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Question 1 of 10
1. Question
1 points
Category: Quantitative Aptitude
There are six consecutive odd numbers in increasing order. If the sum of the squares of the first and the last element is 178, then the average of all the six numbers is?
Correct
Answer : 4.8
Explanation :
Let the numbers be (x-5), (x-3), (x-1), (x+1), (x+3), (x+5)
(x-5)^2 + (x+5)^2 = 178
x^2 = 64
X = 8
Incorrect
Answer : 4.8
Explanation :
Let the numbers be (x-5), (x-3), (x-1), (x+1), (x+3), (x+5)
(x-5)^2 + (x+5)^2 = 178
x^2 = 64
X = 8
Question 2 of 10
2. Question
1 points
Category: Quantitative Aptitude
The ratio between the ages of a father and a son at present is 7:3, respectively. Four years, hence the ratio between the ages of the son and his uncle will be 1:2 respectively. What is the ratio between the present ages of the father and the uncle respectively?
Correct
Answer : 4. Cannot be determined
Explanation:
Let the ages of father and son – 7x, 3x
After four years the age of son = 3x + 4
After four years the age of uncle = 6x + 8
So the present age of uncle = 6x + 4
Ratio of the age of father and uncle = 7x : (6x + 4)
Incorrect
Answer : 4. Cannot be determined
Explanation:
Let the ages of father and son – 7x, 3x
After four years the age of son = 3x + 4
After four years the age of uncle = 6x + 8
So the present age of uncle = 6x + 4
Ratio of the age of father and uncle = 7x : (6x + 4)
Question 3 of 10
3. Question
1 points
Category: Quantitative Aptitude
The average age of Mr. Rakesh and his wife was 24 years when they were married 5 years ago. The average age of Mr.Rakesh, Rakesh’s wife and her daughter who was born during the interval, is 20 years now. How old is the daughter now?
Correct
Answer : 3. 2
Explanation:
Sum of the age of Rakesh and his wife 5 years ago = 24*2 = 48
Total present age of husband and wife is 48 + 10 = 58
Sum of age of Rakesh, Rakesh’s wife and daughter = 60
Age of daughter 60-58 = 2 year
Incorrect
Answer : 3. 2
Explanation:
Sum of the age of Rakesh and his wife 5 years ago = 24*2 = 48
Total present age of husband and wife is 48 + 10 = 58
Sum of age of Rakesh, Rakesh’s wife and daughter = 60
Age of daughter 60-58 = 2 year
Question 4 of 10
4. Question
1 points
Category: Quantitative Aptitude
Total distance between P & Q is ‘x’ km. If the distance travelled along the stream is thrice of the total distance and the distance travelled against the stream is double of the total distance. If the time taken to cover the distance along the stream is 40% less than the time taken to cover the distance against the stream . if a person cover a distance of 21 km in 1 hr 24 min along the stream, then find the rate of current ?
Bala is 20% efficient than Arun. Bala started the work and do it for x days. And then Bala is replaced Arun . And Arun completed the remaining work in x+8 days. Ratio of work done by Bala & Arun is 2:3. In how many days Arun & Bala working together to complete the whole work?
Correct
Answer : 2. 150/11
Explanation:
Efficiency of Arun = 100
Efficiency of Bala = 120
Bala worked for x days.
Arun completed the remaining work in (x + 8) days.
Work done ratio = 2:3
Bala did 2/5 part of the work in x days.
Bala can complete the work = 5x/2
Similarly, Arun’s one day work = 5(x+8)/3 days.
Now,
(5x/2)/[5*(x+8)/3] = 100/20
(5x/2) * 120 = 100 * [5*(x+8)/3]
90x = 50x + 400
40x = 400
x = 10.
Now, Bala completed the whole work= (5*10)/2 = 25 days.
Arun complete whole work = [5*(10+8)/3] = 30 days.
Arun and Bala together to complete the work = (1/25 + 1/30) = 150/11
Incorrect
Answer : 2. 150/11
Explanation:
Efficiency of Arun = 100
Efficiency of Bala = 120
Bala worked for x days.
Arun completed the remaining work in (x + 8) days.
Work done ratio = 2:3
Bala did 2/5 part of the work in x days.
Bala can complete the work = 5x/2
Similarly, Arun’s one day work = 5(x+8)/3 days.
Now,
(5x/2)/[5*(x+8)/3] = 100/20
(5x/2) * 120 = 100 * [5*(x+8)/3]
90x = 50x + 400
40x = 400
x = 10.
Now, Bala completed the whole work= (5*10)/2 = 25 days.
Arun complete whole work = [5*(10+8)/3] = 30 days.
Arun and Bala together to complete the work = (1/25 + 1/30) = 150/11
Question 6 of 10
6. Question
1 points
Category: Quantitative Aptitude
A sum of money is invested in Scheme A at 15% per annum for Simple Interest. After 2 years an amount is received and that amount is invested in scheme B at 20% per annum for compound interest. If the interest received at the end of 2 years is 1430, find the original sum invested?
