Solution:

Let the number of 5 rupee notes be x and the number of 10 rupee notes be y.

From the first condition,
 5x + 10y = 350 ... (i)

From the second condition,
 x = 2y − 10 ... (ii)

Substituting the value of x in (i),

5x + 10y = 350 ... (i)

∴ 5(2y − 10) + 10y = 350

∴ 10y − 50 + 10y = 350

∴ 20y − 50 = 350

∴ 20y = 350 + 50

∴ 20y = 400

\(\displaystyle \therefore y = \frac {400}{20}\)

∴ y = 20 ... (iii)

Substituting the value of y in (ii), we get,

 x = 2y − 10 ... (ii)

∴ x = 2 × 20 − 10

∴ x = 40 − 10

∴ x = 30 ... (iv)

Hence, there are 30 notes of ₹ 5 and 20 notes of ₹ 10.

Solution:

Let, the numerator of that fraction be x and the denominator be y.

From the first condition,
 y = 2x − 1 ... (i)

From the second condition,

\(\displaystyle \frac {x + 1}{y + 1} = \frac {3}{5}\)

By cross multiplication,

   5(x + 1) = 3(y + 1)

∴ 5x + 5 = 3y + 3

∴ 5x − 3y = 3 − 5

∴ 5x − 3y = − 2 ... (ii)

Substituting the value of y from (i) in (ii), we get,

   5x − 3y = −2 ... (ii)

∴ 5x − 3(2x − 1) = − 2

∴ 5x − 6x + 3 = − 2

∴ − x + 3 = − 2

∴ − x = − 2 − 3

∴ − x = − 5

i.e. x = 5 ... (iii)

Substituting the value of x in (i), we get,

  y = 2x − 1 ... (i)

∴ y = 2 × 5 − 1

∴ y = 10 − 1

∴ y = 9 ... (iv)

Hence, the required fraction is \(\displaystyle \frac {5}{9}\).

Solution:

Let, the present age of Priyanka be x years and the present age of Deepika be y years.

From the first condition,
 x + y = 34 ... (i)

From the second condition,
 x − y = 6 ... (ii)

Adding (i) and (ii),

		x	+	y	=		34	... (i)
+		x	−	y	=		6	... (ii)
		2x			=		40

\(\displaystyle \therefore x = \frac {40}{2}\)

∴ x = 20 ... (iii)

Substituting the value of x in (i),

 x + y = 34

∴ 20 + y = 34

∴ y = 34 − 20

∴ y = 14 ... (iv)

∴ Priyanka’s present age is 20 years and Deepika’s present age is 14 years.

Solution:

Let, the number of lions be x and the number of peacocks be y.

From the first condition,
 x + y = 50 ... (i)

From the second condition,
 4x + 2y = 140 ... 

Dividing both sides by 2, we get:
 2x + y = 70 ... (ii)

Subtracting (i) from (ii),

		2x	+	y	=		70	... (ii)
−	⊕	x	⊕	y	=	⊕	50	... (i)
	−		−			−
		x			=		20	... (iii)

Substituting x = 20 in equation (i),

 x + y = 50 ... (i)

∴ 20 + y = 50

∴ y = 50 − 20

∴ y = 30 ... (iv)

∴ There are 20 lions and 30 peacocks in that zoo.

Solution:

Let, Sanjay’s monthly salary be ₹ x and the yearly increment be ₹ y.

From the first condition,

 x + 4y = 4500 ... (i)

From the second condition,

 x + 10y = 5400 ... (ii)

Subtracting equation (i) from equation (ii),

		x	+	10y	=		5400	... (ii)
−	⊕	x	⊕	4y	=	⊕	4500	... (i)
	−		−			−
				6y	=		900

\(\displaystyle \therefore y = \frac {900}{6}\)

∴ y = ₹ 150 ... (iii)

Substituting the value of y in (i), we get,

 x + 4y = 4500 ... (i)

∴ x + 4 × 150 = 4500

∴ x + 600 = 4500

∴ x = 4500 − 600

∴ x = ₹ 3900 ... (iv)

Hence, Sanjay’s monthly salary is ₹ 3900 and his yearly increment is ₹ 150.

Solution:

Let the price of one chair be ₹ x and the price of one table be ₹ y.

From the first condition,
 3x + 2y = 4500 ... (i)

From the second condition,
 5x + 3y = 7000 ... (ii)

Multiplying equation (i) by 3, we get,
 9x + 6y = 13500 ... (iii)

Multiplying equation (ii) by 2, we get,
 10x + 6y = 14000 ... (iv)

Subtracting equation (iv) from equation (iii),

		9x	+	6y	=		13500	... (iii)
−	⊕	10x	⊕	6y	=	⊕	14000	... (iv)
	−		−			−
		− x			=		− 500

i.e. x = 500 ... (v)

Substituting the value of x in (i), we get,

 3x + 2y = 4500 ... (i)

∴ 3 × 500 + 2y = 4500

∴ 1500 + 2y = 4500

∴ 2y = 4500 − 1500

∴ 2y = 3000

\(\displaystyle \therefore y = \frac {3000}{2}\)

∴ y = 1500 ... (vi)

Hence, the price of 2 chairs and 2 tables
= 2x + 2y
= 2 × 500 + 2 × 1500
= 1000 + 3000
= ₹ 4000.

