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ICSE Class 10 Maths Question Paper 2014 Solved
Time Allowed: 2 1/2 hours
Max. Marks: 80
Answers to this paper must be written on the paper provided separately. You will not be allowed to write during the first 15 minutes. This time is to be spent in reading the question paper. The time given at the head of this Paper is the time allowed for writing the answers.
Attempt all questions from Section A and any four questions from Section B. All working, including rough work, must be clearly shown and must be done on the same sheet as the rest of the answer. Omission of essential working will result in loss of marks. The intended marks for questions or parts of questions are given in brackets [ ]
Mathematical tables are provided.
Section-A (40 Marks)
(Attempt all questions from this section)
Question 1.
(a) Ranbir borrows ₹ 20,000 at 12% per annum compound interest. If he repays ₹ 8400 at the end of the first year and ₹ 9680 at the end of the second year, find the amount of loan outstanding at the beginning of the third year. [3]
(b) Find the values of x, which satisfy the inequation
-2\(\frac{5}{6}\) < \(\frac{1}{2}\) – \(\frac{2 x}{3}\) ≤ 2, x ∈ W. Graph the solution set on the number line. [3]
(c) A die has 6 faces marked by the given numbers as shown below:
The die is thrown once. What is the probability of getting
(i) a positive integer.
(ii) an integer greater than -3.
(iii) the smallest integer.
Solution :
(a) Amount borrowed (Loan) = ₹ 20000
Rate of interest = 12% p.a.
Interest due after one year = ₹ \(\frac{20000 \times 12 \times 1}{100}\)
= ₹ 2400
Amount of loan outstanding after one year
= ₹ (20000 + 2400)
= ₹ 22400
Total payment at the end of first year = ₹ 8400
Principal for second year = ₹ (22400 – 8400)
= ₹ 14000
Interest due after second year = ₹ \(\frac{14000 \times 12 \times 1}{100}\) = ₹ 1680
Amount of loan outstanding after second year
= ₹ (14000 + 1680) = ₹ 15680
Total payment at the end of second year = ₹ 9680
Thus, amount of loan outstanding at the beginning of the third year = ₹ (15680 – 9680) = ₹ 6000.
(b) Given inequation is -2\(\frac{5}{6}\) < \(\frac{1}{2}\) – \(\frac{2 x}{3}\) ≤ 2, x ∈ W
⇒ –\(\frac{17}{6}\) < \(\frac{1}{2}\) – \(\frac{2 x}{3}\) ≤ 2
⇒ –\(\frac{17}{6}\) – \(\frac{1}{2}\) < –\(\frac{2 x}{3}\) ≤ 2 – \(\frac{1}{2}\)
⇒ –\(\frac{20}{6}\) < –\(\frac{2 x}{3}\) ≤ \(\frac{3}{2}\)
⇒ -20 < -4x ≤ 9 ⇒ 5 > x ≥ –\(\frac{9}{4}\)
Hence, the required solution set on number line is :
(c)
(i) Here, we have only three positive integers (1, 2, 3)
∴ Probability (getting a positive integer) = \(\frac{3}{6}\) = \(\frac{1}{2}\)
(ii) Now, we have five integers greater than – 3 i.e., -2, -1, 1, 2, 3.
∴ Probability (getting a integer greater than – 3) = \(\frac{5}{6}\)
(iii) Here, we have only one smallest integer i.e., -3
∴ Probability (getting smallest integer) = \(\frac{1}{6}\)
Question 2.
(a) Find x, y if \(\left[\begin{array}{cc}
-2 & 0 \\
3 & 1
\end{array}\right]\left[\begin{array}{c}
-1 \\
2 x
\end{array}\right]\) + 3\(\left[\begin{array}{c}
-2 \\
1
\end{array}\right]\) = 2\(\left[\begin{array}{l}
y \\
3
\end{array}\right]\) [3]
(b) Shahrukh opened a Recurring Deposit Account in a bank and deposited ₹ 800 per month for 1 1/2 years. If he received ₹ 15,084 at the time of maturity, find the rate of interest per annum. [3]
(c) Calculate the ratio in which the line joining A(-4, 2) and B (3, 6) is divided by point P (x, 3). Also find
(i) x
(ii) Length of AP. [4]
Solution:
(a)
Comparing the corresponding entries of equal matrix, we obtain
2y = -4 and 2x = 6
y = -2 and x = 3
(b)
Monthly deposit = ₹ 800
Period of deposit (n) = 1\(\frac{1}{2}\) years or 18 months
Total deposited amount = ₹ 800 × 18
= ₹ 14400
Let rate of interest be r% p.a.
