# Exercise 10

(b) 73

(c) 25

(d) 49

(e) 37

**Ans.e**

Let the original number be l0 x + y

y = 2x + 1....(i)

and (l0y + x) – (10x + y) = 10x + y – 1

or, 9y – 9x = l0x + y – 1

or, l9x – 8y = 1 ...(ii)

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

19x – 8(2x + 1) = 1

or, 19x – 16x – 8 = 1

or, 3x = 9 or, x = 3

So, y = 2 × 3 + 1 = 7

\ original number = 10 × 3 + 7 = 37

(b) 640

(c) 256

(d) 512

(e) None of these

**Ans.d**

**In case I:** Let the no. of children = x.

Hence, total no. of notebooks distributed

^{1}⁄_{8}x.x or ^{x2}⁄_{8}.......(i)

**In case II:** no.of children = ^{x}⁄_{2}

Now, the total no. of notebooks

= 16 × ^{x}⁄_{2} .......(ii)

Comparing (i) & (ii), we get

^{x2}⁄_{8} = 8x

or, x = 64

Hence, total no. of notebooks

= \(\frac{64 \times 64}{8} = 512\)

(b) 20

(c) 30

(d) Cannot be determined

(e) None of these

**Ans.a**

Let the positive integer be x.

Now, x^{2} – 20x = 96

or, x^{2} – 20x – 96 = 0

or, x^{2} – 24x + 4x – 96 = 0

or, x(x – 24) + 4(x – 24) = 0

or, (x – 24)(x + 4) = 0

or, x = 24, – 4

x ≠ –4 because x is a positive integer

(b) 20 km

(c) 8 km

(d) Data inadequate

(e) None of these

**Ans.a**

Distance walked by man

= 2 + 1 + ^{1}⁄_{2} + ^{1}⁄_{4} + ^{1}⁄_{8} + ^{1}⁄_{16}....∞

The above series is in infinity GP.

\(s_{\infty } = \frac{2}{1 - \frac{1}{2}} = 4 \; km\)**Note:** If the series is in GP then

Sum of the Infinity GP = \(\frac{first \;\; term}{\left | 1 - common \;\; ratio \right |}\)

(b) 16

(c) 34

(d) Data inadequate

(e) None of these

**Ans.a**

Let ½ of the no. = 10x + y

and the no. = 10V + W From the given conditions,

W = x and V = y – 1

Thus the no. = 10(y – 1) + x…..(A)

∴ 2(10x + y) = 10(y – 1) + x ⇒ 8y – 19x = 10…(i)

Again, from the question,

V + W = 7 ⇒ y – 1 + x = 7

∴ x + y = 8 …(ii)

Solving equations (i) and (ii), we get x = 2 and y = 6.

∴ From equation (A), Number = 10(y – 1) + x = 52

(b) 2

(c) 7

(d) Cannot be determined

(e) None of these

**Ans.e**

Suppose the two-digit number be 10x + y.

Then we have been given

l0x + y – (10y + x) = 9

⇒ 9x – 9y = 9

⇒ x – y = 1

Hence, the required difference = 1

Note that if the difference between a two-digit number and the number obtained by interchanging the digits is D, then the difference between the two digits of the number = ^{D}⁄_{9}

(b) 100

(c) 125

(d) Cannot be determined

(e) None of these

**Ans.c**

Suppose the number is N.

Then N - ^{3}⁄_{5}N = 50

⇒ ^{2N}⁄_{5} = 50

∴ \(N = \frac{50 \times 5}{2} = 125\)

^{5}⁄

_{8}and if the numerator of the same fraction is increased by 3 and the denominator is increased by I the fraction becomes

^{3}⁄

_{4}. What is the original fraction?

(b)

^{2}⁄

_{7}

(c)

^{4}⁄

_{7}

(d)

^{3}⁄

_{7}

(e) None of these

**Ans.d**

Let the original fraction be ^{x}⁄_{y}.

Then \(\frac{x + 2}{y + 1} = \frac{5}{8}\) or, 8x – 5y = – 11........ (i)

Again, \(\frac{x + 3}{y + 1} = \frac{3}{4}\) or, 4x – 3y = –9........ (ii)

Solving, (i) and (ii) we get x = 3 and y = 7

∴ fraction = ^{3}⁄_{7}

(b) 32

(c) 26

(d) 22

(e) None of these

**Ans.d**

On solving equation we get

x = 7, y = 4, z = 11

(b) 26

(c) 28

(d) 91

(e) None of these

**Ans.e**

Let the number = x

Then, x^{2} + x = 182

or, x^{2} + x – 182 = 0

or, x + 14x – 13x – 182 = 0

or, x(x + 14) – 13(x + 14) = 0

or, (x – 13)(x + 14) = 0

or, x = 13 (negative value is neglected)