| Marks | 5 - 7 | 8 - 10 | 11 - 13 | 14 - 16 | 17 - 19 | 20 - 22 |
| Frequency | 4 | 7 | 26 | 41 | 14 | 8 |
The table above shows the marks obtained by 100 pupils in a test. Find the upper class boundary of the class containing the third quartile.
13.0
16.0
16.5
22.5
Correct answer is C
The third quartile can be found in the (3N4)th position
= (3×1004)th position
This is found in the class 14 - 16.
The upper class boundary = 16.5
-11
-9
-3
4
Correct answer is C
Using the remainder theorem, the remainder when a polynomial ax2+bx+c is divided by (x−a) is equal to f(a).
2x3+3x2+qx−1 divided by (x+2), the remainder = f(−2)
⟹f(−2)=f(1)
f(−2)=2(−23)+3(−22)+q(−2)−1=−16+12−2q−1=−5−2q
f(1)=2(13)+3(12)+q(1)−1=2+3+q−1=4+q
4+q=−5−2q⟹4+5=−2q−q=−3q
q=−3
Find the value of p for which x2−x+p becomes a perfect square.
−12
14
12
1
Correct answer is B
The equation ax2+bx+c is a perfect square if b2=4ac.
x2−x+p
(−1)2=4(1)(p)
1=4p⟹p=14
k=825
k=258
k<258
k>258
Correct answer is C
No explanation has been provided for this answer.
18.75 N
15.75 N
9.50 N
8.66 N
Correct answer is D
F_{1} = (5 N, 060°) = 5 \cos 60 i + 5 \sin 60 j = 2.5 i + \frac{5\sqrt{3}}{2}j
F_{2} = (10 N, 180°) = 10 \cos 180 i + 10 \sin 180 j = -10i
R = F_{1} + F_{2} = -7.5 i + \frac{5\sqrt{3}}{2}j
|R| = \sqrt{(-7.5)^{2} + (\frac{5\sqrt{3}}{2})^{2}}
= \sqrt{56.25 + 18.75} = \sqrt{75}
= 8.66 N