Reasoning Aptitude > Data Interpretation
TABULATION MCQs
Table Charts
Total Questions : 185
| Page 16 of 19 pages
Answer: Option B. -> 2720
Total marks
= 40 × 68
= 2720
Total marks
= 40 × 68
= 2720
Answer: Option D. -> 60
$$\eqalign{
& \text{Average in four subjects,} \cr
& = \frac{56+68+68+48}{4} \cr
& = \frac{240}{4} \cr
& = 60 \cr} $$
$$\eqalign{
& \text{Average in four subjects,} \cr
& = \frac{56+68+68+48}{4} \cr
& = \frac{240}{4} \cr
& = 60 \cr} $$
Answer: Option D. -> D
Here,
A = 820, B = 830, C = 880, D = 700 and E = 905
Hence, The school, which has the maximum number of players is E.
Here,
A = 820, B = 830, C = 880, D = 700 and E = 905
Hence, The school, which has the maximum number of players is E.
Answer: Option B. -> 35
Required difference,
= (200 + 195 + 245 + 200 + 225) - (150 + 200 + 250 + 230 + 200)
= 1065 - 1030
= 35
Required difference,
= (200 + 195 + 245 + 200 + 225) - (150 + 200 + 250 + 230 + 200)
= 1065 - 1030
= 35
Answer: Option B. -> 19$$\frac{4}{9}$$%
$$\eqalign{
& = \left(\frac{175}{900}\times100\right)\% \cr
& = \frac{175}{9}\% \cr
& = 19\frac{4}{9}\% \cr} $$
$$\eqalign{
& = \left(\frac{175}{900}\times100\right)\% \cr
& = \frac{175}{9}\% \cr
& = 19\frac{4}{9}\% \cr} $$
Answer: Option C. -> 50
$$\eqalign{
& x\% \text{ of } 130 = 65 \cr
& \Rightarrow \frac{x\times130}{100} = 65 \cr
& \Rightarrow x = 50 \cr} $$
$$\eqalign{
& x\% \text{ of } 130 = 65 \cr
& \Rightarrow \frac{x\times130}{100} = 65 \cr
& \Rightarrow x = 50 \cr} $$
Answer: Option D. -> 805
Average number of candidates appeared from T all over the years
$$\eqalign{
& = \frac{\left(6.2+6.2+6.4+7.8+9.9+11.8\right)\times100}{6} \cr
& = \frac{48.3\times100}{6} \cr
& = \frac{4830}{6} \cr
& = 805 \cr} $$
Average number of candidates appeared from T all over the years
$$\eqalign{
& = \frac{\left(6.2+6.2+6.4+7.8+9.9+11.8\right)\times100}{6} \cr
& = \frac{48.3\times100}{6} \cr
& = \frac{4830}{6} \cr
& = 805 \cr} $$
Answer: Option A. -> 4 : 7
$$\eqalign{
& \text{Required ratio} \cr
& = \frac{3.2}{5.6} \cr
& = \frac{32}{56} \cr
& = \frac{4}{7} \cr
& = 4:7 \cr} $$
$$\eqalign{
& \text{Required ratio} \cr
& = \frac{3.2}{5.6} \cr
& = \frac{32}{56} \cr
& = \frac{4}{7} \cr
& = 4:7 \cr} $$
Answer: Option A. -> P
Total number of candidates who qualified the test in 2009 and 2010 from various zones are:
P → (4.8 + 5.6) = 10.4
Q → (5.2 + 6.4) = 11.6
R → (6.8 + 7.4) = 14.2
S → (5.2 + 11.4) = 16.6
T → (6.9 + 9.4) = 16.3
Total number of candidates who qualified the test in 2009 and 2010 from various zones are:
P → (4.8 + 5.6) = 10.4
Q → (5.2 + 6.4) = 11.6
R → (6.8 + 7.4) = 14.2
S → (5.2 + 11.4) = 16.6
T → (6.9 + 9.4) = 16.3
Answer: Option B. -> 2007
In zone S, the difference between the appeared candidates and the qualified candidates in various years is given below:
2005 → (4.2 - 2.4) = 1.8
2006 → (7.4 - 3.3) = 4.1
2007 → (8.3 - 5.6) = 2.7
2008 → (9.3 - 6.4) = 2.9
2009 → (11.4 - 5.2) = 6.2
2010 → (14.2 - 11.4) = 2.8
If was lowest in 2005 and second lowest in 2007
In zone S, the difference between the appeared candidates and the qualified candidates in various years is given below:
2005 → (4.2 - 2.4) = 1.8
2006 → (7.4 - 3.3) = 4.1
2007 → (8.3 - 5.6) = 2.7
2008 → (9.3 - 6.4) = 2.9
2009 → (11.4 - 5.2) = 6.2
2010 → (14.2 - 11.4) = 2.8
If was lowest in 2005 and second lowest in 2007
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