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Right choice is (b) Real

Explanation: To check the nature of the roots, the discriminant must be EITHER equal to zero, less than zero or GREATER than zero.

Discriminant = b^2 – 4AC = 10^2 – 4 × 2 × 9 = 100 – 72 = 28

Since discriminant is greater than zero, the roots of the equation are real and distinct.

Right answer is (d) 29.06 km/hr

Best explanation: Let the speed of the metro TRAIN be x km/hr

Time taken to cover 396 km at xkm/hr = \(\frac {396}{x}\)hrs

Time taken to cover 396 km at x+5 km/hr = \(\frac {396}{x+5}\)hrs

∴ \(\frac {396}{x}-\frac {396}{(x+5)}\)=2

\(\frac {x+5-x}{x(x+5)}=\frac {2}{396}\)

\(\frac {5}{x^2+5x}=\frac {1}{198}\)

990=x^2+5x

x^2+5x=990

Adding \(\frac {b^2}{4}\) on both sides, where b=5

x^2 + 5x + \(\frac {5^2}{4}=\frac {5^2}{4}\) + 990

x^2 + 5x + \(\frac {25}{4}=\frac {25}{4}\) + 990

\((x + \frac {5}{2})\)^2=\((\frac {\sqrt {3985}}{2})\)^2

x + \(\frac {5}{2}\) = ±\(\frac {\sqrt {3985}}{2}\)

x = \(\frac {63.12}{2} -\frac {5}{2}=\frac {58.12}{2}\) = 29.06 and x = \(\frac {-63.12}{2} -\frac {5}{2} = \frac {-68.12}{2}\) = -34.06

Since, speed cannot be NEGATIVE hence x=29.06 km/hr

The correct option is (c) \(\FRAC {1}{10}\)

To elaborate: Let the numerator be X.

Denominator of a fraction is 1 more than 9 times the numerator.

Denominator = 1+9x

The fraction is \(\frac {x}{11+9x}\)

Fraction + RECIPROCAL = \(\frac {101}{10}\)

\(\frac {x}{1+9x}+\frac {1+9x}{x}=\frac {101}{10}\)

\(\frac {x^2+(1+9x)^2}{x(1+9x)}=\frac {101}{10}\)

10(x^2+1+81x^2+18x)=101X(1+9x)

10(82x^2+18x+1)=101x+909x^2

820x^2+180x+10=101x+909x^2

89x^2-79x-10=0

89x^2-89x+10x-10x=0

89x(x-1)+10(x-1)=0

(x-1)(89x+10)=0

x=1, \(\frac {-10}{89}\)

The fraction is \(\frac {1}{10}\)

The CORRECT answer is (b) 73, 24

Best explanation: Let the LENGTH of the RECTANGLE be x. The sum of length and breadth is 97.

Breadth will be 97-x.

Area of rectangle = 1752.

length × breadth = 1752

x(97-x)=1752

97x-x^2=1752

x^2-97x+1752=0

x^2-73x-24x+1752=0

x(x-73)-24(x-73)=0

(x-73)(x-24)=0

x=73, 24

Hence, the length is 73 and breadth 24 or length 24, breadth 73.

The correct CHOICE is (b) 10

For EXPLANATION: The n^th term of the AP is 5n+2.

T1=a=5+2=7

T2=5(2)+2=12

d=T2-T1=12-7=5

Sn=295

Sn=\(\FRAC {n}{2}\)(2a+(n-1)d)

295=\(\frac {n}{2}\)(2(7)+(n-1)5)

590=14n+5n^2-5n

5n^2+9n-590=0

5n^2+59n-50n-590=0

n(5n+59)-10(5n+59)=0

(n-10)(5n+59)=0

n=10, \(\frac {-59}{5}\)

Correct option is (c) 58

To explain: Sum of the NUMBERS is 13.

Let one number be x. Other number is 13-x.

