CDS MATHS
NDA MATHS
Arithmetic
Algebra + Vector Algebra
Algebra
Trigonometry
Trigonometry
Determinants and matrices
Geometry
Analytical Geometry
Mensuration
Vector Algebra
Statistics
Statistics and Probability Differential calculus Integral calculus
AFCAT MATHS
TA MATHS
Decimal Fraction
Arithmetic
Simplification
Algebra
Average
Trigonometry
Profit & loss
Geometry
Percentage
Mensuration
Ratio & Proportion
Statistics
Simple Interest
101 MATHS SHORTCUTS 1.
2.
3.
4.
5
6.
7.
Metho Method d to multip multiply ly 2-dig 2-digit it numb numbeer. (i) AB × CD = AC AC / AD + BC BC / BD 35 × 47 = 12 / 21 + 20 / 35 = 12 / 41 / 35 = 1645 (ii) AB × AC AC = A2 / A (B + C) / BC 74 × 76 = 72 / 7(4 + 6) / 4 × 6 = 49 / 70 / 24 = 49 / 70 / 24 = 5624 (iii) (iii) AB × CC = AC / (A (A + B)C / BC = 35 × 44 = 3 × 4 / (3 + 5) × 4 / 5 × 4 = 12 / 32 / 20 = 12 / 32 / 20 = 1540 Metho Method d to to mult multiiply ply 3-di 3-digi gitt no. no. ABC × DEF = AD / AE + BD / AF + BE + CD / BF + CE / CF 456 × 234 = 4 × 2 / 4 × 3 + 5 × 2 / 4 × 4 + 5 × 3 + 6 × 2 / 5 × 4 + 6 × 3 / 6 × 4 = 8 / 12 + 10 / 16 + 15 + 12 1 2 / 20 + 18 / 24 2 3 8 4 = 8 / 2 /4 /4 / 3 / 2 = 106704 If in a series series all number number contains contains repeating repeating 7. To find find their sum, we start start from from the left left multiply 7 by 1, 2, 3, 4, 4, 5 & 6. Look Look at at the example below. 777777 + 77777 + 7777 + 777 + 77 + 7 = ? =7×1/7×2/7×3/7×4/7×5/7×6 = 7 / 14 / 21 / 28 / 35 / 42 = 864192 0.55 0.5555 55 + 0.5 0.555 55 + 0.5 0.55 5 + 0.5 0.5 = ? To find the sum of those number in which one number is repeated after decimal, then first write the number in either increasing in creasing or decreasing order. Then -find the sum by using the below method. 0.5555 + 0.555 + 0.55 + 0.5 =5×4/5×3/5×2/5×1 = 20 / 15 / 10 / 5 = 2.1605 Thos Thosee num numb bers ers whose hose all all dig digit itss are are 3. 3. 2 (33) = 1089 089 Those Those numb number er.. in which hich all all dig digits its are are num numbe berr is is 3 tw two or or mor moree than than 2 tim times es repe repeat ated ed,, to to find find the squar squaree of these number, we repe repeat at 1 and 8 by (n – 1) time. Where n ® Number of times 3 repeated. 2 (333) = 110889 (3333)2 = 11108889 Thos Thosee numb number er who whose se all all digi digits ts are are 9. 2 (99) = 9801 (999)2 = 998001 (9999)2 = 99980001 (99999)2 = 9999800001 Thos Thosee numb number er who whose se all all digi digits ts are are 1. A number whose whose one’s, ten’s, hundred’s digit is 1 i.e., 11, 111, 1111, .... In this we count count number of digits. We write 1, 2, 3, ..... in th their eir square the digit in the number, then write in d ecreasing ecreasing order up to 1. 2 11 = 121 2 111 = 12321 11112 = 1234321
shop.ssbcrack.com
S-2
101 Shortcuts in Quantit ati ve Aptitude
8.
Some Some prope properti rties es of squa square re and and squa square re roo root: t: (i) (i) Complete square square of a no. is possible possible if its last digit is 0, 1, 4, 5, 6 & 9. If last digit digit of a no. is 2, 3, 7, 8 then complete square root root of this no. is not possible. (ii) ii) If last digit digit of a no. is 1, then last digit of its complete complete square square root root is either 1 or 9. (iii iii) If last digit of a no. is 4, then last digit of its complete complete square square root is either either 2 or 8. (iv) (iv) If last digit of a no. is 5 or 0, then last digit of its complete complete square square root root is either 5 or 0. (v) (v) If last last digit of a no. is 6, then last last digit of its complete complete square square root root is either either 4 or 6. (vi) (vi) If last digit digit of a no. is 9, then last digit of its complete complete square square root root is either 3 or 7.
9.
Prime Number : (i) Find the approx square root root of given no. Divide Divide the given no. by by the prime no. less than approx square square root of of no. If given no. is not divisible by any of these prime no. then t he no. is pri me otherwise not. check 359 is a prime number or not. For example : To check Sol. Approx sq. root = 19 Prime no. < 19 are 2, 3, 5, 7, 11, 13, 17 359 is not divisible by any of of these prime nos. So 359 is a prime no. 5001 For example: Is 2 + 1 is prime or n ot?
25001 + 1 2 +1
Þ Reminder = 0,
\
10. 11. 12.
13.
25001 + 1 is not prime. (ii) ii) There are 15 prime prime no. from from 1 to 50. 50. (iii) iii) There are 25 prime prime no. from from 1 to 100. 100. (iv) (iv) There are are 168 prime prime no. no. from 1 to 1000. 1000. n n If a no. no. is is in the the form form of of x + a , then it is divisible by (x + a); if n is odd. If xn ¸ (x – 1), then remainder is alway al wayss 1. n If x ¸ (x + 1) (i) If n is eve even, n, then then rem remaind ainder er is 1. (ii) ii) If n is odd, odd, then then remai remainde nderr is x. x. Value of
P + P + P + ..........¥
=
(ii) Valu Valuee of
P - P - P - ..........¥
=
(iii) ii) Value Value of
P. P. P. ..........¥
(i)
(iv (iv) Value Value of
P P P P P
4P + 1 + 1 2 4P + 1 - 1 2
=P
2n -1) ¸2n ( =P
[Where n ® no. of times P repeated]. val ue of Note: If factors of P are n & (n + 1) type then value 14. 14.
