RS Aggarwal Solutions Chapter 10 Quadratic Equation Exercise 10E Class 10 Maths
Chapter Name  RS Aggarwal Chapter 10 Quadratic Equation 
Book Name  RS Aggarwal Mathematics for Class 10 
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Related Study  NCERT Solutions for Class 10 Maths 
Exercise 10E Solutions
1. The sum of a natural number and its square is 156. Find the number.
Solution
Let the required natural number be x.
According to the given condition,
x + x^{2} = 156
⇒ x^{2} + x – 156 = 0
⇒ x^{2} + 13x – 12x  156 = 0
⇒ x(x + 13) – 12(x + 13) = 0
⇒ (x + 13)(x – 12) = 0
⇒ x + 13 = 0 or x – 12 = 0
⇒ x + 13 = 0 or x – 12 = 0
⇒ x =  13 or x = 12
∴ x = 12 (x cannot be negative)
Hence, the required natural number is 12.
2. The sum of natural number and its positive square root is 132. Find the number.
Solution
Let the required natural number be x.
According to the given condition,
x + √x = 132
Putting √x = y or x = y^{2}, we get
y^{2} + y = 132
⇒ y^{2} + y – 132 = 0
⇒ y^{2} + 12y – 11y – 132 = 0
⇒ y(y + 12) – 11(y + 12) = 0
⇒ (y + 12)(y – 11) = 0
⇒ y + 12 = 0 or y – 11 = 0
⇒ y =  12 or y = 11
∴ y = 11 (y cannot be negative)
Now,
√x = 11
⇒ x = (11)^{2} = 121
Hence, the required natural number is 121.
3. The sum of two natural number is 28 and their product is 192. Find the numbers.
Solution
Let the required number be x and (28 – x).
According to the given condition,
x(28 – x) = 192
⇒ 28x – x^{2} = 192
⇒ x^{2} – 28x + 192 = 0
⇒ x^{2 }– 16x – 12x + 192 = 0
⇒ x(x – 16) – 12(x – 16) = 0
⇒ (x – 12)(x – 16) = 0
⇒ x – 12 = 0 or x – 16 = 0
⇒ 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.
Solution
Let the required two consecutive positive integers be x and (x + 1).
According to the given condition,
x^{2} + (x + 1)^{2} = 365
⇒ x^{2} + x^{2} + 2x + 1 = 365
⇒ 2x^{2} + 2x – 364 = 0
⇒ x^{2} + x – 182 = 0
⇒ x^{2} + 14x – 13x – 182 = 0
⇒ x(x + 14) – 13(x + 14) = 0
⇒ (x + 14)(x – 13) = 0
⇒ x + 14 = 0 or x – 13 = 0
⇒ x =  14 or x = 13
∴ 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 to two consecutive positive odd numbers is 514. Find the numbers.
Solution
Let the two consecutive positive odd numbers be x and (x + 2).
According to the given condition,
x^{2} + (x + 2)^{2} = 514
⇒ x^{2} + x^{2 }+ 4x + 4 = 514
⇒ 2x^{2} + 4x – 510 = 0
⇒ x2 + 2x – 255 = 0
⇒ x2 + 17x – 15x – 255 = 0
⇒ x(x + 17) – 15(x + 17) = 0
⇒ (x + 17)(x – 15) = 0
⇒ x + 17 = 0 or x – 15 = 0
⇒ x =  17 or x = 15
∴ x = 15 (x is a positive odd number)
When x = 15,
x + 2 = 15 + 2 = 17
Hence, the required positive integers are 15 and 17.
6. The sum of the squares of two consecutive even numbers is 452. Find the numbers.
Solution
Let the two consecutive positive even numbers be x and (x + 2).
According to the given condition,
x^{2} + (x + 2)^{2} = 452
⇒ x^{2} + x^{2} + 4x + 4 = 452
⇒ 2x^{2} + 4x – 448 = 0
⇒ x^{2} + 2x – 224 = 0
⇒ x^{2 }+ 16x – 14x – 224 = 0
⇒ x(x + 16) – 14x – 224 = 0
⇒ x(x + 16) – 14(x + 16) = 0
⇒ (x + 16)(x – 14) = 0
⇒ x + 16 = 0 or x – 14 = 0
⇒ x =  16 or x = 14
∴ 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.
Solution
Let the two consecutive positive integers be x and (x + 1).
According to the given condition,
x(x + 1) = 306
⇒ x^{2} + x – 306 = 0
⇒ x^{2} + 18x – 17x – 306 = 0
⇒ x(x + 18) – 17(x + 18) = 0
⇒ (x + 18)(x – 17) = 0
⇒ x + 18 = 0 or x – 17 = 0
⇒ x = 18 or x = 17
∴ x = 17 (x is a positive integers)
When x = 17,
x + 1 = 17 + 1 = 18
Hence, the required integers are 17 and 18.
8. Two natural number differ by 3 and their product is 504. Find the numbers.
Solution
Let the required numbers be x and (x + 3).
According to the question:
x(x + 3) = 504
⇒ x^{2} + 3x – 504 = 0
⇒ x^{2} + (24 – 21)x – 504 = 0
⇒ x^{2} + 24x – 21x – 504 = 0
⇒ x(x + 24) – 21(x + 24) = 0
⇒ (x + 24)(x – 21) = 0
⇒ x + 24 = 0 or x – 21 = 0
⇒ 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 two consecutive multiples of 3 whose product is 648.
Solution
Let the required consecutive multiples of 3 be 3x and 3(x + 1).
According to the given condition,
3x × 3(x + 1) = 648
⇒ 9(x^{2} + x) = 648
⇒ x^{2} + x = 72
⇒ x^{2} + x – 72 = 0
⇒ x^{2} + 9x – 8x – 72 = 0
⇒ x(x + 9) – 8(x + 9) = 0
⇒ (x + 9)(x – 8) = 0
⇒ x + 9 = 0 or x – 8 = 0
⇒ x =  9 or x = 8
∴ x = 8 (Neglecting the negative value)
When x = 8,
3x = 3 × 8 = 24
3(x + 1) = 3 × (8 + 1)
= 3 × 9
= 27
Hence, the required multiples are 24 and 27
10. Find the two consecutive positive odd integer whose products 483.
Solution
Let the two consecutive positive off integers be x and (x + 2).
