BOARD EXAM REVISION TEST 19 (ANSWERS)

CLASS: X : MATHEMATICS
M.M. 40 Marks T.T. 1½ hr
Q1-Q4 of 1 mark, Q5-Q9 of 2 marks, Q10-Q15 of 3 marks & Q16-Q17 of 4 marks
1. Express 5050 as product of its prime factors.
Ans: 5050 = 2 × 5 × 5 × 101 = 2 × 52 × 101

2. If 1 is one zero of the polynomial p(x) = ax2 – 3(a – 1)x – 1, then find the value of a.
Ans: Given, 1 is the zero of the polynomial p(x) = ax2 – 3(a – 1)x – 1
∴ p(1) = 0
⇒ a(1)2 – 3 (a – 1) (1) – 1 = 0
⇒ a – 3a + 3 – 1 = 0
⇒ – 2a + 2 = 0 ⇒ a = 1

3. For what value of k will the consecutive terms 2k + 1, 3k + 3 and 5k – 1 form an A.P.?
Ans: Let 2k + 1, 3k + 3 and 5k – 1 are in A.P.
∴ 2(3k + 3) = 2k + 1 + 5k – 1
⇒ 6k + 6 = 7k ⇒ k = 6

4. For the following distribution:

Marks Below Below Below Below Below Below
10 20 30 40 50 60
No. of Students 3 12 27 57 75 80
Find the modal class is
Ans:
Marks 0 – 10 10 – 20 – 30 – 40 40 – 50 –
20 30 50 60
No. of Students 3 9 15 30 18 5
Highest frequency is 30 which belong to 30 – 40. Hence, Modal class is 30 – 40

5. Prove that 3 + 2√5 is an irrational number

Ans: Let 3 + 2√5 is a rational number such that
3 + 2√5 = a, where a is a rational number
a3
⇒ 2√5 = a – 3 ⇒ 5 
2
a 3
Since ‘a’ is a rational number and 2, 3 are integers, therefore is a rational number
2
⇒ √5 is a rational number which contradict the facts that √5 is an irrational number
Therefore, our assumption is wrong
Hence, 3 + 2√5 is an irrational number
6. Find the number of natural numbers between 101 and 999 which are divisible by both 2 and 5.
Ans: Natural numbers between 101 and 999 which are divisible by both 2 and 5 are 110, 120,
..., 990, which forms an A.P.
Here, a = 110, d = 10, an = 990
Now, an = a + (n – 1)d
⇒ 110 + (n – 1)10 = 990
⇒ (n – 1)10 = 880 ⇒ n – 1 = 88 ⇒ n = 89
Hence, there are 89 numbers between 101 and 999 which are divisible by both 2 and 5.

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7. In the given figure, PA and PB are tangents to the circle with centre O. If ∠APB = 60°, then
find ∠OAB

Ans: PA = PB ( Tangents drawn from external point are equal)

⇒ ∠ABP = ∠BAP = x ( Angles opposite to equal sides are equal)
In ∆APB, 60° + x + x = 180°
⇒ 2x = 120° ⇒ x = 60°
Now, ∠OAP = 90° (∵ Tangent is perpendicular to the radius through the point of contact)
∴ ∠OAB = 90° – 60° = 30°

8. If the vertices of a parallelogram in order are A(1, 2), B(4, y), C(x, 6) and D(3, 5), then find
(x, y).
Ans:

9. A card is drawn at random from a well-shuffled deck of playing cards. Find the probability of
drawing a (i) face card (ii) card which is neither a king nor a red card.
Ans: Total number of outcomes = 52
(i) Number of face cards = 12
∴ Probability of drawing a face card = 12/52 = 3/13
(ii) Number of cards which are neither king nor red = 24
∴ Probability of drawing a card which is neither a king nor a red card = 24/52 = 6/13

10. Find the zeroes of the quadratic polynomial x2 – 2x – 8 and verify the relationship between the
zeroes and the coefficients of the polynomial.
Ans: Given, f(x)=x2 − 2x – 8
The zeroes of f(x) are given by, f(x) = 0
⇒ x2 + 2x − 4x – 8 = 0 ⇒ x (x + 2) − 4(x + 2) = 0 ⇒ (x + 2) (x − 4) = 0
⇒ x = −2 (or) x = 4
Hence, the zeros of f(x) = x2 − 2x − 8 are α = −2 and β = 4
b
α + β = -2 + 4 = 2 = =2
a
c
αβ = -2 x 4= -8 = =-8
a

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11. Amita, Suneha and Raghav start preparingcards for all the persons of an old age home. In
order to complete one card, they take 10, 16 and 20 minutes respectively. If all of them started
together, after what time will they start preparing a new card together?
Ans: Given, Amita, Suneha and Raghav takes 10, 16 and 20 mins. respectively to complete
one card.
Prime factorisation of 10, 20 and 16 is
10 = 2 × 5, 16 = 2 × 2 × 2 × 2 = 24,
20 = 2 × 2 × 5 = 22 × 5
∴ Time after which they start preparing a new card together = LCM (10, 16, 20)
= 24 × 5 = 80 mins.
Therefore, after 80 mins, they will start preparing a new card together.

