Class 10 Maths Full Syllabus Practice Paper 2

PA X Class 10 - Mathematics

1. The HCF of two numbers is 27 and their LCM is 162. If one of the numbers is 54, what is the other number is: [1] a) 36 b) 45 c) 81 d) 9 x y [1]

2. If 1080 = 2 × 3 × 5, then (x - y) is equal to: a) 1 b) 0 c) 6 d) -1

3. If the LCM of a and 18 is 36 and the HCF of a and 18 is 2, then a = [1] a) 1 b) 3 c) 2 d) 4

4. The ratio of HCF to LCM of the least composite number and the least prime number is: [1] a) 2 : 1 b) 1 : 1 c) 1 : 2 d) 1 : 3 2 5 3 2 [1]

5. If two positive integers m and n can be expressed as m = x y and n = x y , where x and y are prime numbers, then HCF(m, n) = 3 2 2 2 a) b) x y x y c) 3 3 d) 2 3 x y x y 2 2 [1]

6. If p and q are co-prime numbers, then p and q are a) not coprime b) odd c) coprime d) even 3 4 [1]

7. LCM of (2 × 3 × 5) and (2 × 5 × 7) is a) 1680 b) 40 c) 1120 d) 560

8. Two tanks contain 504 and 735 litres of milk respectively. Find the maximum capacity of a container which can [1] measure the milk of either tank in exact number of times. a) 21 litres b) 7 litres c) 6 liters d) 42 litres

9. ________ is neither prime nor composite. [1] a) 3 b) 2 c) 1 d) 4 10 . A charitable trust donates 28 different books of Maths, 16 different books of Science and 12 different books of [1] Social Science to poor students. Each student is given maximum number of books of only one subject of their interest and each student got equal number of books. i . Find the number of books each student got. ii . Find the total number of students who got books. a) (i) - (3), (ii) - (10) b) (i) - (4), (ii) - (14) c) (i) - (4), (ii) - (10) d) (i) - (3), (ii) - (15) 11 . 2.35 is [1] a) a rational number b) a natural number c) an integer d) an irrational number – – 2 12 . The number ( √ 3 + √ 5 ) is [1] a) an integer b) an irrational number c) not a real number d) a rational number 13 . The sum of two irrational numbers is always [1] a) a rational number or an irrational number b) a rational number c) an integer d) an irrational number – 14 . If p is a prime number, then √ p is [1] a) Integer b) Rational c) Prime number d) Irrational √ 5 + √ 2 [1] 15 . The number is √ 5 − √ 2 a) an integer b) not a real number c) a rational number d) an irrational number √ √ 16 . If a is a non-zero rational and b is irrational, then a b is: [1] a) a natural number b) a rational number c) an integer d) an irrational number – 17 . 2 √ 3 is [1] a) a whole number b) an irrational number c) an integer d) a rational number 18 . The number 1.732 is [1] a) an integer b) a whole number c) a rational number d) an irrational number – – 19 . The number (5 - 3 √ 5 + √ 5 ) is: [1] a) a whole number b) a rational number c) an integer d) an irrational number – 20 . (2 + √ 2 ) is [1] a) an integer b) a rational number c) A real number d) an irrational number 21 . The graph of y = f(x) is shown in the figure for some polynomial f(x). [1] The number of zeroes of f(x) is a) 6 b) 4 c) 8 d) 5 – 2 22 . The zeros of the polynomial x − √ 2 x − 12 are [1] a) 3, -1 b) 3, 1 – – – – c) 3 √ 2 , −2 √ 2 d) √ 2 , − √ 2 23 . The graph of y = p(x) is shown in the figure for some polynomial p(x). The number of zeroes of p(x) is/are: [1] a) 3 b) 0 c) 1 d) 2 5 3 2 [1] 24 . A polynomial of the form ax + bx + cx + dx + e has atmost ________ zeroes.

a) 5 b) 11 c) 3 d) 7 25 . Which of the following is not the graph of a quadratic polynomial? [1] a) b) c) d) 26 . Which of the following graph has more than three distinct real roots? [1] a) b) c) d) 2 [1] 27 . The zeroes of the quadratic polynomial x + 99x + 127 are a) both positive b) both equal c) one positive and one negative d) both negative 2 [1] 28 . If one root of the polynomial f(x ) = 5x + 13x + k is reciprocal of the other, then the value of k is 1 a) b) 0 6 c) 5 d) 6 2 [1] 29 . The zeroes of the polynomial p(x) = 25x - 49 are:

7 7 7 7 a) , − b) , 5 5 5 5 49 49 49 49 c) − , + d) , 25 25 25 25 2 [1] 30 . For what value of k, the product of zeroes of the polynomial kx - 4x - 7 is 2? 7 2 a) b) − 2 7 1 7 c) − d) − 14 2 2 [1] 31 . If α and β are the zeroes of the polynomial 3x + 4x - 3, then value of αβ is 4 4 a) − b) 3 3 c) -1 d) 1 2 2 [1] 32 . If one zero of the polynomial p(x) = (a + 9)x + 45x + 6a is reciprocal of the other, then the value of a is a) 1 b) 0 c) 2 d) 3 2 [1] 33 . If p(x) = x + 5x + 6, then p(-2) is: a) 8 b) 20 c) 0 d) -8 [1] 2 34 . If the sum of the zeroes of the quadratic polynomial kx + 2x + 3k is equal to their product, then k equals.

