# Frank Solutions for Chapter 7 Problems Based on Quadratic Equation Class 10 ICSE Mathematics

### Exercise 7.1

**1. The sum of the square of the 2 consecutive natural numbers is 481. Find the numbers.**

**Answer**

Let us assume the two consecutive number be P and P + 1

As per the condition given in the question, sum of the square of the 2 consecutive natural numbers is 481, P^{2} + (P + 1)^{2} = 481

Then,

P^{2} + (P + 1)^{2} = 481

We know that, (a + b)^{2} = a^{2} + 2ab + b^{2}

P^{2} + P^{2} + 2P + 1 = 481

⇒ 2P^{2} + 2P + 1 = 481

By transposing we get,

2P^{2} + 2P + 1 – 481 = 0

⇒ 2P^{2} + 2P – 480 = 0

Divide by 2 for both side of each term we get,

2P^{2}/2 + 2P/2 – 480/2 = 0/2

⇒ P^{2} + P – 240 = 0

⇒ P^{2} + 16P – 15P – 240 = 0

Take out common in each terms,

P(P + 16) – 15(P + 15) = 0

⇒ (P + 16) (P – 15) = 0

Equate both to zero,

P + 16 = 0, P – 15 = 0

⇒ P = – 16, P = 15

So, P = 15 ** [because – 16 is not a natural number]**

Then,

P = 15

⇒ P + 1 = 15 + 1 = 16

Therefore, the 2 consecutive numbers are 15 and 16.

**2. ****The sum of 2 numbers is 18. If the sum of their reciprocals is ¼, find the numbers.**

**Answer**

Let us assume the numbers be P and Q.

As per the condition given in the question,

The sum of 2 numbers is 18, P + Q = 18 **…[equation i]**

the sum of their reciprocals is ¼, 1/P + 1/Q = ¼ **…[equation ii]**

Consider equation (i), P + Q = 18

P = 18 – Q

Now, substitute the value of P in equation (ii) we get,

1/(18 – Q) + 1/Q = ¼

⇒ (Q +(18 – Q))/((18 – Q)Q) = ¼

By cross multiplication,

4(Q + 18 – Q) = (18 – Q)Q

⇒ 4(18) = 18Q – Q^{2}

⇒ 72 = 18Q – Q^{2}

Now, transposing we get,

Q^{2} – 18Q + 72 = 0

⇒ Q^{2}_{ }– 12Q – 6Q + 72 = 0

Take out common in each terms,

Q(Q – 12) – 6(Q – 12) = 0

⇒ (Q – 12) (Q – 6) = 0

Equate both to zero,

Q – 12 = 0, Q – 6 = 0

⇒ Q = 12, Q = 6

Therefore, the numbers are 6 and 12.

**3. ****The sum of a number and its reciprocal is 2.9/40. Find the number.**

**Answer**

Let us assume the number be B.

As per the condition given in the question, B + 1/B =

So, B + 1/B = 89/40

(B^{2} + 1)/B = 89/40

Cross multiplication we get,

B^{2} + 1 = (89/40)B

By transposing we get,

B^{2}_{ }+ 1 – (89/40)B = 0

Multiply by 40 for both side of each term we get,

40B^{2} + 40 – 89B = 0

⇒ 40B^{2} – 89B + 40 = 0

⇒ 40B^{2} – 64B – 25B + 40 = 0

Take out common in each terms,

8B(5B – 8) – 5(5B – 8) = 0

⇒ (5B – 8) (8B – 5) = 0

Equate both to zero,

5B – 8 = 0, 8B – 5 = 0

⇒ 5B = 8, 8B = 5

B = 8/5, B = 5/8

Therefore, the numbers are 5/8 and 8/5.

**4.** **A two digit number is such that its product of its digit is 18. When 63 is subtracted from the number, the digits interchange their places. Find the number.**

**Answer**

let us assume two digits be PQ.