According to the given condition,
x(x + 2) = 483
⇒ x^{2} + 2x – 483 = 0
⇒ x^{2} + 23x – 21x – 483 = 0
⇒ x(x + 23) – 21(x + 23) = 0
⇒ (x + 23)(x + 21) = 0
⇒ x + 23 = 0 or x – 21 = 0
⇒ x =  23 or x = 21
∴ 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 the two consecutive positive even integers whose product is 288.
Solution
Let the two consecutive positive even integers be x and (x + 2).
According to the given condition,
x(x + 2) = 288
⇒ x^{2} + 2x – 288 = 0
⇒ x^{2} + 18x – 16x – 288 = 0
⇒ x(x + 18) – 16(x + 18) = 0
⇒ (x + 18)(x – 16) = 0
⇒ x + 18 = 0 or x – 16 = 0
⇒ x =  18 or x = 16
∴ 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 1/2. Find the numbers.
Solution
Let the required natural numbers be x and (9,  x).
According to the given condition,
1/x + 1/(9 – x) = 1/2
⇒ (9 – x + x)/x(9 – x) = 1/2
⇒ 9/(9x – x^{2}) = 1/2
⇒ 9x – x^{2} = 18
⇒ x^{2} – 9x + 18 = 0
⇒ x^{2} – 6x – 3x + 18 = 0
⇒ x(x – 6) – 3(x – 6) = 0
⇒ x – 3 = 0 or x – 6 = 0
⇒ 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 3/10. Find the numbers.
Solution
Let the required natural numbers be x and (15 – x).
According to the given condition,
1/x + 1/(15 – x) = 3/10
⇒ (15 – x + x)/x(15 – x) = 3/10
⇒ 15/(15x – x^{2}) = 3/10
⇒ 15x – x^{2} = 50
⇒ x^{2} – 15x + 50 = 0
⇒ x^{2} – 10x – 5x + 50 = 0
⇒ x(x – 10) – 5(x – 10) = 0
⇒ (x – 5)(x – 10) = 0
⇒ x – 5 = 0 or x – 10 =0
⇒ 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 number is 3 and the difference of their reciprocals is 3/28. Find the numbers.
Solution
Let the required natural numbers be x and (x + 3).
Now, x < x + 3
∴ 1/x > 1/(x + 3)
According to the given condition,
1/x – 1/(x + 3) = 3/28
⇒ (x + 3 – x)/x(x + 3) = 3/28
⇒ 3/(x^{2} + 3x) = 3/28
⇒ x^{2} + 3x = 28
⇒ x^{2} + 3x – 28 = 0
⇒ x^{2} + 7x – 4x – 28 = 0
⇒ x(x + 7) 4(x + 7) = 0
⇒ (x + 7)(x – 4) = 0
⇒ x + 7 = 0 or x – 4 = 0
⇒ 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 5/14. Find the numbers.
Solution
Let the required natural numbers be x and (x + 5).
Now, x < x + 5
∴ 1/x > 1/(x + 5)
According to the given condition,
1/x – 1/(x + 5) = 5/14
⇒ (x + 5 – x)/x(x + 5) = 5/14
⇒ 5/(x^{2 }+ 5x) = 5/14
⇒ x^{2} + 5x = 14
⇒ x^{2} + 5x – 14 = 0
⇒ x^{2} + 7x – 2x – 14 = 0
⇒ x(x + 7) – 2(x + 7) = 0
⇒ (x + 7)(x – 2) = 0
⇒ x + 7 = 0 or x – 2 = 0
⇒ x =  7 or x = 2
∴ x = 2 (7 is not a natural number)
When x = 2,
x + 5 = 2 + 5 = 7
16. The sum of the squares two consecutive multiples of 7 is 1225. Find the multiples.
Solution
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
⇒ 49x^{2 }+ 49(x^{2} + 2x + 1) = 1225
⇒ 49x^{2} + 49x^{2} + 98x + 49 = 1225
⇒ 98x^{2} + 98x – 1176 = 0
⇒ x^{2 }+ x – 12 = 0
⇒ x^{2 }+ 4x – 3x – 12 = 0
⇒ x(x + 4) – 3(x + 4) = 0
⇒ (x + 4)(x – 3) = 0
⇒ x+ 4 = 0 or x – 3 = 0
⇒ x = 4 or x = 3
∴ x = 3 (Neglecting the negative value)
When x = 3,
7x = 7 × 3 = 21
7(x + 1) = 7(3 + 1)
= 7 × 4
= 28.
17. The sum of natural number and its reciprocal is 65/8. Find the number.
Solution
Let the natural number be x.
According to the given condition,
x + 1/x = 65/8
⇒ (x^{2} + 1)/x = 65/8
⇒ 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 or x = 1/8
∴ x = 8 (x is a natural number)
Hence, the required number is 8.
18. Divide 57 into two parts whose product is 680.
Solution
Let the two parts be x and (57 – x).
According to the given condition,
x(57 – x) = 680
⇒ 57x – x^{2} = 680
⇒ x^{2} – 57x + 680 = 0
⇒ x^{2} – 40x – 17x + 680 = 0
⇒ x(x – 40) – 17(x – 40) = 0
⇒ (x – 40)(x – 17) = 0
⇒ x – 40 = 0 or x – 17 = 0
⇒ 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 reciprocal is 3/20.
Solution
Let the two parts be x and (27 – x).
According to the given condition,
1/x + 1/(27 – x) = 3/20
⇒ (27 – x + x)/x(27 – x) = 3/20
⇒ 27/(27x – x^{2}) = 3/20
⇒ 27x – x^{2} = 180
⇒ x^{2} – 27x + 180 = 0
⇒ x^{2} – 15x – 12x + 180 = 0
⇒ x(x – 15) – 12(x – 15) =or 0
⇒ (x – 12)(x – 15) = 0
⇒ x – 12 = 0 or x – 15 = 0
⇒ 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 the larger part exceeds the square of the smaller part by 164.
Solution
Let the larger and smaller parts be x and y, respectively.