Equivalent principal for one month
= \(\frac{n(n+1)}{2}\) × monthly deposit
= ₹ \(\left(\frac{18 \times 19}{2} \times 800\right)\) = ₹ 136800
Interest paid by the bank = \(\frac{136800}{100 \times 12}\) = 114r
Now, total deposited amount + interest amount = amount on maturity
14400 + 114r = 15084
⇒ 114 r = 15084 – 14400
⇒ 114r = 684
⇒ r = \(\frac{684}{114}\) = 6%
Hence, the rate of interest per annum is 6%
(c) Let the line joining A(-4, 2) and B (3, 6) is divided by P(x, 3) in the ratio k : 1
Hence, the required ratio is 1 : 3, x = –\(\frac{9}{4}\) and length of AP is \(\frac{\sqrt{65}}{2}\) units.
Question 3.
(a) Without using trigonometric tables, evaluate
sin2 34° + sin2 56° + 2 tan 18° tan 72° – cot2 30°. [3]
(b) Using the Remainder and Factor Theorem, factorise the following polynomial:
x3 + 10x2 – 37x + 26. [3]
(c) In the figure given below, ABCD is a rectangle. AB = 14 cm, BC = 7 cm. From the rectangle, a quarter circle BFEC and a semicircle DGE are removed. Calculate the area of the remaining piece of the rectangle.
(Take π = 22/7) [4]
Solution:
(a) sin2 34° + sin2 56° + 2 tan 18° tan 12° – cot2 30°
= sin2 34° + sin2 (90° – 34°) + 2 tan 18°. tan (90° – 18°) – (\(\sqrt{3}\))2
= sin2 34° + cos2 34° + 2 tan 18°. cot 18° – 3
[∵ sin (90°- θ) = cos θ, tan (90° – θ) = cot θ]
= 1 + 2 – 3 = 0
(b) Let p(x) = x3 + 10x2 – 37x + 26
Now, factors of 26 are ± 2 and ±13
Put x = 2, we obtain p(2) = (2)3 + 10(2)2 – 37(2) + 26
= 74 – 74 = 0
⇒ (x – 2) is a factor of p(x)
By long division, we obtain
Thus, x3 + 10x2 – 37x + 26 = (x – 2)(x2 + 12x – 13)
= (x – 2) (x2 + 13x – x – 13)
= (x – 2) (x – 1) (x + 13)
(c) Area of rectangle ABCD = 14 × 7 = 98 cm2
Area of quarter circle BFEC = \(\frac{1}{4}\) × \(\frac{22}{7}\) × 7 × 7
= 38.5 cm2
Area of semicircle DGE = \(\frac{1}{2}\) × \(\frac{22}{7}\) × \(\frac{7}{2}\) × \(\frac{7}{2}\)
= 19.25 cm2
Area of the remaining piece of the rectangle = 98 – 38.5 – 19.25
= 40.25 cm2
Question 4.
(a) The numbers 6, 8, 10, 12, 13 and x are arranged in an ascending order. If the mean of the observations is equal to the median, find the value of x. [3]
(b) In the figure, ZDBC 58°. BD is a diameter of the circle. Calculate:
(i) ∠BDC
(ii) ∠BEC
(iii) ∠BAC [3]
(c) Use a graph paper to answer the following questions. (Take 2 cm 1 unit on both axis)
(i) Plot the points A (-4, 2) and B (2, 4)
(ii) A’ is the image of A when reflected in the y-axis. Plot it on the graph paper and write the coordinates of A’.
(iii) B’ is the image of B when reflected in the line AA’. Write the coordinates of B’.
(iv) Write the geometric name of the figure ABA’B’.
(v) Name a line of symmetry of the figure formed. [4]
Solution:
(a) Here, the given observations are 6, 8, 10, 12, 13 and x.
