20 of (a+b)=50 (a-b).find the ratio of a and b
1) A: B: C is in the ratio of 3: 2: 5. How much money will C get out of Rs 1260?
Answer: D Explanation: C's share = [C's ratio/ sum of ratios] * total amount 2) If a: b is 3: 4 and b: c is 2: 5. Find a: b: c.
Answer: C Explanation: The ratio of a: b is 3: 4 Note: To find the ratio in such questions, multiply a to b, then b to b, and then b to c. a: b: c = 3*2: 4*2: 4*5 3) A: B is 1: 2; B: C is 3: 2 and C: D is 1:3. Find A: B: C: D.
Answer: A Explanation: ATQ, Note: To understand the shortcut, remember you need to make the right-hand side missing numbers the same as that of last given number, and for the left-hand side the same is done. i.e., C: D will contain 2: 2 because 2 is the last number on the right side. Or, A: B: C: D Now, multiply vertically and get A: B: C: D. So, A: B: C: D = (1*3*1): (2*3*1): (2*2*1): (2*2*3) 4) 5600 is to be divided into A, B, C, and D in such a way that the ratio of share of A: B is 1: 2, B: C is 3: 1, and C: D is 2: 3. Find the sum of (A and C) and (B and C).
Answer: D Explanation: ATQ, Note: To understand the shortcut, remember you need to make the right-hand side missing numbers the same as that of last given number, and for the right-hand side the same is done. i.e., C: D will contain 2: 2 because 2 is the last number on the right side. Or, A: B: C: D Now, multiply vertically and to get A: B: C: D. So, A: B: C: D = (1*3*2): (2*3*2): (2*1*2): (2*1*3) Now,
the share of A and C = [(A+C)/ (A+B+C+D)] * total amount Solution 2: Find A: B: C: D Now, multiply vertically and get A: B: C: D. So, A: B:
C: D = (1*3*2): (2*3*2): (2*1*2): (2*1*3) Sum of the ratios = 3+6+2+3 = 14, but ATQ, it is 5600 Rs. Now, the share of (A+C) = 1200+800 = 2000 5) The ratio of the total amount distributed in all the males and females as salary is 6: 5. The ratio of the salary of each male and female is 2: 3. Find the ratio of the no. of males and females.
Answer: D Explanation: The total salary of males: the total salary of females = 6:5 To find the
number of men and women, divide the total salary of males and females by salary of each male and female. Ratio and Proportion Aptitude Test Paper 2 Ratio and Proportion Aptitude Test Paper 3 Ratio and Proportion Aptitude Test Paper 4 Ratio and Proportion Concepts Mathematics Part I Solutions Solutions for Class 9 Math Chapter 4 Ratio And Proportion are provided here with simple step-by-step explanations. These solutions for Ratio And Proportion are extremely popular among Class 9 students for Math Ratio And Proportion Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Mathematics Part I Solutions Book of Class 9 Math Chapter 4 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Mathematics Part I Solutions Solutions. All Mathematics Part I Solutions Solutions for class Class 9 Math are prepared by experts and are 100% accurate. Page No 61:Question 1:From the following pairs of numbers, find the reduced form of ratio of first number to second