According to the question:
x + y = 16 ...(i)
2x^{2} = y^{2} + 164 ...(ii)
From (i), we get;
x = 16 – y ....(iii)
From (ii) and (iii), we get:
2(16 – y)^{2} = y^{2 }+ 164
⇒ 2(256 – 32y + y^{2}) = y^{2 }+ 164
⇒ 512 – 64y + 2y^{2} = y^{2 }+ 164
⇒ 512  64y + 348 = 0
⇒ y2 – (58 + 6)y + 348 = 0
⇒ y2 – 58y  6y + 348 = 0
⇒ y(y – 58) – 6(y – 58) = 0
⇒ (y – 58)(y – 6) = 0
⇒ y – 58 = 0 or y – 6 = 0
⇒ y = 6 (∵ 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. Divide two natural numbers, the sum of whose squares is 25 times their sum and also equal to 50 times their difference.
Solution
Let the two natural numbers be x and y.
According to the question:
x^{2} + y^{2} = 25(x + y) ....(i)
x^{2} + y^{2} = 50(x – y) ....(ii)
From (i) and (ii), we get:
25(x + y) = 50(x – y)
⇒ x + y = 2(x – y)
⇒ x + y = 2x – 2y
⇒ y + 2y = 2x – x
⇒ 3y = x ....(iii)
From (ii) and (iii), we get:
(3y)^{2} + y^{2} = 50(3y – y)
⇒ 9y^{2} + y^{2} = 100y
⇒ 10y^{2} = 100y
⇒ y = 10
From (iii), we have:
3 × 10 = x
⇒ 30 = x
Hence, the two natural numbers are 30 and 10.
22. The difference of the squares of two natural numbers is 45. The square of the smaller number is four times the larger number. Find the numbers.
Solution
Let the greater number be x and the smaller number be y.
According to the question,
x^{2 }– y^{2} = 45 ...(i)
y^{2} = 4x ...(ii)
From (i) and (ii), we get:
x^{2} – 4x = 45
⇒ x^{2} – 4x – 45 = 0
⇒ x^{2} – (9 – 5)x – 45 = 0
⇒ x^{2} – 9x + 5x – 45 = 0
⇒ x(x – 9) + 5(x – 9) = 0
⇒ (x – 9)(x + 5) = 0
⇒ x – 9 = 0 or x + 5 = 0
⇒ x = 9 or x =  5
⇒ x = 9 (∵ x is a natural number)
Putting the value of x in equation (ii), we get:
y^{2} = 4 × 9
⇒ y^{2} = 36
⇒ y = 6
Hence, the two numbers are 9 and 16.
23. Three consecutive positive integers are such that the sum of the square of the first and product of the other two is 46. Find the integers.
Solution
Let the three consecutive positive integers be x, x + 1 and x + 2.
According to the given condition,
x^{2} + (x + 1)(x + 2)= 46
⇒ x^{2} + x^{2} + 3x + 2 = 46
⇒ 2x^{2} + 3x – 44 = 0
⇒ 2x^{2} – 11x – 8x – 44 = 0
⇒ x(2x + 11) – 4(2x + 11) = 0
⇒ (2x + 11)(x – 4) = 0
⇒ 2x + 11 = 0 or x – 4 = 0
⇒ x =  11/2 or x = 4
∴ 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 twodigit number is 4 times the sum of its digits and twice the product of digits. Find the number.
Solution
Let the digits at units and tens places be x and y, respectively.
Original number = 10y + x
According to the question:
10y + x = 4(x + y)
⇒ 10y + x = 4x + 4y
⇒ 3x – 6y = 0
⇒ 3x = 6y
⇒ x = 2y ...(i)
Also,
10y + x = 2xy
⇒ 10y + 2y = 2.2y.y [From (i)]
⇒ 12y = 4y^{2}
⇒ y = 3
From (i), we get:
x = 2 × 3 = 6
∴ Original number = 10 × 3 + 6 = 36
25. A twodigit number is such that the product of its digits is 14. If 45 is added to the number, the digit interchange their places. Find the number.
Answer:
Let the digits at units and tens places be x and y, respectively.
∴ xy = 14
⇒ y = 14/x ...(i)
According to the question:
(10y + x) + 45 = 10x + y
⇒ 9y – 9x =  45
⇒ y – x =  5 ...(ii)
From (i) and (ii), we get:
14/x – x =  5
⇒ (14 – x^{2})/x =  5
⇒ 14 – x^{2} =  5x
⇒ x^{2} – 5x – 14 = 0
⇒ x^{2} – (7 – 2)x – 14 = 0
⇒ x^{2} – 7x + 2x – 14 = 0
⇒ x(x – 7) + 2(x  7) = 0
⇒ (x – 7)( x + 2) = 0
⇒ x – 7 = 0 or x + 2 = 0
⇒ x = 7 or x =  2
⇒ x = 7 (∵ the digit cannot be negative)
Putting x = 7 in equation (i), we get:
y = 2
∴ Required number = 10 × 2 + 7 = 27
26. The denominator of a fraction is 3 more than its numerator. The sum of the fraction and its reciprocal is 2.9/10. Find the fraction.
Solution
Let the numerator be x.
∴ Denominator = x + 3
∴ Original number = x/(x + 3)
According to the question:
x/(x + 3) + 1/(x/(x + 3) = 2.9/10
⇒ x/(x + 3) + (x + 3)/x = 29/10
⇒ (x^{2} + (x + 3)^{2})/(x(x + 3)) = 29/10
⇒ (x^{2} + x^{2} + 6x + 9)/(x^{2} + 3x) = 29/10
⇒ (2x^{2} + 6x + 9)/(x^{2} + 3x) = 29/10
⇒ 29x^{2} + 87x = 20x^{2} + 60x + 90
⇒ 9x^{2 }+ 27x – 90 = 0
⇒ 9(x^{2} + 3x – 10) = 0
⇒ x^{2} + 3x – 10 = 0
⇒ x^{2} + 5x – 2x – 10 = 0
⇒ x(x + 5) – 2(x + 5) = 0
⇒ (x – 2)(x + 5) = 0
⇒ x – 2 = 0 or x + 5 = 0
⇒ x = 2 or x =  5 (rejected)
So, number x = 2
denominator x + 3 = 2 + 3 = 5
So, required fraction = 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 1/15. Find the fraction.
Solution
Let the denominator of the required fraction be x.
Numerator of the required fraction = x – 3
∴ Original fraction = (x – 3)/x
If 1 is added to the denominator, then the new fraction obtained is (x – 3)/(x + 1).