P + P + P + .. ....¥
= ( n + 1) and
P - P - P - ....¥
Number of divisors : (i) If N is is any no. no. and and N = an × bm × c p × .... where a, b, c are prime no. No. No. of of divisors divisors of N = (n + 1) (m + 1) (p + 1) .... e.g. Find the no. of divisors of 90000. N = 90000 90000 = 22 × 32 × 52 × 102 = 22 × 32 × 52 × (2 × 5)2 = 24 × 32 × 54 So, the no. of divisors = (4 + 1) (2 + 1) (4 + 1) = 75 (ii) N = an × bm × c p, where a, b, c are prime Then set of co-prime co-prime factors factors of N = [(n + 1) (m + 1) (p + 1) – 1 + nm + mp + pn + 3mnp]
(iii) If N = an × bm × c p..., where a, b & c are prime no. Th en sum of the divisors =
( a n+1 - 1) ( bm+1 - 1) ( cp+1 - 1) ( a - 1) ( b - 1 ) ( c - 1 )
= n.
shop.ssbcrack.com
101 Shortcuts in Quantitati ve Aptitude Aptitude
15.
To find the last digit di git or digit at a t the unit’s place of an. (i)
If the last digit or digit at the unit’ unit ’s place of a is 1, 5 or 6, whatever whatever be the value of n, it will have the same digit at uni t’s place, place, i.e., (.....1 ...1))n (.....5 ...5))n n (... ....6)
(ii)
= (..... ......1 ...1)) ........5 ..5) = (..... = (... .......6)
If the last digit or digit at th e units place of a is 2, 3, 5, 7 or 8, then th e last digit of an depends upon the value of n and an d follows a repeating pat tern in terms of 4 as given below below : last digit of (....2)n
n
last digit of (....3)n
last digit digi t of (....7) (... .7)n
last digit of (....8)n
4x+1
2
3
7
8
4x+2
4
9
9
4
4x+3
8
7
3
2
4x
6
1
1
6
(iii)
If the last digit or digit at the unit’s place of a is either either 4 or 9, then the last digit of an depends upon the th e value of n and an d follows repeating pattern p attern in terms of 2 as given below. below. n
16.
S-3
(i)
last digit of (....4)n
last digit of (....9)n
2x
6
1
2x + 1
4
9
Sum of n nat natural ural numb numbeer
=
(n ) (n (n + 1) 2
(ii) (ii) Sum of n even even number = (n) (n + 1) (iii iii) Sum of of n odd number number = n2 17.
(i)
Sum of sq. sq. of firs irst n nat natur ural al no. no. =
n ( n + 1) ( 2n + 1) 6
(
)
n 4n 2 - 1 (ii) ii) Sum of of sq. sq. of of first first n odd odd natural no. no. =
(iii) iii) Sum of of sq. sq. of of first first n even natural no. =
18.
(i)
Sum Sum of of cube ube of first irst n nat natur ural al no. no. =
3 2n ( n + 1) ( 2n + 1) 3
n 2 ( n + 1) 4
2
é n ( n + 1) ù = ê ú ë 2 û
2
(ii) ii) Sum of of cube cube of of first first n even even natural no. = 2n2 (n + 1)2 (iii) iii) Sum of of cube cube of first n odd natural no. = n2 (2n2 – 1) 19.
(i)
xn – yn is divisible di visible by (x + y) When n is even
(ii) xn – yn is divisible d ivisible by (x – y) When n is either odd or even. 20.
For any any int inteeger ger n, n3 – n is divisible by 3, n5 – n is divisible by 5, n11 – n is divisible by 11, n13 – n is i s divisible by 13.
21.
Some articles related to Divisibility :
(i)
A no. of 3-digits which which is formed formed by repeating repeating a digit 3-times, 3-times, then this no. is divisible divisible by by 3 and 37. e.g., 111, 222, 333, .......
(ii) ii) A no. of 6-digit which is formed formed by by repeating repeating a digit 6-times then this no. is divisible divisible by by 3, 7, 11, 13 and 37. e.g., 111111, 222222, 333333, 444444, .............
shop.ssbcrack.com
S-4
22. 22.
23. 23.
24. 24.
25. 25.
26. 26.
27. 27.
101 Shortcuts in Quantit ati ve Aptitude Divisible by 7 : We use osculator (– 2) for divisibility test.
99995 : 9999 – 2 × 5 = 9989 9989 : 998 – 2 × 9 = 980 980 : 98 – 2 × 0 = 98 Now 98 is divisible by 7, so 99995 is also divisible by by 7. of sum of digit at even p laces and sum of digit at odd places is either 0 or multiple of Divisible by 11 : In a number, if difference of 11, then no. is divisible by 11. For example, 12342 ¸ 11 Sum of even place digit = 2 + 4 = 6 Sum of odd place digit = 1 + 3 + 2 = 6 Difference = 6 – 6 = 0 \ 12342 is divisible by 11. Divisible by 13 : We use (+ 4) as osculator. e.g., 876538 ¸ 13 876538: 8 × 4 + 3 = 35 5 × 4 + 3 + 5 = 28 8 × 4 + 2 + 6 = 40 0 × 4 + 4 + 7 = 11 1 × 4 + 1 + 8 = 13 13 is divisible d ivisible by 13. \ 876538 is also divisible by 13. Divisible by 17 : We use (– 5) as osculator. e.g., 294678: 294678: 29467 29467 – 5 × 8 = 29427 29427 27427: 2942 – 5 × 7 = 2907 2907: 290 – 5 × 7 = 255 255: 25 – 5 × 5 = 0 i s completely divisible by 17. \ 294678 is Divisible by 19 : We use (+ 2) as osculator. e.g: 149264: 4 × 2 + 6 = 14 4 × 2 + 1 + 2 = 11 1 × 2 + 1 + 9 = 12 2×2+1+4=9 9 × 2 + 1 = 19 19 is divisible by 19 di visible by 19. \ 149264 is divisible HCF (Highest Common factor) There are two methods to find the HCF– (a) Factor method (b) Division method (i) For two two no. a and b if a < b, b, then HCF of a and b is always always less than or equal to a . (ii) ii) The greatest greatest number by which which x, y and z completely completely divisible divisible is the HCF of x, y and z. (iii) iii) The greatest number by which x, y, z divisible and gives the remainder a, b and c is the HCF of (x (x –a), (y–b) (y–b) and (z–c). (iv) (iv) The greatest number by which x, y and z divisible and gives same same remainder in each case, that number is HCF HCF of (x–y), (y–z) and (z–x). (v) H.C.F .C.F.. of
28.