Correct
Answer : 2. 2500
Explanation:
P[(1 + 20/100)^2 – 1] = 1430
P = 130 * 25 = Amount of Scheme A
P[RT/100 + 1] = 130 * 25
P[30/100 + 1] = 130 * 25
P = 2500
Incorrect
Answer : 2. 2500
Explanation:
P[(1 + 20/100)^2 – 1] = 1430
P = 130 * 25 = Amount of Scheme A
P[RT/100 + 1] = 130 * 25
P[30/100 + 1] = 130 * 25
P = 2500
Question 7 of 10
7. Question
1 points
Category: Quantitative Aptitude
Suresh lends equal sum of money at the same rate of interest to Arun and Bala. The money lends to Arun becomes twice of the original amount in just four years at simple interest. While Suresh lends to Bala for the first two years at compound interest and for the rest two years at simple interest. If the difference between the amount of Arun and Bala after 4 years is Rs. 3850. What is the amount of money that Suresh lends to each one?
Correct
Answer : 4. Rs. 11200
Explanation:
To Arun
P = P*4*100/R
R = 25%
After 4 years = 2P
To Bala
P(1 + 25/100)^2 = 25P/16
25P/16*2*25/100 = 25P/32
After 4 years = 25P/16 + 25P/32 = 75P/32
75P/32 – 2P = 3850
11P/32 = 3850
P = 11200
Incorrect
Answer : 4. Rs. 11200
Explanation:
To Arun
P = P*4*100/R
R = 25%
After 4 years = 2P
To Bala
P(1 + 25/100)^2 = 25P/16
25P/16*2*25/100 = 25P/32
After 4 years = 25P/16 + 25P/32 = 75P/32
75P/32 – 2P = 3850
11P/32 = 3850
P = 11200
Question 8 of 10
8. Question
1 points
Category: Quantitative Aptitude
A Vessel contain some liter of pure wine. If 25 liter of water added to the vessel, ratio becomes wine to water 12:5, 17 liter of mixture is drawn from the vessel and 10 liter of water is added. Find the new ratio between Wine and Water.
Correct
Answer : 2. 8:5
Explanation:
Wine -> 60-(12/17)*17= 48
Water -> 25-(5/17)*17+10=30
Wine : Water =48/30=8/5
Incorrect
Answer : 2. 8:5
Explanation:
Wine -> 60-(12/17)*17= 48
Water -> 25-(5/17)*17+10=30
Wine : Water =48/30=8/5
Question 9 of 10
9. Question
1 points
Category: Quantitative Aptitude
Ramya gave 1/6th of her salary to mother. From the remaining 2/5 for EMI and 1/6 for rent. She saved the balance in bank. If the difference between the amount given to her mother and bank is 2500, what is her salary?
Correct
Answer :4. Rs. 25000
Explanation:
Salary – “x”
To mother – x/6
x-x/6 = 5x/6
2/5 to EMI & 1/6 to rent
Bank = 5x/6 – (2x/5+x/6) = 4x/15
Difference between the amount given to her mother and bank = 2500
i.e. 4x/15 – x/6 = 2500
x = 25000/-
Incorrect
Answer :4. Rs. 25000
Explanation:
Salary – “x”
To mother – x/6
x-x/6 = 5x/6
2/5 to EMI & 1/6 to rent
Bank = 5x/6 – (2x/5+x/6) = 4x/15
Difference between the amount given to her mother and bank = 2500
i.e. 4x/15 – x/6 = 2500
x = 25000/-
Question 10 of 10
10. Question
1 points
Category: Quantitative Aptitude
The marked price of a Top and leggings are in the ratio 1:2. The shopkeeper gives 50% discount on the Top. If the total discount in the set of the Top and leggings is 30%, the discount offered on the leggings is:
Correct
Answer : 2. 20%
Explanation:
Let the price of Top and leggings be Rs. 100 and Rs. 200 respectively.
Price of set of Top and leggings = Rs. 300.
Selling Price = 300 – 30% of 300 = 210.
Total Discount on Set = 90.
SP of Top alone = 100 – 50% of 100 = 50.
Rs. 50 is the discount on Top then Rs. 40 must be the discount on the leggings.
The discount offered on the leggings = (40*100)/200 = 20%.
Incorrect
Answer : 2. 20%
Explanation:
Let the price of Top and leggings be Rs. 100 and Rs. 200 respectively.
Price of set of Top and leggings = Rs. 300.
Selling Price = 300 – 30% of 300 = 210.
Total Discount on Set = 90.
SP of Top alone = 100 – 50% of 100 = 50.
Rs. 50 is the discount on Top then Rs. 40 must be the discount on the leggings.
The discount offered on the leggings = (40*100)/200 = 20%.
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