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Class 6 R S Aggarwal Maths Solutions

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10. Ratio, Proportion and Unitary Method 10A, 10B, 10C

14. Constructions 14A, 14B

16. Triangles 16A,16B

EXERCISE 10A UNITARY METHOD

 

Q1. cost of 24 oranges = Rs. 33.60

1 = Rs. [less orange less cost]

30 = Rs.

Q2. Cost of 5 bar = Rs. 31

Cost of 1 bar = Rs.
24 =

 

Q3. Cost of 25 m cloth = Rs. 912.50

1m = Rs.

8m = Rs. .

 

Q4. Cost of 9kg rice = Rs. 120.60

Cost of 1 kg rice = Rs.

cost of 50 kg rice= Rs.

 

 

Q5. Weight of 22.5 m rod = 85.5kg.

Wt. of 1m rod =

wt. of 5m rod =

 

Q6. In a day, 18 horses needed to feed = 135 kg.

In a day, 1 horses needed to feed =

In a day, 45 horses needed to feed =

 

Q7. Oils contain in 15 tins = 234 kg

Oils contain in 1 tins =

Oils contain in 10 tins =

Q8. To contains 6000 pens needed 48 boxes.

To contains 6000 pens needed

To contain 1875

 

Q9. Distance covered in 6 hrs = 399km

Dist. covered in 1 hr =

Dist. covered in 16 hrs =

Since, 399 km covered in 6 hrs.

1km

1330 km

 

Q10. Charge to carry 25 tons = Rs. 180.

Charge to carry 1 ton = Rs.

35 ton =

 

Q11. Dist. covered on 12 ltrs of petrol = 222km

1ltr =

22ltrs =

 

Q12. Copper contain in 4.5 g of alloy = 3.5g

1g =

18.9 g =

 

Q13. No. of table can be purchased in Rs. 23520 = 32

No. of table can be purchased in Rs. 1 =

No. of table can be purchased in Rs.51450 = 70.

 

Q14. No. of letters we can buy for Rs. 26.25 = 35

Rs 1 =

Rs.105 =

 

Q15. Time taken by bus to cover 36 km = 40min.

Time taken by bus to cover 1 km =

162km =

Since, in 3 hrs bus travels 162 km

in 1 hrs bus

in 8 hrs bus

 

Q16. 52 x 12 = 624

Cost of 624 pencils = Rs. 499.20

Cost of 1 pencil = Rs.

cost of 650 = .

 

Q17. Cost of 40 x 25 = 1000kg wheat = Rs. 2750.

1kg =Rs.

1750kg = Rs.

 

Q18. In a map 3cm represent distance = 25km

1cm =

7.2cm =

 

Q19. Time taken by a car to cover 60km = 1hr.

1km =

1140km = 1140 = 19hrs.

 

Q20. 1 bag rice is sufficient for 60 person for 3days

1 bag rice is sufficient for 1 person for 3 x 60 days.

1 bag rice is sufficient for 18 person for days.

 

Q21. Provision for 550 men for 28 days

Provision for 1 men for 28 x 550 days

Provision for 700 men for days.

 

Q22. Time taken by 24 workers to build a wall = 15 days

Time taken by 1 workers to build a wall = 15 x 24 days

Time taken by 9 workers to build a wall = days.

 

Q23. In 26 days a work can be finished by 40 men.

In 1 days a work can be finished by 40 x 26 men.

In 16 days a work can be finished by men.

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EXERCISE- 10B RATIO AND PROPORTION

Q1. I] 24 : 36 =

Ii] 45 paise to Rs. 3 = 45 Paise : 300 paise = 3 : 20.

Iii] 4000g : 200 g = 20 : 1.

Iv] 250g : 6000g = 1 : 24.

V] 48 min : 60min = 4 : 5.

Vi] 200cm : 35 cm = 40 : 7.

Vii] 35 x 60 = 2100sec : 45sec =140 : 3.

Viii] 12 : 20 = 3 : 5.