number. (i) 72, 60(ii) 38, 57(iii) 52, 78 Answer: (i) Thus, the reduced form of 72 : 60 is 6 : 5. (ii) Thus, the reduced form of 38 : 57 is 2 : 3. (iii) Thus, the reduced form of 52 : 78 is 2 : 3. Page No 61:Question 2:Find the reduced form of the ratio of the first quantity to second quantity. (i) 700 rs, 308 rs(ii) 14 rs, 12 rs. 40 paise. (iii) 5 litre, 2500 ml (iv) 3 years 4 months, 5years 8 months (v) 3.8 kg, 1900 gm (vi) 7 minutes 20 seconds, 5 minutes 6 seconds. Answer: (i) (ii) 12 rupees 40 paise = 1200 + 40 = 1240 p ∴ 14 rupees : 12 rupees 40 paise = 1400 p : 1240 p = 14001240= 1400÷401240÷40=3531 = 35 : 31 (HCF of 1400 and 1240 = 40) (iii) ∴ 5 litre : 2500 mL = 5000 mL : 2500 mL = 50002500 =5000÷25002500÷2500=21 = 2 : 1 (HCF of 5000 and 2500 = 2500) (iv) 5 years 8 months = 5 years + 8 months = 5 × 12 + 8 = 60 + 8 = 68 months ∴ 3 years 4 months : 5 years 8 months = 40 months : 68 months = 4068=40÷468÷4=1017 = 10 : 17 (HCF of 40 and 68 = 4) (v) ∴ 3.8 kg : 1900 g = 3800 g : 1900 g = 38001900=3800÷19001900÷1900=21 = 2 : 1 (HCF of 3800 and 1900 = 1900) (iv) 5 minutes 6 seconds = 5 minutes + 6 seconds = 5 × 60 + 6 = 300 + 6 = 306 seconds ∴ 7 minutes 20 seconds : 5 minutes 6 seconds = 440 seconds : 306 seconds = 440306= 440÷2306÷2=220153 = 220 : 153 (HCF of 220 and 153 = 2) Page No 61:Question 3:Express the following percentages as ratios in the reduced form. (i) 75 : 100 (ii) 44 : 100 (iii) 6.25% (iv) 52 : 100 (v) 0.64% Answer: (i) (ii) (iii) (iv) (v) Page No 61:Question 4:Three persons can build a small house in 8 days. To build the same house in 6 days, how many persons are required? Answer:Let the number of persons required to build the same house in 6 days be x. The number of persons and the number of days required to build the house are in inverse variation. So, the product of number of persons and the number of days required to build the house is constant. ∴ x × 6 = 3 × 8 ⇒ x = 3×86=246 = 4 Thus, 4 persons are required to build the same house in 6 days. Page No 61:Question 5:Convert the following ratios into percentage. (i) 15 : 25 (ii) 47 : 50 (iii) 710 (iv) 546600 (v) 716 Answer:(i) (ii) (iii) (iv) (v) Page No 61:Question 6:The ratio of ages of Abha and her mother is 2 : 5. At the time of Abha's birth her mothers age was 27 year. Find the present ages of Abha and her mother. Answer:The ratio of the present ages of Abha and her mother is 2 : 5. Let the present age of Abha and her mother be 2x years and 5x years, respectively. ∴ Abha's mother age at the time of Abha's birth = 5x − 2x = 3x years It is given that at the time of Abha's birth her mothers age was 27 year. ∴ 3x = 27 ⇒ x = 9 ∴ Present age of Abha = 2x = 2 × 9 = 18 years Present age of Abha's mother = 5x = 5 × 9 = 45 years Thus, the present age of Abha and her mother is 18 years and 45 years, respectively. Page No 61:Question 7:Present ages of Vatsala and Sara are 14 years and 10 years respectively. After how many years the ratio of their ages will become 5 : 4? Answer:Let after x years, the ratio of their ages will become 5 : 4. Age of Vatsala after x years = (14 + x) years Age of Sara after x years = (10 + x) years ∴14+x10+x=54