Sum of their reciprocals = \(\frac {13}{40}\)

\(\frac {1}{x} + \frac {1}{13-x}=\frac {13}{40}\)

\(\frac {13-x+x}{x(13-x)}=\frac {13}{40}\)

\(\frac {13}{13x-x^2}=\frac {13}{40}\)

\(\frac {1}{13x-x^2}=\frac {1}{40}\)

40=13x-x^2

x^2-13x+40=0

x^2-5x-8x+40=0

x(x-5)-8(x-5)=0

(x-8)(x-5)=0

x=8, 5

The number is 58 or 85.

Correct answer is (a) Imaginary

Explanation: To check the nature of the roots, the discriminant must be either equal to ZERO, LESS than zero or greater than zero.

Discriminant = b^2 – 4AC = – 11^2 – 4 × 5 × 13 = 121 – 260 = – 139

Since discriminant is less than zero, the roots of the EQUATION are imaginary.

The correct choice is (d) 10, -1

The best EXPLANATION: x^2-9X-10=0

Shifting -10 to RHS

x^2-9x=10

Adding \(\frac {b^2}{4}\) on both SIDES, where b=-9

x^2 – 9x + \(\frac {-9^2}{4} = \frac {-9^2}{4}\) + 10

x^2 – 9x + \(\frac {81}{4} = \frac {81}{4}\) + 10

x^2 – 9x + \(\frac {81}{4} = \frac {121}{4}\)

\((x-\frac {9}{2})\)^2=\((\frac {11}{2})\)^2

x – \(\frac {9}{2}\) = ± \(\frac {11}{2}\)

x = \(\frac {11}{2}+\frac {9}{2}=\frac {20}{2}\) = 10 and x = \(\frac {-11}{2} + \frac {9}{2} = \frac {-2}{2}\) = -1

The roots of the equation are 10 and -1.

The CORRECT option is (a) 1

Explanation: SIDES of the right angled triangle is X+2, x+1, x

From Pythagoras THEOREM,

hyp^2=adj^2+base^2

(x+2)^2=(x+1)^2+(x)^2

x^2+4x+4=x^2+1+2x+x^2

4x+4=x^2+1+2x

x^2-2x-3=0

x^2+3x-x-3=0

x(x+3)-1(x+3)=0

(x-1)(x+3)=0

x=1, -3

Since, sides of triangle cannot be negative.

Hence, x=1

Correct choice is (B) TRUE

For explanation I would say: To check the NATURE of the roots, the DISCRIMINANT must be either equal to zero, less than zero or greater than zero.

Discriminant = b^2 – 4ac = -2^2 – 4 × 9 × 5 = 4 – 180 =-176

Since discriminant is less than zero, the roots of the EQUATION are imaginary. Hence, for any real value of x the equation is not true.

Correct option is (d) 15, 16

To explain I would say: Since the pages of books are consecutive numbers, so let the LEFT page number be x. The RIGHT page number will be x+1.

Sum of the squares of the pages is 481

x^2+(x+1)^2=481

x^2+x^2+1+2x=481

2x^2+2x-480=0

x^2+x-240=0

x^2+16x-15x-240=0

x(x+16)-15(x+16)=0

(x-15)(x+16)=0

x=15, -16

Since, page number cannot be negative, so x=15

The two page numbers are 15 and 16.

The CORRECT choice is (a) 8

Explanation: Let the number be X

x+\(\frac {1}{x}=\frac {65}{8}\)

\(\frac {x^2+1}{x}=\frac {65}{8}\)

8(x^2+1)=65x

8x^2+8=65x

8x^2-65x+8=0

8x^2-64x-x+8=0

8x(x-8)-1(x-8)=0

(x-8)(8x-1)=0

x=8, \(\frac {1}{8}\)

The number is 8 or \(\frac {1}{8}\).

Right option is (a) Eera 31, Her sister 15

The explanation: Let the age of Eera be x years. The age of her sister will be 46-x years.

