Preparing for the OSSC CGL exam requires strong proficiency in arithmetic and numerical ability. Focusing on the top 30 multiple-choice questions (MCQs) in this area can significantly enhance your performance. These MCQs cover essential topics such as percentages, ratios, averages, simple and compound interest, profit and loss, time and work, and data interpretation. Practicing these questions helps in understanding key concepts and improves speed and accuracy. Comprehensive preparation with these critical MCQs ensures a solid foundation in arithmetic and numerical ability, crucial for excelling in the OSSC CGL exam
Top 30 Arithmetic and Numerical Ability MCQs for OSSC CGL
- Question: What is the area of a circle with a radius of 7 cm?
A) 154 cm²
B) 144 cm²
C) 132 cm²
D) 162 cm²
Answer: A) 154 cm²
Solution:
The formula for the area of a circle is:
Area=𝜋𝑟^2
Area=πr ^2
Given
r=7 cm, the area is:
Area=𝜋×7^2= =π×49≈3.14×49=153.86≈154 cm - What is the least number which when divided by 12, 15, and 20 leaves a remainder of 4 in each case?
a) 60
b) 64
c) 124
d) 244
Answer: b) 64
Solution:
The least number which leaves a remainder of 4 in each case can be found by calculating the LCM of the divisors and then adding the common remainder.
LCM(12,15,20)=60
The required number = LCM + remainder = 60 + 4 = 64. - A person purchased a chair at₹70 and spent 17 on its repair and 50 paise on cartage. If he sold the chair at 100, then his approximate margin of profit was
(1) 11.25%
(2) 12.50%
(3) 13.30%
(4) 14.30%
Answer:(4) 14.30%
Sol. Profit % = Profit /CP x 100= (100 – (70 +17 +0.50))/(70+17+0.50)×100 = 12.5 /87.5 ×100=14.30% - A hawker gains the selling price of 4 ball-point pens on selling 1 dozen pens. His gain percentage is
(1) 25
(2) 33wh1/3
(3) 50
(4) 661
Answer: (3) 50
Sol. Gain=SP of 4 pens
→ SP of 1 dozen pens – CP of 1 dozen pens
= SP of 4 pens
→ SP of 12 pens – SP of 4 pens {Gain = SP – CP}
= CP of 12 pens
⇒ SP of 8 pens = CP of 12 pens
Gain % = 4/8x 100 = 50% - A person incurs a loss of 6% by selling a watch at 423. In order to get a profit of 6%, the person should sell the watch at
(1) ₹486
(2) ₹477
(3) ₹586
(4) ₹577
Answer:(2) ₹477
Sol. SP₂ = (423 x (100+6))/ (100-6) =423 x 106 /94 = 477 - A reduction of 20% in the price of oranges enables a man to buy 5 oranges more for₹ 10. The price of an orange before reduction was
(1) 20 paise
(2) 40 paise
(3) 50 paise
(4) 60 paise
Answer: (3) 50 paise
Reduced price = (10 x 20 )/(5 x 100) = Rs 2/5 per orange
Original price = (2×100)/(5×80 )=1/2 = 0.50 - If 3 toys are sold at the cost price of 4 toys of the same kind, the profit will be
(1) 25%
(2) 23%
(3) 662%
(4) 33.33%
Answer:(4) 33.33%
Sol .CP of 4 toys= SP of 3 toys
=Profit % = 1/3 x 100 = 33.33% - A shopkeeper marks his goods 20% above cost price, but allows 30% discount for cash. His net loss is
(1) 8%
(2) 10%
(3) 16%
(4) 20%
Answer:(3) 16%
Sol. M% (above CP) = {(d-l)/(100-d) × 100} %
20 = (30-1)/(100-30 )× 100
l= 16% - A student’s mark was incorrectly entered as 83 instead of 63. Due to this error, the average marks of the class increased by 1. The number of students in the class is:
a) 20
b) 10
c) 30
d) 25
Answer: b) 20
Solution:
Let the number of students be 𝑛
Correct sum = Total sum – Incorrect mark + Correct mark
New average = Old average + 1
(Total sum−83+63)/n = Old average
Total sum/n +1= Total sum−20/n
Solving, we get
n=20. - A shopkeeper marks his goods 20% above the cost price and then allows a discount of 10%. Find his gain percent.
a) 8%
b) 10%
c) 12%
d) 15%
Answer: a) 8%
Solution:
Let the cost price be 100.