Now, mean = \(\frac{6+8+10+12+13+x}{6}\) = \(\frac{49+x}{6}\)
(6/2)th observation
Median = \(\frac{+(6 / 2+1)^{\text {th }} \text { observation }}{2}\)
= \(\frac{3^{\text {th }} \text { observation }+4^{\text {th }} \text { Obsrvation }}{2}\)
= \(\frac{10+12}{2}\) = 11
According to given condition, we have \(\frac{49+x}{6}\) = 11
⇒ 49 + x = 66
⇒ x = 66 – 49 = 17
(b) Since BD is a diameter of the circle
∴ ∠BCD = 90°
∠DBC = 58° (given)
∠BDC = 90° – 58° = 32°
Also, BECD is a cyclic quadrilateral.
∴ ∠BEC + ∠BDC = 180°
⇒ ∠BEC + 32° = 180°
⇒ ∠BEC = 180° – 32° = 148°
Again, ∠BAC = ∠BDC = 32
[∠s is the same segment]
Hence, ∠BDC = 32°, ∠BEC = 148° and
∠BAC = 32°.
(c) (i) Points A(-4, 2) and B(2, 4) are plotted in the graph.
(ii) A’ (4, 2)
(iii) B’ (2, 0)
(iv) Kite
(v) AA’ is the line of symmetry of the figure ABA’B’.
Section—B (40 Marks)
(Attempt any four questions from this Section)
Question 5.
(a) A shopkeeper bought a washing machine at a discount of 20% from a wholesaler, the printed price of the washing machine being ₹ 18,000. The shopkeeper sells it to a consumer at a discount of 10% on the printed price. If the rate of sales tax is 8%, find:
(i) the VAT paid by the shopkeeper.
(ii) the total amount that the consumer pays for the washing machine. [3]
(b) If \(\frac{x^2+y^2}{x^2-y^2}\) = \(\frac{17}{8}\) then find the value of:
(i) x : y
(ii) \(\frac{x^3+y^3}{x^3-y^3}\)
(c) In ΔABC, ∠ABC = ∠DAC. AB = 8 cm, AC = 4 cm, AD = 5 cm.
(i) Prove that ΔACD is similar to ΔBCA.
(ii) Find BC and CD.
(iii) Find area of ΔACD : area of ΔABC. [4]
Solution.
(a) Printed price of washing machine = ₹ 18000
Rate of discount = 20%
Sale price of washing machine of wholesaler
= ₹\(\left(18000 \times \frac{80}{100}\right)\) = ₹ 14400
VAT@8% = ₹\(\left(\frac{8}{100} \times 14400\right)\) = ₹ 1152
Sale price of washing machine for shopkeeper
= ₹\(\left(18000 \times \frac{90}{100}\right)\) = ₹ 16200
VAT@ 8% = ₹ \(\left(\frac{8}{100} \times 16200\right)\) = ₹ 1296
Thus,VAT paid by the shopkeeper
= ₹ (1296 – 1152) = ₹ 144
Total amount that the consumer pays for the washing machine = ₹ (16200 + 1296) = ₹ 17496
(b) Given that: \(\frac{x^2+y^2}{x^2-y^2}\) = \(\frac{17}{8}\)
By componendo and dividendo, we obtain
(c) (i) In ΔACD and ΔBCA, we have
∠ACD = ∠BCA [common]
∠DAC = ∠ABC [given)
⇒ ΔACD \(\sim\) ΔBCA
[by AA similarity axiom]
(ii) Since corresponding sides of similar triangles are proportional
(iii)
Question 6.
(a) Find the value of ‘a’ for which the following points A (a, 3), B (2, 1) and C (5, a) are collinear. Hence find the equation of the line. [3]
(b) Salman invests a sum of money in ₹ 50 shares, paying 15% dividend quoted at 20% premium. If his annual dividend is ₹ 600, calculate:
(i) the number of shares he bought.
(ii) his total investment.
(iii) the rate of return on his investment. [3]
(c) The surface area of a solid metallic sphere is 2464 cm2. It is melted and recast into solid right circular cones of radius 3.5 cm and height 7 cm. Calculate:
(i) the radius of the sphere.
(ii) the number of cones recast. (Take π = 22/7) [4]
Solution:
(a) Since the points A(u, 3), B(2,1) and C(5, a) are collinear
∴ ar (Δ ABC) = 0
⇒ \(\frac{1}{2}\)|a(1 – a) + 2(a – 3) + 5(3 – 1)| = 0
⇒ a – a2 + 2a – 6 + 15 – 5 = 0
⇒ 3a – a2 + 4 = 0
a2 – 3a – 4 = 0
(a – 4) (a + 1) = 0
a = 4
a = -1
Thus, points are A(4, 3) , B (2, 1) and C (5, 4) or A(-1, 3), B(2, 1) and C(5, -1).
