 # RS Aggarwal Class 10 Solutions Chapter 10 - Quadratic Equations Ex 10G ( 10.7)

## RS Aggarwal Class 10 Chapter 10 - Quadratic Equations Ex 10G ( 10.7) Solutions Free PDF

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1. The sum of a natural number and its square is 156. Find the number.

Let the required natural number be x.

According to the given condition,

$\Rightarrow$x + x2 = 156

$\Rightarrow$x2 + x -156 =0

$\Rightarrow$x2 + 13x – 12x — 156 = 0

$\Rightarrow$x(x + 13) – 12(x + 13) = 0

$\Rightarrow$ (x + 13)(x — 12) = 0

$\Rightarrow$x + 13 = 0 or x – 12 = 0

$\Rightarrow$x = -13 or x = 12

Therefore, x = 12 (x cannot be negative)

Hence, the required natural number is 12.

2. The sum of a natural number and its positive square root is 132. Find the number.

Let the required natural number be x.

According to the given condition .

x + $\sqrt{x}$= 132

Putting $\sqrt{x}$= y or x = y2. we get

y2 + y = 132

$\Rightarrow$y2+y-132 =0

$\Rightarrow$y2 +12y— 11y— 132 = 0

$\Rightarrow$y(y + 12) — 11(y + 12) = 0

$\Rightarrow$ (y+ 12)(y— 11) = 0

$\Rightarrow$y + 12 =0 or y—11 =0

$\Rightarrow$y = —12 or y = 11

Therefore, y = 11 (y cannot be negative)

Now,

$\sqrt{x}$ = 11

$\Rightarrow$x=(11)2 = 121

Hence, the required natural number is 121.

3. The sum of two natural numbers is 28 and their product is 192. Find the numbers.

Let the required numbers be x and (28 – x).

According to the given condition,

x(28 — x) = 192

$\Rightarrow$28x — x2 = 192

$\Rightarrow$x2 — 28x +192 = 0

$\Rightarrow$ x2 – 16x — 12x + 192 = 0

$\Rightarrow$ x(x — 16) — 12(x — 16) = 0

$\Rightarrow$ (x — 12)(x — 16) = 0

$\Rightarrow$x — 12 = 0 or x — 16 = 0

$\Rightarrow$x = 12 or x = 16

When x = 12,

28 -x=28 – 12=16

When x = 16,

28 – x = 28 – 16 = 12

Hence, the required numbers are 12 and 16.

4. The sum of the squares of two consecutive positive integers is 365. Find the integers.

Let the required two consecutive positive integers be x and (x + 1).

According to the given condition,

x2 +(x + 1)2 = 365

$\Rightarrow$x2+x2+2x+1 = 365

$\Rightarrow$2x2 + 2x— 364 = 0

$\Rightarrow$ x2 + x — 182 = 0,

$\Rightarrow$ x2 + 14x — 13x — 182 = 0

$\Rightarrow$ x(x + 14) —13(x + 14) = 0

$\Rightarrow$ (x + 14)(x — 13) = 0

$\Rightarrow$ x+14 = 0 or x— 13 = 0

$\Rightarrow$ x=-14 or x= 13

Therefore, x= 13 (x is a positive integer)

When x = 13,

x + 1 = 13 + 1 = 14

Hence, the required positive integers are 13 and 14.

5. The sum of the squares of two consecutive positive odd numbers is 514. Find the numbers.

Let the two consecutive positive odd numbers be x and (x + 2).

According to the given condition,

X2 + (x + 2)2 = 514

$\Rightarrow$ x2 + x2 + 4x + 4 = 514

$\Rightarrow$ 2x2 + 4x — 510 = 0

$\Rightarrow$ x2 ± 2x — 255 = 0

$\Rightarrow$ x2 + 17x — 15x — 255 = 0

$\Rightarrow$ x(x + 17) — 15(x + 17) = 0

$\Rightarrow$ (x + 17)(x — 15) =0

$\Rightarrow$ x+17 = 0 or x— 15 = 0

$\Rightarrow$ x=-17 or x=15

Therefore, x = 15 (X is a positive odd number)

When x = 15,

x+2 =15 +2 =17

Hence, the required numbers are 15 and 17.

6. The sum of the squares of two consecutive positive even numbers is 452. Find the numbers.

Let the two consecutive positive even numbers be x and (x + 2).

According to the given condition,

x2 + (x + 2)2 = 452

$\Rightarrow$ x2 + x2 + 4x + 4 = 452

$\Rightarrow$ 2x2 + 4x — 448 = 0

$\Rightarrow$ x2 + 2x — 224 = 0

$\Rightarrow$ x2 + 16x — 14x — 224 = 0

$\Rightarrow$ x(x + 16) —14(x + 16) = 0

$\Rightarrow$ (x +16)(x — 14) = 0

$\Rightarrow$ x+16=0 or x-14=0

$\Rightarrow$ x = —16 or x = 14

Therefore, x = 14 (x is a positive even number)

When

x = 14, x+2 =14 + 2 = 16

Hence, the required numbers are 14 and 16.

7. The product of two consecutive positive integers is 306. Find the integers.

Let the two consecutive positive integers be x and (x + 1).

According to the given condition,

x(x + 1) = 306

$\Rightarrow$ x2 + x – 306 = 0

$\Rightarrow$ x2 + 18x — 17x — 306 = 0

$\Rightarrow$ x(x + 18) — 17(x + 18) = 0

$\Rightarrow$ (x + 18)(x — 17) = 0

$\Rightarrow$ x+18 =0 or x— 17= 0

$\Rightarrow$ x = —18 or x = 17

Therefore, x = 17 (x is a positive integer)

When x + 1 = 17 + 1 = 18

Hence, the required integers are 17 and 18.

8. Two natural numbers differ by 3 and their product is 504. Find the numbers.

Let the required numbers be x and (x + 3).

According to the question:

x(x + 3) = 504

$\Rightarrow$ x2 + 3x = 504

$\Rightarrow$ x2 + 3x — 504 = 0

$\Rightarrow$ x2 + (24 — 21)x — 504 = 0

$\Rightarrow$ x2 + 24x — 21x — 504 = 0

$\Rightarrow$ x(x + 24) — 21(x + 24) = 0

$\Rightarrow$ (x + 24)(x — 21) = 0

$\Rightarrow$ x + 24 = 0 or x — 21 = 0

$\Rightarrow$ x = — 24 or x = 21

If x = — 24, the numbers are — 24 and {(-24 + 3) = — 21}.

If x = 21, the numbers are 21 and {(21 + 3) = 24}.

Hence, the numbers are (-24, – 21) and (21, 24).

9. Find twoconsecutive multiples of 3 whose product is 648.

Let the required consecutive multiples of 3 be 3x and 3(x + 1).

According to the given condition,

3x x 3(x + 1) = 648

$\Rightarrow$ 9 (x2 + x) =648

$\Rightarrow$ x2 + x = 72

$\Rightarrow$ x2 + x – 72 = 0

$\Rightarrow$ x2 + 9x — 8x — 72 = 0

$\Rightarrow$ x(x + 9) — 8(x + 9) = 0

$\Rightarrow$ (x + 9)(x – 8) = 0

$\Rightarrow$ x + 9 = 0 or x — 8 = 0

$\Rightarrow$ x = —9 or x = 8

Therefore, x = 8 (Neglecting the negative value)

When x = 8,

3x = 3 x 8 = 24

3(x + 1) = 3 x (8 + 1) = 3 x 9 = 27

Hence, the required multiples are 24 and 27.

10. Find two consecutive positive odd integers whose product is 483.

Let the two consecutive positive odd integers be x and (x + 2).

According to the given condition,

x(x + 2) = 483

$\Rightarrow$ x2 + 2x — 483 = 0

$\Rightarrow$x2 + 23x — 21x — 483 = 0

$\Rightarrow$x(x + 23) — 21(x + 23) = 0

$\Rightarrow$ (x + 23)(x — 21) = 0

$\Rightarrow$ x+23=0 or x-21=0

$\Rightarrow$ x = —23 or x = 21

Therefore, x = 21 (x is a positive odd integer)

When x= 21,

x+ 2 = 21 +2 = 23

Hence, the required integers are 21 and 23.

11. Find two consecutive positive even integers whose product is 288.

Let the two consecutive positive even integers be x and (x + 2).

According to the given condition,

x(x + 2) = 288

$\Rightarrow$ x2 + 2x — 288 = 0

$\Rightarrow$ x2 + 18x — 16x — 288 = 0

$\Rightarrow$ x(x + 18) — 16(x + 18) =0

$\Rightarrow$ (x + 18)(x — 16) = 0

$\Rightarrow$ x+18 = 0 or x-16 =0

$\Rightarrow$ x = —18 or x = 16

Therefore, x = 16 (x is a positive even integer)

When x = 16,

x+ 2 = 16 + 2 = 18

Hence, the required integers are 16 and 18.

12. The sum of two natural numbers is 9 and the sum of their reciprocals is $\frac{1}{2}$ Find the numbers.

Let the required natural numbers be x and (9 – x).

