A) 1:3 | B) 2:1 |

C) 3:4 | D) 2:3 |

Explanation:

Let, B1 : B2 : B3 = 3x : 4x : 5x

again B1 : B2 : B3 = 5y : 4y : 3y

Since there is increase in no.of oranges in first two baskets only, it means the no. of oranges remains constant in the third basket

5x = 3y

Hence 3x : 4x : 5x

and 5y : 4y : 3y 25x : 20x : 15x

Therfore, increment in first basket = 16

Increment in second basket = 8

Thus, required ratio = 16/8 = 2:1

A) 7 : 12 | B) 8 : 13 |

C) 9 : 4 | D) 2 : 5 |

Explanation:

To get the solution that contains 1 part of milk and two parts of water,

they must be mixed in the ratio as

7x+6x/5y+11y = 1/2

26x = 16y

x/y = 16/26

x/y = 8/13

A) 11 | B) 7 |

C) 9 | D) 13 |

Explanation:

Here the number of cogs is inveresly proportional to number of revolutions

=> More cogs less revolution

=> 6 : 14 :: x : 21

=> (21x6)/14 = 9

A) 9 points | B) 10 points |

C) 11 points | D) 8 points |

Explanation:

a:b = 60:40

a:c = 60:45

c/a x a/b = 45/60 x 60/40 = 45/40 = 90/80

So C gives B => 10 points.

A) 125 | B) 155 |

C) 135 | D) 165 |

Explanation:

Ratio C1:C2 = 3:5 and C2:C3 = 7:11

So C1:C2:C3 = 21 : 35 : 55

Let the strength of three classes are 21x, 35x and 55x respectively, then

Given that 21x + 35x + 55x = 333

=> 111x = 333 or x=3

So strength of the class with highest number of students =55x = 55x3 = 165.

A) 1:3 | B) 2:1 |

C) 3:2 | D) 2:3 |

Explanation:

Let us assume S as number of shirts and T as number of trousers

Given that each trouser cost = Rs.70 and that of shirt = Rs.30

Therefore, 70 T + 30 S = 810

=> 7T + 3S = 81......(1)

T = ( 81 - 3S )/7

We need to find the least value of S which will make (81 - 3S) divisible by 7 to get maximum value of T

Simplifying by taking 3 as common factor i.e, 3(27-S) / 7

In the above equation least value of S as 6 so that 27- 3S becomes divisible by 7

Hence T = (81-3xS)/7 = (81-3x6)/7 = 63/7 = 9

Hence for S, put T in eq(1), we get

S = 81-7(9)/3 = 81-63 / 3 = 18/3 = 6.

The ratio of T:S = 9:6 = 3:2.