Class 7 - Maths - Simple Equations

Exercise 4.1

Question 1:

Complete the last column of the table: Answer:

(i) x + 3 = 0

LHS = x + 3

Put x = 3, we get

LHS = 3 + 3 = 6 ≠ RHS

So, the equation is not satisfied.

(ii) x + 3 = 0

LHS = x + 3

Put x = 0, we get

LHS = 0 + 3 = 3 ≠ RHS

So, the equation is not satisfied.

(iii) x + 3 = 0

LHS = x + 3

Put x = -3, we get

LHS = -3 + 3 = 0 = RHS

So, the equation is satisfied.

(iv) x - 7 = 1

LHS = x - 7

Put x = 7, we get

LHS = 7 - 7 = 0 ≠ RHS

So, the equation is not satisfied.

(v) x - 7 = 1

LHS = x - 7

Put x = 8, we get

LHS = 8 - 7 = 1 = RHS

So, the equation is satisfied.

(vi) 5x = 25

LHS = 5x

Put x = 0, we get

LHS = 5 * 0 = 0 ≠ RHS

So, the equation is not satisfied.

(vii) 5x = 25

LHS = 5x

Put x = 5, we get

LHS = 5 * 5 = 25 = RHS

So, the equation is satisfied.

(viii) 5x = 25

LHS = 5x

Put x = -5, we get

LHS = 5 * (-5) = -25 ≠ RHS

So, the equation is not satisfied.

(ix) m/3 = 2

LHS = m/3

Put m = -6, we get

LHS = -6/3 = -2 ≠ RHS

So, the equation is not satisfied.

(x) m/3 = 2

LHS = m/3

Put m = 0, we get

LHS = 0/3 = 0 ≠ RHS

So, the equation is not satisfied.

(xi) 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 (n = 1)          (b) 7n + 5 = 19 (n = -2)          (c) 7n + 5 = 10 (n = 2)

(d) 4p – 3 = 13 (p = 1)       (e) 4p – 3 = 13 (p = -4)           (f) 4p – 3 = 13 (p = 0)

Answer:

(a) n + 5 = 19 (n = 1)

LHS = n + 5

Put n = 1 in LHS, we get

LHS = 1 + 5 = 6 ≠ RHS

So, n = 1 is not the solution of given equation.

(b)7n + 5 = 19 (n = -2)

LHS = 7n + 5

Put n = -2 in LHS, we get

LHS = 7 * (-2) + 5 = -14 + 5 = -9  ≠ RHS

So, n = -2 is not the solution of given equation.

(a) 7n + 5 = 19 (n = 2)

LHS = 7n + 5

Put n = 2 in LHS, we get

LHS = 7 * 2 + 5 = 14 + 5 = 19 = RHS

So, n = 2 is the solution of given equation.

(d) 4p – 3 = 13 (p = 1)

LHS = 4p - 13

Put p = 1 in LHS, we get

LHS = 4 * 1 - 3 = 4 - 3 = 1 ≠ RHS

So, p = 1 is not the solution of given equation.

(e) 4p – 3 = 13 (p = -4)

LHS = 4p - 13

Put p = -4 in LHS, we get

LHS = 4 * (-4) - 3 = -16 - 3 = -19 ≠ RHS

So, p = -4 is not the solution of given equation.

(f) 4p – 3 = 13 (p = 0)

LHS = 4p - 13

Put p = 0 in LHS, we get

LHS = 4 * 0 - 3 = 0 - 3 = -3 ≠ RHS

So, p = 0 is not the solution of given equation.

Question 3:

Solve the following equations by trial and error method:

(i) 5p + 2 = 17               (ii) 3m – 14 = 4

Answer:

(i) 5p + 2 = 17

Put p = 1 in LHS

5 * 1 + 2 = 5 + 2 = 7 ≠ RHS

Put p = 2 in LHS

5 * 2 + 2 = 10 + 2 = 12 ≠ RHS

Put p = 3 in LHS

5 * 3 + 2 = 15 + 2 = 17 = RHS

Hence, p = 3 is the solution of the given method.

