Mathematics Test

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Mathematics Test - 63

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Weekly Quiz Competition

Question 1
1 / -0

The average of 25 results is 18. The average of first 12 of those is 14 and the average of last 12 is 17. What is the 13th result?

A
74

B
75

C
69

D
78

Solution

Sum of 1st 12 results = 12 × 14
Sum of last 12 results = 12 × 17
13^th result = x (let)
Now,
12 × 14 + 12 × 17 + x = 25 × 18
Or, x = 78
Question 2
1 / -0

Find the number of triangles which can be formed by joining the angular points of a polygon of 8 sides as vertices.

A
56

B
24

C
16

D
8

Solution

A triangle needs 3 points.
And polygon of 8 sides has 8 angular points.
Hence, number of triangle formed,
=⁸C₃
=8×7×6/1×2×3
=56
Question 3
1 / -0

Two dice are thrown simultaneously. What is the probability of getting two numbers whose product is even?

A
1/2

B
3/4

C
3/8

D
5/16

Solution

In a simultaneous throw of two dice, we have n(S) = (6 x 6) = 36
Then, E = {(1, 2), (1, 4), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 2), (3, 4), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 2), (5, 4), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}
∴n (E)=27
∴P (E)=n(E)/n(S)
=27/36
=34
Question 4
1 / -0

If \(a+b+c=6\) and \(a^{2}+b^{2}+c^{2}=38\), then what is the value of

\(a\left(b^{2}+c^{2}\right)+b\left(c^{2}+a^{2}\right)+c\left(a^{2}+b^{2}\right)+3 a b c ?\)

A
3

B
6

C
-6

D
'-3

Solution

\(a+b+c=6\) and \(a^{2}+b^{2}+c^{2}=38\)
As we know, there are three variable but only 2 equation is given so 1 variable will be \(0,\) so
Put \(c=0,\) then
\(a+b=6\) and \(a^{2}+b^{2}=38\)
As we know,
\((a+b)^{2}=a^{2}+b^{2}+2 a b\)
\(\Rightarrow 6^{2}=38+2 a b\)
\(\Rightarrow 2 a b=36-38\)
\(\Rightarrow 2 a b=-2\)
\(\Rightarrow a b=-2 / 2\)
\(\Rightarrow a b=-1\)
Now,
\(a\left(b^{2}+c^{2}\right)+b\left(c^{2}+a^{2}\right)+c\left(a^{2}+b^{2}\right)+3 a b c\)
Put \(c=0,\) then
\(\Rightarrow a b^{2}+b a^{2}\)

\(\Rightarrow a b(a+b)\)
\(\Rightarrow(-1) \times 6\)
\(\Rightarrow(-6)\)
Question 5
1 / -0

Three gentlemen and three ladies are candidates for two vacancies. A voter has to vote for two candidates. In how many ways can one cast his vote?

A
9

B
30

C
36

D
15

Solution

There are 6 candidates and a voter has to vote for any two of them.
So, the required number of ways is,
=⁶C₂
=6!/2!×4!
=15
Question 6
1 / -0

In a party every person shakes hands with every other person. If there are 105 hands shakes, find the number of person in the party.

A
15

B
14

C
21

D
25

Solution

Let n be the number of persons in the party Number of hands shake \(=105\) Total number of hands shake is given by \({ }^{n} C_{2}\) Now, According to the question, \({ }^{n} \mathrm{C}_{2}=105\)
or, \(\frac{n !}{2 ! \times(n-2) !}=105\)
or, \(\frac{n \times(n-1)}{2}=105\)
or, \(n^{2}-n=210\)
or, \(n^{2}-n-210=0\)
or, \(n=15,-14\) But, we cannot take negative value of \(n\)
So, \(n=15\) i.e. number of persons in the party \(=15\)
Question 7
1 / -0

In the next World cup of cricket there will be 12 teams, divided equally in two groups. Teams of each group will play a match against each other. From each group 3 top teams will qualify for the next round. In this round each team will play against each others once. Four top teams of this round will qualify for the semifinal round, where they play the best of three matches. The Minimum number of matches in the next World cup will be:

A
54

B
53

C
38

D
43

Solution

The number of matches in first round,
\(={ }^{6} \mathrm{C}_{2}+{ }^{6} \mathrm{C}_{2}\)
Number of matches in next round,
\(={ }^{6} \mathrm{C}_{2}\)
Number of matches in semifinals,
\(={ }^{4} \mathrm{C}_{2}\)
Total number of matches,
\(={ }^{6} \mathrm{C}_{2}+{ }^{6} \mathrm{C}_{2}+{ }^{6} \mathrm{C}_{2}+{ }^{4} \mathrm{C}_{2}+2\)
\(=53\)
Question 8
1 / -0

If α + β = 90°, then the value of (1 - sin²α)(1 - cos²α) × (1 - cot²β)(1 - tan²β) is?

