# RS Aggarwal Solutions Class 6 Ex 10B

## RS Aggarwal Class 6 Ex 10B Chapter 10

Objective Questions-

Marks the correct answer in each of the following:

Q1: The ratio 92 : 115 in its simplest form is

(a) 23 : 25                   (b) 18 : 23                   (c) 3 : 5                   (d) 4 : 5

Sol:

(d) 4 : 5

92 :115 = $\frac{92 \div 23}{115 \div 23} = \frac{4}{5}$  (As H.C.F. of 92 and 115 is 23)

Q2: If 57 : x :: 51 : 85, then the value of x is

(a) 95                      (b) 76                      (c) 114                    (d) none of these

Sol:

(a) 95

57 : x :: 51 : 85

$\frac{57}{x} = \frac{51}{85}$

$\Rightarrow x = \frac{57 \times 85}{51}$

$\Rightarrow x = \frac{4845}{51} = 95$

Q3: If 25 : 35 :: 45 : x, then the value of x is

(a) 63                     (b) 72                      (c) 54                    (d) none of these

Sol:

(a) 63

25 : .35 :: 45 : x

$\frac{25}{35} = \frac{45}{x}$

$\Rightarrow x = \frac{45 \times 35}{25} = \frac{1575}{25} = 63$

Q4: If 4 : 5 :: x : 35, then the value of x is

(a) 42                      (b) 32                      (c) 28                    (d) none of these

Sol:

(c) 28

4 : 5 :: x : 35

$\Rightarrow \frac{4}{5} = \frac{x}{35}$

$\Rightarrow x = \frac{4 \times 35}{5} = 4 \times 7 = 28$

Q5: If a, b, c, d are in proportion, then

(a) ac = bd             (b) ad = bc               (c) ab = cd            (d) none of these

Sol:

Given :

a, b, c, d are in proportion.

a : b :: c : d

$\Rightarrow \frac{a}{b} = \frac{c}{d}$

$\Rightarrow ad = bc$

Q6: If a, b, c are in proportion, then

(a) $a^{2} = bc$            (b) $b^{2} = ac$              (c) $c^{2} = ab$            (d) none of these

Sol:

(b) $b^{2} = ac$

Given:

a, b, c are in proportion.

a : b :: b : c

Product of means = Product of extremes

$\Rightarrow b^{2} = ac$

Q7: Choose the correct statement:

(a) (5 : 8) > ( 3 : 4)       (b) (5 : 8) < ( 3 : 4)      (c) Two ratios cannot be compared

Sol:

(b) (5 : 8) < ( 3 : 4)

We can write

$5 : 8 = \frac{5}{8} \;\; and \;\; 3 : 4 = \frac{3}{4}$

Making the denominator equal:

$\frac{5}{8} \;\; and \;\; \frac{3 \times 2}{4 \times 2} = \frac{6}{8}$

As 6 > 5, $\frac{5}{8} < \frac{3}{4}$

Q8: If Rs.760 is divided between A and B in the ratio 8 : 11 then B’s share is

(a) Rs. 440                (b) Rs. 320                 (c) Rs. 430                (d) Rs. 330

Sol:

(a) Rs. 440

A : B = 8 : 11

Sum of ratio terms = 8 + 11 = 19

B’s share = $\frac{11}{19} \times 760 = \frac {8360} {19}$ = Rs. 440

Q9: Two numbers are in the ratio 5 : 7 and the sum of these numbers is 252. The larger of these numbers is

(a) 85                       (b) 119                        (c) 105                         (d) 147

Sol:

(d) 174

Ratio = 5 : 7

Let x be any number such that we have:

5x + 7x = 252

$\Rightarrow$ 12 x = 252

$\Rightarrow$ x = $\frac{252}{12} = 21$

Now, 5x = 5 $\times$ 21 = 105

7x = 7 $\times$ 21 = 147

The largest number is 147.

Q10: The side of a triangle are in the ratio 1 : 3 : 5 and its perimeter is 90 cm. The length of its largest side is

(a) 40 cm                  (b) 50 cm                  (c) 36 cm                   (d) 54 cm

Sol:

(b) 50 cm

The sided of the triangle are in the ratio 1 : 3 : 5.