 

Q2. I]

Ii]

Iii]

Iv]

 

Q3. I]

Ii] 21 days : 3days = 7 : 1

Iii] 300cm 5cm : 35cm = 305 : 35 = 61 : 7.

Iv] 48min : 2 x 60 4omin = 48 : 120min 40min = 48 : 160min = 3 : 10.

V] 1000ml 35ml : 270ml = 1035 : 270 =

Vi] 4000g : 2500g = 40 : 25 = 8 : 5

 

Q4. I] 8400 : 5250 =

Ii] 5250 : 8400 = 5 : 8

Iii] 8400 : 13650 =

 

Q5. I]

Ii] Expenditure = income saving = 7650 918 = 6732

7650 : 6732 =

Iii]

 

Q6. Speed = & = 2 : 3.

 

Q7. => 5x = 115 x 3 = 345. x =

 

Q8. => 9x + 5x = 448 => 14x = 448 => x =

 

Q9. Since, in 6 students, no. of pass students = 5

in 1 students, no. of pass students =

in 6 students, no. of pass students = .

Q10. Let the first part be 7x & 2nd part 2x

7x + 2x = 63. x = first part = 7x = 49 & second pare = 2x = 14.

Q11. Let As share = 1x, Bs share 2 & Cs share = 3x.

x + 2x + 3x = 642. => 6x = 642. x =

As share = x = 107, Bs share = 2x = 214 & Cs share = 321.

Q12. Let the two no. be 11x & 12x. ATQ, 11x + 12x = 460.

23x = 460. x =

Hence, 1st no. = 11x = 220 & 2nd no. is 12x = 240. [check 240 + 220 = 460]

Q13. Let the first part be 3x & 2nd part = 4x. ATQ,

3x + 4x = 35cm => 7x = 35cm. x = 5cm.

1st part = 3x = 15cm & 4x = 20cm [check : 20 + 15 = 35cm]

 

Q14. Since, no. of defective bulb out of 10 bulb = 1

no. of defective bulb out of 1 bulb =

no. of defective bulb out of 630 bulb =

 

Q15. LCM of 4 & 16 is 16. Then,

Hence, larger than

 

Q16. Cost of 20 bundle = Rs. 98

Cost of 1 bundle =

Cost of 60 bundle = x 60 = Rs. 288.

Cost of 12 ball pen = 50.40

Cost of 1 ball pen =

Cost of 1 ball pen =

Ratio of price =

 

Q17. 81.9 = 9x => x =

Q18. => 210 = 3x x =

 

Q19. => 2x = 760 x 11 = 8360. x =

income saving = expense

4180 760 = Rs. 3420.

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EXERCISE -10C PROPORTION

 

Q1. I] (T)

Ii] (T)

Iii] (T)

Iv] (F)

V] (T)

Vi] (T)

Vii] (T)

Q2. I] we have 7 : 42 = and 13 : 78 = = 7:42 = 13 : 78.

Hence 7, 42, 13, 78 are in proportion.

Ii] we have 22 : 33 = = and 42 : 63 = 22 : 33 = 42 : 63

Hence 22, 33, 13, 78 are in proportion.

V] we have 72 : 84 = and 186 : 217 = = 72 : 84 = 186 : 217

Hence 72, 84, 186, 217 are in proportion.

Vii] and 46 : 69 69 : 46. Hence 46, 69, 69, 46 are not in proportion.

Q3. Verify the following:

I]

Ii] OR

91 : 104 = 119 : 136

Product of extreme terms = 91 x 136 = 12376

Prod. Of middle terms = 104 x 119 = 12376.

Thus, prod of extreme terms = prod. Of middle term.

 

Q4. I] => =

Ii] => => 3x = 27 x 4 = 108. x =

 

Q5. 32 : 112 :: x : 217

7x = 2 x 217 = 434. x =

 

Q6. 51 : x :: 81 : 108

=> => 3x = 51 x 4 => x =

Hence 1st term = 63 & 2nd term = 68.

 

Q7. I] => => = .

