Numerical Ability Questions - Bank Exams

Numerical Ability Questions – Bank Exams

1. The sum of three consecutive natural numbers each divisible by 3 is 72. What is the largest among them?
1. 25                       2. 26                       3. 27                       4. 30                       5. 26
Ans: (3) 

Let the three consecutive numbers divisible by 3 be { x + (x+3) + (x+6)}

It is given x + (x+3) + (x+6) = 72. Thus the value of x is 21 and the highest of the three is x+6 -> 27

2. The numerator of a non-zero rational number is five less than the denominator. If the denominator is increased by eight and the numerator is doubled, then again we get the same rational number. The required rational number is:
1. 1/8                     2. 4/9                     3. 2/8                     4. 3/8                     5. 5/8
Ans: (4)           Easy. Can be answered from the choices

3. Find the greatest number that will divide 640, 710 and 1526 so as to leave 11, 7, 9 as remainders respectively.
1. 36                       2. 37                       3. 42                       4. 29                       5. 47
Ans: (2) 

Let us assume (640 – 11) is divisible by ‘x’. Then ‘x’ also divides 710-7 and 1526-9

Thus we have ‘x’ that can divide 629, 703 and 1517. The HCF of these three numbers -> 37 is the answer.

4. Jayesh is as much younger than Anil as he is older than Prashant. If the sum of the ages of Anil and Prashant is 48 years, what is Jayesh’s age in years?
1. 29                       2. 30                       3. 24                       4. 25                       5. 28
Ans: (3)

Let J be Jayesh’s age and his difference from Anil and Prashant be ‘x’. Then we have Anil’s age as

J+x and Prashant’s age J-x.

We are given the sum of Anil’s and Prashant’s age as 48.

This gives us an equation (J+x) + (J-x) 48. Solving we get the value of J as 24 ie Jayesh’s age.

5. A bag contains one rupee, 50-paise and 25-paise coins in the ratio 2 : 3 : 5. Their total value is Rs. 114. The value of 50 paise coins is:
1. Rs. 28                2. Rs. 36                3. Rs. 49                4. Rs. 72                5. Rs. 50
Ans: (2)

Let ‘x’ be the rupee coin. Then 50 paise will be x/2 and 25 paise x/4.

The coins are in the ratio of 2 : 3 : 5    thus we have 2x + 3x/2 + 5x/4 = 114. Solving we get

The value of ‘x’as 24. Hence the number of 50 paise coins are 3x/2 -> 36 Rupees.

6. A tempo is insured to the extent of 4/5 of its original value. If the premium on it at the rate of 1.3 percent amounts to Rs. 910, the original value of the tempo is:
1. Rs. 78,000       2. Rs. 78,500       3. Rs. 80,000       4. Rs. 85,000       5. Rs. 87,500
Ans: (5)

At 1.3% the premium amount works out to Rs 910. Hence the insured value of the tempo is

910 / 1.3 -> 700 * 100 -> 70000. This insured value is 4/5 of the original value. Hence the

Original value is 70000 x 5/4 -> Rs 87,500.

7. By selling 45 lemons for Rs. 40, a man loses 20%. How many should he sell for Rs. 24 to gain 20% in the transaction?
1. 19                       2. 18                       3. 24                       4. 22                       5. 23
Ans: (2)

At Rs 40/- the man loses 20%. Hence the cost price is 40 x 100/80 = Rs 50/-. Thus 45 lemons cost price is Rs 50/-. Cost of one lemon is 50/45 -> Rs 10/9.

Selling at Rs 24 gives a profit of 20%. Hence the cost price is 24* 100/120 = Rs 20/-

Since the cost of one lemon is Rs 10/9  the number of lemons to be sold for Rs 20/- will be

20 / 10/9 -> 20 x 9/10 -> 18 lemons.

8. In a ratio, which is equal to 3 : 4, if the antecedent is 12, then the consequent is:
1. 10                       2. 16                       3. 20                       4. 22                       5. 18
Ans: (2)

9. A, B and C are employed to do a piece of work for Rs. 529. A and B together are supposed to do 19/23 of the work and B and C together 8/23 of the work. What amount should A be paid?
1. Rs. 320             2. Rs. 345             3. Rs. 355             4. Rs. 380             5. None of these
Ans: (2)

A+B are to do 19/23 work. Hence C will do 1 – 19/23 -> 4/23 work.

B+C are to do 8/23 work. Hence B will be doing 8/23 – 4/23 -> 4/23 work.

So, A will be doing 19/23 – 4/23 -> 15/23 work. Thus the ratio of work done by

A : B : C  ::  15 : 4 : 4. Total amount agreed for the work is Rs 529.

Thus A’s share of the money will be 529 x 15/23 -> Rs 345.

10. A boy rides his bicycle 10 km at an average speed of 12 km/hr and again travels 12 km at an average speed of 10 km/hr. His average speed for the entire trip is approximately how much?
1. 10.4 km/hr     2. 10.8 km/hr     3. 12 km/hr         4. 14 km/hr         5. 13 km/hr
Ans: (2)

Time taken to travel a distance of 10 km at a speed of 12 kmph is 10/12-> 5/6 hour -> 50 minutes.

Time taken to travel a distance of 12 km at a speed of 10 kmph is 12/10-> 6/5 hours -> 1 hour 12 minutes.

Thus to total time taken to travel a distance of 22 km is 2 hours and 2 minutes -> 122 minutes.

Hence the average speed for the entire journey will be 22/122 * 60 _> 10.8 kmph.

11. From a pack of 52 cards, two cards are drawn together at random. What is the probability of both the cards being kings?
1. 1/19                  2. 27/29                3. 35/256              4. 1/221                5. 35/121
Ans: (4)

There are four cards that are King. Thus the answer is arrived at 4C2/52C2

12. An automobile financier claims to be lending money at simple interest, but he includes the interest every six months for calculating the principal. If he is charging an interest of 10%, the effective rate of interest becomes:
1. 13%                   2. 10.25%             3. 15%                   4. 11%                   5. None of these
Ans: (2)

Apply the formula P (1+R/100)ⁿ – P. Assume P as 100, ‘n’ as two since the interest is compounded twice in a year and R half of 10% ie 5%.

  

13. A towel, when bleached, was found to have lost 20% of its length and 10% of its breadth. The percentage of decrease in area is:
1. 11%                   2. 12%                   3. 20%                   4. 28%                   5. 33%
Ans: (4)

Assume the original measurement of the Towel as Length 100 and Breadth 50. Area is 500 Sq. units

After bleaching length shrunk by 20% and breadth by 10%. So, the new length and breadth are

80 and 45. The new area is 80 * 45 = 360 sq. units. Thus the reduction in area is 500 – 360 = 140 sq. units.

The percentage of reduction in area is 140 x 100/500 -> 28%