As per the condition given in the question,

Product of 2 digits is 18, PQ = 18 **…[equation (i)]**

63 is subtracted from the number, the digits interchange their places,

PQ – 63 = QP **…[equation (ii)]**

Now, assume two-digit number be PQ, means P = 10p (as it comes in tens digit)

Then,

PQ – 63 = QP

⇒ 10p + q – 63 = 10q + p

By transposing we get,

9p – 9q – 63 = 0

Divide both side by 9,

P – Q – 7 = 0 **…[equation (iii)]**

So,

PQ =18

⇒ P = 18/Q

Substitute the value of P in equation (iii),

(18/Q) – Q – 7 = 0

⇒ 18 – Q^{2} – 7Q = 0

⇒ Q^{2}+ 7Q – 18 = 0

⇒ Q^{2} + 9Q – 2Q – 18 = 0

⇒ Q(Q + 9) – 2(Q + 9) = 0

⇒ Q + 9 = 0, Q – 2 = 0

⇒ Q = -9, Q = 2

Therefore, Q = 2 **…[because value of Q can’t be negative]**

Then, P = 18/Q

P = 18/2

⇒ P = 9

Therefore, the number is 92.

**5.** **A two digit number is such that the product of the digits is 14. When 45 is added to the number, then the digit are reversed. Find the number.**

**Answer**

Let us assume the two digits be MN

As per the condition given in the question,

Product of 2 digits is 18, MN = 14 **…[equation (i)]**

45 is added to the number, then the digit are reversed,

MN + 45 = NM **…[equation (ii)]**

So, from equation (i) possible combinations are 27 and 72.

Then, from equation (ii) it is clear that number MN is less than NM, we are adding 45 to MN which makes it equal to NM.

Therefore, the number 27 is added by 45, we get a number 72.

72 is a number where the digits are interchanged.

Therefore, the number is 27.

**6. ****Divide 25 into two parts such that twice the square of the larger part exceeds thrice the square of the smaller part by 29.**

**Answer**

Let us assume the two numbers be A and B, B being the bigger number.

As per the condition given in the question,

A + B = 25 **…[equation (i)]**

2B^{2} = 3A^{2} + 29 **…[equation (ii)]**

Now, consider the equation (i), A + B = 25

B = 25 – A

Then, substitute the value of B is equation (ii),

2(25 – A)^{2} = 3A^{2} + 29

We know that,

(a – b)^{2} = a^{2} – 2ab + b^{2}

⇒ 2(25^{2} – (2 × 25 × A) + A^{2}) = 3A^{2} + 29

⇒ 1250 + 2A^{2} – 100A = 3A^{2} + 29

⇒ A^{2} + 100A – 1221 = 0

⇒ A^{2} – 11A + 111A – 1221 = 0

Take out common in each terms,

A(A – 11) + 111(A – 11) = 0

⇒ (A – 11) (A + 111) = 0

Equate both to zero,

A – 11 = 0, A + 111 = 0

⇒ A = 11, A = -111

Therefore, A = 11** …[because A can’t be negative]**

So,

B = 25 – A

= 25 – 11

= 14

**7.**** The sum of the square of 2 positive integers is 208. If the square of larger number is 18 times the smaller number, find the numbers.**

**Answer**

Let us assume the two numbers be, P and Q.

As per the condition given in the question,

P^{2} + Q^{2} = 208 **…[equation (i)]**

⇒ Q^{2} = 18P **…[equation (ii)]**

Consider the equation (i),

P^{2} + Q^{2} = 208

⇒ Q^{2} = 208 – P^{2}

Now, substitute the value of Q^{2} in equation (ii) we get,

208 – P^{2} = 18P

By transposing we get,

P^{2} + 18P – 208 = 0

⇒ P^{2} + 26P – 8P – 208 = 0

Take out common in each terms,

P(P + 26) – 8(P + 26) = 0

⇒ (P + 26) (P – 8) = 0

Equate both to zero,

P + 26 = 0, P – 8 = 0

⇒ P = – 26, P = 8

Therefore, P = 8 **…[because P can’t be negative]**

So,

Q^{2} = 18P

⇒ Q^{2} = 18 × 8

⇒ Q^{2} = 144

⇒ Q = √144

⇒ Q = 12

**8. ****The difference of the square of two natural numbers is 45. The square of the smaller number is 4 times the larger number. Determine the numbers.**

**Answer**

Let us assume the two numbers be, P and Q, Q being the bigger number.

