| **Answers & Solutions ** | | 1 . | Answer : Option B | Explanation : | Reqd ratio = $980 over 1120$ = $7over 8 $ = 7 : 8 | | | 2 . | Answer : Option D | Explanation : | Average no.of students in college A = $5610 \over 6$ = 935
Average no.of students in college C = $5490 \over 6$ = 915
Reqd difference = 935 - 915 = 20 | | | 3 . | Answer : Option D | Explanation : | Total number of students in College B = 5810
Total number of students in College D = 5598
Reqd difference = 5810 – 5598 = 212
| | | 4 . | Answer : Option D | Explanation : | Average no.of students in College E = $5880\over 6$ = 980 | | | 5 . | Answer : Option A | Explanation : | | | | 6 . | Answer : Option D | Explanation : | For 2 years CI at 8%
= 8 + 8 + $8 \times 8 \over 100$
= 16.64%
CI = 3980 x $16.64 \over 100$ = 662.2 $\cong$ $665$ | | | 7 . | Answer : Option D | Explanation : | $a\over 6$ = $b\over 10$ = $c\over 4$ = 1
a = 6k , b = 10k, c = 4k
$a + b + c \over c$ = $(6 + 10 + 4)k\over 4k$ = 5 | | | 8 . | Answer : Option C | Explanation : | 7x + 8y = 53 ... (i)
5x + 12y = 63 ... (ii)
On solving the equations, we get
x = 3, y = 4
y – x = 4 – 3 = 1
| | | 9 . | Answer : Option D | Explanation : | Time = $14 \over 5$ + $14 \over 10$ = $42 \over 10$h = 4.2h | | | 10 . | Answer : Option A | Explanation : | Let the numbers be x, (x + 1), (x + 2) and (x + 3).
Then, x + (x + 3) = 103
or, x = 50
Hence, product of B and C
= (50 + 1) (50 + 2) = 51 × 52 = 2652
| | | | |

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