2 2 a) b) − 3 3 1 1 c) − d) 3 3 35 . If am = bl and bn ≠ cm, then the system of equations [1] ax + by = c Ix + my = n a) Has no solution. b) Has a unique solution. c) Has infinitely many solutions. d) May or may not have a solution. 36 . When L and L are coincident, then the graphical solution of system of linear equation have [1] 1 2 a) infinite number of solutions b) no solution c) unique solution d) one solution 37 . If a pair of linear equations has infinitely many solutions, then the lines representing them will be [1] a) parallel b) always coincident c) intersecting or coincident d) always intersecting 38 . For what value of k, do the equations [1] 3x – y + 8 = 0 and 6x – ky = –16 represent coincident lines?

1 a) b) –2 2 1 c) − d) 2 2 39 . Graphically, the pair of equations 6x - 3y + 10 = 0, 2x - y + 9 = 0 represents two lines which are [1] a) coincident b) Intersect at two points c) parallel d) intersect at a point 40 . The ratio of a 2-digit number to the sum of digits of that number is 4 : 1. If the digit in the units place is 3 more [1] than the digit in the tens place, then what is the number? a) 63 b) 36 c) 24 d) 40 41 . Which equation satisfies the data given in the table? [1] x -1 0 1 2 y -3 -1 1 3 a) y = 2x - 1 b) y = x - 2 c) y = x + 1 d) y = 3x - 3 – x+y x-y [1] 42 . If 2 = 2 = √ 8 then the value of y is 1 a) b) 0 2 3 c) d) 1 2 o o o o [1] 43 . In a cyclic quadrilateral ABCD, if ∠ A = (2x - 1) , ∠ B = (y + 5) , ∠ C = (2y + 15) and ∠ D = (4x - 7) , then the value of ∠ C is o o a) b) 65 55 c) o d) o 115 125 44 . The sum of the digits of a two-digit number is 15. The number obtained by interchanging the digits exceeds the [1] given number by 9. The number is a) 87 b) 96 c) 69 d) 78 45 . In △ ABC, if ∠ C = 3 ∠ B = 2( ∠ A + ∠ B), then ∠ C = [1] a) o b) o 120 60 o o c) d) 150 90 46 . If 29x + 37y = 103 and 37x + 29y = 95 then [1] a) x = 2, y = 1 b) x = 3, y = 2 c) x = 1, y = 2 d) x = 2, y = 3 47 . The sum of the digits of a two-digit number is 12. The number obtained by interchanging the two digits exceeds [1] the given number by 18. Find the number.

a) 58 b) 57 c) 75 d) 85 o [1] 48 . In △ ABC, if ∠ C = 50° and ∠ A exceeds ∠ B by 44 , then ∠ A = a) o b) o 40 43 c) o d) o 67 87 49 . If ∠ A and ∠ B are complementary angles and ∠ A is x, then which equation can be used to find ∠ B which is [1] denoted by y? a) y = (180° - x) b) y = (90° - x) c) y = (x + 180°) d) y = (90° + x) 50 . 5 years hence, the age of a man shall be 3 times the age of his son while 5 years earlier the age of the man was 7 [1] times the age of his son. The present age of the man is a) 47 years b) 50 years c) 40 years d) 45 years 51 . A and B are friends. A is elder to B by 5 years. B’s sister C is half the age of B while A’s father D is 8 years [1] older than twice the age of B. If the present age of D is 48 years, then find the present ages of A, B and C respectively.

a) 40 years, 20 years, 15 years b) 20 years, 15 years, 10 years c) 25 years, 20 years, 10 years d) 50 years, 25 years, 20 years 52 . The cost of a notebook is twice the cost of a pen. If the cost of a notebook is ₹ x and that of a pen is ₹ y, then a [1] linear equation in two variables to represent the given condition is ________. a) x - 2y = 0 b) 2x - y = 0 c) 2x + y = 0 d) x + 2y = 0 53 . Sum of two numbers is 80 and their difference is 36. Find the numbers. [1] a) 44, 36 b) 40, 40 c) 58, 22 d) 52, 28 54 . A part of monthly expenses of a family on milk is fixed which is ₹ 700 and remaining varies with quantity of [1] milk taken extra at the rate of ₹ 25 per litre. Taking quantity of milk required extra as x litres and total expenditure on milk as ₹ y, write a linear equation from the above information.

