Ncert Solutions For Class 10 Maths Ex 3.7

Ncert Solutions For Class 10 Maths Chapter 3 Ex 3.7

Question 1:

The ages of Reuben and Alex are at a difference of 3 yrs. Reuben’s father Thomas is twice as old as Reuben and Alex is twice as old as his sister Ann. The ages of Ann and Thomas are at a difference of 30 yrs. What is the present age of Reuben and Alex?

Solution:

The age difference between Reuben and Alex is 3 yrs.

Either Alex is 3 yrs older than that of Reuben or Reuben is 3 yrs older than Alex. From both the cases we find out that Reuben’s father’s age is 30 yrs more than that of Ann’s age.

Let the ages of Reuben and Alex be A and B respectively.

Therefore, the age of Thomas = 2 x A = 2A yrs.

And the age of Alex’s sister Ann B/2 yrs

By using the information that is given,

Case (i)

When Reuben is older than that of Alex by 3 yrs then A – B = 3 – – – – – – – – (1)

2AB2=30$2A-\frac{B}{2}=30$

4A – B = 60 – – – – – – – – – – – (2)

By subtracting the equations (1) and (2) we get,

3A = 60 – 3 = 57

A=573=19$A=\frac{57}{3}=19$

Therefore, the age of Reuben = 19 yrs

And the age of Alex is 19 – 3 = 16 yrs.

Case (ii)

When Alex is older than Reuben,

B – A = 3 – – – – – – – – – (1)

2AY2=30$2A-\frac{Y}{2}=30$

4A – B = 60 – – – – – – – – – (2)

Adding the equation (1) and (2) we get,

3A = 63

A = 21

Therefore, the age of Reuben is 21 yrs

And the age of Alex is 21 + 3 = 24 yrs.

Question 2:

Sangam says, “Give me Rs100, bro! I’ll become two times richer than you” Reuben replies, “If you give me Rs 10, I’ll become six times richer than you” What is the capital amount of Sangam and Reuben.

[Hint: A + 100 = 2 ( B – 100) , B + 10 =6(A- 10)]

Solution:
Let Sangam have Rs A with him and Reuben have Rs B with him.

Using the information that is given we get,

A + 100 = 2(B – 100) A + 100 = 2B – 200

Or A – 2B = -300 – – – – – – – (1)

And

6(A – 10) = ( B + 10 )

Or 6A – 60 = B + 10

Or 6A – B = 70 – – – – – – (2)

When equation (2) is multiplied by 2 we get,

12A – 2B = 140 – – – – – – – (3)

When equation (1) is subtracted from equation (3) we get,

11A = 140 + 300

11A = 440A

=> 40

Using A =40 in equation (1) we get,

40 – 2B = -300

40 + 300 = 2B

2B = 340

B = 170

Therefore, Sangam had Rs 40 and Reuben had Rs 170 with them.

Question 3:

A train travelling at a uniform speed covers a certain distance. If the train would have travelled 10km/hr faster, then it would have reached 2 hours prior to the scheduled time. And if the train were slower by 10km/hr, it would have taken 3 hours more than the scheduled time. Find the distance covered by the train.

Solution:

Let the speed of the train be A km/hr and the time taken by the train to travel a distance be N hours and the distance to travel be X hours.

Speed of the train: DistancetravelledbythetrainTimetakentotravelthatdistance$\frac{Distance\, travelled\, by\, the\, train}{Time\, taken\, to\,travel\, that\, distance}$

A=N(distance)X(time)$A = \frac{N(distance)}{X(time)}$

Or, N = AX – – – – – – – – – – – (1)

Using the information that is given, we get:

(A+10)=X(N2)$(A+10)=\frac{X}{(N-2)}$

(A + 10) (N – 2) = X

DN + 10N – 2A – 20 = X

By using the equation (1) we get,

– 2A + 10N = 20 – – – – – – – – – – (2)

(A10)=X(N+3)$(A-10)=\frac{X}{(N+3)}$

(A – 10) (N + 3) = X

AN – 10N  + 3A – 30 = X

By using the equation (1) we get,

3A – 10N = 30 – – – – – – – – – (3)

Adding equation (2) and equation (3) we get,

A = 50

Using the equation (2) we get,

( -2) x (50) + 10N = 20

-100 +10N = 20

=> 10N = 120

N = 12hours

From the equation (1) we get,

Distance travelled by the train, X = AN

=> 50 x 12

=> 600 km

Hence, the distance covered by the train is 600km.