(ii) 3m – 14 = 4

Put m = 4 in LHS

3 * 4 – 14 = 12 – 14 = -2 ≠ RHS

Put m = 5 in LHS

3 * 5 – 14 = 15 – 14 = 1 ≠ RHS

Put m = 6 in LHS

3 * 6 – 14 = 18 – 14 = 4 = RHS

Hence, m = 6 is the solution of the given method.

Question 4:

Write equations for the following statements:

(i) The sum of numbers x and 4 is 9.

(ii) 2 subtracted from y is 8.

(iii) Ten times a is 70.

(iv) The number b divided by 5 gives 6.

(v) Three-fourth of t is 15.

(vi) Seven times m plus 7 gets you 77.

(vii) One-fourth of a number x minus 4 gives 4.

(viii) If you take away 6 from 6 times y, you get 60.

(ix) If you add 3 to one-third of z, you get 30.

Answer:

(i) The sum of numbers x and 4 is 9.

=> x + 4 = 9

(ii) 2 subtracted from y is 8.

=> y – 2 = 8

(iii) Ten times a is 70.

=> 10a = 70

(iv) The number b divided by 5 gives 6.

=> b/5 = 6

(v) Three-fourth of t is 15.

=> 3t/4 = 15

(vi) Seven times m plus 7 gets you 77.

=> 7m + 7 = 77

(vii) One-fourth of a number x minus 4 gives 4.

=> x/4 – 4 = 4

(viii) If you take away 6 from 6 times y, you get 60.

=> 6y – 6 = 60

(ix) If you add 3 to one-third of z, you get 30.

=> z/3 + 3 = 30

Question 5:

Write the following equations in statement form:

(i) p + 4 = 15                (ii) m – 7 = 3             (iii) 2m = 7             (iv) m/5 = 3

(v) 3m/5 = 6                (vi) 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

7 subtracted from m is 3.

(iii) 2m = 7

Two times m is 7.

(iv) m/5 = 3

The number m is divided by 5 gives 3.

(v) 3m/5 = 6

Three-fifth of the number m is 6.

(vi) 3p + 4 = 24

Three times p plus 4 gets 25.

(vii) 4p – 2 = 18

If you take away 2 from 4 times p, you get 18.

(viii) p/2 + 2 = 8

If you added 2 to half is p, you get 8.

Question 6:

Set up an equation in the following cases:

(i) Irfan says that he has 7 marbles more than five times the marbles Parmit has. Irfan has 37 marbles.

(Tale m to be the number of Parmit’s marbles.)

(ii) Laxmi’s father is 49 years old. He is 4 years older than three times Laxmi’s age. (Take  axmi’s age to be y years.)

(iii) The teacher tells the class that the highest marks obtained by a student in her class are twice the lowest marks plus 7.

The highest score is 87. (Take the lowest score to be l. )

(iv) In an isosceles triangle, the vertex angle is twice either base angle. (Let the base angle be B in degrees.

Remember that the sum of angles of a triangle is 1800.)

Answer:

(i) Let m be 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

Now, 2b + b + b = 1800

=> 4b = 1800                         [Angle sum property of triangle]

Exercise 4.2

Question 1:

Give first the step you will use to separate the variable and then solve the equations:

(a)  x – 1 = 0             (b) x + 1 = 0               (c) x - 1 = 5              (d) x + 6 = 2

(e) y – 4 = -7             (f) y – 4 = 4               (g) y + 4 = 4             (h) y + 4 = -4

Answer:

(a)  x – 1 = 0

=> x – 1 + 1 = 0 + 1           [Adding 1 on both sides]

=> x = 1

(b) x + 1 = 0

=> x + 1 – 1 = 0 – 1          [Subtracting 1 on both sides]

=> x = -1

(c) x - 1 = 5

=> x – 1 + 1 = 5 + 1          [Adding 1 on both sides]

=> x = 6

(d) x + 6 = 2

=> x + 6 – 6 = 2 – 6          [Subtracting 6 on both sides]

=> x = -4

(e) y – 4 = -7

=> y – 4 + 4 = -7 + 4        [Adding 4 on both sides]

=> y = -3

(f) y – 4 = 4

=> y – 4 + 4 = 4 + 4          [Adding 4 on both sides]

=> y = 8

(g) y + 4 = 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 = -8

Question 2:

Give first the step you 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

(e) 8y = 36           (f) z/3 = 5/4             (g) a/5 = 7/15          (h) 20t = -10