Now, equation of line AC is
y – 3 = \(\frac{1-3}{2-4}\)(x – 4)
⇒ y – 3 = \(\frac{-2}{-2}\)(x – 4)
⇒ y – 3 = x – 4
⇒ x – y – 1 = 0
(b) Dividend on 1 share of ₹ 50 = 15% of ₹ 50 = ₹ \(\frac{15}{100}\) × 50
= ₹ 7.5
Annual dividend = ₹ 600
∴ Number of shares bought = \(\frac{600}{7.5}\) = 80
His total investment = ₹ (60 × 80)
[∵ a share of ₹ 50 is quoted at 20% premium]
= ₹ 4800
Rate of return on investment = \(\frac{600}{4800}\) × 100
= 12.5%
(c) Surface area of metallic sphere = 2464 cm2
4πr2 = 2464
r2 = \(\frac{2464 \times 7}{4 \times 22}\) = 196
∴ r = 14 cm
Let the total number of cones recasted be n, then n × volume of a cone = volume of solid metallic sphere
n × \(\frac{1}{3} \pi r^2\) × h = \(\frac{4}{3} \pi r^3\)nri 3 3
n × 3.5 × 3.5 × 7 = 4 × 14 × 14 × 14
n = \(\frac{4 \times 14 \times 14 \times 14}{3.5 \times 3.5 \times 7}\)
= 128
Hence, the radius of the sphere is 14 cm and number of cones recasted is 128.
Question 7.
(a) Calculate the mean of the distribution given below using the short cut method.
| Marks | 11-20 | 21-30 | 31-40 | 41-50 | 51-60 | 61-70 | 71-80 |
| No. of students | 2 | 6 | 10 | 12 | 9 | 7 | 4 |
[3]
(b) In the figure given below, diameter AB and chord CD of a circle meet at R PT is a tangent to the circle at T. CD = 7.8 cm, PD = 5 cm, PB = 4 cm. Find:
(i) AB.
(ii) the length of tangent PT. [3]
(c) Let A = \(\left[\begin{array}{cc}
2 & 1 \\
0 & -2
\end{array}\right]\), B = \(\left[\begin{array}{cc}
4 & 1 \\
-3 & -2
\end{array}\right]\), and C = \(\left[\begin{array}{cc}
-3 & 2 \\
-1 & 4
\end{array}\right]\). Find A2 + AC – 5B.
Solution:
(a)
Let a (assumed mean) = 45.5
Mean (\(\overline{\mathrm{X}}\)) = a + \(\frac{\Sigma f_i u_i}{\Sigma f_i}\) × h
= 45.5 + \(\frac{7}{50}\) × 10
= 45.5 + 1.4 = 46.9
(b) Here, two chords, AB and CD,
when produced, meet at P
∴ PA × PB = PC × PD
= PA × 4 = (7.8 + 5) × 5
⇒ PA × 4 = 64
⇒ PA = 16
⇒ AB + PB = 16
⇒ AB + 4 = 16
⇒ AB = 12 cm
Now, ABP is secant line and PT be a tangent
∴ PT2 = PA × PB
⇒ PT2 = 16 × 4
⇒ PT = \(\sqrt{16 \times 4}\) = 8 cm
(c)
A2 + AC – 5B
Question 8.
