Medical Classes 10th CBSE{SA I} - Learn Maths Online … CBSE ... 100–120 120–140 140–160...
Transcript of Medical Classes 10th CBSE{SA I} - Learn Maths Online … CBSE ... 100–120 120–140 140–160...
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10th CBSE{SA – I}
Mathematics
Solution Visits: www.pioneermathematics.com/latest_updates
Mock Paper With
Blue Print of Original Paper on Latest Pattern
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10th CBSE First Term {SA- I}
Blue Print
Topic/Unit MCQs SA(1) SA(II) LA Total
Number System 2(2) 1(2) 2(6) – 5(10)
Algebra 2(2) 2(4) 2(6) 2(8) 8(20)
Geometry 1(1) 2(4) 2(6) 1(4) 6(15)
Trigonometry 4(4) 1(2) 2(6) 2(8) 9(20)
Statistics 1(1) 2(4) 2(6) 1(4) 6(15)
Total 10(10) 8(16) 10(30) 6(24) 34(80)
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General instructions:
Time: 3hrs. M: M: 90
(i) All questions are compulsory.
(ii) The question paper consists of 34 questions divided into sections A, B, C and D.
Section-A comprises of 8 questions of 1 mark each, section-B comprises of 6
questions of two marks each, section-C comprises of 10 questions of three marks
each and section-D comprises of 10 questions of four marks each.
(iii) Question numbers 1 to 8 in section-A are multiple choice questions where you
are required to select one option out of the given four.
(iv) There is no overall choice. However, internal choice has been provided in 1
question of two marks, 3 questions of three marks each and two questions of four
marks each. You have to attempt only one of the alternatives in all such questions.
(v) Use of calculator is not permitted.
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Section-A
Questions number 1 to 8 carry one marks each
1. The pair of equation x = 2 and x = –4 has :
(a) infinitely many solutions (b) no solution
(c) two solutions (d) one solution
Sol : (b)
2. If sides of two similar triangles are in ratio 4 : 9, then area of these triangles are in the
ratio
(a) 2 : 3 (b) 4 : 9 (c) 81 : 16 (d) 16 : 81
Sol : (d)
Ratio of Areas = 2
4
9 =
16
81
3. The x-coordinate of the point of intersection of more than and less then ogive is :
(a) mode (b) mean (c) median (d) Variance
Sol : (c)
4. If LCM (54, 336) = 3024, then HCF (54, 336) is:
(a) 54 (b) 6 (c) 336 (d) 36
Sol : (b)
L.C.M × H.C.F
=Product of two numbers
= 54 × 336 = 3024 × H.C.F
H.C.F = 6
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5. Value of 5 tan2A–5sec2A is :
(a) 1 (b) 0 (c) –5 (d) 5
Sol : (c)
5 tan2A – 5 × (1 + tan2A) = 5 × (tan2A – 1 – tan2A)
= –5
6. If 2 sin2A = 3 , then A equal to :
(a) 90o (b) 60o (c) 45o (d) 30o
Sol : (d)
sin 2A = 3
2Sin2A = Sin600 A = 300
7. If sec A = q
p, then value of
1
p cos A is:
(a) 1
p (b)
1
q (c)
1
pq (d)
p
q
Sol : (b)
1 1 p 1cos A
p p q q
8. If a positive integer n is divided by 2, then the remainder can be :
(a) 1 or 2 (b) 1, 2, or 3 (c) 0 or 1 (d) 2
Sol : (c)
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Section-B
Questions number 9 to 14 carry two marks each
9. In the given figures, find measure of X .
Sol :
In PQR and zyx
PQ 4.2 1
ZY 8.4 2
PR 3 3 1
ZX 26 3
QR 7 1
YX 14 2
PQ PR QR
ZY ZX YX
PQR ZYX (By SSS similarity criteria)
PRQ ZXY ..(1) (By CPST)
In PQR
PQR + PRQ + RPQ = 1800 (Angle sum property of )
600 + 700 + PRQ = 1800
0PRQ 50 ..(2)
From (1) & (2)
0ZXY X 50
10. Is 7 × 11 ×13+13 a composite number? Justify your answer.
Sol :
7 × 11 × 13 + 13
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= 13(7 × 11 × 1 + 1)
= 13(77 + 1)
= 13(78)
= 13 × 2 × 13 × 3
As, the no.7 × 11 × 13 + 13 is having more than 2 prime factors,
The given no. is a composite no.
