MENTAL MATH TRICKS FOR DIVISION- BASICS FOR EXAMS

Quantitative Aptitude and Data Interpretation sections of various competitive exams call for long, time-intensive calculations. The time you spend here could mean the difference between selection and disqualification. Calculations can be made easier using divisibility tests and mental math tricks for division specially. You can even use these mental tricks of maths for division to eliminate options. These tricks will be useful for all competitive exams including IBPS Clerk, SBI Clerk, IBPS PO, SBI PO, IBPS RRB, SSC CGL, CAT, XAT, Placement Aptitude, NICL AO, LIC AAO, SBI Associate Clerk, and SBI Associate PO.

Note: In every divisibility test, please assume that 0 is divisible by all divisors.

Divisibility Test for 2

A number is divisible by 2 if the last digit is divisible by 2 i.e. last digit is 0, 2, 4, 6 or 8.
E.g. 864 is divisible by 2 since the last digit 4 is divisible by 2.

Divisibility Test for 3

A number is divisible by 3 if the sum of the digits of the number is divisible by 3.
E.g. 861 is divisible by 3 since the sum of the digits is 15 (8 + 6 + 1 = 15), and 15 is divisible by 3.

Divisibility Test for 4

A number is divisible by 4 if the number formed by the last two digits is divisible by 4.
E.g. 504 is divisible by 4 since 04 is divisible by 4.

Divisibility Test for 5

A number is divisible by 5 if the last digit is either 0 or 5.
E.g. 425 is divisible by 5 since the last digit is 5.

Divisibility Test for 6

A number is divisible by 6 if it is divisible by 2 AND it is divisible by 3.
E.g. 186 is divisible by 6 since it is divisible by 2 AND it is divisible by 3.

Divisibility Test for 7

To find out if a number is divisible by 7 or not, follow these steps:

1. Separate the last digit from the rest of the number. Let us call the rest of the number the truncated number. The truncated number has one less digit than the original number or the previous truncated number.

2. Double the last digit and subtract it from the truncated number.

3. Check if this result is sufficiently small so that you can immediately say if this is divisible by 7. If it is divisible by 7, then so was the original number. If it is not divisible by 7, then neither was the original number.

4. If the number is still too large to visually check if it is divisible, apply this rule over and over again as necessary.

E.g. Check 6132. The last digit is 2 and the truncated number is 613. Twice of 2 is 4. So subtract 4 from the truncated number 613 i.e. 613 – 4 = 609. Again, the last digit is now 9, and the truncated number is 60. Twice of 9 is 18. Subtract it from the truncated number 60, i.e. 60 – 18 = 42. Now 42 is small enough to check visually. We know that 42 is divisible by 7, so we can tell that 6132 is divisible by 7 also.

NOTE: Explaining this step is long. But actually using it is a very short and time saving method. This method is especially useful in Geometry and Mensuration problems where the value of π plays an important role.

Divisibility Test for 8

A number is divisible by 8 if the number formed by the last three digits is divisible by 8.

E.g. 8120 is divisible by 8 since 120 is divisible by 8.

Divisibility Test for 9

A number is divisible by 9 if the sum of its digits is divisible by 9. If the sum is large, you can once again add the digits and check if the new sum is divisible by 9.
E.g. 27549 is divisible by 9 since the sum of the digits is 27 (2 + 7 + 5 + 4 + 9 = 27), and 27 is divisible by 9.

Divisibility Test for 10

A number is divisible by 10 if the last digit is 0.
E.g. 1760 is divisible by 10 since the last digit is 0.

Divisibility Test for 11

Method 1:- The (sum of the odd place digits) – (sum of the even place digits) is divisible by 11.

E.g. 963391
Sum of odd place digits = 9 + 3 + 9 = 21
Sum of even place digits = 6 + 3 + 1 = 10
Difference = 21 – 10 = 11 which is divisible by 11.
So, yes, 963391 is divisible by 11.

Method 2: Subtract the last digit from the remaining truncated number. If the result is divisible by 11, then so was the first number. Apply this rule over and over again as necessary.
E.g. 963391 → 96339 – 1 = 96338 → 9633 – 8 = 9625 → 962 – 5 = 957 → 95 – 7 = 88, which is divisible by 11. So yes, 963391 is divisible by 11.

Divisibility Test for 13

To find out if a number is divisible by 13 or not, follow these steps:

1. Separate the last digit from the rest of the number. Let us call the rest of the number the truncated number. The truncated number has one less digit than the original number or the previous truncated number.
2. Multiply the last digit by 4 and add it to the truncated number.
3. Check if this result is sufficiently small so that you can immediately say if this is divisible by 13. If it is divisible by 13, then so was the original number. If it is not divisible by 13, then neither was the original number.
4. If the number is still too large to visually check if it is divisible, apply this rule over and over again as necessary.