Answer:

(a) 3l = 42

=> 3l/3 = 42/3                       [Dividing both sides by 3]

=> l = 12

(b) b/2 = 6

=> (b/2) * 2 = 6 * 2              [Multiplying both sides by 2]

=> b = 12

(c) p/7 = 4

=> (p/7) * 7 = 4 * 7              [Multiplying both sides by 7]

=> p = 28

(d) 4x = 25

=> 4x/4 = 25/4                     [Dividing both sides by 4]

=> x = 25/4

(e) 8y = 36

=> 8y/8 = 36/8                         [Dividing both sides by 8]

=> y = 9/2

(f) z/3 = 5/4

=> (z/3) * 3 = (5/4) * 3           [Multiplying both sides by 3]

=> z = 15/4

(g) a/5 = 7/15

=> (a/5) * 5 = (7/15) * 5        [Multiplying both sides by 5]

=> a = 35/15                            [35 and 15 are dividing by 5]

=> a = 7/3

(h) 20t = -10

=> 20t/20 = -10/20                 [Dividing both sides by 20]

=> t = -1/2

Question 3:

Give first the step you will use to separate the variable and then solve the equations:

(a) 3n – 2 = 46         (b) 5m + 7 = 17          (c) 20p/3 = 40              (d) 3p/10 = 6

Answer:

(a) 3n – 2 = 46

=> 3n – 2 + 2 = 46 + 2          [Adding 2 on both sides]

=> 3n = 48

=> 3n/3 = 48/3                     [Dividing both sides by 3]

=> n = 16

(b) 5m + 7 = 17

=> 5m + 7 – 7 = 17 – 7             [Subtracting 7 on both sides]

=> 5m = 10

=> 5m/5 = 10/5                        [Dividing both sides by 5]

=> m = 2

(c) 20p/3 = 40

=> (20p/3) * 3 = 40 * 3           [Multiplying both sides by 3]

=> 20p = 120

=> 20p/20 = 120/20               [Dividing both sides by 20]

=> p = 6

(d) 3p/10 = 6

=> (3p/10) * 10 = 6 * 10       [Multiplying both sides by 10]

=> 3p = 60

=> 3p/3 = 60/6                       [Dividing both sides by 6]

=> p = 10

Question 4:

Solve the following equation:

(a) 10p = 100    (b) 10p + 10 = 100     (c) p/4 = 5    (d) –p/3 = 5    (e) 3p/4 = 6    (f) 3s = -9

(g) 3s + 12 = 0   (h) 3s = 0      (i) 2q = 6    (j) 2q – 6 = 0     (k) 2q + 6 = 0     (l) 2q + 6 = 12

Answer:

(a) 10p = 100

=> 10p/10 = 100/10                       [Dividing both sides by 10]

=> p = 10

(b) 10p + 10 = 100

=> 10p + 10 – 10 = 100 – 10        [Subtracting 10 on both sides]

=> 10p = 90

=> 10p/10 = 90/10                       [Dividing both sides by 10]

=> p = 9

(c) p/4 = 5

=> (p/4) * 4 = 5 * 4                      [Multiplying both sides by 4]

=> p = 20

(d) –p/3 = 5

=>  (–p/3) * (-3) = 5 * (-3)          [Multiplying both sides by (-3)]

=> p = -15

(e) 3p/4 = 6

=> (3p/4) * 4 = 6 * 4                    [Multiplying both sides by 4]

=> 3p = 24

=> 3p/3 = 24/3                             [Dividing both sides by 3]

=> p = 8

(f) 3s = -9

=> 3s/3 = -9/3                                [Dividing both sides by 3]

=> s = -3

(g) 3s + 12 = 0

=> 3s + 12 – 12 = 0 – 12               [Subtracting 12 on both sides]

=> 3s = -12

=> 3s/3 = -12/3                             [Dividing both sides by 3]

=> s = -4

(h) 3s = 0

=> 3s/3 = 0/3                                [Dividing both sides by 3]

=> s = 0

(i) 2q = 6

2q/2 = 6/2                                    [Dividing both sides by 2]

=> q = 3

(j) 2q – 6 = 0

=> 2q – 6 + 6 = 0 + 6                   [Adding 6 on both sides]

