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Question 1 of 10

1. Question

1 points

Category: Quantitative Aptitude

The average weight of all the 11 players of RCB is 49 kg. If the average of first six lightest weight players of RCB is 49 kg and that of the six heaviest players of RCB is 52 kg. The average weight of the player which lies in the sixth position in the list of players when all the 11 players of RCB are arranged in the order of increasing or decreasing weights.

Correct

Answer – 4. 67 kg
Explanation:
Average of First six players = 49 * 6 = 294
Average of Last six players = 52 * 6 = 312; Average of all players = 49 * 11 = 539
Average weight of sixth player = 294 + 312 – 539 = 67

Incorrect

Answer – 4. 67 kg
Explanation:
Average of First six players = 49 * 6 = 294
Average of Last six players = 52 * 6 = 312; Average of all players = 49 * 11 = 539
Average weight of sixth player = 294 + 312 – 539 = 67

Question 2 of 10

2. Question

1 points

Category: Quantitative Aptitude

A container which contains a mixture of Sulfuric acid and water in ratio 13:4. 25.5 litres of mixture is taken out from the container and 2.5 litres of pure water and 5 litres of Sulfuric acid is added to the mixture. If resultant mixture contains 25% water, what was the initial quantity of mixture in the container before the replacement in litre?

Correct

Answer – 2. 68 litre
Explanation :
Quantity of Sulfuric acid = 13x
Quantity of water = 4x
Total = 17x
Resultant Mixture = 17x – 25.5 + 2.5 + 5 = 17x – 18
Resultant water = 4x – 25.5 * (4/17) + 2.5 = 4x – 3.5
Resultant mixture contains 25% water
(17x – 18)*25/100 = 4x – 3.5
x = 4
Initial quantity = 17*4 = 68

Incorrect

Answer – 2. 68 litre
Explanation :
Quantity of Sulfuric acid = 13x
Quantity of water = 4x
Total = 17x
Resultant Mixture = 17x – 25.5 + 2.5 + 5 = 17x – 18
Resultant water = 4x – 25.5 * (4/17) + 2.5 = 4x – 3.5
Resultant mixture contains 25% water
(17x – 18)*25/100 = 4x – 3.5
x = 4
Initial quantity = 17*4 = 68

Question 3 of 10

3. Question

1 points

Category: Quantitative Aptitude

The present age of Rohini is one-fourth that of her father. After 6 years, the father’s age will be twice the age of Kajol. If Kajol celebrated fifth birthday 8 year ago, What is the Present age of Rohini?

Correct

Answer – 3. 8 years
Explanation :
Kajol’s present age = 8 + 5 = 13
Kajol’s age after 6 years = 13 + 6 = 19
Kajol’s father age = 2 * 19 = 38
Father’s present age = 32
Rohini’s present age = 32 / 4 = 8

Incorrect

Answer – 3. 8 years
Explanation :
Kajol’s present age = 8 + 5 = 13
Kajol’s age after 6 years = 13 + 6 = 19
Kajol’s father age = 2 * 19 = 38
Father’s present age = 32
Rohini’s present age = 32 / 4 = 8

Question 4 of 10

4. Question

1 points

Category: Quantitative Aptitude

Initially, a shopkeeper had “Y” bangles. A customer bought 10% of bangles from “Y” then another customer bought 20% of the remaining bangles after that one more customer purchased 25% of the remaining bangles. Finally, shopkeeper is left with 270 bangles in his shop. How many bangles were there initially in his shop?

Ritesh sells his bike to Mukesh at a loss of 20% who subsequently sells it to Sharma at a profit of 25%. Sharma after finding some defect in the bike, returns it to Mukesh but could recover only Rs.4.50 for every Rs. 5 he had paid. Find the amount of Sharma’s loss if Ritesh had paid Rs.50,000 for the bike?

Syndicate Bank lent Rs. 10,000 to Dinesh 7% SI for 10 years. Meanwhile, the government implemented a scheme due to which interest rate reduced by 2%. By this Dinesh paid Rs.16,000 in total. Then after how many years after Dinesh took the loan, the government introduced the scheme?