a) -25x + y = 700 b) 20x + 10y = 300 c) 20x + y = 500 d) x + 25y = 900 55 . Two numbers whose sum is 12 and the absolute value of whose difference is 4 are the roots of the equation [1] ________. a) 2 b) 2 2x - 6x + 7 = 0 2x - 24x + 43 = 0 c) 2 d) 2 x - 12x + 30 = 0 x - 12x + 32 = 0 2 2 [1] 56 . 5x + 8x + 4 = 2x + 4x + 6 is a a) constant b) cubic equation c) quadratic equation d) linear equation 2 [1] 57 . Value of k for which x = 2 is a solution of the equation 5x - 4x + (2 + k) = 0, is a) 10 b) -10 c) -14 d) 14 2 [1] 58 . If y = 1 is one of the solutions of the quadratic equation py + py + 3 = 0, then the value of p is:

a) -3 b) -2 3 c) − d) 2 2 2 [1] 59 . If one root of the equation 2x + ax + 6 = 0 is 2 then a = ? 7 a) 7 b) 2 −7 c) d) -7 2 60 . The hypotenuse of a right triangle is 6m more than twice the shortest side. The third side is 2m less than the [1] hypotenuse. The representation of the above situation in the form of a quadratic equation is 2 2 2 2 2 2 a) b) (2x - 6) = x - (2x - 4) (2x + 6) = x - (2x + 4) c) 2 2 2 d) 2 2 2 (2x + 6) + x = (2x + 4) (2x + 6) = x + (2x + 4) 2 [1] 61 . If x = 3 is a solution of the equation 3x + (k - 1)x + 9 = 0 then k = ?

a) 11 b) -11 c) 13 d) -13 2 [1] 62 . If one root of the equation 2x + kx + 4 = 0 is 2, then the other root is a) 1 b) -1 c) 6 d) -6 63 . The denominator of a fraction is one more than twice the numerator. If the sum of the fraction and its reciprocal [1] 16 is 2 , find then fraction. 21 7 3 a) b) 3 7 3 4 c) d) 4 3 2 [1] 64 . If one root of 5x + 13x + k = 0 be the reciprocal of the other root then the value of k is a) 2 b) 5 c) 1 d) 0 2 [1] 65 . The least positive value of k, for which the quadratic equation 2x + kx - 4 = 0 has rational roots, is – a) √ 2 b) 2 – c) ±2 d) ±2 √ 2 2 [1] 66 . If one root of the equation a(b - c)x + b(c - a)x + c(a - b) = 0 is 1, then the other root is ________.

c ( a − b ) a ( b − c ) a) b) a ( b − c ) b ( c − a ) b ( c − a ) a ( b − c ) c) d) a ( b − c ) c ( a − b ) 2 [1] 67 . The equation x - 8x + k = 0 has real and distinct roots if a) k = 8 b) k 16 2 2 2 2 2 [1] 68 . If (a + b ) x + 2(ac + bd) x + c + d = 0 has no real roots, then a) ab = cd b) ac = bd c) ad ≠ bc d) ad = bc 2 [1] 69 . The discriminant of the equation (2a + b) x = x + 2ab is ________ a) 2 b) 2 (2a + b) (2a - b) c) 2 d) 2 (2a + b ) (2a - b ) 2 2 [1] 70 . If 2 is a root of the equation x + ax + 12 = 0 and the quadratic equation x + ax + q = 0 has equal roots, then q = a) 8 b) 16 c) 20 d) 12 71 . A train travels 360km at a uniform speed. If the speed had been 5 km/hr more, it would have taken 1 hour less [1] for the same journey, then the actual speed of the train is a) 45 km/hr b) 48 km/hr c) 40 km/hr d) 36 km /hr 72 . Rohan’s mother is 26 years older than him. The product of their ages 3 years from now will be 360, then [1] Rohan’s present age is a) 8 years b) 10 years c) 6 years d) 7 years 73 . If I had walked 1 km per hour faster, I would have taken 10 minutes less to walk 2 km. Then the rate of my [1] walking is a) 6 km /hr b) 3 km/hr c) 8 km/hr d) 4 km/hr 74 . A takes 10 days less than the time taken by B to finish a piece of work. If both A and B together can finish the [1] work in 12 days, then the time taken by B to finish the work is a) 20 days b) 25 days c) 30 days d) 28 days 75 . The angry Arjun carried some arrows for fighting with Bheeshma. With half the arrows, he cut down the arrows [1] thrown by Bheeshma on him and with six other arrows he killed the rath driver of Bheeshma. With one arrow each, he knocked down respectively the rath, flag and bow of Bheeshma. Finally, with one more than four times the square root of arrows, he laid Bheeshma unconscious on an arrow bed. The total number of arrows that Arjun had, is a) 100 b) 120 c) 80 d) 96

Solution

PA X Class 10 - Mathematics 1 .
(c) 81 Explanation: Let the two numbers be x and y. It is given that: x = 54 HCF = 27 LCM = 162 We know, x × y = HCF × LCM ⇒ 54 × y = 27 × 162 ⇒ 54y = 4374 4374 ⇒ ∴ y = = 81 54 2 .
(b) 0 Explanation: 3 3 1080 = 2 × 3 × 5 On comparing x = 3, y = 3 x - y = 3 - 3 = 0 3 .
(d) 4 Explanation: LCM (a, 18) = 36 HCF (a, 18) = 2 We know that the product of numbers is equal to the product of their HCF and LCM. Therefore, 18a = 2(36) 2(36) a = 18 a = 4 4 .
(c) 1 : 2 Explanation:

Least composite number is 4 and the least prime number is 2. LCM (4, 2) = 4 HCF (4, 2) = 2 The ratio of HCF to LCM = 2 : 4 or 1 : 2. 5 . 2 2
(b) x y Explanation: 2 5 3 2 2 x y = y (x y ) 3 3 2 2 x y = x(x y ) 2 2 Therefore HCF (m, n) is x y 6 .
(c) coprime Explanation: We know that the co-prime numbers have no factor in common, or, their HCF is 1. 2 2 Thus, p and q have the same factor with exponent 2 each. which again will not have any common factor. 2 2 Thus we can conclude that p and q are co-prime numbers.

