Mathematics Test

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Mathematics Test - 1

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Weekly Quiz Competition

Question 1
1 / -0

For acute angle θ, find value of sec²θ − tan²θ

A
1

B
2

C
3

D
0

Solution
Question 2
1 / -0

The average weight of 45 passenger on board an aircraft is 50 kg. If the weight of 5 members of the crew is added, the average is reduced by half kilogram . What is the average weight of the crew members?

A
45 kg

B
48 kg

C
58 kg

D
68 kg

Solution

Total weight of 45 passenger = 45 x 50 = 2250 kg

Total weight of 45 passenger and 5 crews = 50 x 49.5 = 2475 kg

Total weight of 5 crews = 2475-2250 = 225kg

Average weight of 5 crews = 225/5 = 45kg.
Question 3
1 / -0

A gardener had a number of shrubs to plant in rows . At first he tried to plant 8, then 12 and then 16 in a row but he always had 3 shrubs left with him . On trying 7 shrubs he was left with none.Find the total number of shrubs.

A
149

B
159

C
156

D
147

Solution

L.C.M of 8,12,16 = 48

Now, 48 x 1 + 3 = 51 -not divisible by 7

48 x 2 + 3 = 99 - not divisible by 7

48 x 3 + 3 = 147 - not divisible by 7

Required number = 147
Question 4
1 / -0

If tan ⁡α/3 = 1/2, then the value of tan ⁡α + cot ⁡α is:

A
11/23

B
12/23

C
121/46

D
125/22

Solution
Question 5
1 / -0

A box contains 5 black and 4 white balls. A ball is drawn at random and its colour is noted. The ball is then put back in the box along with two additional balls of its opposite colour. If a ball is drawn again from the box, then the probability that the ball drawn now is black, is

A
7/11

B
5/11

C
53/99

D
48/99

Solution

P(B) = 59⋅511 + 49 − 711 = 5399
Question 6
1 / -0

Let a, b, c, d and e be distinct positive numbers. If a, b, c and 1/c, 1/d, 1/e both are in A.P. and b, c, d are in G.P. then

A
a, c, e are in G.P.

B
a, b, e are in G.P.

C
a, b ,e are in A.P.

D
a, c, e are in A.P.

Solution
Question 7
1 / -0

A code word of length 4 consists of two distinct consonants in the English alphabet followed by two digits from 1 to 9, with repetition allowed in digits. If the number of code words so formed ending with an even digit is 432 k, then k is equal to :

A
7

B
5

C
49

D
35

Solution

Number of consonants = 21

Number of given digits = 9

so total number formed = 21 × 20 × 9 × 4 = 432k

⇒ k = 35
Question 8
1 / -0

If f is a function of real variable x satisfying f (x + 4) – f (x + 2) +f(x) = 0, then f is a periodic functionwith period:

A
6

B
8

C
10

D
12

Solution

We have , f(x+4)-f(x+2)+f(x)=0 for all x .

Replacing x by x +2, we get

f(x+6)-f(x+4)+f(x+2)=0 for all x

Adding (i) and (ii) ,we get

f(x+6)+f(x)=0 …….(iii)

Replacing x by x+6 , we get

f(x+12)+f(x+6)=0 ....(iv)

From (iii) and (iv), we obtain

f(x+12)=f(x) for all x

Therefore, f(x) is periodic with period 12.
Question 9
1 / -0

One of the two events, A and B must occur. If P (A) = (2/3) P(B), the odds in favour of B are:

A
1 : 2

B
3 : 2

C
2 : 3

D
3 : 5

Solution
Question 10
1 / -0

Find the least number of five digits which is exactly divisible by 12, 15 and 18.

A
1080

B
10080

C
10025

D
11080

Solution

The least number of 5 digits is 10000.

L.C.M. of 12, 15 and 18 is 180.

On dividing 10000 is 100.

=> 10000 + 180- 100 = 10080

10080 is divisible by 180.