Toggle navigation
Home
Study Room
AP Eamcet Engineering Online Mock Exams
AP Eamcet Medical Online Mock Exams
Exams
Features
Pricing
Login
Register
Study Room
Study Room
Arithmetic Ability
Lessons
Alligation or Mixture
Area
Average
Banker's Discount
Boats and Streams
Calendar
Chain Rule
Clocks
Compound Interest
Decimal Fractions
HCF and LCM
Height and Distance
Logarithms
Numbers
Odd Man Out and Series
Partnership
Percentage
Permutations and Combinations
Pipes and Cistern
Probability
Problems on Ages
Problems on Numbers
Problems on Trains
Profit and Loss
Races and Games
Ratio and Proportion
Simple Interest
Simplification
Square Roots and Cube Roots
Stocks and Shares
Time and Distance
Time and Work
True Discount
Volume and Surface Area
Goto
Arithmetic Ability
Problems on Ages
Take Exam
01
02
03
04
05
06
07
08
09
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
Previous
67 / 93
Next
Q.
Ten years ago A was half of B in age. If the ratio of their present ages is 3 : 4, what will be the total of their present ages
A.
15
B.
25
C.
35
D.
45
View Answer
Correct choice: C
Explanation:
Let A's age 10 years ago = x years. Then, B's age 10 years ago = 2x years. (x + 10) / (2x+ l0) = 3/4 => x = 5. So, the total of their present ages =(x + 10 + 2x + 10) = (3x + 20) = 35 years.
Previous
67 / 93
Next