Question 4:

The students of a class are made to stand in rows and if there are 3 students that are extra in a row, there would be 1 row lesser. And if 3 students are lesser in a row, there would be 2 more rows. Find the number of students in the class.

Solution:

Let the number of rows be A and the number of students in a row be B.

Total number of students:

= Number of rows x Number of students in a row

=AB

Using the information that is given,

First Condition:

Total number of students = (A – 1) ( B + 3)

Or AB = ( A – 1 )(B + 3) = AB – B + 3A – 3

Or 3A – B – 3 = 0

Or 3A – Y = 3 – – – – – – – – – – – – – (1)

Second condition:

Total Number of students = (A + 2 ) ( B – 3 )

Or AB = AB + 2B – 3A – 6

Or 3A – 2B = -6 – – – – – – – – – (2)

When equation (2) is subtracted from (1)

(3A – B) – (3A – 2B) = 3 – (-6)

-B + 2B = 3 + 6B = 9

By using the equation (1) we get,

3A – 9 =3

3A = 9+3 = 12

A = 4

Number of rows, A = 4

Number of students in a row, B = 9

Number of total students in a class => AB => 4 x 9 = 36

Question 5:

In a ΔZXC, B=2(A + B) and C=3. What are the three angles?

Solution:

Given,

∠C = 3 ∠B = 2(∠B + ∠A)

∠B = 2 ∠A+2 ∠B

∠B=2 ∠A

∠A – ∠B= 0- – – – – – – – – – – –  (i)

We know, the sum of all the interior angles of a triangle is 180O.

Thus, ∠ A +∠B+ ∠C = 180O

∠A + ∠B +3 ∠B = 180O

∠A + 4 ∠B = 180O– – – – – – – – – – – – – – -(ii)

Multiplying 4 to  equation (i) , we get

8 ∠A – 4 ∠B = 0- – – – – – – – – – – – (iii)

Adding equations (iii) and (ii) we get

9 ∠A = 180O

∠A = 20O

Using this in equation (ii), we get

20O+ 4∠B = 180O

∠B = 40O

3∠B =∠C

∠C = 3 x 40 = 120O

Therefore, ∠A = 20O

∠B=40O

∠C = 120O

Question 6:

Represent the equations 5x–y = 5 and 3x – y =3 in a graph. What are the coordinates of the vertices of the triangles formed by the y-axis and these lines?

5x – y = 5

=> y =5x – 5

Its solution table will be.

 X 2 1 0 Y 5 0 -5

Also given,3x – y = 3

y = 3x – 3

Its solution table will be.

 X 2 1 0 Y 3 0 -3

The graphical representation of these lines will be as follows:

From the above graph we  can see that the triangle formed is ∆ABC by the lines and the y axis. Also the coordinates of the vertices are A(1,0) ,  C(0,-5) and B(0,-3)

Question 7:

Find the solutions of the following set of linear equations.

(i) x/a –y/b = 0 , ax + by = a2+b2

(ii) by + ax= c,  ay + bx = 1+ c

(iii) px + qy = p-q, qx – py = p+q

(iv) 152x -378y =-74

-378x + 152y = -604

(v) (a + b) y+(a-b)x = a 2− 2ab – b2

(x + y)(a+b) = a 2+ b2

Solution:

(i) x/a – y/b = 0

=>  bx − ay = 0 ……. (i)

ax + by = a 2 + b 2 …….. (ii)

Multiplying a and b to equation (i) and (ii) respectively, we get

b2x − aby = 0 …………… (iii)

a2x + aby = a 3 + ab3 …… (iv)

Adding equations (iii) and (iv), we get

b2x + a 2x = a 3 + ab2

x (b2 + a2 ) = a (a2 + b2 ) x = a

Using equation (i), we get

b(a) − ay = 0

ab − ay = 0

ay = ab,

y = b

(ii)Given,

ax + by= c…………………(i)

bx + ay = 1+ c………… ..(ii)