11. Solve: 3 2 2 5
0 and 19x y x y
Sol :
3 20
x y ..(1)
2 519
x y ..(2)
Put 1 1
a & bx y
in (1) & (2)
3a – 2b = 0 ..(3)
2a + 5b = 19 ..(4)
Multiply (3) & (4) by 5 & 2 respectively,
(3a – 2b = 0) × 5
15a – 10b = 0 ..(5)
& (2a + 5b = 19) × 2
4a + 10b = 38 …(6)
Adding (5) & (6)
15a 10b 0
4a 10b 38
19a 38 a 2
Put = a = 2 in (2)
3(2) – 2b = 0
6 – 2b = 0
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–2b = – 6
b = 3
1 1
xa 2
& 1 1
yb 3
12. The following distribution give the daily income of 50 workers of a factory:
Daily income in Rs. 100–120 120–140 140–160 160–180 180–200
Number of workers 12 14 8 6 10
Write the above distribution as “more than type” cumulative frequency distribution.
Sol :
Daily income (in
Rs)
More than type
distribution
fi Cf
100–120
120–140
140–160
160–180
180–200
More than 100
More than 120
More than 140
More than 160
More than 180
12
14
8
6
10
50
38
24
16
10
Total 50
13. If 1
and are zeroes of polynomial 4x2 –2x + (k –4). Find k.
Sol :
If f(x) = 4x2 – 2x + (k – 4)
where, 1
x ,
According to relationship between zeroes & coefficients of a polynomial,
1 c
a
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k 41
4
4 = k – 4
k = 8
Value of k = 8.
14. If tan 2A = cot (A –18o), where 2A is an acute angle, find the value of A.
Sol :
0tan 2A cot A 18
0cot 90 2A
= cot (A – 18)
90 – 2A = A – 18
90 + 18 = 3A
3A = 108
A = 360
Or
If is an acute angle and sin = cos , find the value of 3 tan2 + 2sin2 – 1.
Sol :
sin cos …(1) (Given)
2 23 tan 2sin 1
=
222 2 2
2
sin3 2 sin sin cos
cos 2 2[ sin cos 1]
=
2
2 2 2sin3 2 sin sin cos sin cos from(1)
cos
= 2 23 sin sin sin cos
= 3
Value of 2 23tan 2sin 1 3
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Section-C
Questions number 15 to 24 carry three marks each
15. A survey regarding the height (in cm) of 50 girls of class X of a school was conducted and
the following data was obtained:
Height(in cm) 120–130 130–140 140–150 150–160 160–170 Total
Number of girls 2 8 12 20 8 50
Find the mode of the data.
Sol :
Model class = 150 – 160
f1 = 20
f0 = 12
f2 = 8
h = 10
l = 150
Mode = 1 0
1 0 2
f fl h
2f f f
= 20 12
150 1040 12 8
= 8
150 1020
= 150 + 4
= 154 cm
16. Prove that : cot A cos A cosecA 1
cot A cos A cosecA 1.
Sol :
To prove : cot A cos A cosec A 1
cot A cos A cosec A 1
L. H. S. = cot A cos A
cot A cos A
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=
cos Acos A
sin Acos A
cos Asin A
=
cos A sin A cos A
sin Acos A sin A cos A
sin A
= cos A sin A cos A
cos A sin A cos A
= cos A 1 sin A
cos A 1 sin A
=
11
cosec A1
1cosec A
=
cosec A 1
cosec Acosec A 1
cosec A
= cosec A 1
cosec A 1 = R. H. S.
