ANSWER KEY
QUE
NO.
1
2
3
4
5
6
7
8
Conventional Mathematical Techniques
5X + 4 = 19
100 – X = 45
2,3,7 & 8
1,4,5,6,9 & 0
8,000
Vedic Mathematical Techniques
ANSWER
5X + 4 = 19
100 – X = 45
2,3,7 & 8
1,4,5,6,9 & 0
8,000
83 X 57 = 83
x 57
4150
+ 581
4731
2515 ÷ 99
2 5
99 2 5 1 5
AhI,waJy = 2515
198
Waajk = 99
535
Waagk5 = 25
495
xe8 = 40
40
(5xy2 – 3)( –4xy)
= (5xy2 – 3) × (–4xy)
= 5xy2 × (–4xy) – 3× (–4xy)
= 5 × (−4) x2y3 + 12xy
= (−
−20) x2y3 + 12xy
(5xy2 – 3) × ( – 4xy ) = (–20)x2y3 + 12xy.
83 X 57 = 83
x 57
4731
2515 ÷ 99
1
1
1
1
1
TOTAL
MARKS
1
1
1
1
1
2
2
2
2
2
2
AhI,waJy = 2515
Waajk = 99
Waagk5 = 25
xe8 = 40
99 2 5 1 5
01
0 2
05
(1004I
A>tr)
25 40
(5xy2 – 3)( –4xy)
4xy
5xy
20x2y3
3
12xy
2
(5xy – 3)( –4xy) = (– 20)x2y3 + 12xy.
2
301
MARKS
9
10
11
8x + 5 = 61
∴ 8x = 61 – 5
∴ 8x = 56
∴ x = 56
8
∴ x= 7
mud\l = ½a.
½a 1,000
mud\t = 1 v8R
Vyajno dr = 10 %
½a.
½a. 10
½a 100 nu> 1 v8Rnu> Vyaj = ½a
½a.
½a 1,000 nu> 1 v8Rnu> Vyaj = ( ? )
∴ Vyaj = 1,000 x 10 x 1
100 x 1
Vyaj = 100 ½a.
½a
mud\l = ½a.
½a 4,000
mud\t = 3 v8R
Vyajno dr = 12 %
½a.
½a. 12
½a 100 nu> 1 v8Rnu> Vyaj = ½a
½a.
½a 4,000 nu> 3 v8Rnu> Vyaj = ( ? )
∴ Vyaj = 4,000 x 12 x 3
100 x 1
Vyaj = 40 x 12 x 3
Vyaj = 1440 ½a.
½a
Vyajmud\l = mudl
\ + Vyaj
= 4,000 + 1,440
Vyajmud\l = 5,440 ½a.
½a
8x + 5 = 61
∴ 8x = 61 – 5
∴ 8x = 56
∴ x = 56
8
∴ x= 7
P = ½a.
½a 1,000
R = 10 %
1
1
2
1
N = 1 v8R
1
a
2
P = ½a.
½a 4,000
R = 12 %
N = 3 v8R
a
1
a
1
AHI>,
302
2
12
13
14
2aro ke 7eya pase x ½ipya 0e.
quxI pase 7eya krta car g`a ½ipya 0e.
te4I quxI pase 4x ½ipya 4ay.
b>neno srva5o 110 ½ipya 0e.
smIkr` Sv½p
Sv½p :
∴ x + 4x = 110
∴ 5x = 110
∴ x = 110
5
∴ x = 22
∴ 7eyane x = 22 ½ipya m5e.Ane
quxIne 4x = 4 x 22 = 88 ½a.m5e.
(67)2 = 67 x 67
67
x 67
4020
+ 469
4489
2
(67) = 4489.
1 il3r = 1,000 imlIil3r
8 il3r = ( ? ) imlIil3r
= 8 x 1,000 imlIil3r
= 8,000 imlIil3r
4elInI s>Qya = 8,000 imlIil3r
200 imlIil3r
= 40 4elIAo wrI xkay.
2aro ke 7eya pase x ½ipya 0e.
te4I quxI pase 4x ½ipya 4ay.
b>neno srva5o 110 ½ipya 0e.