Correct

Answer – 3. 5 years
Explanation :
6000 = 10000(7*x+5*(10-x))/100
x = 5

Incorrect

Answer – 3. 5 years
Explanation :
6000 = 10000(7*x+5*(10-x))/100
x = 5

Question 7 of 10

7. Question

1 points

Category: Quantitative Aptitude

A Minivan travelled to a place B from A, the first 50 km at 10 kmph faster than the usual speed, but it returned the same distance at 10 kmph slower than usual speed. If the total time taken by the Minivan is 12 hours, then how many hours will travel at the faster speed?

Correct

Answer – 3. 2 hours
Explanation :
Total time taken,
[50/(x-10)] + 50/(x +10)] = 12 hours.
By solving the equation, we get
x = 15
Time is taken by the Minivan at faster speed = 50/(15+10) = 2 hours.

Incorrect

Answer – 3. 2 hours
Explanation :
Total time taken,
[50/(x-10)] + 50/(x +10)] = 12 hours.
By solving the equation, we get
x = 15
Time is taken by the Minivan at faster speed = 50/(15+10) = 2 hours.

Question 8 of 10

8. Question

1 points

Category: Quantitative Aptitude

In a firecrackers company, there are equal number of women and children. Women work for 8 hours a day and children for 6 hours a day. During festival time, the work load goes up by 50%. The government rule does not allow children to work for more than 8 hours a day. If they are equally efficient and the extra work is done by women, then extra hours of work put in by women every day are?

Correct

Answer – 1. 5
Explanation :
Let extra hours a day are x.
According to the formula,
(M1D1T1) / W1 = (M2D2T2) / W2
⇒ [1 x 1 x (8 + 6)] / 1 = [1 x 1 x (8 + 8 + x)] / 3/2
⇒ (3/2) x 14 = 16 + x
⇒ 21= 16 + x
∴ x = 21 – 16 = 5

Incorrect

Answer – 1. 5
Explanation :
Let extra hours a day are x.
According to the formula,
(M1D1T1) / W1 = (M2D2T2) / W2
⇒ [1 x 1 x (8 + 6)] / 1 = [1 x 1 x (8 + 8 + x)] / 3/2
⇒ (3/2) x 14 = 16 + x
⇒ 21= 16 + x
∴ x = 21 – 16 = 5

Question 9 of 10

9. Question

1 points

Category: Quantitative Aptitude

The ratio of the age between Arun and Bala is 6:5, and the age of Chitra and Daniel is 9/10 times that of Bala. Age of Faruq is less than Arun but greater than Bala. The ratio of age of Bala & Preeti is 2:3, also age of A is 3 years less than Preeti .What is the ratio of age of Arun and Faruq, if all the ages are in integers ?

Correct

Answer – 4. 12 : 11
Explanation :
B = 5/6A
C = D = 9/10B
B = 3/2E
E – A =3
A/E = ⅘ => E = 5/4A
E-A = 5A/4 – A = 3
A = 12, E = 15, B =10, C = D = 9 and F = 11
A : F = 12 : 11

Incorrect

Answer – 4. 12 : 11
Explanation :
B = 5/6A
C = D = 9/10B
B = 3/2E
E – A =3
A/E = ⅘ => E = 5/4A
E-A = 5A/4 – A = 3
A = 12, E = 15, B =10, C = D = 9 and F = 11
A : F = 12 : 11

Question 10 of 10

10. Question

1 points

Category: Quantitative Aptitude

James and Dinesh have to clear their respective loans by paying 3 equal annual installments of Rs. 30000 each. James pays @ 10% per annum of simple interest while Dinesh pays 10% per annum compound interest. What is the difference in their payments?

Correct

Answer – 1. Rs. 300
Explanation :
James = SI
1st year=100
2nd = 100+10% of 100=110
3rd = 110+10% of 100=120
Total = 330
Dinesh = CI
1st yr = 100
2nd yr =100+10% of 100=110
3rd year=110+10% of 110=121
Total = 331
Diff = 331-330=1%
30000*1%=300

Incorrect

Answer – 1. Rs. 300
Explanation :
James = SI
1st year=100
2nd = 100+10% of 100=110
3rd = 110+10% of 100=120
Total = 330
Dinesh = CI
1st yr = 100
2nd yr =100+10% of 100=110
3rd year=110+10% of 110=121
Total = 331
Diff = 331-330=1%
30000*1%=300

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