According to the given condition,

$\frac{1}{x}+ \frac{1}{9-x}$ = $\frac{1}{2}$

$\Rightarrow$ $\frac{9-x+x}{x(9-x)}$= $\frac{1}{2}$

$\Rightarrow$ $\frac{9}{9x-x^{2}}$= $\frac{1}{2}$

$\Rightarrow$ 9x — x2 = 18

$\Rightarrow$ x2 – 9x+18=0

$\Rightarrow$ x2 — 6x — 3x + 18 = 0

$\Rightarrow$ x(x — 6) — 3(x — 6) = 0

$\Rightarrow$ (x — 3)(x — 6) = 0

$\Rightarrow$ x — 3 = 0 or x — 6 = 0

$\Rightarrow$ x = 3 or x = 6

When x = 3,

9-x=9-3=6

When x = 6,

9-x=9-6=3

Hence, the required natural numbers are 3 and 6.

13. The sum of two natural numbers is 15 and the sum of their reciprocals is $\frac{3}{10}$ Find the numbers.

Let the required natural numbers be x and (15 – x).

According to the given condition,

$\frac{1}{x}$+ $\frac{1}{15-x}$ = $\frac{3}{10}$

$\frac{15-x+x}{x(15-x)}$ = $\frac{3}{10}$

$\frac{15}{15x-x^{2}}$ = $\frac{3}{10}$

$\Rightarrow$15x — x2 = 50

$\Rightarrow$x2 — 15x + 50 = 0

$\Rightarrow$x2 — 10x — 5x + 50 = 0

$\Rightarrow$x(x — 10) — 5(x — 10) = 0

$\Rightarrow$ (x — 5)(x — 10) = 0

$\Rightarrow$x — 5 = 0 or x — 10 = 0

$\Rightarrow$x = 5 or x = 10

When x = 5,

15-x=15 -5=10

When x = 10,

15-x=15-10=5

Hence, the required natural numbers are 5 and 10.

14. The difference of two natural numbers is 3 and the difference of their reciprocals is $\frac{3}{28}$. Find the numbers.

Let the required natural numbers be x and (x + 3).

Now, x <x + 3

$\frac{1}{x}$ > $\frac{1}{x+3}$

According to the given condition,

$\Rightarrow$ $\frac{1}{x}$$\frac{1}{x+3}$=$\frac{3}{28}$

$\Rightarrow$ $\frac{x+3-z}{x(x+3)}$ = $\frac{3}{28}$

$\Rightarrow$X2 + 3x = 28

$\Rightarrow$x2+ 3x— 28=0

$\Rightarrow$x2 + 7x — 4x — 28 = 0

$\Rightarrow$x(x + 7) — 4(x + 7) = 0

$\Rightarrow$ (x + 7)(x —4) = 0

$\Rightarrow$x+7=0 or x-4=0

$\Rightarrow$x = -7 or x = 4

x = 4 (-7 is not a natural number)

When x = 4,

x+3=4+3=7

Hence the required natural numbers are 4 and 7.

15. The difference of two natural numbers is 5 and the difference of their reciprocals is $\frac{5}{14}$ . Find the numbers.

Let the required natural numbers be x and (x + 5).

Now, x <x + 5

$\frac{1}{x}$>$\frac{1}{x+5}$=$\frac{5}{14}$

According to the given condition,

$\frac{1}{x}$$\frac{1}{x+5}$=$\frac{5}{14}$

$\Rightarrow$ $\frac{x+5-x}{x(x+5)}$=$\frac{5}{14}$

$\Rightarrow$ $\frac{5}{x^{2}+5x}$ = $\frac{5}{14}$

$\Rightarrow$x2+5x=14

$\Rightarrow$ x2 +5x-14=0

$\Rightarrow$ x2 +7x – 2x— 14= 0

$\Rightarrow$ x(x + 7) — 2(x + 7) = 0

$\Rightarrow$ (x + 7)(x — 2) = 0

$\Rightarrow$ x+7=0 or x-2=0

$\Rightarrow$ x= -7 or x= 2

Therefore, x = 2 (-7 is not a natural number)

When x=2,

x+5=2+5=7

Hence,the required natural numbers are 2 and 7.

16. The sum of the square of two Consecufive multiples of 7 is 1225. Find the multiples.

Let the required consecutive multiples of 7 be 7x and 7(x + 1).

According to the given condition,

(7x)2 + [7(x + 1)]2 = 1225

$\Rightarrow$ 49x2 + 49 (x2 + 2x + 1) = 1225

$\Rightarrow$ 49x2 + 49x2 + 98x + 49 =1225

$\Rightarrow$ 98x2 + 98x — 1176 = 0

$\Rightarrow$ x2 +x-12 = 0

$\Rightarrow$ x2 -4x — 3x —12 = 0

$\Rightarrow$ x(x + 4) — 3(x + 4) = 0

$\Rightarrow$ (x + 4)(x — 3) = 0

$\Rightarrow$ x+4 = 0 or x-3 = 0

$\Rightarrow$ x = —4 or x= 3

Therefore, x = 3 (Neglecting the negative value)

When x = 3,

7x=7 x 3=21

7(x +1)=7(3+1)=7 x 4=28

Hence, the required multiples are 21 and 28.

17. The sum of a natural number and its reciprocal is $\frac{65}{8}$. Find the number.

Let the natural number be x.

According to the given condition,

x + $\frac{1}{x}$ = $\frac{65}{8}$

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

$\Rightarrow$ 8x2 – 8 = 65x

$\Rightarrow$ 8x2 — 65x + 8 = 0

$\Rightarrow$ 8x2 — 64x — x -I- 8 = 0

$\Rightarrow$ 8x(x — 8) — 1(x — 8) = 0

$\Rightarrow$ (x — 8)(8x — 1) = 0

$\Rightarrow$ x — 8 = 0 or 8x — 1 = 0

$\Rightarrow$ x = 8 or x = $\frac{1}{8}$

Therefore, x = 8 (x is a natural number)

Hence, the required number is 8.

18. Divide 57 into two parts whose product is 680.

Let the two parts be x and (57 – x).

According to the given condition,

x(57 — x) = 680

$\Rightarrow$ 57x — x2 = 680

$\Rightarrow$x2 – 57x + 680 = 0

$\Rightarrow$x2 — 40x — 17x + 680 = 0

$\Rightarrow$x(x — 40) — 17(x — 40) = 0

$\Rightarrow$ (x — 40)(x — 17) = 0

$\Rightarrow$x — 40 = 0 or x — 17 = 0

$\Rightarrow$x = 40 or x = 17

When x = 40,

57 – x = 57 – 40 = 17

When x = 17,

57 – x = 57 – 17 = 40

Hence, the required parts are 17 and 40.

19. Divide 27 into two parts such that the sum of their reciprocals is $\frac{3}{20}$.

Let the two parts be x and (27 – x).

According to the given condition,

$\Rightarrow$ $\frac{1}{x}$+ $\frac{1}{27-x}$= $\frac{3}{20}$

$\Rightarrow$ $\frac{27-x+x}{x(27-x) }$= $\frac{3}{20}$

$\Rightarrow$ $\frac{27}{27x-x^{2}}$= $\frac{3}{20}$

$\Rightarrow$27x—x2 = 180

$\Rightarrow$x2 — 27x + 180 = 0

$\Rightarrow$x2 — 15x — 12x + 180 = 0

$\Rightarrow$x(x — 15) — 12(x — 15) = 0

$\Rightarrow$ (x — 12)(x — 15) = 0

$\Rightarrow$x — 12 = 0 or x — 15 = 0

$\Rightarrow$x = 12 or x = 15

When x = 12,

27 -x=27 – 12 =15

When x = 15,

27 -x=27 – 15=12

Hence, the required parts are 12 and 15.

20. Divide 16 into two parts such that twice the square of e larger part exceeds the square of the smaller part by 164.

Let the larger and smaller parts be x and y, respectively.

According to the question:

x + y = 16 …(i)

2x2 = y2 + 164 …(ii)

From (i), we get:

x =16 — y …(iii)

From (ii) and (iii), we get:

$\Rightarrow$2(16 — y)2 = y2 + 164

$\Rightarrow$2(256 — 32y + y2) = Y2 + 164

$\Rightarrow$512 — 64y + 2y2 = y2 + 164

$\Rightarrow$y2 — 64y + 348 = 0

$\Rightarrow$y2 — (58 + 6)y + 348 = 0

$\Rightarrow$y2 — 58y — 6y + 348 = 0

$\Rightarrow$y(y — 58) — 6(y — 58) = 0

$\Rightarrow$ (y — 58)(y — 6) = 0

$\Rightarrow$y — 58 =0 or y — 6=0

y = 6 (Because y < 16)

Putting the value of y in equation (iii), we get :

x = 16 — 6 = 10

Hence, the two natural numbers are 6 and 10.

21. Find two natural numbers, the sum of whose squares is 25 times their sum and also equal to 50 times their difference.

