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Maths Ratio and Proportion Test.20

**Quiz Instructions:**

- There will be 20 multiple choice question in this online test.
- Answer of the questions will change randomly each time you start this test.
- Practice this test at least 3 times if you want to secure High Marks.
- At the End of the Test you can see your Test score and Rating.

1 / 20

**What is the ratio of Faheem’s salary to Imran’s salary to Naveed’s salary if Faheem makes 80,000 rupees ,Imran 70,000 rupees and Naveed makes 50,000 rupees ?**

2 / 20

**Two numbers are in the ratio 5:4 and their difference is 10.What is the largest number?**

3 / 20

**In what ratio should the profit be divided if M, N, O invests capital in ratio 2:3:5 and their timing of their investments are in the ratio 4:5:6.**

P1:P2:P3 = (2*4):(3*5):(5*6)

= 8 : 15 : 30

P1:P2:P3 = (2*4):(3*5):(5*6)

= 8 : 15 : 30

4 / 20

**A bag contains 50 paise, 1 rupee and 2 rupee coins in the ratio 2:5:8. If the total amount is Rs. 352, find the total number of coins in the bag.**

Let the number of 50p, 1 rupee and 2 rupee coins be 2x,5x and 8x. 0.5*(2x)+5x+2*8x = 352 22x=352 x=16 Total no. of coins

= 2x+5x+8x

= 15x = 240

Let the number of 50p, 1 rupee and 2 rupee coins be 2x,5x and 8x. 0.5*(2x)+5x+2*8x = 352 22x=352 x=16 Total no. of coins

= 2x+5x+8x

= 15x = 240

5 / 20

**Three quantities X, Y and Z are such that XY=kZ where k is constant. Initially, X was at 4 and X:Y:Z was 2:3:4. If the value of Y is changed to 12 and Z is kept constant, find the value of X.**

X:Y:Z=2:3:4

X=4

Then, Y=6, Z=8 XY=kZ 4*6=k*8 8k=24 k=3

Y=12, Z=8

X*12=3*8

X=2

X:Y:Z=2:3:4

X=4

Then, Y=6, Z=8 XY=kZ 4*6=k*8 8k=24 k=3

Y=12, Z=8

X*12=3*8

X=2

6 / 20

**Find the mean proportion between 5 and 45.**

5:x::x:45 x²=45*5 = 225 x=15

5:x::x:45 x²=45*5 = 225 x=15

7 / 20

**If X:Y = 2:3, Y:Z = 4:5. Find X:Y:Z.**

If A:B = m1:n1 ; B:C = m2:n2

Then, A:B:C = (m1*m2):(n1*m2):(n1*n2)

Therefore, X:Y:Z = (2*4):(3*4):(3*5)

= 8:12:15

If A:B = m1:n1 ; B:C = m2:n2

Then, A:B:C = (m1*m2):(n1*m2):(n1*n2)

Therefore, X:Y:Z = (2*4):(3*4):(3*5)

= 8:12:15

8 / 20

**The ratio of two numbers a and b is 3:5. If 2 is added to both the numbers, the ratio becomes 2:3. Find b.**

a:b = 3:5 5a = 3b (a+2):(b+2) = 2:3 3(a+2) = 2(b+2) 3a+6 = 2b+4 3*(3b/5) = 2b-2 9b = 10b-10 b =10

a:b = 3:5 5a = 3b (a+2):(b+2) = 2:3 3(a+2) = 2(b+2) 3a+6 = 2b+4 3*(3b/5) = 2b-2 9b = 10b-10 b =10

9 / 20

**The plan for a rectangular area is drawn to a scale of 1:500. If the length and breadth in the plan is 30 cm and 20 cm, find the actual area.**

Actual length

= 30*500 cm Actual breadth

= 20*500 cm Actual area

= 30*20*500*500 = 15000 m²

Actual length

= 30*500 cm Actual breadth

= 20*500 cm Actual area

= 30*20*500*500 = 15000 m²

10 / 20

**The length and breadth of a rectangle are in the ratio 5:4. Find its perimeter if its area is 180 cm²**

Let the length and breadth be 5x cm and 4x cm.

5x*4x=180

20x²=180

x²=9,x=3 Length = 15cm Breadth = 12cm Perimeter

= 2(l+b)

= 2*27

= 54 cm

Let the length and breadth be 5x cm and 4x cm.

5x*4x=180

20x²=180

x²=9,x=3 Length = 15cm Breadth = 12cm Perimeter

= 2(l+b)

= 2*27

= 54 cm

11 / 20

**In what ratio should the profit of Rs.8000 be divided if X starts a business with an investment of Rs. 20000, Y invests Rs.7500 for 4 months and Z invests Rs.15000 after 3 months from the start of the business.**

Let the profit of X be P1, that of Y be P2 and of Z be P3.