Hence proved.
17. In the given figure, oACB 90 and CD AB . Prove that 2
2
BC BD
AC AD
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Sol:
ADC ~ ACB
AC AD CD
AB AC BC
2AC AD.AB ……….(1)
CDB ~ ACB
CD BC BD
CA AB BC
2BC AB.BD ……….(2)
Equation (2)/(1) we get
2
2
BC AB.BD
AD.ABAC
2
2
BC BD
ADAC
Or
In the given figure, if AD BC, prove that AB2 +CD2 = BD2 +AC2.
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Sol :
Given : In ABC, AD BC
To Prove : AB2 + CD2 = BD2 + AC2
Proof : In ADC, AD CD (Given)
AC2 = CD2 + AD2 (By Pythagoras theorem)
AD2 = AC2 – CD2 ..(1)
In ADB, AD BD (Given)
AB2 = AD2 + BD2 (By Pythagoras theorem)
AD2 = AB2 – BD2 ..(2)
From (1) & (2)
AC2 – CD2 = AB2 – BD2
AC2 + BD2 = AB2 + CD2
AB2 + CD2 = BD2 + AC2
Hence, proved.
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18. In the given figure, the line segment XY is parallel to AC of a ABC and it divides the
triangle into two parts of equal area. FindAX
AB.
Sol :
Let Area of BXY = k
Area of BAC = 2k
In XBY and ABC
XBY = ABC (Common angle)
BXY = BAC (Corresponding angles)
BXY ~ BAC by AA similarity criteria
2Area of BXY BX
Area of BAC AB
2K BX
2k AB
BX 1
AB 2
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AB AX 1
AB 2
2AB 2AX AB
AX 2 1
AB 2
2 2 1
2 2=
AX
AB
AX 2 2
AB 2
19. Prove that: 3
3
sin 2sintan
2cos cos
Sol :
To prove : 3
3
sin 2 sintan
2 cos cos
Proof : = 3
3
sin 2 sin
2 cos cos
= 2
2
sin 1 2 sin
cos 2cos cos
= 2
2
1 2 sintan
2 1 sin 1
= 2
2
1 2 sintan
2 2 sin 1
= 2
2
1 2 sintan
1 2 sin
= tan = R. H. S.
Hence, proved.
20. Find the mean of the following data:
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Or
Find the missing frequency for the given data if mean of distribution is 52.
Wages (in Rs.) 10–20 20–30 30–40 40–50 50–60 60–70 70–80
No. of workers 5 3 4 f 2 6 13
Sol :
Class interval Frequency (fi) xi fixi
100-120
120-140
140-160
160-180
180-200
12
14
8
6
10
110
130
150
170
190
1320
1820
1200
1020
1900
if 50 i if x 7260
Mean = i i
i
x f
f
= 7260
50
= 145.2
Class- interval 100–120 120–140 140–160 160–180 180–200
Frequency 12 14 8 6 10
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Or
Sol :
Wages in Rs. No. of workers xi di = xi – 45 fidi
10-20 20-30 30-40 40-50 50-60 60-70 70-80
5 3 4 f 2 6
13
15 25 35 45 55 65 75
–30 –20 –10
0 10 20 30
–150 –60 –40
0 20
120 390
if 33 f i if d 280
Mean = A + i i
i
f d
f
52 = 45 + 280
33 f
7 = 280
33 f
231 + 7f = 280
7f = 280 – 231
f = 7
21. Prove that 2 – 5 is an irrational number.
Sol :
Let us assume the 2 5 is a rational number. Then, according to Euclid’s division
lemma, use can find two co-prime integers, such them,
a2 5
b
a5 2
b
2b a5
b
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As, a & b are integers, this implies, (2b – a)/b is a rational no.