Let the two natural numbers be x and y.

According to the question:

x2 + y2 = 25(x + y) … (i)

x2 + y2 = 50(x — y) . . . (ii)

From (i) and (ii), we get:

25(x + y) = 50(x — y)

$\Rightarrow$x + y = 2(x — y)

$\Rightarrow$x + y = 2x — 2y

$\Rightarrow$y + 2y = 2x — x

$\Rightarrow$3y = x … (iii)

From (ii) and (iii), we get :

(3y)2 + y2 = 50(3y — y)

$\Rightarrow$9y2+ y2 = 100y

$\Rightarrow$10y2 = 100y

$\Rightarrow$y = 10

From (iii), we have :

3 x 10 = x

$\Rightarrow$30 = x

Hence, the two natural numbers are 30 and 10.

22. The difference of the squares of two natural numbers is 45. The squat of the smaller number is four times the larger number. Find the numbers.

Let the greater number be x and the smaller number be y.

According to the question: x2 — y2 = 45 … (i)

y2 = 4x … (ii)

From (i) and (ii), we get:

x2 — 4x = 45

$\Rightarrow$x2 — 4x — 45 = 0

$\Rightarrow$x2 — (9 — 5)x — 45 = 0

$\Rightarrow$x2 — 9x + 5x — 45 = 0

$\Rightarrow$x(x — 9) + 5(x — 9) =0

$\Rightarrow$ (x — 9)(x + 5) =0

$\Rightarrow$x — 9=0 or x + 5 = 0

$\Rightarrow$x = 9 or x = —5

$\Rightarrow$x = 9 (Because x is a natural number)

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

y2 = 4 x 9

y2 = 36

y = 6

Hence, the two numbers are 9 and 6.

23. Three consecutive positive integers are such that the sum of the square of the first and the product of the other two is 46. Find the integers.

Let the three consecutive positive integers be x, x + 1 and x + 2.

According to the given condition,

$\Rightarrow$x2 + (x+1)(x+ 2) =46

$\Rightarrow$x2+x2+3x+2=46

$\Rightarrow$2x2 + 3x -44 =0

$\Rightarrow$2x2 + 11x – 8x -44= 0

$\Rightarrow$x(2x +11) -4(2x+11) = 0

$\Rightarrow$ (2x +11)(x -4) = 0

$\Rightarrow$2x+11=0 or x-4=0

$\Rightarrow$x = –$\frac{11}{2}$ or x =4

Therefore, x= 4 (x is a positive integer)

When x = 4,

x+1=4+1=5

x+2=4+2=6

Hence, the required integers are 4, 5 and 6.

24. A two-digit number is 4 times the sum of its digits and twice the product of its digits. Find the number.

Let the digits at units and tens places be x and y, respectively.

Therefore, Original number = 10y + x

According to the question:

$\Rightarrow$10y + x = 4(x + y)

$\Rightarrow$10y + x = 4x + 4y

$\Rightarrow$3x — 6y = 0

$\Rightarrow$3x = 6y

$\Rightarrow$x = 2y …. (i)

Also,

10y + x = 2xy

10y + 2y = 2.2y. y [ From (i)]

12y = 4y2

y = 3

From (i), we get:

x = 2 x 3= 6

Therefore, Original number =10 x 3 + 6 = 36

25. A two-digit number is such that the product of its digits is 14. If 45 is added to the number, the digits interchange their places. Find the number.

Let the digits at units and tens places be x and y, respectively.

Therefore, xy = 14

$\Rightarrow$y= $\frac{14}{x}$ ….(i)

According to the question:

(10y + x) + 45 = 10x + y

$\Rightarrow$9y — 9x = — 45

$\Rightarrow$y — x = —5 …(ii)

From (i) and (ii), we get:

$\frac{14}{x}$– X = – 5

$\Rightarrow$ $\frac{14-x2}{x}$ =- 5

$\Rightarrow$14 — x2 = —5 x

$\Rightarrow$x2 — 5x — 14 = 0

$\Rightarrow$x2 — (7 — 2)x — 14 = 0

$\Rightarrow$x2 — 7x + 2x — 14 = 0

$\Rightarrow$x(x — 7) + 2(x — 7) =0

$\Rightarrow$ (x — 7)(x + 2) =0

$\Rightarrow$x — 7 = 0 or x+ 2=0

$\Rightarrow$x = 7 or x= —2

$\Rightarrow$x = 7 (Because the digit cannot be negative)

Putting x = 7 in equation (i), we get:

y = 2

Therefore, Required number = 10 x 2 + 7 = 27

26. The denominator of a fraction is 3 more than its numerator. The sum of 9 the fraction and its reciprocal is 2$\frac{9}{10}$. Find the fraction.

Let the numerator be x.

Therefore, Denominator = x + 3

Therefore, Original number = $\frac{x}{x+3}$

According to the question:

$\frac{x}{x+3} + \frac{1}{\frac{x}{x+3}}$= $2\frac{9}{10}$

$\Rightarrow$ $\frac{x}{x+3}$+$\frac{x+3}{x}$=$\frac{29}{10}$

$\Rightarrow$ $\frac{x^{2}+(x+3) ^{2}}{x(x+3)}$ = $\frac{29}{10}$

$\Rightarrow$ $\frac{x}{x+3}$+$\frac{x+3}{x}$=$\frac{29}{10}$

$\Rightarrow$ $\frac{x^{2}+x^{2}+6x+9}{x^{2}+3x}$=$\frac{29}{10}$

$\Rightarrow$29x^{2} + 87x = 20x^{2} + 60x + 90

$\Rightarrow$9x2 + 27x — 90 = 0

$\Rightarrow$9(x2 + 3x — 10) =0

$\Rightarrow$x2 + 3x — 10 = 0

$\Rightarrow$x2 + 5x — 2x — 10 = 0

$\Rightarrow$x(x + 5) — 2(x + 5) = 0

$\Rightarrow$ (x — 2)(x + 5) = 0

$\Rightarrow$x — 2 = 0 or x + 5 = 0

$\Rightarrow$x = 2 or x = — 5 (rejected)

So, numerator = x = 2

denominator — x+ 3= 2+ 3= 5

So, required fraction = $\frac{2}{5}$

27. The numerator of a fraction is 3 less than its denominator. If 1 is added to the denominator, the fraction is decreased by $\frac{1}{15}$. Find the fraction.

Let the denominator of the required fraction be x.

Numerator of the required fraction = x – 3

Original fraction = $\frac{x-3}{x}$

If 1 is added to the denominator, then the new fraction obtained is $\frac{x-3}{x+1}$.

According to the given condition,

$\frac{x-3}{x+3}$=$\frac{x-3}{x}$$\frac{1}{15}$

$\Rightarrow$ $\frac{x-3}{x}$$\frac{x-3}{x+1}$ = $\frac{1}{15}$

$\Rightarrow$ $\frac{(x-3)(x+1)-x(x-3)}{x(x+1)}$=$\frac{1}{15}$

$\Rightarrow$ $\frac{ x^{2}-2x-3-x^{2}+3x }{ x^{2}+x}$= $\frac{1}{15}$

$\Rightarrow$ $\frac{x-3}{x^{2}+x}$=$\frac{1}{15}$

$\Rightarrow$x2+x=15x-45

$\Rightarrow$x2-14x+45=0

$\Rightarrow$x2 — 9x — 5x +45 = 0

$\Rightarrow$x(x — 9) — 5(x — 9) = 0

$\Rightarrow$ (x — 5)(x — 9) = 0

$\Rightarrow$x — 5 = 0 or x — 9 = 0

$\Rightarrow$x = 5 or x = 9

When x=5

$\frac{x-3}{x}$=$\frac{5-3}{5}$=$\frac{2}{5}$

When x=9

$\frac{x-3}{x}$=$\frac{9-3}{9}$=$\frac{6}{9}$=$\frac{2}{3}$

Hence, the required fraction is $\frac{2}{5}$.

28. The sum of a number and its reciprocal is 2$\frac{1}{30}$. Find the number.

Let the required number be x.

According to the given condition,

X+$\frac{1}{X}$ = 2$\frac{1}{30}$

$\Rightarrow$ $\frac{x^{2}+1}{x}$=$\frac{61}{30}$

$\Rightarrow$30x2 + 30 = 61x

$\Rightarrow$30x2 — 61x + 30 = 0

$\Rightarrow$30x2 — 36x — 25x + 30 = 0

$\Rightarrow$6x(5x — 6) — 5(5x — 6) = 0

$\Rightarrow$ (5x — 6)(6x — 5) = 0

$\Rightarrow$5x-66 =0 or 6x-5=0

$\Rightarrow$x =$\frac{6}{5}$ or x= $\frac{5}{6}$

Hence, the required number is $\frac{5}{6}$ or $\frac{6}{5}$.

29. A teacher on attempting to arrange the students for mass drill in the form of a solid square found that 24 students were left. When he increased the size of the square by one student, he found that he was short of 25 students. Find the number of students.

Let there be x rows.