∴ x + 4x = 110
∴ 5x = 110
∴ x = 110
5
∴ x = 22
∴ quxI pase 4x = 4 x 22 = 88 ½a.m5e.
1
1
2
(67)2
(67)2 =
=
=
2
(67) =
D(6) / D(67) / D(7)
368449
4489.
4489.
8 il3r = 8,000 imlIil3r
1
1
2
1
4elInI s>Qya = 8,000 imlIil3r
200 imlIil3r
= 40 4elIAo wrI xkay.
1
303
2
15
7y2 + 3y - 4 ma>4I y2 - 5y + 8 bad kro.
Jvab ::- (7y2 + 3y - 4) – (y2 - 5y + 8)
= 7y2 + 3y – 4 – y2 + 5y – 8
= 7y2 – y2 + 3y + 5y – 4 – 8
= 6y2 + 8y – 12
A4va
2
(5x + 4y) nu> sadu>½p Aapo.
jvab ::- (5x + 4y)2 = (5x)2 + 2 (5x) (4y) + (4y)2
= 25x2 + 40xy + 16y2.
16
(9m2 – 7)(2m + 1) nu> sadu>½p Aapo.
Jvab ::- (9m2 – 7)(2m + 1)
= 9m2 (2m + 1) – 7 (2m + 1)
= 18m3 + 9m2 – 14m – 7
(9m2 – 7)(2m + 1) = 18m3 + 9m2 – 14m – 7
A4va
Jvab ::(x – 7 + 4x2) + (8 – 2x2 + 4x) + (7x – 5 +3x2)
= x – 7 + 4x2 + 8 – 2x2 + 4x + 7x – 5 + 3x2
= 4x2 – 2x2 + 3x2 + x + 4x + 7x – 7 + 8 – 5
= 5x2 + 12x – 4.
y1
y0
y2
p/4m pdavlI
7
3
4
bI+ pdavlI
1
5
8
6
8
12
(7y2 + 3y – 4 ) – ( y2 – 5y + 8 ) = 6y2 + 8y – 12.
A4va
(5x + 4y)2
jvab ::- (5x + 4y)2 = D(5x) / D(5x)(4y) / D(4y)
= 25x2 + 40xy + 16y2.
9m2
7
2m
18m3
14m
1
9m2
7
(9m2 – 7)(2m + 1) = 18m3 + 9m2 – 14m – 7.
A4va
x2
X
x0
p/4m pdavlI
4
1
7
bI+ pdavlI
2
4
8
+I+
I+ pdavlI
3
7
5
5
12
4
2
2
(x – 7 + 4x ) + (8 – 2x + 4x) + (7x – 5 + 3x2)
= 5x2 + 12x – 4.
304
1
1
2
A4va
A4va
1
1
2
2
1
3
A4va
2
1
3
17
18
2aro ke phelI s>Qya = x
bI+ k/imk s>Qya = x + 1 Ane
+I+
I+ k/imk s>Qya = x + 2 4ay.
+`
` k/imk s>Qyano srva5o 87 4ay 0e.
te4I smIkr` Sv½p
Sv½p :
∴x + x + 1 + x + 2 = 87
∴ 3x + 3 = 87
∴ 3x = 87 – 3
∴ 3x = 84
∴ x = 84
3
∴ x = 28
∴ phelI s>Qya = x = 28 4ay.
l>b6n 3a>kInI l>ba[ = 50 semI.
l>b6n 3a>kInI pho5a[ = 40 semI.
l>b6n 3a>kInI }>ca[ = 90 semI.
l>b6n 3a>kInu> 6nf5 = l> x p. x }>
= lxbxh
= 50 x 40 x 90
= 1,80,000 6n semI
l>b6n 3a>kInu> 6nf5 = 1,80,000 6n semI
1,000 6n semI = 1 il3r
1,80,000 6n semI = ( ? ) il3r
= 1,80,000 il3r
1,000
= 180 il3r
pa`InI
pa`InI 3a>kIma> 180 il3r pa`I smay.
2aro ke phelI s>Qya = x
bI+ k/imk s>Qya = x + 1 Ane
+I+
I+ k/imk s>Qya = x + 2 4ay.