⇒56+4x=50+5x⇒5x-4x=56-50⇒x=6 Page No 61:Question 8:The ratio of present ages of Rehana and her mother is 2 : 7. After 2 years, the ratio of their ages will be 1 : 3. What is Rehana's present age ? Answer:Let the present ages of Rehana and her mother be 2x years and 7x years, respectively. After 2 years, Age of Rehana = (2x + 2) years Age of Rehana's mother = (7x + 2) years It is given that after 2 years, the ratio of their ages will be 1 : 3. ∴2x+27x+2=13⇒6x+6=7x+2⇒7x-6x=
6-2⇒x=4 Thus, the present age of Rehana is 8 years. Page No 63:Question 1:Using the property ab = akb k , fill in the blanks substituting proper numbers in the following. (i) 57 = ....28 = 35.... = . ...3.5 (ii) 914= 4.5.... = ....42 = ....3.5 Answer: (i) Now, 57 = 5×47×4
= 5×77×7 = 5×0.57×0.5⇒57 = 20
28 = 3549 = 2.53.5 Now, 914=9×0.514×0.5=9×314×3=9×0.2514×0.25 ⇒914= 4.57 = 2742 = 2. 253.5 Page No 63:Question 2:Find the following ratios. (i) The ratio of radius to circumference of the circle. (ii) The ratio of circumference of circle with radius r to its area. (iii) The ratio of diagonal of a square to its side, if the length of side is 7 cm. (iv) The lengths of sides of a rectangle are 5 cm and 3.5 cm. Find the ratio of its perimeter to area. Answer: (i) ∴ Circumference of the circle = 2πr units Radius of the circle : Circumference of the circle = r : 2πr = r2πr=12π = 1 : 2π Thus, the ratio of radius to cirumference of the circle is 1 : 2π. (ii) ∴ Circumference of the circle = 2πr units Area of the circle = πr2 square units Circumference of the circle : Area of the circle = 2πr : πr2 = 2πrπr2=2 r = 2 : r Thus, the ratio of circumference of circle with radius r to its area is 2 : r. (iii) ∴ Length of diagonal of the square = 2 × Side of the square = 72 cm Length of diagonal of the square : Side of the square = 72 cm : 7 cm = 727=21 = 2 : 1 Thus, the ratio of diagonal of the square to its side is 2 : 1. (iv) Breadth of the rectangle, b = 3.5 cm ∴ Perimeter of the rectangle = 2(l + b) = 2 × (5 + 3.5) = 2 × 8.5 = 17 cm Area of the rectangle = l × b = 5 × 3.5 = 17.5 cm2 Perimeter of the rectangle : Area of the rectangle = 17 : 17.5 = 1717.5=170175=170÷5175÷5=3435 = 34 : 35 Thus, the ratio of perimeter to area of the rectangle is 34 : 35. Page No 63:Question 3:Compare the following pairs of ratios. (i) 53, 37 (ii) 35 57, 63125 (iii) 518 , 17121 (iv) 80 48 , 4527 (v) 9.25.1 , 3.47.1 Answer: (i) ∴53<37 (ii) ∴ 3557<63125 (iii) ∴ 518>17121 (iv) ∴8048=4527 (v) ∴9.25.1>3.47.1 Page No 63:Question 4:(i) □ ABCD is a parallelogram. The ratio of ∠A and ∠B of this parallelogram is 5 : 4. Find the measure of ∠B. (ii) The ratio of present ages of Albert and Salim is 5 : 9. Five years hence ratio of their ages will be 3 : 5. Find their present ages. (iii) The ratio of length and breadth of a rectangle is 3 : 1, and its perimeter is 36 