Exercise 2.5

1. Solve the following equations.

i. n5 - 57 = 23

Solution:

n5 - 57 + = 23 +

n5 = 23 + 57

Finding a common denominator for 23 and 57 (which is ):

n5 = 2×73×7 + 5×37×3

n5 = +

n5 = 14+1521

n5 =

n5 × = 2921 ×

n = 29×521

n =

Therefore, the solution is n = 14521.

ii. x3 - x4 = 14

Solution:

4x123x12 - 3x124x12 = 14

4x−3x12 = 14

= 14

x12 × = 14 ×

x =

Therefore, the solution is x = 168.

iii. z2 + z3 - z6 = 8

Solution:

Finding a common denominator for the fractions (which is ):

3z6z6 + 2z64z6 - z6 = 8

3z+2z−z6 = 8

= 8

Simplifying: = 8

2z3 × = 8 ×

2z =

2z2 = 242

z =

Therefore, the solution is z = 12.

iv.2p3 - p5 = 1123

Solution:

1123 = 11×3+23 =

- = 353

Finding a common denominator (which is ): 10p155p15 - 3p152p15 = 353

= 353

7p15 × = 353 ×

7p =

7p7 = 1757

p =

Therefore, the solution is p = 25.

v. 9 14 = y - 1 13

Solution:

9 14 = 9×4+14 =

1 13 = 1×3+13 =

= y -

374 + 43 = y

Finding a common denominator(which is ):

37×312 + 4×412 = y

+ = y

= y

Therefore, y = 12712.

vi. x2 - 45 + x5 + 3x10 = 15

Solution:

Simplifying the left side: 5x−456x−45 = 15

10 × 10x−810 = 10 × 15

10x - 8 =

10x - 8 + 8 = 2 + 8

10x =

10x10 = 1010

x =

vii. x2 - 14 = x3 + 12

Solution:

x2 - x3 = 12 + 14

Finding a common denominator for the x terms (which is ) and for the constant terms (which is ):

() - () = 2436 + 14

x6 = 34

x6 × 6 = 34 × 6

x =

x =

Therefore, the solution is x = 92.

viii. 2x−33x+2 = −23

Solution:

2x−33x+2 × (3x + 2) = −23 × (3x + 2)

= −233x+2

× (2x - 3) = × −233x+2

3(2x - 3) = (3x + 2)

- = +

6x - 9 + 6x = -6x - 4 + 6x

- 9 = -4

12x - 9 + 9 = -4 + 9

12x =

12x12 = 512

x =

Therefore, the solution is x = 512.

ix. 8p−57p+1 = −24

Solution:

−24 =

8p−57p+1 = −12

8p−57p+1 × (7p + 1) = −12 × (7p + 1)

8p - 5 = −127p+1

× (8p - 5) = × −127p+1

2(8p - 5) = (7p + 1)

- 10 = -7p - 1

16p - 10 + 7p = -7p - 1 + 7p

- 10 = -1

23p - 10 + 10 = -1 + 10

23p =

23p23 = 923

p =

Therefore, the solution is p = 923.

x. 7y−25 = 6y−511

Solution:

7y + 25 - 6y = 6y - 511 - 6y

y + 25 =

y + 25 - 25 = -511 - 25

y = -511 - 25

Finding a common denominator for −511 and −25 (which is ):

y = -

y =

Therefore, the solution is y = −4755.

xi. x+56 - x+19 = x+34

x + 56 - x - 19 = x + 34

56 - 19 = x + 34

Finding a common denominator for 56 and 19 (which is ):

1518 - 218 = x + 34

= x + 34

1318 - 34 = x

Finding a common denominator for 1318 and 34 (which is ):

() - () = x

= x

Therefore, the solution is x = −136.

xii. 3t+116 - 2t−37 = t+38 + 3t−114

Solution:

Finding a common denominator for all the fractions: the least common multiple of 16, 7, 8, and 14 is .

So,

7×3t+1112 - 16×2t−3112 = 14×t+3112 + 8×3t−1112

Since all fractions now have the same denominator, we can equate the numerators:

7(3t + 1) - 16(2t - 3) = 14(t + 3) + 8(3t - 1)

+ 7 - + = 14t + + +

+ 55 = 38t +

55 = + 34

= 49t

t =

t =

Therefore, the solution is t = 37.

2. What number is that of which the third part exceeds the fifth part by 4?

Solution:

Let 'x' represent the unknown number.

The third part of the number:

The fifth part of the number:

The third part exceeds the fifth part by 4, so:

x3 - x5 = 4

Finding a common denominator for the fractions (which is ):

() - () = 4

Combine the fractions:

= 4

Multiply both sides by 15:

2x15 × 15 = 4 × 15

2x =

Divide both sides by 2:

2x2 = 602

x =

Therefore, the number is 30.

3. The difference between two positive integers is 36. The quotient when one integer is div> ided by other is 4. Find the integers.

Solution:

Let 'x' be the larger integer and 'y' be the smaller integer.

The difference between the two integers is 36: - =

The quotient when one integer is divided by the other is 4: xyyx = 4

Solve for x in terms of y from the second equation: x =

Substitute this value of x into the first equation: - =

Simplify and solve for y: = 36

y =

Substitute the value of y back into the equation x = 4y to find x: x = 4 ×

x =

Therefore, the two integers are 48 and 12.

4. The numerator of a fraction is 4 less than the denominator. If 1 is added to both its numerator and denominator , it becomes 1/2. Find the fraction.

Solution:

Let 'x' represent the denominator.

