A club consists of members whose ages are in A.P. The common difference being 3 months. If the youngest member of the club is just 7 years old and the sum of the ages of all the members is 250, then number of members in the club are
Answer: C Let n be the number of members in the club Then, 250 = n/2 [2*7 + (n-1)*3/12] => n= 25.
Q. No. 8:
Students of a class are made to stand in rows. If 4 students are extra in each row, there would be 2 rows less. If 4 students are less in each row, there would be 4 more rows. The number of students in the class is
Answer: D Let number of rows be x and number of students in each row be n. Then, total number of students = x*n Again, (n+4)(x-2) = (x-4)(n-4) = xn => n=12 and x=8 Number of students = 12*8 = 96.
Q. No. 9:
A man arranges to pay off a debt of Rs 3600 by 40 annual installments which are in A.P. When 30 of the installments are paid he dies leaving one-third of the debt unpaid. The value of the 8th installment is
Answer: C Let the first installment be 'a' and the common difference between any two consecutive installments be 'd' Using the formula for the sum of an A.P S= n/2[2a + (n-1)d] We have, 3600 = 40/2[2a+ (40-1)d] => 180 = 2a + 39d........(i) and 2400 = 30/2[2a + (30-1)d] => 160 = 2a +29d..........(ii) On solving both the equations we get d=2 and a =51 Value of 8th installment = 51 + (8-1)2 = Rs 65.
Q. No. 10:
A father with 8 children takes 3 children at a time to the zoological garden, as often as he can without taking the same 3 children together more than once. Then
A :
number of times he will go to zoological garden is 56.
B :
number of times each child will go to the zoological garden is 21.
C :
number of times a particular child will not go to the zoological garden is 35.
Answer: D The number of times the father would go to the zoological garden = Number of ways of selection of 3 children taken at a time = 8C3 = 56. Number of times a child will go to the zoological garden = Number of times he is accompanished by two other = 1* 7C2 =21 => Number of times a child will not go to the zoological garden = 56-21 =35.
Q. No. 11:
A farmer has decided to build a wire fence along one straight side of his property. For this, he planned to place several fence posts at 6m intervals, with posts fixed at both ends of the side. After he bought the posts and wire, he found that the number of posts he had bought was 5 less than required. However, he discovered that the number of posts he had bought would be just sufficient if he spaced them 8m apart. What is the length of the side of his property and how many posts did he buy?
Answer: D Let the length of the side of the property be x m and y be the number of posts bought. When the space between polls is 8m, The number of poles = x/8 + 1 =y.........(i) When space between poles is 6m, The number of poles = x/6 + 1 = y+5......(ii) On solving both of the equations we get, x= 120m and y =16
Q. No. 12:
Ram's age was square of number last year and it will be cube of a number next year. How long must he wait before his age is again a cube of a number.