(a) The compound interest, calculated yearly, on a certain sum of money for the second year is ₹ 1320 and for the third year is ₹ 1452. Calculate the rate of interest and the original sum of money. [3]
(b) Construct a ΔABC with BC = 6.5 cm, AB = 5.5 cm, AC = 5 cm. Construct the incircle of the triangle. Measure and record the radius of the incircle. [3]
(c) (use a graph paper for this question.) The daily pocket expenses of 200 students in a school are given below:
| Pocket expenses (in ₹) | Number of students (frequency) |
| 0-5
5-10 10-15 15-20 20-25 25-30 30-35 35-40 |
10 14 28 42 50 30 14 12 |
Draw a histogram representing the above distribution and estimate the mode from the graph .[4]
Solution:
(a) Compound interest for second year = ₹ 1320
Compound interest for third year = ₹ 1452
S.I. on ₹ 1320 for one year = ₹ (1452 – 1320)
= ₹ 132
Principal = ₹ 1320,
Time = 1 year
∴ R = \(\frac{\mathrm{I} \times 100}{\mathrm{P} \times \mathrm{T}}\) = \(\frac{132 \times 100}{1320 \times 1}\) = 10%
Let principal be ₹ 100,Interest for first year
= \(\frac{100 \times 10 \times 1}{100}\) = ₹ 10
Interest for second year = \(\frac{110 \times 10 \times 1}{100}\) = ₹ 11
If interest for second year is ₹ 11, then, principal = ₹ 100
If interest for second year is ₹ 1320, then, principal = ₹ \(\left(\frac{100}{11} \times 1320\right)\) = ₹ 12000
Hence, the rate of interest is 10% p.a. and the original sum of money is ₹ 12000.
(b) Steps of construction :
(i) Construct ΔABC, such that AB = 5.5 cm, BC = 6.5 cm and AC = 5 cm.
(ii) Draw the angle bisectors BO and CO of ∠B and ∠C and let they intersect in O.
(iii) Draw OE ⊥ BC, intersecting BC is E.
(iv) With O as centre and OE as radius draw the required incricle of ΔABC. Radius of incircle = 1.5 cm
(c)Required histogram is as given below :
Clearly, from the graph estimated mode is ₹ 22.
Question 9.
(a) If (x – 9) : (3x + 6) is the duplicate ratio of 4 : 9, find the value of x. [3]
(b) Solve for x using the quadratic formula. Write your answer correct to two significant figures.
(x – 1)2 – 3x + 4 = 0
(c) A page from the savings bank account of Priyanka is given below:
If the interest earned by Priyanka for the period ending September, 2006 is ₹ 175, find the rate of interest. [4]
Solution:
(a) Duplicate ratio of 4 : 9 is (x – 9): (3 + 6).
∴ \(\frac{4^2}{9^2}\) = \(\frac{x-9}{3 x+6}\)
⇒ \(\frac{16}{81}\) = \(\frac{x-9}{3 x+6}\)
⇒ 48x + 96 = 81x – 729
⇒ 81x – 48x = 729 + 96
⇒ 33x = 825
⇒ x = \(\frac{825}{33}\) = 25
(b) Given quadratic equation is
(x – 1)2 – 3x + 4 = 0
x2 – 2x + 1 – 3x + 4 = 0
x2 – 5x + 5 = 0
⇒ x = \(\) = \(\)
= \(\frac{5 \pm 2.236}{2}\) = \(\frac{7.236}{2}\), \(\frac{2.764}{2}\)
= 3.618, 1.382
x = 3.62,1.38 upto two significant figures
(c) Minimum qualifying balance for April, 2006 = ₹ 6000
May, 2006 = ₹ 7000
June, 2006 = ₹ 10000
July, 2006 = ₹ 6000
August, 2006 = ₹ 6000
September, 2006 = ₹ 7000
Total = ₹ 42000
Interest = ₹ 175, Time = \(\frac{1}{12}\) year
∴ R = \(\frac{\mathrm{I} \times 100}{\mathrm{P} \times \mathrm{T}}\) = \(\frac{175 \times 100 \times 12}{42000 \times 1}\) = 5% p.a
Question 10.
(a) A two digit positive number is such that the product of its digits is 6. 1f 9 is added to the number, the digits interchange their places. find the number. [4]
(b) The marks obtained by 100 students in a Mathematics test are given below :
| Marks | No. of students |
| 0-10 | 3 |
| 10-20 | 7 |
| 20-30 | 12 |
| 30-40 | 17 |
| 40-50 | 23 |
| 50-60 | 14 |
| 60-70 | 9 |
| 70-80 | 6 |
| 80-90 | 5 |
| 90-100 | 4 |
Draw an ogive for the given distribution on a graph sheet.
Use a scale of 2 cm = 10 units on both axis.
Use the ogive to estimate the:
(i) median.
(ii) lower quartile.
(iii) number of students who obtained more than 85% marks in the test.