12. In the figure XY and X'Y' are two parallel tangents to a circle with centre O and another
tangent AB with point of contact C interesting XY at A and X'Y' at B, what is the measure of
∠AOB.

Ans: Join OC. Since, the tangents drawn to a circle from an external point are equal.
∴ AP = AC

In Δ PAO and Δ AOC, we have:

AO = AO [Common]
OP = OC [Radii of the same circle]
AP = AC
⇒ Δ PAO ≅ Δ AOC [SSS Congruency]
∴ ∠PAO = ∠CAO = ∠1
∠PAC = 2 ∠1 ...(1)
Similarly ∠CBQ = 2 ∠2 ...(2)
Again, we know that sum of internal angles on the same side of a transversal is 180°.
∴ ∠PAC + ∠CBQ = 180°
⇒ 2 ∠1 + 2 ∠2 = 180° [From (1) and (2)]
⇒ ∠1 + ∠2 = 180°/2 = 90° ...(3)
Also ∠1 + ∠2 + ∠AOB = 180° [Sum of angles of a triangle]
⇒ 90° + ∠AOB = 180° ⇒ ∠AOB = 180° − 90°
⇒ ∠AOB = 90°.

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13. How many terms of the A.P. 18, 16, 14, ... be taken so that their sum is zero?
Ans: Let a be the first term and d be the common difference of A.P.
n
Sum of n terms, S n  [2a  (n  1)d ]
2
n
 [(2  18)  (n  1)(2)]  0
2
Here, a = 18, d = 16 – 18 = –2
⇒ n[18 – n + 1] = 0 ⇒ n = 19 ( n ≠ 0)
∴ Sum of 19 terms of the A.P. is zero.

14. Two dice are thrown at the same time. What is the probability that the sum of the two numbers
appearing on the top of the dice is
(i) 8? (ii) 7? (iii) less than or equal to 12?
Ans: (i) Number of outcomes with sum of the numbers 8 = 5
∴ Required Probability = 5/36
(ii) Number of outcomes with sum of the numbers 7 = 6
∴ Required Probability = 6/36 = 1/6
(iii) Number of outcomes with sum of the numbers less than or equal to 12 = 36
∴ Required Probability = 36/36 = 1

15. Find the ratio in which the point (–3, k) divides the line segment joining the points (–5, –4)
and (–2, 3). Also find the value of k.
Ans: Let the ratio be m : 1 then we have
m1 x2  m2 x1 m  (2)  (1)(5)
x  3 
m1  m2 m 1
 3(m  1)  2m  5  3m  3  2m  5  m  2
m y  m2 y1 2  (3)  1(4) 64
Now, k  1 2 k k
m1  m2 2 1 3
2
k
3

16. The median of the following data is 868. Find the values of x and y, if the total frequency is
100
Class Frequency
800 – 820 7
820 – 840 14
840 – 860 x
860 – 880 25
880 – 900 y
900 – 920 10
920 – 940 5
Ans:
Class Frequency Frequency
800 – 820 7 7
820 – 840 14 21
840 – 860 x x + 21
860 – 880 25 x + 46
880 – 900 y x + y + 46
900 – 920 10 x + y + 56
920 – 940 5 x + y + 61

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 From table, we have x + y + 61 = 100 ⇒ x + y = 100 – 61 ⇒ x + y = 39
Here, median = 868, therefore median class is 860 – 880
So, l = 860, cf = x + 21, f = 25, h = 20, n/2 = 50
n 
 2  cf   50  ( x  21) 
Now, Median  l    h   868  860    20 
 f   25 
 
 50  x  21)  29  x
 868  860    4  8  4
 5  5
 40  (29  x )4  29  x  10  x  29  10  19
 y  39  19  20

17. Prove that “If a line is drawn parallel to one side of a triangle to intersect the other two sides
in distinct points, the other two sides are divided in the same ratio.”
AD 3
In ∆ABC, DE||BC and  if EA = 6.6cm, then find AC using the above theorem.
DB 1
Ans: For Theorem:
Given, To Prove, Constructions and Figure – 1 ½ mark
Proof – 1 ½ mark
To find the value of AC = 8.8 cm – 1 mark
AD AE 3 6.6
    EC  2.2cm
DB EC 1 EC
 AC  AE  EC  6.6  2.2  8.8cm

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