CBSE Class 10 Maths Previous Question Paper 2018 Solution
Section A
1) If x = 3 is one root of the quadratic equation x2 – 2kx – 6 = 0, then find the value of k.
solution:
x=3 is one root of the equation
∴9-6k-6=0
->k=1/2
2) What is the HCF of smallest prime number and the smallest composite number ?
solution:
The required numbers are 2 and 4.
HCF of 2 and 4 is 2.
3) Find the distance of a point P(x, y) from the origin.
solution:
OP=√x2+y2
4) In an AP, if the common difference (d) = –4, and the seventh term (a7 ) is 4, then find the first term.
solution:
a+6(-4)=4
->a=28
5)What is the value of (cos2 67° – sin2 23°) ?
solution:
∴ cos 67°=sin 23 °
∴ cos267°-sin223=0
6) Given D ABC ~ D PQR, if AB/ PQ = 1/ 3 , then find ar D ABC /ar D PQR
solution:
ar△ABC/ar△PQR=AB2/PQ2
=(1/3)2=1/9
Section B
7) Given that √2 is irrational, prove that (5 + 3√ 2) is an irrational number.
solution:
Let us assume 5+3√2 is a rational number.
∴ 5+3√2=p/q where q≠0 and q are integers.
->√2=p-5q/3q
->√2 is a rational number as RHS is rational
This contradicts the given fact that √2 is rational.
Hence 5+3√2 is an irrational number.
8) In Fig. 1, ABCD is a rectangle. Find the values of x and y.
solution:
AB=DC and BC=AD
-> x+y=30
and x-y=14
solving to get x=22 and y=8.
9) Find the sum of first 8 multiples of 3.
solution:
S=3+6+9+12…+24
=3(1+2+3+…+8)
=3×(8×9/2)
=108
10) Find the ratio in which P(4, m) divides the line segment joining the points A(2, 3) and B(6, –3). Hence find m.
solution:
Let AP:PB=k:1
∴6k+2/k+1=4
->k=1, ratio is 1:1
Hence m =-3+3/2=0
11) Two different dice are tossed together. Find the probability :
(i) of getting a doublet
(ii) of getting a sum 10, of the numbers on the two dice.
solution:
Total number of possible outcomes=36
(i) Doublets are (1,1)(2,2)(3,3)(4,4)(5,5)(6,6)
Total number of doublets=6
∴Prob (getting a doublet)=6/36 or 1/6
(ii)Favourable outcomes are(4,6)(5,5)(6,4)i.e.,3
∴Prob(getting a sum10)=3/36 or 1/12
12) An integer is chosen at random between 1 and 100. Find the probability that it is :
(i) divisible by 8.
(ii) not divisible by 8.
solution:
Total number of outcomes=98
(i) Favourable outcomes are 8,16,24,…96 i.e.,12
∴Prob(integer is divisible by 8)=12/98 or 6/49
(ii) Prob(integer is not divisible by 8)=1-6/49
=43/49
Section C
13) Find HCF and LCM of 404 and 96 and verify that HCF ´ LCM = Product of the two given numbers.
solution:
404=2×2×101=22×101
96=2×2×2×2×2×3=25×3
HCF of 404 and 96=22=4
LCM of 404 and 96=101×25×3=9696
HCF×LCM=4×9696=38784
Also 404×96=38784
Hence HCF×LCM=Product of 404 and 96.
14) Find all zeroes of the polynomial (2x 4 – 9x 3 + 5x 2 + 3x – 1) if two of its zeroes are (2 + √3) and (2 – √3).
Solution:
p(x)=2x4-9x3+5x2+3x-1
2+√3 and 2-√3 are zeroes of p(x)
∴p(x)=(x-2-√3)×(x-2+√3)×g(x)
=(x2-4x+1)g(x)
(2x4-9x3+5x2+3x-1) ÷(x2-4x+1)=2x2-x-1
∴g(x)=2x2-x-1
=(2x+1)(x-1)
Therefore other zeroes are x=-1/2 and x=1
∴Therefore all zeroes are 2+√3,2-√3,-1/2 and 1
15) If A(–2, 1), B(a, 0), C(4, b) and D(1, 2) are the vertices of a parallelogram ABCD, find the values of a and b. Hence find the lengths of its sides.
Solution:
ABCD is a parallelogram
∴diagonals AC and BD bisect each other Therefore
Mid point of BD is same as mid point of AC
->(a+1/2,2/2)=(-2+4/2,b+1/2)
->a+1/2=1 and b+1/2=1
->a=1,b=1, therefore length of sides are √10 units each.
OR
If A(–5, 7), B(–4, –5), C(–1, –6) and D(4, 5) are the vertices of a quadrilateral, find the area of the quadrilateral ABCD.