4ay.
smIkr` Sv½p
Sv½p : x + x + 1 + x + 2 = 87
∴ 3x + 3 = 87
∴ 3x = 87 – 3
∴ 3x = 84
∴ x = 84
3
∴ x = 28
∴ phelI s>Qya = x = 28 4ay.
l = 50 semI.
b = 40 semI l>b6n 3a>kInu> 6nf5 =
h = 90 semI.
=
=
( 1000 6n semI.=
=
I. 1 lI3r)
lI3r
lbh
50 x 40 x 90
1,80,000 6n semI
1,80,000 il3r
1,000
= 180 lI3r
1
1
1
3
1
1
1
3
305
19
2 7056
2 3528
2 1764
2 882
3 441
3 147
7
49
7
7
1
7056 =
7056 = 2x2x2x2x3x3x7x7
= 22 x 22 x 32 x 72
= ( 2 x 2 x 3 x 7)2
= ( 84)2
√7056 = √(84)2 = 84
70 / 56
Aekmno A>k
4 ke 6
70 4I nanI
s>Qya
84 ke 86
pU`RvgR s>Qya
= 64 = 82
84
852 = 7225 4I AapelI s>Qya nanI 0e.
86
√7056 = 84
1
1
1
20
mud\l = ½a.
½a 5,110
mud\t = 200 idvs
Vyajno
Vyajno dr = 10 %
½a.
½a 100 nu> 365 idvsnu> Vyaj = ½a.
½a 10
½a.
½a 5,110 nu> 200 idvsnu> Vyaj = ( ? )
∴ Vyaj = 5,110 x 200 x 10
100 x 365
Vyaj = 280 ½a.
½a
Vyajmud\l = mudl
\ + Vyaj
= 5,110 + 280
Vyajmud\l = 5,390 ½a.
½a
3
P = ½a.
½a 5,110
1
R = 10 %
N = 200 idvs
a
AHI>,
1
1
306
3
21
(1)
77 x 73 =
=
=
77 x 73 =
(75 + 2) (75 – 2)
(75)2 – (2)2
5625 – 4
5621.
(2) (8xy – 5p)2 = (8xy)2 – 2 (8xy) (5p) + (5p)2
= 64x2y2 – 80pxy + 25p2
77 x 73 = 77 / +7
77 / +3
80 / 21
x 7
560 / 21
23 x 27 = 5621.
(2) (8xy – 5p)2
jvab ::- (8xy – 5p)2 = D(8xy) / D(8xy)(5p) / D(5p)
= 64x2y2 – 80pxy + 25p2
A4va
A4va
Jvab ::(9x2y3+ 3xy2 – 2x2y + 4) – (7 – 4xy2 – 8x2y +
3xy3– 4x2y3) + (7x2y – 4xy2 + 5xy – 4)
= 9x2y3+ 3xy2 – 2x2y + 4 – 7 + 4xy2 + 8x2y
– 3xy3 + 4x2y3 + 7x2y – 4xy2 + 5xy – 4
= 9x2y3 +4x2y3 – 3xy3 + 4xy2 – 4xy2 – 2x2y
+ 8x2y + 7x2y + 5xy + 4 – 7 – 4
= 13x2y3 – 3xy3 + 3xy2 + 13x2y + 5xy – 7.
22
Sam6n qaDanI l>ba[ = 8 mI3r = 800 semI.
Sam6n qaDanu> 6nf5 = ( l )3
= ( 800 )3
= 800 x 800 x 800
= 51,20,00,000 6n semI
Sam6n qaDanu> 6nf5 = 51,20,00,000 6n semI
l>b6n [>>3nI l>ba[ = 32 semI
(1)
x2y3
p/4m pdavlI
9
bI+ pdavlI
4
+I+
I+ pdavlI
0
13
x2 y
2
8
7
13
xy2 x y3
3
0
4
3
4
0
3
3
xy
0
0
5
5
x0y0
4
7
4
7
(9x2y3+ 3xy2 – 2x2y + 4) – (7 – 4xy2 – 8x2y +
3xy3– 4x2y3) + (7x2y – 4xy2 + 5xy – 4)
= 13x2y3 – 3xy3 + 3xy2 + 13x2y + 5xy – 7.