According to the given condition,
(x – 3)/(x + 1) = (x – 3)/x – 1/15
⇒ (x – 3)/x – (x – 3)/(x + 1) = 1/15
⇒ {(x – 3)(x + 1) – x(x – 3)}/{x(x + 1)} = 1/15
⇒ (x^{2} – 2x – 3 – x^{2} + 3x)/(x^{2} + x) = 1/15
⇒ (x – 3)/(x^{2} + x) = 1/15
⇒ x^{2} + x = 15x – 45
⇒ x^{2} – 14x + 45 = 0
⇒ x^{2 }– 9x – 5x + 45 = 0
⇒ x(x – 9) – 5(x – 9) = 0
⇒ (x – 5)(x – 9) = 0
⇒ x – 5 = 0 or x – 9 = 0
⇒ x = 5 or x = 9
When x = 5,
(x – 3)/x = (5 – 3)/5 = 2/5
When x = 9,
(x – 3)/x = (9 – 3)/9 = 6/9 = 2/3
(This fraction is neglected because this does not satisfies the given condition.)
Hence, the required fraction is 2/5.
28. The sum of a number and its reciprocal is 2.1/30. Find the number.
Solution
Let the number be x.
According to the given condition,
x + 1/x = 2.1/30
⇒ (x^{2} + 1)/x = 61/30
⇒ 30x^{2} + 30 = 61x
⇒ 30x^{2} – 61x + 30 = 0
⇒ 30x^{2} – 36x – 25x + 30 = 0
⇒ 6x(5x – 6) – 5(5x – 6) = 0
⇒ (5x – 6)(6x – 5) = 0
⇒ 5x – 6 = 0 or 6x – 5 = 0
⇒ x = 6/5 or x = 5/6
Hence, the required number is 5/6 or 6/5.
29. A teacher on attempting to arrange the students for mass drill in the form of 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.
Solution
Let there be x rows.
Then, the number o students in each row will also be x.
∴ Total number of students = (x^{2} + 24)
According to the question:
(x + 1)^{2 }– 25 = x^{2} + 24
⇒ x^{2 }+ 2x + 1 – 25 – x^{2} – 24 = 0
⇒ 2x – 48 = 0
⇒ 2x = 48
⇒ x = 24
∴ Total number of students = 24^{2} + 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.
Solution
Let the total number of students be x.
According to the question:
300/x – 300/(x + 10) = 1
⇒ (300(x + 10)  300x)/(x(x + 10)) = 1
⇒ (300x + 3000 – 300x)/(x^{2} + 10x) = 1
⇒ 3000 = x^{2} + 10x
⇒ x^{2} + 10x – 3000 = 0
⇒ x^{2} + (60 – 50)x = 3000 = 0
⇒ x^{2} + 60x – 50x – 3000 = 0
⇒ x(x + 60) – 50(x + 60) = 0
⇒ (x + 60)(x – 50) = 0
⇒ 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 marks less in English, the product of the marks would have been 360. Find his marks in two subjects separately.
Solution
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
⇒ (x + 3)/(40 – x  4) = 360 [From (i)]
⇒ (x + 3)(36 – x) = 360
⇒ 36 – x^{2} + 108 – 3x = 360
⇒ 33x – x^{2} – 252 = 0
⇒ x^{2} + 33x – 252 = 0
⇒ x^{2} – 33x – 252 = 0
⇒ x^{2} – (21 + 12)x + 252 = 0
⇒ x^{2} – 21x – 12x + 252 = 0
⇒ x(x – 21) – 12(x – 21) = 0
⇒ (x – 21)(x – 12) = 0
⇒ 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.
32. Some students planned a picnic. the total budget for food was ₹ 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?
Solution
Let x be the number of students who planned a picnic.
∴ Original cost of food for each member = ₹ 2000/x
Five students failed to attend the picnic, So, (x – 5) students attended the picnic.
∴ New cost of food for each member = ₹ 2000/(x – 5)
According of the given condition,
₹ 2000/(x – 5)  ₹ 2000/x = ₹ 20
⇒ (2000x – 2000x + 10000)/x(x – 5) = 20
⇒ 10000/(x^{2} – 5x) = 20
⇒ x^{2} – 5x = 500
⇒ x^{2} – 5x – 500 = 0
⇒ x^{2} – 25x + 20x – 500 = 0
⇒ x(x – 25) + 20(x – 25) = 0
⇒ (x – 25)(x + 20) = 0
⇒ x – 25 = 0 or x + 20 = 0
⇒ x = 25 or x =  20
∴ x = 25 (Number of student cannot be negative)
Number of students who attended the picnic = x – 5 = 23 – 5 = 20
Amount Paid by each student for the food = ₹ 2000/(25 – 5) = ₹ 2000/20
= ₹ 100
33. If the price of a book is reduced by ₹ 5, a person can buy 4 more books for ₹ 600. Find the original price of the book.
Solution
Let the original price of the book be ₹ x.
∴ Number of books bought at original price for ₹ 600 = 600/x
If the price of a book is reduced by ₹ 5, then the new price of the book is ₹(x – 5).
∴ Number of books bought at reduced price for ₹ 600 = 600/(x – 5)
According to the given condition,
600/(x – 5) – 600/x = 4
⇒ (600x – 600x + 3000)/x(x + 5) = 4
⇒ 3000/(x^{2 }– 5x) = 4
⇒ x^{2} – 5x = 750
⇒ x^{2} – 5x – 750 = 0
⇒ x^{2} – 30x + 25x – 750 = 0
⇒ x(x – 30) + 25(x – 30) = 0
⇒ x – 30 = 0 or x + 25 = 0
⇒ x = 30 or x =  25
∴ x = 30 (Price cannot be negative)
Hence, the original price of the book is ₹ 30.
34. A person on tour has ₹ 10800 for his expenses. If he extends his tour by 4 days, he has to cut down his daily expenses by ₹ 90. Find the original duration of the tour.
Solution
Let the original duration of the tour be x days.
∴ Original daily expenses = ₹ 10,800/x
If he extends his tour by 4 days, then his new daily expenses = ₹ 10,800/(x + 4)
According to the given condition,
₹ 10,800/x = ₹ 10,800/(x + 4) = ₹ 90
⇒ (10800x + 43200 – 10800x)/x(x + 4) = 90
⇒ 43200/(x^{2} + 4x) = 90
⇒ x^{2} + 4x = 480
⇒ x^{2} + 4x – 480 = 0
⇒ x^{2} + 24x – 20x – 480 = 0
⇒ x(x + 24) – 20 – 480 = 0
⇒ x(x + 24) – 20(x + 24) = 0
⇒ (x + 24)(x – 20) = 0
⇒ x + 24 = 0 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.
