Percentage - Study Mode
[#281] If x% of y is equal to z, what percent of z is x ?
Correct Answer
(D) $$frac{{{{100}^2}}}{y}$$ %
Explanation
Solution: x% of y = z ⇒ $$frac{x}{100}$$y = z ⇒ $$frac{x}{z}$$ = $$frac{100}{y}$$ ∴ Required percentage $$eqalign{
& = left( {frac{x}{z} imes 100}
ight)\% cr
& = left( {frac{{100}}{y} imes 100}
ight)\% cr
& = left( {frac{{{{100}^2}}}{y}}
ight)\% cr} $$
[#282] A part of Rs. 9600 is invested at a 5% annual return, while the remainder is invested at a 3% annual return. If the annual income from both portions is the same, what is the total income from the two investments ?
Correct Answer
Rs. 320
Explanation
Solution: Let the sum invested at 5% be Rs. x and that invested at 3% be Rs. (9600 - x) Then, 5% of x = 3% of (9600 - x) ⇒ 5x = 3 (9600 - x) ⇒ 8x = 28800 ⇒ x = 3600 Hence, total income : = 5% of x + 3% of (9600 - x) = Rs. (5% of 3600 + 3% of 6000) = Rs. (180 + 180) = Rs. 360
[#283] Of the 50 researchers in a work group, 40% will be assigned to Team A and the remaining 60% to Team B. However, 70% of the researchers prefer team A and 30% prefer Team B. What is the least possible number of researchers who will not be assigned to the team they prefer ?
Correct Answer
(A) 15
Explanation
Solution: Number of researchers who prefer Team A = 70% of 50 = 35 Number of researchers who prefer Team B = (50 - 35) = 15 Number of researchers assigned to Team A = 40% of 50 = 20 Number of researchers assigned to Team B = (50 - 20) = 30 To find the least possible number of researchers who will not be assigned to the team they prefer, we assume that the maximum number of researchers get the team they prefer. So,Number of researchers who are assigned to the team they prefer = 20 (Team A) + 15 (Team B) = 35 ∴ Required number = (50 - 35) = 15
[#284] A fraction in reduced form is such that when it is squared and then its numerator is reduced by $$33frac{1}{3}$$ % and denominator is reduced to 20%, its result is twice the original fraction. The sum of numerator and denominator is:
Correct Answer
(A) 8
Explanation
Solution: Let the fraction be $$frac{{ ext{x}}}{{ ext{y}}}$$
When fraction is squared its numerator is reduced by $$33frac{1}{3}$$ and denominator is reduced by 20% $$eqalign{
& { ext{According}},{ ext{to}},{ ext{question,}} cr
& {left( {frac{x}{y}}
ight)^2} imes frac{{33left( {frac{1}{3}}
ight)\% }}{{20\% }} = 2left( {frac{x}{y}}
ight) cr
& { ext{Or}},,{left( {frac{x}{y}}
ight)^2} imes frac{{left( {frac{2}{3}}
ight)}}{{left( {frac{1}{5}}
ight)}} = 2left( {frac{x}{y}}
ight) cr
& { ext{Or}},,frac{x}{y} = frac{3}{5} cr
& { ext{Sum of numerator and denominator is}} cr
& left( {x + y}
ight) = 3 + 5 cr
& ,,,,,,,,,,,,,,,,,,, = 8 cr} $$
[#285] A student appeared in the Mock CAT. The test paper contained 3 sections namely QA, DI and VA. The percentage marks in all VA was equal to the average of percentage marks in all the 3 sections. Coincidentally, if we reverse the digit of the percentage marks of QA we get the percentage marks of DI. The percentage marks in VA scored by student could be:
Correct Answer
(C) 66
Explanation
Solution: Let the percentage marks in QA =(10a + b)% Let the percentage marks in DI = (10b + a)% Let the percentage marks in VA = x% Now, according to the question, we have, $$eqalign{
& frac{{left( {10{ ext{a}} + { ext{b}}}
ight) + { ext{x}} + left( {10{ ext{b}} + { ext{a}}}
ight)}}{3} = { ext{x}} cr
& 11{ ext{a}} + 11{ ext{b}} + { ext{x}} = 3{ ext{x}} cr
& { ext{or, x}} = frac{{11left( {{ ext{a}} + { ext{b}}}
ight)}}{2} cr} $$ Clearly, we can see that the percentage of the VA section will be a multiple of 11 So, required answer will be 66