7 .
(a) 1680 Explanation: LCM = Product of greatest power of each prime factor involved in the numbers 4 = 2 × 3 × 5 × 7 = 16 × 3 × 5 × 7 = 1680 8 .
(a) 21 litres Explanation: 3 2 2 We have, 504 = 2 × 3 × 7 and 735 = 3 × 5 × 7 . ∴ H.C.F. (504, 735) = (3 × 7) = 21 ∴ Capacity of the container = 21 litres. 9 .
(c) 1 Explanation: 1 is neither prime nor composite. A prime is a natural number greater than 1 that has no positive divisors other than 1 and itself e.g. 5 is prime because 1 and 5 are its only positive integers factors but 6 is composite because it has divisors 2 and 3 in addition to 1 and 6.

10 .
(b) (i) - (4), (ii) - (14) Explanation: i . H.C.F. (28, 16, 12) = 2 × 2 = 4 ∴ Number of books each student got = 4 28 ii . Number of students who got Maths books = = 7 4 16 Number of students who got Science books = = 4 4 12 Number of students who got Social Science books = = 3 4 ∴ Total number of students who got books = 7 + 4 + 3 = 14. 11 .
(a) a rational number Explanation: p It can be expressed in form q 235 2.35 = 100 so, 2.35 is a rational number 12 .
(b) an irrational number Explanation:

– – 2 – 2 – 2 – – ( √ 3 + √ 5 ) = ( √ 3 ) + ( √ 5 ) + 2 × √ 3 × √ 5 − − = 3 + 5 + 2 √ 15 − − = 8 + 2 √ 15 − − – – Here, √ 15 = √ 3 × √ 5 – – – – 2 Since √ 3 and √ 5 both are an irrational number. Therefore, ( √ 3 + √ 5 ) is an irrational number. 13 .
(a) a rational number or an irrational number Explanation: The sum of two irrational numbers can be either a rational number or an irrational number. – – – – e.g 5 √ 3 + 3 √ 2 = 5 √ 3 + 3 √ 2 sum is irrational – – (2 + 6 √ 7 ) + ( - 6 √ 7 ) = 2 sum is rational Hence sum can be either rational or irrational 14 .

(d) Irrational Explanation: – – √ p is an irrational number because the square root of every prime number is an irrational number. (for example √ 3 is an irrational number) 15 .
(d) an irrational number Explanation: √ 5 + √ 2 √ 5 − √ 2 √ 5 + √ 2 √ 5 + √ 2 = × √ 5 − √ 2 √ 5 + √ 2 2 ( √ 5 + √ 2 ) = 2 2 ( √ 5 ) − ( √ 2 ) 2 2 ( √ 5 ) + ( √ 2 ) +2× √ 5 × √ 2 = 5−2 5+2+2 √ 10 = 3 7+2 √ 10 = 3 − − – – √ √ √ Here 10 = 2 × 5 – – Since √ 2 and √ 5 both are an irrational number √ 5 + √ 2 Therefore, is an irrational number.

√ 5 − √ 2 16 .
(d) an irrational number Explanation: If possible let a √ b be rational. p Then a √ b = , where p and q are non-zero integers, having no common factor other than 1. q p √ Now, a b = q p ⇒ √ b = ... (i) aq But, p and aq are both rational and aq ≠ 0 p ∵ is rational. aq √ Therefore, from eq. (i), it follows that b is rational. The contradiction arises by assuming that a √ b is rational. Hence, a √ b is irrational. 17 .
(b) an irrational number Explanation: an irrational number 18 .
(c) a rational number Explanation:

Clearly, 1.732 is a terminating decimal. Hence, it is a rational number. 19 .
(d) an irrational number Explanation: an irrational number 20 .
(d) an irrational number Explanation: – (2 + √ 2 ) is an irrational number. If it is rational, then the difference of two rational is rational. – – ∴ (2 + √ 2 ) − 2 = √ 2 = irrational, which is a contradiction. – √ Hence, (2 + 2 ) , is an irrational number. 21 .
(d) 5 Explanation: Graph of f(x) intersect the x-axis at 5 times. hence, No. of zeroes of f(x) = 5 22 .