Solution:
Area of quad ABCD=Ar△ABD+ Ar△BCD
Area of △ABD=1/2|(-5)(-5-5)+(-4)(5-7)+(4)(7+5)|=
=53 sq units
Area of △BCD=1/2|(-4)(-6-5)+(-1)(5+5)+4(-5+6)|
=19 sq units
Hence area of quad,ABCD=53+19=72 sq units
16) A plane left 30 minutes late than its scheduled time and in order to reach the destination 1500 km away in time, it had to increase its speed by 100 km/h from the usual speed. Find its usual speed.
solution:
Let the usual speed of the plane be x km/hr
∴1500/x-1500/x+100=30/60
->x2+100x-300000=0
->x2+600x-500x-300000=0
->(x+600)(x-500)=0
x≠-600,
∴x=500
Speed of plane=500 km/hr
17) Prove that the area of an equilateral triangle described on one side of the square is equal to half the area of the equilateral triangle described on one of its diagonal.
Solution:
Let the side of the square be’a’ units
∴AC2=a2+a2=2a2
->AC=√2 a units
Area of equilateral △BCF=√3/4a2 sq.u
Area of equilateral △ACE=√3/4(√2a)2=√3/2a2sq.u
->Area △BCF=1/2Ar△ACE
OR
If the area of two similar triangles are equal, prove that they are congruent.
Solution:
Let △ABC-△PQR
ar △ABC/ ar△PQR=AB2/PQ2=BC2/QR2=AC2/PR2
Given ar △ABC= ar△PQR
->AB2/PQ2=1=BC2/QR2=AC2/PR2
->AB=PQ,BC=QR,AC=PR
->Therefore △ABC=△PQR.(sss congruence rule)
18) Prove that the lengths of tangents drawn from an external point to a circle are equal.
Solution:
Correct given, To prove Figure, Construction correct proof
19) If 4 tan q = 3, evaluate (4sin θ-cos θ+1/4sin θ + cos θ -1)
Solution:
4 tan θ=3
-> tan θ=3/4
=> Sin θ=3/5 and cos θ=4/5
∴4sin θ-cos θ+1/4sin θ -1=(4×3/5-4/5+1)/(4×3/5+4/5-1)
=13/11
or
If tan 2A = cot (A – 18°), where 2A is an acute angle, find the value of A.
Solution:
tan 2A=cot(A-18°)
=90 °-2A=A-18 °
=>3A=108 °
=>A=36 °
20) Find the area of the shaded region in Fig. 2, where arcs drawn with centres A, B, C and D intersect in pairs at mid-points P, Q, R and S of the sides AB, BC, CD and DA respectively of a square ABCD of side 12 cm. [Use p = 3.14]
Solution:
Radius of each are drawn=6cm
Area of one quadrants=(3.14) ×36/4
Area of four quadrants=3.14×36=113.04cm2
Area of square ABCD=12×12=144cm2
Hence Area of shaded region=144-113.04
=30.96cm2
21) A wooden article was made by scooping out a hemisphere from each end of a solid cylinder, as shown in Fig. 3. If the height of the cylinder is 10 cm and its base is of radius 3.5 cm. Find the total surface area of the article.
solution:
Total surface Area of article=CSA of cylinder +CSA of 2 hemispheres CSA of cylinder =2 πrh
=2×22/7×3.5×10
=220cm2
Surface Area of two hemispherical scoops=4×22/7×3.5×3.5
=154cm2
Total surface Area of article=220+154
=374 cm2
OR
A heap of rice is in the form of a cone of base diameter 24 m and height 3.5 m. Find the volume of the rice. How much canvas cloth is required to just cover the heap ?
Solution:
Radius of conical heap=12m
Volume of rice=1/3×22/7×12×12×3.5m3
=528 m3
Area of canvas cloth required= πr/
l=√122+(3.5)2=12.5m
Area of canvas required=22/7×12×12.5
=471.4m2
22) The table below shows the salaries of 280 persons :
Salary( in thousand Rs) |
No. of persons(f) |
5-10 |
49 |
10-15 |
133 |
15-20 |
63 |
20-25 |
15 |
25-30 |
6 |
30-35 |
7 |
35-40 |
4 |
40-45 |
2 |
45-50 |
1 |
Calculate the median salary of the data.
Solution:
Salary( in thousand Rs)
Salary( in thousand Rs) |
No. of persons(f) |
cf |
5-10 |
49 | 49 |
10-15 | 133 |
182 |
15-20 |
63 | 245 |
20-25 | 15 |
260 |
25-30 |
6 | 266 |
30-35 | 7 |
273 |
35-40 |
4 | 277 |
40-45 | 2 |
279 |
45-50 |
1 |
280 |
N/2=280/2=140
Median class is 10-15
Median=l+h/f(N/2-C)
=10+5/133(140-49)
=10+5×91/133
=13.42
Median salary is Rs 13.42 thousand or Rs 13420(approx)
Section D
23) A motor boat whose speed is 18 km/hr in still water takes 1hr more to go 24 km upstream than to return downstream to the same spot. Find the speed of the stream.
solution:
Let the speed of stream be x km/hr
∴The speed of the boat upstream=(18-x)km/hr
and speed of the boat downstream=(18+x)km/hr
As given in the question,
(24/18-x)-(24/18+x)=1
=>x2+48x-324=0
=>(x+54)(x-6)=0
x≠-54, ∴x=6
∴Speed of the stream=6km/hr
OR
A train travels at a certain average speed for a distance of 63 km and then travels at a distance of 72 km at an average speed of 6 km/hr more than its original speed. If it takes 3 hours to complete total journey, what is the original average speed ?