Then, the number of students in each row will also be x.

Total number of students = (x2 + 24)

According to the question:

(x + 1)2 — 25 = x2 + 24

$\Rightarrow$x2 + 2x + 1 — 25 — x2 — 24 = 0

$\Rightarrow$2x — 48 = 0

$\Rightarrow$2x = 48 x = 24

Total number of students = 242 + 24 = 576 + 24 = 600

30. 300 apples are distributed equally among a certain number of students. Had there been 10 more students, each would have received one apple less. Find the number of students.

Let the total number of students be x.

According to the question :

$\frac{300}{x}$$\frac{300}{x+10}$=1

$\Rightarrow$ $\frac{300(x+10)-300x}{x(x+10)}$=1

$\Rightarrow$ $\frac{300x+3000-300x}{x^{2}+10x}$=1

$\Rightarrow$3000=x2 + 10x

$\Rightarrow$x2 + 10x – 3000=0

$\Rightarrow$x2 + (60 — 50)x – 3000 = 0

$\Rightarrow$x2 + 60x — 50x — 3000 = 0

$\Rightarrow$x(x + 60) — 50(x + 60) = 0

$\Rightarrow$ (x + 60)(x — 50) = 0

$\Rightarrow$x= 50 or x = —60

x cannot be negative; therefore, the total number of students is 50.

31. In a class test, the sum of Kamal’s marks in mathematics and English is 40. Had he got 3 marks more in mathematics and 4 marks less in English, the product of the marks would have been 360. Find his marks in two subjects separately.

Let the marks of Kamal in mathematics and english be x and y, respectively.

According to the question :

x + y = 40 … (i)

Also,

(x + 3)(y — 4) = 360

$\Rightarrow$ (x + 3)(40 — x — 4) = 360 [From(i)]

$\Rightarrow$ (x + 3)(36 — x) = 360

$\Rightarrow$36x — x2 + 108 — 3x = 360

$\Rightarrow$33x — x2 — 252 = 0

$\Rightarrow$x2 ± 33x — 252 = 0

$\Rightarrow$x2 – 33x + 252 = 0

$\Rightarrow$X2 – (21 + 12)x + 252 = 0

$\Rightarrow$X2 – 21x — 12x + 252 =0

$\Rightarrow$x(x — 21) — 12(x — 21) =0

$\Rightarrow$ (x — 21)(x — 12) 0

$\Rightarrow$x = 21 or x = 12

If x = 21,

y = 40 — 21 = 19

Thus, Kamal scored 21 and 19 marks in mathematics and english, respectively.

If x = 12,

y = 40 — 12 = 28

Thus, Kamal scored 12 and 28 marks in mathematics and english, respectively.

Q32. Some students planned a picnic. The total budget for food was rs 2000. But, 5 students failed to attend the picnic and thus the cost for food for each member increased by 20. How many students attended the picnic and how much did each student pay for the food?

Ans:

Let x be the number of the students who planned a picnic

Original cost of the food for each member = $\frac{2000}{x}$

Five students failed to ttend the picnic .So (x-5) students attended the picnic.

New cost of the food for each member =$\frac{2000}{x-5}$

According to the given condition

$\frac{2000}{x-5}$$\frac{2000}{x}$ =20

=$\frac{2000x-2000x+10000}{x(x-5)}=20$

=$\frac{10000}{x^{2} -5x}=20$

=x2 -5x=500

= x2 -5x-500=0

= x2 -25x+20x-500=0

=x-25=0 or x+20=0

x=25 or x=25 or x=-20

x=25

Q33. If the price of a book is reduced by 5, a person can buy 4 more books for rs 600. Find the original price of the book.

Sol:

Let the original price of the book be x

Therefore number of the books bought at the original price for 600= $\frac{600}{x}$

If the price of the book is reduced by the 5 the new price of the book is (x-5)

Number of the books bought at the reduced price for 600=$\frac{600}{x-5}$

According to the given condition

$(\frac{600}{x-5})-(\frac {600}{x})=4$

$\frac{600x-600x+3000}{x(x-5)}=4$

= x2 -5x=750

=x2 -5x-750=0

=x2 -30x+25x-750=0

=x(x-30)+25(x-30)=0

=(x-30)(x+25)=0

X=30 or x=-25

X=30 (Price cannot be negative)

Hence the original price of the book is 30

Q34. A person on tour has rs 10800 for his expenses. If he extends his tour by 4 days, he has to cut down his daily expenses by rs 90. Find the original duration of the tour.

Sol:

Let the original duration of the tour be x days.

Original daily expenses =$\frac{10800}{x}$

If he extends his tour by 4 days, then his new daily expenses=$\frac{10800}{x+4}$

According to the given condition,

$(\frac{10800}{x})-(\frac{10800}{x+4})=90$

=$\frac{43200}{x^{2}+4x}=90$

x2+ 4x= 480

x2+ 4x – 480 = 0

x2 + 24x – 20x – 480 = 0

x(x + 24) – 20(x + 24) = 0

(x + 24)(x – 20) = 0

x+24 = 0 or x-20 = 0

x = -24 or x = 20

x = 20 (Number of days cannot be negative)

Hence, the original duration of the tour is 20 days.

Q35. In a class test, the sum of the marks obtained by P in mathematics and science is 28. Had he got 3 more marks in mathematics and 4 marks less in science, the product of marks obtained in the two subjects would have been 180. Find the marks obtained by him in the two subjects separately.

Let the marks obtained by P in mathematics and science be x and (28 – x), respectively.

According to the given condition,

(x + 3)(28 – x – 4) = 180

(x + 3)(24 -x) = 180

= – x2 + 21x + 72 = 180

= x2 – 21x + 108 = 0

x2 – 12x – 9x + 108 = 0

x(x -12) – 9(x -12) = 0

(x – 12)(x – 9) = 0

x – 12 = 0 or x – 9 = 0

x = 12 or x = 9

When x = 12,

28 -x=28 – 12=16

When x = 9,

28 -x=28 -9=19

Hence, he obtained 12 marks in mathematics and 16 marks in science or 9 marks in mathematics and 19 marks in science.

Q36. A man buys a number of pens for 180. If he had bought 3 more pens for the same amount, each pen would have cost him rs 3 less. How many pens did he buy?

Let the total number of pens be x.

According to the question :

$(\frac{80}{x})-(\frac{80}{x+4}=1)$

$(\frac{80(x+4)-80x}{x(x+4)}=1)$

320 = x2+ 4x

x2 + 4x -320 = 0

x2 + (20 – 16)x – 320 = 0

x2 + 20x – 16x -320 =0

x(x + 20) — 16(x + 20) = 0

(x + 20)(x -16) = 0

x= – 20 or x = 16

The total number of pens cannot be negative; therefore, the total number of pens is 16.

Q37. A dealer sells an article for 75 and gains as muchper cent as the cost price of the article. Find the cost price of the article.

Let the cost price of the article be x.

Gain percent = x%

According to the given condition,

x + $\frac{x}{100}\times x$ = 75

x2 + 100X = 7500

x2 ± 100X – 7500 = 0

x2 -I- 150x — 50x — 7500 = 0

x(x + 150) -50(x + 150) = 0

(x – 50)(x + 150) = 0

x – 50 =0 or x + 150 = 0 x = 50 or

x = -150

x = 50 (Cost price cannot be negative)

(Cost price + Gain = Selling price)

Hence, the cost price of the article is 50.

PROBLEMS ON AGES

Q38. One year ago, a man was 8 times as old as his son. Now, his age is equal to the square of his son’s age. Find their present ages.

Let the present age of the son be x years.

Present age of the man = x2 years One year ago,

Age of the son = (x – 1) years Age of the man = (x2 – 1) years

According to the given condition,

Age of the man = 8 x Age of the son

x2 – 1 = 8(x – 1)

x2 – 1 = 8x – 8

x2 -8x+7=0

x2 – 7x – x +7=0

x(x – 7) – 1(x – 7) = 0

(x -1)(x – 7)=0

x – 1 = 0 or x – 7 = 0

x = 1 or x = 7

(Man’s age cannot be 1 year)

Present age of the man = 72 years = 49 years

Q39. The sum of the reciprocals of Meena’s ages (in years) 3 years ago and . I -. 5 years hence is -3 Find her present age.

Sol:

Let the present age of Meena be x years.

Meena’s age 3 years ago = (x – 3) years

Meena’s age 5 years hence = (x + 5) years

According to the given condition,

$(\frac{1}{x-3})+(\frac{1}{x+5})= \frac{1}{3}$

$\frac{x+5+x-3}{(x-3)(x+5)}= \frac{1}{3}$

x2 ± 2x – 15 = 6x + 6

x2 – 4x – 21 = 0

x2– 7x + 3x — 21 = 0

x(x – 7) + 3(x – 7) = 0

(x – 7)(x + 3) = 0 x – 7 = 0

or x + 3 = 0 x = 7 or x = -3 x = 7 (Age cannot be negative)

Hence, the present age of Meena is 7 years.

