1.

Find the mean proportional between 49 & 81?

**Answer: Option 'C'**

**Formula = √a×b
A = 49 and B = 81
√49×81 = 7 × 9 = 63**

2.

A sum of money is to be distributed among A, B, C, D in the proportion of 7 : 4 : 6 : 5. If C gets Rs. 3000 more than D, what is C's share?

**Answer: Option 'B'**

**Given, Money among A, B, C, D in the proportion of 7 : 4 : 6 : 5
Let the shares of A, B, C and D be Rs. 7x, Rs. 4x, Rs. 6x and Rs. 5x respectively.
Then, if C gets Rs. 3000 more than D
=> 6x = 3000 + 5x
=> 6x - 5x = 3000
**

3.

Two numbers are in the ratio 3 : 5. If 9 is subtracted from each, the new numbers are in the ratio 12 : 23. The smaller number is:

**Answer: Option 'B'**

**Let the numbers be 3x and 5x
Then (3x-9)/(5x-9) = 12/23 => 23(3x-9)=12(5x-9) => 9x = 99 => x = 11
Therefore the smaller number is 3x = 3 x 11 = 33**

4.

Two number are in the ratio 3 : 5. If 9 is subtracted from each, the new numbers are in the ratio 12 : 23. The smaller number is:

**Answer: Option 'A'**

**Let the numbers be 3x and 5x.
Then, 3x - 9 / 5x - 9 = 12/3
=> 23×( 3x-9) = 12× (5x - 9)
=> 69x - 207 = 60 x - 108
=> 9x = 99
=> x = 11
Therefore, The smaller number
= (3 x 11)
= 33.**

5.

The ratio of the number of boys and girls in a college is 7 : 8. If the percentage increase in the number of boys and girls be 20% and 10% respectively, what will be the new ratio?

**Answer: Option 'A'**

**Originally, let the number of boys and girls in the college be 7x and 8x respectively.
Their increased number is (120% of 7x) and (110% of 8x)
=> (120 / 100 x 7x ) and (110/ 100 x 8x )
=> 42x / 5 and 44x / 5
Therefore, The required ratio = ( 42 x/5 : 44x /5 ) = 42 : 44 =**

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