Mathematics MCQs and Answers: Mathematics MCQs are very important for OPSC, OSSC, OSSSC & Other State Exams. Aspirants who are willing to apply for the various Government exams in 2022 must go through the topics of Mathematics for competitive exams, as Mathematics is a key part of the syllabus.
Q1. The value of x + x(xx) when x = 2 is:
(a) 10,
(b) 16,
(c) 18,
(d) 36,
(e) 64
Q2. Mr. Jones sold two pipes at $1.20 each. Based on the cost, his profit one was 20% and his loss on the other was 20%. On the sale of the pipes, he:
(a) broke even,
(b) lost 4 cents,
(c) gained 4 cents,
(d) lost 10 cents,
(e) gained 10 cents
Q3. The distance light travels in one year is approximately 5,870,000,000,000 miles. The distance light travels in 100 years is:
(a) 587 × 108 miles,
(b) 587 × 1010 miles,
(c) 587 × 10-10 miles,
(d) 587 × 1012 miles,
(e) 587 × 10-12 miles
Q4. A man has $ 10,000 to invest. He invests $ 4000 at 5 % and $ 3500 at 4 %. In order to have a yearly income of $ 500, he must invest the remainder at:
(a) 6 % ,
(b) 6.1 %,
(c) 6.2 %,
(d) 6.3 %,
(e) 6.4 %
Q5. Jones covered a distance of 50 miles on his first trip. On a later trip he traveled 300 miles while going three times as fast. His new time compared with the old time was:
(a) three times as much
(b) twice as much,
(c) the same,
(d) half as much,
(e) a third as much
Q6. If (0.2)x = 2 and log 2 = 0.3010, then the value of x to the nearest tenth is:
(a) -10.0,
(b) -0.5,
(c) -0.4,
(d) -0.2,
(e) 10.0
Q7. If 102y = 25, then 10-y equals:
(a) -1/5,
(b) 1/625,
(c) 1/50,
(d) 1/25,
(e) 1/5
Q8. The fraction (5x-11)/(2×2 + x – 6) was obtained by adding the two fractions A/(x + 2) and B/(2x – 3). The values of A and B must be, respectively:
(a) 5x, -11,
(b) -11, 5x,
(c) -1, 3,
(d) 3, -1,
(e) 5, -11
Q9. The sum of the three numbers is 98. The ratio of the first to the second is 2/3, and the ratio of the second to the third is 5/8. The second number is:
(a) 15
(b) 20
(c) 30
(d) 32
(e) 33
Q10.A sum fetched a total simple interest of Rs 1200 at the rate of 7.5%/yr in 4 years. What is the sum (in Rs)?
(a) 4000
(b) 6000
(c) 8000
(d) 7500
(e) 8500
Q11.A sum of Rs 400 becomes Rs 448 at simple interesting 2 years. In how many years will the sum of Rs 550 amount to Rs 682 at the same rate?
(a) 2
(b) 3
(c) 3.5
(d) 4
Q12. The simple interest on a sum of money for 10 years is Rs 3130. If the principal becomes 5 times after 5 years, then what will be the total interest (in Rs) obtained after 10 years?
(a) 6260
(b) 7825
(c) 9390
(d) 15650
Q13.A person invested a total sum of Rs 7900 in three different schemes of simple interest at 3%, 5% and 8% per annum. At the end of one year he got same interest in all three schemes. What is the money (in Rs) invested at 3%?
(a) 2900
(b) 3500
(c) 4000
(d) 5600
Q14. A, B and C invested amounts in the ratio 3 : 4 : 5 respectively. If the schemes offered compound interest at the rate of 20% per annum, 15% per annum and 10% per annum respectively, then what will be the ratio of their amounts after 1 year?
(a) 3 : 15 : 25
(b) 6 : 6 : 5
(c) 36 : 46 : 55
(d) 12 : 23 : 11
Q15. The difference between the compound interest compounding half yearly for 1 year and the simple interest for 1 year on a certain sum of money lent out at 8% per annum is Rs 64. What is the sum (in Rs)?
(a) 40000
(b) 42000
(c) 44000
(d) 44800
Answer:
S1. Ans: (a)
Sol:x + x(xx)
Put the value of x = 2 in the above expression we get,2 + 2(22)= 2 + 2(2 × 2)
= 2 + 2(4)
= 2 + 8
= 10
S2. Ans.(d) lost 10 cents,
Sol: The selling price of the first pipe = $1.20
Profit = 20%
Let’s try to find the cost price of the first pipe
CP = Selling price – Profit
CP = 1.20 – 20% of CP
CP = 1.20 – 0.20CP
CP + 0.20CP = 1.20
1.20CP = 1.20
CP = 1.201.20
CP = $ 1
Selling price of the Second pipe = $1.20
Loss = 20%
Let’s try to find the cost price of the second pipe
CP = Selling price + Loss
CP = 1.20 + 20% of CP
CP = 1.20 + 0.20CP
CP – 0.20CP = 1.20
0.80CP = 1.20
CP = 1.200.80
CP = $1.50
Therefore, total cost price of the two pipes = $1.00 + $1.50 = $2.50
And total selling price of the two pipes = $1.20 + $1.20 = $2.40
Loss = $2.50 – $2.40 = $0.10
Therefore, Mr. Jones loss 10 cents.