Let x be any number such that the sides are 1x cm, 3x cm and 5x cm.

x + 3x + 5x = 90

$\Rightarrow$ 9x = 90

$\Rightarrow x = 10$

First side = x = 10 cm

Second side = 3 $\times$ 10 = 30 cm

Third side = 5x = 5 $\times$ 10 = 50 cm.

The length of the largest side is 50 cm.

Q11: The ratio of boys and girls in a school is 12 : 5. If the number of girls is 840, the total strength of the school is

(a) 1190                 (b) 2380                      (c) 2856                     (d) 2142

Sol:

(c) 2856

Ratio of boys and girls = 12 : 5

Let x be any number such that the number of boys and girls are 12x and 5x respectively.

Number of girls = 840

5x = 840

$\Rightarrow x = \frac{840}{5} = 168$

Number of boys = 12x = 12 $\times$ 168 = 2016

Number of girls = 840

Total strength of the school = 2016 + 840 = 2856

Q12: If the cost of 12 pens is Rs. 138, then the cost of 14 such pens is

(a) Rs. 164             (b) Rs. 161               (c) Rs. 118.30          (d) Rs. 123.50

Sol:

(b) Rs. 161

Cost of 12 pens = Rs. 138

Cost of 1 pen = Rs. $\frac{138}{12}$

Cost of 14 pens = Rs. $14 \times \frac{138}{12} = Rs. \;\frac{1932}{12} = Rs. \; 161$

Q13: If 24 workers can build a wall in 15 days, how many days will 8 workers take to build a similar wall?

(a) 42 days           (b) 45 days             (c) 48 days              (d) none of these

Sol:

(b) 45 days

Time taken by 24 workers to build a wall = 15 days

Time taken by 1 worker to build a wall = $24 \times 15 = 360$ days   ( Less worker will take more time)

Time taken by 8 worker to build a wall = $\frac{360}{8} = 45$ days

Q14: If 40 men can finish a piece of work in 26 days, how many men will be required to finish it in 20 days?

(a) 52                          (b) 31                          (c) 13                          (d) 65

Sol:

(a) 52

Number of men required to finish the work in 26 days = 40

Number of men required to finish it in 1 day = $40 \times 26 = 1040$ men

Number of men required to finish it in 20 days = $\frac{1040}{20} = 52$

Q15: In covering 111 km , a car consumes 6 L of petrol. How many kilometers will it go in 10 L of petrol?

(a) 172 km                (b) 185 km                 (c) 205 km                 (d) 266.5 km

Sol:

(b) 185 km

Distance covered in 6 L of petrol = 111 km

Distance covered in 1 L of petrol = $\frac{111}{6}$ km

Distance covered in 10 L of petrol = $\frac{111}{6} \times 10 = \frac{1110}{6} = 185$

Q16: In a fort, 550 men had provision for 28 days. How many days will it last for 700 men?

(a) 22 days          (b) $35 \frac{7}{11}$ days        (c) 34 days          (d) none of these

Sol:

(a) 22 days

Number of days for which 550 men had provision = 28 days

Number of days for which 1 man had provision = $28 \times 550 = 15400$ days

Number of days for which 700 men had provision = $\frac{15400}{700} = 22$ days.

Q17: The angles of a triangle are in the ratio 3 : 1 : 2. The measure of the largest angle is

(a) $30^{\circ}$                (b) $60^{\circ}$                      (c) $90^{\circ}$                (d) $120^{\circ}$

Sol:

(c) $90^{\circ}$

Ratio of the angles of a triangle is 3 : 1 : 2.