[>3onI s>Qya = sm6n qaDanu> 6nf5
l>b6n [>3nu> 6nf5
= 800 x 800 x 800 6n semI
32 x 20 x 10 6n semI
= 800 x 100
= 80,000 [>3o bnavI xkay.
307
1
1
2
1
1
2
A4va
A4va
1
2
1
1
1
4
23
l>b6n [>>3nI pho5a[ = 20 semI
l>b6n [>3> nI }>ca[ = 10 semI
l>b6n [>>3nu> 6nf5 = l x b x h
= 32 x 20 x 10
= 6,400 6n semI
l>b6n [>>3nu> 6nf5 = 6,400 6n semI
[>3onI s>Qya = sm6n qaDanu> 6nf5
l>b6n [>3nu> 6nf5
= 800 x 800 x 800 6n semI
32 x 20 x 10 6n semI
= 25 x 40 x 80
1,250
= 80,000 [>3o bnavI xkay.
3364
676
3364 = 2 x 2 x 29 x 29
2 3364
= 22 x 292
= (2 x 29)2
2 1682
29 841
= (58)2
29
29
1
2 676
676 = 2 x 2 x 13 x 13
= 22 x 132
2 338
13 169
= (2 x 13)2
13 13
= (26)2
1
√3364
676
2
= √ (58)
(26)2
= 58
26
1
1
4
3364
676
33 /6 4
2 ke 8
52 ke 58
6/76
Aekmno A>k
s>Qya
33 4I nanI pU`RvgR
s>Qya = 25 = 52
52
552 = 3364 4I AapelI
58
s>Qya mo3I 0e.
√3364
= √ (58)2
676
(26)2
308
4 ke 6
24 ke 26
Aekmno A>k
s>Qya
6 4I nanI pU`RvgR s>Qya
= 4 = 22
24
252 = 625 4I AapelI
26
s>Qya mo3I 0e.
= 58
26
1
2
1
4
24
25
mud\l = ½a.
½a 25,000
mud\t = 30 mas
mas
Vyajno dr = 15 %
½a.
nu> 12 masnu
½a 100
masnu> Vyaj = ½a.
½a 15
½a.
½a 25,000 nu> 30 masnu
masnu> Vyaj = ( ? )
∴ Vyaj = 25,000 x 30 x 15
100 x 12
= 125 x 75
Vyaj = 9375 ½a.
½a
Vyajmud\l = mudl
\ + Vyaj
= 25,000 + 9375
Vyajmud\l = 34,375 ½a.
½a
p/kaxwa[Ae mud\tne
tne A>te 34,375 ½a.
½a cUkvva pDe.
2aro ke phelI bekI s>Qya = x
bI+ bekI s>Qya = x + 2 Ane
+I+
I+ bekI s>Qya = x + 4 4ay.
+`e
`ey k/imk bekI sQyaaono srva5o 156 4ay 0e.
∴smIkr` Sv½p
Sv½p : x + x + 2 + x + 4 = 156
∴ 3x + 6 = 156
∴ 3x = 156 – 6
∴ 3x = 150
∴ x = 150
∴ x = 50
3
∴ phelI bekI s>Qya = x = 50,
bI+ bekI s>Qya = x + 2 = 50 + 2 = 52
Ane +I+
I+ bekI s>Qya = x + 4 = 50 + 4 = 54 4ay.
∴ sO4I mo3I bekI s>Qya = x + 4 = 54 4ay.
1
P = ½a.
½a 25,000
N = 30 mas
mas
R = 15%
a
AHI>,
1
1
½a
1
4
2aro ke phelI bekI s>Qya = x
bI+ bekI s>Qya = x + 2 Ane
+I+
I+ bekI s>Qya = x + 4 4ay.
∴smIkr` Sv½p
Sv½p : x + x + 2 + x + 4 = 156
∴ 3x + 6 = 156
∴ 3x = 156 – 6
∴ 3x = 150
∴ x = 150
3
∴ x = 50
∴ phelI bekI s>Qya = x = 50,
bI+ bekI s>Qya = x + 2 = 50 + 2 = 52
Ane +I+
I+ bekI s>Qya = x + 4 = 50 + 4 = 54 4ay.
∴ sO4I mo3I bekI s>Qya = x + 4 = 54 4ay.
309
1
1
1
1
4