Quiz 1

Given ratio of number of candidates is 4:5:6
Let the number of candidates for 3 exams be 4k, 5k and 6k respectively.
After increasing, number of candidates become (140% of 4k), (160% of 5k) & (185% of 6k)
That is, (140x4k)/100, (160x5k)/100 and (185x6k)/100= 56k/10, 80k/10 and 111k/10
Now, the required new ratio is: 56k/100 : 80k/10 : 111k/10= 56 : 80 : 111

A : B = 5 : 7, B : C = 6 : 11 = (6*7/6) : (11*7/6) = 7 : 77/6
A : B : C = 5 : 7 : 77/6 = 30 : 42 : 77.

Quantity of Milk = 60*(2/3) = 40 liters
Quantity of water = 60-40 = 20 liters
As per question we need to add water to get quantity 2:1
=> 40/(20+x) = 1/2
=> 20 + x = 80
=> x = 60 liters

Let the third number be x.
Then, the first number =120x/100 = 6x/5
second number=150/100*6x/5=9x/5
The ratio of two numbers is=6x/5:9x/5=2 : 3

Formula = Fourth propotional = (b × c)/a
A = 0.2 , B = 0.12 and C = 0.3
(0.12 × 0.3)/0.2
0.036/0.2 = 0.18

a/d = (3/4)*(7/9)*(5/7) => 5/12

Let the original salaries of Ravi and Sumit be Rs. 2x and Rs. 3x respectively.
Then,
(2x+4000) / (3x+4000) = 40 / 57
57 × (2x + 4000) = 40 × (3x+4000)
6x = 68,000
3x = 34,000
Sumit's present salary = (3x + 4000) = Rs.(34000 + 4000) = Rs. 38,000

Video solution:

Step (i) Let x be the number of boys and y be the number of girls.
Given total number of boys and girls = 100
x+y=100
Step (ii) A boy gets Rs. 3.60 and a girl gets Rs. 2.40
The amount given to 100 boys and girls = Rs. 312
3.6x+2.4y=312
Step (iii)
Solving (i) and (ii)
3.6x+3.6y=360 multiplying by 3.6
3.6x+2.4y1.20yy=312=48=40
The number of girls is 40.

Let the shares of A, B, C and D be Rs. 5x, Rs. 2x, Rs. 4x and Rs. 3x respectively.Then, 4x - 3x = 1000
x = 1000.
B's share = Rs. 2x = Rs. (2 x 1000) = Rs. 2000.

4x+7x=11x. We know that the sum of the quantities, 11, represents a fraction of the total number of party goers, so our answer MUST be a multiple of 11.The only multiple of 11 in our choices is C. 66.

Given ratio of number of candidates is 4:5:6
Let the number of candidates for 3 exams be 4k, 5k and 6k respectively.
After increasing, number of candidates become (140% of 4k), (160% of 5k) & (185% of 6k)
That is, (140x4k)/100, (160x5k)/100 and (185x6k)/100= 56k/10, 80k/10 and 111k/10
Now, the required new ratio is: 56k/100 : 80k/10 : 111k/10= 56 : 80 : 111

A : B = 5 : 7, B : C = 6 : 11 = (6*7/6) : (11*7/6) = 7 : 77/6
A : B : C = 5 : 7 : 77/6 = 30 : 42 : 77.

Quantity of Milk = 60*(2/3) = 40 liters
Quantity of water = 60-40 = 20 liters
As per question we need to add water to get quantity 2:1
=> 40/(20+x) = 1/2
=> 20 + x = 80
=> x = 60 liters

Let the third number be x.
Then, the first number =120x/100 = 6x/5
second number=150/100*6x/5=9x/5
The ratio of two numbers is=6x/5:9x/5=2 : 3

Formula = Fourth propotional = (b × c)/a
A = 0.2 , B = 0.12 and C = 0.3
(0.12 × 0.3)/0.2
0.036/0.2 = 0.18

a/d = (3/4)*(7/9)*(5/7) => 5/12

Let the original salaries of Ravi and Sumit be Rs. 2x and Rs. 3x respectively.
Then,
(2x+4000) / (3x+4000) = 40 / 57
57 × (2x + 4000) = 40 × (3x+4000)
6x = 68,000
3x = 34,000
Sumit's present salary = (3x + 4000) = Rs.(34000 + 4000) = Rs. 38,000

Video solution:

Step (i) Let x be the number of boys and y be the number of girls.
Given total number of boys and girls = 100
x+y=100
Step (ii) A boy gets Rs. 3.60 and a girl gets Rs. 2.40
The amount given to 100 boys and girls = Rs. 312
3.6x+2.4y=312
Step (iii)
Solving (i) and (ii)
3.6x+3.6y=360 multiplying by 3.6
3.6x+2.4y1.20yy=312=48=40
The number of girls is 40.

Let the shares of A, B, C and D be Rs. 5x, Rs. 2x, Rs. 4x and Rs. 3x respectively.Then, 4x - 3x = 1000
x = 1000.
B's share = Rs. 2x = Rs. (2 x 1000) = Rs. 2000.

4x+7x=11x. We know that the sum of the quantities, 11, represents a fraction of the total number of party goers, so our answer MUST be a multiple of 11.The only multiple of 11 in our choices is C. 66.