cm. Find the length and breadth of the rectangle. (iv) The ratio of two numbers is 31 : 23 and their sum is 216. Find these numbers. (v) If the product of two numbers is 360 and their ratio is 10 : 9, then find the numbers. Answer: (i) Let the measure of ∠A and ∠B be 5x and 4x, respectively. Now, ∠A + ∠B = 180º (Adjacent angles of a parallelogram are supplementary) ∴ 5x + 4x = 180º ⇒ 9x = 180º ⇒ x = 20º ∴ Measure of ∠B = 4x = 4 × 20º = 80º Thus, the measure of ∠B is 80º. (ii) 5 years hence, Age of Albert = (5x + 5) years Age of Salim = (9x + 5) years It is given that five year hence, the ratio of their ages will be 3 : 5. ∴5x+59x+5=35⇒25x+25=27
x+15⇒27x-25x=25-15⇒2x=10⇒x=5 Present age of Salim = 9x = 9 × 5 = 45 years Thus, the present age of Albert is 25 years and the present age of Salim is 45 years. (iii) Perimeter of the rectangle = 36 cm ∴ 2(Length + Breadth) = 36 cm ⇒ 2(3x + x) = 36 ⇒ 2 × 4x = 36 ⇒ 8x = 36 ⇒ x = 4.5 ∴ Length of the rectangle = 3x = 3 × 4.5 = 13.5 cm Breadth of the rectangle = x = 4.5 cm Thus, the length and breadth of the rectangle is 13.5 cm and 4.5 cm, respectively. (iv) Sum of the two numbers = 216 ∴ 31x + 23x = 216 ⇒ 54x = 216 ⇒ x = 21654 = 4 ∴ One number = 31x = 31 × 4 = 124 Other number = 23x = 23 × 4 = 92 Thus, the two numbers are 92 and 124. (v) Product of the two numbers = 360 ∴ 10x × 9x = 360 ⇒ 90x2 = 360 ⇒ x2 = 4 ⇒ x = 2 ∴ One number = 10x = 10 × 2 = 20 Other number = 9x = 9 × 2 = 18 Thus, the two numbers are 18 and 20. Page No 64:Question 5:If a : b = 3 : 1 and b : c = 5 : 1 then find the value of (i) a315b2c 3 (ii) a27bc Answer:a : b = 3 : 1 ∴ab=31 ⇒a=3 b .....(1) b : c = 5 : 1 ∴bc=51 ⇒b=5c .....(2) From (1) and (2), we have a = 3 × 5c = 15c (i) Page No 64:Question 6:If 0.04×0.4×a = 0.4×0.04×b then find the ratio ab. Answer:
Page No 64:Question 7:( x + 3) : ( x + 11) = ( x- 2) : ( x + 1) then find the value of x. Answer:
Page No 70:Question 1:If ab= 73 then find the values of the following ratios. (i) 5a + 3b5a - 3b (ii) 2a2 + 3b22a2 - 3b2 (iii) a3 - b3 b3 (iv) 7a + 9b7a - 9b Answer:
Page No 70:Question 2:If 15a2 + 4b215a2 - 4b2 = 477 then find the values of following ratios. (i) ab (ii) 7a - 3b7a + 3b (iii) b2 - 2a2b2 + 2a2 (iv) b3 - 2a3 b3 + 2a3 Answer: (i) Applying componendo and dividendo, we get 15a2+4b2+15a2-4b215a2
+4b2-15a2-4b2=47+747-7⇒15a2+4b2
+15a2-4b215a2+4b2-15a2+4b2=5440⇒30a2
8b2=2720⇒a2b2=27×820×30=925⇒ab=
925=35 (iv) Page No 70:Question 3:If 3a + 7 b3a - 7b= 43 then find the value of the ratio 3a2 - 7b23a2 + 7b2 Answer:3a+7b3a-7b=43 Applying componendo and dividendo, we get
3a+7b+3a-7b3a+7b-3a-7b=4+34-3⇒3a+7b+3a-7b3a+7b-3a+7b=71⇒6a14b=7
⇒ab=14×76=493 .....
1 3a2-7b23a2+7b2=3a2b
2-7b2b23a2b2+7b2b2
=3×ab2-73×ab2+7=3×
4932-73×4932+7
Using 1=2401-2132401+213 Page No 70:Question 4:Solve the following equations. (i) x2 + 12x -203x - 5 = x2 + 8x + 122x + 3 (ii) 10x2 + 15x + 635x2 - 25x + 12= 2x + 3x - 5 (iii) 2x + 12 + 2x - 122x + 1 2 - 2x - 12= 178 (iv) 4x + 1 + x + 34x + 1 -x + 3= 41 (v) 4x + 1 2 + 2x + 324x2 + 12x + 9 = 6136 (vi) 3x - 4 3 - x + 1 3 3x - 4 3 + x + 1 3 = 61189 Answer: (i) Multiplying both sides by 14 , we get x2+12x-2012x-20=x2+8x+128x+12 Using dividendo, we get x2+12x-20-12x-2012x-20=x2+8x+12-8