e a c , and f b d
=
H.C.M. of (a, c, e) L.C.M. of (b, d, f)
LCM (Least Common Multiple) There are two methods to find the LCM– (a) Factor method (b) Division method (i) For For two numbers numbers a and b if a < b, b, then L.C.M. L.C.M. of of a and b is more more than or or equal equal to b. (ii) (ii) If ratio between between two two numbers is a : b and and their H.C.F. H.C.F. is x, then their L.C.M. = abx.
shop.ssbcrack.com
x
(iii iii)
If ratio ratio betw between een two two numbers numbers is a : b and and their L.C.M. L.C.M. is x, then their H.C.F. H.C.F. =
(iv) (iv) (v)
The smallest smallest number number which which is divisible by x, y and z is L.C.M. of x, y and z. The smallest number which is divided by by x, y and z give remainder remaind er a, b and c, but (x – a) = (y – b) b) = (z – c) c) = k, then number is (L.C.M. of (x, y and z) – k). The smallest smallest number number which which is divided divided by x, y and and z give remainder remainder k in each each case, case, then number number is (L.C.M. (L.C.M. of x, y and and z) + k.
(vi) (vi)
e a c , and f b d
(vii) L.C.M .C.M.. of of
=
ab
L.C.M. of (a, c, e) H.C.F. of (b, d, f)
(vii (viii) i) For two numbers numbers a and b – LCM × HCF = a × b (ix) ix) If a is the H.C.F. of of each each pair from n numbers and L is L.C.M., L.C.M., then product of n numbers = an–1.L
29.
Algebra Identities: (i) (a + b)2 + (a – b) 2 = 2 (a 2 + b2) (iii) a3 + b3 = (a + b) (a2 – ab + b2) (v) a4 + a2 + 1 = (a2 + a + 1) (a 2 – a + 1)
(vii)
2 2 ( a + b) - ( a - b )
ab
(ii) (a + b)2 – (a – b)2 = 4ab (iv) a3 – b3 = (a – b) (a2 + ab + b2) (vi) If a + b + c = 0, then a3 + b3 + c3=3abc
=4
2 2 ( a + b) + ( a - b) =2 a 2 + b2
(viii)
k b e h k - j = ( a + d + g - j) + æç + + - ö÷ l èc f i lø
(ix)
b e h a +d +g c f i
(x)
If a + b + c = abc, abc, then then
æ 2 a ö æ 2 b ö æ 2c ö æ 2a ö æ 2 b ö æ 2c ö ç ÷+ç ÷+ç ÷ = ç ÷ .ç ÷ .ç ÷ and è 1 - a 2 ø è 1 - b2 ø è 1 - c 2 ø è 1 - a 2 ø è 1 - b 2 ø è 1 - c2 ø
æ 3a - a3 ö æ 3b - b3 ö æ 3c - c3 ö çç ÷+ç ÷ +ç ÷ 2 ÷ ç 2 ÷ ç 2 ÷ 1 3 a 1 3 b 1 3 c è ø è ø è ø 30.
=
æ 3a - a 3 ö æ 3b - b3 ö æ 3c - c3 ö ç ÷ .ç ÷ .ç ÷ è 1 - 3a 2 ø è 1 - 3b2 ø è 1 - 3c2 ø
If a1x + b1y = c1 and a2x + b2y = c2, then (i)
If
(iii)
If
a1 a2 a1 a2
¹ =
b1 b2
, one solution.
b1 b2
¹
c1 c2
(ii)
If
a1 a2
=
b1 b2
=
c1 c2
, Infinite man y solutions. solutions.
, No solution 1
1
31.
If a and b are roots of ax 2 + bx + c = 0, then
32.
If a and b are roots of ax2 + bx + c = 0, then (i) One root root is zero zero if c = 0. 0. (ii) ii) Both Both roots roots zero zero if b = 0 and c = 0. (iii) iii) Roots are reciprocal to each other, if c = a. (iv) (iv) If both both roots roots a and b are positive, then sign of a and b are opposite and sign of c and a are same. (v) If both oth root rootss a and b are negative, then sign of a, b and c are same. b
( a + b) = -
a
c
, ab =
a
a
and
b
are roots of cx2 + bx + a = 0
, then
a -b - b = ( a +b + b ) 2 - 4 ab
(
a 4 + b4 = a 2 +b + b2
)
2
2
- 2a2 b2 = é( a + b) 2 - 2ab ù - 2 ( ab ) 2 ë û
shop.ssbcrack.com
33. 33.
Arithmetic Progression: (i) If a, a + d, a + 2d, 2d, ..... are are in A.P. A.P.,, then, then, nth term term of A.P. A.P. an = a + (n – 1)d
Sum of n terms of this A.P. = Sn
n
n
= éë 2a + ( n - 1) d ùû = [ a + l] where l = last term 2 2 a = first term d = common difference
a+b
(ii) A.M. = 34. 34.