P1:P2:P3 = 20000*12 : 7500*4 : 15000*9 = 240 : 30 : 135 = 80 : 10 : 45

= 16 : 2 : 9

Let the profit of X be P1, that of Y be P2 and of Z be P3.

P1:P2:P3 = 20000*12 : 7500*4 : 15000*9 = 240 : 30 : 135 = 80 : 10 : 45

= 16 : 2 : 9

12 / 20

**M:N = 3:4, N:O = 5:7 and O:P = 7:9. Then M:P =________?**

M/P = M/N * N/O * O/P = ¾ * 5/7 * 7/9 = 5/ 12

M/P = M/N * N/O * O/P = ¾ * 5/7 * 7/9 = 5/ 12

13 / 20

**The perimeter of a triangle is 81 cm. If the sides are in the ratio 4:3:2, find the length of the largest side.**

Let the three sides be 4x,3x and 2x. 4x+3x+2x = 81 9x = 81 x = 9 Length of the largest side

= 4x = 36 cm

Let the three sides be 4x,3x and 2x. 4x+3x+2x = 81 9x = 81 x = 9 Length of the largest side

= 4x = 36 cm

14 / 20

**The ratio of boys and girls in a class was 11:13. 8 girls and 4 boys join the class from this academic year and the ratio becomes 4:5. Find the strength of the class.**

Let the number of boys and girls in the class be 11x and 13x. 4 boys and 8 girls join the class. Strength of the class

= 11x+13x+4+8

= 24x+12 (11x+4):(13x+8) = 4:5 5*(11x+4)=4*(13x+8) 55x+20=52x+32 3x=12 x=4 24x+12=108

Let the number of boys and girls in the class be 11x and 13x. 4 boys and 8 girls join the class. Strength of the class

= 11x+13x+4+8

= 24x+12 (11x+4):(13x+8) = 4:5 5*(11x+4)=4*(13x+8) 55x+20=52x+32 3x=12 x=4 24x+12=108

15 / 20

7x+8x=90 15x=90 x=6 Number of boys=7x=42

7x+8x=90 15x=90 x=6 Number of boys=7x=42

16 / 20

**In a partnership for a business, Changaz invests Rs.6000 for complete year & Nasir invests Rs.3000 for 6 months. What is Nasir’s share if they earn Rs.240 as profit?**

If A invests amount C1 for T1 time and his share of profit is P1, and B invests amount C2 for T2 time and his share of profit is P2, then, C1 * T1 / C2 * T2 = P1/P2

If P is the Nasir’s share of profit, then Changaz gets (240 – P)

Therefore, 6000 * 12 / 3000 * 6 = (240 – P) / P = 72/ 18 = 4

4P = (240 – P)

5P = 240

P = 48

If A invests amount C1 for T1 time and his share of profit is P1, and B invests amount C2 for T2 time and his share of profit is P2, then, C1 * T1 / C2 * T2 = P1/P2

If P is the Nasir’s share of profit, then Changaz gets (240 – P)

Therefore, 6000 * 12 / 3000 * 6 = (240 – P) / P = 72/ 18 = 4

4P = (240 – P)

5P = 240

P = 48

17 / 20

**If x% of 200=y% of 250, find y:x.**

(x/100)*200 = (y/100)*250 y/x = 20/25 = 4/5 y:x=4:5

(x/100)*200 = (y/100)*250 y/x = 20/25 = 4/5 y:x=4:5

18 / 20

**A bag contains 1 rupee, 2 rupee and 5 rupee coins amounting to Rs.264. If the ratio of the number of these coins is 3:5:4, find the number of 1 rupee coins.**

Let the number of 1 rupee, 2 rupee and 5 rupee coins be 3x, 5x and 4x. (1*3x)+(2*5x)+(5*4x)=264 3x+10x+20x=264 33x=264 x=8 No. of 1 rupee coins

= 3x = 24

Let the number of 1 rupee, 2 rupee and 5 rupee coins be 3x, 5x and 4x. (1*3x)+(2*5x)+(5*4x)=264 3x+10x+20x=264 33x=264 x=8 No. of 1 rupee coins

= 3x = 24

19 / 20

**Divide Rs.2000 into two shares in the ratio 3:2.**

3x+2x=2000 5x=2000 x=400 3x=1200 2x=800

3x+2x=2000 5x=2000 x=400 3x=1200 2x=800

20 / 20

**The ratio of two numbers is 5:9. If each number is decreased by 5, the ratio becomes 5:11. Find the numbers.**

Let the two numbers be 5x and 9x.

(5x-5)/(9x-5) = 5:11

(5x-5)*11 = (9x-5)*5

55x – 55 = 45x – 25

10x = 30

x = 3

Therefore, the numbers are 15 and 27.

Let the two numbers be 5x and 9x.

(5x-5)/(9x-5) = 5:11

(5x-5)*11 = (9x-5)*5

55x – 55 = 45x – 25

10x = 30

x = 3

Therefore, the numbers are 15 and 27.