– –
(c) 3 √ 2 , −2 √ 2 Explanation: – – – 2 2 x − √ 2 x − 12 = x − 3 √ 2 x + 2 √ 2 x − 12 – – – – – = x ( x − 3 √ 2 ) + 2 √ 2 ( x − 3 √ 2 ) = ( x − 3 √ 2 )( x + 2 √ 2 ) – – ∴ x = 3 √ 2 or x = −2 √ 2 23 .
(b) 0 Explanation: 0 24 .
(a) 5 Explanation: Since, degree of given polynomial is 5, 5 3 2 so ax + bx + cx + dx + e has atmost 5 zeroes. 25 .
(d) Explanation: The shape of a quadratic polynomial is either upward or downward U - shaped curve i.e., an upward or downward parabola. Also, the graph of the quadratic equation cuts the X - axis at the most at two points, but in fig it cuts the X - axis at three points.

∴ fig is not the graph of a quadratic polynomial. 26 .
(b) Explanation: For more than three distinct real roots the graph must cut x-axis at least four times. 27 .
(d) both negative Explanation: 2 As the Discriminant of the given quadratic polynomial x + 99x + 127 is more than Zero. ∴ Both the zeros are negative. 28 .
(c) 5 Explanation: 2 The Given polynomial is f ( x ) = 5 x + 13 x + k . Product of roots = k/5 k 1 = 5 ⇒ k = 5 7 7 29 .
(a) , − 5 5 Explanation: 2 p(x) = 25x - 49 = 0 = (5x - 7)(5x + 7) = 0 7 −7 ∴ x = and 5 5 30 .

7
(d) − 2 Explanation: c Product of zeros = a (−7) 2 = k −7 k = ( ) 2 31 .
(c) -1 Explanation: -1 32 .
(d) 3 Explanation: 1 Let one zero be β then the other zero will be α c 6 a 1 Since αβ = ⇒ α × = a α 2 a +9 6 a ⇒ 1 = 2 a +9 2 ⇒ 6a = a + 9 2 ⇒ a - 6a + 9 = 0 ⇒ (a - 3)(a - 3) = 0 a - 3 = 0 and a - 3 = 0 ⇒ a = 3 and a = 3 33 .
(c) 0 Explanation: 2 p(-2) = (-2) + 5(-2) + 6 p(-2) = 4 - 10 + 6 = 0 34 . 2
(b) − 3 Explanation: −2 3 k −2 −2 α + β = αβ ⇒ = ⇒ = 3 ⇒ k = k k k 3 35 .
(a) Has no solution. Explanation:

We have, ax + by - c and lx + my = n a b c Now, = ≠ (given) l m n ∴ The given system of equations has no solution. 36 .
(a) infinite number of solutions Explanation: When L & L are co-incident, 1 2 a 1 b 1 c 1 ⇒ = = a 2 b c 2 2 ⇒ infinite number solution. 37 .
(b) always coincident Explanation: equation has infinite many solutions if a b c 1 1 1 = = a c 2 b 2 2 i.e. always co-incident. 38 .
(d) 2 Explanation: Condition for coincident lines is - a /a = b /b = c /c …(i) 1 2 1 2 1 2 Given lines are, 3x - y + 8 = 0 and 6x - ky + 16 = 0; Comparing with the standard form, gives a = 3, b = - 1, c = 8; 1 1 1 a 2 = 6, b 2 = - k, c 2 = 16; 3 1 8 and, from Eq. (i), = = 6 k 16 1 1 = k 2 So, k = 2 39 .

(c) parallel Explanation: Given: a 1 = 6, a 2 = 2, b 1 = -3, b 2 = -1, c 1 = 10 and c 2 =9 a = 6, a = 2, b = −3, b = −1, c = 10 and c = 9 1 2 1 2 1 2 a 1 6 3 b 1 −3 3 c 1 10 Here = = , = = , = a 2 2 1 b −1 1 c 2 9 2 c 1 10 but = c 2 9 a b c 1 1 1 ∵ = ≠ a 2 b 2 c 2 Therefore, the lines are parallel. 40 .
(b) 36 Explanation: Let the digit at units place be x and the digit at tens place e be y, then the number = 10y + x 10 y + x 4 Now, according to the question, = y + x 1 ⇒ 10y + x = 4y + 4x ⇒ 6y = 3x ⇒ x = 2y ...(i) Also, x = 3 + y ⇒ 2y = 3 + y [From (i)] ⇒ y = 3 and x = 6 ∴ Required number = 36 41 .
(a) y = 2x - 1 Explanation:

y = 2x - 1 42 .
(b) 0 Explanation: 3 3 x+y x-y 3/2 2 = 2 = 2 ⇒ x + y = and x - y = . So, by adding above two equations we get and x= y = 0 2 2 43 . o
(c) 115 Explanation: o Since the sum of the opposite angles of a cyclic quadrilateral is 180 o ∴ ∠ A + ∠ C = 180 o ⇒ 2x - 1 + 2y + 15 = 180 o ⇒ x + y = 83 ... (i) o And ∠ B + ∠ D = 180 o ⇒ y + 5 + 4x - 7 = 180 o ⇒ 4x + y = 182 ... (ii) Subtracting eq. (ii) from eq. (i), o we get -3x = -99 o ⇒ x = 33 Putting the value of x in eq. (i), o o we get 33 + y = 83 o ⇒ y = 50 o o o ∴ ∠ C = (2y + 15) = (2 × 50 + 15) = 115 44 .