[Q A.M. = Arithmetic mean]
2
Geometric Progression: (i) G.P. ® a, ar, ar 2,......... Then, nth term of G.P. an = ar n–1
(
) ,r >1
a r n -1 Sn
=
=
( r - 1)
a (1 - r n ) (1 - r )
S¥ = 1 - r (ii) 35.
, r <1 [where r = common common ratio, a = first term]
G.M. = ab
If a, b, b, c are are in in H.P H.P., .,
1 1 1 , , are in A.P. a b c
nth term of H.M. =
th
H.M. =
1 n
term of A.P.
2ab a +b
Note : Relation between A.M., G.M. and H.M. (i) A.M. .M. × H.M H.M.. = G.M G.M..2 (ii) ii) A.M. A.M. > G.M G.M.. > H.M H.M.. A.M. ® Arithmetic Mean G.M. ® Geometric Mean H.M. ® Harmonic Mean
36.
(i)
Average rage of firs irst n nat natur ural al no. no. =
n +1 2
(ii) ii) Average Average of first first n even even no. no. = (n + 1) (iii iii) Average Average of of first first n odd odd no. = n 37.
(i)
Aver Averag agee of of sum sum of squar squaree of of firs firstt n natur natural al no. no. =
(ii) ii) Average Average of of sum of of square square of first first n even no. =
(iii) iii)
( n + 1) ( 2n + 1) 6
2 ( n + 1) ( 2n + 1)
æ 4n 2 - 1 ö Average of of sum of of square of of first odd no. = çç 3 ÷÷ è ø
3
shop.ssbcrack.com
38.
(i)
Ave Average rage of cub cubee of of fir first st n natur natural al no. no. =
n ( n + 1)
2
4
(ii) ii) Average of cube of first n even natural no. = 2n(n + 1)2 (iii) iii) Average of cube of of first first n odd natural no. = n(2n2 – 1) m ( n + 1)
39.
Avera Average ge of first first n multip multiple le of of m =
40.
(i) If average average of of some some observati observations ons is x and a is added added in each each observ observations, ations, then new average average is is (x + a). (ii) ii) If average average of some some observations observations is x and a is subtracted subtracted in each observations, then new average average is (x (x – a). (iii) iii) If average average of some some observations observations is x and each observations observations multiply by a, then n ew average average is ax.
2
(iv) (iv) If average average of some some observations observations is x and each observations observations is divided by a, then new average average is
(v) If ave average rage of n1 is A1, & average of n2 is A2, then Average of (n 1 + n2) is
42.
n1 + n 2
a
.
and
n1A1 - n 2 A 2
Average of (n 1 – n2) is 41.
n1A1 + n 2 A 2
x
n1 - n 2
When a person person is included or exclude excluded d the group, then age/weight age/weight of that person = No. of of persons persons in group × (Increase (Increase / Decrease) in average ± New average. For example : In a class average age of 15 students is 18 yrs. When the age of teacher is included their average increased by 2 yrs, then find th e age of teacher. Sol. Age of teacher = 15 × 2 + (18 + 2) = 30 + 20 = 50 yrs. When two or more than two persons persons included included or excluded excluded the group, then then average age of of included included or excluded excluded person person is
=
No. of person ´ ( Increase / Decrease ) in average average ± New average ´ ( No. of person included or excluded ) No. of included or person
weight of 13 students is 44 k kg. g. After including two new students their average weight becomes 48 kg, the n For example : Average weight find th e average weight of two new students. Sol. Average weight of two new students =
13 ´ ( 48 - 44 ) + 48 ´ 2 2
=
13 ´ 4 + 48 ´ 2 2
=
52 + 96 2
= 74 kg 2xy
43.
If a person person travels travels two two equal equal distances distances at a speed speed of x km/h and y km/h, km/h, then average speed speed =
44.
If a person person travels travels three three equal equal distances distances at a speed speed of x km/h, y km/h km/h and z km/h, then average speed speed =
45.
(i)
If
a K1
=
b
=
K2
For example: If
P
c K 3
=
=.... , then
Q
3 4 Sol. P = 3, Q = 4, R = 7 Then
(ii)
If
a1 a2
P + Q+R R
=
a2 a3
=
a3 a4
=
=
=
a + b + c + .... c
=
K1 + K 2 + K3 + ..... K 3
R P + Q + R , then find 7 R
3+ 4 + 7 7
a4 a5
= ....
=2 an a n +1
= K , then a1 : an + 1 = (K)n
x+y
km/h 3xyz xy + yz + zx
km/h.
shop.ssbcrack.com
46.
A number number added or subtracte subtracted d from from a, b, c & d, so that they are in proportion proportion = a (
ad
bc
d)
(b
c)
For example : When a number should be subtracted from 2, 3, 1 & 5 so that they are in proportion. Find that number. 2 ´ 5 - 3 ´1 Sol. Req. No. = 2 + 5 - 3 + 1 ( ) ( )
47. 47.
=
10 - 3 7-4
=
7 3
If X part of A is equal equal to Y part of B, then A : B = Y : X. For example: If 20% of A = 30% of B, then fin d A : B. 30%
Sol. A : B =
48.
20%
3
= = 3 : 2 2
When Xth part of P, Y th part of Q and Zth part of R are equal, then find A : B : C. Then, A : B : C = yz : zx : xy
49.
A can can do do a/b a/b part part of work ork in t1 days and c/d part of work in t2 days, then
50.
(i)
t1 a/b
=
t2 c/d
If A is is K time times ef effici ficient ent than B, Then Then T(K T(K + 1) = Kt KtB
(ii) ii) If A is K times times effici efficient ent than B and takes t days days less less than B Then T = 51. 51.
