a2zinterviews

31.	Find a number such that when 15 is subtracted from 7 times the number, the results is more than twice the number?
	A.  2
	B.  3
	C.  4
	D.  5

Solution

Let the number be x. = Then, 7x-15 = 2x+10 = 5x = 25 ‹=›x = 5. Hence the required number is 5.

32.	54 is to be divided into two parts such that the sum of 10 times the first and 22 times the second is 780. The bigger part is
	A.  24
	B.  30
	C.  32
	D.  34

Solution

Let the two parts be (54 - x) and x. Then, 10(54-x)+22x=780 = 12x = 240 ‹=›x = 20. Therefore Bigger part = (54-x) = 34.

33.	A number is doubled and 9 is added. If the resultant is trebled, it becomes 75. What is that number?
	A.  5
	B.  8
	C.  9
	D.  10

Solution

Let the number be x. Then, 3(2x+9) = 75 ‹=› 2x+9 = 25 ‹=›2x = 16 ‹=›x = 8.

34.	The difference between a two digit number and the number obtained by interchanging the positions of its digits is 36. What is the difference between the two digits of that number?
	A.  2
	B.  3
	C.  4
	D.  8

Solution

Let the ten's digit be x and units digit be y. Then, (10x+y) - (10y+x)= 36 ‹=›9(x - y) = 36 ‹=›x - y = 4.

35.	A number consists of 3 digits whose sum is 10. The middle digit is equal to the sum of the other two and the number will be increased by 99 if its digits are reversed. The number is
	A.  145
	B.  185
	C.  253
	D.  370

Solution

Let the number be x. Then 2x = 10 or x = 5. So, the number is either 253 or 352. Since the number increases on reversing the digits, so the hundred's digit is smaller than the units digit. Hence, required number = 253.