5 is also a rational number. But it contradicts the fact the 5 is irrational. This
contradiction has arisen due to our incorrect assumption that 2 5 is a rational
number.
2 5 is an irrational number.
22. If one zeros of polynomial p(x) = 3x2 – 8x + 2k + 1 is seven times of the other, then find
the zeroes and the value of k.
Sol :
p(x) = 3x2 – 8x + (2k + 1)
x = , 7
b
7a
8 88
3 3
1
3
1
x3
..(1)
7x 7
3 ..(2)
And, c 2k 1
7a 3
1 7 2k 1
3 3 3 (From (1) & (2))
7 = 6k + 3
6k = 4
k = 2
3
Value of k = 2
3
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23. A boat goes 30 km upstream and 44 km downstream in 10 hours. In 13 hours, it can go 40
km upstream and 55 km downstream. Find the speed of the stream and that of the boat in
still water.
Sol :
Speed of stream = y km/hr
Speed of boat = x km/hr
5 = D
T
Case I
30 4410
x y x y
Case II
40 5513
x y x y
Let 1 1
a, bx y x y
1 1
a , b5 11
on solving x = 8 km/hr y = 3 km/hr
Or
Solve by cross multiplication method: ax + by =a –b; bx – ay = a + b
Sol :
ax by a b
bx ay a b
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2 2 2 2
x y
ab b a ab ab b a ab
= 2 2
1
a b
2 2 2 2 2 2
x y 1
a b a b a b
I II III
Equating I & III
2 2 2 2
x 1
a b a b
x = 1
Equating II & III
2 22 2
y 1
a ba b
y = – 1
Values of x = 1 & y = – 1
24. Evaluate: sinA cosA –sin A cos(90 A)cos A cos A sin(90 A)sin A
sec(90 A) cosec(90 A).
Sol :
= sin A sin A cos A cos A cos A sin A
sin A cos Acosec A sec A
= sin A cos A - sin3A cos A – cos3A sin A
= sin A cos A – sin A cos A (sin2A + cos2A)
= sin A cos A – sin A cos A = 0
= 0
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Section-D
Questions number 25to 34 carry Four marks each
25. Draw ‘more than ogive’ and less than ogive’ for the following distribution and hence
obtain the median.
Class-interval 5–10 10–15 15–20 20–25 25–30 30–35 35–40
Frequency 2 12 2 4 3 4 3
Sol:
Less then type of distribution
Class interval Less than type distribution
f.i c.f
5–10 Less than 10 2 2
10–15 Less than15 12 14
15–20 Less than20 2 16
20–25 Less than 25 4 20
25–30 Less than30 3 23
30–35 Less than35 4 27
35–40 Less than40 3 30
We will plot the points (10, 2); (15, 14); (20, 16); (25, 20); (30, 23) (35, 27); (40, 30)
More than type of distribution
Class interval More than type distribution
f.i c.f
5–10 More than 5 2 30
10–15 More than10 12 28
15–20 More than15 2 16
20–25 More than 20 4 14
25–30 More than25 3 10
30–35 More than30 4 7
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35–40 Less than35 3 3
Total 30
We will plot the points (5, 30), (10, 28), (15, 16), (20, 14), (25, 10), (30, 7), (35, 3)
26. Evaluate:2 2 2 o 2 o 2 o
2 o 2 o 2 o 2 o
sin sin (90 ) 3cot 30 sin 54 sec 36
3 sec 61 cot 29 2 cosec 65 tan 25.