(i) (i)
Kt K
2
-1
or or
t K -1
, tB
=
t K -1
= ktA
If a cistern cistern takes X min to be filled filled by by a pipe pipe but but due to to a leak, it takes takes Y extra minutes minutes to be filled, filled, then the time taken taken by leak
æ X 2 + XY ö min ç Y ÷÷ è ø
to empty the cistern = ç
(ii) ii) If a leak empty a cistern cistern in X hours. A pipe which admits Y litres per hour water water into the cistern cistern and now cistern is emptied emptied in Z hours, th en capacity of cistern is =
æ X+Y+Zö ç Z - X ÷ litres. è ø
(iii) (iii) If two pipes A and B fill a cistern cistern in x hours h ours and y hours. A pipe is also an outlet C. If all the th e three pipes are opened opened together,
é ù xyT ú ë yT + xT - xy û
the tank full in T hours. Th en the time taken by C to empty the full tank is = ê 52.
(i)
from A to B is If t1 and t2 time taken t o travel from A to B and B to A, with speed a km/h and b km/h, then distan ce from
æ ab ö ÷ èa+bø
æ ab ö ÷ èa-bø
d = ( t1 + t 2 ) ç
d = ( t1 - t 2 ) ç
æ t1t 2 ö ÷ è t1 - t 2 ø
d = ( a - b) ç
(ii) ii) If Ist part of of distance is covered covered at the speed speed of of a in t1 time and the th e second part is covered at the speed of b in t2 time, then the average speed =
æ at 2 + bt1 ö çè t + t ÷ø 1 2
shop.ssbcrack.com
53.
Simple Fraction
Their Percentage
1
Simple Fraction
100%
1 8
1
1 9
1
11.11%
33.3%
3
1
1
25% 25%
4
10
1
1
20% 20%
5
11 1
1
16.67%
6 1
12
10% 10%
9.09%
8.33%
14.28%
7
(i)
12.5%
50% 50%
2
54.
Their Percentage
æ è
aö æ a %ö = ÷ more than B, then B is ç ÷ less than A. b ø èa+b ø
æ è
a a ö % ÷ more than A = ö÷ less than B, then B is æç b ø è a -b ø
If A is ç x%
(ii) If A is ç x%
if a > b, we take a – b if b > a, we take b – a. 55.
æ b - a ´100 ö % If pric pricee of a article article increase increase from from ` a a to ` b, b, then its expenses decrease by ç ÷ so that expendit ure will be same. è b ø
56.
Due to increase increase/decre /decrease ase the price price x%, A man purchase purchase a kg more in ` y, y, then Per kg increase or decrease =
Per kg starting price = `
æ xy ö ç 100 ´ a ÷ è ø
xy
( 100 ± x) a
shop.ssbcrack.com
S-10
57.
101 Shortcuts in Quantit ati ve Aptitude
For For two two artic articles, les, if price price:: Ist
IInd
Overall
Increase (x%)
Increase (y%)
Increase
Increase (x%)
Decrease (y%)
æ x - y - xy ö % ç ÷ 100 ø è
æ x + y + xy ö % ç ÷ 100 ø è
If +ve (Increase) If –ve (Decrease) Decrease (x%)
æ è
Decrease ç x + y -
Decrease (y%)
Increase (x%)
Decrease (x%)
Decrease
ö% ÷ 100 ø xy
æ x2 ö çç ÷÷ % 100 è ø
58.
æ x2 ö If the side side of a square square or radius radius of a circle is x% increase/dec increase/decrease rease,, then its area increase/dec increase/decrease rease = çç 2x ± 100 ÷÷ % è ø
59.
If the side side of a square, square, x% increase/de increase/decreas creasee then x% its perimete perimeterr and diagonal increase increase/decre /decrease. ase.
60.
(i)
æ 100 ± R ö è 100 ÷ø
t
If popula populatio tion n P increas increase/d e/dec ecrea rease se at r% rate rate,, then after after t year yearss popula populatio tion n = Pç
(ii) ii) If population population P increase increase/decre /decrease ase r 1% first year, r 2% increase/decrease second year and r 3% increase/decrease third year,
æ è
then after 3 years population = P ç1 ±
r ö r ö æ öæ 1 ± 2 ÷ ç1 ± 3 ÷ ÷ ç 100 ø è 100 ø è 100 ø r1
If increase we use (+), if decrease we use (–) 61. 61.
If a man spend x% of this income on on food, y% y% of remaining on rent and z% of of remaining remaining on cloths. If he has ` P P remaining, then total income of man is =
P ´ 100 ´ 100 ´100
(100 - x ) (100 - y ) (100 - z )
[Note: We can use this table for area increase/decrease increase/decrease in mensuration for rectangle, triangle an d parallelogram].
62.
If CP of x things thin gs = SP of y things, then Profit/Loss =
63.
éx - y ù ê y ´ 100ú % ë û
If +ve, Profit; If –ve, Loss If afte afterr selli elling ng x thin gs P/L is equal to SP of y things, then P/L =
y
( x ± y)
é Profit = - ù ê Loss = + ú ë û
´ 100
shop.ssbcrack.com
101 Shortcuts in Quantitative Aptitude
64.
65.
S-11
If CP of of two two articles articles are same, same, and they sold sold at Ist
IInd
Overall
(x%) Profit
(y%) Profit
æ x + y ö% ç 2 ÷ Profit è ø
(x%) Profit
(y%) Loss
æ x - y ö % ìProfit, ifif x > y ç 2 ÷ íLoss, if x < y è ø î
(x%) Loss
(y%) Loss
æ x + y ö % Loss oss ç 2 ÷ è ø
(x%) Profit
(y%) Loss
No profit, no loss
If SP of of two two articles articles are same same and they sold at Ist
IInd
Overall
Profit (x%)
Loss(x%)
æ x2 ö Loss ç % ç 100 ÷÷ è ø
Profit (x%)
Loss (y%)
æ 100 ( x - y ) - 2xy ö then Pr of ofit % é 2 10 ù ì If + ve, th 100 + x ) ( 100 - y ) ç ÷ % or ê ( 100 ú % í 200 + x - y t hen Loss% è 200 + x - y ø ë û î If - ve, th
é P + D ´100ù % úû ë100 - D
66.
