PSC Maths Questions and Answers
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Are you searching for PSC Exam Preparation tips? Then you are in the right place. Here I’m sharing you some shortcut methods in maths you should know if you are preparing for PSC examination. Try to solve more Questions and Answers in this method.
Clock based problems are one of the frequently asked Questions in most of the competitive
exam.To solve these problems, it is always better to understand some of the basic principles and the types of problems that get asked. In this post I hereby explained simple tricks and some simple formulas for solving clock based problems, In every competitive exams clock Questions are categorized in to two ways.
1. Problems in angles.
2. Problems on incorrect clocks.
Before we actually start solving problems on angles, we need to know couple of basic facts clear:
==>Speed of the hour hand = 0.5 degrees per minute (dpm)
==>Speed of the minute hand = 6 dpm
==>At ‘n’ o’ clock, the angle of the hour hand from the vertical is 30n.
The questions based upon these could be of the following types;
Example : 1
What is the angle between the hands of the clock at 7:20 At 7 o’ clock, the hour hand is at 210 degrees from the vertical.
In 20 minutes,
Hour hand = 210 + 20*(0.5) = 210 + 10 = 220 {The hour hand moves at 0.5 dpm}
Minute hand = 20*(6) = 120 {The minute hand moves at 6 dpm}
Difference or angle between the hands = 220 – 120 = 100 degrees .
Method : 2
Example :2
Find the reflex angle between the hands of a clock at 05.30?
The above problem are solved by the bellow formula:
Angle between X and Y =|(X*30)-((Y*11)/2)|
Angle between hands at 5:30
Step 1: X=5 , Y=30
Step 2: 5*30=150
Step 3: (30*11)/2 = 165
Step 4: 165-150=15
Thus, angle between hands at 5:30 is 15 degrees.
Example : 3
At what time 3&4’o clock in the hands of clock together.
Approximately we know at 03:15 hands of the clock together. So 15*60/55=16.36 min.
In 1 minute it covers 360/60 = 6°/ minute. Also, as the hour hand covers just one part out of the given 12 parts in one hour. This implies it covers 30° in 60 min. i.e. ½° per minute. This implies that the relative speed of the minute hand is 6 - ½ = 5 ½ degrees. We will use the concept of relative speed and relative distance while solving problems on clocks.
==>The hands are in the same straight line when they are coincident or opposite to each other.
==>When the two hands are at a right angle, they are 15-minute spaces apart. In one hour, they will form two right angles and in 12 hours there are only 22 right angles. It happens due to right angles formed by the minute and hour hand at 3’o clock and 9'o clock.
==>When the hands are in opposite directions, they are 30-minute spaces apart.
==>If both the hour hand and minute hand move at their normal speeds, then both the hands meet after 5/11 minutes.
Percent is derived from a phrase in Latin “per cent um” which means per hundred. It is a ratio with base (denominator) 100. It evolved as a concept so that there can be a uniform platform for comparing different values.
To express x% as a fraction, divide it by 100 ⇒ x% = x/100
To express a fraction as %, multiply it by 100 ⇒ x/y = [(x/y) × 100] %
x% of y is given by (y × x/100 )
Point to remember for faster Calculation;
1 = 100%
1 /2= 50%
1 /3= 33.33%
1/4 = 25%
1/5 = 20%
1 /6= 16.66%
1/7 = 14.28%
1/8 = 12.5%
1/9 = 11.11%
1/10 = 10%
1/11 = 9.09%
1/12 = 8.33%
If ‘M’ is x% of ‘N’ and ‘P’ is y% of ‘N’ then ‘M’ is (x/y) * 100% of ‘P’.
If the sides of the triangle, rectangle, square, circle, rhombus etc is
(i) Increased by a%. Its area is increased by 2a+(a^2/100)
(ii)If decreased b%. Its areas is decreased by, -2b+(b^2/100)
The population of a town is ‘P’. It increased by x% during 1st year, increased by y% during 2nd year and again increased by z% during 3rd year. The population after 3 years will be, P *[(100+x)/100] * [(100+y)/100] * [(100+z)/100]
exam.To solve these problems, it is always better to understand some of the basic principles and the types of problems that get asked. In this post I hereby explained simple tricks and some simple formulas for solving clock based problems, In every competitive exams clock Questions are categorized in to two ways.