x+128x+12⇒x2+12x-20-12x+2012x-20=x2+8x+12-
8x-128x+12⇒x212x-20=x28x+12 If x ≠ 0, then x2 ≠ 0. Dividing both sides by x2, we get 112x-20=18x+12⇒12x-20=8x+
12⇒12x-8x=20+12⇒4x=32⇒x=8 (ii) 10 x2+15x+6310x2+15x=5x2-25x+125x2-25x Using dividendo, we get 10x2+15x+63-10x2+15x10x2+15x=5x2-25
x+12-5x2-25x5x2-25x⇒6310x2+15x=125x2
-25x⇒635x2x+3=125xx-5⇒632x+3=
12x-5 (iii) Applying componendo and dividendo, we get 2x+
12 +2x-12+2x+12 -2x-12
2x+12 +2x-12-2x+12 -2x
-12=17+817-8⇒22x+12 22x-12
=259⇒2x+12x-1=259=53 (iv) Applying componendo and dividendo, we get 4x+1+x+3
+4x+1-x+34x+1+x+3-4x+1-x+3=
4+14-1⇒24x+12x+3=53⇒4x+1x+3
=53 4x+1x+3=532=259⇒36x+9=
25x+75⇒36x-25x=75-9⇒11x=66⇒x=6 (v) 4x+1
2+2x+32-2x+322x+32=61-3636⇒
4x+122x+32=2536 4x+12x+3=2536
=56⇒24x+6=10x+15⇒24x-10x=15-6⇒14x=9⇒x=
914 (vi) Applying componendo and dividendo, we get 3x-43
-x+13+3x-43+x+133x-43
-x+13-3x-43+x+13=61+18961-189
⇒23x-43-2x+13=250-128⇒3x-4
3x+13=12564 3x-4x+1=125643⇒
3x-4x+1=5433=54⇒12x-16=5x+5⇒12x-5x=16+5⇒7x=21⇒x=3 Page No 73:Question 1:Fill in the blanks of the following (i) x7 = y3 = 3x + 5y.......... = 7x - 9y......... (ii) a3 = b 4 =c7 = a -2b+3c....... = ..........6 - 8 + 14 Answer: (i) x7=y3=7×x7×7=
-9×y-9×3⇒x7=y3=7x
49=-9y-27=7x-9y49+-27
Theorem of equal ratios⇒x7=y3=7x-9y22 (ii) a3=b4=c7=2×a2×3=-2
×b-2×4=2×c2×7⇒a3=b4=c7=2a6=-2b-8=2c14=2a+
-2b+2c6+-8+14 Theorem
of equal ratios⇒a3=b4=c7=
2a-2b+2c6-8+14 Page No 73:Question 2:5 m - n= 3m + 4n then find the values of the following expressions. (i) m2+ n2m2- n2 (ii) 3m + 4 n3m - 4n Answer:
Page No 73:Question 3:(i) If a(y+z) = b(z+x) = c(x+y) and out of a, b, c no two of them are equal then show that, y - za b - c= z - x b c - a = x - yc a - b (ii) If x3x - y - z = y3y - z - x = z3z - x - y and x + y + z ≠ 0 then show that the value of each ratio is equal to 1. (iii) If ax + byx + y = bx +azx + z = ay + bzy + z and x + y + z ≠ 0 then show that a + b2. (iv) If y + za = z + xb = x + yc then show that xb + c - a = yc + a - b = za + b - c . (v) If 3x - 5y5z + 3y = x + 5zy - 5x = y - zx - z then show that every ratio = xy. Answer: (i) ⇒a=ky+z, b=kz+x, c=kx+y∴ab-c=
ky+zkz+x-kx+y⇒ab-c=k2y+z
x+y-z-xz+xx+y⇒y-zab-c=k
2x+yy+zz+x ....
.1 z-xbc-a=k2x+yy+zz+x
.....2x-yca-b
=k2x+yy+zz+x
.....3 y-zab-c=z-x bc-a =x-yca-b (ii) ⇒x3x-y-z=y3y-z-x=z3z-x-y=x+y+zx+ y+z ⇒x3x-y-z=y3y-z-x=z3z-x-y=1 (iii) ⇒ax+byx+y=bx+azx+z=ay+bzy+z=
a+bx+a+by+a+bz2x+2y+2z⇒ax+byx+y
=bx+azx+z=ay+bzy+z=a+bx+y+z2x+y
+z⇒ax+byx+y=bx+azx+z=ay+bzy+z=
a+b2 ay +z=bz+x=cx+y (By invertendo) ay+z=bz+x=c
x+y=b+c-az+x+x+y-y-z Theorem of equal ratios⇒ ay+z=bz+x=cx+y=b+c-a2x
.....1 ay+z=bz+x=cx
+y=c+a-bx+y+y+z-z-x Theorem of equal
ratios⇒ ay+z=bz+x=cx+y=c+a-b2y
.....2 ay+z=bz+x
=cx+y=a+b-cy+z+z+x-x-y Theorem of
equal ratios⇒ ay+z=bz+x=cx+y=a+b-c2z
.....3 b+c-a2x=c