Hence 27, 36, 48 are in proportion.

Iv] 36 : 90 and 90 : 225

= => and

 

Q8. 9 : x :: x : 49 => => x2 = 441 => x == 21.

 

Q9. 48 : 36 : : 36 : x

: : => 4x = 36 x 3 = 108 => x =

 

Q10.

=> 8x = 120 => x =

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EXERCISE -21A CONCEPT OF PERIMETER AND AREA

Perimeter of a rectangle = 2(length + breadth) = 2(l + b)units.

Perimeter of square = (4 x side) = (4a) units.

1dm = 10 cm, 1m = 100cm.

Q1. I] Given, l = 16.8 cm, b = 6.2 cm. we know,

Perimeter of a rectangle = 2(length + breadth) = 2 (16.8 + 6.2)

= 2(23) = 46 cm.

Ii] Perimeter of a rectangle =2(8.6 + 2.45)m = 2 x 11.05 = 22.1 m.

Iii] length = 6m 5dm = 600cm 50cm = 650 cm, b = 4m 6dm = 460 cm.

Perimeter of a rectangle = 2( 650 + 460)cm = 2 x 1110 = 2220 cm = 22.2m.

 

Q2. Perimeter of a rectangular field = 2(24 + 18)m = 2 x 42 = 84m.

Cost of fencing at Rs. 6.25 per meter = 84 x 6.25 = Rs. 525.

Q3. Let the length & breadth of a rectangular field be be 5y & 3y respectively.

ATQ, 2 (5y + 3y ) = 64m. => 2(8y) = 64 => 8y = 32 => y =

Hence, length = 5y = 5 x 4 = 20m. & breadth = 3y = 3 x 4 = 12m.

Q4. Given, b = 23m, l = ?. we know, Perimeter of a rectangle = 2(length + breadth)

Perimeter = 2(23 +y)

ATQ, cost of fencing = 1606, & rate = 14.60 per metre.

Now, length of rectangular field = perimeter =

2(23 + y) = 110 => 23 + y = y = 55 23 = 32.

Q5. Let the length & breadth of a rectangular field be be 3y & 2y respectively.

Perimeter of a rectangle = 2(length + breadth)

Perimeter = 2(3y + 2y) = 2(5y)

ATQ, cost of fencing = Rs. 520. & rate = 6.50 per m.

Perimeter of a rectangular field = Perimeter = 0m.

80 = 2 x 5y => 5y = 40. => y = 8.

Now, length = 3y = 3 x 8 = 24, & breadth = 2y = 2 x 8 = 16m.

 

Q6. I] Given, side of the square = 3.8 cm, we know, Perimeter of square = (4 x side).

  Perimeter of square = (4 x 3.8) = 15.2 cm.

Ii] Given, side of the square = 2.6 m, we know, Perimeter of square = (4 x side).

Perimeter of square = (4 x 2.6) = 10.4 m

Iii] side = 1m 5dm = 100cm 50cm = 150 cm.

Given, side of the square = 150 cm, we know, Perimeter of square = (4 x side)cm.

Perimeter of square = (4 x 150)cm =600cm = 6m.

Q7. Cost of fencing of square field = Rs. 432 & rate = 15 per m.

perimeter of square field = .

4 x side = 28.8m => side =

Q8. perimeter of square field = 4 x side = 4 x 21 = 84m.

Let the length & breadth of a rectangular field be be 4y & 3y respectively.

Perimeter of a rectangular field = 2 (4y + 3y) = 2(7y).

ATQ, 84 = 2(7y) => 7y = 42 => y =

length = 4y = 4 x 6 = 24m, & breadth = 3y = 3 x 6 = 18m.

Q9. Perim. Of a rectangle = 2(5y + 4y)

72 = 2(9y) => 9y = 36 => y = 4. length = 5y = 5 x 4 = 20cm.

Ans (c) 20cm.

Q10. Perim. = . side =

Ans.- (c) 20m.

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EXERCISE 21B CIRCUMFERENCE OF A CIRCLE

I] The perimeter of a circle is called its circumference.