(d) 78 Explanation: Let us assume the tens and the unit digits of the required number be x and y respectively ∴ Required number = (10x + y) According to the given condition in the question, we have x + y = 15 .....(i) By reversing the digits, we obtain the number = (10y + x) ∴ (10y + x) = (10x + y) + 9 10y + x - 10x - y = 9 9y - 9x = 9 y - x = 1 .....(ii) Now, on adding (i) and (ii) we get: 2y = 16 16 ∴ y = = 8 8 Putting the value of y in (i), we get: x + 8 = 15 x = 15 - 8 x = 7 ∴ Required number = (10x + y) = 10 × 7 + 8 = 70 + 8 = 78 o 45 .
(a) 120 Explanation:

o Since ∠ A + ∠ B + ∠ C = 180 ... (i) ∠ C = 3 ∠ B = 2( ∠ A+ ∠ B) 3 ∠ B = 2( ∠ A+ ∠ B) 3 ∠ B - 2 ∠ B = 2 ∠ A ∠ B = 2 ∠ A B ∠ A = ∠ 2 from (i), B o ∠ + ∠ B + 3 ∠ B = 180 2 B o 9 ∠ = 180 2 o ∠ B = 40 ∠ C = 3 ∠ B o ∠ C = 3 × 40 = 120 46 .
(c) x = 1, y = 2 Explanation: 29x + 37y=103 .......(i) 37x+29y=95 .........(ii) Adding (i) and (ii), we get 66 (x + y) = 198 ⇒ x + y = 3. Subtracting (ii) from (i), we get 8 (y - x) = 8 ⇒ y - x = 1. Solve above equations we get x = 1, y = 2 47 .
(b) 57 Explanation: Let the units and tens digits in the number be y and x respectively.

So, the number be 10x + y. According to the question, x + y = 12 ...(i) Also, 10x + y + 18 = 10y + x ⇒ 9x - 9y = -18 ⇒ x - y = -2 ...(ii) Solving (i) and (ii), we get x = 5 and y = 7 ∴ Required number is 57. 48 . o
(d) 87 Explanation: Let x and y be the measures of ∠ A and ∠ B respectively. o Now, ∠ A + ∠ B + ∠ C = 18 [By angle sum property] o o o ⇒ x + y + 50 = 180 [Given, ∠ C = 50 ] o ⇒ x + y = 130 ...(i) o o Also, ∠ A - ∠ B = 44 ⇒ x - y = 44 ...(ii) Adding (i) and (ii), we get o o o 2x = 174 ⇒ x = 87 ⇒ ∠ A = 87 49 .

(b) y = (90° - x) Explanation: We have given, ∠ A + ∠ B = 90° ⇒ x + y = 90° ⇒ y = (90° - x) 50 .
(c) 40 years Explanation: Let us assume the present age of men be x years Also, the present age of his son be y years According to question, after 5 years: (x + 5) = 3 (y + 5) x + 5 = 3y + 15 x - 3y = 10 …(i) Also, five years ago: (x - 5) = 7 (y - 5) x - 5 = 7y - 35 x - 7y = - 30 …(ii) Now, on subtracting (i) from (ii) we get: - 4y = - 40 y = 10 Putting the value of y in (i), we get x - 3 × 10 = 10 x - 30 = 10 x = 10 + 30 x = 40 ∴ The present age of men is 40 years 51 .

(c) 25 years, 20 years, 10 years Explanation: Let the present ages of A, B, C and D are x, y, z and t respectively. Since, present age of D = t = 48 years. According to question, x = y + 5 1 z = y 2 f = 2y + 8 From (iii), 48 = 2y + 8 ⇒ From (iii), 48 = 2y + 8 1 From (ii), z = x 20 = 10 years 2 From (i), x = 20 + 5 = 25 years So, present ages of A, B and C are 25 years, 20 years and 10 years respectively. 52 .
(a) x - 2y = 0 Explanation: According to question, 2 x Cost of pen = Cost of notebook ⇒ 2y = x ⇒ x - 2y = 0 53 .

(c) 58, 22 Explanation: Let the two numbers be x and y. Then, x + y = 80 ...(i) Also, x - y = 36 ...(ii) or y - x = 36 ...(iii) a . If x - y = 36, from (i) and (ii), we have 2x = 116 ⇒ x = 58 and y = 22 b . If y - x = 36, from (i) and (iii), we have 2y = 116 ⇒ y = 58 and x = 22 ∴ Numbers are 58 and 22. 54 .
(a) -25x + y = 700 Explanation: Since, x litres is the extra quantity of milk and y be total expenditure on milk. ∴ Required linear equation is, 700 + 25x = y ⇒ y - 25x = 700 or -25x + y = 700 55 .