As per the condition given in the question,

Q^{2} – P^{2} = 45 **…[equation (i)]**

⇒ P^{2} = 4Q **…[equation (ii)]**

Consider the equation (i), Q^{2} – P^{2} = 45

P^{2} = Q^{2} – 45

Now, substitute the value of P^{2} in equation (ii) we get,

Q^{2} – 45 = 4Q

By transposing we get,

Q^{2} – 4Q – 45 = 0

⇒ Q^{2} – 9Q + 5Q – 45 = 0

Take out common in each terms,

Q(Q – 9) + 5(Q – 9) = 0

⇒ (Q – 9) (Q + 5) = 0

Equate both to zero,

Q – 9 = 0, Q + 5 = 0

⇒ Q = 9, Q = -5

Therefore, Q = 9 **…[because Q can’t be negative]**

So,

P^{2} = 4Q

⇒ P^{2} = 4 × 9

⇒ P^{2} = 36

⇒ P = √36

⇒ P = 6

**9. ****A two digit number is four times the sum and 3 times the product of its digits, find the number.**

**Answer**

Let us assume the two numbers be, P and Q,

Then, P = 10P **…[because it comes in tens digit]**

As per the condition given in the question,

PQ = 4(P + Q)

⇒ 4(P + Q) = 10P +Q **…[equation (i)]**

⇒ PQ = 3(PQ) = 10P + Q **…[equation (ii)]**

Consider the equation (i),

4(P + Q) = 10P +Q

⇒ 4P + 4Q = 10P + Q

⇒ 10P – 4P = 4Q – Q

⇒ 6P = 3Q

⇒ P = 3Q/6

⇒ P = Q/2

Now, substitute the value of P in equation (ii) we get,

3(PQ) = 10P + Q

⇒ (3Q/2) × Q = (10Q/2) + Q

⇒ 3Q^{2}/2 = 5Q + Q

⇒ 3Q^{2}/2 = 6Q

By cross multiplication we get,

3Q^{2} = 12Q

By simplification,

Q = 4

So, P = Q/2 = 2

Therefore, the number is 24.

** 10. Two natural numbers differ by 4. If the sum of their square is 656, find the numbers.**

**Answer**

Let us assume the two numbers be, P and Q.

As per the condition given in the question,

P^{2} + Q^{2} = 656 **…[equation (i)]**

⇒ P – Q = 4

⇒ P = 4 + Q **…[equation (ii)]**

Now, substitute the equation (ii) in equation (i) we get,

(4 + Q)^{2} + Q^{2} = 656

We know that, (a + b)^{2} = a^{2} + 2ab + b^{2}

(4^{2}+ (2 × 4 × Q) + Q^{2}) + Q^{2} = 656

⇒ 16 + 8Q + Q^{2} + Q^{2} = 656

⇒ 16 + 8Q + 2Q^{2} = 656

By transposing we get,

2Q^{2} + 8Q + 16 – 656 = 0

⇒ 2Q^{2} + 8Q – 640 = 0

Divide both side by 2 we get,

2Q^{2}/2 +8Q/2 – 640/2 = 0/2

⇒ Q^{2} + 4Q – 320 = 0

⇒ Q^{2} + 20Q – 16Q – 320 = 0

Take out common in each terms,

Q(Q + 20) – 16(Q + 20) = 0

⇒ (Q + 20) (Q – 16) = 0

Equate both to zero,

Q + 20 = 0, Q – 16 = 0

⇒ Q = – 20, Q = 16

⇒ Q = 16 **…[because Q can’t be negative]**

So, P = 4 + Q

⇒ P = 4 + 16

⇒ P = 20

Therefore, the numbers are 16 and 20.

**11.** **The sum of the square of 2 consecutive odd positive integers is 290. Find them.**

**Answer**

let us assume the two consecutive odd positive number be P and P + 2

As per the condition given in the question, P^{2} + (P + 2)^{2} = 290

Then,

P^{2} + (P + 2)^{2} = 290

We know that, (a + b)^{2} = a^{2} + 2ab + b^{2}

⇒ P^{2} + P^{2} + 4P + 4 = 290

⇒ 2P^{2} + 4P + 4 = 290

By transposing we get,

2P^{2} + 4P + 4 – 290 = 0

⇒ 2P^{2} + 4P – 286 = 0

Divide by 2 for both side of each term we get,

2P^{2}/2 + 4P/2 – 286/2 = 0/2

⇒ P^{2} + 2P – 143 = 0

⇒ P^{2} + 13P – 11P – 143 = 0

Take out common in each terms,

P(P + 13) – 11(P + 13) = 0

⇒ (P + 13) (P – 11) = 0

Equate both to zero,

P + 13 = 0, P – 11 = 0

⇒ P = –13, P = 11

So, P = 11 **…[because –16 is a negative even integer]**

Then,

P = 11

⇒ P + 2 = 11 + 2 = 11

Therefore, the 2 consecutive positive odd integers are 11 and 13.