Multiplying a to equation (i) and  b to equation (ii), we obtain

a2x + aby = ac ………………… (iii)

b2x + aby = b + bc …………… (iv)

Subtracting equation (iv) from equation (iii),

(a 2 − b 2 ) x = ac − bc– b

x=c(ab)ba2b2$x = \frac{c(a-b)-b}{a^{2}-b^{2}}$

From equation (i), we obtain

ax +by = c

a{c(ab)b)a2b2}+by=c$a\left \{ \frac{c(a-b)-b)}{a^{2}-b^{2}} \right \}+by =c$ ac(ab)aba2b2+by=c$\frac{ac(a-b)-ab}{a^{2}-b^{2}}+by=c$ by=cac(ab)aba2b2$by = c – \frac{ac(a-b)-ab}{a^{2}-b^{2}}$ by=a2cb2ca2c+abc+aba2b2$by = \frac{a^{2}c-b^{2}c-a^{2}c+abc+ab}{a^{2}-b^{2}}$ by=abcb2c+aba2b2$by = \frac{abc-b^{2}c+ab}{a^{2}-b^{2}}$ y=c(ab)+aa2b2$y = \frac{c(a-b)+a}{a^{2}-b^{2}}$

(iii)

px + qy = p − q ……………… (i)

qx − py = p + q ……………… .. (ii)

Multiplying p to equation (1)  and q to equation (2), we get

p2x + pqy = p2 − pq ………… (iii)

q2x − pqy = pq + q2 ………… (iv)

Adding equation (iii) and equation (iv),we get

p2x + q2 x = p2  + q2

(p2 + q2 ) x = p2 + q2

x=p2+q2p2+q2=1$x = \frac{p^{2}+q^{2}}{p^{2}+q^{2}} =1$

From equation (i), we get

p(1) + qy = p – q

qy = − q y = − 1

(iv) 152x − 378y = − 74

76x − 189y = − 37

x=(189y-137)/76………(i)

− 378x + 152y = − 604

− 189x + 76y = − 302 ………….. (ii)

Using the value of x in equation (ii), we get

189(189y3776)+76y=302$-189\left ( \frac{189y-37}{76} \right ) + 76y = -302$

− (189) 2 y + 189 × 37 + (76) 2 y = − 302 × 76

189 × 37 + 302 × 76 = (189) 2 y − (76) 2y

6993 + 22952 = (189 − 76) (189 + 76) y

29945 = (113) (265) y

y = 1

Using equation (i), we get

x =(189-37)/76

x=152/76 =2

(v)

(a + b) y + (a – b) x = a2− 2ab − b2 …………… (i)

(x + y)(a + b)  = a 2 + b2

(a + b) y + (a + b) x  = a 2 + b 2 ………………… (ii)

Subtracting equation (ii) from equation (i), we get

(a − b) x − (a + b) x = (a 2 − 2ab − b 2 ) − (a2 + b2 )

x(a − b − a − b) = − 2ab − 2b2

− 2bx = − 2b (b+a)

x =  b + a

Substituting this value in equation (i), we get

(a + b)(a − b)  +y (a + b)  = a2− 2ab – b2

a2 − b2 + y(a + b)  = a2− 2ab – b2

(a + b) y = − 2ab

y=-2ab/(a+b)

Question 8:

A cyclic quadrilateral ABDC is given below, find its angles.

Solution:

It is know that  the sum of the opposite angles of a cyclicquadrilateral is 180o

Thus, we have

∠C +∠A = 180

4y + 20− 4x = 180

− 4x + 4y = 160

x − y = − 40 ……………(1)

And, ∠B + ∠D = 180

3y − 5 − 7x + 5 = 180

− 7x + 3y = 180 ………..(2)

Multiplying 3 to equation (1), we get

3x − 3y = − 120 ………(3)

Adding equation (2) to equation (3), we get

− 7x + 3x = 180 – 120

− 4x = 60

x = −15

Substituting this value in equation (i), we get

x − y = − 40

-y−15 = − 40

y = 40-15

= 25

∠A = 4y + 20 = 20+4(25)  = 120°

∠B = 3y − 5 = − 5+3(25)  = 70°

∠C = − 4x = − 4(− 15) = 60°

∠D =  5-7x

∠D=  5− 7(−15) = 110°