=> 2q = 6

=> 2q/2 = 6/2                              [Dividing both sides by 2]

=> q = 3

(k) 2q + 6 = 0

=> 2q + 6 - 6 = 0 – 6                    [Subtracting 6 on both sides]

=> 2q = -6

=> 2q/2 = -6/2                             [Dividing both sides by 2]

=> q = -3

(l) 2q + 6 = 12

=> 2q + 6 - 6 = 12 – 6                  [Subtracting 6 on both sides]

=> 2q = 6

=> 2q/2 = 6/2                               [Dividing both sides by 2]

=> q = 3

Exercise 4.3

Question 1:

Solve the following equations:

(a) 2y + 5/2 = 37/2       (b) 5t + 28 = 10        (c) a/5 + 3 = 2       (d) q/4 + 7 = 5       (e) 5x/2 = 10

(f) 5x/2 = 25/4              (g) 7m + 19/2 = 13   (h) 6z + 10 = -2     (i) 3l/2 = 2/3       (j) 2b/3 – 5 = 3

Answer:

(a) 2y + 5/2 = 37/2

=> 2y = 37/2 – 5/2

=> 2y = (37 - 5)/2

=> 2y = 32/2

=> 2y = 16

=> y = 16/2

=> y = 8

(b) 5t + 28 = 10

=> 5t = 10 – 28

=> 5t = -18

=> t = -18/5

(c) a/5 + 3 = 2

=> a/5 = 2 – 3

=> a/5 = -1

=> a = -1 * 5

=> a = -5

(d) q/4 + 7 = 5

=> q/4 = 5 – 7

=> q/4 = -2

=> q = -2 * 4

=> q = -8

(e) 5x/2 = 10

=> 5x = 10 * 2

=> 5x = 20

=> x = 20/5

=> x = 4

(f) 5x/2 = 25/4

=> 5x = (25/4) * 2

=> 5x = 50/4

=> 5x = 25/2

=> x = 25/(5 * 2)

=> x = 25/10

=> x = 5/2                               [25 and 10 are divided by 5]

(g) 7m + 19/2 = 13

=> 7m = 13 – 19/2

=> 7m = (13 * 2 - 19)/2

=> 7m = (26 - 19)/2

=> 7m = 7/2

=> m = 7/(2 * 7)

=> m = 7/14

=> m = 1/2                            [7 and 14 are divided by 7]

(h) 6z + 10 = -2

=> 6z = -2 – 10

=> 6z = -12

=> z = -12/6

=> z = -2

(i) 3l/2 = 2/3

=> 3l = (2/3) * 2

=> 3l = 4/3

=> l = 4/(3 * 3)

=> l = 4/9

(j) 2b/3 – 5 = 3

=> 2b/3 = 3 + 5

=> 2b/3 = 8

=> 2b = 8 * 3

=> 2b = 24

=> b = 24/2

=> b = 12

Question 2:

Solve the following equations:

(a) 2(x + 4) = 12             (b) 3(n - 5) = 21              (c) 3(n - 5) = -21            (d) 3 - 2(2 - y) = 7

(e) -4(2 - x) = 9               (f) 4(2 - x) = 9                 (g) 4 + 5(p - 1) = 34       (h) 34 - 5(p - 1) = 4

Answer:

(a) 2(x + 4) = 12

=> 2x + 2*4 = 12

=> 2x + 8 = 12

=> 2x = 12 – 8

=> 2x = 4

=> x = 4/2

=> x = 2

(b) 3(n - 5) = 21

=> 3n – 3 * 5 = 21

=> 3n – 15 = 21

=> 3n = 21 + 15

=> 3n = 36

=> n = 36/3

=> n = 12

(c) 3(n - 5) = -21

=> 3n – 3 * 5 = -21

=> 3n – 15 = -21

=> 3n = -21 + 15

=> 3n = -6

=> n = -6/3

=> n = -2

(d) 3 - 2(2 - y) = 7

=> -2(2 - y) = 7 – 3

=> -2(2 - y) = 4

=> -2 * 2 + 2 y = 4

=> -4 + 2y = 4

=> 2y = 4 + 4

=> 2y = 8

=> y = 8/2

=> y = 4

(e) -4(2 - x) = 9

=> -4 * 2 + 4x = 9

=> -8 + 4x = 9

=> 4x = 9 + 8

=> 4x = 17

=> x = 17/4

(f) 4(2 - x) = 9

=> 4 * 2 – 4x = 9

=> 8 – 4x = 9

=> -4x = 9 – 8

=> -4x = 1

=> 4x = -1

=> x = -1/4

(g) 4 + 5(p - 1) = 34

=> 4 + 5p – 5 * 1 = 34

=> 4 + 5p – 5 = 34

=> 5p – 1 = 34

=> 5p = 34 + 1

=> 5p = 35

=> p = 35/5

=> p = 7

(h) 34 - 5(p - 1) = 4

=> 34 – 5p + 5 * 1 = 4

=> 34 – 5p + 5 = 4

=> 39 – 5p = 4

=> -5p = 4 – 39

=> -5p = -35

=> 5p = 35

=> p = 35/5

=> p = 7

Question 3:

Solve the following equations:

(a) 4 = 5(p - 2)              (b) -4 = 5(p - 2)            (c) -16 = -5(2 - p)              (d)10 = 4 + 3(t + 2)

(e) 28 = 4 + 3(t + 5)     (f) 0 = 16 + 4(m - 6)

Answer:

(a) 4 = 5(p - 2)

=> 4 = 5p – 5 * 2

=> 4 = 5p – 10

=> 5p = 4 + 10

=> 5p = 14

=> p = 14/5

(b) -4 = 5(p - 2)

=> -4 = 5p – 5 * 2

=> -4 = 5p – 10

=> 5p = 10 – 4

=> 5p = 6

=> p = 6/5

(c) -16 = -5(2 - p)

=> -16 = -5 * 2 + 5p

=> -16 = -10 + 5p

=> 5p = -16 + 10

=> 5p = -6

=> p = -6/5

(d) 10 = 4 + 3(t + 2)

=> 10 = 4 + 3t + 3 * 2

=> 10 = 4 + 3t + 6

=> 10 = 10 + 3t

=> 3t = 10 – 10

=> 3t = 0

=> t = 0/3

=> t = 0

(e) 28 = 4 + 3(t + 5)

=> 28 = 4 + 3t + 3 * 5

=> 28 = 4 + 3t + 15

=> 28 = 3t + 19

=> 28 - 19 = 3t

=> 3t = 9

=> t = 9/3

=> t = 3

(f) 0 = 16 + 4(m - 6)

=> 0 = 16 + 4m – 4 * 6

=> 0 = 16 + 4m - 24

=> 0 = 4m – 8

=> 4m = 8

=> m = 8/4

=> m = 2

Question 4:

(a) Construct 3 equations starting with x = 2.

(b) Construct 3 equations starting with x = -2.

Answer:

(a) Construct 3 equations starting with x = 2.

(i) x = 2

Multiplying both sides by 10, we get

10x = 10 * 2

=> 10x = 20

Adding 2 on both sides, we get

10x + 2 = 20 + 2

=> 10x + 2 = 22

(ii) x = 2

Multiplying both sides by 5, we get

5x = 5 * 2

=> 5x = 10

Subtracting 2 on both sides, we get

5x - 3 = 10 - 3

=> 5x - 3 = 7

(iii) x = 2

Dividing both sides by 5, we get

x/5 = 2/5

(b) 3 equations starting with x = -2.

(i) x = -2

Multiplying both sides by 3, we get

3x = -2 * 3

=> 3x = -6

(ii) x = -2

Multiplying both sides by 3, we get

3x = -2 * 3

=> 3x = -6

Adding 7 on both sides, we get

3x + 7 = -6 + 7

=> 3x + 7 = 1

(iii) x = -2

Multiplying both sides by 3, we get

3x = -2 * 3

=> 3x = -6

Adding 10 on both sides, we get

3x + 10 = -6 + 10

=> 3x + 10 = 4

Exercise 4.4

Question 1:

Set up equations and solve them to find the unknown numbers in the following cases:

(a) Add 4 to eight times a number; you get 60.

(b) One-fifth of a number minus 4 gives 3.

(c) If I take three-fourth of a number and add 3 to it, I get 21.

(d) When I subtracted 11 from twice a number, the result was 15.