Q40. The sum of the ages of a boy and his brother is 25 years, and the product of their ages in years is 126. Find their ages

Let the present ages of the boy and his brother be x years and (25 – x) years.

According to the question :

x(25 – x) = 126

25x – x2= 126

x2– (18 + 7)x + 126 = 0

x2 – 18x – 7x + 126 = 0

x(x – 18) – 7(x – 18) = 0

(x -18)(x – 7) =0

x – 18 = 0 or x – 7 = 0 x= 18 or x = 7 x = 18

Present age of the boy cannot be less than his brother) If x = 18,

we have : Present ages of the boy =18

years Present age of his brother = (25 — 18) years = 7 years

Thus, the present ages of the boy and his brother are 18 years and 7 years, respectively.

Q41. The product of Tanvy’s age (in years) 5 years ago and her age 8 years later is 30. Find her present age.

Let the present age of Tanvy be x years.

According to the question :

(x -5)(x + 8) = 30

x2+ 3x – 40 = 30

x2+ 3x – 70 = 0

x2 + (10 – 7)x – 70 = 0

x2+ 10X – 7x – 70 = 0

x(x + 10) – 7(x + 10) = 0

(x + 10)(x – 7) = 0

x + 10 = 0 or x – 7 = 0

x= – 10 or x = 7 x = 7 ( … Age cannot be negative) Thus , the present age of Tanvy is 7 years.

Q42: Two years ago, a man’s age was three times the square of his son’s age. In three years time, his age will be four times his son’s age. Find their present ages.

Let son’s age 2 years ago be x years. Then,

Man’s age 2 years ago = 3 x2 years …

Son’s present age = (x + 2) years

Man’s present age = (3 x2 + 2) years In three years time,

Son’s age = (x + 2 + 3) years = (x + 5) years Man’s age = (3 x2 + 2 + 3) years = (3×2 + 5) years

According to the given condition,

Man’s age = 4 x Son’s age

3 x2 + 5 = 4(x + 5)

3 x2+5=4x+20

3 x2 – 4x – 15 = 0

3 x2 – 9x + 5x – 15 = 0

3x(x – 3) + 5(x – 3) = 0

(x – 3)(3x + 5) = 0

x-3=11 or 3x+5=0 x=3 or x=-1 x = 3 (Age cannot be negative)

Son’s present age = (x + 2) years = (3 + 2) years = 5 years

Man’s present age = (3×2 + 2) years = (3 x 9 + 2) years = 29 years

Q43) A truck covers a distance of 150 km at a certain average sped and then covers another 200 km at an average speed which is 20 km per hour than the first speed. If the truck covers the total distance in 5 hours, find the first speed of the truck.

Ans. 43) Let the first speed of the truck x km/h.

Therefore, Time taken to cover 150 km = $\frac{ 150 }{ x }h$ $(Time = \frac{ Distance }{ Speed })$

New speed of the truck = ( x + 20 ) km/h

Therefore, Time taken to cover 200 km = $\frac{ 200 }{ x + 20 }h$

According to the given condition,

Time taken to cover 150 km + Time taken to cover 200 km = 5 h

Therefore, $\frac{ 150 }{ x } + \frac{ 200 }{ x + 20 }$ = 5

$\frac{ 150x + 3000 + 200x }{ x(x + 20) }$ = 5

350x + 3000 = 5 ($x^{ 2 }$ + 20x )

350x + 3000 = $5x^{ 2 }$ + 100x

$5x^{ 2 }$ – 250x – 3000 = 0

$x^{ 2 }$ – 50x – 600 = 0

$x^{ 2 }$ – 60x + 10x – 600 = 0

x ( x – 60 ) + 10 ( x – 60 ) = 0

( x – 60 ) ( x + 10 ) = 0

x = 60 or x = -10

Therefore, x = 60 ( Speed cannot be negative )

Hence, the first speed of the truck is 60km/h.

Q44) While boarding an aeroplane, a passenger got hurt. The pilot showing promptness and concern, made arrangements to hospitalise the injured and so the plane started late by 30 minutes. To reach the destination, 1500 km away, in time, the pilot increased the speed by 100 km/hour. Find the original speed of the plane.

Do you appreciate the values shown by the pilot, namely promptness in providing help to the injured and his efforts to reach in time?

Ans. 44) Let the original speed of the plane be x km/hr.

Therefore, Actual speed of the plane = (x + 100) km/hr

Distance of the journey = 1500 km

Time taken to reach the destination at original speed = $\frac{ 1500 }{ x }h$

Time taken to reach the destination at actual speed = $\frac{ 1500 }{ x + 100 }h$

According to the given condition,

Time taken to reach the destination at original speed = Time taken to reach the destination at actual speed + 30 min

Therefore, $\frac{ 1500 }{ x } = \frac{ 1500 }{ x + 100 } + \frac{ 1 }{ 2 }$ (30 min = $\frac{ 30 }{ 60 }h = \frac{ 1 }{ 2 }h$ )

$\frac{ 1500 }{ x } – \frac{ 1500 }{ x + 100 } = \frac{ 1 }{ 2 }$

$\frac{ 1500x + 150000 – 1500x }{ x( x + 100) } = \frac{ 1 }{ 2 }$

$\frac{ 150000 }{ x^{ 2 } + 100x } = \frac{ 1 }{ 2 }$

$x^{ 2 }$ + 100x = 300000

$x^{ 2 }$ + 100x – 300000 = 0

$x^{ 2 }$ + 600x – 500x – 300000 = 0

x ( x + 600 ) – 500 ( x + 600 ) = 0

(x + 600)(x – 500) = 0

x + 600 = or x = 500

Therefore, x = 500 (Speed cannot be negative )

Hence, the original speed of the plane is 500 km/hr.

Yes, we appreciate the values shown by the pilot, namely promptness in providing help to the injured and his efforts to reach in time. This reflects the caring nature of the pilot and his dedication to the work.

Q45) A train covers a distance of 480 km at a uniform speed. If the speed had been 8 km/hr less than it would have taken 3 hours more to cover the same distance. Find the usual speed of the train.

Ans. 45) Let the usual speed of the train be x km/hr.

Therefore, Reduced speed of the train = ( x – 8 ) km/ hr

Total distance to be covered = 480 km

Time taken by the train to cover the distance at usual speed = $\frac{ 480 }{ x }h$ $(Time = \frac{ Distance }{ Speed })$

Time taken by the train to cover the distance at reduced speed = $\frac{ 480 }{ x – 8 }h$

According to the given condition,

Time taken by the train to cover the distance at reduced speed = Time taken by the train to cover the distance at usual speed + 3 h

Therefore, $\frac{ 480 }{ x – 8 } = \frac{ 480 }{ x } + 3$

$\frac{ 480 }{ x – 8 } – \frac{ 480 }{ x } = 3$

$\frac{ 480x – 480x + 3840 }{ x(x – 8)} = 3$

$\frac{ 3840 }{ x^{ 2 } – 8x } = 3$

$x^{ 2 }$ – 8x = 1280

$x^{ 2 }$ – 8x – 1280 = 0

$x^{ 2 }$ – 40x + 32x – 1280 = 0

x (x – 40) + 32 ( x – 4) = 0

x – 40 = 0 or x + 32 = 0

Therefore, x = 40 (Speed cannot be negative)

Hence, the usual speed of the train is 40 km/hr.

Q46) A train travels at a certain average speed for a distance of 54 km and then travels a distance of 63 km at an average speed of 6 km/hr more than the first speed. If it takes 3 hours to complete the total journey, what is its first speed?

Ans. 46) Let the first speed of the train be x km/hr

Therefore, Time taken to cover 54 km = $\frac{ 54 }{ x }h$ $(Time = \frac{ Distance }{ Speed })$

New speed of the train = (x + 6) km/hr

Therefore, Time taken to cover 63 km =$\frac{ 63 }{ x + 6 }h$

According to the given condition,

Time taken to cover 54 km + Time taken to cover 63 km = 3 h

Therefore, $\frac{ 54 }{ x } + \frac{ 63 }{ x + 6 } = 3$

$\frac{ 54x + 324 + 63x }{ x(x + 6)} = 3$

117x + 324 = 3 ($x^{ 2 }$ + 6x )

117x + 324 = $3x^{ 2 }$ + 18x

$3x^{ 2 }$ – 99x – 324 = 0

$x^{ 2 }$ – 33x – 108 = 0

$x^{ 2 }$ – 36x + 3x – 108 = 0

x (x – 36) + 3 (x + 3) = 0

x – 36 = 0 or x + 3 = 0

x = 36 or x = -3

Therefore, x = 36 (Speed cannot be negative )

Hence, the first speed of the train is 36 km/hr.

Q47) A train travels 180 km at a uniform speed. If the speed had been 9km/hr more, it would have taken 1 hour less for the same journey. Find the speed of the train.

