CBSE CLASS 10 MATHEMATICS QUESTIONS AND ANSWERS PART 3

CBSE CLASS 10 MATHEMATICS QUESTIONS AND ANSWERS PART 3

Answers are given in Yellow

1. ABC and EDF are two similar triangles such that AB=5 cm, AC=7 cm, DF=15 cm and DE=12 cm, then what is the sum of the remaining sides of the triangles?

The remaining sides are BC and EF
Since the triangles are similar, the sides are proportional.
AB/ED = BC/DF = AC/EF
5/12 = BC/15 gives BC = 25/4= 6.25 cm
5/12 = 7/EF gives EF=84/5 = 16.8 cm
BC+EF= 6.25+16.8 = 23.05 cm

2. During conversion of a solid from one shape to another, is there any change in the volume of the new shape?

No. The volume of the new shape remains unaltered.

3. If sinx + cosx=√2 cosx, then find the value of tanx (x not equal to 90)

sinx + cosx=√2 cos x
Dividing by cosx throughout, we get
tanx + 1 = √2
Therefore tanx = √2-1

4. Observation of some data are x/5, x, x/3, 2x/3, x/4, 2x/5 and 3x/4 where x>0. If the median of the data is 4, then what is the value of x?

Rearranging the terms in the ascending order, we get x/5, x/4, x/3, 2x/5, 2x/3, 3x/4, x.
Since there are 7 terms, the middle most term that is the 4th term will be the median.
Given median=4
Therefore 2x/5= 4
Which gives x=10

5. In a frequency distribution, the mid value of a class is 10 and the width of the class is 6. Find the lower limit of the class.

Let L be the lower limit and U be the upper limit

U-L= 6 --> equation 1
(L+U)/2 = 10
L+U = 20 --> equation 2
Solving equations 1 and 2,
L=7

6. If X, M, Z denote mean, median and mode of a data and X:M = 9:8, then find the ratio of M:Z

The empirical formula is
3 median = mode + 2 mean
3 M = Z + 2X
Given X:M = 9:8
So X=9M/8
3M = Z + 2 (9 M/8)
M(3-(9/4)) = Z
M(3/4) = Z
M/Z = 4/3
M:Z = 4:3

7. Find the area of a sector of a circle of radius 14 cm and subtending an angle of 45° at the centre.

Area of the sector = (45/360) x (22/7) x 14 x 14
Simplifying, we get
Area of the sector as 77 sq.cm

8. Find the smallest natural number by which 1200 should be multiplied so that the square root of the product is a rational number.

Expressing 1200 as a product of its prime factors, we get
1200 = 2 x 2 x 2 x 2 x 5 x 5 x 3
So to get a product such that its square root is a rational number, we should multiply 1200 by 3.
1200x3= 3600
√3600 = 60

9. Find the values of a, b and c such that a, 10, b, c, 31 are in AP

Let us take a, 10, b
10-a = b-10
gives a+b= 20 -->equation 1
10, b, c
b-10=c-b
gives 2 b - c=10 -->equation 2
b, c, 31
c-b= 31-c
2 c - b = 31 --> equation 3
Solving equations 2 and 3, we get
b=17, c=24
Substitute b=17 in equation 1, we have
a=3

10. A game consists of tossing a coin 3 times and noting the outcome each time. If getting the same result in all the tosses is a success, find the probability of losing the game.

Sample space in the event of tossing a coin 3 times is
HHH, HHT, HTH, THH, HTT,THT, TTH, TTT
Total number of possible outcomes is 8
Given that getting the same result (HHH, TTT) is a success
P(success) = 2/8 =1/4
Therefore
P(losing the game) = 1 - (1/4) = 3/4

11. If five times the fifth term of an AP is equal to eight times its eighth term, show that its 13th term is zero.

a₅ = a + 4 d
a₈ = a + 7 d
5 a₅ = 8 a₈
5 a + 20 d = 8 a + 56 d
-3 a = 36 d
a = -12 d

Therefore a₁₃ = a + 12 d = -12 d + 12 d = 0

12. Show that any positive even integer can be written in the form 6 q, 6 q +2 or 6 q + 4, where q is an integer.

Let a be any positive integer and b= 6

By Euclid's division algorithm,
a= 6 q+r for some integer q>=0.
So, r can take the values 0, 1, 2, 3, 4, 5.
When r=0,
a=6 q is even
When r=2,
a=6 q + 2 is even
When r= 4,
a=6 q + 4 is even

Therefore any positive even integer is of the form 6 q, 6 q +2, 6 q + 4.

13. Find A and B if sin(A+2 B) = √3/2 and cos(A+4 B) = 0 , where A and B are acute angles.

sin(A+2 B) = √3/2
sin (A + 2 B ) = sin 60°
A+ 2 B = 60 -->equation 1
cos(A+4 B) = 0
cos (A + 4 B ) = cos 90°
A + 4 B = 90 -->equation 2
Solving equations 1 and 2, we get
A=30° and B=15°

14. A horse is tethered to one corner of a rectangular field of dimensions 70 m x 52 m , by a rope of length 21 m. How much area of the field can it graze?

Area of the field = Length x Breadth
= 70 x 52 = 3640 sq.m
Area of the quadrant (since the field is rectangular)
= (90/360) x (22/7) x 21 x 21
= 693/2 = 346.5 sq m
Grazing area = 3640 - 346.5 = 3293.5 sq.m

15. If 3 x+ 4 y : x + 2 y = 9 : 4, then find the value of 3 x + 5 y : 3 x - y

4 ( 3 x+ 4 y ) = 9 ( x + 2 y )
12 x + 16 y = 9 x + 18 y
12 x - 9 x = 18 y - 16 y
3x = 2 y

Substituting this, we get
3 x + 5 y : 3 x - y
2 y + 5 y : 2 y - y
7 y : y
7 : 1