After After D% D% discount, discount, require requiress P% profit, profit, then total total increase increase in C.P.= C.P.= ê
67.
M.P. .P. = C.P C.P ×
68.
Profit % =
69.
(i)
( 100 + P ) ( 100 - D )
( M.P. - C.P.) ´ 100 C.P.
For discount r1 % and r 2%, successive discount
éæ 100 + r1 ö æ 100 + r2 ö æ 100 + r3 ö ù = êç ÷ç ÷ç ÷ - 1ú ´ 100 ëè 100 ø è 100 ø è 100 ø û
(ii (ii)) For For disco discount unt r 1%, r 2% and r 3%, successive discount
If
P = Principal,
R = Rate per annum,
T = Time in in ye years, rs,
SI = Si Simple inte nterest,
A = Amount 70.
PRT
(i)
SI =
(ii)
A = P + SI = P ê1 +
100
é ë
RT ù
100 úû
éæ 100 + r1 ö æ 100 + r2 ö æ 100 + r3 ö ù = êç ÷ø çè 100 ÷ø èç 100 ÷ø - 1ú ´ 100 è 1 0 0 ë û
shop.ssbcrack.com
71.
If P = Pri Princi ncipa pal, l, A = Amou Amount nt in n years, R = rate of interest per ann um.
é ë
A = P ê1 +
72.
(i)
R ù
100 úû
n
, interest payable annually
n¢ R ¢ ù é A = P ê1 + , interest payable half-yearly ë 100 úû R¢ = R/2, n¢ = 2n 4n
R ù é (ii) A = P ê1 + , interest payable quarterly; ë 400 úû
73.
74.
(i)
é1 + R ù êë 400 úû is the yearly growth factor;
(ii)
é1 – R ù th e yearly decay decay factor or depreciation factor.. êë 400 úû is the 3
When When time is fractio fraction n of a yea year, r, say say 4
4
, years, then,
é 3 R ù ê 4 ú é Amount = P ê1 + ´ ê1 + ú ë 100 ûú ë 100 û R ù
4
75.
L O R I CI = Amoun mountt – Princ Principa ipall = P MF G1 + J - 1P
76.
When Rates Rates are differe different nt for for differe different nt years, years, say R 1, R 2, R 3% for 1st, 2nd & 3rd years respectively, respectively, then, th en,
77.
Amount
n
MNH
100 K
PQ
é R ù é R ù é R ù = P ê1 + 1 ú ê1 + 2 ú ê1 + 3 ú ë 100 û ë 100 û ë 100 û
In general, interest is consider ed to be SIMPLE SIMPLE unless otherwise stated.
78.
(i)
Sum Sum of of all all the the ext exter erio iorr angl anglee of of a poly polygo gon n = 360 360°°
(ii) ii) Each exterio exteriorr angle of of a regular regular polygon polygon =
360° n
(iii iii) Sum of of all the interior interior angles of a poly polygon gon = (n – 2) × 180° (iv) (iv) Each interior angle of a regular regular polygon polygon =
(v) (v) No. of diagonals of a polygo polygon n=
79. 79.
n ( n - 3) 2
( n - 2) n
´ 180°
, n ® no. of sides.
(vi) (vi) The ratio of sides a polygon polygon to the diagonals of a polygon polygon is 2 : (n – 3) (vii) (vii) Ratio of interior angle an gle to exterior exterior angle of a regular polygon polygon is (n – 2) : 2 Properties of triangle: (i) When one one side is extended extended in any direction, direction, an angle is formed formed with with another side. side. This is called called the exterior exterior angle. angle. There are six exterior angles of a triangle. (ii) ii) Interior Interior angle + corre correspo sponding nding exterio exteriorr angle = 180°. 180°.
shop.ssbcrack.com
(iii) ii)
An exterio exteriorr angle = Sum of the other other two two interior interior oppos opposite ite angles angles..
(iv) (iv) (v) (v)
Sum of of the lengths of of any any two two sides sides is greater greater than the length length of third side. Diffe Differenc rencee of of any any two two sides sides is less less than than the third third side. side.
(vi) (vi) (vii (vii))
Side opposite opposite to the greatest angle is greatest and vice versa. A triangl trianglee must must have at at least least two acute acute angle angles. s. Triangles on on equal bases bases and betwee between n the same parallels parallels have equal areas. areas.
(vii (viii) i) If a, b, c denote the sides of a triangle then (i) if c2 < a2 + b2, Triangle is acute angled. (ii) if c2 = a2 + b2, Triangle is right angled. (iii) if c2 > a2 + b2, Triangle is obtuse angled. (ix) If 2 triangles triangles are equiangul equiangular, ar, their correspo corresponding nding sides are are proportio proportional. nal. In triangles ABC and XYZ, XYZ, if
ÐA = ÐX, ÐB = ÐY, ÐC = ÐZ, then AB XY
•
(i)
=
AC XZ
=
BC YZ
A
..
In DABC, ÐB = 90° BD ^ AC
\ BD × AC = AB × BC 1
=
1
1
+
(ii)
BD 2
(iii)
BD2 = AD × DC
(x)
AB2
D
BC 2
C B The perpendic perpendiculars ulars drawn drawn from from vertice verticess to oppos opposite ite sides sides (calle (called d altitudes) altitudes) meet meet at a point called calledOrthocentre of the triangle.
(xi)
The line drawn drawn from a vertex of a triangle to the opposite opposite side such that it bisects bisects the side is called theMedian of the triangle. triang le. A median bisects the area of the tri angle. (xii) When a vertex vertex of a triangle is joined joined to the midpoint of the opposite opposite side, side, we get a median. The point of intersectio intersection n of the medians is called the Centroid of the triangle. tri angle. The centroid divides any median in the rati o 2 : 1. (xiii) (xiii) Angle Angle Bisector Bisector Theore Theorem– m– In the figure if AD is the an gle bisector (interior) of Ð BAC. Then, A
B
1. 2.
