Ratio - Study Mode

[#1] The students of three classes are in the ratio 4 : 6 : 9. If 12 students are increased in each class the ratio changes to 7 : 9 : 12. Then the total number of students in the three classes before the increase is = ?
Correct Answer

(B) 76

Explanation

Solution: Let class A, B and C then Number of student ratio in
A : B : C = 4 : 6 : 9 = 4x : 6x : 9x And Now 12 Student join in all Class and their ratio become A : B : C = 7 : 9 : 12 = 7x : 9x : 12x ∴ in A → 7x - 4x = 12 , In B → 9x - 6x = 12 and In C →12x - 9x = 12 i.e. x = 4 Total number of Student before new student are = 4 × 4 + 6 × 4 + 9 × 4 = 76

[#2] Two vessels A and B contain milk 8 : 5 and 5 : 2 respectively. The ratio in which these two mixtures be mixed to get a new mixture containing $$69frac{3}{{13}}$$ % milk is =?
Correct Answer

(D) 2 :7

Explanation

Solution: Milk in 1 litre mix. in A = $$frac{8}{{13}}$$ Milk in 1 litre mix. in B = $$frac{5}{{7}}$$ Milk in 1 litre of final mix. = $$frac{{900}}{{13}} imes frac{1}{{100}} imes 1 = frac{9}{{13}}$$ By the rule of alligation, we have : ∴ Required ratio $$ = frac{2}{{91}} : frac{1}{{13}} = 2 : 7$$

[#3] In two types of stainless steel the ratio of chromium and steel are 2 : 11 and 5 : 21 respectively. In what proportion should the two types be mixed so that the ratio of chromium to steel in the mixed type becomes 7 : 32 = ?
Correct Answer

(C) 1 : 2

Explanation

Solution: Given: ⇒ Let the alloys be mixed in the ratio of $$x:y$$ xa0 (Assumption) ⇒ In 1st alloy, chromium = $$frac{2x}{13}$$ ⇒ In 1st alloy, steel = $$frac{11x}{13}$$ ⇒ In 2nd alloy, chromium = $$frac{5y}{26}$$ ⇒ In 2nd alloy, steel = $$frac{21y}{26}$$ ⇒ The ratio in which 2 alloys must be mixed to get a new alloy with a ratio of chromium and steel be 7 : 32 =? Now we have, $$eqalign{
& left( {frac{{2x}}{{13}} + frac{{5y}}{{26}}}
ight) : left( {frac{{11x}}{{13}} + frac{{21y}}{{26}}}
ight) = 7 : 32 cr
& Rightarrow frac{{left( {frac{{2x}}{{13}} + frac{{5y}}{{26}}}
ight)}}{{ left( {frac{{11x}}{{13}} + frac{{21y}}{{26}}}
ight) }} = frac{7}{{32}} cr
& Rightarrow frac{{frac{{4x + 5y}}{{26}}}}{{ frac{{22x + 21y}}{{26}} }} = frac{7}{{32}} cr
& Rightarrow frac{{4x + 5y}}{{22x + 21y}} = frac{7}{{32}} cr
& Rightarrow 128x + 160y = 154x + 147y cr
& Rightarrow 154x - 128x = 160y - 147y cr
& Rightarrow 26x = 13y cr
& Rightarrow frac{x}{y} = frac{{13}}{{26}} cr
& herefore frac{x}{y} = frac{1}{2} cr} $$ ∴ Ratio in which 2 alloys must be mixed to get a new alloy with a ratio of chromium and Steel to be 7 : 32 is 1 : 2

[#4] One - fourth of sixty percent of a number is equal to two - fifths of twenty percent of another number. What is the respective ratio of the first number to the second number ?
Correct Answer

4 : 7

Explanation

Solution: Let the numbers be x and y $$eqalign{
& { ext{Then,}} cr
& = frac{1}{4}{ ext{ of }}left( {60\% { ext{ of }}x}
ight) cr
& = frac{2}{5}{ ext{ of }}left( {20\% { ext{ of }}y}
ight) cr
& Rightarrow left( {frac{1}{4} imes frac{{60}}{{100}} imes x}
ight) = left( {frac{2}{5} imes frac{{20}}{{100}} imes y}
ight) cr
& Rightarrow frac{{3x}}{{20}} = frac{{2y}}{{25}} cr
& Rightarrow frac{x}{y} = frac{2}{{25}} imes frac{{20}}{3} = frac{8}{{15}} cr
& Rightarrow x:y = 8:15 cr} $$

[#5] The total number of boys in a school is 16% more than the total number of girls in the school. What is the respective ratio of the total number of boys to the total number of girls in the school ?
Correct Answer

25 : 21

Explanation

Solution: Let the number of girls be x Then, $$eqalign{
& { ext{Number of boys}} cr
& = 116\% { ext{ of }}x = frac{{29}}{{25}}x cr
& herefore { ext{Required ratio}} cr
& = frac{{29}}{{25}}x:x cr
& = 29:25 cr} $$