Ans. 47) Let the speed of train be x km/hr

Distance = 180 km

So, time = 180/x

When speed is 9 km/hr more, time taken = 180 / (x + 9)

According to the given information:

$\frac{ 180 }{ x } – \frac{ 180 }{ x + 9 } = 1$

$\frac{ 180 (x + 9 – x) }{ x(x + 9) } = 1$

180 * 9 = x (x + 9)

1620 = $x^{ 2 }$ + 9x

$x^{ 2 }$ + 45x – 36x – 1620 = 0

x (x + 45) – 36 (x + 45) = 0

(x – 36)(x + 45) = 0

x = 36 or -45

But x being speed cannot be negative.

So, x = 36

Hence, the speed of the train is 36 km/hr.

Q48) A train covers a distance of 90 km at a uniform speed. Had the speed been 15 km/hr more, it would have taken 30 minutes less for the journey. Find the original speed of the train.

Ans. 48) Let the original speed of the train be x km/hr.

According to the question:

$\frac{ 90 }{ x } – \frac{ 90 }{ (x + 15) } = \frac{ 1 }{ 2 }$

$\frac{ 90 ( x + 15 ) – 90x }{ x(x + 15) } = \frac{ 1 }{ 2 }$

$\frac{ 90x + 1350 – 90x }{ x^{ 2 } + 15x } = \frac{ 1 }{ 2 }$

$\frac{ 1350 }{ x^{ 2 } + 15x } = \frac{ 1 }{ 2 }$

2700 = $x^{ 2 }$ + 15x

$x^{ 2 }$ + (60 – 45)x – 2700 = 0

$x^{ 2 }$ + 60x – 45x – 2700 = 0

x(x + 60) – 45(x + 60) = 0

(x + 60) (x – 45) = 0

x = -60 or x = 45

x cannot be negative

Therefore, the original speed of train is 45 km/hr.

Q49) A passenger train takes 2 hours less for a journey of 300 km if its speed is increased by 5 km/hr from its usual speed. Find its usual speed.

Ans. 49) Let the usual be x km/hr.

According to the question:

$\frac{ 300 }{ x } – \frac{ 300 }{ (x + 5) } = 2$

$\frac{ 300( x + 5) – 300x }{ x ( x + 5) } = 2$

$\frac{ 300x + 1500 – 300x }{ x^{ 2 } + 5x } = 2$

1500 = 2 ($x^{ 2 }$ + 5x)

1500 = $2x^{ 2 }$ + 10x

$x^{ 2 }$ + 5x – 750 = 0

$x^{ 2 }$ + ( 30 – 25 ) x – 750 = 0

$x^{ 2 }$ + 30x – 25x – 750 = 0

x (x + 30) – 25 ( x + 30 ) = 0

( x + 30 ) ( x – 25 ) = 0

x = -30 or x = 25

The usual speed cannot be negative;

Therefore. the speed is 25 km/hr.

Q50) The distance between Mumbai and Pune is 192 km. Travelling by the Deccan Queen, it takes 48 minutes less than another train. Calculate the speed of the Deccan Queen if the speeds of the two trains differ by 20 km/hr.

Ans. 50) Let the speed of the Deccan Queen be x km/hr.

According to the question:

Speed of another train = ( x – 20 ) km/hr

Therefore, $\frac{ 192 }{ x -20 } – \frac{ 192 }{ x } = \frac{ 48 }{ 60 }$

$\frac{ 4 }{ x -20 } – \frac{ 4 }{ x } = \frac{ 1 }{ 60 }$

$\frac{ 4x -4( x -20 ) }{ (x – 20)x } = \frac{ 1 }{ 60 }$

$\frac{ 4x – 4x + 80 ) }{ x^{ 2 } – 20x } = \frac{ 1 }{ 60 }$

$\frac{ 80 }{ x^{ 2 } – 20x } = \frac{ 1 }{ 60 }$

$x^{ 2 }$ – 20x = 4800

$x^{ 2 }$ – 20x – 4800 = 0

$x^{ 2 }$ – ( 80 – 60 ) x – 4800 = 0

$x^{ 2 }$ – 80x + 60x – 4800 = 0

x ( x – 80 ) + 60 ( x + 60 ) = 0

x = 80 or x = – 60

The value of x cannot be negative :

Therefore , the original speed of Deccan Queen is 80 km/hr.

Question 51: A motor boat whose speed in still water is 18 km/hr, takes 1 hour more to go 24 km upstream than to return to the same spot. Find the speed of the stream.

Solution 51:

Let the speed of the stream be x km/hr.

Given:

Speed of the boat = 18 km/hr

Speed downstream = (18 + x) km/hr

Speed upstream = (18 – x) km/hr

Therefore, $\frac {24} {\left (18 – x \right)} – \frac {24} {\left (18 + x \right)} = 1$

$\Rightarrow \frac {1} {\left ( 18 – x \right )} – \frac{1} { \left ( 18 + x \right )} = \frac {1} {24}$

$\Rightarrow \frac {18 + x – 18 + x} {\left ( 18 – x \right )\left ( 18 + x \right )} = \frac {1} {24}$

$\Rightarrow \frac {2x} {18^{2} – x^{2}} = \frac {1} {24}$

$\Rightarrow 324 – x^{2} = 48x$

$\Rightarrow 324 – x^{2} – 48x = 0$

$\Rightarrow x^{2} + 48x – 324 = 0$

$\Rightarrow x^{2} + \left ( 54 – 6 \right )x – 324 = 0$

$\Rightarrow x^{2} + 54x – 6x – 324 = 0$

$\Rightarrow x\left ( x + 54 \right ) – 6\left ( x + 54 \right ) = 0$

$\Rightarrow \left ( x + 54 \right ) \left ( x – 6 \right ) = 0$

$\Rightarrow x = -54\; or \; x = 6$

The value of x cannot be negative; therefore, the speed of the stream is 6 km/hr.

Question 52: The speed of the boat in still water is 8 km/hr. It can go 15 km upstream and 22 km downstream in 5 hours. Find the speed of the stream.

Solution:

Speed of the boat in still water = 8 km/hr

Let the speed of the stream be x km/hr

Speed upstream = (8 – x) km/hr

Speed downstream = (8 + x) km/hr

Time taken to go 22 km downstream = $\frac {22} { \left ( 8 + x \right ) } hr$

Time taken to go 15 km upstream = $\frac {15} { \left ( 8 – x \right )} hr$

According to the question:

$\Rightarrow \frac {22} { \left ( 8 + x \right )} + \frac {15} { \left ( 8 – x \right )} = 5$

$\Rightarrow \frac {22} { \left ( 8 + x \right )} + \frac {15} { \left ( 8 – x \right )} -5 = 0$

$\Rightarrow \frac { 22\left ( 8 – x \right ) + 15(8 + x) -5 \left ( 8 – x \right )\left ( 8 + x \right )} {\left ( 8 – x \right ) \left ( 8 + x \right) } = 0$

$\Rightarrow 176 – 22x + 120 + 15x – 320 + 5x^{2} = 0$

$\Rightarrow 5x^{2} – 7x – 24 = 0$

$\Rightarrow 5x^{2} – \left ( 15 – 8 \right )x – 24 = 0$

$\Rightarrow 5x^{2} – 15x + 8x – 24 = 0$

$\Rightarrow 5x\left ( x – 3 \right ) – 8\left ( x – 3 \right ) = 0$

$\Rightarrow \left ( x – 3 \right )\left ( 5x – 8 \right ) = 0$

$\Rightarrow x – 3 = 0 \; or \; 5x – 8 = 0$

$\Rightarrow x = 3 \; or \; x = \frac {8} {5}$

$\Rightarrow x = 3 \; \; \; \; \; \; \; \left ( Because \; speed\; cannot\; be\; a\; fraction \right )$

Speed of the stream = 3 km/hr

Question 53: A motor boat whose speed is 9 km/hr in still water, goes 15 downstream and comes back in a total time of 3 hours 45 minutes. Find the speed of the stream.

Solution:

Let the speed of the stream be x km/hr.

Downstream speed = (9 + x) km/hr

Upstream speed = (9 – x) km/hr

Distance covered downstream = distance covered upstream = 15 km

Total time taken = 3 hr 45 min = $\left ( 3 + \frac {45} {60} \right ) minutes = \frac {225} {60} minutes = \frac {15} {4} minutes$

Therefore, $\frac {15} { \left ( 9 + x \right )} + \frac {15} { \left ( 9 – x \right ) } = \frac {15} {4}$

$\Rightarrow \frac {1} { \left ( 9 + x \right ) } + \frac {1} { \left ( 9 – x \right ) } = \frac {1} {4}$

$\Rightarrow \frac { 9 – x + 9 + x} { \left ( 9 + x \right ) \left ( 9 – x \right ) } = \frac {1} {4}$

$\Rightarrow \frac {18} { 9^{2} – x^{2}} = \frac {1} {4}$

$\Rightarrow \frac {18} { 81 – x^{2} } = \frac {1} {4}$

$\Rightarrow 81 – x^{2} = 72$

$\Rightarrow 81 – x^{2} – 72 = 0$

$\Rightarrow – x^ {2} + 9 = 0$

$\Rightarrow x^ {2} = 9$

$\Rightarrow x = 3 \; or\; x = -3$

The value of x cannot be negative; therefore, the speed of the stream is 3 km/hr.