E.g. 12675 → 1267 + 20 = 1287 → 128 + 28 = 156 → 15 + 24 = 39, which is divisible by 13. So yes, 12675 is divisible by 13.

Divisibility Test for 17

To find out if a number is divisible by 17 or not, follow these steps:

1. Separate the last digit from the rest of the number. Let us call the rest of the number the truncated number. The truncated number has one less digit than the original number or the previous truncated number.
2. Multiply the last digit by 5 and subtract it from the truncated number.
3. Check if this result is sufficiently small so that you can immediately say if this is divisible by 17. If it is divisible by 17, then so was the original number. If it is not divisible by 17, then neither was the original number.
4. If the number is still too large to visually check if it is divisible, apply this rule over and over again as necessary.

E.g. 21165 → 2116 – 25 = 2091 → 209 – 5 = 204 → 20 – 20 = 0, which is divisible by 17. So yes, 21165 is divisible by 17.

Divisibility Test for 19

Add two times the last digit to the remaining leading truncated number. If the result is divisible by 19, then so was the first number. Apply this rule over and over again as necessary.

E.g. 185117 → 18511 + 2 × 7 = 18525 → 1852 + 2 × 5 = 1862 → 186 + 2 × 2 = 190, which is divisible by 19. So yes, 185117 is divisible by 19.

Divisibility Test for Composites

A number is divisible by a composite if it is also divisible by all the prime factors.

E.g. 157905 is divisible by 33 if it is divisible by 3 AND by 11.

Hope you liked our effort and we are sure that these mental math tricks for division will help you a lot in exams. Well, there are already a number of books related to the basics of division but they look messy. Here I have complied only important tricks of division that everyone should know. Yes you are free to learn more and more because nothing is enough if we talk about competitive exam. Good Luck!

Previously asked mathematics question solved

Previously asked maths questions from numbers for competitive exams

This type of questions often seen in competitive exams, try it...

1. Difference of 2 number is 11 and their product is 60, find the sum of numbers ?

If you try to solve this question without using equation, it will consume too much time.. so use this equation..

(X + Y)2 = (X - Y)2 + 4 XY

We have X-Y = 11 and XY = 60 

hence

(X + Y)2 = 112 + 4 x 60

(X + Y)2 = 121 + 240

(X + Y)2 = 361

X + Y = 19

Shortcuts for solving Compound Interest Problems

Compound Interest related problems are common in almost all competitive Exams. In this post we are discussing with you some of the handy shortcuts formulas that can be used to solve Compound Interest related Problems. In compound interest, the interest for each period is added to the principle before interest is calculated for the next period. With this method the principle grows as the interest is added to it. This method is mostly used in investments such as savings account and bonds etc. 

The Basic Formula used for solving Compound Interest Problems is:

If A = Amount

P = Principle

C.I. = Compound Interest

T = Time in years

R = Interest Rate Per Year 

Shortcut methods

Shortcut 1: If rate of interest is R1% for first year, R2% for second year and R3% for third year, then:

Let’s find it out with an example:

1) Find the total amount after three years on Rs 1000 if the compound interest rate for first year is 4%, for second year is 5% and for third year is 10%.

P = 1000, R1 = 4%, R2 = 5% and R3 = 10%

Shortcut 2: If principle = P, Rate = R% and Time = T years then

a) If the interest is compounded annually:

b) If the interest is compounded half yearly (two times in year):

c) If the interest is compounded quarterly (four times in year): 

Shortcut 3: If difference between Simple Interest and Compound Interest is given.

a) If the difference between Simple Interest and Compound Interest on a certain sum of money for 2 years at R% rate is given then:

Example: If the difference between simple interest and compound interest on a certain sum of money at 10% per annum for 2 years is Rs 2 then find the sum.

b) If the difference between Simple Interest and Compound Interest on a certain sum of money for 3 years at R% is given then:

Shortcut 4: If sum A becomes B in T1 years at compound interest, then after T2 years

Example:

Qn. Rs 1000 becomes 1100 after 4 years at certain compound interest rate. What will be the sum after 8 years?

Here A = 1000, B = 1100, T1 = 4, T2 = 8

Jammu and Kashmir Circle Mathematics Answer Key Exam held on 18.05.2014

Jammu and Kashmir Circle Mathematics Answer Key Exam held on 18.05.2014

Thanks to Norboo Thinless (Question Paper Shared)

Part - B (Mathematics)

                     

Jammu and Kashmir Circle -19 Booklet Series :B

Q 26. The average temperature of the first 3 days of a week is 26.5 C and that of the next 3 days is 29 C. If the weekly average is 27.4 C, what is the temperature of the day:

A. 26.8 C

B. 24.8 C

C.25.3 C

D.25. C

ANS:C

Q 27 Find the Volume of a cube where longest diagonal is 2_/3 cm:

A. 8cm3

B. 4 cm3

C. 16 cm3

D. 2 cm3

ANS:A

Q 28 A sells a bicycle to B at a profit of 20% B sells to C at a profit of 25%. If C pays Rs. 225 for it, what did a pay for it?