1. Problems in angles.
2. Problems on incorrect clocks.
Problems in angles
Method : 1Before we actually start solving problems on angles, we need to know couple of basic facts clear:
==>Speed of the hour hand = 0.5 degrees per minute (dpm)
==>Speed of the minute hand = 6 dpm
==>At ‘n’ o’ clock, the angle of the hour hand from the vertical is 30n.
The questions based upon these could be of the following types;
Example : 1
What is the angle between the hands of the clock at 7:20 At 7 o’ clock, the hour hand is at 210 degrees from the vertical.
In 20 minutes,
Hour hand = 210 + 20*(0.5) = 210 + 10 = 220 {The hour hand moves at 0.5 dpm}
Minute hand = 20*(6) = 120 {The minute hand moves at 6 dpm}
Difference or angle between the hands = 220 – 120 = 100 degrees .
Method : 2
Example :2
Find the reflex angle between the hands of a clock at 05.30?
The above problem are solved by the bellow formula:
Angle between X and Y =|(X*30)-((Y*11)/2)|
Angle between hands at 5:30
Step 1: X=5 , Y=30
Step 2: 5*30=150
Step 3: (30*11)/2 = 165
Step 4: 165-150=15
Thus, angle between hands at 5:30 is 15 degrees.
Example : 3
At what time 3&4’o clock in the hands of clock together.
Approximately we know at 03:15 hands of the clock together. So 15*60/55=16.36 min.
Basic Concept of Clocks
A clock is a complete circle having 360 degrees. It is divided into 12 equal parts i.e. each part is 360/12 = 30°. As the minute hand takes a complete round in one hour it covers 360° in 60 min.In 1 minute it covers 360/60 = 6°/ minute. Also, as the hour hand covers just one part out of the given 12 parts in one hour. This implies it covers 30° in 60 min. i.e. ½° per minute. This implies that the relative speed of the minute hand is 6 - ½ = 5 ½ degrees. We will use the concept of relative speed and relative distance while solving problems on clocks.
Some facts about clocks :
==>Every hour, both the hands coincide once. In 12 hours, they will coincide 11 times. It happens due to only one such incident between 12 and 1'o clock.==>The hands are in the same straight line when they are coincident or opposite to each other.
==>When the two hands are at a right angle, they are 15-minute spaces apart. In one hour, they will form two right angles and in 12 hours there are only 22 right angles. It happens due to right angles formed by the minute and hour hand at 3’o clock and 9'o clock.
==>When the hands are in opposite directions, they are 30-minute spaces apart.
==>If both the hour hand and minute hand move at their normal speeds, then both the hands meet after 5/11 minutes.
Percentage
Percentage is the another type question topic ,in the maths syllabus percentage is compulsory topic, just follow the rules and study basic rule in this topic definitely you can score high marks in this area. now we can discuss some percentage related Questions and answers.Percent is derived from a phrase in Latin “per cent um” which means per hundred. It is a ratio with base (denominator) 100. It evolved as a concept so that there can be a uniform platform for comparing different values.
To express x% as a fraction, divide it by 100 ⇒ x% = x/100
To express a fraction as %, multiply it by 100 ⇒ x/y = [(x/y) × 100] %
x% of y is given by (y × x/100 )
Point to remember for faster Calculation;
1 = 100%
1 /2= 50%
1 /3= 33.33%
1/4 = 25%
1/5 = 20%
1 /6= 16.66%
1/7 = 14.28%
1/8 = 12.5%
1/9 = 11.11%
1/10 = 10%
1/11 = 9.09%
1/12 = 8.33%
Shortcuts
If X’s income is a% more than Y’s income, the Y’s income is less than X’s income by [ a / (100+a)] * 100%If ‘M’ is x% of ‘N’ and ‘P’ is y% of ‘N’ then ‘M’ is (x/y) * 100% of ‘P’.
If the sides of the triangle, rectangle, square, circle, rhombus etc is
(i) Increased by a%. Its area is increased by 2a+(a^2/100)
(ii)If decreased b%. Its areas is decreased by, -2b+(b^2/100)
The population of a town is ‘P’. It increased by x% during 1st year, increased by y% during 2nd year and again increased by z% during 3rd year. The population after 3 years will be, P *[(100+x)/100] * [(100+y)/100] * [(100+z)/100]
Multiplying by Powers of 2
It simplifies multiplication if a number in the equation is a power of 2, meaning it works for 2, 4, 8, 16 and so on.
Here’s what to do: For each power of 2 that makes up that number, double the other number.