The PRODUCT of their AGES is 465.

x(46-x)=465

46x-x^2=465

x^2-46x+465=0

x^2-46x=-465

Adding \(\frac {b^2}{4}\) on both sides, where b=-46

x^2 – 46x + \(\frac {-46^2}{4}=\frac {-46^2}{4}\) – 465

x^2 – 46x + \(\frac {2116}{4}=\frac {2116}{4}\) – 465

x^2 – 46x + \(\frac {2116}{4}=\frac {256}{4}\)

\((x-\frac {46}{2})\)^2=\((\frac {16}{2})\)^2

x-\(\frac {46}{2}\)=±\(\frac {16}{2}\)

x = \(\frac {16}{2}+\frac {46}{2}=\frac {62}{3}\) = 31 and x = \(\frac {-16}{2}+\frac {46}{2}=\frac {30}{2}\) = 15

The ages of Eera and her sister are 31 and 15 RESPECTIVELY or 15 and 31 years.

Correct option is (a) 100

Best explanation: LET the number be x. It square ROOT will be √x

Sum of the number and its square root is 110.

x+√x=110

Let √x=y, x=y^2

The EQUATION becomes,

y^2+y=110

Adding \(\frac {b^2}{4}\) on both sides, where b=1

y^2 + y + \(\frac {1^2}{4}=\frac {1^2}{4}\) + 110

y^2 + y + \(\frac {1}{4}=\frac {1}{4}\) + 110

y^2 + y + \(\frac {1}{4}=\frac {441}{4}\)

\((y+\frac {1}{2})\)^2 = \((\frac {21}{2})\)^2

y + \(\frac {1}{2}\) = ±\(\frac {21}{2}\)

y = \(\frac {21}{2}-\frac {1}{2}=\frac {20}{2}\) = 10 and y = \(\frac {-20}{2}-\frac {1}{2}=\frac {-21}{2}\) = -10.5

x = y^2 = 10^2 = 100 and x = y^2 = -10.5^2 = 110.25

The numbers are 100 and 110.25.

Correct choice is (b) -3 < k < 3

The best I can explain: Roots are imaginary. ∴ b^2 – 4ac < 0

(2K)^2 – 4(9)(1) < 0

4k^2 – 36 < 0

k^2 – 9 < 0

k^2 < 9

k < ±3

-3 < k < 3

The correct choice is (c) The roots of the equation are real, DISTINCT and irrational

Easiest explanation: Roots are real. ∴ b^2 – 4ac ≥ 0

5^2 – 4(1)(1)

25 – 4 = 21 which is GREATER than 0. Hence, the discriminant of the equation is greater than ZERO, so roots are real.

Right choice is (C) 73

The explanation is: Let the units place of the two digit number be x and the tens place be y.

Product of the digits of the places = 21

xy=21

y=\(\FRAC {21}{x}\)

Now, the number will be 10y+x

If 36 is subtracted from the number the digits interchange their places.

New number = 10x+y

10y+x-36=10x+y

9y-9x=36

y-x=4

Now, y=\(\frac {21}{x}\)

\(\frac {21}{x}\)-x=4

21-x^2=4x

x^2+4x-21=0

x^2+7x-3x-21=0

x(x+7)-3(x+7)=0

(x+7)(x-3)=0

x = -7, 3

The number is 73.

Right OPTION is (b) 10 < a < 36

The explanation is: The roots of both the EQUATIONS are real.

Discriminant of 5x^2 + ax + 5 : b^2 – 4ac = a^2 – 4 × 5 × 5 = a^2 – 100

Since, roots are real; discriminant will be greater than 0.

a^2 ≥ 100

a ≥ ±10

Now, discriminant of x^2 – 12x + a : b^2 – 4ac = -12^2 – 4 × 1 × a = 144 – 4a

Since, roots are real; discriminant will be greater than 0.

144 ≥ 4a

a ≤ \(\frac {144}{4}\) = 36

For both the equations to have real roots the value of a must lie between 36 and 10.

Correct answer is (c) 20

For explanation: Roots are EQUAL. ∴ b^2 – 4ac = 0

-(2K)^2 – 4(k – 4)(k + 5) = 0

4k^2 – 4(k^2 – 4k + 5K – 20) = 0

4k^2 – 4(k^2 + k – 20) = 0

4k^2 – 4k^2 – 4k + 80 = 0

-4k =– 80

k = \(\frac {-80}{-4}\) = 20

Right answer is (C) L=5, b=7

Best explanation: Let the length of the rectangle be l cm and breadth be b cm.