2
(d) x - 12x + 32 = 0 Explanation: Let the two roots be a and b, then a + b = 12 ...(i) and a - b = 4 ...(ii) ⇒ a = 8 and b = 4 (from (i) and (ii)) 2 ∴ Required equation is x - 12x + 32 = 0 56 .
(c) quadratic equation Explanation: 2 2 Given: 5x + 8x + 4 = 2x + 4x + 6 2 2 ⇒ 5x - 2x + 8x - 4x + 4 - 6 2 ⇒ 3x + 4x - 2 = 0 Here, the degree is 2, therefore it is a quadratic equation. 57 .
(c) -14 Explanation: x = 2 is solution 2 p(2) = 5(2) - 4(2) + (2 + k) 0 = 20 - 8 + 2 + k k = -14 58 . 3
(c) − 2 Explanation:

If y = 1 is solution p(1) = 0 2 p(1) + p(1) + 3 = 0 p + p + 3 2p = -3 −3 p = 2 59 .
(d) -7 Explanation: 2 One root of the equation 2 x + ax + 6 = 0 is 2 i.e. it satisfies the equation 2 2(2) + 2a + 6=0 8 + 2 a + 6=0 2a = - 14 a = - 7 60 . 2 2 2
(d) (2x + 6) = x + (2x + 4) Explanation: Let the shortest side of a right angled triangle be x meters. Then according to question, its hypotenuse will be (2x + 6) meters and, the third side will be (2x + 6 -2) = (2x + 4) meters. 2 2 2 Now, using Pythagoras theorem, (Hypotenuse) = (Base) + (Perpendicular) 2 2 2 ⇒ (2x + 6) = x + (2x + 4) 61 .

(b) -11 Explanation: 2 3 x + ( k − 1) x + 9 = 0 x = 3 is a solution of the equation means it satisfies the equation Put x = 3, we get 2 3(3) + (k - 1) 3 + 9 = 0 27 + 3 k - 3 + 9 = 0 27 + 3 k + 6 = 0 3 k = - 33 k = - 11 62 .
(a) 1 Explanation: 2 Let α and β be the roots of quadratic equation 2x + kx + 4 = 0 in such a way that α = 2 Here, a = 2, b = k and c = 4 Then, according to question sum of the roots − b α + β = a − k 2 + β = 2 − k β = - 2 2 − k −4 β = 2 And the product of the roots c α ⋅ β = a 4 = 2 = 2 − k −4 Putting the value of β = in above 2 − k −4 2 × = 2 2 (-k - 4) = 2 k = -4 - 2 = -6 − k −4 Putting the value of k in β = 2 −(6)−4 β = 2 6−4 = 2 2 = 2 β = 1 Therefore, value of other root be β = 1 63 .

3
(b) 7 Explanation: x Let the fraction be . y According to the question, y = 2x + 1 ...(i) y x 16 58 Also, + = 2 = y x 21 21 x 2 x +1 58 ⇒ + = [From (i)] 2 x +1 x 21 2 2 x +4 x +1+4 x 58 ⇒ = x (2 x +1) 21 2 2 ⇒ 105x + 84x + 21 = 116x + 58x ⇒ (x - 3)(11x + 7) = 0 7 ⇒ x = 3, or x = − (Not possible) 11 ∴ y = 7 3 ∴ Required fraction = 7 64 .
(b) 5 Explanation: 1 2 Let the roots of the equation ( 5 x + 13 x + k = 0 ) be α and α c Product of the roots = a 1 k ⇒ α × = α 5 k ⇒ 1 = 5 ⇒ k = 5. 65 .
(b) 2 Explanation:

2 Dividing the equation by the coefficient of x i.e., 2 we got 2 kx x + - 2 = 0 2 2 2 k k ( x + ) − - 2 = 0 4 16 2 2 k k +32 ( x + ) = 4 16 2 k +32 Hence for rational roots, has to be a perfect square. 16 2 k +32 36 6 We get a perfect square at k = ±2 for ( ) i.e., which becomes upon removing the square 16 16 4 2 k +32 81 9 We get a perfect square at k = ±7 for ( ) i.e., which becomes upon removing the square 16 16 4 Hence the least positive value of k is 2. c ( a − b ) 66 .
(a) a ( b − c ) Explanation:

Given equation is 2 a(b - c)x + b(c - a)x + c(a - b) = 0 Let α be the other root, then c ( a − b ) Product of roots = α × 1 = a ( b − c ) c a − b ⇒ α = ( ) a b − c 67 .
(b) k 0 2 b - 4ac > 0 2 (-8) - 4(1)(k) > 0 64 - 4k > 0 64 > 4k 64 ( ) > k 4 16 > k 68 .
(c) ad ≠ bc Explanation: 2 2 2 2 2 (a + b ) x + 2(ac + bd)x + c + d = 0 2 2 2 2 Here A = a + b , B = 2(ac + bd), C = c + d 2 2 2 2 2 2 D = B − 4AC = [2(ac + bd)] − 4(a + b ) (c + d ) 2 2 2 2 2 2 2 2 2 2 2 2 = 4[a c + b d + 2abcd]−4[a c + a d + b c + b d ] 2 2 2 2 2 2 2 2 2 2 2 2 = 4a c + 4b d + 8abcd − 4a c −4a d - 4b c - 4b d 2 2 2 2 = −4a d − 4b c + 8abcd 2 2 2 2 = −4(a d + b c − 2abcd) 2 = −4(ad − bc) ∵ Roots are not real ∴ D < 0 2 2 ∴ −4(ad − bc) < 0 ⇒ (ad − bc) < 0 ⇒ ad − bc < 0 or ad ≠ bc 69 .