S3. Ans.D.
Sol: The distance the light travels in 100 years is:
5,870,000,000,000 × 100 miles.
= 587,000,000,000,000 miles.
= 587 × 1012 miles.
S4.Ans: (e)
Sol:Income from $ 4000 at 5 % in one year = $ 4000 of 5 %.
= $ 4000 × 5/100.
= $ 4000 × 0.05.
= $ 200.
Income from $ 3500 at 4 % in one year = $ 3500 of 4 %.
= $ 3500 × 4/100.
= $ 3500 × 0.04.
= $ 140.
Total income from 4000 at 5 % and 3500 at 4 % = $ 200 + $ 140 = $ 340.
Remaining income amount in order to have a yearly income of $ 500 = $ 500 – $ 340.
= $ 160.Total invested amount = $ 4000 + $ 3500 = $7500.
Remaining invest amount = $ 10000 – $ 7500 = $ 2500.
We know that, Interest = Principal × Rate × Time
Interest = $ 160,
Principal = $ 2500,
Rate = r [we need to find the value of r],
Time = 1 year.
160 = 2500 × r × 1.
160 = 2500r
160/2500 = 2500r/2500 [divide both sides by 2500]
0.064 = r
r = 0.064
Change it to a percent by moving the decimal to the right two places r = 6.4 %
Therefore, he invested the remaining amount $ 2500 at 6.4 % in order to get $ 500 income every year.
S5. Ans. (b)
Sol:Let speed of the 1st trip x miles / hr. and speed of the 2nd trip 3x / hr.
We know that
Speed = Distance/Time.
Or, Time = Distance/Speed.
So, times taken to covered a distance of 50 miles on his first trip = 50/x hr.
And times taken to covered a distance of 300 miles on his later trip = 300/3x hr.
= 100/x hr.
So we can clearly see that his new time compared with the old time was: twice as much.
S6. Ans.(c)
Sol:(0.2)x = 2.
Taking log on both sides
log (0.2)x = log 2.
x log (0.2) = 0.3010, [since log 2 = 0.3010].
x log (2/10) = 0.3010.
x [log 2 – log 10] = 0.3010.
x [log 2 – 1] = 0.3010,[since log 10=1].
x [0.3010 -1] = 0.3010, [since log 2 = 0.3010].
x[-0.699] = 0.3010.
x = 0.3010/-0.699.
x = -0.4306….
x = -0.4 (nearest tenth)
S7. Ans. (e)
Sol:102y = 25
(10y)2 = 52
10y = 5
1/10y = 1/5
10-y = 1/5
S8 Ans.(D)
Sol:Partial Fraction
130Save
S9 Ans. (C)
Sol:Let the three numbers be x, y and z.
Sum of the numbers is 98.
x + y + z = 98………………(i)
The ratio of the first to the second is 2/3.
x/y = 2/3.
x = 2/3 × y.
x = 2y/3.
The ratio of the second to the third is 5/8.
y/z = 5/8.
z/y = 8/5.
z = 8/5 × y.
z = 8y/5.
Put the value of x = 2y/3 and z = 8y/5 in (i).
2y/3 + y + 8y/5 = 98
49y/15 = 98.
49y = 98 × 15.
49y = 1470.
y = 1470/49.
y = 30 .Therefore, the second number is 30.
S10. Ans.(a)
Sol. SI= (p x r x t)/100
1200= (p x 4 x 7.5)/100
p=4000
S11. Ans. (d)
Sol. 2 year interest = 448 – 400 = 48
1 year interest = 24
Rate = 24/400 × 100 = 6%
682 – 550 = 550×6×t/100
1320 = 55 × 6 × t
t = 4 year
S12. Ans. (C)
Sol. (P×r×10)/100=3130
A/Q, (P×r/100×5)+(5P×r/100×5)=P×r/100(5+25)
3130×30/100=9390
S13. Ans. (c)
Sol. let Amount x,y,z
Interest
=3% of x = 5% of y = 8% of z =K
x : y : z= k/3 : k /5 : k/8
=40 : 24 : 15
Money Invested at 3%=40 / 79 × 7900 =4000Rs
S14. Ans. (C)
Sol. Let amount invested by A, B and C be Rs. 300, 400 and
500 respectively for 1 year.
Amount under compound interest
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P (1+ r/100)t
So A will get 360 B will get 460 and C will get 550
Ratio of their amount is 360 : 460 : 550 or 36 : 46 : 55
S15. Ans. (a)
Sol. Let the given sum be 100x
R = 8 and T = 1 year
C I = [P (1 + R/200)2T – 1]
Solving we get 204x / 25
S I = P*R*T / 100
Solving we get 8x
Difference between C I & S I
( 204x / 25 ) – (8x) = 64
Solving for x we get 400 so value = 400*100 = 40000
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