Solution:
Let the original average speed of train be x km/hr
Therefore 63/x+72/x+6=3
=>x2-39x-126=0
=>(x-42)(x+3)=0
x≠-3
x=42
Original speed of train is 42 km/hr
24) The sum of four consecutive numbers in an AP is 32 and the ratio of the product of the first and the last term to the product of two middle terms is 7 : 15. Find the numbers.
solution:
Let the four consecutive terms of the A.P.be
a-3d,a-d,a+d,a+3d
By given conditions
(a-3d)+(a-d)+(a+d)+(a+3d)=32
=>4a=32
=>a=8
and (a-3d)(a+3d)/(a-d)(a+d)=7/15
=>8a2=128d2
=>d2=4
=>d=≠2
Numbers are 2,6,10,14,10,6,2.
25) In an equilateral D ABC, D is a point on side BC such that BD = 1 /3 BC. Prove that 9(AD)2 = 7(AB)2
solution:
Draw AE⊥BC
△AEB=△AEC(RHS Congruence Rule)
BE=EC=1/2BC=1/2AB
Let AB=BC=x
Now BE=x/2 and DE=BE-BD
=x/2-x/3
=x/6
Now AB2=AE2+BE2….(1)
and AD2=AE2+DE2….(2)
From(1) and (2) AB2-AD2=BE2-DE2
=>X2-AD2=(X/2)2-(X/6)2
=>AD2=X2-X2/4+X2/36
=>AD2=(28/36)X2
=>9AD2=7AB2
OR
Prove that, in a right triangle, the square on the hypotenuse is equal to the sum of the squares on the other two sides.
solution:
Given, to prove, Construction and Figure
Correct proof
26) Draw a triangle ABC with BC = 6 cm, AB = 5 cm and ÐABC = 60°. Then construct a triangle whose sides are 3 /4 of the corresponding sides of the D ABC.
solution:
Correct construction of △ABC
Correct construction of similar to △ABC.
27) Prove that : sin A – 2 sin3 A/ 2 cos3 A – cos A = tan A.
solution:
LHS=sinA-2sin3A/2cos3A-cosA
=sinA(1-2sin2A)/cos A(2cos2A-1)
=sinA(1-2(1-cos2A))/cosA(2cos2A-1)
=tanA(2cos2A-1)/(2cos2A-1)
=tanA=RHS
28) (i) The area of the metal sheet used to make the bucket. (ii) Why we should avoid the bucket made by ordinary plastic ? [Use p = 3.14]
solution:
Here, r1=15cm,r2=5cm and h=24cm
(i) Area of metal sheet=CSA of the bucket+area of lower end= πl(r1+r2)+ πr22
Where l=√242+(15-5)2=26cm
Surface area of metal sheet=3.14(26×20+25)cm2
=1711.3cm2
we should avoid use of plastic because it is non-degradable of similar value.
29) As observed from the top of a 100 m high light house from the sea-level, the angles of depression of two ships are 30° and 45°. If one ship is exactly behind the other on the same side of the light house, find the distance between the two ships. [Use √3 = 1.732]
solution:
Let AB be the tower and ships are at point C and D.
tan 45°=AB/BC
=>AB/BC=1
=>AB=BC
Also tan 30°=1/√3=AB/BC+CD
=>1/√3=AB/AB+CD
=>AB+CD=√3AB
=>CD=AB(√3-1)
=100×(1.732-1)
=73.2m
30) The mean of the following distribution is 18. Find the frequency f of the class 19 – 21.
Class | 11-13 | 13-15 | 15-17 | 17-19 | 19-21 | 21-23 | 23-25 |
Frequency | 3 | 6 | 9 | 13 | f | 5 | 4 |
solution:
class |
x | f |
fx |
11-13 |
12 | 3 | 36 |
13-15 | 14 | 6 |
84 |
15-17 |
16 | 9 | 144 |
17-19 | 18 | 13 |
234 |
19-21 |
20 | f | 20f |
21-23 | 22 | 5 |
110 |
23-25 |
24 | 4/40+f |
96/704+20f |
Mean=18=704+20f/40+f
=>720+18f=704+20f
=>f=8
or
The following distribution gives the daily income of 50 workers of a factory :
Daily Income (in₹) | 100-120 | 120-140 | 140-160 | 160-180 | 180-200 |
Numbers of workers | 12 | 14 | 8 | 6 | 10 |
solution:
Cumulative frequency distribution table of less than type is
Daily income |
Cumulative frequency |
Less than 100 |
0 |
Less than 120 |
12 |
Less than 140 |
26 |
Less than 160 |
34 |
Less than 180 |
40 |
Less than 200 |
50 |
CBSE Class 10 Previous Question Paper 2018 Solution
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