A
1

B
-1

C
0

D
2

Solution

\(\left(1-\sin ^{2} \alpha\right)\)
\(\left(1-\cos ^{2} \alpha\right)\)
\(\times\left(1-\cot ^{2} \beta\right)\)
\(\left(1-\tan ^{2} \beta\right)\)
\(\Rightarrow\left(\cos ^{2} \alpha\right)\left(\sin ^{2} \alpha\right)\)
\(\times\left(\operatorname{cosec}^{2} \beta\right)\)
\(\left(\sec ^{2} \beta\right)\)
Put \(\rightarrow \beta=45^{\circ}\)
\(\Rightarrow \cos ^{2} 45^{\circ}\)
\(\cdot \sin ^{2} 45^{\circ}\)
\(\cdot \operatorname{cosec}^{2} 45^{\circ}\)
\(\cdot \operatorname{sce}^{2} 45^{\circ}\)
\(\Rightarrow 1\)
\(2 \cdot \frac{1}{2} \cdot 2 \cdot 2\)
\(\Rightarrow 1\)

Alternate:
\(\left(1-\sin ^{2} \alpha\right)\)
\(\left(1-\cos ^{2} \alpha\right)\)
\(\times\left(1-\cot ^{2} \beta\right)\)
\(\left(1-\tan ^{2} \beta\right)\)
\(\Rightarrow\left(\cos ^{2} \alpha\right)\left(\sin ^{2} \alpha\right)\)
\(\times\left(\operatorname{cosec}^{2} \beta\right)\)
\(\left(\sec ^{2} \beta\right)\)
\(\Rightarrow \cos ^{2}\left(90^{\circ}-\beta\right)\)
\(\sin ^{2} \alpha\)
\(\cdot \operatorname{cosec}^{2} \beta\)
\(\sec ^{2}\left(90^{\circ}-\alpha\right.\)
\(\Rightarrow \sin ^{2} \beta\)
\(\cdot \operatorname{cosec}^{2} \beta\)
\(\cdot \sin ^{2} \alpha\)
Question 9
1 / -0

A tank is filled by three pipes with uniform flow. The first two pipes operating simultaneously fill the tank in the same time during which the tank is filled by the third pipe alone. The second pipe fills the tank 5 hours faster than the first pipe and 4 hours slower than the third pipe. The time required by the first pipe is:

A
6 hours

B
10 hours

C
15 hours

D
30 hours

Solution

Suppose, first pipe alone takes \(x\) hours to fll the tank.
Then,
Second and third pipes will take \((x-5)\) and \((x-9)\) hours respectively to fill the tank.

\(\therefore \frac{1}{x}+\frac{1}{(x-5)}=\frac{1}{(x-9)}\)
\(\Rightarrow \frac{x-5+x}{x(x-5)}=\frac{1}{(x-9)}\)
\(\Rightarrow(2 x-5)(x-9)=x(x-5)\)
\(\Rightarrow x^{2}-18 x+45=0\)
\((x-15)(x-3)=0\)
\(\Rightarrow x=15 \quad \text { [neglecting } x=3]\)
Question 10
1 / -0

If log₁₀7 = a, then log₁₀(1/70) is equal to

A
- (1 + a)

B
(1 + a)^-1

C
a/10

D
1/10a

Solution

\(\log _{10}\left(\frac{1}{70}\right)\)
\(=\log _{10} 1-\log _{10} 70\)
\(=-\log _{10}(7 \times 10)\)
\(=-\left(\log _{10} 7+\log _{10} 10\right)\)
\(=-(a+1)\)

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Re-Attempt Weekly Quiz Competition

Mathematics Test - 63

Result

Mathematics Test - 63

Score

-

Rank

-

SHARING IS CARING

Question 1

74

75

69

78

Solution

Question 2

56

24

16

8

Solution

Question 3

1/2

3/4

3/8

5/16

Solution

Question 4

3

6

-6

'-3

Solution

Question 5

9

30

36

15

Solution

Question 6

15

14

21

25

Solution

Question 7

54

53

38

43

Solution

Question 8

1

-1

0

2

Solution

Question 9

6 hours

10 hours

15 hours

30 hours

Solution

Question 10

- (1 + a)

(1 + a)-1

a/10

1/10a

Solution

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(1 + a)^-1