Ii] circumference of a circle = 2. Where .=

Iii] radius (r) = = .

Q1. I] Given, r = 28cm, we know, circumference of a circle = 2

= 2 x x 28 = 176cm.

Ii] Given, r = 10.5cm, we know, circumference of a circle = 2

= 2 x x 10.5 =

Q2. I] Given, d = 14cm, then, r = we know, circumference of a circle = 2

= 2 x x 7 = 44cm.

Ii] Given, d = 35cm, then, r = we know, circumference of a circle = 2

= 2 x x 17.5 =

Q3. circumference of a circle = 2 => 176 = 2 x x r => r =

Q4. circumference of a circle = 2 => 2 x x r => r =

d = r x 2 = 24cm.

Q5. Radius of the wheel = r =

Circumference of the wheel = 2 x x = 242 cm.

In 1 revolution the wheel covers dist. equal to its circumference.

dist. covered by the wheel in 1 revolution = 242 cm.

500 = 500 x 242 = 121000cm = 1210m.

Q6. Radius of the wheel = r =

COC = 2 x x 35cm = 220cm = 2.20m.

No. of revolution required to cover 2.20m = 1

1m =

1.65km = 1650m =

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EXERCISE 21D AREA OF RECTANGLE

I] Area of rectangle = (length x breadth)sq units

Ii] Area of square = (side)2sq units.

Q1. Find the area of rectangle:

I] Area of rectangle = (length x breadth)cm2 = 46 x 25 = 1150cm2.

Ii] = 9 x 6 = 54m2.

Iii] = 14.5 x 6.8 = 98.6m2.

Iv] = 3.5 x 2 = 7km2.

Q2. Area of square = (side)2 = (12)2 = 144cm2.

Q3. 2m 25cm = 2.25m & 1m 20 cm = 1.20m. Area = 2.25 x 1.20 = 2.7m2.

Q4. 30m 75cm = 30.75m, 80cm = 0.80m. Area = (30.75 x 0.80)m = 24.6m2.

Cost t Rs. 20/m = 24.6 x 20 = Rs. 492.

Q5. Area of sheet paper = (324 x 172)cm = 55728 cm2.

Area of 1 envelope = 18cm x 12cm = 216cm2.

No. of envelope can be made =

Q6. Area of the room = 12.5m x 8m = 100m2. Area of square carpet = 82 = 64m2.

Now, 100 64 = 36m2.

Q7. Area of lane = 150m x 9m = 1350m2 = 13500000cm2.

Area of brick = 22.5cm x 7.5 cm = 168.75cm2.

No. of bricks required =

 

Q8. Area of the room = 13m x 9m = 117m2.

Required carpet = (75cm = 0.75m)

Cost of carpeting at the rate of Rs. 6.50/m = 156 x 6.50 = Rs. 1014.

 

Q9. Let the length and breadth of rectangular park are 5y & 3y respectively.

Given, perimeter = 128m. Area of park = ?. 5y x 3y = 15y2 -----(1)

Perimeter of park = 2(l + b)

  128 = 2(5y + 3y) = 2(8y) => 8y = => y =

From (1) area of park = 15y2 = 15 x (8)2 = 15 x 64 = 960m2.

Q10. (|) Draw the line OP ||to AE

In rectangular area APDE :

Area =( 10 x 1)m = 10 m2

In rectangular area OPBC:

Area = ( 8 x 2 )m = 16 m2

Total area = (10 + 16)m2 = 26 m2.

(||) QD = AB DC => 9m 7.5m = 1.5 m

(a) area AQDP =(AQ x QD) = 1m x 1.5 m = 1.5 m2

(b) Area PBCD = (DC X BC)= 7.5 m x 1m = 7.5m2

HQ = HA QA = ( 10 1 )m = 9m

(c) Area QDOH = QD X OH = 9m x 1.5m = 13.5 m2

EF = HG RE = (12 1.5)m = 10.5 m

(d) Area OEFG = (GE X EF) = 10.5 x 2 = 21m2

Total area = 1.5 + 7.5+ 13.5 + 21 = 43.5m2

(|||) AF = BC ED = (9 7.5)m =1.5m

FE = AB DC = (12 2)m = 10 m

AF = ED = 1.5m

(a)    area AFEG = AF X FE = (1.5 x 10)m = 15m2

Area DCBG = DC X BG = (2 x 9)m = 18m2

Total area = 15 +18 = 33m2.

