Mathematics Test 159

Result

Mathematics Test 159

Score
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Weekly Quiz Competition

Question 1
4 / -1

There are several tea cups in the kitchen, some with handles and other without handles. The number of ways of selecting one cup without a handle and one cup with a handle is exactly 36. Then the minimum possible numbers of cups in the kitchen is equal to

A
20

B
12

C
10

D
28

Solution

Let us assume that no. of cups with handle is n and without handle is m. No. of ways of selecting one cup each is ^mC₁ . ⁿC₁ = 36, mn = 36
Question 2
4 / -1

The sum of values of such that the equation (x + 10) (x – a) + 5 = 0 has integral roots

A
–4

B
-16

C
-20

D
0

Solution

(x + 10) (x – a) = –5

Case (i) Either x + 10 = 5, x – a = –1

x = 5, –5 – a = –1

a = –4

or x + 10 = –1, x – a = 5

x = – 11, –11 – a = 5

a = –16

Case (ii) x + 10 = –5, x – a = 1

or x + 10 = 1, x – a = –5 it gives same values of 'a'.

Sum of values of 'a' is –16 – 4 = –20
Question 3
4 / -1

for x, y, z > 0 then the least value of (x – 1)(y – 1)(z – 1) is

A
3

B
6

C
0

D
8

Solution
Question 4
4 / -1

The number of points in square |x| + |y| = 2 which lie on the curve y = x + cosx and at which the tangent to the curve is parallel to the x-axis is

A
0

B
2

C
4

D
8

Solution
Question 5
4 / -1

Let f(x) = max{|x|, x – [x]}, where [.] denotes greatest integer function, then is equal to

A
1/4

B
3/4

C
5/4

D
7/4

Solution
Question 6
4 / -1

Length of the line segment joining the feet of the perpendicular drawn from the point (3, 4) on the pair of lines x² – 4xy + y² = 0 is

A
5√3 / 2

B
5√3

C
2√3

D
√3

Solution
Question 7
4 / -1

Let z₁, z₂, z₃ be the complex numbers such that |z₁ – 1| = |z₂ – 1| = |z₃ – 1| and then z₂(z₂ – 1) –z₃(z₂ + 1) + (z₃ + 1) (z₃ – 1) =

A
0

B
1

C
2

D
-2

Solution
Question 8
4 / -1

A
0

B
1

C
2

D
More than 2

Solution
Question 9
4 / -1

The curve satisfying the differential equation (2x²y – 2y⁴)dx + (2x³ + 3xy³)dy = 0 and passing through (1, 1) is given by then the value of m + n is

A
5

B
4

C
3

D
6

Solution
Question 10
4 / -1

The real no. x and y satisfying (x – 4)² + (y – 5)² = 7² then the minimum value of

A
1

B
2

C
3

D
4

Solution