**12. Three consecutive natural numbers are such that the square of the first increased by the product of other two gives 154. Find the numbers.**

**Answer**

let us assume the three consecutive natural number be P – 1, P and P + 1

As per the condition given in the question, (P – 1)^{2} + (P) (P + 1) = 154

Then,

(P – 1)^{2} + P^{2} + P = 154

We know that, (a – b)^{2} = a^{2} – 2ab + b^{2}

P^{2} – 2P + 1 + P^{2} + P = 154

⇒ 2P^{2} – P + 1 = 154

By transposing we get,

2P^{2} – P + 1 – 154 = 0

⇒ 2P^{2} – P – 153 = 0

⇒ 2P^{2} – 18P – 17P – 153 = 0

Take out common in each terms,

2P(P – 9) – 17(P – 9) = 0

⇒ (P – 9) (2P – 17) = 0

Equate both to zero,

P – 9 = 0, 2P – 17 = 0

⇒ P = 9, 2P = 17

⇒ P = 9, P = 17/2

So, P = 9 **…[because 9 is a natural number]**

Then,

P = 9

⇒ P – 1 = 9 – 1 = 8

⇒ P + 1 = 9 + 1= 10

Therefore, the three consecutive natural number are 8, 9 and 10.

**13. The sum of the square of two numbers is 233. If one of the numbers is 3 less than twice the other number. Find the numbers.**

**Answer**

let us assume the two numbers be y, 2y – 3

As per the condition given in the question, y^{2} + (2y – 3)^{2} = 233

Then,

y^{2} + (2y – 3)^{2} = 233

We know that, (a – b)^{2} = a^{2} – 2ab + b^{2}

y^{2} + 4y^{2} – 12y + 9 = 233

⇒ 5y^{2} – 12y + 9 = 233

By transposing we get,

5y^{2} – 12y + 9 – 233= 0

⇒ 5y^{2} – 12y – 224 = 0

⇒ 5y^{2} – 40y + 28y – 224 = 0

Take out common in each terms,

5y(y – 8) + 28(y – 8) = 0

⇒ (y – 8) (5y + 28) = 0

Equate both to zero,

y – 8 = 0, 5y + 28 = 0

⇒ y = 8, 5y = – 28

⇒ y = 8, y = – 28/5

So, y = 8 **…[because 8 is a natural number]**

Then,

y = 8

⇒ 2y – 3 = 2(8) – 3

= 16 – 3

= 13

Therefore, the 2 numbers are 8 and 13.

**14. ****Find two natural numbers which differ by 3 and whose squares have the sum of 117.**

**Answer**

let us assume the two natural numbers be y, y – 3

As per the condition given in the question, y^{2} + (y – 3)^{2} = 117

Then,

y^{2} + (y – 3)^{2} = 117

We know that, (a – b)^{2} = a^{2} – 2ab + b^{2}

y^{2} + y^{2} – 6y + 9 = 117

⇒ 2y^{2} – 6y + 9 = 117

By transposing we get,

2y^{2} – 6y + 9 – 117 = 0

⇒ 2y^{2} – 6y – 108 = 0

Divide by 2 for both side of each term we get,

2y^{2}/2 – 6y/2 – 108/2 = 0/2

⇒ y^{2} – 3y – 54 = 0

⇒ y^{2} – 9y + 6y – 54 = 0

Take out common in each terms,

y(y – 9) + 6(y – 9) = 0

⇒ (y – 9) (y + 6) = 0

Equate both to zero,

y – 9 = 0, y + 6 = 0

⇒ y = 9, y = – 6

⇒ y = 9, y = – 6

So, y = 9 **…[because 9 is a natural number]**

Then,

y = 9

⇒ y – 3 = 9 – 3 = 6

Therefore, the 2 natural numbers are 6 and 9.