35. 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.
Solution
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
⇒ x^{2} – 21x + 72 = 180
⇒ x^{2} – 21x + 108 = 0
⇒ x^{2} – 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.
36. 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 ₹ 3 less. How many pens did he buy?
Solution
80/x  80/(x + 4) = 1
⇒ (80(x + 4) – 80x)/x(x + 4) = 1
⇒ (80 + 320 – 80x)/(x^{2 }+ 4x) = 1
⇒ 320 = x^{2} + 4x
⇒ x^{2 }+ 4x – 320 = 0
⇒ x^{2} + (20 – 16)x – 320 = 0
⇒ x^{2} + 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.
37. A dealer sells an article for ₹ 75 and gains as much per cent as the cost priced of the article. Find the cost price of the article.
Solution
Let the cost price of the article be x
∴ Gain percent = x%
According to the given condition,
₹ x + ₹(x/100 × x) = ₹ 75 (Cost price + Gain = Selling price)
⇒ (100x + x^{2})/100 = 75
⇒ x^{2} + 100x = 7500
⇒ x^{2} + 100x – 7500 = 0
⇒ x^{2} + 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)
Hence, the cost price of the article is ₹ 50.
38. One year ago, 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.
Solution
Let the present age of the son be x years.
∴ Present age of the man = x^{2} years
One year ago,
Age of the son = (x – 1) years
Age of the man = (x^{2} – 1) years
According to the given condition,
Age of the man = 8 × Age of son
∴ x^{2} – 1 = 8(x – 1)
⇒ x^{2 }– 1 = 8x – 8
⇒ x^{2} – 8x + 7 = 0
⇒ x^{2} – 7x – x + 7 = 0
⇒ x(x – 7) – 1(x – 7) = 0
⇒ x = 1 or x = 7
∴ x = 7 (Man’s age cannot be 1 year)
Present age of the son = 7 years
Present age of the man = 7^{2} years = 49 years.
39. The sum of reciprocals of Men’s ages (in years) 3 years ago and 5 years hence 1/3. Find her present ages.
Solution
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,
1/(x – 3) + 1/(x + 5) = 1/3
⇒ (x + 5 + x – 3)/{(x – 3)(x + 5)} = 1/3
⇒ (2x + 2)/(x^{2} + 2x – 15) = 1/3
⇒ x^{2} + 2x – 15 = 6x + 6
⇒ x^{2} – 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.
40. 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.
Solution
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 – x^{2} = 126
⇒ x^{2} – (18 – 7)x + 126 = 0
⇒ x^{2} – 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 ages 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.
41. The product of Tanvy ’s age (in years) 5 years ago and her age is 8 years later is 30. Find her present age.
Solution
Let the present age of Meena be x years.
According to the question:
(x – 5)(x + 8) = 30
⇒ x^{2} + 3x – 40 = 30
⇒ x^{2} + 3x – 70 = 0
⇒ x^{2} + (10 – 7)x – 70 = 0
⇒ x^{2} + 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 Meena is 7 years.
42. Two years ago, 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.
Solution
Let son’s age 2 years ago be x years, Then,
Man’s age 2 years ago = 3x^{2} years
∴ Sons present age = (x + 2) years
Man’s present age = (3x^{2} + 2) years
In three years time,
Son’s age = (x + 2 + 3) years = (x + 5) years
Man’s age = (3x2 + 2 + 3) years = (3x^{2} + 5) years
According to the given condition,
∴ 3x^{2} + 5 = 4(x + 5)
⇒ 3x^{2} + 5 = 4x + 20
⇒ 3x^{2} – 4x – 15 = 0
⇒ 3x^{2} – 9x + 5x – 15 = 0
⇒ 3x(x – 3) + 5(x – 3) = 0
⇒ (x – 3)((3x + 5) = 0
⇒ x – 3 = 0 or 3x + 5 = 0
⇒ x = 3 or x =  5/3
∴ x = 3 (Age cannot be negative)
Son’s present age = (x + 2) years = (3 + 2) years = 5 years
Man’s present age = (3x^{2} + 2) years = (3 × 9 + 2) = 29 years
43. A truck covers a distance of 150 km at a certain average speed and then covers another 200 km at an average speed which is 20 km per hour more than the first speed. If the truck covers the total distance in 5 hours, find the first speed of the truck.
Solution
Let the first speed of the truck be x km/h.
∴ Time taken to cover 150 km = (150/x).h (Time = distance/Speed)
New speed of the truck = (x + 20)km/h
∴ Time taken to cover 200 km = 200/(x + 20).h
According to the given condition,
Time to cover 150 km + Time taken to cover 200 km = 5 h
∴ 150/x + 200/(x + 20) = 5
⇒ (150x + 3000 + 200x)/x(x + 20) = 5
⇒ 350x + 3000 = 5(x^{2} + 20x)
⇒ 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
∴ x = 60 (Speed cannot be negative)
Hence, the first speed of the truck is 60 km/h.
44. While boarding an aeroplane, a passengers got hurt. The pilot showing promptness and concern, made arrangements to hospitalize 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?
Solution
Let the original speed of the plane be x km/h.
∴ Actual speed of the plane = (x + 100) km/h
Distance of the journey = 1500 km
Time taken to reach the destination at original speed = (1500/x).h (Time = Distance/speed)
Time taken to reach the destination at actual speed = 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
∴ 1500/x = 1500/(x + 100) + 1/2 (30 min = 30/60.h = 1/2h)
⇒ 1500/x  1500/(x + 100) = 1/2
⇒ (1500x + 150000 – 1500x)/x(x + 100) = 1/2
⇒ 150000/(x^{2 }+ 100x) = 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 = 0 or x – 500 = 0
⇒ x =  600 or x = 500
∴ x = 500 (Speed cannot be negative)
Hence, the original speed of the plane is 500 km/h.
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.
45. 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.
Solution
Let the usual speed of the train be x km/h.