Or
Prove that :1 cos A sin A 1 sin A
sin A cos A 1 cos A
Sol:
2 2 2
2 2 2 2 2
sin cos 3 3 sin 54
3 sec 61 tan 61 cos 36 2 sec 25 tan 25
= 2
2
9 sin 90 361
3 1 cos 36 2 1
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= 2
2
1 9 cos 36
3 2 cos 36
= 1 9
3 2
= 2 27
6
= 25
6
Or
To prove : 1 cos A sin A 1 sin A
sin A cos A 1 cos A
L.H.S. = 1 sin A cos A 1 sin A cos A
sin A cos A sin A cos A 1
2 2
2 2
1 sin A cos A
sin A cos A 1
2 2
2 2
1 sin 2sin A cos A
sin A cos A 2sin Acos A 1
2 2sin A sin A 2sin A
1 1 2sin Acos A
2sin A sin A 1
2sin Acos A
sin A 1R.H.S.
cos A Hence proved
27. If tan A + sin A = m and tan A – sin A = n, Show that m2 –n2 = 4 mn .
Sol:
Given : sin A tan A m & tan A sin A n
To prove :
2 2m n 4 mn
Proof:
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2 2m n L.H.S.
=2 2
sin A tan A tan A sin A
= 2 2 2 2sin A tan A 2sin A tan A tan A sin A 2sin A tan A
= 4sinA tanA
R.H.S. = 4 mn
= 4 tan A sin A tan A sin A
= 2 24 tan A sin A
=2
2
2
sin A4 sin A
cos A
=2 2 2
2
sin A sin Acos A4
cos A
=2
2
2
1 cos A4 sin A
cos A
=2
2
2
sin A4 sin A
cos A
= 2 24 sin A tan A
= 4sin A tan A L.H.S. Hence proved
28. If the median of the distribution given below is 32.5. find x and y.
Class interval 0–10 10–20 20–30 30–40 40–50 50–60 60–70 Total
Frequency x 5 9 12 y 3 2 40
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Sol:
Class interval f.i c.f
0–10 x x
10–20 5 x+5
20–30 9 x+14
30–40 12 x+26
40–50 y x+ y+26
50–60 3 x+ y+29
60–70 2 x+ y+31
40
x+ y + 31 = 40
x + y = 9
Median = 32.5 = 30 + 20 x 14
1012
x = 3, x + y = 9
3 + y = 9
y = 6
29. In the given figure, AB||PQ||CD, AB = x units, CD = y units and PQ = z units,
prove that, 1 1 1
x y z
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Or
In an equilateral triangle ABC, D is a point on BC, such that BD=1
BC3
. Prove that 9 AD2 = 7
AB2.
Sol :
ABD ~ PQD
x z
l m m
xm = z (l + m)
x m = zl + zm
xm – zm = zl
m (x – z) = zl
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m z
l x z ..(1)
BPQ ~ BCP
z l
y l m
z(l + m) = yl
zl + mz = yl
mz = yl – zl
mz = l(y – z)
m y z
l z ..(2)
From (1) & (2)
z y z
x 2 z 2z y z x z
z2 = xy – yz – xz + z2
xy = yz +zx xy yz zx
xyz xyz xyz
1 1 1
2 x y
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Or
Sol:
Given : Equilateral ABC , BD=1
BC3
To prove : 9AD2 = 7AB2
Construction: Draw AM BC.
Proof: BD = 1
3BC ……(1)(Given)
BM = 1
2BC …(2) [Altitude of an equilateral is also its median]
DM = BM –BD
DM = BC BC
2 3 [From (1) and (2)]
DM = 3BC 2BC BC
6 6 ….(5)
In ADM,AM DM
AD2 = AM2 + DM2 (By Pythagoras theorem)
AM2 = AD2 –DM2 …..(3)
Similarly, in ABM,
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AM2 = AB2 –BM2 …(4)
From (3) and (4)
AD2 –DM2 = AB2–BM2
AD2 –AB2 = DM2 –BM2
AD2 –AB2 = 2 2
BC BC
6 2 (from (2) and (5)
AD2 –AB2 = 2 2BC BC
36 4
=2 2BC 9BC
36
=28BC
36
AD2 –AB2 = 22BC
9
AD2 = 2 22BC 9AB
9
9AD2 = –2AB2 +9AB2
9AD2 = 7AB2 Hence proved
30. Prove that in a right angle triangle, the square of the hypotenuse is equal to the sum of the
squares of other two sides.