D
C
AB/AC = BD/DC. AB x AC – BD x DC = AD2 .
(xiv (xiv)) Midpo Midpoint int Theore Theorem m– (xv)
In a trian gle, the line joining th e mid points of two two sides is parallel to the third side and half of it. Basic Basic Propo Proportio rtional nality ity Theore Theorem m A line parall p arallel el to any one side of a triangle tria ngle divides divi des the other two sides proportionally. If DE DE is parallel to BC, then A D
E
B
AD BD
C
=
AE AB AB EC¢ AD AD
=
AC AD , AE DE
=
AB BC
and so on.
shop.ssbcrack.com
80. 80.
Properties of circle –
(i)
Only Only one circl circlee can can pass pass through through three three giv given en poi points. nts.
(ii) ii)
There is one and and only one tangent tangent to the circle circle passing passing through through any point point on the circle circle..
(iii iii) (iv) (iv)
From any exterior exterior point point of the circle, two tangents can can be drawn on to the circle. circle. The lengths lengths of two two tangents tangents segment segment from from the exterio exteriorr point point to the circle, circle, are equal. equal.
(v) (v)
The tangent tangent at any point of of a circle circle and the the radius radius through through the point point are perpe perpendicu ndicular lar to each each other other..
(vi) (vi)
When two two circles circles touch touch each other, other, their centres & the point of of contact contact are collinear collinear..
(vii (vii)) If two two circles touch externally externally,, distance betwee between n centres = sum of radii. (vii (viii) i) If two circles touch touch i nternally, nternally, distance between between centres = difference difference of radii (ix)
Circles Circles with with same same centre centre and diffe different rent radii radii are concentric circles.
(x)
Point Pointss lying lying on on the sam samee circ circle le are are call called ed concyclic points.
(xi)
Measure Measure of an arc means means measur measuree of of central central angle. m(minor arc) + m(major arc) = 360°.
(xii)
Angle Angle in a semic semicirc ircle le is a right angle angle..
(xiii) Only one one circle can pass through three given (xxv) If ON is ^ from the th e centre O of a circle to a ch ord AB, then AN = NB.
O
A
B
N
( from centre bisects chord)
(xv (xv)
If N is the midpoint midpoint of a chord chord AB of of a circle with centre O, then ÐONA = 90°. (Converse,
from centre bisects chord)
(xvi (xvi)) Two congruent figures have equal areas but the converse converse need not be true. (xvi (xvii) i) A diagonal of a parallelogram divides it into two triangles of of equal area. (xvi (xviii ii)) Parallelograms on the same base and between between the same parallels ar e equal in area.
•
(xix)
Triangles on the same same bases bases and betwee between n the same parallels are equal in area.
(xx)
If a triangle and a parallelogram parallelogram are on the same same base and between between the same parallels, then the area of the triangle is equal to the half of the parallelogram.
If PT is is a tang tangeent to the the cir circcle, le, the then n OP OP2 = PT2 = OT2 T
P
•
O
If PT is tang tangeent and and PA PAB is is sec secant ant of a circ circle le,, then then PT PT2 = PA.PB T
P
O A B
101 Shortcuts in Quantitati ve Aptitude Aptitude
•
S-15
If PB & PD are two secant secant of of a circle circle,, then PA.PB PA.PB = PC.PD PC.PD B
A P
•
D
C
If two two circ circles les touch touch exte externally rnally,, then distance distance betw betwee een n their their centre centress = (r (r 1 + r 2)
B
A r 1
•
r 2
If two two circles circles touch touch interna internally lly,, then then distanc distancee betw betwee een n their their centre centress = r 1 – r 2 where r 1 > r 2.
A
81.
(i)
Area of of triangle =
1 2
× base × altitude
(ii) ii) Area Area of triangle triangle using using heron’ heron’ss form formula ula = 82.
S / S - a (S (S - b) (S - c) , where S =
a+b+c 2
In an an equilat equilateeral triang triangle le with with side side a, then 4A 3
83.
B
=
4h
2
3
=
P
2
9
where A ® Area of triangle
=a
2
P ® Perimeter h ® Height
In an isosc isoscel elees triang triangle le PQR PQR ar D PQR =
Height =
b 4
4a
2
P
- b2 a
a
4a 2 - b 2 2 Q
b
R
84.
1 (i) Are Area of D = bc SinP where ÐP = ÐQPR 2 (ii) Area rea of of D = (iii) ii) Area Area of D =
85. 85.
CosP =
b 2
2 1 2
ab SinR SinR
+ c2 - a 2
a2
Sine Rule :
b
c
ac SinQ SinQ
2bc
CosR =
86. 86.
1
P
Q
, CosQ =
a 2 + c2 - b 2 2ac
a
R
,
+ b2 - c2 2ab
a SinP
=
b SinQ
=
c SinR Square
Area rea of squ square are
87. 87.
88.
=
Perim Perimet eter er of squa square re 4
=
Diag Diagon onal al of squar squaree 4
In a circ circle le wit with h radiu radiuss r. r. A C
=
D 4
where A - Area of circle
r O
C - Circumference of circle D - Diameter of circle 89.
= sid side of squa square re
If q = 60°, ar D AOB =
3 2 r 4
If If q = 90°, ar D AOB =
1 2 r 2
1 2 If q , ar D AOB = r 2
r
q
r B
A
æ q ö . cos æ q ö ÷ ç2÷ è 2ø è ø
2 sinq = r sin ç
90.
(i)
A circle circle with with large largest st area area insc inscrib ribed ed in a right right angle angle triangle triangle,, then then r =
A
r B
C
2 ´ area of of
DABC . Perimeter of DABC
(ii) ii) If ABC ABC is an equilateral equilateral triangle with with side a, then Area Area of of circle circle =
pa 2 12
A
r C
B (iii iii) If ABC ABC is an equilateral equilateral triangle with with side a, then area of circle circle =
pa 2 3
.
r
(iv) If DABC is an equilateral triang le, and two circles with with radius r adius r and R, th en
r R
pr 2 1 = = and 2 pR 2 4 1
A
r R B
C
(
)
(v) (v) Three equal equal circle circle with with radius r and an equilateral equilateral triangle ABC, ABC, then area of shaded region region = 2 3 - p .