Total distance to be covered = 480 km
Time taken by the train to cover the distance at usual speed = 480/x h (Time = Distance/Speed)
Time taken by the train to cover the distance at reduced speed = 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
∴ 480/(x – 8) = 480/x + 3
⇒ 480/(x – 8) = 480/x = 3
⇒ (480x – 480x + 3840)/(x(x – 8) = 3
⇒ 3840/(x^{2 }– 8x) = 3
⇒ x^{2} – 8x = 1280
⇒ x^{2} – 8x – 1280 = 0
⇒ x(x – 40) + 32(x – 40) = 0
⇒ (x – 40) (x + 32) = 0
⇒ x – 40 = 0 or x + 32 = 0
⇒ x = 40 or x =  32
∴ x = 40 (Speed cannot be negative)
Hence, the usual speed of the train is 40 km/h.
46. A train travels at a certain average speed for a distanced 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 ?
Solution
Let the first speed of the train be x km/h.
Time taken to cover 54 km = (54/x).h (Time = Distance/Speed)
New speed of the train = (x + 6) km/h
∴ Time taken to cover 63 km = 63/(x + 6).h
According to the given condition,
Time taken to cover 54 km + Time taken to cover 63 km = 3 h
∴ 54/x + 63/(x + 6) = 3
⇒ (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 – 324 = 0
⇒ x^{2} – 33x – 108 = 0
⇒ x^{2 }– 36x + 3x – 108 = 0
⇒ x(x – 36) + 3(x – 36) = 0
⇒ (x – 36)(x + 3) = 0
⇒ x  36 = 0 or x + 3 = 0
⇒ x = 36 or x =  3
∴ x = 36 (Speed cannot be negative)
Hence, the first speed of the train is 36 km/h.
47. A train travels 180 km at a uniform speed. If the speed had been 9 km/hr more, it would have taken 1 hour less for the same journey. Find the speed of the train.
Solution
36 km/hr
48. A train travels 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.
Solution
Let the original speed of the train be km/hr.
According to the question:
90/x – 90/(x + 15) = 1/2
⇒ {90(x + 15) – 90x}/x(x + 15) = 1/2
⇒ (90x + 1350 – 90x)/(x^{2} + 15x) = 1/2
⇒ 1350/(x^{2} + 15x) = 1/2
⇒ 2700 = x^{2} + 15x
⇒ x^{2} + (60 – 45)x – 2700 = 0
⇒ x^{2} + 60x – 45x – 2700 = 0
⇒ x(x + 60) – 45x(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.
49. 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.
Solution
Let the usual speed x km/hr.
According to the question:
300/x – 300/(x + 5) = 2
⇒ {300(x + 5) – 300x}/x(x + 5) = 2
⇒ (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.
50. 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 two train differ by 20 km/hr.
Solution
Let the speed of the Deccan Queen be x km/hr.
According to the question:
Speed of another train = (x – 20)km/hr
∴ 192/(x – 20) – 192/x = 48/60
⇒ 4/(x – 20) – 4/x = 1/60
⇒ 4x – 4(x – 20)/(x – 20)x = 1/60
⇒ (4x – 4x + 80)/(x^{2} – 20x) = 1/60
⇒ (4x – 4x + 80)/(x^{2} – 20x) = 1/60
⇒ 80/(x^{2} – 20x) = 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 – 80) = 0
⇒ (x – 80)(x + 60) = 0
⇒ x = 80 or x =  60
The value of x cannot be negative; therefore, the original speed of Deccan Queen 180 km/hr.
51. A motor boat whose speed in still water is 178 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
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
∴ 24/(18 – x) – 24/(18 – x) = 1
⇒ 1/(18 – x) – 1/(18 + x) = 1/24
⇒ (18 + x – 18 + x)/(18 – x)(18 + x) = 1/24
⇒ 2x/(18^{2} – x^{2}) = 1/24
⇒ 324 – x^{2} = 48x
⇒ 324 – x^{2} – 48x = 0
⇒ x^{2} + 48x – 324 = 0
⇒ x^{2} + (54 – 6)x – 324 = 0
⇒ x^{2} + 54x – 6x – 324 = 0
⇒ x(x + 54) – 6(x + 54) = 0
⇒ (x + 54)(x – 6) = 0
⇒ x =  54 or x = 6
The value of x cannot be negative; therefore, the speed of the stream is 6 km/hr.
52. The speed of a boat in still water is 8 km/hr. it can go 15 km upstream and 22 km downstream is 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 = 22/(8 + x)hr
Time taken to go 15 km upstream = 15/(8 – x) hr
According to the question:
⇒ 22/(8 + x) + 15/(8 – x) = 5
⇒ 22/(8 + x) + 15/(8 – x) – 5 = 0
⇒ {22(8 – x) + 15(8 + x) – 5(8 – x)(8 + x)}/{(8 – x)(8 + x)} = 0
⇒ 176 – 22x + 120 + 15x – 320 + 5x^{2} = 0
⇒ 5x^{2} – 7x – 24= 0
⇒ 5x^{2} – (15 – 8)x – 24 = 0
⇒ 5x^{2} – 15x + 8x – 24 = 0
⇒ 5x(x – 3) – 8(x – 3) = 0
⇒ (x – 3)(5x – 8) = 0
⇒ x – 3 = 0 or 5x – 8 = 0
⇒ x = 3 or x = 8/5
⇒ x = 3 (∵ Speed cannot be a fraction)
∴ Speed of the stream = 3 km/hr
53. A motorboat whose speed is 9 km/hr in still water, goes 15 km 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 hours 45 minutes = (3 + 45/60) minutes = 225/60 minutes = 15/4 minutes
∴ 15/(9 + x) + 15/(9 – x) = 15/4
⇒ 1/(9 + x) + 1/(9 – x) = 1/4
⇒ (9 – x + 9 + x)/(9 + x)(9 – x) = 1/4
⇒ 18/(9^{2} – x^{2}) = 1/4
⇒ 18/(81 – x^{2}) = 1/4
⇒ 81 – x^{2} = 72
⇒ 81 – x^{2} – 72 = 0
⇒  x^{2} + 9 = 0
⇒ x^{2} = 9
⇒ x = 3 or x =  3
The value of x cannot be negative; therefore, the speed of the stream is 3 km/hr.