The price of 2 tables and 2 chairs is ₹ 4000.

Solution:

Let, the digit in the unit’s place be x and the digit in the ten’s place be y.

From the first condition,
 x + y = 9 ... (i)

Now, the original number
= 10y + x

And the number obtained by interchanging the digits
= 10x + y

From the second condition,
 (10x + y) − (10y + x) = 27

∴ 10x + y − 10y - x = 27

∴ 9x − 9y = 27

∴ 9(x − y) = 27

∴ x − y = 3 ... (ii)

Adding equation (i) and equation (ii),

		x	+	y	=		9	... (i)
+		x	−	y	=		3	... (ii)
		2x			=		12

\(\displaystyle \therefore x = \frac {12}{2}\)

∴ x = 6 ... (iii)

Substituting the value of x in (i), we get,

 x + y = 9 ... (i)

∴ 6 + y = 9

∴ y = 9 − 6

∴ y = 3 ... (iv)

Hence, the digit in the unit’s place is 6 and the digit in the ten’s place is 3.

Therefore, the required two-digit number is 36.

Solution:

\(\angle \)A + \(\angle \)B + \(\angle \)C = 180° ... (Theorem)

But, \(\angle \)A = \(\angle \)B + \(\angle \)C ... (Given)

∴ \(\angle \)A + \(\angle \)A = 180°

∴ 2\(\angle \)A = 180°

∴ \(\displaystyle \angle A = \frac {180}{2}\)

∴ \(\angle \)A = 90° ... (i)

∴ \(\angle \)B + \(\angle \)C = 90° ... (ii)

Also, the ratio of measures of \(\angle \)B and \(\angle \)C is 4 : 5.

∴ \(\displaystyle \frac {\angle B}{\angle C} = \frac {4}{5}\)

By cross multiplication,

   5\(\angle \)B = 4\(\angle \)C

∴ 5\(\angle \)B − 4\(\angle \)C = 0 ... (iii)

Multiplying equation (ii) by 4, we get,

   4\(\angle \)B + 4\(\angle \)C = 360° ... (iv)

Adding equation (iii) and equation (iv),

		5\(\angle \)B	−	4\(\angle \)C	=		0	... (iii)
+		4\(\angle \)B	+	4\(\angle \)C	=		360°	... (iv)
		9\(\angle \)B			=		360°

\(\displaystyle \therefore \angle B = \frac {360}{9}\)

∴ \(\angle \)B = 40° ... (v)

Substituting the value of \(\angle \)B in (ii), we get,

 \(\angle \)B + \(\angle \)C = 90° ... (ii)

∴ 40° + \(\angle \)C = 90°

∴ \(\angle \)C = 90° − 40°

∴ \(\angle \)C = 50° ... (vi)

Hence, the measures of the angles of that triangle are 90°, 40° and 50°.

Solution:

Let, the length of the smaller part be x cm and the length of the larger part be y cm.

From the first condition,
 x + y = 560 ... (i)

From the second condition,
 2x = \(\displaystyle \frac {1}{3}\) y

Multiplying both sides by 3, we get,

6x = y ... (ii)

i.e. y = 6x ... (ii)

Substituting the value of y in (i), we get,

  x + y = 560 ... (i)

∴ x + 6x = 560

∴ 7x = 560

∴ \(\displaystyle \therefore x = \frac {560}{7}\)

∴ x = 80 cm ... (iii)

Substituting the value of x in (i), we get,

 x + y = 560 ... (i)

∴ 80 + y = 560

∴ y = 560 − 80

∴ y = 480 cm ... (iv)

∴ The length of the larger part is 480 cm.

Solution:

Let, the number of correct answers be x and the number of incorrect answers be y.

From the first condition,
 x + y = 60 ... (i)

From the second condition,
 2x - y = 90 ... (ii)

Adding equation (i) and equation (ii),

		x	+	y	=		60	... (i)
+		2x	−	y	=		90	... (ii)
		3x			=		150

\(\displaystyle \therefore x = \frac {150}{3}\)

∴ x = 50 ... (iii)

Substituting the value of x in (i), we get,

 x + y = 60 ... (i)

∴ 50 + y = 60

∴ y = 60 − 50

∴ y = 10 ... (iv)

Yashwant got 10 questions wrong.