+a-b2y=a+b-c2z⇒b+c-ax=c+a-b
y=a+b-cz⇒xb+c-a=yc+a-b=za+b-c
By invertendo Page No 74:Question 4:Solve (i) 16x2 - 20x + 98x2 + 12x + 21 = 4x - 52x + 3 (ii) 5y2 + 40y - 12 5y + 10y2 - 4 = y + 81 + 2y Answer: (i) If x = 0, then 16×0-20×0+98×0+12×0+21=4×0-52×0+3 ⇒ 9 21=-53, which is not true. So, x = 0 is not a solution of the given equation. Now, 16x2-20x+98x2+12x+21=
4x-52x+3=4x4x-54x2x+3⇒16x2-20x+
98x2+12x+21=4x-52x+3=16x2-20x+9-4x4x-5
8x2+12x+21-4x2x+3 Theorem of equal ratios
⇒16x2-20x+98x2+12x+21=4x-52x+3=16x2-20
x+9-16x2+20x8x2+12x+21-8x2-12x⇒16x2-20x+98
x2+12x+21=4x-52x+3=921 (ii) If y = 0, then 5×0+40×0-125×0+10×0-4= 0+81+2×0 ⇒ -12-4=81, which is not true. So, y = 0 is not a solution of the given equation. Now,
5y2+40y-125y+10y2-4=y+81+2y=5yy+85y
1+2y⇒5y2+40y-125y+10y2-4=y+81+2y=
5y2+40y-12-5yy+85y+10y2-4-5y1+2y
Theorem of equal ratios⇒5y2+40y-125y+10y
2-4=y+81+2y=5y2+40y-12-5y2-40y5y+10y2-4-5
y-10y2⇒5y2+40y-125y+10y2-4=y+81+2y=
-12-4 Page No 77:Question 1:Which number should be subtracted from 12, 16 and 21 so that resultant numbers are in continued proportion? Answer:Le the number x be subtracted from 12, 16 and 21 so that resultant numbers are in continued proportion. ∴ (12 − x), (16 − x), (21 − x) are in continued proportion. ⇒12-x16-
x=16-x21-x⇒16-x2=12-x×21-x⇒256+x
2-32x=252-12x-21x+x2⇒256-32x=252-33x⇒33x-32x=252-256
⇒x=-4 Page No 77:Question 2:If (28-x) is the mean proportional of (23-x) and (19 -x) then find the vaue of x. Answer:It is given that (28 − x) is the mean proportional of (23 − x) and (19 − x). ∴28-x2=23-x×19
-x⇒784+x2-56x=437-23x-19x+x2⇒784-56x=437-42x⇒56
x-42x=784-437⇒14x=347⇒x=34714 Page No 77:Question 3:Three numbers are in continued proportion, whose mean proportional is 12 and the sum of the remaining two numbers is 26, then find these numbers. Answer:Let the remaining two numbers be x and y. So, the numbers x, 12, y are in continued proportion. ∴ xy = (12)2 = 144 .....(1) Also, x + y = 26 .....(2) Solving (1) and (2), we get x+144x=26⇒x2+144=26x⇒x2-26x+144
=0⇒x2-18x-8x+144=0⇒xx-18-8x-18=0⇒x
-8x-18=0 ⇒ x = 8 or x = 18 When x = 8, y = 26 − 8 = 18 [Using (2)] When x = 18, y = 26 − 18 = 8 Thus, the numbers are 8, 12, 18 or 18, 12, 8. Page No 77:Question 4:If (a + b + c) (a- b + c) = a2 + b2 + c2 show that a, b, c are in continued proportion. Answer:
Page No 77:Question 5:If ab= bc and a,b,c > 0 then show that, (i) (a + b + c) (b - c) = ab- c2 (ii) (a2 + b2) (b2 + c2 ) = (ab + bc)2 (iii) a2 + b2ab = a + c b Answer:ab=bc⇒b2=ac .....1 (i) Page No 77:Question 6:Find mean proportional of x + yx - y , x2 - y2 x2y2 Answer:If b is the mean proportional of a and c, then b2=ac or b=ac. Mean proportional of x+yx-y and x2-y2x2y2 = x+yx-y×x2-y2x2y2=x+yx-y×x -yx+yx2y2=x+y2xy2= x+yxy2=x+yxy Page No 77:Question 1:Select the appropriate alternative answer for the following questions. (i) If 6 : 5 = y : 20 then what will be the value of y ? (A) 15(B) 24(C) 18(D) 22.5 (ii) What is the ratio of 1 mm to 1 cm ? (A) 1 : 100(B) 10 : 1(C) 1 : 10(D) 100 : 1 (iii * ) The ages of Jatin, Nitin and Mohasin are 16, 24 and 36 years respectively. What is the ratio of Nitin’s age to Mohasin’s age ? (A) 3 : 2 (B) 2 : 3(C) 4 : 3(D) 3 : 4