A
C
B
91.
ABCD ABCD is a square placed placed inside a circle with side side a and radius of circle r, then A
a
B r a
a
D
a
C
area of square area of circle
=
7 11
2
r
2
S-18
101 Shortcuts in Quantit ati ve Aptitude
92.
Diago Diagonal nal of a cub cubee =
93.
Diago Diagonal nal of a cub cuboi oid d = l 2 + b2
94.
For tw two cube ubes A1 A2
95. 95. •
96.
=3
v1 v2
=
a1 a2
3 × side
=
d1
d2
+ h 2 ; where l ® Length, b ® breadth, h ® height
where A1, A2 ® Area of cubes v1, v2 ® Volume a1, a2 ® Sides d 1, d 2 ® Diagonals Units of Measurement Mea surement of Area and Volume The inter-relations inter-relationships hips betwe between en various various units units of measureme measurement nt of length, area and and volume volume are listed listed below below for for ready reference reference:: Length 1 Centimetre (cm) = 10 milimetre (mm) 1 Decimetre (dm) = 10 centimetre 1 Metre (m) = 10 dm = 100 cm = 1000 mm 1 Decametre (dam) = 10 m = 1000 cm 1 Hectometre (hm) = 10 dam = 100 m 1 Kilometre (km) = 1000 m = 100 dam = 10 hm 1 Myriametre = 10 kilometre Area 1 cm cm2 = 1 cm × 1 cm = 10 mm × 10 mm = 100 mm2 1 dm2 = 1 dm × 1 dm = 10 cm × 10 cm = 100 cm2 1 m2 = 1 m × 1 m = 10 dm × 10 dm = 100 dm2 1 dam2 or or 1 are = 1 dam × 1dam = 10 m × 10 m = 100 m2 1 hm2 = 1 h ectare = 1 hm × 1 hm = 100 m × 10000m2 = 100 dm2 1 km2 = 1 km × 1 km = 10 hm × 10 hm = 100 hm2 or 100 hectare Volume 1 cm3 = 1 ml = 1 cm × 1 cm × 1 cm = 10 mm × 10 10 mm × 10 mm = 1000 mm3 1 litre = 1000 ml = 1000 cm3 1 m3 = 1 m×1 m×1m = 100 cm × 10 100 cm × 100 cm cm = 106 cm3 = 1000 1000 litr litree = 1 kil kilo ometre etre 3 3 3 3 3 1 dm = 1000 cm , 1m = 1000 dm , 1 km = 109 m3 If a, b, c are the edges of a cuboid, then The The lo longest ngest diag diagon onal al
=
a 2 + b 2 + c2
(i) If the the height of of a cubo cuboid id is zero zero it beco becomes mes a rect rectangle angle.. (ii) ii) If “a” “a” be the edge edge of of a cube cube,, then (iii) ii) The longes longestt diagonal diagonal = aÖ3 97.
Volume lume of of pyra pyram mid =
1 ´ Base Area ´ height (H) 3 2
98.
(i)
A 1 F l1 I = G J If A1 & A2 denote the areas of two similar similar figures an d l1 & l2 denote their corresponding linear measures, then A 2 H l2 K
(ii)
V1 F l1 I = G J If V1 & V2 denote the volumes of two two similar solids and l1, l2 denote their corresponding linear measures, then V2 H l2 K
3
(iii iii) The rise or or fall of of liquid level level in a container =
Total volume of objects submerged or taken out Cross sectional area of container
101 Shortcuts in Quantitati ve Aptitude Aptitude
99.
S-19
If a largest possibl possiblee cube cube is inscribed inscribed in a sphere of radius ‘a’ cm, then (i)
the edge edge of the cube cube =
2a 3
.
(ii) ii) If a largest largest possible possible sphere sphere is inscribed inscribed in a cylinder cylinder of radius radius ‘a’ cm and height height ‘h’ cm, then for h > a, •
the the ra radiu dius of the sphe phere = a and and
•
the radius =
h (for a > h) 2
(iii iii) If a largest possible possible sphere is inscribed inscribed in a cone of of radius ‘a’ cm and slant height equal to the diameter of the base, base, then •
the rad radius ius of the sphere here =
a 3
.
(iv) (iv) If a largest largest possible possible cone cone is inscribed in a cylinder cylinder of radius ‘a’ cm and height ‘h’ cm, then the radius of the cone cone = a and height = h.
(v) (v) If a largest largest possible possible cube cube is inscribed inscribed in a hemisphere hemisphere of radius ‘a’ cm, cm, then the edge of of the cube cube = a
2 . 3
S-20
101 Shortcuts in Quantit ati ve Aptitude
100. In any quadril quadrilate ateral ral (i) (i)
Area =
1 1 ´ one diagonal × (sum of perpendiculars to it from opposite vertices) = × d (d 1 + d 2) 2 2
(ii) ii) Area Area of a cycli cyclicc quadrila quadrilateral teral = bs - agbs - bgbs - cgbs - dg where a, b, c, d are sides of quadrilatera l and s = semi perimeter =
a+b+c+d 2
101. 101. If length, length, breadth & height of a three dimensional dimensional figure increase/decrease by x%, y% and z%, then Change Chan ge in area =
éæ 100 ± x ö æ 100 ± y ö ù êç 100 ÷ ç 100 ÷ - 1ú ´ 100% øè ø û ëè
Change Chang e in Volume Volume =
éæ 100 ± x ö æ 100 ± y ö æ 100 ± z ö ù êç 100 ÷ ç 100 ÷ ç 100 ÷ - 1ú ´ 100% øè øè ø û ëè
