Numerical Ability
 A container originally contains 10 litres of pure spirit. From this container 1 litre of spirit is replaced with 1 litre of water. Subsequently, 1 litre of the mixture is again replaced with 1 litre of water and this processes is repeated one more time. How much spirit is now left in the container?

 7.58 litres
 7.84 litres
 7 litres
 7.29 litres
 7.58 litres

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10 10  1 ^{3} = 10 9 ^{3} 10 10 = 729 1000 ∴ 729 × 1 = 7.29 litres 1000
Correct Option: D
10 10  1 ^{3} = 10 9 ^{3} 10 10 = 729 1000 ∴ 729 × 1 = 7.29 litres 1000

If Log(P) = 1 Log(Q) = 1 Log(P), then which of the following options is TRUE ? 2 3

 P^{2} = Q^{3}R^{2}
 Q^{2} = PR
 Q^{2} = R^{3}P
 R = P^{2}Q^{2}
 P^{2} = Q^{3}R^{2}

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log P = 1 log Q 2 log P = 1 log (R) = k 3
∴ P = b^{k} , Q = b^{2k} , R = b^{3k}
Now , Q^{2} = b^{4k} = b^{3k}b^{k} = PRCorrect Option: B
log P = 1 log Q 2 log P = 1 log (R) = k 3
∴ P = b^{k} , Q = b^{2k} , R = b^{3k}
Now , Q^{2} = b^{4k} = b^{3k}b^{k} = PR
 Hari (H), Gita (G), Irfan (I) and Saira (S) are siblings (i.e. brothers and Sisters). All were born on 1st January. The age difference between any two successive siblings (that is born one after another) is less than 3 years. Given the following facts: 1. Hari's age + Gita's age > lrfan's age + Saira's age. 2. The age difference between Gita and Saira is 1 year. However, Gita is not the oldest and Saira is not the youngest. 3. There are no twins. In what order were they born (oldest first)?

 HSIG
 SGHI
 IGSH
 IHSG
 HSIG

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(i) H + G > I + S
⇒ I – H < G – S ...(i)
(ii) G – S = 1 ...(ii)
From equations (i) and (ii), we get
I – H < 1
⇒ H > 1
Hence order is SGHI because G is not oldest and S is not youngest.Correct Option: B
(i) H + G > I + S
⇒ I – H < G – S ...(i)
(ii) G – S = 1 ...(ii)
From equations (i) and (ii), we get
I – H < 1
⇒ H > 1
Hence order is SGHI because G is not oldest and S is not youngest.
 Given digits 2, 2, 3, 3, 3, 4, 4, 4, 4 how many distinct 4 digit numbers greater than 3000 can be formed?

 50
 51
 52
 54
 50

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Case 1: First Digit = 3 ⇒ 3...
Rest 3 digits may be combination of
234 → 3!223 → 3! 2! 224 → 3! 2! 332 → 3! 2! 334 → 3! 2!
442 → 3! 2! 443 → 3! 2!
444 → 1
∴ Total combinations = 25
Case 2: First Digit = 4 ⇒ 4...
Rest 3 digits may be combination of
234 → 6
223 → 3
224 → 3
332 → 3
334 → 3
442 → 3
443 → 3
444 → 1
333 → 1
∴ Total combinations = 26
∴ Required answer = 25 + 26 = 51Correct Option: B
Case 1: First Digit = 3 ⇒ 3...
Rest 3 digits may be combination of
234 → 3!223 → 3! 2! 224 → 3! 2! 332 → 3! 2! 334 → 3! 2!
442 → 3! 2! 443 → 3! 2!
444 → 1
∴ Total combinations = 25
Case 2: First Digit = 4 ⇒ 4...
Rest 3 digits may be combination of
234 → 6
223 → 3
224 → 3
332 → 3
334 → 3
442 → 3
443 → 3
444 → 1
333 → 1
∴ Total combinations = 26
∴ Required answer = 25 + 26 = 51
 5 skilled workers can build a wall in 20 days; 8 semiskilled workers can build a wall in 25 days; 10 unskilled workers can build a wall in 30 days. If a team has 2 skilled, 6 semiskilled and 5 unskilled workers, how long will it take to build the wall?

 20 days
 18 days
 16 days
 15 days
 20 days

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5 skilled workers build wall in 20 days
∴ 1 skilled worker build wall in 20 × 5 daysHence in 1 day,part of work done by skilled work = 1 100 Similarly in 1 day part of work done by semiskilled workers = 1 25 × 8 and in 1 day part of work done by unskilled worker = 1 30 × 10 So part of work done in 1 day by 2 skilled, 6 semiskilled and 5 unskilled = 2 + 6 + 5 = 1 100 200 300 15
So work done by given workers in days = 15
Correct Option: D
5 skilled workers build wall in 20 days
∴ 1 skilled worker build wall in 20 × 5 daysHence in 1 day,part of work done by skilled work = 1 100 Similarly in 1 day part of work done by semiskilled workers = 1 25 × 8 and in 1 day part of work done by unskilled worker = 1 30 × 10 So part of work done in 1 day by 2 skilled, 6 semiskilled and 5 unskilled = 2 + 6 + 5 = 1 100 200 300 15
So work done by given workers in days = 15