(iv) 24 Bananas were distributed between Shubham and Anil in the ratio 3 : 5, then how many bananas did Shubham get ? (A) 8(B) 15 (C) 12(D) 9 (v) What is the mean proportional of 4 and 25 ? Answer: (i) ∴65=y20⇒y=6×205=24 Hence, the correct answer is option (B). (ii) = 1 mm : 1 cm = 1 mm : 10 mm (1 cm = 10 mm) = 1 : 10 Hence, the correct answer is option (C). (iii) Age of Mohasin = 36 years ∴ Ratio of Nitin's age to Mohansin's age = Age of Nitin : Age of Mohasin = 24 years : 36 years =2436=24÷12
36÷12 HCF of 24 and 36=12=23 Thus, the ratio of Nitin's age to Mohasin's age is 2 : 3. Hence, the correct answer is option (B). (iv) Let the number of bananas received by Shubham and Anil be 3x and 5x, respectively. ∴ 3x + 5x = 24 ⇒ 8x = 24 ⇒ x = 3 ∴ Number of bananas received by Shubham = 3x = 3 × 3 = 9 Thus, Shubham gets 9 bananas. Hence, the correct answer is option (D). (v) =4×25=100=10 Hence, the correct answer is option (C). Page No 78:Question 2:For the following numbers write the ratio of first number to second number in the reduced form. (i) 21, 48 (ii) 36, 90 (iii) 65, 117 (iv) 138, 161(v) 114, 133 Answer: (i) Thus, the reduced form of 21 : 48 is 7 : 16. (ii) Thus, the reduced form of 36 : 90 is 2 : 5. (iii) Thus, the reduced form of 65 : 117 is 5 : 9. (iv) Thus, the reduced form of 138 : 161 is 6 : 7. (v) Thus, the reduced form of 114 : 133 is 6 : 7. Page No 78:Question 3:Write the following ratios in the reduced form. (i) Radius to the diameter of a circle. (ii) The ratio of diagonal to the length of a rectangle, having length 4 cm and breadth 3 cm. (iii) The ratio of perimeter to area of a square, having side 4 cm. Answer: (i) Diameter of the circle = 2 × Radius of the circle = 2r units ∴ Ratio of radius to diameter of a circle = Radius of the circle : Diameter of the circle = r : 2r = 1 : 2 (ii) Breadth of the rectangle, b = 3 cm Now, Diagonal of the rectangle = l2+b2=32+42=9+16=25 = 5 cm ∴ Ratio of diagonal to the length of the rectangle = Diagonal of the rectangle : Length of the rectangle = 5 cm : 4 cm = 5 : 4 (iii) Perimeter of the square = 4 × Side of the squre = 4 × 4 cm = 16 cm Area of the squre = (Side of the square)2 = (4 cm)2 = 16 cm2 ∴ Ratio of perimeter to area of the square = Perimeter of the square : Area of the squre = 16 : 16 = 1 : 1 Page No 78:Question 4:Check whether the following numbers are in continued proportion. (i) 2, 4, 8 (ii) 1, 2, 3 (iii) 9, 12, 16 (iv) 3, 5, 8 Answer:The numbers a, b, c are in continued proportion if ab=bc. (i) (ii) (iii) (iv) Page No 78:Question 5:a, b, c are in continued proportion. If a = 3 and c = 27 then find b. Answer:It is given that 3, b, 27 are in continued proportion. ∴3b=b27⇒b2=3×27⇒b2=81⇒b=81=9 Page No 78:Question 6:Convert the following ratios into percentages.. (i) 37 : 500 (ii) 58 (iii) 2230 (iv) 516 ( v) 1441200 Answer: (i) (ii) (iii) (iv) (v) Page No 78:Question 7:Write the ratio of first quantity to second quantity in the reduced form. (i) 1024 MB, 1.2 GB [(1024 MB = 1 GB)] (ii) 17 Rupees, 25 Rupees 60 paise (iii) 5 dozen, 120 units (iv) 4 sq.m, 800 sq.cm (v) 1.5 kg, 2500 gm Answer: (i) = 1024 MB : 1.2 × 1024 MB =10241.2×1024
