1. The sum of three consecutive integers is 102. Find the lowest of the three?
Answer - Option D
Explanation -
Three consecutive numbers can be taken as (P - 1), P, (P + 1).
So, (P - 1) + P + (P + 1) = 102
3P = 102 => P = 34.
The lowest of the three = (P - 1) = 34 - 1 = 33.
2. The sum of the two digits of a number is 10. If the number is subtracted from the number obtained by reversing its digits, the result is 54. Find the number?
Answer - Option B
Explanation -
Any two digit number can be written as (10P + Q), where P is the digit in the tens place and Q is the digit in the units place.
P + Q = 10 ----- (1)
(10Q + P) - (10P + Q) = 54
9(Q - P) = 54
(Q - P) = 6 ----- (2)
Solve (1) and (2) P = 2 and Q = 8
The required number is = 28
3. The sum of three consecutive even numbers is 42. Find the middle number of the three?
Answer - Option A
Explanation -
Three consecutive even numbers (2P - 2), 2P, (2P + 2).
(2P - 2) + 2P + (2P + 2) = 42
6P = 42 => P = 7.
The middle number is: 2P = 14.
4. The sum of the present ages of two persons A and B is 60. If the age of A is twice that of B, find the sum of their ages 5 years hence?
Answer - Option C
Explanation -
A + B = 60, A = 2B
2B + B = 60 => B = 20 then A = 40.
5 years, their ages will be 45 and 25.
Sum of their ages = 45 + 25 = 70.
5. The ratio of the present ages of P and Q is 3:4. 5 years ago, the ratio of their ages was 5:7. Find the their present ages?
A. 30, 40
B. 25, 30
C. 50, 60
D. 20, 40
Answer - Option A
Explanation -
Their present ages be 3X and 4X.
5 years age, the ratio of their ages was 5:7, then (3X - 5):(4X - 5) = 5:7
X = 35 - 25 => X = 10.
Their present ages are: 30, 40.