(e) Munna subtracts thrice the number of notebooks he has from 50, 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 will get 8.

(g) Answer thinks of a number. If he takes away 7 from 5/2 of the number, the result is 11/2.

Answer:

(a) Let the number be x.

According to question,

8x + 4 = 60

=> 8x = 60 – 4

=> 8x = 56

=> x = 56/8

=> x = 7

So, the number is 7

(b) Let the number by y.

According to question,

y/5 – 4 = 3

=> y/5 = 3 + 4

=> y/5 = 7

=> y = 7 * 5

=> y = 35

So, the number is 35

(c) Let the number be z.

According to question,

3z/4 + 3 = 21

=> 3z/4 = 21 – 3

=> 3z/4 = 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 + 11

=> 2x = 26

=> x = 26/2

=> x = 13

So, the number is 13

(e) Le the number be m.

According to question,

50 – 3m = 8

=> -3m = 8 – 50

=> -3m = -42

=> 3m = 42

=> m = 42/3

=> m = 14

So, the number is 14

(f) Let the number be n.

According to question,

(n + 19)/5 = 8

=> n + 19 = 8 * 5

=> n + 19 = 40

=> n = 40 – 19

=> n = 21

So, the number is 21

(g) Let the number be x.

According to question,

5x/2 – 7 = 11/2

=> 5x/2 = 11/2 + 7

=> 5x/2 = (11 + 7 * 2)/2

=> 5x/2 = (11 + 14)/2

=> 5x/2 = 25/2

=> 5x = 25

=> x = 25/5

=> x = 5

So, the number is 5

Question 2:

Solve the following:

(a) The teacher tells the class that the highest marks obtained by a student in her class are twice the lowest marks plus 7.

The highest score is 87. What is the lowest score?

(b) In an isosceles triangle, the base angles are equal. The vertex angle is 400. What are the base angles of the triangle?

(Remember, the sum of three angles of a triangle is 1800.)

(c) Sachin scored twice as many runs as Rahul. Together, their runs fell two short of a double century. How many runs did each one score?

Answer:

(a) Let the lowest mark be y.

According to question,

2y + 7 = 87

=> 2y = 87 – 7

=> 2y = 80

=> y = 80/2

=> y = 40

Hence, the lowest score is 40.

(b) Let the base angle of the triangle be b.

Given, a = 400, b = c

Since a + b + c = 1800          [Angle sum property of a triangle]

=> 400 + b + b = 1800

=> 400 + 2b = 1800

=> 2b = 1800 – 400

=> 2b = 1400

=> b = 1400/2

=> b = 700

Thus, the base angles of the isosceles triangle are 700 each. (c) Let the score of Rahul be x runs and Sachin’s score is 2x.

According to question,

x + 2x = 198

=> 3x = 198

=> x = 198/3

=> x = 66

So, 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 five times the marbles Parmit has. Irfan has 37 marbles. How many marbles does Parmit have?

(ii) Laxmi’s father is 49 years old. He is 4 years older than three times Laxmi’s age. What is Laxmi’s age?

(iii) People of Sundergram planted a total of 102 trees in the village garden. Some of the trees were fruit trees.

The numbers of non-fruit trees were two more than three times the number of fruit trees. What was the number of fruit trees planted?

Answer:

(a) Let the number of marbles Permit has be m.

According to question,

5m + 7 = 37

=> 5m = 37 – 7

=> 5m = 30

=> m = 30/5

=> m = 6

Thus, Permit has 6 marbles.

(b) Let the age of Laxmi be y years.

Then the age of her father = 3y + 4 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.

Then the number of non-fruit trees = 3t + 2

According to question,

t + 3t + 2 = 102

=> 4t + 2 = 102

=> 4t = 102 – 2

=> 4t = 100

=> t = 100/4

=> t = 25

Thus, the number of fruit trees is 25.

Question 4:

Solve the following riddle:

I am a number,

Tell my identity!

Take me seven times over,

And add a fifty!

To reach a triple century,

You still need forty!

Answer:

Let the number be n.

According to question,

7n + 50 + 40 = 300

=> 7n + 90 = 300

=> 7n = 300 – 90

=> 7n = 210

=> n = 210/7

=> n = 30

Thus, the required number is 30.

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