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
=> x = 3000
C's share = Rs. 6x = Rs. (6 x 3000) = Rs. 18,000
Answer is 18,000
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 = 21 : 22