2
(b) (2a - b) Explanation: 2 (2a + b)x = x + 2ab 2 x - (2a + b)x + 2ab = 0 2 D = b - 4ac 2 D = [-(2a + b)] - 4 × 1 × 2ab 2 2 D = 4a + b + 4ab - 8ab 2 2 D = 4a + b - 4ab 2 D = (2a - b) 70 .
(b) 16 Explanation: 2 2 is root equation x + ax + 12 = 0 2 ∴ (2) + a × 2 + 12 = 0 ⇒ 4 + 2a + 12 = 0 ⇒ 2a + 16 = 0 −16 ⇒ a = = −8 2 2 and given that roots of x + ax + q = 0 are equal. 2 ∴ b - 4ac = 0 2 2 ⇒ a − 4 q = 0 ⇒ (−8 ) − 4 q = 0 ⇒ 64 − 4 q = 0 ⇒ 4 q = 64 64 ⇒ q = = 16 4 ∴ q = 16 71 .
(c) 40 km/hr Explanation:

Let the actual speed of the train be x km/hr 360 Time taken to cover 360 km at this speed = hrs. x 360 Time taken to cover 360 km at the increased speed = hrs. x +5 360 360 According to condition, − = 1 x x +5 1 1 ⇒ 360 [ − ] = 1 x x +5 x +5− x ⇒ 360 [ ] = 1 x ( x +5) 5 ⇒ 360 [ ] = 1 x ( x +5) 2 ⇒ x + 5x - 1800 = 0 2 ⇒ x + 45x - 40x - 1800 ⇒ x(x + 45) - 40(x + 45) = 0 ⇒ (x - 40)(x + 45) = 0 ⇒ x - 40 = 0 and x + 45 = 0 ⇒ x = 40 km/hr and x = -45 km/hr [But x = -45 is not possible] Therefore, the actual speed of the train is 40 km/hr.

72 .
(d) 7 years Explanation: Let Rohan’s present age be x years. Then Rohan’s mother age will be (x + 26) years. And after 3 years their ages will be (x + 3) and (x + 29) years. According to question, (x + 3)(x + 29) = 360 2 ⇒ x + 29x + 3x + 87 = 360 2 ⇒ x + 32x - 273 = 0 2 ⇒ x + 39x + 7x - 273 = 0 ⇒ x(x + 39) -7(x + 39) = 0 ⇒ (x - 7)(x + 39) = 0 ⇒ (x - 7) = 0 and x + 39 = 0 ⇒ x = 7 and x = -39 [x = -39 is not possible] Therefore, Rohan’s present is 7 years 73 .
(b) 3 km/hr Explanation: Let the rate of my walking be x km/h 2 ∴ Time taken to cover 2 km at the rate of x km/h = hrs x New rate = (x + 1) km/h 2 ∴ Time taken to cover 2 km at new rate = hrs x +1 2 2 10 According to question, − = x x +1 60 1 1 1 ⇒ − = x x +1 12 x +1− x 1 ⇒ = x ( x +1) 12 1 1 ⇒ = 2 12 x + x 2 ⇒ x + x - 12 = 0 2 ⇒ x + 4x - 3x - 12 = 0 ⇒ x(x + 4) -3(x + 4) = 0 ⇒ (x + 4)(x - 3) = 0 ⇒ (x + 4) = 0 and x - 3 = 0 ⇒ x = -4 [not possible] and x = 3 Therefore, the rate of my walking is 3 km/h.

74 .
(c) 30 days Explanation: Let B takes x days to do the work, then A takes (x - 10) days to do it. 1 1 ∴ Work done by B in 1 day = and work done by A in x x −10 1 1 1 According to question, + = x x −10 12 x −10+ x 1 ⇒ = 12 x ( x −10) 2 ⇒ x - 10x = 24x - 120 2 ⇒ x - 34x + 120 = 0 2 ⇒ x - 30x - 4x + 120 = 0 ⇒ x(x - 30) -4(x - 30) = 0 ⇒ (x - 30)(x - 4) = 0 ⇒ x - 30 = 0 and x - 4 = 0 ⇒ x = 30 and x = 4 [x = 4 is not possible] Therefore, B can finish the work in 30 days. 75 .
(a) 100 Explanation: Let Arjun had x arrows.

According to question, x − − + 6 + 3 + 4 x + 1 = x √ 2 − − x ⇒ 10 + 4 x = √ 2 − − ⇒ 20 + 8 √ x = x − − ⇒ 8 x = x - 20 √ 2 ⇒ 64x = x - 40x + 400 2 ⇒ x - 104x + 400 = 0 2 ⇒ x - 100x - 4x + 400 = 0 ⇒ x(x - 100) - 4(x - 100) = 0 ⇒ (x - 100)(x - 4) = 0 ⇒ x - 100 = 0 and x - 4 = 0 ⇒ x = 100 and x = 4 [which is not possible] Therefore, Arjun had 100 arrows.