Sol:
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Given : 0ABC, BAC 90 ; AD BC
To prove : BC2 = AB2 +AC2
Proof: According to a theorem, if a is drawn from the right of a to its hypotenuse,
then the two s formed are similar to each other & to the whole .
BAD ACD BCA ……(1)
BAD BCA From (1)
AB BC
BD AB (by CPST)
AB2 = BC . BD ……..(2)
ACD BCA (from (1))
AC BC
CD AC
AC2 = BC . CD …….(2)
Adding (1) & (2)
AC2 +AB2 = BC .BD +BC .CD
= BC(BD + CD)
= BC × BC
= BC2
AC2 +AB2 = BC2 Hence proved
31. Use Euclid's division lemma to show that the cube of any positive integer is of the form
9m, 9m +1 or 9m + 8.
Sol:
Let a be any +ve integer and b = 3, the according to Euclid’s division lemma, a can be
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written as,
a = 3q +r, whose, 0 r<3.
a is of the form 3q, 3q+ 1&3q +2.
Case I: a = 3q
cubing both sides
a3 = (3q)3
= 27 q3
= 9(3q3)
=9m, where m = 9q3
a3 is of the of 9m.
Case II: a = 3q + 1
cubing both sides
a3 = (3q +1)3
= 27 q3 + 1 + 27 q2 +9q
= 9(3q3 + 3 q2 +q) +1
= 9m +1, where m = 3q3 +3q2 +q
a3 is of the form 9m +1.
Case III: a = 3q +2
cubing both sides
a3 = (3q+2)3
= 27q3 +8 +54q2 + 36q
= 9(3q3 +6q2 +4q) + 8
= 9m +8, where m = 3q3 + 6q2 +4q
a3 is of the form 9m +8
32. Draw the graph of 2x + y = 6 and 2x – y + 2 = 0. Shade the region bounded by
these lines with x axis. Find the area of the shaded region.
Sol:
2x + y = 6
2x – y =–2
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Area of 1
b h2
= 214 4 8 cm
2
33. Find other zeroes of the polynomial p(x) = 2x4– 21x3 + 49x2 – 10x –20, if two of its zeroes
are 5 ± 5 .
Sol:
P(x) = 2x4 –213 + 49x2 – 10x –20
x = 5 5
x –5 5 = 0
( x –5) – 5 =0 …….(1)
x = 5 5
x –5 + 5 = 0
(x –5) + 5 = 0 ……….(2)
Multiplying (1) & (2)
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(x – 5) 2 – 2
5 =0
x2 + 25 –10x – 5 = 0
f(x) = x2 –10x +20 = 0
2
2 4 3 2
4 3 2
3 2
3 2
2
2
2x x 1
x –10x 20 2x 21x 49x 10x 20
2x 20x 40x
x 9x 10x
x 10x 20x
x 10x 20
x 10x 20
0
g(x) = 2x2 – x – 1 = 0
= 2x2 – 2x + x + 1 = 0
= 2x ( x –1) + 1(x – 1)
= (2x + 1) ( x–1) = 0
x = 1
,1& 5 5 , 5 52
Ans.
34. Let days taken by 1 women = x
Let days taken by 1 man = y
2 5 1
x y 4 ……(1)
&
3 6 1
x y 3 ……(2)
Put 1
x = a &
1b
y in (1) & (2)
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2a +5b = 1
4 ……(3)
3a + 6b = 1
3
a 2b 1/9 2
2a + 4b = 2/9 …..(4)
Subtracting (3) from (4) we get
2a + 5b = ¼
2a + 4b = 2/9
Put b = 1/36 in (3)
2a + 51 1
36 4
72a + 5 = 36
94
72a = 4
a = 4 1
72 18
x = 1
18 daysa
1
y 36 daysb
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