The bar chart shows the distribution of marks in a class test. How many students took the test?
15
20
25
30
Correct answer is B
Number f students that took the test = \(\sum f\). Where f is the frequencies
= 2 + 5 + 0 + 3 + 4 + 0 + 0 + 2 + 3 + 1 + 2 = 20
From the figure, calculate TH in centimeters
\(\frac{5}{\sqrt{3} + 1}\)
\(\frac{5}{\sqrt{3} - 1}\)
\(\frac{5}{\sqrt{3}}\)
\(\frac{\sqrt{3}}{5}\)
Correct answer is B
TH = tan 45o, TH = QH
\(\frac{TH}{5 + QH}\) = tan 30o
TH = (b + QH) tan 30o
QH = 56 (5 + QH) \(\frac{1}{\sqrt{3}}\)
QH(1 - \(\frac{1}{\sqrt{3}}\)) = \(\frac{5}{\sqrt{3}}\)
QH = \(\frac{5\sqrt{3}}{\sqrt{3}} - \frac{1}{\sqrt{3}}\)
= \(\frac{5}{\sqrt{3} - 1}\)
6cm
7cm
8cm
9cm
Correct answer is C
Since area of square PQRS = 100cm2
each lenght = 10cm
Also TUYS : XQVU = 1 : 16
lengths are in ratio 1 : 4, hence TU : UV = 1: 4
Let TU = x
UV = 1: 4
hence TV = x + 4x = 5x = 10cm
x = 2cm
TU = 2cm
UV = 8cm
But TU = SY and UV = YR
YR = 8cm
7 \(\sqrt{3}\)cm
12.9cm
\(\sqrt{87}\)cm
7cm
Correct answer is B
SQ2 + OS2 = OQ2 + 52 = 142
SQ2 = 142 - 52
196 - 25 = 171
ST2 + TQ2 = SQ2
ST2 + 22 = 171
ST2 = 171 - 4
= 167
ST = \(\sqrt{167}\)
= 12.92 = 12.9cm
In the diagram, QPS = SPR, PR = 9cm. PQ = 4cm and QS = 3cm, find SR.
6\(\frac{3}{4}\)cm
3\(\frac{3}{8}\)cm
4\(\frac{3}{8}\)cm
2\(\frac{3}{8}\)cm
Correct answer is A
Using angle bisector theorem: line PS bisects angle QPR
QS/QP = SR/PR
3/4 = SR/9
4SR = 27
SR = \(\frac{27}{4}\)
= 6\(\frac{3}{4}\)cm
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