All types of Mathematics MCQs for PPSC, FPSC, NTS, and all government exam.
If the height of a pole is 2√3 meters and the length of its shadow is 2 meters, find the angle of elevation of the sun.
A. 50°
B. 60°
C. 70°
D. 80°
Let AB be the pole and AC be its shadow. Let the angle of elevation, ∠ACB = θ. Then, AB = 2√3m, AC = 2 m. tan θ = AB/AC = 2√3/2 = √3 θ = 60°. So, the angle of elevation is 60°
The angle of elevation of the sun, when the length of the shadow of a tree is √3 times the height of the tree, is:________?
A. 30°
B. 45°
C. 60°
D. 90°
Let AB be the tree and AC be its shadow.
Let ∠ACB = θ.
Then, AC/AB = √3
Cot θ = √3
θ = 30°
From point P on level ground, the elevation angle of a tower’s top is 30°. If the tower is 100 m high, the distance of point P from the foot of the building is:_________?
A. 149 m
B. 156 m
C. 173 m
D. 200 m
Let AB be the tower. Then, ∠APB = 30° and AB = 100 m, AB/AP = tan 30° = 1/√3 AP = (AB X √3)= 100√3 m. = (100 X 1.73) m = 173 m.
What will be the compound interest on a sum of Rs. 25,000 after 3 years at 12% p.a.?
A. Rs. 9000.30
B. Rs. 9720
C. Rs. 10123.20
D. Rs. 10483.20
Amount = [25000 * (1 + 12/100)3]
= 25000 * 28/25 * 28/25 * 28/25 = Rs. 35123.20
C.I. = (35123.20 – 25000) = Rs. 10123.20
A man saves Rs. 200 at the end of each year and lends the money at 5% compound interest. How much will it become at the end of 3 years?
A. Rs. 565.25
B. Rs. 635
C. Rs. 662.02
D. Rs. 666.50
Amount = [200(1 + 5/100)3 + 200(1 + 5/100)2 + 200(1 + 5/100)]
= [200 * 21/20(21/20 * 21/20 + 21/20 + 1)] = Rs. 662.02
Hafeez invested Rs. 15000 at 10% per annum for one year. If the interest is compounded half-yearly, then the amount received by Hafeez at the end of the year will be?
A. Rs. 16,500
B. Rs. 16,525.50
C. Rs. 16,537.50
D. Rs. 18,150
P = Rs. 15000; R = 10% p.a. = 5% per half-year; T = 1 year = 2 half-year
Amount = [15000 * (1 + 5/100)2] = (15000 * 21/20 * 21/20) = Rs. 16537.50
A cricketer has a certain average for 10 innings. In the eleventh inning, he scored 108 runs, thereby increasing his average by 6 runs. His new average is_____________?
A. 48 runs
B. 52 runs
C. 55 runs
D. 60 runs
Let the average for 10 innings be x. Then,
(10x + 108)/11 = x + 6
= 11x + 66 = 10x + 108
= x = 42.
New average = (x + 6) = 48 runs
The average age of a husband, wife, and their child 3 years ago was 27 years and that of a wife and the child 5 years ago was 20 years. The present age of the husband is___________?
A. 35 years
B. 30 years
C. 40 years
D. None of these
27×3+3×3=90 husband, wife and child
20×2+5×2=50 wide n child
Hsbnd age= 90-50
= 40 yrs
The arithmetic mean of the scores of a group of students in a test was 52. The brightest 20% secured a mean score of 80 and the dullest 25% a mean score of 31. The mean score of the remaining 55% is___________?
A. 45
B. 50
C. 51.4 approx.
D. 54.6 approx.
Let the required means score be x. Then,
20 * 80 + 25 * 31 + 55 * x = 52 * 100
= 1600 + 775 + 55x = 5200
= 55x = 2825
= x = 565/11 = 51.4.
The average salary of all the workers in a workshop is Rs. 8000. The average salary of 7 technicians is Rs. 12000 and the average salary of the rest is Rs. 6000. The total number of workers in the workshop is__________________?
A. 20
B. 21
C. 22
D. 23
Let the total number of workers be x. Then,
8000x = (12000 * 7) + 6000(x – 7)
= 2000x = 42000
= x = 21
The average age of students in a class is 15.8 years. The average age of boys in the class is 16.4 years and that of the girls is 15.4 years. The ratio of the number of boys to the number of girls in the class is___________?
A. 1:2
B. 2:3
C. 3:4
D. 3:5
The average of women and child workers in a factory was 15%yr. The average age of all the 16 children was 8yr and the average age of women workers was 22 yrs if ten women workers were married then the number of unmarried women workers was________?
A. 16
B. 12
C. 8
D. 6
Let the number of women workers be x
According to the question 22*x+16*8 = 15(16+x)
=> 22x+128 =240 +15x
=> 22x-15x = 240 -128
=> 7x=112
Therefore, x=112/7 =16
Therefore, Unmarried women workers =(16-10) = 6