(iv) number of students who did not pass in the test if the pass percentage was 35. [6]
Solution:
(a) Let one’s place digit be x
∴ Ten’s place digit = \(\frac{6}{x}\)
Original number = 10 × \(\frac{6}{x}\) + x = \(\frac{60}{x}\) + x
Number obtained by interchanging their digits
= 10x + \(\frac{6}{x}\)
According to question, we obtain \(\frac{6}{x}\) + x + 9 = 10x + \(\frac{6}{x}\)
⇒ 60 + x2 + 9x = 10x2 + 6
⇒ 9x2 – 9x – 54 = 0
⇒ x2 – x – 6 = 0
⇒ (x – 3) (x + 2) = 0
⇒ x = 3 or x = -2
[Rejecting -ve value, ∵ given no. is positive]
Thus, one’s place digit is 3 and ten’s place digit is \(\frac{6}{3}\)
i.e., 2.
Hence, the required number is 23.
(b) Cummulative frequency distribution table is as below:
Plot the points (0, 0), (10, 3), (20, 10), (30, 22), (40, 39), (50, 62), (60, 76), (70, 85),
(80, 91), (90, 96) and (100, 100). Join them free hand to get the required ogive. Now, from the above graph, we obtain
(i) median = 45
(ii) lower quartile = 32
(iii) number of students who obtained more than 85% marks in the test = 100 – 94 = 6
(iv) number of students who did not pass in the test if the pass percentage was 35 = 30
Question 11.
(a) In the figure given below, O is the centre of the circle. AB and CD are two chords of the circle. OM is perpendicular to AB and ON is perpendicular to CD. AB = 24 cm, OM = 5 cm, ON = 12 cm. Find the:
(i) radius of the circle.
(ii) length of chord CD. [3]
(b) Prove the identity : (sin θ + cos θ) (tan θ + cot θ) = sec θ + cosec θ. [3]
(c) An aeroplane at an altitude of 250 m observes the angle of depression of two boats on the opposite banks of a river to be 45° and 60° respectively. Find the width of the river. Write the answer correct to the nearest whole number. [4]
Solution:
(a) We know that perpendicular drawn from the centre to the chord, bisects the chord.
∴ AM = MB = 12 cm and
CN = ND = \(\frac{1}{2}\) CD
OM = 5 cm, ON = 12 cm
Join AO and CO.
In rt. ∠ed ΔAMO, by Pythagoras Theorem, we have
AO2 = AM2 + OM2 = 122 + 52
= 144 + 25 = 169
AO = 13 cm
Thus, radius of the circle is 13 cm.
Now, in rt. ∠ed ΔCNO, we obtain
CN2 = CO2 – ON2 = 132 – 122 = 169 – 144 = 25 CN = 5 cm
∴ CD = CN + ND
= CN + CN
= 2CN = 2 × 5 = 10 cm
(b) L.H.S. = (sin θ + cos θ) (tan θ + cot θ)
= (sin θ + cos θ) \(\left(\frac{\sin \theta}{\cos \theta}+\frac{\cos \theta}{\sin \theta}\right)\)
= (sin θ + cos θ) \(\left(\frac{\sin ^2 \theta+\cos ^2 \theta}{\cos \theta \sin \theta}\right)\)
= \(\frac{\sin \theta+\cos \theta}{\cos \theta \sin \theta}\) = \(\frac{\sin \theta}{\cos \theta \sin \theta}\) + \(\frac{\cos \theta}{\cos \theta \sin \theta}\)
= sec θ + cosec θ
= R.H.S.
(c) Let AB be the altitude of an aeroplane, P and Q are the positions of two boats on either sides of AB, such that AB = 250 m, ∠P = 45°, ∠Q = 60°
In rt. ∠ed ΔPBA, we have
\(\frac{A B}{P B}\) = tan 45°
⇒ \(\frac{250}{\mathrm{~PB}}\) = 1 ⇒ PB = 250 m
In rt. ∠ed ΔQBA, we have
\(\frac{A B}{B Q}\) = tan 60°
⇒ \(\frac{250}{\mathrm{BQ}}\) = \(\sqrt{3}\)
⇒ BQ = \(\frac{250}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}}\)
= \(\frac{250 \sqrt{3}}{3}\) m
Thus, PQ (width of the river) = PB + BQ
= 250 + \(\frac{250 \sqrt{3}}{3}\)
= \(\left(\frac{750+250 \sqrt{3}}{3}\right) \mathrm{m}\)
= 394.33 m