**15.** **A two digit number is such that the product of its digit is 8. When 18 is subtracted from the number, the digits interchange its place. Find the numbers.**

**Answer**

Let us assume the two numbers be, P and Q,

Then, P = 10P **…[because it comes in tens digit]**

As per the condition given in the question,

PQ = 8 **…[equation (i)]**

⇒ 10P + Q – 18 = 10Q + P

⇒ 10P – P + Q – 10Q – 18 = 0

⇒ 9P – 9Q – 18 = 0

Divide by 2 for both side of each term we get,

9P/9 – 9Q/9 – 18/9 = 0/9

⇒ P – Q – 2 = 0 **…[equation (ii)]**

Consider the equation (i), PQ = 8

P = 8/Q

Now, substitute the value of P in equation (ii) we get,

8/Q – Q – 2 = 0

By simplification,

8 – Q^{2} – 2Q = 0

⇒ Q^{2} + 2Q – 8 = 0

⇒ Q^{2} + 4Q – 2Q – 8 = 0

Take out common in each terms,

Q(Q + 4) – 2(Q + 4) = 0

⇒ (Q + 4) (Q – 2)

Equate both to zero,

Q + 4 = 0, Q – 2= 0

⇒ Q = –4, Q = 2

So, Q = 2

From equation (i), PQ = 8

2P = 8

⇒ P = 8/2

⇒ P = 4

Therefore, the number is 42.

**16. A two digit number is such that the product of the digit is 12. When 36 is added to the number, the digits interchange their places. Find the numbers. **

**Answer**

Let this two digit number be xy which means x = 10x (as it comes in tens digit).

**17. A two digit number is 4 times the sum of its digit and twice the product of its digit. Find the number.**

**Answer**

Let this two digit number be xy. Which means x = 10x (as it comes in tens digit).

**18. Three years ago, a man was 5 times the age of his son. Four years hence, he will be thrice his son’s age. Find the present ages of the man and his son.**

**Answer**

Let the present age of the man be M years and his sons age be S years.

**19. The sum of the ages of a father and his son is 45 years. Five years ago, the product of their ages was 124. Determine their present ages.**

**Answer**

Let the present age of the man be M years and his son’s age be S years.

**20. The present age of the mother is square of her daughter’s present age. 4 years hence, she will be 4 times as old as her daughter. Find their present ages.**

**Answer**

Let the present age of the Mother be M years and her daughter age be D years.

**21. The product of a girls age five years ago and her age 3 years later is 105. Find her present age. **

**Answer**

Let the present age of the girl be G year. Then, as per the question description,

**22. The hypotenuse of a right-angled triangle is 17 cm. If the smaller side is multiplied b 5 and the larger side is doubled, the new hypotenuse will be 50 cm. Find the length of each side of the triangle.**

**Answer**

Let hypotenuse = h, and other sides by x and y (x bigger than y). As per the question,

**23. The hypotenuse of a grassy land in the shape of a right triangle is 1 m more than twice the shortest side. If the third side is 7 m more than the shortest side, find the sides of the grassy land.**

**Answer**

Let hypotenuse = h, and other sides by x and y (x bigger than y). As per the question,

^{2}. find the hypotenuse.**Answer**

Area of a triangle = (Height × Base)/2

**25. The area of right-angled is 600 cm ^{2}. If the base of the triangle exceeds the altitude by 10 cm, find the dimensions of the triangle. **

**Answer**

Area of a triangle = (Height × Base)/2

**26. The area of the isosceles triangle is 60 cm ^{2}, and the length of each one of its equal side is 13 cm. Find its base.**

**Answer**

**Answer**

Perimeter = a + b + h where a and b are sides and h is hypotenuse. Here h = 25.

**28. A farmer wishes to grow a 100m ^{2} rectangular vegetable garden. Since he was with him only 30 m barbed wire, he fences 3 sides of the rectangular garden letting the compound of his house to act as the 4^{th} side. Find the dimensions of his garden. **

**Answer**

Area of rectangular garden with sides a and b = 100.

a × b = 100

**29. The perimeter of a rectangular field is 82 m and its area is 400 m ^{2}, find the dimension of the rectangular field.**

**Answer**

Area of rectangular garden with sides a and b = 400 = ab

**30. There is a square field whose side is 44m. A square flower bed is prepared in its centre leaving a gravel path all around the flower bed. The total cost of laying the flower bed and graving the path at Rs 2.75 and Rs 1.5 per square metre, respectively, is Rs 4,904. Find the width of the gravel path.**

**Answer**

Let the side of flower bed be a and that of gravel path b.

**31. The length of a hall is 5 m more than the breadth. If the area of the floor of the hall is 84 m ^{2}, find the length and breadth of the hall.**

**Answer**

Let the breadth of the hall be a. Then length = a + 5

**32. A takes 6 days less than the time taken by B to finish a piece of work. If both A and B work together they can finish in 4 days. Find the time taken by B to finish the work.**

**Answer**

Let B take ‘c’ days to finish the work. Hence, A will take ‘c – 6’ days.