A. 175 B. 125 C. 200 D.150

ANS:D

Q 29. Two train of 400 meters and 475 meters in length run at the speed of 45km/h and X km/h respectively in opposite directions on parallel tracks. The time taken by them to across each other is 42 seconds, then find the value of X:

A.28km/h B.30km/h C. 32km/h D. 30km/h

ANS:D

Q 30. X borrowed Rs. 500 from Y@ 5% p.a. simple interest. If X paid Rs.50 as interest, what was the tenure of this loan?

A. 2years B. 212 C. 3 years D. 4 years

ANS:D

Q 31 At an examination 65% of the total candidates were successful. The total number of failures was 420. Determine the number of students who appeared for the exam:

A. 567 B. 639 C. 1000 D. 1200

ANS:D

Q. 32 A sum of Rs. 550 was taken as loan. This is to be paid back in two equal installments. If the rate of interest be 20% compounded annually, then the value of each installment is:

A. Rs. 421 B. 396 C. 360 D. 350

Q .33 A man cycles at the rate of 15.6km/h. How many meters does he cover in 2 minutes?

A. 31.2 B. 260 C. 520 D. 5200

ANS:B

Q. 34 The value of ((.013) ^2+(.007) ^2)/((.013) ^2-.13×.007+(.007) ^2 )

A. .006

B. .02

C. .0091

D. .00185

Q. 35 If 3712 of a number is 900, then 6212% of the number will be:

A. 1200 B. 1350 C. 1500 D. 450

ANS:C

Q. 36 A and B can do a piece of work in 18 days; B and C in 24 days, C and A in 36 days. In how many days can they do it all working together:

A. 12 B. 13 C. 16 D. 26

ANS:C

Q. 37 The average price per kg of rice at 10 different places was Rs. 4085. After a week, the price per kg was increased by 20 paisa at 3 places and decreased by 10 paisa at one place. The new average of price per kg is:

A. 4.88 B. 4.86 C. 4.90 D. 5.35

ANS:C

Q. 38 The average marks of 30 students in a class is 45. While checking two errors were noticed. after correction, one student got 45 marks more and other 15 marks less. What is the corrected average?

A. 47 B. 48. C. 46 D. 50

ANS:C

Q. 39 In a group of 15 persons, 7 persons have studies German, 8 have studies French and 3 persons have not studies either. How many persons of these have studies both German and French?

A. 4 B. 6 C. 5 D. 3

Q. 40 If the radius of a right circular cylinder is decreased by 50% and its height is increased by 60% its volume will be decreased by:

A. 60% B. 50% C. 55% D. None of these

ANS:A

Q 41 Given that √(1225 ) =35; then √(12.25 )+√0.1225+√0.0 01225 is equal to:

A. .385

B. 3.885

C. 38.85

D. 3.585

ANS:B

Q. 42 Ramu spend 75% of his income. His income is increased by 20% he has increased his Expenditure by 10%. His saving are increased by:

A.40% B. 50% C. 30% D. 25%

ANS:D

Q. 43 If the length of a rectangle is increased by 20% and width of the same rectangle decreased by 20% then the area of the rectangle:

A. Decreases by 20%

B. Decreases by 4%

C. Is unchanged

D. Increases by 20%

ANS:B

Q. 44 If a clock started at noon then the angle traversed by the hour hand at 5.45 PM will be

A. 112 ½*
B. 120*
C. 172 ½*
D. none of these
ANS:C

Q. 45 The sum of two numbers is 42 and their product is 437. Then the absolute difference between the numbers is:

A. 3 B. 4 C. 5 D. 7
ANS:B

Q. 46 A does 80% of a work in 20 days. He then calls in B and together they finish the remaining work in 3 days. How long B alone would take to do the whole work:

A.23 days B. 37 days C.37 1/2days D. 40 days


Q. 47 If 6 men and 8 boys can do a piece of work in 10 days while 26 men and 48 boys can do the same in 2 days, the time taken by 15 men and 20 boys in doing the same type of work will be:

A. 4 days B. 5 days C. 6days D. 3 days
ANS:A

Q.48 The ratio between the speeds of two train is 7:8. If the second train runs 400 km in 4 hours, then the speed of the first train is;

70km/h B. 75km/h C. 84km/ h D. 87.5km/h
ANS:D

Q. 49 When the circumference of a circle decreases from 3π to π, its area decreases by:

A. 1623 % B. 6623 % C.888/9 % D.1212 %
ANS:B