When the index of an exponential expression with a positive base is doubled, then the expression increases by 700%. If one of the values that the base can not have is X which of the following is not a possible value of P?
Answer: D Let the expression be an. => a2n = 8an => n = loga 8. a can be 5, 4 and 8 but it cannot be 1.
Q. No. 26:
Abhishek had a certain number of Re1 coins, Rs 2 coins and Rs 10 coins. If the number of Re 1 coins he had is six times the number of Rs 2 coins Abhishek had, and the total worth of his coins is Rs 160, find the maximum number of Rs 10 coins Abhishek could have had.
Answer: A If the Abhishek had x Re 1, y Rs 2 coins and z Rs 10 coins, the total value of coins he had = x(1)+y(2)+z(10) = x+2y+10z = 160. Since, 6y =x Thus , 8y + 10z =160 i.e 8y is a multiple of 10 i.e y=5 or y=10 i.e (x,y,z) = (30,5,12) or (60,10,18) Thus, the maximum value of 'z' is 12.
Q. No. 27:
In an A.P, the 12th term is 7 times the 2nd term and the 8th term is 3 more than 10 times the first term. What is the 5th term of the G.P whose first term is the first term of A.P and whose common ratio is equal to the common difference of the A.P.
Answer: A Let the progression be a, a+d, a+2d.......... => a+11d = 7a +7d => 6a = 4d => 3a =2d Also, a+7d = 10a+3 => 7d = 9a+3 => 7d = 6d+3 => d=3, a=2 The GP is 2, 2(3), 2(9),2(27), 2(81) The firth term is 162.
Q. No. 28:
The digits of a three number are in AP. If the number is subtracted from the number formed by reversing its digits, the result is 396. What could be the original number?
Answer: B The difference between 3-digit number and its reverse is 99 times the difference between its extreme (hundred and units) digits. As the first difference is 396, the second is 4. Further as the digits are in AP and the hundred's digits is less than the unit's digit, we have following possibilities 135, 246, 357, 468, 579...
Q. No. 29:
Three persons Abhishek, Dishant and Prashant were born on different days in the same year. If the date and month of birth of Abhishek, Dishant and Prashant are numerically equal, then what could be the minimum difference in the ages of youngest and oldest in days?
Answer: C To have the minimum differences in the ages of the oldest and youngest. One of them should be born in February. Also the other two should be born in March and April of January and March. Difference between the youngest and the oldest is the number of days from 2/2 to 4/4 = (26+31+4) = 61 or, 1/1 to 3/3 = 30+28+3 = 61 days.
Q. No. 30:
The total age of some 7 years old and some 5 years old children is 60 years. If I have to select a team from these children such that their total age is 48 years, In how many ways can it be done?
Answer: C Let 'a' children of 7 years and 'b' children of 5 years be taken. Then 7a+5b =48. This is possible only when x=4 and b=4. Hence only one combination is possible.