Question 54: 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 work in 12 days, find the time taken by B to finish the work.

Solution:

Let B takes x days to complete the work.

Therefore, A will take (x – 10) days.

Therefore, $\frac {1} {x} + \frac {1} {\left ( x – 10 \right )} = \frac {1} {12}$

$\Rightarrow \frac {\left ( x – 10 \right ) + x} {x \left ( x – 10 \right )} = \frac {1} {12}$

$\Rightarrow \frac { 2x – 10 } {x^{2} – 10x } = \frac {1} {12}$

$\Rightarrow x^ {2} – 10x = 12 \left ( 2x – 10 \right )$

$\Rightarrow x^ {2} – 10x = 24x – 120$

$\Rightarrow x^ {2} – 10x = 24x – 120$

$\Rightarrow x^ {2} – \left ( 30 + 4 \right ) x + 120 = 0$

$\Rightarrow x^ {2} – 30x – 4x + 120 = 0$

$\Rightarrow x \left ( x – 30 \right ) – 4 \left ( x – 30 \right ) = 0$

$\Rightarrow \left ( x – 30 \right ) \left ( x – 4 \right ) = 0$

$\Rightarrow x = 30 \; or \; x = 4$

Number of days to complete the work by

B cannot be less than by A; therefore, we get:

x = 30

Thus, B completes the work in 30 days.

Question 55: Two pipes running together can fill a cistern in 3 1/13 minutes. If one pipe take 3 minutes more than the other to fill it, find the time in which each pipe would fill the cistern.

Solution:

Let one pipe fill the cistern in x mins.

Therefore, the other pipe will fill the cistern in (x + 3) mins.

Time taken by both, running together, to fill the cistern = $3 \frac {1} {13} mins = \frac {40} {13} mins$

Part filled by one pipe in 1 min = 1/x

Part filled by the other pipes in 1 min = 1/ x + 3

Part filled by both pipes, running together, in 1 in = $\frac {1} {x} + \frac {1} {x + 3}$

Therefore, $\frac {1} {x} + \frac {1} {x + 3} = \frac {1} { \frac {40} {13}}$

$\Rightarrow \frac { \left ( x + 3 \right ) + x } {x\left ( x + 3 \right ) } = \frac {13} {40}$

$\Rightarrow \frac { 2x + 3 } { x^ {2} + 3x} = \frac {13} {40}$

$\Rightarrow 13x^{2} + 39x = 80x + 120$

$\Rightarrow 13x^{2} – 41x – 120 = 0$

$\Rightarrow 13x^{2} – \left ( 65 – 24 \right )x – 120 = 0$

$\Rightarrow 13x^{2} – 65x + 24x – 120 = 0$

$\Rightarrow 13x\left ( x – 5 \right ) + 24\left ( x – 5 \right ) = 0$

$\Rightarrow \left ( x – 5 \right ) \left ( 13x + 24 \right ) = 0$

x – 5 = 0 or 13x + 24 = 0

x = 5 or x = -24/13

x = 5 (because the speed cannot be in negative fraction)

Thus, one pipe will take 5 mins and the other will take {(5 + 3) = 8} mins to fill the cistern.

Question 56:

Two pipes running together can fill tank in $11\frac{1}{9}$ minutes. If one pipe takes 5 minutes more than the other to fill the tank separately, find the time in which each pipe would fill the tank separately.

Let the time taken by one pipe to fill the tank be x minutes.

Time taken by the other pipe to fill the tank = (x + 5) min

Suppose the volume of the tank be V.

Volume of the tank filled by one pipe in x minutes = V

Volume of the tank filled by one pipe in 1 minute = Vx

Volume of the tank filled by one pipe in 1119 minutes = Vx * 1119

=Vx*1009

Similarly,

Volume of the tank filled by the other pipe in 1119 minutes = Vx+5×1119

=Vx+5×1009

Now,

Volume of the tank filled by one pipe in 1119 minutes + Volume of the tank filled by the other pipe in 1119 minutes = V

.:Vlx + 1x+ 5x 1009 = V1x+1x+5=9100

=> x+5+xxx+5=9100

=> 2x+5×2+5x=9100

x = 20 (Time cannot be negative)

Time taken by one pipe to fill the tank = 20 min

Time taken by other pipe to fill the tank = (20+5) => 25min

Question 57:

Two water taps together can fill a tank in 6 hours. the tap of larger diameter takes 9 hours less than the smaller one to fill the tank separately. Find the time in which each tap can separately fill the tank.

Let the tap of smaller diameter fill the tank in x hours.

Time taken by the tap of larger diameter to fill the tank = (x – 9) h

Suppose the volume of the tank be V.

Volume of the tank filled by the tap of smaller diameter in x hours = V

Volume of the tank filled by the tap of smaller diameter in 1 hour = Vx

Volume of the tank filled by the tap of smaller diameter in 6 hours = Vx*6

Similarly,

Volume of the tank filled by the tap of larger diameter in 6 hours = Vx-9×6

Now,

Volume of the tank filled by the tap of smaller diameter in 6 hours + Volume of the tank filled by the tap of larger diameter in 6 hours = V

V1x + 1x – 9 x 6=V

=> 1x + 1x – 9 = 16

=> x – 9 + xxx – 9 =16

=> 2x-9×2 – 9x =16

=> 12x – 54 = x2- 9x

=> x2-21x+54=0

=> x2-18x-3x+54=0

=> x = 18 or x = 3

For x = 3, time taken by the tap of larger diameter to fill the tank is negative which is not possibe.

x = 18

Time taken by the tap of smaller diameter to fill the tank = 18 h

Time taken by the tap of larger diameter to fill the tank = (18 – 9) = 9 h

Hence, the time taken by the taps of smaller and larger diameter to fill the tank is 18 hours and 9 hours, respectively.

Question 58:

The length of a rectangle is twice the breadth and its area is 288$cm^{2}$. Find the dimensions of the rectangle.

Let the length and breadth of the rectangle be 2x m and x m, respectively.

According to the question:

2x * x = 288

=> 2$x^{2}$ = 144

=> x = 12 or x = -12

=> x = 12 (since, x cannot be negative)

Length = 2 x 12 = 24

Question 59:

The length of a rectangle field is three times its breadth. If the area of the field be 147 sq metres, find the length of the field.

Let the length and breadth of the rectangle be 3x m and x m, respectively.

According to the question:

3x * x = 3$x^{2}$ = 147

=>$x^{2}$ = 49

x = 7 or x = -7

x = 7 (x cannot be negative)

Length = 3 x 7 = 21m

Question 60:

The length of a hall is 3 meters more than its breadth. if the area of the hall is 238 sq metres, calculate its length and breadth

Let the breadth of the rectangular hall be x metre.

Therefore, the length of the rectangular hall will be (x + 3) metre

According to the question:

x(x + 3) = 238

=> $x^{2}$ + 3x = 238

=> $x^{2}$ + 3x – 238 = 0

=> $x^{2}$ + (17 – 14)x – 238 = 0

=> x(x + 17) – 14(x + 17) = (x + 17)

(x – 14) = 0

=> x = -17 or x = 14 But the value x cannot be negative.

Therefore, the breadth of the hall is 14.

Question 61:

The perimeter of a rectangular plot is 62m and its area is 228 sq metres. Find the dimensions of the plot.

Let the length and breadth of the rectangular plot be x and y meter, respectively.

Therefore, we have : Perimeter = 2(x + y) = 62

(i) and Area = xy = 228

=> y = 228x

Putting the value of y in (i),

we get => 2(x + 228x) = 62

=> x + 228x = 31

=> $x^{2}$ + 228 = 31x

=> $x^{2}$ – 31x + 228 = 0

=> $x^{2}$ – (19 + 12)x + 228 =0.

=> (x – 19) (x – 12) = 0

=> x = 19 or x = 12

if x=19 m, y=22819 = 12 m

Therefore, the length and breadth of the plot are 19m and 12m, respectively.

Question 62:

A rectangular field is 16m long and 10m wide. There is a path of uniform width all around it, having an area of 120 sq metres. Find the width of the path.

Let the width of the path be X metre.

Length of the field including the path = 16 + x + x = 16 + 2x

Breadth of the field including the path = 10 + x +x => 10+2x

Now,

(Area of the field including path) – (Area of the field excluding path)= Area of the path

=> (16 + 2x) (10 + 2x) – (16 x 10) = 120

=> 160 + 32x + 20x + 4×2 – 160 = 120

4 x 2 + 52x – 120 = 0

$x^{2}$ + 13x – 30 = 0

=> $x^{2}$ + (15 – 2)x + 30 = 0

$x^{2}$ + 15x – 2x + 30 = x(x + 15) – 2(x + 15) = (x – 2)

(x + 15) = 0

=> x – 2 = 0 or

x + 15 = 0

=> x = 2 or x = -15

=> x = 2 (Width cannot be negative)

Thus, the width of the path is 2m.