54. A take 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.
∴ 1/x + 1/(x – 10) = 1/12
⇒ {(x – 10) + x}/{x(x – 10)} = 1/12
⇒ (2x – 10)/(x^{2} – 10x) = 1/12
⇒ x^{2} – 10x = 24x – 120
⇒ x^{2} – 34x + 120 = 0
⇒ x^{2} – (30 + 4)x + 120 = 0
⇒ x^{2} – 30x – 4x + 120 = 0
⇒ x(x – 30) – 4(x – 30) = 0
⇒ (x – 30)(x – 4) = 0
⇒ x = 30 or x = 4
Number of days to complete the work by B cannot be less than that by A; therefore, we get: x = 30
Thus, B completes the work in 30 days.
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 fills 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 = 3.1/13 mins = 40/13 mins
Part filled by one pipe in 1 min = 1/x
Part filled by the other pipe in 1 min = 1/(x + 3)
Part filled by the both pipes, running together, in 1 min = 1/x + 1/(x + 3)
∴ 1/x + 1/(x + 3) = (1/40)/13
⇒ {(x + 3) + x}/{x(x + 3)} = 13/40
⇒ (2x + 3)/(x^{2} + 3a) = 13/40
⇒ 13x^{2} + 39x = 80x + 120
⇒ 13x^{2} – 41x – 120 = 0
⇒ 13x^{2} – (65 – 24)x – 120 = 0
⇒ 13x^{2} – 65x + 24x – 120 = 0
⇒ 13x(x – 5) + 24(x – 5) = 0
⇒ (x – 5)(13x + 24) = 0
⇒ x – 5 = 0 or 13x + 24 = 0
⇒ x = 5 or x = 24/13
⇒ x = 5 (∵ Speed cannot be a negative fraction)
Thus, one pipe will take 5 mins and other will take {(5 + 3) = 8} mins to fill the cistern.
56. Two pipes running together can fill a tank in 11.1/9 minutes. If on 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.
Solution
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 = V/x
⇒ Volume of the tank filled by one pipe in 11.1/9 minutes = V/x × 11.1/9 = V/x × 100/9
Similarly,
Volume of the tank filled by the other pipe in 11.1/9 minutes = V/(x + 5) × 11.1/9
= V/(x + 5) × 100/9
Now,
Volume of the tank filled by one pipe in 11.1/9 minutes + Volume of the tank filled by the other pipe in 11.1/9 minutes = V
∴ V(1/x + 1/(x + 5)) × 100 = V
⇒ 1/x + 1/(x + 5) = 9/100
⇒ (x + 5 + x)/(x(x + 5)) = 9/100
⇒ (2x + 5)/(x^{2} + 5x) = 9/100
⇒ 200x + 500 = 9x^{2} + 45x
⇒ 9x^{2} – 155x – 500 = 0
⇒ 9x^{2} – 180x + 25x – 500 = 0
⇒ 9x(x – 20) + 25(x – 20) = 0
⇒ (x – 20)(9x + 25) = 0
⇒ x – 20 = 0 or 9x + 25 = 0
⇒ x = 20 or x =  (25/9)
∴ 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) = 25 min
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 which each tap can separately fill the tank.
Solution
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 = V/x
⇒ Volume of the tank filled by the tap of smaller diameter in 6 hours = V/x × 6
Similarly,
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
∴ V (1/x + 1/(x – 9) × 6 = V
⇒ 1/x + 1/(x – 9) = 1/6
⇒ (x – 9 + x)/x(x – 9) = 1/6
⇒ (2x – 9)/(x^{2} – 9x) = 1/6
⇒ 12x – 54 = x^{2} – 9x
⇒ x^{2} – 21x + 54 = 0
⇒ x^{2} – 81x – 3x + 54 = 0
⇒ x(x – 18) – 3(x – 18) = 0
⇒ (x – 18) = 0 or x – 3 = 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 possible.
∴ x = 18
Time taken by the tap of smaller diameter to fill the tank = 18h
Time taken by the tap of larger diameter to fill the tank = (18 – 9) = 9h
Hence, the time taken by the taps of smaller and larger diameter to fill the tank is 18 hours and 9 hours, respectively.
58. The length of rectangle is twice its breadth and its areas is 288 cm and 288 cm2. Find the dimensions of the rectangle.
Solution
Let the length and breadth of the rectangle be 2x m and x m, respectively.
According to the question:
2x × x = 288
⇒ 2x^{2} = 288
⇒ x^{2} = 144
⇒ x = 12 or x =  12
⇒ x = 12 (∵ x cannot be negative)
∴ Length = 2 × 12 = 24m
Breadth = 12 m
59. The length of a rectangular field is three times its breadth. If the area of the filed be 147 sq. meters, find the length of the field.
Solution
Let the length and breadth of the rectangle be 3x m and x m, respectively.
According to the question:
3x × x = 147
⇒ 3x^{2} = 147
⇒ x^{2} = 49
⇒ x = 7 or x =  7
⇒ x = 7 (∵ x cannot be negative)
∴ Length = 3 × 7
= 21 m
Breadth = 7 m
60. The length of a hall is 3 meter more than its breadth. If the area of the hall is 238 sq metres, calculate its length and breadth.
Solution
Let the breadth of the rectangular hall be x meter.
Therefore, the length of the rectangular hall will be (x + 3) meter.
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^{2 }+ 17x – 14x – 238 = 0
⇒ x(x + 17) – 14(x + 17) = 0
⇒ (x + 17)(x – 14) = 0
⇒ x =  17 or x = 14
But the value of x cannot be negative.
Therefore, the breadth of the hall is 14 meter and the length is 17 meter.
61. The perimeter of a rectangular plot is 62 m and its area is 288 sq meters. Find the dimension of the plot.
Solution
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 = 228/x
Putting the value of y in (i), we get
⇒ 2(x + 228/x) = 62
⇒ x + 228/x = 31
⇒ (x^{2} + 228)/x = 31
⇒ x^{2} + 228 = 31x
⇒ x^{2} – 31x + 228 = 0
⇒ x^{2} – (19 + 12)x + 228 = 0
⇒ x^{2} – 19x – 12x + 228 = 0
⇒ x(x – 19) – 12(x – 19) = 0
⇒ (x – 19)(x – 12) = 0
⇒ x = 19 or x = 12
If x = 19 m, y = 228/19 = 12 m
Therefore, the length and breadth of the plot are 19 m and 12 m respectively.
62. A rectangular field in 16 m long and 10 m wide. There is a path of uniform width all around it, having an area of 120 m^{2}. Find the width of the path
Solution
Let the width of the path be x m.