Let x be any number such that the three angles are $(3x)^{\circ} , (1x)^{\circ} \;\; and \;\; (2x)^{\circ}$

We know that the sum of the angles of a triangle is $180^{\circ}$

3x + x + 2x = $180^{\circ}$

$\Rightarrow 6x = 180$

$\Rightarrow x = 30$

Therefore, $(3x)^{\circ} = (3 \times 30)^{\circ} = 90^{\circ}$

$(x)^{\circ} = (1 \times 30)^{\circ} = 30^{\circ}$

$(2x)^{\circ} = (2 \times 30)^{\circ} = 60^{\circ}$

The measure of the largest angle is $90^{\circ}$

Q18: Length and breadth of a rectangle field are in the ratio 5 : 4. If the width of the field is 36 m , what is its length?

(a) 40 m                     (b) 45 m                          (c) 54 m                      (d) 50 m

Sol:

(b) 45 m

Length : Breadth = 5 : 4

Let x be any real number such that the length and the breadth are 5x and 4x respectively.

Now, 4x = 36

x = 0

Length = 5x = 45 m

Q19: If a bus covers 195 km in 3 hours and a train covers 300 km in 4 hours, then the ratio of their speed is

(a) 13 : 15                 (b) 15 : 13                 (c) 13 : 12                (d) 12 : 13

Sol:

(a) 13 : 15

Speed = $\frac{Distance}{Time}$

Speed of the bus = $\frac{195}{3}$ = 65 km/h

Speed of the train = $\frac{300}{4}$ = 75 km/h

Ratio = $\frac{65}{75} = \frac{65 \div 5}{75 \div 5} = \frac{13}{15} = 13 : 15$

Q20: If the cost of 5 bars of soap is Rs. 82.50, then the cost of one dozen such bars is

(a) Rs. 208              (b) Rs. 192                (c) Rs. 198                 (d) Rs. 204

Sol:

(c) Rs. 198

Cost of 5 bars of soap = Rs. 82.50

Cost of 1 bar of soap = $\frac{82.50}{5} = Rs. \; 16.5$

Cost of 12 bars of soap = $16.5 \times 12 = Rs. \; 198$

Q21: If the cost of 30 packets of 8 pencils each is Rs. 600, what is the cost of 25 packets of 12 pencils each?

(a) Rs. 725            (b) Rs. 750              (c) Rs. 480                (d) Rs. 720

Sol:

(b) Rs. 750

Cost of 30 packets of 8 pencils each = Rs. 600

Cost of 1 packet of 8 pencils = $\frac{600}{30} = Rs. \;\; 20$

Cost of 1 pencil = Rs. $\frac{20}{8}$

Cost of 1 packet of 12 pencils = $12 \times \frac{20}{8} = \frac{240}{8} = Rs. \;\; 30$

Cost of 25 packets of 12 pencils each = 25 $\times$ 30 = Rs. 750

Q22: A rail journey of 75 km costs Rs. 215. How much will a journey of 120 km cost?

(a) Rs. 344              (b) Rs. 324           (c) Rs. 268.75         (d) None of these

Sol:

(a) Rs. 344

Cost of rail journey of 75 km = Rs. 215

Cost of rail journey of 1 km = Rs. $\frac{215}{75}$

Cost of rail journey of 120 km = $120 \times \frac{215}{75} = \frac{25800}{75} = Rs. \;\; 344$

Q23: The 1st, 2nd and 4th items of a proportion are 12, 21 and 14 respectively. Its third term is

(a) 16                       (b) 18                           (c) 21                        (d) 8

Sol:

(d) 8

Let the third term be x.,

Then, we have :

12 : 21 :: x : 14

We know:

Product of means = Product of extremes

21x = 12 $\times$ 14

$\Rightarrow 21x = 168$

$\Rightarrow x = \frac{168}{21} = 8$

The third term is 8.

Q24: 10 boys can dig a pitch in 12 hours. How long will 8 boys take to do it?

(a) 9 h 36 min            (b) 15 h              (c) 6 h 40 min             (d) 13 h 20 min

Sol:

(b) 15 h

Time taken by 10 boys to dig a pitch = 12 hours

Time taken by 1 boy to dig a pitch = 12 $\times$ 10 = 120 hours

Time taken by 8 boys to dig a pitch = $\frac{120}{8} = 15$ hours

#### Practise This Question

Find the estimated product of 67 × 62 by rounding off.