Numerator =

Express the fraction: Fraction = x−4xxx−4

Express the new fraction after adding 1 to both numerator and denominator:

New Fraction = x−4+1x+1 = x−3x+1x−2x+2

So:

x−3x+1 =

× (x - 3) = × (x + 1)

- = +

Solve for x: 2x - x = 1 + 6

x =

Find the numerator:

Numerator = x - 4 = 7 - 4 =

Form the original fraction: =

Therefore, the fraction is 37.

5. Find three consecutive numbers such that if they are divided by 10, 17, and 26 respectively, the sum of their quotients will be 10.

Solution:

Let the three consecutive numbers be n, , and .

Express the quotients when the numbers are divided by 10, 17, and 26 respectively:

n+117n17

n+226n26

So,

n10 + n+117 + n+226 = 10

Finding a common denominator for the fractions: The least common multiple of 10, 17 and 26 is .

221n2210 + 130n+12210 + 85n+22210 = 10

221n+130n+130+85n+1702210 = 10

436n372n + =

436n = 21800

n = 21800436

n =

Find the other two numbers:

n + 1 = 50 + 1 =

n + 2 = 50 + 2 =

Therefore, the three consecutive numbers are 50, 51, and 52.

6. In class of 40 pupils the number of girls is three-fifths of the number of boys. Find the number of boys in the class.

Solution:

Let 'b' represent the number of boys.

Number of girls = × b

Thus, Number of boys + Number of girls =

b + 35b = 40

() + 3b5 = 40

= 40

8b =

b = 2008

b =

Therefore, the number of boys in the class is 25.

7. After 15 years , Mary’s age will be four times of her present age. Find her present age.

Solution:

Let 'x' represent Mary's present age.

Mary's age after 15 years:

x + 15 =

15 = 4x - x

15 =

x = 153

x =

Therefore, Mary's present age is 5 years.

8. Aravind has a kiddy bank. It is full of one-rupee and fifty paise coins. It contains 3 times as many fifty paise coins as one rupee coins.The total amount of the money in the bank is 35.How many coins of each kind are there in the bank?

Solution:

Let 'x' represent the number of one-rupee coins.

Number of fifty paise coins =

Value of one-rupee coins = 1 × x =

Value of fifty paise coins = × 3x =

Value of one-rupee coins + Value of fifty paise coins =

x + 1.5x = 35

Combine the 'x' terms: = 35

x = 352.5

x =

Number of fifty paise coins = 3x = 3 × 14 =

Therefore, there are one-rupee coins and fifty paise coins in the bank.

9. A and B together can finish a piece of work in 12 days. If ‘A’ alone can finish the same work in 20 days , in how many days B alone can finish it?

Solution: A's work per day =

Find the fraction of work A and B together do in one day:

(A + B)'s work per day =

Find the fraction of work B does in one day:

B's work per day = (A + B)'s work per day - A's work per day

B's work per day = 112 - 120

Find a common denominator for the fractions (which is ):

B's work per day = 560 - 360

B's work per day =

B's work per day =

Find the number of days B takes to complete the work alone:

Number of days for B = 1B's work per day

Number of days for B = 1130

Number of days for B =

Therefore, B alone can finish the work in 30 days.

10. If a train runs at 40 kmph it reaches its destination late by 11 minutes . But if it runs at 50 kmph it is late by 5 minutes only. Find the distance be covered by the train.

Solution: Let 'd' be the distance and 't' be the scheduled time in hours.

Time taken at 40 kmph = hours

Time taken at 50 kmph = hours

Converting the delay into hours: 11 minutes = hours

5 minutes = hours

So, d40 = t + 1160 (1)

d50 = t + 560 (2)

d40 - d50 = t+1160 - t+560

d40 - d50 =

d40 - d50 =

Finding a common denominator for the fractions (which is ):

() - () = 110

= 110

d = 110 ×

d =

Therefore, the distance to be covered by the train is 20 km.

11. One fourth of a herd of deer has gone to the forest. One third of the total number is grazing in a field and remaining 15 are drinking water the bank of a river. Find the total number of deer.

Solution:

Let 'x' be the total number of deer.

Deer in forest =

Deer grazing =

Deer drinking water =

Deer in forest + Deer grazing + Deer drinking water = Total number of deer

14x + 13x + 15 = x

Combine the fractions on the left side. The common denominator is :

() + () + 15 = x

() + 15 = x

15 = x - 7x12

15 = () - 7x12

15 =

15 × = 5x

= 5x

x = 1805

x =

Therefore, the total number of deer is 36.

12. By selling a radio for Rs. 903, a shop keeper gains 5%. Find the cost price of the radio.

Solution:

Let 'c' represent the cost price of the radio.

Selling Price = Cost Price + (Gain Percentage × Cost PriceProfit )

Selling Price = + ( × )

Selling Price = c

= 1.05c

c = 9031.05

c =

Therefore, the cost price of the radio is ₹ 860.

13. Sekhar gives a quarter of his sweets to Renu and then gives 5 sweets to Raji. He has 7 sweets left. How many sweets did he have to start with?

Solution:

Let 'x' be the number of sweets Sekhar had to start with.

Sweets to Renu =

Sweets remaining after Renu = x - 14x = ()x

Sweets to Raji =

Sweets left =

Sweets remaining after Renu - Sweets to Raji = Sweets left

34x - 5 = 7

34x = 7 + 5

34x =

x = 12 × 43

x =

Therefore, Sekhar had 16 sweets to start with.