**33. A takes 10 days less than B to finish a piece of work. If both A and B work together, they can finish the work in 12 days. Find the time taken by B to finish the work.**

**Answer**

Let B take ‘c’ days to finish the work. Hence, A will take ‘c – 10’ days.

**34. A train travels a distance of 300 kms at a constant speed. If the speed of the train is increased by 10 km/hour, the journey would have taken 1 hour less. Find the original speed of the train.**

**Answer**

Let the speed of the train be S.

**35. A fast train takes 3 hours less than a slow train for a journey of 600 kms. If the speed of the slow train is 10 km/hr less than the fast train, find the speed of the train.**

**Answer**

Let the speed of the slow train be S, Hence speed of the fast train = S + 10

**36. An ordinary train taken 3 hours less for a journey of 360 kms when its speed is increased by 10km/hr. Find the usual speed of the train.**

**Answer**

Let the usual speed of the train be S, Hence speed of the train when speed is increased S + 10

**37. The time taken by a person to cover 150 km was 2.5 hrs more than the time taken in the return journey. If he returned at a speed of 10km/h more than the speed when going, find his speed per hour in each direction.**

**Answer**

Let the speed of the person be S, hence return speed of the person = S + 10

**38. The speed of a boat in still water is 15 km/hr. It can go 30 km upstream and return downstream to the original point in 4 hours 30 minutes. Find the speed of the stream.**

**Answer**

Let the speed of the stream be S km/hr. So in upstream, boat speed will be 15 - S (Against the water flow) and downstream will be S + 15 (towards the water flow and hence speed is added).

**39. The speed of the boat in still water is 11 km/hr. It can go 21 km upstream and 12 km downstream in 3 hours. Find the speed of the stream.**

**Answer**

Let the speed of the stream be S km/hr. So, in upstream, boat speed will be 11-S (Against the water flow) and downstream will be S + 11 (Towards the water flow and hence speed is added).

**40. In a flight of 600 km, an aircraft was slowed due to bad weather. Its average speed for the trip was reduced by 200 km/hr and the time of flight increased by 30 minutes. Find the duration of the flight.**

**Answer**

Let the normal speed of the aircraft be S, and the time taken be t. Distance = 600 km

**41. A plane left 40 minutes late due to bad weather and in order to reach its destination, 1600 kms away in time, it had to increase its speed by 40 km/hr from its usual speed. Find the usual speed of the plane. **

**Answer**

Let the usual speed of the aircraft be S, and time taken be t.

**42. An aeroplane takes 1 hour less for a journey of 1200 km, if its speed is increased by 100 km/hr, from its usual sped. Find the usual speed.**

**Answer**

Let the usual speed of the aircraft be S, and time taken be t. Distance = 1200 km

**43. A piece of cloth cost Rs 5000. If the cost price of the cloth was Rs 5 less per meter, 50 m more of the cloth would have been purchased. Find the cost price per meter of cloth and length of the cloth purchased.**

**Answer**

Let the cloth meters purchase initially was a meters.

**44. A scholarship account of Rs 75,000 was distributed equally among a certain number of students. Had there been 10 students more, each would have got Rs 250 less. Find the original number of persons. **

**Answer**

Let the original number of students be S.

**45. A shopkeeper buys a number of books for Rs 740. If the has bought 5 more books for the same amount, each book would have cost Rs 4 less. How many books did he buy ?**

**Answer**

Let the number of books be b.

**46. One fourth of herd of camel was seen in the forest. Twice the square root of herd has gone to mountains and the remaining 15 camels were seen on the bank of a river. Find the total number of camel.**

**Answer**

Let the total number of camel be a.

**47. The angry Arjun carried some arrows for fighting with Bheeshm. With half the arrows, he cut down the arrows thrown by Bheeshm on him and with six other arrows, he killed the chariot driver of Bheeshm. With one arrow each, he knocked down respectively the chariot, the flag and the bow of Bheeshm. Finally, with one more than 4 times the square root of arrows, he laid below Bheeshm unconscious on the arrow bed. Find the total number of arrows Arjun had. **

**Answer**

Let the total number of arrows be a.

**48. Out of a group of Swans, 3 and half times the square root of the total number are playing on the shore of a pond. The two remaining ones are swinging in water. Find the total number of swans.**

**Answer**

Let the total number of swans be a.