Perimeter of rectangle = 24

2(l+b)=24

l+b=12

l=12-b(1)

Now, the area of rectangle is 35 cm^2

l×b=35(2)

SUBSTITUTING (1) in (2)

(12-b)b=35

12b-b^2=35

b^2-12b+35=0

b^2-12b=-35

Adding \(\frac {b^2}{4}\) on both sides, where b=-12

b^2 – 12b + \(\frac {-12^2}{4}=\frac {-12^2}{4}\) – 35

b^2 – 12b + \(\frac {144}{4}=\frac {144}{4}\) – 35

b^2 – 12b + \(\frac {144}{4}=\frac {4}{4}\)

\((b-\frac {12}{2})\)^2=\((\frac {2}{2})\)^2

b-\(\frac {12}{2}\)=±\(\frac {2}{2}\)

b = \(\frac {2}{2}+\frac {12}{2}=\frac {14}{2}\) = 7 and b = \(\frac {-2}{2}+\frac {12}{2}=\frac {10}{2}\) = 5

The length of the rectangle is l=12-b=12-7=5 and l=12-5=7

The length and breadth of the rectangle are 7 and 5 or 5 and 7.

Right answer is (a) 40, 42

For explanation: Let one NUMBER be x. The other number is x+2

Sum of the squares of the NUMBERS is 3364.

x^2+(x+2)^2=3364

x^2+x^2+4x+4=3364

2x^2+4x-3360=0

x^2+2x-1680=0

x^2+42x-40x-1680=0

x(x+42)-40(x+42)=0

(x+42)(x-40)=0

x=40, -42

Since we only need positive numbers.

Hence, x=40

The TWO numbers are 40, 42.

Right OPTION is (b) 75

Explanation: Sum of the NUMBERS is 12.

Let one NUMBER be x. Other number is 12-x.

Sum of their SQUARES = 74

x^2+(12-x)^2=74

x^2+144+x^2-24x=74

2x^2-24x+144-74=0

2x^2-24x+70=0

x^2-12x+35=0

x^2-7x-5x+35=0

x(x-7)-5(x-7)=0

(x-7)(x-5)=0

x=7, 5

The number is 57 or 75.

Correct option is (c) 1200

The best I can explain: LET the length of shorter side be x.

Length of longer side = x+10

Length of diagonal = x+20

Here, ∆BDC forms a right-angled triangle. HENCE by Pythagoras Theorem,

BD^2=BC^2+DC^2

(x+20)^2=x^2+(x+10)^2

x^2+40x+400=x^2+x^2+20x+100

40x+400=x^2+20x+100

x^2-20x-300=0

x^2-20x=300

Adding \(\frac {b^2}{4}\) on both sides, where b=-20

x^2 – 20x + \(\frac {-20^2}{4}=\frac {-20^2}{4}\) + 300

x^2 – 20x + \(\frac {400}{4}=\frac {400}{4}\) + 300

x^2 – 20x + \(\frac {400}{4}=\frac {1600}{4}\)

\((x-\frac {20}{2})\)^2 = \((\frac {40}{2})\)^2

x – \(\frac {20}{2}\) = ±\(\frac {40}{2}\)

x = \(\frac {40}{2}+\frac {20}{2}=\frac {60}{2}\) = 30 and x = \(\frac {-40}{2}+\frac {20}{2}=\frac {-20}{2}\) = -10

Since length cannot be negative, hence x = 30

The length of the other side is x + 10 = 30 + 10 = 40

Area of RECTANGLE = 40 × 30 = 1200 units

Correct answer is (B) 15.28, 20.28

Explanation: Let the sides of two squares be x and y.

Their AREAS will be x^2 and y^2 respectively.