Marked price = 120.
Selling price after 10% discount = 120 – 12 = 108.
Gain percent:(108−100)/100 ×100=8% - A and B together can complete a task in 12 days. B and C together can do it in 16 days. A and C together can do it in 24 days. How long will A, B, and C together take to complete the task?
a) 8 days
b) 10wh 2/3 days
c) 12 days
d) 14 days
Answer: b) 10wh 2/3 days
Solution:
Let the work done by A, B, and C per day be
=1/A,1/B, 1/C
From the given information
2( 1/A+1/B+ 1/C)=( 1/12+1/16+ 1/24)
=(4+3+2)/48=9/48=3/16
So( 1/A+1/B+ 1/C)=3/32
A+B+C=8. - 10^7.5*10^4.5?10^2=10^_?
(1) 10
(2) 6
(3) 8.5
(4) 9.5
Answer:(1) 10 - 36^2-24^2=?
(1) 840
(2) 640
(3) 720
(4) 960
Answer:(3) 720 - 32×15-1850/37=?
(1) 295
(2) 326
(3) 410
(4) 430
Answer:(4) 430 - 11{2/9}*36/101=?
(1) 16
(2) 18
(3) 4
(4) 7
Answer:(3) 4 - 1087.05+187.005+87.005=?
(1) 1361.066
(2) 1361.556
(3) 1361.055
(4) 1361.06
Answer:(4) 1361.06 - Root of _+187=220
(1) 1089
(2) 529
(3) 23
(4) 33
Answer:(1) 1089 - (27*2^3 /18)/6=?
(1) 3
(2) 2
(3) 92
(4) 8
Answer:(2) 2 - 80% of 135+?% of 750= 228
(1) 25
(2) 20
(3) 16
(4) 24
Answer:(3) 16 - 632/8+11^2=?
(1) 180
(2) 185
(3) 200
(4) 199
Answer:(3) 200 - ? of7/8 of 1/3 of 2/7=12
(1) 108
(2) 144
(3) 132
(4) 136
Answer:(2) 144 - 10101.001+1001.01+101.1 = ?
(1) 12121.011
(3) 12203.101
(2) 11111.111
(4) 11203.111
Answer:(4) 11203.111 - 2310/14/5=?
(1) 33
(2) 825
(4) 233
(3) 35
Answer:(1) 33 - 198×15+30=?
(1) 98
(2) 19
(3) 96
(4) 99
Answer:(4) 99 - 7.8+5.4×8.2=?
(1) 52.08
(2) 108.24
(3) 48.05
(4) 102.05
Answer:(1) 52.08
10whole2/3 - 5 students are to be seated on 5 seats. In how many different ways can they be seated?
(1) 120
(2) 20
(3) 60
(4) 72
(5) None of these
Answer:(1) 120 - If 2x-3y = 1 and 3x + y = 18, then what is the value of x – y?
(1)-2
(2) 2
(3)3
(4) 3
Answer:(2) 2 - Question: If 12 men can complete a work in 15 days, how many days will 10 men take to complete the same work?
A) 18 days
B) 20 days
C) 15 days
D) 12 days
Answer: A) 18 days
Solution:
The total amount of work can be represented as
W=12 men×15 days=180 man-days.
For 10 men, let the number of days required be
10 men×D days=180 man-days
Solving for
𝐷=180/10=18 days - Question: A train 150 meters long is running at a speed of 54 km/hr. How long will it take to pass a pole?
A) 10 seconds
B) 12 seconds
C) 8 seconds
D) 15 seconds
Answer: B) 10 seconds
Solution:
First, convert the speed from km/hr to m/s:
54 km/hr=54×1000/3600 m/s=15 m/s
The time t taken to pass the pole is:
𝑡=Length of the train/Speed=150 m/15 m/s=10 seconds - Question: If the ratio of two numbers is 3:4 and their sum is 70, what are the numbers?
A) 30 and 40
B) 35 and 35
C) 21 and 49
D) 24 and 46
Answer: A) 30 and 40
Solution:
Let the numbers be 3x and 4x.
Their sum is given as:
3x+4x=70
So,
7x=70
x=10
Thus, the numbers are:
3x=3×10=30
4x=4×10=40