Q11. Area of square = (64m)2 = 4096m2

Area of rectangular plot =70m x 58m = 4060m2

Difference in area = 4096 4060 = 36m2

Hence, the grater plot is square plot by 36m2.

[Perimeter of square = (4 x sides) = 4 x 64 = 256m

Perimeter of rectangular plot = 2( l + b) = 2(70 +58) = 2 x 128 = 256m]

 

Q12. Given cost of cultivating a rectangular area = Rs. 1833

=>Rate = 1.50 per minute

Area = = 1222m2 Given, breadth = 26m, l = ?. Area = length x breadth.

  1222 = length x 26 length = =47

Perimeter of rectangular area = 2( l + b) = 2(47 + 26) = 2 x 73 = 146m

  Cost of fencing the rectangular area at the rate of Rs. 7.50 m2 = 146 x 7.50

= Rs. 1095.

 

Q13. Given , length = 62 m 60 cm= 62. 60m

Breadth = 25m 40cm = 25.40m.

Area of the play ground = 62.60 x 25.40 = 1590.04m2.

Cost of turfing at the rate of Rs.2.50/m2 = 1590.04 x 2.50 = Rs. 3975.10

Perimeter of play ground = 2(l + b) = 2(62.60 + 25.40) = 2 x 88 = 176m.

Since, a man walks 2m he takes 1 sec

1m

176m sec.

For three times he will take = 88 x 3 = 264 sec = 4 min 24 sec.

Q14. Area = length x width

540 = 36 x width => width = = 15.

Perimeter = 2(l + b) = 2(36 + 15) = 2 x 51 = 102 cm. ----------------------------------------------------------------X--------------------------------------------

EXERCISE 22 DATA HANDLING

Q1. Define :

I] Data: an information in the form of numerical figures is known as data.

Ex. - In Science subject marks obtained by 5 students in class test :- 30, 45, 47, 32, 43.

Ii] Raw data: Data obtained in original form is known as raw data.

Iii] Array: Arranging the numbers in an ascending or a descending order is called array.

Iv] Tabulation of data: Arranging the data in the form of a table is called tabulation of data.

V] Observation : Each number in a data is called observation.

Vi] Frequency of an observation : The no. of times a particular observation occurs is called its frequency.

Vii] Statistics : it is the science which deals with the collection, presentation, analysis and interpretation of numerical data.

Q6. I] Data means information in the form of numerical figures.

Ii] Data obtained in the original form is called raw data.

Iii] Arranging the numerical figures in an ascending or a descending order is called an array.

Iv] the no of times a particular observation occurs is called its frequency.

V] Arranging the data in the form of a table is called tabulation.

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EXERCISE 24 BAR GRAPH

Q1. I] it gives information about the marks obtained by a student in an examination.

Ii] The student is poorest in science subject.

Iii] the student is best in mathematics.

Iv] in Hindi and maths he got more than 40 marks.

Q2. I] The bar graph gives the information about the number of members in each of the 60 families of a colony.

Ii] 10 families have 3 members each.

Iii] 5 couples have no child.

Iv] A family of 4 members is most common.

Q3. I] In the 2nd week production was maximum.

Ii] in the 4th week production was minimum.

Iii] The average production is 720 per week.

Iv] 2400 cycles were produced in the first three weeks.

Q4. I] The bar graph shows the different modes of transport to a school used by 51 students of a locality.

Ii] Bicycle is used by maximum number of students.

Iii] 14 students use bus for going to school.

Iv] 37 students of the locality do not use bus for going to school.

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