Mathematics questions with solutions
1. A regular polygon has each of its angles as 160⁰. What is the number of sides of the polygon?
1. A regular polygon has each of its angles as 160⁰. What is the number of sides of the polygon?
a) 36
b) 9
c) 18
d) 20
Solution
Number of sides of the polygon » n
»180(n-2)/n
160 »180(n-2)/n
Cross multiply
160n » 180n-360
Collect like terms
360 »180n-160n
360 » 20n
Divide both sides by 20
n » 18 (C)
b) 9
c) 18
d) 20
Solution
Number of sides of the polygon » n
»180(n-2)/n
160 »180(n-2)/n
Cross multiply
160n » 180n-360
Collect like terms
360 »180n-160n
360 » 20n
Divide both sides by 20
n » 18 (C)
2. A girl walks 30m from a point p on a bearing of 040⁰ to a point Q. She then walks 30m on a bearing of 140⁰ to a point R. The bearing of R from P is?
2. A girl walks 30m from a point p on a bearing of 040⁰ to a point Q. She then walks 30m on a bearing of 140⁰ to a point R. The bearing of R from P is?
a) 90⁰
b) 50⁰
c) 45⁰
d) 40⁰
Solution
From the figure above, the bearing of R from P » 40⁰ + ø
Triangle PQR In isosceles,
Therefore, angle P = angle R = ø
Ø » (180-80)/2 » 50⁰
The bearing of R from P
Ø » 40⁰ + 50⁰
» 90⁰ [A]
3. How many different three digits number can be formed using the integers 1 to 6 if no integer occurs twice in a number?
a) 24
b) 120
c) 60
d) 48
Solution
The number of three digit number from integer 1 to 6
6P3 » n!/(n-r)!
6!/(6-3)!
6!/3!
6×5×4×3!/3!
120 [b]
4. In how many different ways can the letters of the word GEOLOGY be arranged in order?
a) 720
b) 1260
c) 2620
d) 5040
Solution
The word GEOLOGY has 7 letters » 7!
‘0’ occurs twice (2!)
‘G’ occurs twice = (2!)
The rest appears only ‘once’
In how many different ways can the letters be arranged in order
7!/(2!2!)
7×6×5×4×3×2!/(2!2!)
» 7×6×5×2×3
1260 [b]
5. A cyclist rode for 30 minutes at xkm/hr and due to a breakdown he had to push the bike for 2hrs at x-5km/hr. If the total distance covered is less than 60km, what is the range of the value of x?
a) X<14
b) X<20
c) X<29
d) X<28
Solution
Distance covered in the first part » speed × timetaken
X × 30/60
½xkm
Distance covered in the second part » 2(x-5)km
When distance covered is less than 60km
Therefore, 2(x-5)2 + ½x < 60
4x-20 + ½x < 60
4x-20 + x < 120
5x<140
X<28 [d]
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