Question 63:

The sum of the areas of two squares is 640 square metres. If the difference in their perimeters be 64m, find the sides of the two squares.

Let the length of the side of the first and the second square be x and y, respectively.

According to the question : $x^{2}$+ $y^{2}$= 640 …(i)

Also,4x – 4y = 64

=> x – y = 16

=> x = 16 + y

Putting the value of x in (i), we get;

$x^{2}$ + $y^{2}$ = 640

=> $(16 + y)2 +y^{2}$ = 640

=> 2$y^{2}$ + 32y – 384 = 0

=> $y^{2}$+ 16y – 192 = 0

=>$y^{2}$ + (24 – 8)y – 192 =0

=> $y^{2}$ + 24y – 8y – 192 = 0

=> y(y + 24) – 8(y + 24) = (y + 24) (y – 8) = 0

=> y = -24 or y = 8

y = 8 (Side cannot be negative)

x = 16 + y => 16 + 8 = 24 m

Thus, the sides of the squares are 8m and 24m.

Question 64:

The length of a rectangle is thrice as long as the side of a square. The side of the square is 4cm more than the width of the rectangle. Their areas being equal, find their dimensions.

Let the breadth of rectangle be x cm.

According to the question:

Side of the square = (x + 4) cm Length of the rectangle = {3+(x+4)}cm

It is given that the areas of the rectangle and square are same.

=> 3(x + 4) * x =$(x + 4)^{2}$

=> 3×2 + 12x = $(x + 4)^{2}$

=> 3$x^{2}$ + 12x = $y^{2}$ + 8x + 16

=> $x^{2}$ + 4x – 16 = x2 + 2x – 8 = x2 + (4 – 2)x – 8 = $x^{2}$ + 4x – 2x – 8 =0

=> x(x + 4) – 2(x + 4) = (x + 4) (x – 2) = 0

=> x = -4 or x = 2

=> x = 2 (The value of x cannot be negative)

Thus, the breadth of the rectangle is 2cm and length is {3(2+4) = 18}cm. Also, the side of the square is 6 cm.

Question 65:

A farmer prepares a rectangle vegetable garden of area 180 sq metres. With 39 metres of barbed wire, he can fence the three sides of the garden, leaving one of the longer sides unfenced. Find the dimensions of the garden.

Let the length and breadth of the rectangular garden be x and y metre, respectively.

Given :

xy = 180 sq m

(i) and 2y + x = 39

=> x = 39 – 2y

Putting the value of x in (i), we get:

(39 – 2y)y = 180

=> 39y – 2$y^{2}$ = 180

=> 39y – 2$y^{2}$ – 180 = 0

=> 2$y^{2}$ – 39y + 180 = 0

=> 2$y^{2}$ – (24 + 15)y + 180 = 2$y^{2}$ – 24y – 15y + 180 = 2y(y – 12) – 15(y – 12) = (y – 12)

(2y-15)=0

=> y=12 or y = 152 = 7.5

If y = 12, x = 39 – 24 = 15

If y = 7.5, x = 39 – 15 = 24Thus, the length and breadth of the garden are (15m and 12m) or (24m and 7.5m), respectively.

Question 66:

The area of a right triangle is 600 sq centimeter. If the base of the triangle exceeds the altitude by 10cm, find the dimensions of the triangle.

Let the altitude of the triangle be x cm.

Therefore, the base of the triangle will be (x + 10) cm.

Area of triangle = 12 * (x + 10) = 600

=> x(x + 10) = 1200

=> $x^{2}$ + 10x – 1200 = 0

=> $x^{2}$ + (40 – 30)x – 1200 = 0 x2 + 40x – 30x – 1200 = 0

=> x(x + 40) – 30(x + 40) = (x + 40) (x – 30) = 0

=> x = -40 or x = 30

=> x = 30 [Altitude cannot be negative]

Thus, the altitude and base of the triangle are 30 cm and (30 + 10 = 40) cm, respectively.

Hypotenuse2 = Altitude2 + Base2

=> $Hypotenuse^{2}$ = $30^{2}$ + $40^{2}$

=> $Hypotenuse^{2}$ = 900 + 1600 = 2500

=> $Hypotenuse^{2}$ = $50^{2}$

Hypotenuse = 50

Thus, the dimensions of the triangle are:

Hypotenuse = 50cm.

Altitude = 30cm.

Base = 40 cm.

Question 67:

The area of a right-angled triangle is 86 sq metres. If the base is three times the altitude, find the base.

Let the altitude of the triangle be x m.

Therefore, the base will be 3x m.

Area of a triangle = 12 x Base x Altitude

12 * 3x * x = 96 (Area = 96 sq m)

=> $x^{2}$ = 64

=> x = ±8

The value of X cannot be negative.

Therefore, the altitude and base of the triangle are 8 m and (3 x 8 = 24 m), respectively.

Question 68:

The area of the right angled triangle is 165 sq metres. Determine its base and altitude if the latter exceeds the former by 7 metres.

Let the base be x m.

Therefore, the altitude will be x + 7 m.

Area of a triangle = 12 x Base x Altitude

12 * x * (x + 7) = 165

=> $x^{2}$ + 7x = 330

=> $x^{2}$ + 7x – 330 = 0

=> $x^{2}$ + (22 – 15)x – 330 = 0.1

=> x(x + 22) – 15(x + 22) = 0

(x + 22)(x – 15) = 0

=> x = -22 or x = 15

The value of x cannot be negative.

Therefore, the base is 15 m and the altitude is {(15 + 7) = 22 m}.

Question 69:

The hypotenuse of a right-angled triangle is 20 metres. If the difference between the lengths of the other sides be 4 metres, find the other sides.

Let one side of the right-angled triangle be x m and the other side be x + 4 m.

On applying Pythagoras theorem, we have:

202 = $(x+4)^{2}$ + $x^{2}$

=> 400 = $x^{2}$ + 8x + 16 + $x^{2}$

=> 2×2 + 8x – 384 = $x^{2}$ + 4x – 192 = 0

=> $x^{2}$ + (16 – 12)x – 192 = 0

=> $x^{2}$ + 16x – 12x – 192 = 0

=> x(x + 16) – 12(x + 16) = 0

=> (x + 16)(x – 12) = 0

=> x = -16 or x = 12

The value of x cannot be negative.

Therefore, the base is 12 m and the other side is {(12 + 4) = 16 m}.

Question 70:

The length of the hypotenuse of a right angled triangle exceeds the length of the base by 2cm and exceeds twice the length of the altitude by 1cm. Find the length of each side of the triangle.

Let the base and altitude of the right-angled triangle be x and y cm, respectively.

Therefore, the hypotenuse will be (x + 2) cm.

$(x+2)^{2}$= $y^{2}$ + $x^{2}$

(i)Again, the hypotenuse exceeds twice the length of the altitude by 1 cm.

h = (2y + 1)

=> x + 2 = 2y + 1

=> x =2y -1

Putting the value of x in (i), we get:

$(2y – 1 + 2)^{2}$= $y^{2}$ + $(2y-1)^{2}$

=> (2y + 1)2 = $y^{2}$ + 4$y^{2}$ – 4y + 4$y^{2}$ + 4y + 1 = 5$y^{2}$ – 4y + –$y^{2}$ + 8y = 0

=>$y^{2}$ – By = 0

=> y(y – 8) = 0

=> y = 8 x = 16 – 1 = 15

h = 16 + 1 = 17 cm

Thus, the base, altitude and hypotenuse of the triangle are 15 cm, 8 cm and 17 cm, respectively.

Question 71:

The hypotenuse of a right-angled triangle is 1 metre less than twice the shortest side. If the third side is 1 metre more than the shortest side, find the sides of the triangle.

Let the shortest side be x m.

Therefore, according to the question:

Hypotenuse = 2x – 1 m

Third side = x + 1 m

On applying Pythagoras theorem, we get:

$(2x – 1)^{2}$ = $(x+1)^{2}$ + $x^{2}$

=> 4×2 – 4x + 1 = $x^{2}$ + 2x + 1 + $x^{2}$

=> 2×2 – 6x = 2x(x – 3) = 0

=> x = 0 or x = 3

The length of the side cannot be 0; therefore, the shortest side is 3 m. Therefore,

Hypotenuse = 2 x 3 – 1 = 5 m

Third side = (3 + 1) = 4 m.

### Key Features of RS Aggarwal Class 10 Solutions Chapter 10 –  Quadratic Equations Ex 10G ( 10.7)

• It provides easier solutions for tough and difficult questions.
• It is considered as one of the best useful resource for revising the entire syllabus.
• Students are advised to practice RS Aggarwal maths solution to master in the subject of maths.
• It will improve your speed and accuracy.

#### Practise This Question

In Drosophila, the red-eye character is dominant over white eye character. When a homozygous red-eyed individual is crossed with a homozygous white-eyed individual, and individuals of F1 generation are intercrossed, 12 individuals are produced. White-eyed individuals of these will be: