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Q 1 - On the off chance that a:b=2:3 and b:c=5:7, discover a:c.

**Answer - B**

**Explanation**

We have a/b = 2/3 and b/c = 5/7 So a/c = (a/b*b/c) = (2/3*5/7) = 10/21 So its demonstrate that a:c = 10:21

Q 2 - On the off chance that a:b=2:3 and b:c=5:7, discover a:b:c.

**Answer - A**

**Explanation**

Here a/b = 2/3 and b/c = 5:7 = 3/5*5:3/5*7 = 3:21/5. So a:b=2:3 and b:c=3:21/5 So a:b:c = 2:3:21/5 = 10:15:21.

Q 3 - On the off chance that 4a = 5b and 8b = 9c, find a:b:c.

**Answer - A**

**Explanation**

4a = 5b => a/b=5/4 and 8b = 9c => b/c=9/8 So a:b = 5:4 and b:c = 9:8 = (4/9)(9):(4/9)(8) = 4:32/9 => a:b:c = 5:4:32/9 = 45:36:32. Hence,a:b:c = 45:36:32.

Q 4 - On the off chance that a/8 = b/9 = c/12, find a:b:c.

**Answer - B**

**Explanation**

Let a/8 = b/9 =c/12 =k. Then a=8k ,b=9k and c=12k. So a:b:c = 8k:9k:12k =8:9:12. Hence,a:b:c = 8:9:12.

Q 5 - In the event that a:b =1:3, b:c = 5:7 and c:d = 9:8 ,find a:b:c:d.

**Answer - C**

**Explanation**

We have a:b = 1:3, b:c = 5:7 and c:d = 9:8 => a:b = 5:15, b:c = 15:21, c:d =(21/9)*9 : (21/9)*8 => a:b = 5:15, b:c = 15:21, c:d = 21:56/3 => a:b:c:d =5:15:21:56/3 = 15:45:63:56 Consequently, a:b:c:d = 15:45:63:56

**Answer - A**

**Explanation**

Here (5x+3y)/(5x-3y) = 3/1 => 5x+3y = 15x-9y => 10x = 12y => x/y = 12/10 = 6/5 So x:y =6:5

**Answer - D**

**Explanation**

Given x/y = 5/3 Dividing numerator and denominator by y. (8x-5y)/(8x+5y) = {8(x/y) - 5}/{8(x/y) + 5} = {8*(5/3)-5}/{8*(5/3)+5} = (40-15)/(40+15) = 25/55 = 5/11 So (8x-5y):(8x+5y)= 5:11

**Answer - D**

**Explanation**

Let 4:5::12:x. => 4*x = (5*12) => x = 5*12/4 = 15 So the fourth relative to 4,5,12 is 15.

**Answer - B**

**Explanation**

Third relative to 9 and 12 is equivalent to fourth corresponding to 9,12 and 12. Give it a chance to be x at that point => 9:12::12:x => 9x = 12*12 => x = 12*12/9 =16 So the third relative is 16.

**Answer - B**

**Explanation**

Mean relative somewhere around 49 and 64 is 49*64 = (7*8) = 56.

Q 11 - An aggregate of rs. 391 has been divided between a,b,c in the proportion 1/2 :2/3:3/4 , discover the offer of each.

**Answer - A**

**Explanation**

We have a:b:c=1/2:2/3:3/4= 6:8:9. A share = (391*6/23) = 102 rs. B offer = (391*8/23) = 136 rs. C offer = (391*9/23) = 153 rs.

Q 12 - A sack contain one rupee, fifty paisa and 25 paisa in the proportion of 8:9:11, if the aggregate cash of the pack is 122, discover the no. of coins of every sorts.

**Answer - A**

**Explanation**

Let the quantity of one rupee, 50-p and 25-p coins be 8x, 9x and 11x individually. At that point, 8x + 9x/2 + 11x/4 =122 => 32x + 18x + 11x = 488 => 61x =488 => x = 8 No. of one rupee coins = 8*8= 64 No. of 50-p coins =9*8= 72 No. of 25-p coins =11*8 =88

Q 13 - A blend contains liquor and water in the proportion 4:3, if 7 liter of water is added to the blend, the proportion of liquor and water gets to be 3:4. Discover the amount of liquor in the blend.

**Answer - B**

**Explanation**

Let the amount of liquor and water be 4x liter and 3x liter separately. At that point , 4x/3x + 7 = 3/4 => 16x = 9x+21 => 7x = 21 so estimation of x is 3 Amount of liquor in the blend is = 4*3 =12 liters.

Q 14 - In a collection, the no. of understudy considering expressions, trade and science in the proportion of 4:7:9. On the off chance that the no. of understudy in expressions of the human experience, business and science be expanded by 30%, 20% and 40%. What will be the new proportion?

**Answer - A**

**Explanation**

Let the no. of understudy in expressions, business and science be 4x,7x and 9x individually. Presently they are 130% of 4x, 120 % of 7x and 140 % of 9x. Required proportion = (130/100*4x): (120/100*7x) (140/100*9x) =26x/5:42x/5:63x/5 =26:42:63.

Q 15 - The expense of assembling an auto is comprised of three items: cost of material, work and overheads. In a year, the expense of these things were in the proportion 4:3:2.Next year, the expense of material rose by 10%,cost of work expanded by 8% however the overheads lessened by 5%.Find the increment for every penny in the auto's cost.

**Answer - A**

**Explanation**

Let the expense of material, work and over head be rs. 4x, 3x and 2x separately. At that point aggregate expense =9x rs . New cost= {(110% of 4x) + (108% of 3x) +(90% of 2x)} ={(110/100*4x)+(108/100*3x)+(90/100*2x)} = (22x/5 + 81x/25 + 9x/5) = (110x+81x+45x)/25= 236x/25 Increment = {(236x/25)-9x} = 11x/25 Increase%= (11x/25)*(1/9x)*100 % = 44/9 %

Q 16 - The proportion of no. of young men to that of the young ladies in a school is 3:2 .if 20% of young men and 25% of young ladies are grant holders, discover the % of the individuals who are not grant holders.

**Answer - B**

**Explanation**

Let the no. of young men be 3x and the no. of young ladies 2x. Aggregate no. = 5x No. of the individuals who are not grant holders = (80% of 3x)+(75% of 2x) = (80/100 * 3x) + ( 75/100 * 2x) = (12x/5 + 3x/2) = 39x/10 Required % = (39x/10)*(1/5x)*100 % = 78%

Q 17 - An and B together have rs.1210 with them. In the event that 4/15 of A sum is equivalent to 2/5 of B sum, what amount of sum does B have?

**Answer - A**

**Explanation**

Let (4/15)a = (2/5)b = x then a = 15x/4 and b = 5x/2 So. 15x/4 + 5x/2 =1210 => 15x + 10x = 4840 => 25x = 4840 => x=193.6 So. B = (5/2*193.6) = 484 Henceforth B has Rs. 484.

Q 18 - In the event that (x+y): (x-y)= 4:1,then (x^{2}+y^{2}): (x^{2}-y^{2})=?

**Answer - A**

**Explanation**

(x + y)/(x - y)= 4/1 => x + y = 4x-4y => 3x = 5y => x/y = 5/3 Now (x^{2}+y^{2})/ (x^{2}-y^{2})= {(x/y)^{2}+1}/ { (x/y )^{2}-1 } = {(5/3)^{2}+1} / {(5/2)^{2}-1} = 34/16 = 17/8

Q 19 - In the event that (4x^{2}-3y^{2}) :( 2x^{2}+5y^{2})= 12:19 , then x:y=?

**Answer - B**

**Explanation**

(4x^{2}-3y^{2})/ (2x^{2}+5y^{2}) = 12/19 =>76x^{2}-57y^{2}= 24x^{2}+60y^{2}=> 52x^{2}= 117y^{2}=> x^{2}/y^{2}= 117/52 = 9/4 => (x/y)^{2}=(3/2)^{2}=> x/y = 3/2. => x:y = 3:2

**Answer - A**

**Explanation**

As x^{2}+4y^{2}= 4xy => x^{2}+ 4y^{2}- 4xy = 0 => (x-2y)^{2}= 0 => x-2y=0 => x = 2y => x/y = 2/1. => x:y = 2:1.

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