Q. 1: The smallest integer n for which 4^n > 17^19 holds, is closest to?
1. 33
2. 37
3. 39
4. 35
Given , 4^n > 17^19
Or 16^(n/2) > 17^19
Now from the given options, the only possible value of n is 39 as for other
values n/2 will be less than 19 and 16 < 17.
Correct answer 3. 39
let the concentration of A, B & C are a% , b% & c% respectively .
Then (a+2b+3c )/600=20/100=1/5
a+2b+3c=120——1)
And (3a+2b+c )/600=30/100=3/10
3a+2b+c=180——-2)
By eq 1)/ eq 2) , (a+2b+3c )/(3a+2b+c )=2/3
3a+6b+9c=6a+4b+2c
(2b+7c)/a=3
So correct answer : 2. 1 :3
Q. 3: The area of a rectangle and the square of its perimeter are in the ratio 1 : 25. Then the lengths of the shorter and longer sides of the rectangle are in the ratio
1. 1:4
2. 2:9
3. 1:3
4. 3:8
Given ratio of areas of rectangle and square = 1:25 = 4∶ 100=(1×4):(10×10)
Thus possible ratio of perimeter =(1+4)/(10+10) = 5/20= 1∶4
Q. 4: The scores of Amal and Bimal in an examination are in the ratio 11 : 14. After an appeal, their scores increase by the same amount and their new scores are in the ratio 47 : 56. The ratio of Bimal’s new score to that of his original score is
1. 5 : 4
2. 8 : 5
3. 4 : 3
4. 3 : 2
let their scores were 11x and 14x and it increase by n then
(11x+n)/(14x+n)=47/56
616x+56n=658x+47n
42x=9n
n = 42x/9
So Bimal’s new score = 14x + 42x/9 = 168x/9
So required ratio = (168x/9)/14x = 4∶ 3
Q. 5: From a rectangle ABCD of area 768 sq cm, a semicircular part with diameter AB and area 72π sq cm is removed. The perimeter of the leftover portion, in cm, is
1. 80 + 16π
2. 86 + 8π
3. 82 + 24π
4. 88 + 12π
Given area of semicircle = 72π
Or (πr^2)/2 = 72π
r = 12
So AB = 24
Thus 24×AD = 768
AD = 32
So Required perimeter = AD+CD+CB+πr = 32+24+32+12π = 88+12π
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