Sum of their areas is 625 m^2

x^2+y^2=625(1)

Their perimeters will be 4x and 4y

Difference in their perimeters is 20 m

4x-4y=20

x-y=5

x=5+y(2)

Substituting (2) in (1) we get,

(y+5)^2+y^2=625

y^2+10y+25+y^2=625

2y^2+10y+25=625

2y^2+10y-620=0

y^2+5y=310

Adding \(\FRAC {b^2}{4}\) on both sides, where b=5

y^2 + 5y + \(\frac {5^2}{4}=\frac {5^2}{4}\) + 310

y^2 + 5y + \(\frac {25}{4}=\frac {25}{4}\) + 310

y^2 + 5y + \(\frac {25}{4}=\frac {1265}{4}\)

\((y+\frac {5}{2})\)^2 = \((\frac {\sqrt {1265}}{2})\)^2

y+\(\frac {5}{2}\) = ±\(\frac {\sqrt {1265}}{2}\)=±\(\frac {35.56}{2}\)

y = \(\frac {35.56}{2}-\frac {5}{2}=\frac {30.56}{2}\) = 15.28 and y = \(\frac {-35.56}{2}-\frac {5}{2}=\frac {-40.56}{2}\) = -20.28

Since, length cannot be negative HENCE, y=15.28

Now, x=5+y=5+15.28=20.28

The CORRECT option is (a) True

The best explanation: Roots are equal.∴ b^2 – 4ac = 0

(9p)^2 – 4(8)(15) = 0

81p^2 – 480 = 0

81p^2 = 480

p = ± \(\sqrt {\frac {480}{81}}\) = ± \(\frac {4\sqrt {30}}{9}\)

Correct option is (B) 11.205 m

For explanation I would say: Let the width of the path be X metres.

Length of the field including the path = 50+2x m

Breadth of the field including the path = 14+2x m

Area of the field including the path = (50+2x)(14+2x) m^2

Area of the field excluding the path = 50×14=700 m^2

∴ area of the path = [(50+2x)(14+2x)]-700

∴ [(50+2x)(14+2x)]-700=121

700+100x+28x+4x^2-700=121

4x^2+128x-121=0

x^2+42x=\(\frac {121}{4}\)

Adding \(\frac {b^2}{4}\) on both sides, where b=42

x^2 + 42x + \(\frac {42^2}{4}=\frac {42^2}{4}+\frac {121}{4}\)

x^2 + 42x + \(\frac {1764}{4}=\frac {1764}{4}+\frac {121}{4}\)

x^2 + 42x + \(\frac {1764}{4}=\frac {1885}{4}\)

\((x+\frac {21}{2})\)^2=\((\frac {\sqrt {1885}}{2})\)^2

x + \(\frac {21}{2}\) = ±\(\frac {\sqrt {1885}}{2}\) = ±\(\frac {43.41}{2}\)

x = \(\frac {43.41}{2}-\frac {21}{2}=\frac {22.41}{2}\) = 11.205 and x = \(\frac {-43.41}{2}-\frac {21}{2}=\frac {-64.41}{2}\) = -32.20

Since, the length cannot be negative; therefore, x=11.205 m

The CORRECT option is (d) 4

Easiest EXPLANATION: There are FOUR methods of solving quadratic equations namely, factorization, completing the SQUARE method, USING quadratic formula and using square roots.

The correct answer is (a) base = 32 cm, ALTITUDE = 42 cm

Explanation: Let the altitude of the triangle be x cm.

The base will be x-10 cm.

The area of triangle is 500 cm^2

\(\frac {1}{2}\) × base × altitude = 672

\(\frac {1}{2}\) × (x-10) × x = 672

x^2-10X=1344

Adding \(\frac {b^2}{4}\) on both sides, where b=-10

x^2 – 10x + \(\frac {-10^2}{4}=\frac {-10^2}{4}\) + 1344

x^2 – 10x + \(\frac {100}{4}=\frac {100}{4}\) + 1344

x^2 – 10x + \(\frac {100}{4}=\frac {5476}{4}\)

\((x-\frac {10}{2})\)^2=\((\frac {74}{2})\)^2

x-\(\frac {10}{2}\)=±\(\frac {74}{2}\)

x = \(\frac {74}{2}+\frac {10}{2}=\frac {84}{2}\) = 42 and x = \(\frac {-74}{2}+\frac {10}{2}=\frac {-64}{2}\) = -32

Since, length cannot be NEGATIVE, hence x=42 cm

Base will be x-10=42-10=32 cm