=11.2=1012=56 (ii) 25 Rupees 60 paise = 25 Rupees + 60 paise = 2500 paise + 60 paise = 2560 paise ∴ Ratio of 17 Rupees to 25 Rupees 60 paise = 1700 paise : 2560 paise =17002560=1700÷202560÷20
HCF of 1700 and 2560=20=85128 (iii) = 5 × 12 units : 120 units = 60 units : 120 units
=60120=12 (iv) 4 m2 = 4 × 10000 cm2 = 40000 cm2 ∴ Ratio of 4 m2 to 800 cm2 = 40000 cm2 : 800 cm2 =40000
800=50 (v) ∴ Ratio of 1.5 kg to 2500 g = 1500 g : 2500 g =15002500=35 Page No 78:Question 8:If ab= 23 then find the values of the following expressions. (i) 4a + 3b3b (ii) 5a2 + 2b25a2 - 2b2 (iii) a3 + b3 b3 (iv) 7b - 4a7b + 4a Answer: (i) 4a+3b3
b=8+99⇒4a+3b3b=179 5a2+2b25a2-2b2=20+1820-18
⇒5a2+2b25a2-2b2=382⇒5a2+2b2
5a2-2b2=19 a3+b3b3
=8+2727⇒a3+b3b3=3527 Applying invertendo, we get ba=32 ⇒7b4a=7×
34×2=218 7b+4a7b-4a=21+821-8⇒7b+4a7b-4a=2913 7b-4a7b+4a=1329 Page No 78:Question 9:If a, b, c, d are in proportion , then prove that (i) 11a2 + 9ac11b2 + 9bd = a2 + 3acb2 + 3bd (ii) a2 + 5c2 b2 + 5d2= ab (iii) a2 + ab + b2a 2 - ab + b2= c2 +cd+ d2c2 -cd+ d2 Answer:It is given that a, b, c, d are in proportion. ∴ab=cd=k⇒a=bk, c=dk 11a2+9ac11b2+9bd=a2+3acb2+3bd (ii) a2+5c2b2+5d2=ab (iii) a2+ab+b2a2-ab+b2=c2+cd+d2c2-cd +d2 Page No 79:Question 10:If a, b, c are in continued proportion , then prove that (i) aa + 2b= a - 2ba - 4c (ii) bb + c = a - b a - c Answer:a, b, c are in continued proportion. ∴ab=bc=k⇒a=bk, b=ck⇒a=bk=ck×k=ck2 aa+2b=a- 2ba-4c (ii) bb+c=a-ba-c Page No 79:Question 11:Solve : 12x2 + 18x + 4218x2 + 12x + 58= 2x + 33x + 2 Answer:12x2+18x+4218x2 +12x+58=2x+33x +2 If x = 0, then 12×0+18×0+4218×0+12×0+58=2×0+33×0+2 ⇒4258=32, which is not true. So, x = 0 is not a solution of the given equation. Now, 12x2+18x+4218x2 +12x
+58=2x+33x+2=6x2x+36x3x+2⇒12x2
+18x+4218x2 +12x+58=2x+33x+2=12x2+18x+42-
6x2x+318x2 +12x+58-6x3x+2
Theorem of equal ratios⇒12x2+18x+4218x2 +12x+58=
2x+33x+2=12x2+18x+42-12x2-18x18x2 +12x+58-18x
2 -12x⇒12x2+18x+4218x2 +12x+58=2x+33x+
2=4258 Page No 79:Question 12:If 2x - 3y3z + y= z - yz - x = x + 3z2y - 3x then prove that every ratio = xy. Answer:
Page No 79:Question 13:(13 * ) If by + czb2 + c2= cz +ax c2 + a2= ax + bya2 + b2 then prove that xa= yb= zc . Answer:by+czb2 +c2=cz+ax c2+a2=ax+
bya2+b2⇒by+czb2 +c2=cz+ax c2
+a2=ax+bya2+b2=by+cz+cz+ax+ax+by
b2 +c2+c2+a2+a2+b2
Theorem of equal ratios⇒by+czb2 +c2=c
z+ax c2+a2=ax+bya2+b2=2ax+2by+2cz2
a2+2b2+2c2⇒by+czb2 +c2=cz+ax
c2+a2=ax+bya2+b2=2ax+by+cz2a2+b
2+c2⇒by+czb2 +c2=cz+ax c2+a2
=ax+bya2+b2=ax+by+cza2+b2+c2 by+cz-ax+by+czb2 +c2-a2+
b2+c2=cz+ax-ax+by+cz c2+a2-
a2+b2+c2=ax+by-ax+by+cza2+b2
-a2+b2+c2=ax+by+cza2+b2+c2 View NCERT Solutions for all chapters of Class 9 |