> due north = 40 – 19
=> northward = 21
So, the number is 21
(grand) Let the number exist x.
According to question,
5x/2 – 7 = xi/2
=> 5x/two = 11/ii + 7
=> 5x/2 = (11 + vii * 2)/2
=> 5x/2 = (eleven + xiv)/ii
=> 5x/ii = 25/two
=> 5x = 25
=> ten = 25/five
=> x = 5
So, the number is 5
Question 2:
Solve the post-obit:
(a) The teacher tells the grade that the highest marks obtained by a student in her form are twice the everyman marks plus vii.
The highest score is 87. What is the everyman score?
(b) In an isosceles triangle, the base angles are equal. The vertex angle is 400. What are the base of operations angles of the triangle?
(Call up, the sum of three angles of a triangle is 1800.)
(c) Sachin scored twice as many runs every bit Rahul. Together, their runs savage ii short of a double century. How many runs did each ane score?
Respond:
(a) Let the everyman mark be y.
Co-ordinate to question,
2y + seven = 87
=> 2y = 87 – vii
=> 2y = lxxx
=> y = lxxx/two
=> y = 40
Hence, the everyman score is 40.
(b) Let the base of operations angle of the triangle be b.
Given, a = twoscore0, b = c
Since a + b + c = 1800 [Angle sum property of a triangle]
=> twoscore0 + b + b = 1800
=> 400 + 2b = 1800
=> 2b = 1800 – 400
=> 2b = 1400
=> b = 1400/two
=> b = 700
Thus, the base angles of the isosceles triangle are seventy0 each.
(c) Permit the score of Rahul exist x runs and Sachin's score is 2x.
According to question,
x + 2x = 198
=> 3x = 198
=> 10 = 198/iii
=> ten = 66
And then, the score of Rahul = 66 runs
and Score of Sachin = 2 * 66 = 132 runs.
Question 3:
Solve the following:
(i) Irfan says that he has 7 marbles more than 5 times the marbles Parmit has. Irfan has 37 marbles. How many marbles does Parmit take?
(ii) Laxmi'south father is 49 years one-time. He is four years older than iii times Laxmi'southward age. What is Laxmi's historic period?
(iii) People of Sundergram planted a total of 102 trees in the hamlet garden. Some of the trees were fruit trees.
The numbers of non-fruit trees were ii more than iii times the number of fruit copse. What was the number of fruit copse planted?
Answer:
(a) Let the number of marbles Allow has exist grand.
Co-ordinate to question,
5m + seven = 37
=> 5m = 37 – 7
=> 5m = 30
=> m = thirty/5
=> m = vi
Thus, Permit has half dozen marbles.
(b) Let the age of Laxmi be y years.
Then the age of her father = 3y + iv years
According to question,
3y + 4 = 49
=> 3y = 49 – 4
=> 3y = 45
=> y = 45/3
=> y = 15
Hence, the age of Laxmi is 15 years.
(c) Let the number of fruit trees be t.
So the number of non-fruit trees = 3t + 2
According to question,
t + 3t + two = 102
=> 4t + 2 = 102
=> 4t = 102 – 2
=> 4t = 100
=> t = 100/four
=> t = 25
Thus, the number of fruit trees is 25.
Question 4:
Solve the following riddle:
I am a number,
Tell my identity!
Have me seven times over,
And add a fifty!
To reach a triple century,
You still need forty!
Answer:
Let the number be n.
Co-ordinate to question,
7n + fifty + 40 = 300
=> 7n + ninety = 300
=> 7n = 300 – 90
=> 7n = 210
=> n = 210/7
=> due north = 30
Thus, the required number is 30.
Source: https://www.examfear.com/cbse-ncert-solution/Class-7/Maths/Simple-Equations/solutions.htm
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