∴ 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 × 10) = 120
⇒ 160 + 32x + 20x + 4x^{2} – 160 = 120
⇒ 4x^{2} + 52x – 120 = 0
⇒ x^{2 }+ 13x – 30 = 0
⇒ x2 + (15 – 2)x + 30 = 0
⇒ x^{2} + 15x – 2x + 30 = 0
⇒ x(x + 15) – 2(x + 15) = 0
⇒ (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 2 m.
63. The sum of the areas of two squares is 640 m^{2}. If the difference in their perimeter be 64 m, find the sides of the two square.
Solution
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
⇒ 256 + 32y + y^{2 }+ y^{2} = 640
⇒ 2y^{2} + 32y – 384 = 0
⇒ x^{2} + 16y – 192 = 0
⇒ y^{2} + (24 – 8)y – 192 = 0
⇒ y^{2} + 24y – 8y – 192 = 0
⇒ y(y + 24) – 8(y + 24) = 0
⇒ (y + 24) (y – 8) = 0
⇒ y =  24 oy = 8
∴ y = 8 (∵ Side cannot be negative)
∴ x = 16 + y = 16 + 8 = 24 m
Thus, the sides of the square are 8 m and 24 m.
64. The length of a rectangle is thrice as long as the side of a square. The side of the square is 4 cm more than the width of the rectangle. Their areas being equal, find the dimensions.
Solution
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}
⇒ 3x^{2} + 12x = (x + 4)^{2}
⇒ 3x^{2} + 12x = x^{2} + 8x + 16
⇒ 2x^{2} + 4x – 16 = 0
⇒ x^{2} + 2x – 8 = 0
⇒ x^{2} + (4 – 2)x – 8 = 0
⇒ x^{2} + 4x – 2x – 8 = 0
⇒ x(x + 4)  2(x + 4) = 0
⇒ (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 2 cm and length is {3(2 + 4) = 18} cm.
Also, the side of the square is 6 cm.
65. A farmer prepares rectangular garden of area 180 sq meters. With 39 meters 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.
Solution
Let the length and breadth of the rectangular garden be x and y meter, 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
⇒ 39 – 2y^{2} = 180
⇒ 39y – 2y^{2} – 180 = 0
⇒ 2y^{2} – 39y + 180 = 0
⇒ 2y^{2} – (24 + 15)y + 180 = 0
⇒ 2y^{2} – 24y – 15y + 180 = 0
⇒ 2y(y – 12) – 15(y – 12) = 0
⇒ (y – 12)(2y – 15) = 0
⇒ y = 12 or y = 15/2 = 7.5
If y = 12, x = 39 – 24 = 15
If y = 7.5, x = 39 – 15 = 24
Thus, the length and breadth of the garden are (15 m and 12 m) or (24 m and 7.5 m), respectively.
66. The area of a right triangle is 600 cm^{2}. If the base of the triangle exceeds the altitude by 10 cm, find the dimensions of the triangle.
Solution
Let the altitude of the triangle be x cm.
Therefore, the base of the triangle will be (x + 10) cm
Are a of triangle = 1/2.x (x +10) = 600
⇒ (x + 10) = 1200
⇒ x^{2} + 10x – 1200 = 0
⇒ x^{2} + (40 – 30)x – 1200 = 0
⇒ x^{2} + 40x – 30x – 1200 = 0
⇒ x(x + 40) – 30(x + 40) = 0
⇒ (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) cm, respectively.
(Hypotenuse)^{2} = (Altitude)^{2} + (base)^{2}
⇒ (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 = 50 cm
Altitude = 30 cm
Base = 40 cm
67. The area of rightangled triangle is 96 sq meters. If the base is three times the altitude, find the base.
Solution
Let the altitude of the triangle be x m.
Therefore, the base will be 3x m.
Area of a triangle = 1/2 × Base × Altitude
∴ 1/2 × 3x × x = 96 (∵ Area = 96 sq m)
⇒ x^{2}/2 = 32
⇒ 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 × 8 = 24 m), respectively.
68. The area of rightangled triangle is 165 sq meters. Determine its base and altitude if the latter exceeds the former by 7 metres.
Solution
Let the base be x m.
Therefore, the altitude will be (x + 7) m.
Area of a triangle = 1/2 × base × Altitude
∴ 1/2 × x × (x + 7) = 165
⇒ x^{2} + 7x = 330
⇒ x^{2} + 7x – 330 = 0
⇒ x^{2} + (22 – 15)x – 330 = 0
⇒ x^{2} + 22x – 15x – 330 = 0
⇒ 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}
69. The hypotenuse of a rightangled triangle is 20 meters. If the difference between the lengths of the other sides be 4 metres, find the other sides.
Solution
Let one side of the rightangled triangle be x m and the other side be (x + 4) m.
On applying Pythagoras theorem, we have:
20^{2} = (x + 4)^{2} + x^{2}
⇒ 400 = x^{2} + 8x + 16 + x^{2}
⇒ 2x^{2} + 8x – 384 = 0
⇒ x^{2} + 4x – 192 = 0
⇒ x^{2} + (16 – 12)x – 192 = 0
⇒ x^{2} + 16x – 12x – 192 = 0
⇒ x^{2}(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}.
70. The length of the hypotenuse of a rightangled triangle exceeds the length of the base by 2 cm and exceeds twice the length of the altitude by 1 cm. Find the length of each side of the triangle.
Solution
Let the base and altitude of the rightangled 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} + 4y^{2} – 4y + 1
⇒ 4y^{2 }+ 4y + 1 = 5y^{2} – 4y + 1
⇒  y^{2} + 8y = 0
⇒ y^{2} – 8y = 0
⇒ y(y – 8) = 0
⇒ y = 8 cm
∴ x = 16 – 1 = 15 cm
∴ h = 16 + 1 = 17 cm
Thus, the base, altitude and hypotenuse of the triangle are 15 cm, 8 cm and 17 cm, respectively.
71. The hypotenuse of a rightangled triangle is 1 meter less than twice the shortest side. If the third side 1 meter more than the shortest side, find the side, find the sides of the triangle.
Solution
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}
⇒ 4x^{2} – 4x +1 = x^{2} + 2x +1 + x^{2}
⇒ 2x^{2} – 6x = 0
⇒ 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 = (2×3 – 1) = 5 m
Third side = (3 + 1) = 4 m