| Largest amount in dollars that an one student could have | 300 |
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Solution
Total amount = $600. Three students having no savings.
∴ 7 students have a total saving of $600.
∴ Maximum amount in $ that one student can have = 600 - (25 × 5 + 130) = 600 - (125 + 130) = $345
| The product of integers from -5 to 3 | The product of integers from -4 to 4 |
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Solution
Each column has 'zero' as one of the integers.
∴ Product in both cases is zero.
| Smallest number reater than 100 in the seuecence | 120 |
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Solution
The terms are 1,2,5,14,41,122 122 >120
| 4⁄5 of x | 0 |
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Solution
\(\frac{2x}{3}=\frac{3x}{4}\) 2⁄3 ≠ 3⁄4
∴ x = 0 ∴ Col. A = 0
| pq | p + q |
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Solution
Plug actual nos.-
(1.1) (1.2) = 1.32 → Col. A
1.1 + 1.2 = 2.3 → Col. B
Addition of 2 fractions is always more than its multiplication.
| (m + 1)(m -1) | 28/42 |
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Solution
Col. A = m2 = 17 ∴ m2 - 1 = 16
Col. B = 28/24 = 24 = 16
| Number of prime factor of p | Number of prime factor of 2p |
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Solution
Each factor of p is a factor of 2p. But 2 is a factor only of 2p. Since p is an odd positive integer, 2 is not the factor of p.
∴2p has more prime factors than p
Shaded region above is bounded by 2 semicircles and two sides of square PQRS. Perimeter of shaded region is 4 + 2π.
| Area of shaded region | 1 |
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Solution
PQ + RS + Zπr = perimeter of shaded region.
∴ 2PQ + 2πr = 4 + 2π ∴ r = 1,PQ = 2
Area of shaded region = area of square - area of circle
= (4 - π) <1
| Number of males at picnic | Number of females at picnic |
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Solution
Let the no. of males and females be 3x and 2x respectively.
Col. A = 20% of 3x = 0.6x
Col. B = 30% of 2x = 0.6x
| a+b | 67 |
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Solution
If 'a' and 'b' are integers, the greatest value for a + b = 65.
But they need not be integers a = 1⁄10, b = 640
∴ a + b = 640.1