EG:- 9 x 16 is the same thing as 9 x (2 x 2 x 2 x 2) or 9 x 24.. Students can therefore double 9 four times to reach the answer:
- 9 x 24.
- 18 x23
- 36 x 22
- 72 x 2
- 144
Multiplication using Split and Merge Method
This method is the most effective method for multiplication. We usually don’t do mistakes when we multiply a number with less than 10. We do mistakes only when the number is big.
1) What is 63 x 15?
It took 20 seconds for me. For you it may take less than that.
= 63 (10 + 5) = 630 + 315 = 945. its much easy method guyz.
In competitive exams, every second counts and even a 0.1 mark can change your future.
2) What is 81 x 19?
= 81 (20 - 1). It don’t need to be only addition. It can be subtraction also. The ultimate aim is to reduce the time by using simple calculation.
= 1620 – 81 = 1539
3) What is 131 x 26
= 26 (100 + 30 + 1) = 2600 + 780 + 26 = 3406. (You can split the number to n
number of times)
4) What is 147 x 150
You can split this to 150 (100 + 40 + 4 + 3) OR 147 (100 + 50) = 14700 + 7350
friends always use this simple way to find the answer.
5) What is 1005 x 106
Traditional Method will surely take 30 plus seconds. But follow this tricky method to crack your exam, and earn more time to solve other questions.
= 106 (1000 + 2 + 2 + 1)
I have split the number to many numbers for your understanding. It can be just 106 (1000 + 5).
6) What is 1256 x 516
= 1256 (500 + 10 + 3 + 3)
(It may look big. But once you practice this method, you will understand how simple and effective it is)
7)Multiplication with 5 - Into 10 By 2 Method (x 10/2)
When you have to multiply any number with 5, First multiply with 10 and
divide by 2.
• 2 x 5 = 20 /2 = 10
• 77 x 5 = 770/2 = 385
• 1876 x 5 = 18760/2 = 9380
• 978672 x 5 = 9786720/2 = 4893360 (You can directly write the answer)
8) What is 1082 x 107
= 1082 (100 + 5 + 2)
= 108200 + [10820/2] + 2164
= 108200 + 5410 + 2164
= 115774
Here I write this many steps only for your understanding. you can directly write the answer in single step.
9) What is 103 x 97
Here rather than using split and merge method, we can apply another Formula.
(a + b) (a - b) = a2 – b2
(100 + 3) (100 - 3) = 1002 – 32 = 10000 – 9 = 9991.
==> What is 53 x 47
= (50 + 3) (50 - 3) = 2500 – 9 = 2491. It just took 5 seconds.
==> What is 163 x 157
= (160 + 3) (160 - 3) = 25600 – 9 = 25591
Multiplication with 11
1) 21 x 11 = ?
Leave the traditional method. Shortcut method is write 2 and 1 at the first and last position. (Always add from the left) i.e. 2___1 Now add 2 and 1. That is 3.
Answer is 231. (You got answer in one second).
2) 25 x 11 = 2 ___ 5 = 275
3) 32 x 11 = 3 ___2 = 352
4) 253 x 11 = 2 _______ 3 = 2783 (Just adding the adjacent numbers).
5) 531 x 11 = 5 ____1 = 5841
6) 277 x 11 = 2____7
Here 2 followed by (2 + 7) then (7 + 7) then 7
= 29 (14) 7 (Carry the 1 to the left)
= 3047 is the answer.
7) 9879 x 11 = 9_____9
= 9 (9 + 8) (8 + 7) (7 + 9) 9
= 9 (17) (15) (16) (9) (Add the carry over number to the left)
= (9+1) (7+1) (5+1) (6)(9)
= 108669
8) 1387 x 11 = 1 ______ 7
= 1 (1+3) (3+8) (8+7) 7
= 1 (4) (11) (15) 7
= 1(4 +1) (1+1) (5) (7)
= 15257
9) 2587 x 11 = 2______7 = 27(13)(15)7 = 28457
k) 3768912 x 11 = 3_______2 = 3(10)(13)(14)(17)(10)(3)2 = 41458032
Practice Problems:
1. 98763 x 11
2. 9659213 x 11
3. 8621098 x 11
4. 55598 x 11
5. 127409 x 11
friends the above problems are try to solve your self
PSC General Knowledge Questions and Answer https://a2zgkinfo.blogspot.com/2017/12/psc-general-knowledge-questions-and.html
PSC exam English Grammer Qustions and answer https://a2zgkinfo.blogspot.com/2017/12/psc-exam-english-grammer.html
